Measure Theorey and Functional Analysisebooks.lpude.in/arts/ma_mathematics/year_2/DMTH505...Unit 9: Measure Spaces 100 Unit 10: Measurable Functions 106 Unit 11: Integration 121 Unit

DMTH505

Measure�Theorey�and�Functional�Analysis�

MEASURE THEORY ANDFUNCTIONAL ANALYSIS

Copyright © 2012 AnuradhaAll rights reserved

Produced & Printed byEXCEL BOOKS PRIVATE LIMITED

A-45, Naraina, Phase-I,New Delhi-110028

forLovely Professional University

Phagwara

Sr. No Description

1 Differentiation and Integration: Differentiation of monotone functions,

Functions of bounded variation,

2 Differentiation of an integral, Absolute continuity

3 Spaces, Holder, Minkowski inequalities, Convergence and Completeness

4 Bounded linear functional on the Lp spaces, Measure spaces, Measurable

Functions, Integration

5 General Convergence Theorems, Signed Measures, Radon-Nikodym

theorem.

6 Banach spaces: Definition and some examples, Continuous linear

transformations, The Hahn-Banach theorem

7 The natural imbedding of N in N**, The open mapping theorem, The

closed graph theorem,

8 The conjugate of an operator, The uniform boundedness theorem, The

uniform boundedness theorem, Hilbert spaces : The definition and some

simple properties

9 Orthogonal complements, Orthonormal Sets, The conjugate space H*, The

Adjoint of an Operator, Self Adjoint Operators

10 Normal and Unitary Operators, Projections, Finite dimensional spectral

theory : the spectrum of an operator on a finite dimensional

Hilbert space, the Spectral theorem

SYLLABUS

Measure Theory and Functional AnalysisObjectives: This course is designed for the of analysis of various types of spaces like Banach Spaces, Hilbert Space, etc. and also

CONTENTS

Unit 1: Differentiation and Integration: Differentiation of Monotone Functions 1

Unit 2: Functions of Bounded Variation 11

Unit 3: Differentiation of an Integral 27

Unit 4: Absolute Continuity 36

Unit 5: Spaces, Hölder 50

Unit 6: Minkowski Inequalities 65

Unit 7: Convergence and Completeness 72

Unit 8: Bounded Linear Functional on the Lp-spaces 82

Unit 9: Measure Spaces 100

Unit 10: Measurable Functions 106

Unit 11: Integration 121

Unit 12: General Convergence Theorems 141

Unit 13: Signed Measures 158

Unit 14: Radon-Nikodym Theorem 166

Unit 15: Banach Space: Definition and Some Examples 174

Unit 16: Continuous Linear Transformations 183

Unit 17: The Hahn-Banach Theorem 201

Unit 18: The Natural Imbedding of N in N** 211

Unit 19: The Open Mapping Theorem 217

Unit 20: The Closed Graph Theorem 225

Unit 21: The Conjugate of an Operator 231

Unit 22: The Uniform Boundedness Theorem 238

Unit 23: Hilbert Spaces: The Definition and Some Simple Properties 244

Unit 24: Orthogonal Complements 253

Unit 25: Orthonormal Sets 262

Unit 26: The Conjugate Space H* 273

Unit 27: The Adjoint of an Operator 283

Unit 28: Self Adjoint Operators 294

Unit 29: Normal and Unitary Operators 305

Unit 30: Projections 316

Unit 31: Finite Dimensional Spectral Theory 329

LOVELY PROFESSIONAL UNIVERSITY 1

Unit 1: Differentiation and Integration: Differentiation of Monotone Functions

NotesUnit 1: Differentiation and Integration: Differentiation of Monotone Functions

CONTENTS

Objectives

Introduction

1.1 Differentiation and Integration

1.1.1 Lipschitz Condition

1.1.2 Lebesgue Point of a Function

1.1.3 Covering in the Sense of Vitali

1.1.4 Four Dini's Derivatives

1.1.5 Lebesgue Differentiation Theorem

1.2 Summary

1.3 Keywords

1.4 Review Questions

1.5 Further Readings

Objectives

After studying this unit, you will be able to:

Understand differentiation and integration

Describe Lipschitz condition and Lebesgue point of a function

State Vitali’s Lemma and understand its proof.

Explain four Dini’s derivatives and its properties

Describe Lebesgue differentiation theorem.

Introduction

Differentiation and integration are closely connected. The fundamental theorem of the integralcalculus is that differentiation and integration are inverse processes. The general principle maybe interpreted in two different ways:

1. If f is a Riemann integrable function over [a, b], then its indefinite integral i.e.

F : [a, b] R defined by F (x) = x

a

f (t) dt is continuous on [a, b]. Furthermore if f is

continuous at a point xo [a, b], then F is differentiable thereat and F (xo) = f (xo).

2. If f is Riemann integrable over [a, b] and if there is a differentiable function F on [a, b] suchthat F f(x) for x [a, b], then

x

a

f (t) dt = F (x) – F (a) [a x b].

2 LOVELY PROFESSIONAL UNIVERSITY

Measure Theory and Functional Analysis

Notes 1.1 Differentiation and Integration

1.1.1 Lipschitz Condition

Definition: A function f defined on [a, b] is said to satisfy Lipschitz condition (or Lipschitzianfunction), if a constant M > 0 s.t.

|f (x) – f (y)| M |x – y|, x, y [a, b].

1.1.2 Lebesgue Point of a Function

Definition: A point x is said to be a Lebesgue point of the function f (t), if

x n

h 0 x

1Lim |f(t) f (x)|dt 0.h

1.1.3 Covering in the Sense of Vitali

Definition: A set E is said to be covered in the sense of Vitali by a family of intervals (may be open,closed or half open), M in which none is a singleton set, if every point of the set E is contained insome small interval of M i.e., for each x E, and > 0, an interval I M s.t. x I and (I) .

The family M is called the Vitali Cover of set E.

Example: If E = {q : q is a rational number in the interval [a, b]}, then the family qiI

where qi

1 1I q , qi i

, i N is a vitali cover of [a, b].

Vitali's Lemma

Let E be a set of finite outer measure and M be a family of intervals which cover E in the sense ofVitali; then for a given > 0, it is possible to find a finite family of disjoint intervals {Ik, k = 1, 2,… n} of M, such that

n

kk 1

m * E I < .

Proof: Without any loss of generality, we assume that every interval of family M is a closedinterval, because if not we replace each interval by its closure and observe that the set of endpoints of I1, I2, …… In has measure zero.

[Due to this property some authors take family M of closed intervals in the definition of Vitali’scovering].

Suppose 0 is an open set containing E s.t. m* (0) < m* (E) + 1 < we assume that each interval inM is contained in 0, if this can be achieved by discarding the intervals of M extending beyond 0and still the family M will cover the set E in the sense of Vitali.

Now we shall use the induction method to determine the sequence <Ik : k = 1, 2, … n> of disjointintervals of M as follows:



NotesLet I1 be any interval in M and let 1 be the supremum (least upper bound of the lengths of theintervals in M disjoint from I1 (i.e., which do not have any point common with I1).

Obviously 1 < as 1 m (0) < .

Now we choose an interval I2 from M, disjoint from I1, such that 2 11(I ) .2

Let 2 be the

supremums of lengths of all those intervals of M which do not have any point common with I2

or I2 obviously 2 < .

In general, suppose we have already chosen r intervals I 1, I2, … Ir (mutually disjoint). Let r bethe supremums of the length of those intervals of M which do not have any point in common

with ii 1

I (i.e., which do not meet any of the intervals I1, I2, … Ir. Then r m (0) < .

Now if E is contained in r

ii 1

I , then Lemma established. Suppose r

ii 1

I E . Then we can find

interval Ir+1 s.t. r 1 r1(I )2

which is disjoint from I1, I2, … Ir.

Thus at some finite iteration either the Lemma will be established or we shall get an infinite

sequence <Ir> of disjoint intervals of M s.t. r 1 r1(I )2

and r < , n = 1, 2, 3 ….

Note that < r > is a monotonically decreasing sequence of non-negative real numbers.

Obviously, we have that r rr 1i 1

I 0 ( ) m(0) hence for any arbitrary > 0, we can

find an integer N s.t.

rr N 1

1(I )5

.

Let a set F = N

rr 1

I .

The lemma will be established if we show that m* (F) < . For, let x F, then x N

rr 1

I x is an

element of E not belonging to the closed set N

rr 1

I an interval I in M s.t. x I and (I) is so

small that I does not meet the N

rr 1

I , i.e.

I Ir = , r = 1, 2, … N.

Therefore we shall have N n 1(I) 2 (I ) as by the method of construction we take

n 1 N1I2

.



Notes It also I IN+1 = , we should have N 1I . Further if the interval I does not meet any of theintervals in the sequence <Ir>, we must have

rI , r

which is not true as r r 12 (I ) 0 as r .

I must meet at least one of the intervals of the sequence <Ir>. Let p be the least integer s.t. Imeets Ip. Then p > N and p 1 p(I) 2 (I ). Further let x I as well x Ip, then the distance of x

from the mid point of Ip is at most

p p p p1 1 5(I) ( ) 2 (I ) (I ) (I )2 2 2

Thus if Ip is an interval having the same mid point as Ip but length 5 times the length of Ip, i.e. p p(J ) 5 (I ) . Then x Jp also.

Thus for every x F, an integer p > N s.t. x Jp

and p p(J ) 5 (I ) . Also

pp N 1

F J

p pp N 1 p N 1

m * (F) (J ) 5 (I ) 55

and hence the Lemma holds good.

1.1.4 Four Dini's Derivatives

The usual condition of differentiability of a function f (x) is too strong. Here we are studying thefunctions under slightly weaker condition (measurability). So why we define four quantities,called as Dini’s Derivatives, which may be defined even at the points where the function is notdifferentiable.

1. D+ f (x) = n 0

f(x h) f (x)Limh

, called upper right derivative

2. D+ f (x) = h 0

f (x h) f (x)Limh , called lower right derivative

3. D– f (x) = h 0

f (x h) f (x)Limh

,

or h 0

f (x h) f (x)Limh

, called upper left derivative

4. D– f (x) = h 0

f (x h) f (x)Limh

or h 0

f (x h) f (x)Limh

, called lower left derivative



Notes

Notes

1. D+f (x) D+ f (x) and D f(x) D f(x)

If D+ f (x) = D+ f (x), then we conclude that right hand derivative of f (x) exists at thepoint x and denoted by f (x+). Similarly if D– f (x) = D– f (x), we say that f (x) is leftdifferentiable at x and denote this common value by f (x–).

2. The function is said to be differentiable at x if all the four Dini’s derivatives are equalbut different than , i.e. if

D+ f (x) = D+ f (x) = D– f (x) = D– f (x)

and their common value is denoted by f (x).

Properties of Dini's Derivatives

1. Dini’s derivatives always exist, may be finite or infinite for every function f.

2. D+ (f + g) D+ f + D+ g with similar properties for the other derivatives.

3. If f and g are continuous at a point ‘x’, then

D+ (f . g) (x) f (x) D+ g (x) + g (x) D+ f (x).

4. D+ f (x) = – D+ (– f (x))

and D– f (x) = – D– (– f (x)).

5. If f is a continuous function on [a, b] and one of its derivatives (say D +) is non-negative on(a, b). Then f is non-decreasing on [a, b] i.e.

f (x) f (y) whenever x y, y [a, b].

6. If f is any function on an interval [a, b], then the four derivatives if exist are measurable.

1.1.5 Lebesgue Differentiation Theorem

Statement: Let f : [a, b] R be a finite valued monotonically increasing function, then f isdifferentiable. Also f : [a, b] R is L-integrable and

b

af (x) dx f (b) f (a) .

Proof: Define a sequence <fn> of non-negative functions, where fn : [a, b] R such that,

fn (x) = 1n f x f (x) , x [a, b]n … (1)

and set f (x) = f (b), for x b.

By hypothesis, f : [a, b] R is an increasing function, therefore fn : [a, b] R is also an increasingfunction and hence integrable in the Lebesgue sense.

Again from (i) we have

nnLim f (x) =

1/n 0

f {x (1/n)} f (x)Lim , x [a, b](1/n) ,

= f (x), a.e.



Notes Thus, the sequence <fn> converges to f (x), a.e.

Using Fatou’s Lemma, we have

b

af (x) dx

b

nn aLim inf f (x) dx … (ii)

Againb

nn aLim inf f (x) dx =

b

n a

1Lim inf n f x f(x) dxn

= b b

n a a

1Lim inf n f x dx f(x) dxn

Putting t = x + (1/n), we get

b

a

1f x dxn =

b (1/n) b (1/n)

a (1/n) a (1/n)f (t) dt f (x) dx

[By the first property of definite integrals]

b


b (1/n) b

n a (1/n) aLim inf n f (x) dx f (x) dx

= b (1/n) a (1/n)

n b aLim inf n f (x) dx f (x) dx … (iii)

Now extend the definition of f by assuming

f (x) = f (b), x [b, b + 1/n].

b (1/n)

bf (x) dx =

b (1/n)

b

1f (b) dx f (b)n

Also f (a) f (x), for x 1a, an

, therefore

a (1/n)

af (x) dx

a (1/n )

a

1f (a) dx f (a)n

–a (1/n)

af (x) dx 1 f (a)

n

(iii) b


b (1/n) a (1/n)

n b aLim inf n f(b) dx f(x) dx

n

1 1Lim inf n f (b) f (a) f (b) f (a)n n

Thus from (ii), we get

b

af (x) dx f (b) – f (a)

f (x) is integrable and hence finite a.e. thus f is differentiable a.e.



NotesExample: Let f be a function defined by f (0) = 0 and f (x) = x sin (1/x) for x 0. Find

D+ f (0), D+ f (0), D– f (0), D– f (0).

D+ f (0) = h oh o

1h sin 0f(0 h) f(o) hLim Limh h

= h o

1 1Lim sin 1, as 1 sin 1h h

Also D+ f (0) = h o h o

f(0 h) f(0) 1Lim Lim sin 1h h

D– f (0) = h oh o

1( h)sin 0f(0 h) f(0) hLim Lim0 h h

= h o

1Lim sin 1h

and D– f (0) = h o h o

f(0 h) f(0) 1Lim Lim sin 1h h

Theorem: Let x be a Lebesgue point of a function f (t); then the indefinite integral

F (x) = F (a) + x

af (t) dt

is differentiable at each point x and F (x) = f (x).

Proof: Given that x is a Lebesgue point of f (t), so that

x h

h o x

1Lim f(t) f(x) dth

= 0 … (i)

Nowx h

x

1 f (x) dth

= x h

x hx

x

1 1f (x) 1 dt f (x) [t]h h

= 1 f (x).h f (x)h

Thus f (x) = x h

x

1 f (x) dth

… (ii)

Also F (x + h) – F (x) = x h x

a af (t) dt f (t) dt

= x x h x x h

a x a xf (t) dt f (t) dt f (t)dt f (t) dt

F(x h) F(x)h =

x h

x

1 f (t)dth … (iii)



Notes From (ii) and (iii) we have

F(x h) F(x) f(x)h

= x h x h

x x

1 1f (t)dt f (x)dth h

= x h

x

1 f (t) f (x) dth

x h

x

1 f (t) f (x) dth

h o

F(x h) F(x)Lim f(x)h

x h

h o x

1Lim f(t) f (x) dt 0h

[Using (i)]

orh o


0 … (iv)

Since modulus of any quantity is always positive, therefore

h o


0 … (v)

Combining (iv) and (v), we obtain

h o

F(x h) F(x)Lim f(x)h = 0

h o

F(x h) F(x)Limh

= f (x)

F (x) = f (x).

Theorem: Every point of continuity of an integrable function f (t) is a Lebesgue point of f (t).

Proof: Let f (t) be integrable over the closed interval [a, b] and let f (t) be continuous at thepoint xo.

f (t) is continuous at t = xo implies that > 0, a > 0 such that,

|f (t) – f (xo) | < , whenever |t – xo| < .

x h x ho o

ox xo o

|f (t) f(x )|dt dt h whenever|h| .

x ho

oxo

1 |f (t) f(x )|dth

… (i)

Now h 0 0. So from (i), we have

x ho

oh o xo

1Lim |f (t) f(x )|dt 0h … (ii)

Nowx ho

oh o xo

1Lim |f (t) f(x )|dth

x ho

oh o xo

1Lim |f (t) f(x )|dt 0h [Using (ii)]



Notesor

x ho

oh o xo

1Lim |f (t) f(x )|dt 0h [ Modulus of any quantity is always non-negative]

i.e.x ho

oh o xo

1Lim |f (t) f(x )|dt 0h .

This shows that xo is a Lebesgue point of f (t).

1.2 Summary

A function f defined on [a, b] is said to satisfy Lipschitz condition if a constant M > 0 suchthat

| f (x) – f (y) M |x – y|, x, y [a, b].

A point x is said to be a Lebesgue point of the function f (t), if

x h

h o x

1Lim f(t) f (x) dt 0h

= 0

Let E be a set of finite outer measure and M be a family of intervals which cover E in thesense of Vitali; then for a given > 0, it is possible to find a finite family of disjointintervals

{Ik, k = 1, 2, …, n} of M, such that

n

kk 1

m * E I .

Lebesgue differentiation theorem: Let f : [a, b] R be a finite valued monotonicallyincreasing function, then f is differentiable. Also

f : [a, b] R is L-integrable and

b

a

f (x) dx f (b) f (a) .

1.3 Keywords

Dinni’s Derivatives: These are the ways to define the quantities to judge themeasurability of the functions even at the points where it is not differentiable.

Fundamental Theorem of the Integral: The fundamental theorem of the integral calculus is thatdifferentiation and integration are inverse processes.

Measurable functions: An extended real valued function f defined over a measurable set E is saidto be measurable in the sense of Lebesgue if set

E (f > a) = {x E : f (x) > a} is measurable for all extended real numbers a.

Vitali's Lemma: Let E be a set of finite outer measure and M be a family of intervals which coverE in the sense of Vitali; then for a given > 0, it is possible to find a finite family of disjointintervals {Ik, k = 1, 2, … n} of M, such that

n

kk 1

m * E I < .



Notes 1.4 Review Questions

1. If the function f assumes its maximum at c, show that D+ f (c) 0 and D– f (c) 0.

2. Give an example of functions such that D+ (f + g) D+ f + D+g.

3. Find the four Dini’s derivatives of function f : [0, 1] R

such that f (x) = 0, if x 0, if x Q and f (x) = 1, if x Q.

4. Evaluate the four Dini’s derivative at x = 0 of the function f (x) given below:

f (x) = 2 2

2 2

1 1ax sin bx cos , x 0x x1 1px sin qx cos , x 0x x

and f (0) = 0, given that a < b, p < q.

5. Every point of continuity of an integrable function f (t) is a Lebesgue point of f (t). Elucidate.


Books J. Yeh, Real Analysis: Theory of Measure and Integration

Bartle, Robert G. (1976). The Elements of Real Analysis (second edition ed.)

Online links www.solitaryroad.com/c756.html

www.public.iastate.edu/.../Royden_Real_Analysis_Solutions.pdf


Unit 2: Functions of Bounded Variation

NotesUnit 2: Functions of Bounded Variation

CONTENTS

Objectives

Introduction

2.1 Functions of Bounded Variation

2.1.1 Absolute Continuous Function

2.1.2 Monotonic Function

2.1.3 Functions of Bounded Variation – Definition

2.1.4 Theorems and Solved Examples

2.2 Summary

2.3 Keywords



Objectives


Define absolute continuous function.

Define monotonic function.

Understand functions of bounded variation.

Solve problems on functions of bounded variation.

Introduction

Functions of bounded variation is a special class of functions with finite variation over aninterval. In Mathematical analysis, a function of bounded variation, also known as a BV function,is a real-valued function whose total variation is bounded: the graph of a function having thisproperty is well behaved in a precise sense. Functions of bounded variation are precisely thosewith respect to which one may find Riemann – Stieltjes integrals of all continuous functions.

In this unit, we will study about absolute continuous function, Monotonic function and functionsof bounded variation.

2.1 Functions of Bounded Variation


A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for anarbitrary 0 , however small, a, 0, such that

n n

r r r rr 1 r 1

f b – f a ,wherever b a ,



Notes where 1 1 2 2 n na b a b ... a b i.e., a i’s and b i’s are forming finite collection

i ia ,b : i 1,2,...,n of pair-wise disjoint (non-overlapping) intervals (or of disjoint closed

intervals).

Obviously, every absolutely continuous function is continuous.

Notes

If a function satisfies r rf b – f a , even then it is absolutely continuous.

The condition n

r rr 1

b a , means that total length of all the intervals must be

less than .

2.1.2 Monotonic Function

Recall that a function f defined on an interval I is said to be monotonically non-increasing, iff

x y f x f y , x,y I

and monotonically non-decreasing, iff

x y f x f y , x,y I

Also f is said to be strictly decreasing, iff

x y f x f y

and strictly increasing, iff

x y f x f y

2.1.3 Functions of Bounded Variation – Definition

Let f be a real-valued function defined on [a,b] which is divided by means of points

0 1 2 na x x x ... x b.

Then the set 0 1 2 nP x ,x ,x ,...,x is termed as subdivision or partition of [a,b].

Let us take n 1b

r 1 ra r 0V f,P f x f x , and

b b

a aV f,P sup V f,P for all possible subdivisions P of

[a,b]. (This is called total variation of f over [a,b] and also denoted by b

aT f ).



NotesIf

b

aV f is finite, then f is called a function of bounded variation or function of finite variation

over [a,b].

Set of all the functions of bounded variation on [a,b] is denoted by BV [a,b].

Notes

If f is defined on R, then we define

a

a aV f lin V f .

Some important observations about the functions of bounded variations.

Let f: [a,b] R and P be any subdivision of [a,b]. Then:

(i)x

af x f a V f ,x [a,b]

(ii)a

aV f 0

(iii)b b

1 2 1 2a aP P V f,P V f,P , where 1P and 2P are any two subdivisions of [a,b].

(iv)b b

a aV f,P V f , for all subdivisions P of [a,b].

(v) For each 0, however small, at least one subdivision P’ of [a,b] such that

b b

a aV f,P' V f .

(vi)b

aV f 0.

(vii)b c

a aa b c V f V f .


Theorem 1: A monotonic function on [a,b] is of bounded variation.

Proof: Divide the interval [a,b] by means of points

0 1 2 na x x x ... x b.

without any loss of generality, we can take f(x) as increasing function on [a,b]. Since if f is adecreasing function, –f is an increasing function and so by taking –f = g, we see that g is anincreasing function and so we are allowed to consider only increasing functions. Thus

r r 1 r r 1x x f x f x

r 1 r f x – f x 0

r 1 r r 1 r f x – f x f x – f x ...(i)



NotesNow

n 1 n 1

r 1 r r 1 rr 0 r 0

V f x – f x f x – f x [using (i)]

n 0V f x f x f b f a .

Now f is monotonic f b and f a are finite quantities.

V a finite quantity independent of the mode subdivision. Hence f is of bounded variation.

Notes

If f is a monotonic function on [a,b], then

b

aT f f b f a

Theorem 2: Let V, P, N denote total, positive and negative variations of a bounded function f on[a,b]; then prove that

V = P+N, and P–N= f(b)–f(a).

Proof: Let the interval [a,b] be divided by means of points

0 1 2 na x x x ... x b.

n 1

r 1 rr 0

v f x – f x

If P denotes the sum of those differences r 1 rf x – f x which are +n for positive and –n fornegative, then obviously,

v = p + n, f(b) – f(a) = p – n ...(i)

Let P supp,V supv,N supn, ...(ii)

where suprema are taken over all subdivisions of [a,b]. From (i), we have

v = 2p + f(a) – f(b), ...(iii)

v = 2n + f(b) – f(a). ...(iv)

Taking supremum in (iii) and (iv) and using (ii), we get

V = 2P + f(a) – f(b), ...(v)

V = 2N + f(b) – f(a). ...(vi)

By adding and subtracting, (v) and (vi) give

V = P+N and f(b) – f(a) = P–N.

Theorem 3: If f1 and f2 are non-decreasing functions on [a,b], then f1–f2 is of bounded variation on[a,b].

Proof: Let f = f1 – f2 defined on [a,b].

Then for any partition 0 1 nP a x ,x ,...,x b , we have



Notesi i 1 1 i 1 i 1 2 i 2 i 1f x – f x f x – f x f x – f x

1 1 2 2f b – f a f b – f a

as f1 and f2 are monotonically increasing.

b

1 2 2 1aV f f b f b f a f a , which is a finite quantity.

b

aV f and hence f is of bounded variation.

Theorem 4: If f BV [a,b] and c a,b , then f BV [a,c] and f BV [c,b] . Also

b c b

a a cV f V f V f

Proof: Since f BV[a,b] and [a,c] [a,b] we get

c b

a aV f V f

f BV [a,c] and similarly f BV [c,b] .

Now if P1 and P2 are any subdivisions of [a,c] and [c,b] respectively, then 1 2P P P is asubdivision of [a,b].

c b b b

1 2a c a aV f,P V f,P V f,P V f .

But P1 and P2 are any subdivisions. So taking supremums on P1 and P2, we get

c b b

a c aV f V f V f . ...(1)

Now let 0 1 2 nP a x ,x ,x ,...,x b be a subdivision of [a,b] and r 1 rc x ,x

1 0 1 2 r 1P x ,x ,x ,...,x ,c and

2 r r 1 r 2 nP c,x ,x ,x ,...,x are the subdivisions of [a,c] and [c,b] respectively.

Nowr 1 nb

i i 1 r r 1 i i 1a i 1 i r 1V f,P f x f x f x f x f x f x

r 1 n

i i 1 r r 1 i i 1i 1 i r 1

f x f x f x f c f c f x f x f x

r 1 n

i i 1 r 1 r i i 1i 1 i r 1

f x f x f c f x f x f c f x f x

c b c b

1 2a c a cV f,P V f,P V f V f

(i) and (ii) b c b

a a cV f V f V f . ...(ii)



Notes

Notes

This theorem enables us to define a new function (called variation function) say

x

aV x V f , x [a,b] .

If x > y in [a,b], then y yx

a a xV f V f V f .

i.e. v(y) = v(x) + y

xV f .

v x is an increasing function.

If 1 2 na c c ... c b,then

c cb b1 2

a a c c1 nV f V f V f ... V f

Corollary:

f BV[a,b] f BV[a,c],

f BV[c,b] for each c [a,b].

Theorem 5: If a function f of bounded variation in [a,b] is continuous at c [a,b], then the function

defined by v(x) = x

aV f , is also continuous at x = c and vice versa.

Proof: Suppose f is continuous at x = c. Hence for arbitrary /2 0, we can find a 1 such that

1 1a c x c or x c f x f c /2 ...(i)

Also we know by remark (v) after the definition (2.1.3), for above , we can get a subdivision

P = 0 1 2 na x ,x ,x ,...,x c of [a,c]

s.t. c c

a aV f V f,P

2...(ii)

Now choosing positive 1 n 1min[ ,c x ], we get that for any x such that c x c , we also

have n 1 nx x x .

(ii)n 1c

r r 1 n n 1a r 1V f f x f x f x f x

2

n 1

r r 1 n n 1r 1

f x f x f x f x f x f x2



Notes

n 1

r r 1 n 1 nr 1

f x f x f x f x f x f x2

x

aV f + f c f x

2

c x

a aV f V f , [by (i)]

c x

a a0 V f V f ,

x s.t. c x c,we have v c v x

x c 0lt v x = v c .

v x is continuous on the left at x = c.

Similarly considering the partition of [c,b], one can show that v(x) is right continuous also atx = c and hence v(x) is also continuous at x = c.

Converse of the above Theorem

If v(x) is continuous at x c [a,b] so is f also at x – c.

Proof: Since v(x) is continuous at x = c, for arbitrary small 0, a 0 such that

v x v c ,x c ,c ...(i)

Now let c x c . Then by Note (ii) of Theorem 4, we get

x c x

a a cV f V f V f

x

cv x v c V f

x

cv x v c V f f x f c

f x f c v x v c [by (i)] ...(ii)

Similarly, we can show that f c f x ,if c x c. ...(iii)

(ii) and (iii) show that f(x) is also continuous at x = c.

Theorem 6: Let f and g be functions of bounded variation on [a,b] ; then prove that f+g, f-g, fg and

f/g g x 0, x and cf are functions of bounded variation, c being constant.

Proof:

(i) Set f + g = h, then

r 1 r r 1 r 1 r rh x h x f x g x f x g x

where a = x0 < x1< x2<...<xn= b



Notesr 1 r r 1 rf x f x g x g x

r 1 r r 1 rf x f x g x g x .

n 1 n 1 n 1

r 1 r r 1 r r 1 rr 0 r 0 r 0

h x h x h x h x g x g x .

orb b b

a a aV h V f V g .

Now by hypothesis, f, g are functions of bounded variations.

b b

a aV f and V g are finite.

b

aV h a finite quantity.

Hence h = f + g is of bounded variation in [a,b].

(ii) Let h = f – g. Then as above,

r 1 r r 1 r r 1 rh x h x f x f x g x g x .

b b b

a a aV h V f V g

b

aV h a finite quantity.

Hence h = f – g is of bounded variation in [a,b].

(iii) Let h(x) = f(x).g(x). Then

r 1 r r 1 r 1 r rh x h x f x .g x f x .g x

r 1 r 1 r r 1 r r 1 r rf x .g x f x g x f x g x f x g x

r 1 r 1 r r r 1 rg x f x f x f x g x g x .

r 1 r 1 r r r 1 rg x f x f x f x . g x g x .

Let A = sup f x : x [a,b] ,

B = sup g x : x [a,b] ,

r 1 r r 1 r r 1 rh x h x B. f x f x A. g x g x .

n 1 n 1 n 1

r 1 r r 1 r r 1 rr 0 r 0 r 0

h x h x B. h x h x A g x g x .



Notesi.e.

b b b

a a aV h B.V f A.V g .

= a finite quantity.

Hence h(x) = f(x).g(x) is of bounded variation in [a,b].

(iv) First, we shall show that 1/g is of bounded variation, where g x 0, x [a,b].

Now, g x 0, x [a,b]

1 1 0, x [a,b].g x

Again, we observe that

r r 1r r 12

r 1 r r r 1

g x g x1 1 1 g x g xg x g x g x .g x

n 1 n 1

r r 12r 0 r 0r 1 r

1 1 1 g x g xg x g x

b b

2a a

1 1V V g a finite quantity.g

Hence 1g is of bounded variation in [a,b].

Now f and 1g are of bounded variation in [a,b].

f. 1g

is of bounded variation in [a,b] [by case (iii)]

fg

is of bounded variation in [a,b].

(v) Do yourself. Note that b b

a aV cf c V f .

Notes

Since BV [a,b] is closed for all four algebraic operations, it is a linear space.

Theorem 7: Every absolutely continuous function f defined on [a,b] is of bounded variation.

Proof: Since f is absolutely continuous on [a,b]; for 1, a 0



Notess.t.

n

i ii 1

f b f a 1,

whenever n

i ii 1

b a ,

and 1 1 2 2 n na a b a b ... a b b.

Now consider another subdivision of [a,b] or say refinement of P by adjoining some additionalpoints to P in such a way that all the intervals can be divided into r parts each of total length less

than .

Let the r sub-intervals be [c0,c1], [c1,c2],...,[cr-1,cr] such that

a = c0, cr = b and (ck+1–ck) < , K 0,1,2,...,(r 1)

Obviously, i 1 i 1 i 1 i k k 1i

f x – f x 1,where x ,x . [c ,c ]

orck 1

ckV f 1, [Using (i)]

Hencec c cb 1 2 r

a c c c10 r 1V f V f V f ... V f 1 1 1 ... 1 r finite quantity.

Hence, f is of bounded variation.

Notes

Converse of above theorem is not necessarily true. These exists functions of boundedvariation but not absolutely continuous.

Theorem 8: Jordan Decomposition Theorem

A function f is of bounded variation, if and only if it can be expressed as a difference of twomonotonic functions both non-decreasing.

Proof: Let f be the function of f :[a,b] R.

Case I. f BV[a,b]. Then we can write

f = v – (v – f), ...(i)

so that f x v x v x f x ,x [a,b].

Now if x, y [a, b] such that x < y, then by the remark (ii) of theorem 4, we get

y yx

a a xV f V f V f .

y

xv y v x V f 0

v x v y and hence v is a non-decreasing function on [a,b].



NotesAgain, if x<y in [a,b], then as above

y

xv y v x V f f y f x f y f x

v y f y v x f x v f y v f x

v f is also a non-decreasing function on [a,b].

Thus (i) shows that f is expressible as a difference of two monotonically non-decreasing functions.

Case II. Set g (x) and h (x) be increasing functions such that f(x) = g(x) – h(x).

Divide the closed interval [a,b] by means of points

a = x0 < x1< x2<...< xn=b.

Let n 1

r 1 rr 0

V f x f x

Now, we have that

r 1 r r 1 r 1 r r

r 1 r r r 1

r 1 r r r 1

r 1 r r 1 r

f x f x g x h x g x h x

g x g x h x h x

g x g x h x h x

g x g x h x h x

Now, g(x) and h(x) are monotonically increasing functions, so that r 1 rg x g x 0

r 1 rand h x h x 0

r 1 r r 1 rg x g x g x g x

and r 1 r r 1 rh x h x h x h x .

Hence r 1 r r 1 r r 1 rf x f x g x g x h x h x

n 1 n 1 n 1

r 1 r r 1 r r 1 rr 0 r 0 r 0

f x f x g x g x h x h x

Now n 1

r 1 r 1 0 2 1 n n 1r 0

g x g x g x g x g x g x ..... .... g x g x

n 0

n 0

g x g x

g b g a x b,x a

Similarly, n 1

r 1 rr 0

h x h x h b h a .

Hence n 1

r 1 rr 0

f x f x g b g a h b h a .



Notessince f is finite in [a,b] Now g b ,g a h b ,h a are finite numbers.

n 1

r 1 rr 0

f x h x

b

aV f .

f is a function of bounded variation. Alternatively, since g (x) and h(x) are both non-decreasing,so by theorem 3, g(x) – h(x) and hence f(x) is of bounded variation.

Corollary: A continuous function is of bounded variation iff it can be expressed are as a differenceof two continuous monotonically increasing functions. It follows from the results of Theorems5 and 8.

Theorem 9: An indefinite integral is a function of bounded variation, i.e. if f L[a,b] and F x is

indefinite integral of f x i.e. F (x) = x

a

f t dt, then F BV[a,b].Also show that

xb

aa

V f f .

Proof: Since f L[a,b], also f L[a,b].

Let P = ix : i 0,1,2,...,n be a subdivision of the interval [a,b]. Then

x xi i 1n n

i i 1r 0 i 1 a a

F x F x f f

x xi in n

i 1 i 1x xi 1 i 1

f f

b

a

f .

bb

aa

f BV[a,b] and V f,p f .

Further above result is true for any subdivision of P of [a,b]. Therefore taking supremum, we get

b b

a a

f f .



NotesExample: A function f of bounded variation on [a,b] is necessarily bounded on [a,b] but

not conversely.

Solution: If x [a,b], then x b

a af x f a V f V f

b b

a aV f f x f a V f

b b

a af a V f f x V f f a

f x is bounded on [a,b]

For the converse, define the function f on [0,1] by

0,if x 0f x

x.sin ,if 0 x 1x

since 0 x 1 and 1 sin 1,x

the function f is obviously bounded. Now consider the

partition

2 2 2 2 2P= 0, , ,..., , , ,1 of [0,1]2n+1 2n–1 7 5 3

Where n N. Then we get

1

0

2 2 2 2V f,P f f 0 ... f f f 1 f2n 1 3 5 3

n2 2 2 21 0 ... 1 .1 0 1

2n 1 3 5 3

2 2 2 2...

2n 1 3 5 3

1 1 14. ... .3 5 2n 1

But we know that series 1

2n 1 is divergent. Therefore letting n we get that

1 1

0 n 0V f lt V f,P

f is not of bounded variation.



NotesExample: Show that the function

xsin if 0 x 2f x is continuousx

0,if x 0

without being of bounded variation.

or

show that there exists a continuous function without being of bounded variation.

Solution: We know that x 0lt f x 0 f 0

f x is continuous but not of bounded variation (see converse of above example.)

Hence the result.

Problem: Show that if f exists and is bounded on [a, b], then f BV [a, b].

Solution: According to given, let |f | M on [a, b].

Then for any Xi – 1, xi [a, b], we get

i i 1i i 1 i i 1

i i 1

f(x ) f(x ) M |f(x ) f(x )| M(x x )x x

for any partition P of [a, b],

b

i i 1aV(f) M (x x ) M(b a)

f B V [a, b].

Problem: Show that the function f defined as

p 1f(x) x sin for 0 x 1, f(o) 0, p 2.x

is of bounded variation [0, 1].

Solution: Note that RF (0) =

p

h o

1(0 h) sin 0hLim

h

= (p 1)

h o

1Lim h sin 0h

and

p

h o

1( h) sin 0hLf (0) lim 0

h



Notesf (0) = 0 and f (x) = p p 1

2

1 1 1x cos px sinx x x

f (x) = xp–2 1 1pxsin cosx x , for 0 < x 1

f (x) is bounded for 0 x 1.

According to above problem, f BV [0, 1].

2.2 Summary

A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if foran arbitrary 0, however small, a, 0,s.t.

n n

r r r rr 1 r 1

f b f a whenever b a ,

where 1 1 2 2 n na b a b ... a b

A function f defined on an interval I is said to be monotonically non-increasing, iff

, , .x y f x f y x y I

and monotonically non-decreasing, iff x > y f(x), f(y) x, g I.

Let n 1

b

r 1 rar 0

V f,P f x f x , and b b

a aV f Sup V f,P for all possible subdivisions P of

[a,b]. If b

aV f is finite, then f is called a function of bounded variation over [a,b].

2.3 Keywords

Absolute Continuous Function: A real valued function f defined on [a, b] is said to be absolutelycontinuous on [a, b], if for an arbitrary > 0, however small, a, > 0, such that

n n

r r r rr 1 r 1

|f(b ) f(a )| , wherever (b a )

where a1 < b1 a2 < b2 … an < bn i.e. a1’s and b1’s are forming finite collection {(ai, bi) : i = 1, 2,…, n} of pair-wise disjoint intervals.

Continuous: A continuous function is a function f : X Y where the pre-image of every openset in Y is open in X.

Disjoint: Two sets A and B are said to be disjoint if they have no common element, i.e. A B .



Notes Monotonic Decreasing Function: A monotonic decreasing function is a function that eitherdecreases or remains the same, never increases i.e. a function f(x) such that f(x2) f(x1) for x2 > x1.

Monotonic Function: A monotonic function is a function that is either a monotonic increasing ormonotonic decreasing.

Monotonic Increasing Function: A monotonic increasing function is a function that eitherincreases or remains the same, never decreases i.e. a function f(x) such that f(x2) f(x1) for x2 > x1.


1. Show that sum and product of two functions of bounded variation is again a function ofbounded variation.

2. Show that the function f defined on [0,1] by

xxcos for 0 x 1f x 2

0 for x 0

is continuous but not of bounded variation on [0,1].

3. Show that the function f defined on [0,1] as f(x) = xsinx for x > 0, f(0)=0 is continuous but

is not of bounded variation on [0,1].

4. Define a function of bounded variation on [a,b]. Show that every increasing function on[a,b] is of bounded variation and every function of bounded variation on [a,b] isdifferentiable on [a,b].

5. Show that a continuous function may not be of bounded variation.

6. Show that a function of bounded variation may not be continuous.

7. If f is a function such that its derivative f’ exists and is bounded. Then prove that thefunction f is of bounded variation.


Books Halmos, Paul (1950), Measure Theory, Van Nostrand and Co.

Kolmogorov, Andrej N.; Fomin, Sergej V. (1969). Introductory Real Analysis, NewYork: Dovers Publications.

Online links www.ams.org

www.whitman.edu/mathematics/SeniorProjectArchive/.../grady.pdf


Unit 3: Differentiation of an Integral

NotesUnit 3: Differentiation of an Integral

CONTENTS

Objectives

Introduction

3.1 Differentiation of an Integral

3.2 Summary

3.3 Keyword



Objectives


Define differentiation of an integral.

Solve problems related to it.

Introduction

If f is an integrable function on [a, b], we define its indefinite integral to be the function F definedon [a, b] by

F (x) = x

a

f (t) dt

Here, it is shown that the derivative of the indefinite integral of an integrable function is equalto the integrand almost everywhere. We begin by establishing some lemmas.

3.1 Differentiation of an Integral

If f is an integrable function on [a, b] then f is integrable on any interval [a, x] [a, b]. Thefunction F given by

F (x) = x

a

f (t) dt c ,

where c is a constant, called the indefinite integral of f.

Lemma 1: If f is integrable on [a, b] then the indefinite integral of f namely the function F on

[a, b] given by F (x) = x

a

f (t) is a continuous function of bounded variation on [a, b].

Proof: Let xo be any point of [a, b].



Notes

Then oF(x) F(x ) = xx o

a a

f (t) dt f (t) dt

= x a

a xo

f (t) dt f (t) dt

= a x

x ao

f (t) dt f (t) dt

= x

xo

f (t) dt

x

xo

|f(t)|dt

But f is integrable on [a, b]

|f| is integrable on [a, b]

[Since we know that measurable function f is integrable over iff |f| is integrable over E]

Given > 0, > 0 such that for every measurable set A [a, b] with m (A) < , we have

A

|f| by theorem, “if f is a non-negative function which is integrable over a set E, then

given > 0, there is a > 0 such that for every set A E with m (A) < , A

f .”

x

xo

|f(t)|dt < , for |x – xo| < .

|F(x) – F (xo)| = x x

x xo o

f (t) dt |f (t)|dt <

whenever |x – xo| < .

|F(x) – F(xo)| < wherever |x – xo| <

F is continuous at xo and hence in [a, b].

Now we shall show that F is a function of bounded variation.

Let P = {a = xo < x1 < x2 < … < xn = b} be a partition of [a, b].

Then

n

i i 1i 1

F(x ) F(x ) = xin

i 1 xi 1

f (t) dt



Notes

xin

i 1 xi 1

|f(t)|dt

= b

a

|f(t)|dt

baT (F)

b

a

|f(t)|dt

But |f| is integrable therefore.

b

a

|f|dt

baT (F) <

F BV [a, b]

Hence the Proof.

Theorem 1: Let f be an integrable on [a, b].

If x

a

f(t)dt 0 x [a, b] then f = 0 a.e. in [a, b].

Proof: Let if possible, f 0 a.e. in [a, b].

Let f (t) > 0 on a set E of positive measure, then there exists a closed set F E with m (F) > 0.

Let A = (a, b) – F.

Then A is an open set.

Nowb

a A F

f (t) dt f (t) dt

Butb

a

f (t) dt 0

b

A F

f (t) dt 0

A F

f (t) dt f (t) dt 0

A F A F

f (t) dt f (t) dt 0 f (t) dt f (t) dt



Notes But f (t) > 0 on F with m (F) > 0 implies

F

f (t) dt 0

Therefore A

f (t) dt 0

Now, A being as open set, it can be expressed as a union of countable collection {(an, bn)} ofdisjoint open intervals as we know that an open set can be expressed as a union of countablecollection of disjoint open intervals.

Thusbn

nA an

f (t) dt f (t) dt

ButA

f (t) dt 0

bn

n an

f (t) dt 0

bn

an

f (t) dt 0 for some n

either an

a

f (t) dt 0

Orbn

a

f (t) dt 0

In either case, we see that if f is positive on a set of positive measure, then for some x [a, b] wehave

x

a

f (t) dt 0 .

Similarly if f is negative on a set of positive measure we have

x

a

f (t) dt 0 .

But it leads to the contradiction of the given hypothesis. Hence our supposition is wrong.

f = 0 a.e. in [a, b].

Hence the proof.



NotesTheorem 2: First fundamental theorem of calculus statement: If f is bounded and measurable on

[a, b] and F (x) = x

a

f (t) dt + F (a), then F (x) = f (x) a.e. in [a, b].

Proof: Since every indefinite integral is a function of bounded variation, therefore F (x) is afunction of bounded variation over [a, b]. Thus F (x) can be expressed as a difference of twomonotonic functions and since every monotonic function has a finite differential coefficient atevery point of a set of non-zero measure, therefore F (x) has a finite differential coefficient a.e. in[a, b]. Now F is given to be bounded;

|f| M (say) … (1)

Let fn (x) = F(x h) F(x)h

.

with h = 1x

.

Then |fn (x)| = 1 (F(x h) F(x)h

= x h x

a a

1 f (t) dt f (t) dth

= x h a

a x


= a x h

x a


= x h

x

1 f (t) dth

But |f| M

|fn(x)| x h

x

M Mdt (x h x)h h

|fn(x)|M (h)h

|fn(x)| M

Since fn (x) F (x) a.e.,



Notes then the bounded convergence theorem implies thatx

a

F (x) dx = x

xha

lim F (x) dx

= x

h 0a

1lim [F(x h) F(x)]dxh

= x h a h

h 0x a

1 1lim [F(x)dx F(x) dxh h

= F (x) – F (a)

= x

a

f (t) dt, by hypothesis

x x

a a

F(x) f (t) dt F(a) F(x) F(a) f (t) dt

orx

a

[F (t) f(t)]dt 0, x

F (x) – f (x) = 0 a.e. in [a, b]

Hence F (x) = f(x) a.e. in [a, b] by the theorem, “If f is integrable on [a, b] and x

a

f (t) dt 0, x [a, b]

then f = 0 a.e. in [a, b]”.

Hence F (x) = f (x) a.e. in [a, b].

Hence the proof.

Theorem 3: If f is an integrable function on [a, b] and if F(x) = x

a

f (t) dt + F(a) then F (x) = f (x) a.e.

in [a, b].

Proof: Without loss of generality, we may assume that f (x) 0 x

Let us define a sequence {fn} of functions

fn : [a, b] R, where

fn(x) = f(x) if f(x) n,

n if f(x) n

Clearly, each fn is bounded and measurable function and so, by the theorem,

Let f be a bounded and measurable function defined on [a, b]. If F(x) = x

a

f (t) dt + F(a), then F (x)

= f(x) a.e. in [a, b]”, we havex

n n

a

d f f (x) a.e.dx



NotesAlso, f – fn > 0 n, and therefore, the function Gn defined by

x

n n

a

G (x) (f f )

is an increasing function of x, which must have a derivative almost everywhere by Lebesguetheorem and clearly, this derivative must be non-negative.

Since Gn (x) =

x

n

a

(f f )

= x x

n

a a

f (t ) dt f (t ) dt

x

a

f (t ) dt = x

n n

a

G (x) f (t ) dt

Now the relation

F (x) = x

a

f (t ) dt F(a) becomes

F (x) = x

n n

a

G (x) f (t ) dt F(a) ,

F (x) = n nG (x) f (x) a.e.

fn(x) a.e. n.

since n is arbitrary, we have

F (x) f(x) a.e.

b

a

F (x) dxb

a

f (x) dx … (1)

Also by the Lebesgue’s theorem, i.e. “Let f be an increasing real-valued function defined on[a, b].

Then f is differentiable a.e. and the derivative f is measurable.

andb

a

f (x) dx f (b) – f (a)”, we have

b

a

F (x) dx F (b) – F (a) … (2)



Notes

But F (x) = b

a

f (t) dt F(a)

F(b) – F(a) = b

a

f (x) dx

Therefore (2) becomesb

a

F (x) dxb

a

f(x) dx … (3)

From (1) and (3), we getb

a

F (x) dx = b

a

f(x) dx

b b

a a

F (x) dx f(x) dx = 0

b

a

F (x) f(x) dx = 0

since F (x) – f(x) 0 a.e., which gives that

F (x) – f(x) = 0 a.e. and

so F (x) = f(x) a.e.

3.2 Summary

If f is an integrable function on [a, b] then f is integrable on any interval [a, x] [a, b]. Thefunction F given by

F (x) = x

a

f (t) dt c ,

where c is a constant, called the indefinite integral of F.

Let f be an integrable on [a, b]. If x

a

f (t) dt 0 x [a,b] then f = 0 a.e. in [a, b].

3.3 Keyword

Differentiation of an Integral: If f is an integrable function on [a, b] then f is integrable on anyinterval [a, x] [a, b]. The function F given by

F (x) = x

a

f (t) dt c ,

where c is a constant, called the indefinite integral of f.



Notes3.4 Review Questions

1. If f is an integrable function on [a, b] and if F (x) = x

a

f (t) dt F(a) then check whether

F (x) = f (x) is absolute continuous function in [a, b] or not.

2. If F is an absolutely continuous function on [a, b], then prove that F (x) = x

a

f (t) dt C where

f = F a.e. on [a, b] and C is constant.


Books Flanders, Harley. Differentiation under the Integral Sign

Frederick S. Woods, Advanced Calculus, Ginn and Company

David V. Widder, Advanced Calculus, Dover Publications Inc., New Edition(Jul 1990).

Online links www.physicsforums.com > Mathematics > Calculus & Analysis

www.sp.phy.cam.ac.uk/~ alt 36/partial diff.pdf



Notes Unit 4: Absolute Continuity

CONTENTS

Objectives

Introduction

4.1 Absolute Continuity



4.2 Summary

4.3 Keywords



Objectives


Define Absolute Continuous function.

Solve problems on absolute continuity

Understand the proofs of related theorems.

Introduction

It may happen that a continuous function f is differentiable almost everywhere on [0,1], itsderivative f’ is Lebesgue integrable, and nevertheless the integral of f’ differs from the incrementof f. For example, this happens for the Cantor function, which means that this function is notabsolutely continuous. Absolute continuity of functions is a smoothness property which isstricter than continuity and uniform continuity.

4.1 Absolute Continuity


A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if for anarbitrary 0, however small, a, 0, such that

n

r rr 1

f b f a whenever n

r rr 1

b a ,

where 1 1 2 2 n na b a b ... a b i.e. a i’s and b i’s are forming finite collection

i ia ,b : i 1,2,...,n of pair-wise disjoint intervals.

Obviously, every absolutely continuous function is continuous.


Unit 4: Absolute Continuity

Notes

Notes

1. If a function satisfied r rf b f a , even then it is absolutely continuous.

2. The condition n

r rr 1

b a means that total length of all the intervals must be less

than .


Theorem 1: Every absolutely continuous function f defined on [a,b] is of bounded variation.

Proof: Since f is absolutely continuous on [a,b]; for 1, a 0 such that

n

i ir 1

f b f a 1,

whenevern

i ir 1

b a ,

and 1 1 2 2 n na a b a b ... a b b.

Now consider another subdivision of [a,b] or say refinement of P by adjoining some additionalpoints to P in such a way that all the intervals can be divided into r parts each of total length lessthan .

Let the r-sub-intervals be 0 1 1 2 r 1 rc ,c , c ,c ,..., c ,c such that

0 r k+1 ka c ,c b and c c , k 0,1,2,..., r 1

Obviously, i 1 ii

f x f x 1,

where i 1 i k k 1x ,x c ,c

or ck 1

ckV f 1,

Hence c c cb 1 2 r

a c c c0 1 r 1V f V f V f ... V f 1 1 ... 1 r finite quantity.

Hence f is of bounded variation.



Notes

Notes

Converse of above theorem is not necessarily true. There exists functions of boundedvariation but not absolutely continuous.

Theorem 2: Let f(x) and g(x) be absolutely continuous functions, then prove that f x g x and

f x .g x are also absolutely continuous functions. Hence show that f x if g x 0, xg x

is also

absolutely continuous function.

Proof: Given f x and g x are absolutely continuous functions on the closed interval [a,b],

therefore for each 0 , there exists 0 such that

n

r rr 1

f b f a and

n

r rr 1

g b g a ,

whenever n

r rr 1

b a , for all the points 1 1 2 2 n na ,b ,a ,b ,...,a ,b such that

1 1 2 2 n na b a b ... a b .

(i) We have, n n n

r r r r r r r rr 1 r 1 r 1

f b g b f a g a f b f a g b g a .

Now if n

r rr 1

b a ,then

n n

r r r rr 1 r 1

f b f a and g b g a .2 2

n

r r r rr 1

f b g b f a g a + = ,2 2

whenever n

r rr 1

b a .

This show that f x g x are also absolutely continuous functions over [a,b].



Notes

(ii) We have n

r r r rr 1

f b g b f a g a

n

r r r r r r r rr 1

f b g b f b g a f b g a f a g a

n

r r r r r rr 1

f b g b g a g a f b f a

n n

r r r r r rr 1 r 1

f b g b g a g a f b f a

n n

r r r r r rr 1 r 1

f b g b g a g a f b f a .

Now every absolutely continuous function is bounded therefore f(x) and g(x) are boundedin the closed interval [a,b].

Let 1 2f x K , g x K , x [a,b].

Then we have

n

r r r r 1 2 1 2r 1

f b g b f a g a K K K K ,

Whenever n

r rr 1

b a .

Setting 1 2K K *,

We have n

r r r rr 1

f b g b f a g a = *,

Whenever n

r rr 1

b a ,

where 1 1 2 2 n na b a b ... a b ;

Product of two absolutely continuous functions is also absolutely continuous.

(iii) We have g x 0 x [a,b] ; therefore

g x ,where 0, x [a,b].



Notes

Now, n n

r r2

r r r rr 1 r 1

g a g b1 1 = < ,g b g a g b g a

Whenever n

r r 2r 1

b a . Setting = *,we get

n

r rr 1

1 1 < *.g b g a

This show that 1g x

is absolutely continuous function over [a,b].

Now f(x), 1g x

are absolutely continuous.

1f x .g x

is absolutely continuous.

f xg x

is also absolutely continuous over [a,b].

Hence the theorem is true.

Note

By Theorem 1, its remark and above theorem it follows that set of all absolutely continuousfunctions on [a,b] is a proper subspace of the space BV [a,b] of all functions of boundedvariation on [a,b].

Theorem 3: If BV[a,b], then f is absolutely continuous on [a,b], iff the variation function

x

av x V f is absolutely continuous on [a,b].

Proof: Case I: Given v(x) is absolutely continuous.

For arbitrary >0, 0 s.t.

n n

r r r rr 1 r 1

v b v a < ,whenever b a .

Also, we know that x

af x f a V f v x



Notesn n

r r r rr 1 r 1

f b f a = f b f a f a f a

n

r rr 1

f b f a f a f a

Now taking supremum over all collections of Pi of [ai, bi] for i = 2,...,n, we get

n bi

air 1

V f .

But b a bi i i

a a aiV f V f V f

b b ai i i

a a aiV f V f V f

bi

i iaiV f v b v a

n

i ii 1

v b v a <

v x is absolutely continuous.

Theorem 4: A necessary and sufficient condition that a function should be an indefinite integralis that it should be absolutely continuous.

Proof: Condition is sufficient.

Let f(x) be an absolutely continuous function over the closed interval [a,b].

Therefore f is of bounded variation and hence we can express f(x) as

f(x) = f1(x) – f2(x)

where f1(x) and f2(x) are monotonically increasing functions and hence both are differentiable.

n n

r r r rr 1 r 1

v b v a < whenever b a

f is also absolutely continuous on [a,b].

Case II: Given f is absolutely continuous on [a,b].

for a given 0, a 0 s.t.

n

i ii 1

f b f a < , ...(i)



Notesfor every finite collection i iP a ,b , i 1,2,...,n of pairwise disjoint sub-intervals of [a,b] such

that n

i ii 1

b a .

Now, let i ii k 1 ki iP x ,bx ,k 1,...,m be a finite collection of non-overlapping intervals of the

interval [ai,bi].

Then the collection 1 , : 1,2,..., , 1,...,i ik k i ix bx i n k m is a finite collection of non-overlapping

sub-intervals of [a,b] such that

mn nii ik k 1 i i

i 1 k 1 i 1

x x b a .

and hence by (i), mn i

i ik k 1

i 1 k 1

f x f x .

Hence f’(x) exists and ' '1 2f ' x f x f x

b

1 2 1 2

a

f x f b f b f a f a ,

f x is integrable also.

Now let F(x) be an definite integral of f’(x) i.e.

F(x) = F(a) + x

a

f t dt, x [a,b] ...(ii)

Using fundamental theorem of integral calculus,

We get

F’(x) = f’(x)

or F(x) = f(x) + constant (say c) ...(iii)

From (ii), we have F(a) = f(a),

Using this in (iii), we get c = 0 and hence F (x) = f(x).

Thus every absolutely continuous function f(x) is an indefinite integral of its own derivative.

Condition is necessary: Let f(x) be an indefinite integral of f(x) defined on the closed interval[a,b], so that

x

a

F x f t dt f a , x [a,b] and f(x) is integrable over [a,b].

Corresponding to arbitrary small 0, let >0 be such that if A

m(A) ,then f ,



NotesNow select 2n real numbers such that

i i 2 2 3 n na b a b a ... a b

such that A = n

n

i i i ii 1i 1

U a ,b and b a .

Then b ai in n

i ii 1 i 1 a a

F b F a f f

b bi in n

i 1 i 1a a Ai i

F f f .

Thus, we have shown that for arbitrary small n

i ii 1

0, a 0 s.t. b a .

n

i ii 1

F b F a .

F is absolutely continuous.

Thus every indefinite integral is absolutely continuous.

Theorem 5: If a function f is absolutely continuous in an interval [a,b] and if f’(x) = 0. a.e. in [a,b],then f is constant.

Proof: Let c [a, b] be arbitrary. If we show that f(c) = f(a), then the theorem will be proved.

Let E = x a, c : f '(x) 0 .

since c is arbitrary, therefore set E a,c . This implies any x E f ' x 0.

Let , > 0 arbitrary. Now f ' x 0, x E an arbitrary small interval x,x h a,c

such that f x h f x

f x h f x h.h

This implies that corresponding to every x E, an arbitrary small closed interval x,x hcontained in [a,c] s.t.

f x h f x h.

Thus the interval x,x h , x E, over E in Vitali’s sense. Thus by Vitali’s Lemma, we candetermine a finite number of non-overlapping intervals Ik, where

k k kI x ,y k 1,2,3,...,n

such that this collection covers all of E except for a set of measure less than 0 where is pre-assigned number which corresponds to occurring in the definition of absolute continuity of f.



Notes Suppose k k 1x x ; then adjoining the points y0, xn+1.

We have 0 1 1 2 2 n n n 1a y x y x y ... x y x c.

Now since f is absolutely continuous, therefore for above subdivision of [a,c], we have

n n

k 1 k k 1 kk 0 k 0

f x f y , whenever x y .

(i)n n

k k k kk 1 k 1

f y f x n y x n c a .

Now n n

k 1 k k kk 0 k 1

f c f a f x f y f y f x

n n

k 1 k k kk 0 k 1

f x f y f y f x

n c a

But ,n and hence n c a are arbitrary small positive numbers. So letting 0,n 0

We get f(c) = f(a)

f x is a constant function.

Corollary: If the derivatives of two absolutely continuous functions are equivalent, then thefunctions differ by a constant.

Proof: Let f and g be two absolutely continuous functions and f’ = g’ f g ' 0 by above

theorem f – g = constant and hence the result.

Example: If f is an absolutely continuous monotone function on [a,b] and E a set ofmeasure zero, then show that f (E) has measure zero.

Proof: Let the function f be monotonically increasing. By the definition of absolute continuity of

f, for 0, 0 and non-overlapping intervals n n nI a ,b such that

n n n nb a f b f a

or n nf b f a

Now,nE [a,b] E I

n nf E f I f I

n n nm * f E m * f I f x f x ,



Noteswhere nf x and nf x are the maximum and maximum values of f(x) in the interval [an,bn].

Also note that n n n nx x b a

m * f E , being arbitrary.

m * f E 0 m f E 0.

Example: Give an example which is continuous but not absolutely continuous.

Solution: Consider the function f : F R, where F is the Cantor’s ternary set.

Let k1 2 3 kk

K 1

xx F x x x x ... ,x 0 or 23

Define kk kk

K 1

r 1f x , where r x .2 2

= 0. r1, r2, r3.....

This function is continuous but not absolutely continuous.

(i) Note that this function is constant on each interval contained in the complement of theCantor’s ternary set.

For, let (a,b) be one of the countable open intervals contained in F c. Then in ternarynotation,

a = 0.a1a2...an–1 0 2 2 2

and b = 0.a1a2...an-1 2 0 0 0,

where ai = 0 or 2, for i n 1.

i1 2 n 1 i

af a 0.r ,r ,...,r 0 1 1 1 1 ...,where r ,2

1 2 n 1f b 0.r ,r ,...,r 1 0 0 0 0 ...

But in binary notation

1 2 n 1 1 2 n 10.r ,r ,...,r 0 1 1 1 1 ... 0.r ,r ,...,r 1 0 0 0 0 ...

f a f b .

Thus, we extend the function f overall of the set [0,1] instead of F by definingcf x f b , x a,b F . Thus, the Cantor’s function is defined over [0,1] and maps it

onto [0,1].

It is clearly a non-decreasing function.



Notes (ii) To show that f(x) is a continuous function. Note that if c', c'' F, then we have

1 2 3i i

1 2 3

c' 0. 2p 2p 2p ...each p ,q 0 or =1

c'' 0. 2q 2q 2q ...

If n

1c' c'' ,3

then pi = qi, for 1 i n 1 and hence

n

1f c' f c''2 ...(i)

as n ,c' c'', f c' f c'' ,

Hence if 0 nc F and c is a sequence in F such that n 0c c ,when n , then

n 0f c f c ,when n .

Now let 0x [0,1] and let nx be a sequence in [0,1] such that n 0x x as n .

Case I: Let c0 0x F x I,say a,b F

n nx I and hence f x f x f a

and hence n 0f x f x as n .

Case II: Let 0x F. Now for each n such that n n n n 0x F,set x c and hence f x f x .

If cnx F, then an open int erval I F .

(i) if n 0x x , then set cn as the upper end point of I.

(ii) If 0 nx x , then set cn as the lower end point of I.

n 0in any case f x f x as n .

But the sequence nx was any sequence satisfying the stated conditions.

f is a continuous function.

(iii) To show f(x) is not absolutely continuous. Note that f’(x) = 0 at each cx F .

f ' x exists and is zero on [0,1] and is summable on [0,1].

We know that for f(x) to be absolutely continuous, we must have

x

0

f x f ' x dx f 0 .

Particularly, we must have

1

0

f 1 f 0 f ' x dx.



NotesBut f(1) – f(0) = 1 and

x

0

f ' 1 dx 0 as f ' x 0

x

0

f 1 f 0 f ' 1 dx 0 as f ' x 0

f x is not absolutely continuous.

Theorem 6: Prove that an absolutely continuous function on [a, b] is an indefinite integral.

Proof: Let f(x) be an absolutely continuous function in a closed interval [a, b] so that f (x) &

x

a

f (t) dt exists finitely x [a, b].

Let F(x) be an indefinite integral of f (x), so that

F(x) = x

a

f(a) f (t) dt, x [a,b] … (1)

We shall prove that F(x) = f(x).

Since an indefinite integral is an absolutely continuous function.

Therefore F(x) is absolutely continuous in [a, b].

Then from (1),

F (x) = f (x) a.e.

d [F(x) f(x)]dx

= 0.

Integrating, we get

F(x) – f(x) = c (constant) … (2)

Taking x = a in (1), we get

F(a) = a

a

f(a) f (t) dt

F(a) – f(a) = 0

or F(x) – f(x) = 0 for x = a

Then from (2), we get c = 0.

Thus (2) reduces to

F(x) – f(x) = 0 a.e.

F(x) = f(x) a.e.

which shows that f(x) is indefinite integral of its own derivative.



Notes 4.2 Summary

A real-valued function f defined on [a,b] is said to be absolutely continuous on [a,b], if foran arbitrary 0, however small, a, 0, such that

n n

r r r rr 1 r 1

f b f a , whenever b a .

Every absolutely continuous function is continuous.

Every absolutely continuous function f defined on [a,b] is of bounded variation.

4.3 Keywords

Absolute Continuity of Functions: Absolute continuity of functions is a smoothness propertywhich is stricter than continuity and uniform continuity.

Absolute Continuous Function: A real-valued function f defined on [a,b] is said to be absolutelycontinuous on [a,b], if for an arbitrary 0, however small, a, 0, such that

n

r rr 1

f b f a whenever n

r rr 1

b a ,

where 1 1 2 2 n na b a b ... a b i.e. a i’s and b i’s are forming finite collection

i ia ,b : i 1,2,...,n of pair-wise disjoint intervals.


1. Define absolute continuity for a real variable. Show that f(x) is an indefinite integral, if Fis absolutely continuous.

2. If f,g: [0,1] R are absolutely continuous, prove that f + g and fg are also absolutelycontinuous.

3. Show that the set of all absolutely continuous functions on an interval I is a linear space.

4. If g is a non-decreasing absolutely continuous function on [a,b] and f is absolutely continuouson [g(a), g(b)], show that fog is also absolutely continuous on [a,b].

5. If f is absolutely continuous on [a,b] and f ' x 0 for almost all x [a,b], show that f is

non-decreasing on [a,b].


Books Krishna B Athreya, N Soumendra Lahiri, Measure Theory and Probability Theory,Springer (2006).



NotesAole Nielsen, An Introduction to Integration and Measure Theory, Wiley-Interscience,(1997).

N.L. Royden, Real Analysis (third ed.), Collier MacMillan, (1988).

Online links dl.acm.org

mrich.maths.org



Notes Unit 5: Spaces, Hölder

CONTENTS

Objectives

Introduction

5.1 Spaces, Hölder

5.1.1 Lp-spaces

5.1.2 Conjugate Numbers

5.1.3 Norm of an Element of Lp-space

5.1.4 Simple Version of Hölder's Inequality

5.1.5 Hölder's Inequality

5.1.6 Riesz-Hölder's Inequality

5.1.7 Riesz-Hölder's Inequality for 0 < p < 1

5.2 Summary

5.3 Keywords



Objectives


Understand Lp-spaces, conjugate numbers and norm of an element of Lp-space

Understand the proof of Hölder’s inequality.

Introduction

In this unit, we discuss an important construction, which is extremely useful in virtually allbranches of analysis. We shall study about Lp-spaces and Hölder’s inequality.

5.1 Spaces, Hölder

5.1.1 LP-Spaces

The class of all measurable functions f (x) is known as Lp-spaces over [a, b], if Lebesgue –integrable over [a, b] for each p exists, 0 < p < , i.e.

b

p

a

|f| dx , (p 0)

and is denoted by Lp [a, b].


Unit 5: Spaces, Hölder

Notes

Note The symbol Lp is used for such classes when limits of integration are known andmentioning of interval is not necessary.

5.1.2 Conjugate Numbers

Let p, q be any two n on-negative extended real numbers s.t. 1 1 1p q

, then p, q are called

(mutually) conjugate numbers.

Obviously, 2 is self-conjugate number.

Also if p 2, then q 2. Further, if p = , then q = 1 1, are conjugate numbers.

Note Non-negativity p 1, q 1.

5.1.3 Norm of an Element of LP-space

The p-norm of any f Lp [a, b], denoted by f p, is defined as

f p =

1b p

p

a

|f| , 0 < p < .

Theorem 1: If f Lp [a, b] and g f, then g Lp [a, b].

Proof: Let be any positive real number.

{x [a, b] : g (x) > } = {x [a, b] : < g (x) f (x)} ( g f)

= {x [a, b] : f (x) > }

Again f Lp [a, b]

f is measurable over [a, b].

{x [a, b] : f (x) > } is a measurable set.

{x [a, b] : g (x) > } is a measurable set.

g is a measurable function over [a, b]

Again since g (x) f (x), x [a, b]

b b

p p

a a

|g| dx |f| dx ( |f|p L [a, b])

orb

p

a

|g| dx

Thus |g|p L [a, b].



Notes Thus we have proved that g is a measurable function over [a, b] such that

|g|p L [a, b]

Hence g Lp [a, b]

Theorem 2: If f Lp [a, b], p > 1, then f L [a, b]

Proof: f Lp [a, b] f is measurable over [a, b]

Let A1 = {x [a, b] : |f (x) 1}

and A2 = {x [a, b] : |f (x) 1}

Then [a, b] = A1 A2 and A1 A2 =

Using countable additive property of the integrals, we haveb

a

|f|dx = A A1 2

|f|dx |f|dx … (i)

Now |f (x)| 1, x A1

|f| |f|p on A1 as p > 1

A1

|f|dx p

A1

|f| dx as f Lp [a, b] … (ii)

Now |f (x) |<|, x A2

Using first mean value theorem, we get

2

A2

|f|dx m (A ) = A finite quantity … (iii)

Combining (ii) and (iii) and making use of (i), we get

b

a

|f|dx <

Thus f is a measurable function over [a, b], such that

b

a

|f|dx <

|f| L [a, b] and hence f L [a, b].

Theorem 3: If f Lp [a, b], g Lp [a, b]; then f + g Lp [a, b]

Proof: Since f, g Lp [a, b] f, g are measurable over [a, b]

f + g is measurable over [a, b]

Let A1 = {x [a, b] : |f (x)| |g (x)|}

and A2 = {x [a, b] : |f (x)| < |g (x)|}

Then [a, b] = A1 A2 and A1 A2 =



NotesTherefore p p p

A A1 2

|f g| dx |f g| |f g| dx .

Again, |f + g|p (|f| + |g|)p (|g| + |g|)p on A2 and (|f| + |f|)p on A1

2P |g|p on A2 and 2p |f|p on A1

Integrating, we have

p

A1

|f g| p p

A1

2 |f|

and p

A2

|f g| p p

A2

2 |g|

Since f, g Lp [a, b] p p

A A1 2

|f| and |g|

p p

A A1 1

|f g| and |f g| [by (i) and (ii)]

b

p

a

|f g| dx f + g Lp [a, b]

5.1.4 Simple Version of Hölder's Inequality

Lemma 1: Let p, q > 1 be such that 1 1 1p q

, and let u and v be two non-negative numbers, at

least one being non-zero. Then the function f : [0, 1] R defined by

f (t) = ut + 1

q qv(1 t ) , t [0, 1],

has a unique maximum point at

s =

1p q

p p

uu v

… (1)

The maximum value of f is

t [0 , 1]max f(t) =

1p p p(u v ) … (2)

Proof: If v = 0, then f (t) = tu, t [0, 1] (with u > 0), and in this case, the Lemma is trivial.

Likewise, if u = 0, then

f (t) = 1

q qv(1 t ) , t [0, 1] (with v > 0), … (3)



Notes and using the inequality

1q q(1 t ) 1, t (0, 1],

We immediately get f (t) < f (0), t (0, 1],

and the Lemma again follows.

For the remainder of the proof we are going to assume that u, v > 0.

Obviously f is differentiable on (0, 1) and the solutions of the equation (3)

f (t) = 0

Let s be defined as in (1), so under the assumption that u, v > 0, we clearly have 0 < s < 1.

We are going to prove first that s is the unique solution in (0, 1) of the equation (3).

We have

f (t) = 1 1

q q q 11u v (1 t ) q tq

… (4)

= q

q

tu v , t (0, 1)1 t

so the equation (3) reads

q

q

tu v1 t

= 0.

Equivalently, we have

1q p

q

t1 t

= u/v,

q

q

t1 t

= (u/v)p,

t2 = p p

p p p

(u/v) u ,1 (u/v) u v

Having shown that the “candidates” for the maximum point are 0, 1 and s let us show that s is theonly maximum point.

For this purpose, we go back to (4) and we observe that f is also continuous on (0, 1).

Sincet 0lim f (t) = u > 0 and

t 1lim f (t) = –

and the equation (3) has exactly one solution in (0, 1), namely s, this forces

f (t) > 0 t (0, s)



Notesand f (t) < 0 t (s, 1).

This means that, f is increasing on [0, s] and decreasing on [s, 1], and we are done.

The maximum value of f is then given by

t [0 , 1]max f(t) = f (s),

and the fact that f (s) equals the value in (2) follows from an easy computation.

5.1.5 Hölder's Inequality

Statement: Let a1, a2, …, an, b1, b2, …, bn be non-negative numbers. Let p, q > 1 be real number

with the property 1 1 1p q

. Then

n

j jj 1

a b

1 1n np q

p qj j

j 1 j 1

a b … (5)

Moreover, one has equality only when the sequences p p1 na , , a and q q

nb , , b are proportional.

Proof: The proof will be carried on by induction on n. The case n = 1 is trivial.

Case n = 2.

Assume (b1, b2) (0, 0). (otherwise everything is trivial).

Define the number

r = 11

q q q1 2

b

b b.

Notice that r [0, 1] and we have

21

q q q1 2

b

b b=

1q q1 r

Notice also that, upon dividing by 1

q q q1 2b b , the desired inequality

a1 b1 + a2 b2 1 1

q q q qp q1 2 1 2a a b b … (6)

reads

a1 r + a2 1

q q1 r 1

q q p1 2a a … (7)

It is obvious that this is an equality when a1 = a2 = 0. Assume (a1, a2) (0, 0), and set up thefunction.

f (t) = 1

q q1 2a t a 1 t , t [0, 1].



Notes We now apply Lemma (1) stated above, which immediately gives us (7).

Let us examine when equality holds.

If a1 = a2 = 0, the equality obviously holds, and in this case (a1, a2) is clearly proportional to (b1,b2). Assume (a1, a2) (0, 0).

Again by Lemma (1), we know that equality holds in (7), exactly when

r =

1p q1

p p1 2

aa a

that is 11

q q q1 2

b

b b=

1p q1

p p1 2

aa a

,

or equivalently

q1

q q1 2

bb b

= p1

p p1 2

aa a

.

Obviously this forces

q1

q q1 2

bb b =

q1

p p1 2

aa a ,

so indeed p p1 2a , a and q q

1 2b , b are proportional.

Having proven the case n = 2, we now proceed with the proof of:

The implication: Case n = k case n = k + 1, start with two sequences (a1, a2, …, ak, ak+1) and(b1, b2, …, ak, bk+1).

Define the numbers

a =

1k p

pj

j 1

a and b =

1k q

qj

j 1

b .

Using the assumption that the case n = k holds, we have

k 1

j jj 1

a b

1 1k kp q

p qj j

j 1 j 1

a b + ak+1 bk+1

= ab + ak+1 bk+1 … (8)

Using the case n = 2, we also have

ab + ak+1 bk+1 1 1

p p q qp qk 1 k 1a a b b

=

1 1k 1 k 1p q

p qj j

j 1 j 1

a b , … (9)

so combining with (8) we see that the desired inequality (5) holds for n = k + 1.



NotesAssume now we have equality. Then we must have equality in both (8) and in (9).

On one hand, the equality in (8) forces p p p q q q1 2 k 1 2 ka , a , , a and b , b , , b to be proportional (since

we assume the case n = k). On the other hand, the equality in (9) forces p pk 1a , a and q q

k 1b , b to

be proportional (by the case m = 2). Since

k kp p q q

j jj 1 j 1

a a and b b ,

it is clear that p p p p1 2 k k 1a , a , , a , a and q q q q

1 2 k k 1b , b , , b , b are proportional.

5.1.6 Riesz-Hölder's Inequality

Statement: Let p and q be conjugate indices or exponents (numbers) and f Lp [a, b], g Lq [a, b];then show that

(i) f g L[a, b]

(ii) p qfg f g i.e.

1 1p qp q|fg| |f| |g|

with equality only when |f|p = |g|q a.e. for some non-zero constants and .

Lemma: If A and B are any two non-negative real numbers and 0 < < 1, then

A B1– A + (1 – ) B, with equality when A = B.

Proof: If either A = 0 or B = 0, then the result is trivial.

Let A > 0, B > 0

Consider the function

(x) = x – x, where 0 x < and 0 < < 1

1d xdx

and 2

12

d ( 1)xdx

.

Now solving ddx

= 0, we get x = 1.

Also at x = 1, 2

2

ddx

< 0 as 0 < < 1.

By calculus, (x) is maximum at x = 1, so

(x) (1) i.e. x – x l – . … (1)

Now, putting x = AB

, we get

A AB B

A1 or A B 1B



Notes Or A B1– – A (1 – ) B or

A B1– A + B (1 – ) … (2)

Obviously equality holds good only for x = 1, i.e. only when A = B.

Proof of Theorem

Note that when p = 1, q = , the proof of theorem is obvious. Let us assume that 1 < p < and1 < q < .

Now set = 1p

; p > 1 < 1

Therefore1q = 1 –

1 1 1p q

Putting these values of and 1 – in (2), we get

1 1p qA B A B

p q… (3)

If one of the functions f (x) and g (x) is zero a.e. then the theorem is trivial. Thus, we assume thatf 0, g 0 a.e. and hence the integrals

b b

p q

a a

|f| dx and |g| dx

are strictly positive and hence f p > 0, g q > 0.

Set f (x) = p q

g(x)f(x) , g(x)f g

and1pA = |f(x)|,

1qB g(x) .

Then (3) gives

|f(x) g(x)| qp g(x)f(x)

p q

Integrating, we get

b

a

|f(x) g(x)|dxb b

p p

a a

1 1|f(x)| dx |g(x)| dxp q

= b b qp

b b

p qa a

a a

|g(x)|1 |f(x)| 1dx dxp q

|f| dx |g| dx



Notes

=

b b

p q

a ab b

p q

a a

|f| dx |g| dx1 1p q

|f| dx |g| dx

= 1 1 1p q

.

Henceb

a

|f(x) g(x)|dx 1.

Putting the values of f (x) and g (x), we get

b

p qa

|f(x) g(x)|dxf g

1 or fg f p g q … (4)

Now f Lp [a, b], g Lq [a, b]

b b

p q

a a

|f| dx and |g| dx

p qf and g

Therefore, from (4), we have

fg 1 < fg L [a, b]

Also the equality will hold when A = B

i.e. |f (x)|p = |g (x)|q, a.e.

i.e.p

p

p

|f|iff

= q

q

q

|g|g

, a.e.

or if q pq

g |f| = p qp

f |g| , a.e.

or if we have got some non-zero constants ,

|f|p = |g|q, a.e.

Hence the theorem.

5.1.7 Riesz-Hölder's Inequality for 0 < p < 1

If 0 < p < 1 and p and q are conjugate exponents, and f Lp and g Lq, then

qp q|fg| f g , provided |g| 0 .

(In this case, the inequality is reversed than that of the case for 1 p < .)

Proof: Conjugacy of p, q 1 1 1p q



Notes 1 11q p

p 1 pq

pp 1q

If we take 1pP

and p 1P Q

, then 1 1 1p Q

and since 1 10 p 1 1 P 10 P

,

i.e. 1< P < < and also 1 p 11 p 0 1Q q Q

as 0 < p < 1 Q > 1.

P, Q are conjugate numbers with 1 < P < .

If we take |fg| = Fp and |g|q = GQ.

Then fg = 1 qp Q

g|f | |g| = 1 q pp Q qp|f| |g|

= |f|p.

f, g are non-negative measurable functions s.t.

Also f Lp and g Lq.

Applying the Hölder’s inequality for P, Q to the functions f and g, we get

|FG| F P G Q

p|f| 1 1P QP Q|F| |G| as |fg| = fg = |f|p

p|f|pp

qq|fg| |g|

1pp|f|

1qq|fg| |g|

1pp|f| 1

qq

|fg|

|g|, provided q|g| 0

|fg|1 1p qp q

p q|f| |q| f g



NotesTheorem 4: SCHWARZ or CAUCHY-SCHWARZ INEQUALITY statement: Let f and g be squareintegrable, i.e.

f, g L2[a, b]; then fg L [a, b] and fg f 2 g 2.

Proof: Let x [a, b] be arbitrary, then

[|f(x)| – |g(x)|]2 0

or 2|f(x)|.|g(x)| |f(x)2 + |g(x)|2.

On integrating, we get

b

a

2 |f(x)g(x)|dx b b

2 2

a a

|f(x) dx |g(x)| dx … (i)

Now f, g L2[a, b] f and g are measurable over [a, b] and b b

2 2

a a

|f(x)| dx , |g(x)| dx .

Using in (i), we get b

a

|f(x) g(x)|dx

Thus fg L [a, b].

Let a R be arbitrary. Then

( |f| + |g|)2 0

b

2

a

( |f| |g|) 0

or b b b

2 2 2

a a a

|f| dx 2 |fg|dx |g| dx 0

Write A = b b b

2 2

a a a

|f| dx, B 2 |fg|dx, C |g| dx

Then we have 2A + B + C 0 … (ii)

Now, if A = 0, then f(x) = 0 a.e. in [a, b] and hence B = 0 and both sides of the inequality to beproved are zero. Thus when A = 0, the inequality is trivial.

Again, let A 0. Writing = – B

2A in (ii), we get

2B BA B C 02A 2A

.

which gives B2 4AC.



Notes Now putting the values of A, B, C in last inequality, we have

2b

a

4 |fg|dx b b

2 2

a a

4 |f| |g|

orb

a

|f(x) (g(x)|dx 1/2 1/2b b

2 2

a a

|f(x)| |g(x)|

or fg f 2 g 2.

Note: The above theorem is a particular case of Hölder’s inequality.

Example: Let f, g be square integrable in the Lebesgue sense then prove f + g is alsosquare integrable in the Lebesgue sense, and f + g 2 f 2 + g 2.

Solution: By hypothesis f2 L [a, b], g2 L [a, b].

f2, g2 L [a, b] fg L [a, b]. [by Schwarz inequality]

Again (f + g)2 = f2 + g2 + 2fg L [a, b].

Hence (f + g) is square integrable, again, we have

b

2

a

(f g) = b b b

2 2

a a a

f g 2 fg

1 /2 1 /2b b b b

2 2 2 2

a a a a

f g 2 f g (by Schwarz inequality)

=

21 /2 1 /2b b

2 2

a a

f g

1 /2b

2

a

(f g)1 /2 1 /2b b

2 2

a a

f g

or f + g 2 f 2 + g 2.

Example: Prove that f + g 1 f 1 + g 1.

Solution: We know that |f + g| |f| + |g|.

Integrating both the sides.

|f g| |f| |g|

f + g 1 f 1 + g 1.



Notes5.2 Summary

The class of all measurable functions f (x) is known as Lp – space over [a, b], if Lebesgue-integrable over [a, b] for each p exists, 0 < p < , i.e.

b

p

a

|f| dx , (p 0)

The p-norm of any f Lp [a, b], denoted by f p, is defined as

1b p

pp

a

f |f| , 0 p

Let p, q > 1 be such that 1 1 1p q

, and let u and v be two non-negative numbers, at least

one being non-zero. Then the function f : [0, 1] R defined by

1q qf(t) ut v 1 t , t [0, 1],

has a unique maximum point at1

p q

p p

usu v

Let p and q be conjugate indices or exponents and f Lp [a, b], g Lq [a, b], then it is evidentthat

(i) f, g L [a, b]

(ii) fg f p g q i.e.

1 1p qp q|fg| |f| |g|

5.3 Keywords

Conjugate Numbers: Let p, q be any two n on-negative extended real numbers s.t. 1 1 1p q

, then

p, q are called (mutually) conjugate numbers.

Hölder's Inequality: Let a1, a2, …, an, b1, b2, …, bn be non-negative numbers. Let p, q > 1 be real

number with the property 1 1 1p q

. Then

n

j jj 1

a b

1 1n np q

p qj j

j 1 j 1

a b

Moreover, one has equality only when the sequences p p1 na , , a and q q

nb , , b are proportional.



Notes LP-Spaces: The class of all measurable functions f (x) is known as Lp-spaces over [a, b], if Lebesgue– integrable over [a, b] for each p exists, 0 < p < , i.e.

b

p

a

|f| dx , (p 0)


p-norm: The p-norm of any f Lp [a, b], denoted by f p, is defined as

f p =

1b p

p

a

|f| , 0 < p < .


1. If f and g are non-negative measurable functions, then show that in Hölder’s inequality,equality occurs iff some constants s and t (not both zero) such that sfp + tgq = 0.

2. State and prove Hölder’s Inequality.


Books G.H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge University Press,(1934)

L.P. Kuptsov, Hölder inequality, Springer (2001)

Kenneth Kuttler, An Introduction of Linear Algebra, BRIGHAM Young University,2007

Online links www.m–hiKari.com

www.math.Ksu.edu

www.tandfonline.com


Unit 6: Minkowski Inequalities

NotesUnit 6: Minkowski Inequalities

CONTENTS

Objectives

Introduction

6.1 Minkowski Inequalities

6.1.1 Proof of Minkowski Inequality Theorems

6.1.2 Minkowski Inequality in Integral Form

6.2 Summary

6.3 Keywords



Objectives


Define Lp-space, conjugate numbers and norm of an element of Lp-space.

Understand Minkowski inequality.

Solve problems on Minkowski inequality.

Introduction

In mathematical analysis, the Minkowski inequality establishes that the L p spaces are normedvector spaces. Let S be a measure space, let 1 p and let f and g be elements of Lp (s). Thenf + g is in Lp (s), we have the triangle inequality

f + g p f p + g p

with equality for 1 < p < if and only if f and g are positively linearly dependent, i.e. f = g forsome 0. In this unit, we shall study Minkowski’s inequality for 1 p < and for 0 < p < 1. Weshall also study almost Minkowski’s inequality in integral form.

6.1 Minkowski Inequalities

Here, the norm is given by:

f p = 1/p

p|f| d

if p < , or in the case p = by the essential supremum

f = ess supx S |f (x)|.

The Minkowski inequality is the triangle inequality in Lp(S). In fact, it is a special case of themore general fact



Notes f p =

g 1q

sup |fg|d , 1/p + 1/q = 1

where it is easy to see that the right-hand side satisfies the triangular inequality.

Like Holder's inequality, the Minkowski inequality can be specialized to sequences and vectorsby using the counting measure:

1/p 1/p 1/pn n np p p

k k k kk 1 k 1 k 1

|x y | |x | |y |

for all real (or complex) numbers x1, ..., xn, y1, ..., yn and where n is the cardinality of S (thenumber of elements in S).

Thus, we may conclude that

If p > 1, then Minkowski's integral inequality states that

1 /p 1 /p 1 /pb b bp p p

a a a|f(x) g(x)| dx |f(x)| dx |g(x)| dx

Similarly, if p >1 and ak, bk > 0, then Minkowski's sum inequality states that

1/p 1/p 1/pn n np p p

k k k kk 1 k 1 k 1

|a b | |a | |b |

Equality holds iff the sequences a1, a2, ... and b1, b2, ... are proportional.

6.1.1 Proof of Minkowski Inequality Theorems

Theorem 1: State and prove Minkowski inequality. If f and g Lp (1 p < ), then f + g Lp and

f + g p f p + g p.

or

Let 1 p . Prove that for every pair f, g Lp {0, 1}, the function f + g Lp {0, 1} and that f + g p f p + g p. When does equality occur?

Or

Suppose 1 p . Prove that for any two functions f and g in Lp [a, b]

1 1 1b b bp p p

p p p

a a a

|f g| |f| dx |g| dx

Proof: When p = 1, the desired result is obvious.

If p = , then

|f| f a.e.

|g| g a.e.



Notes|f + g| f + |g|

f + g a.e.

f + g f + g

Hence the result follows in this case also. Thus, we now assume that 1 < p < .

Since Lp is a linear space, f + g Lp.

Let q be conjugate to p, then 1 1 1p q

.

Now (f + g) Lp

(f + g)p/q Lq

Since 1 1 1 1 1 p 11 1p q q p q p

qq p 1 p(p 1) p, |f g| |f g|

and therefore |f + g|p–1 Lp (f + g)p/q Lp because p – 1 = pq

.

On applying Hölder’s inequality for f and (f + g)p/q, we get

pq|f||f g| dx

11 p pp qp q|f| ) dx |f g| dx

orp

q|f||f g| dx1 1

p qp p|f| ) dx |f g| dx … (1)

Since g Lp, therefore interchanging f and g in (1), we get

pq|g||f g| dx

1 1p q

p p|g| ) dx |f g| dx … (2)

Adding, we get

p pq q|f||f g| dx |g||f g| dx

1 1 1p p q

p p p|f| dx |g| dx |f g| dx … (3)

Now |f + g|p = p 1f g f g

But 1 1 1p q

p p1 p p 1q q

|f + g|p = p

qf g f g

p

q|f| |g| |f g|



Notesor |f + g|p

p pq q|f||f g| |g||f g|

Integrating, we get

p|f g| dx p p

q q|f||f g| |g||f g| dx … (4)

Using (3), relation (4) becomes

p|f g| dx 1 1

p p 1p p p q|f| dx |g| dx |f g| dx

Dividing each term by 1

p q|f g| dx , we get

11 qp|f g| dx

1 1p p

p p|f| dx |g| dx

But 1 1 1p q

1 11q p

So1

pp|f g| dx

1 1p p

p p|f| dx |g| dx

or f + g p f p + g p

Hence the proof.

Note Equality hold in Minkowski’s inequality if and only if one of the functions f and gis a multiple of the other.

Theorem 2: Minkowski’s inequality for 0 < p < 1. If 0 < p < 1 and f, g are non-negative functionsin Lp, then

f + g p f p + g p.

Proof: For this proceed as in theorem Minkowski’s inequality and applying the Hölder’sinequality for 0 < p < 1 for the functions f Lp and (f + g)p/q Lq, we get

p/q|f||f g| 1 p 1 q

p p/q q|f| |f g| )

p/q|f||f g| 1 p 1 q

p p|f| |f g| … (i)

Also g Lp, proceeding as above, we get

p/q|g||f g| 1 p 1 q

p p|g| dx |f g| … (ii)



NotesAdding these two,

p/q|f g| (|f| |g|) 1 1 1

p p pp p q|f| |g| |f g| … (iii)

Also 1 1 1p q

p1 pq

|f + g|p = p p

q q|f g||f g| |f| |g| |f g| , as f 0, g 0

p|f g| = 1

q1 1p p pp p|f| |g| |f g|

Dividing by 1

qp|f g| , we get

1 (1/q)p|f g|

p pf g

1/qp|f g| f p + g p

f + g p f p + g p

6.1.2 Minkowski Inequality in Integral Form

Statement: Suppose f : × is Lebesgue measurable and 1 p < . Then

1/p

h(x, y) dy dx 1/p

ph(x, y) dx dy

Proof: By an approximation argument we need only consider h of the form

N

j jj 1

h(x,y) f (x)1 f (y), (x,y) ,

where N is a positive integer, f j is Lebesgue measurable, and F j Ln, j = 1, … Ni and Fi Fj = if1 i < j N. We use Minkowski’s inequality to estimate

1/p

h(x,y) dy dx =

1/ppN N 1/p

pj j j i

j 1 j 1dx

F f (x) F f (x) dx

But

1/pp

h(x,y) dx dy = N N1/p 1/p

p pj

F Fi jj 1 j 1

|h(x,y)| dx |f (x)| dx



NotesExample: If <fn> is a sequence of functions belonging to L2(a, b) and also f L2 (a, b) and

Lim fn – f 2 = 0, then prove that

b b

2 2n

a a

f dx Lim f dx

Solution: By Minkowski’s inequality, we get

n 2 2f f fn – f 2

n 2Lim f f Lim fn – f 2 = 0

n 2Lim f f = 0 Lim fn 2 = f 2

1/2 1/2b b

2 2n

a a

Lim (f ) dx f dx b b

2 2n

a a

Lim f dx f dx .

6.2 Summary

The class of all measurable function f (x) is known as Lp space over [a, b], if Lebesgueintegrable over [a, b] for each p exists, 0 < p < .

If f and g Lp (1 p ), then f + g Lp and f + g p f p + g p.

6.3 Keywords

Lp-space: The class of all measurable functions f (x) is known as Lp-space over [a, b], if Lebesgue-integrable over [a, b] for each exists, 0 < p < , i.e.,

b

p

a

|f| dx , (p 0)


Minkowski Inequality in Integral Form: Suppose f : × is Lebesgue measurable and 1 p < . Then

1/p

h(x, y) dy dx 1/p

ph(x, y) dx dy

Minkowski Inequality: Minkowski inequality establishes that the Lp spaces are normed vectorspaces. Let S be a measure space, let 1 p and let f and g be elements of Lp (s). Then f + g is inLp (s), we have the triangle inequality

f + g p f p + g p

with equality for 1 < p < if and only if f and g are positively linearly dependent.




1. If f, g are square integrable in the Lebesgue sense, prove that f + g is also square integrableand

f + g 2 f 2 + g 2.

2. If | < p < , then show that equality can be true, iff there are non-negative constants and, such that f = g.


Books Books: Stein, Elias (1970). Singular Integrals and Differentiability Properties ofFunctions. Princeton University Press.

Hardy, G.H.; Littlewood, J.E.; Polya, G. (1952). Inequalities, CambridgeMathematical Library (second ed.). Cambridge: Cambridge University Press.

Online links Mathworld.wolfram.com>Calculus and Analysis>Inequalities

Planet math.org/Minkowski In-equality.html



Notes Unit 7: Convergence and Completeness

CONTENTS

Objectives

Introduction

7.1 Convergence and Completeness

7.1.1 Convergent Sequence

7.1.2 Cauchy Sequence

7.1.3 Complete Normed Linear Space

7.1.4 Banach Space

7.1.5 Summable Series

7.1.6 Riesz-Fischer Theorem

7.2 Summary

7.3 Keywords



Objectives


Understand convergence and completeness.

Understand Riesz-Fischer theorem.

Solve problems on convergence and completeness.

Introduction

Convergence of a sequence of functions can be defined in various ways, and there are situationsin which each of these definitions is natural and useful. In this unit, we shall start with thedefinition of convergence and Cauchy sequence and proceed with the topic completeness of Lp.

7.1 Convergence and Completeness

7.1.1 Convergent Sequence

Definition: A sequence <xn> in a normal linear space X with norm . is said to converge to anelement x X if for arbitrary > 0, however small, n0 N such that xn – x < , n > n0.

Then we write nnlim x x .


Unit 7: Convergence and Completeness

Notes7.1.2 Cauchy Sequence

Definition: A sequence <xn> in a normal linear space (X, . ) is said to be a Cauchy sequence if forarbitrary > 0, n0 N s.t.

xn – xm < , n, m n0.


Definition: A normed linear space (X, . ) is said to be complete if every Cauchy sequence <xn> init converges to an element x X.

7.1.4 Banach Space

Definition: A complete normed linear space is also called Banach space.

7.1.5 Summable Series

Definition: A series nn 1

u in N1 is said to be summable to a sum u if u N1 and nnlim S u , where

Sn = u1 + u2 + … + un

In this case, we write

nn 1

u u .

Further, the series nn 1

u is said to be absolutely summable if nn 1

u .

7.1.6 Riesz-Fischer Theorem

Theorem: The normed Lp-spaces are complete for (p 1).

Proof: In order to prove the theorem, we shall show that every Cauchy sequence in Lp [a, b] spaceconverges to some element f in Lp-space. Let <fn> be one of such sequences in Lp-space. Then forgiven > 0, a natural number n0, such that

m, n n0 fm – fn p < ,

since is arbitrary therefore taking 12

, we can find a natural number n1 such that

for all m, n n1 fm – fn p < 12

Similarly, taking k

1 , k N,2

we can find a natural number nk, such that

for all m, n nk fm – fn p < k

12



NotesIn particular, m > nk fm – nk

f < k

12 .

Obviously n1 < n2 < n3 … < nk < …

i.e. <nk> is a monotonic increasing sequence of natural numbers.

Set gk = nkf , then from above, we have

g2 – g1 p = n n2 2 p

1f f2 ,

g3 – g2 p = n n 23 2 p

1f f2

,

… … … … …

… … … … …

gk + 1 – gk p = n n kk 1 k p

1f f2

.

… … … … …

… … … … …

Adding these inequalities, we get

k 1 k pk 1

g g < kk 1

1 12

… (i)

Thus kk 1 pk 1

g g is convergent. Define g such that

g (x) = 1 kk 1 pk 1

g (x) g g if R.H.S. is convergent … (ii)

and g (x) = , if right hand side is divergent.

Now,b

p

a

|g(x)| dx =

1b n p

p

1 kk 1 dxnk 1a

lim |g (x) g g

or g p = n

1 kk 1p pnk 1

lim g g g (By Minkowski’s inequality)

= g1 p + kk 1 pk 1

g g < g1 p + 1, [by (i)]

g p < g Lp [a, b].

Let E = {x [a, b] : g (x) = }.



NotesNow we define a function f such that

f (x) = 0, x E

and f (x) = g1 (x) + kk 1k 1

g g , for x [a, b] but x E,

or f (x) = m 1

1 kk 1mk 1

lim g g g , for x E

= mmlim g (x)

Thus f (x) = 0, for x E and

f (x) = mmlim g (x) for x E.

f (x) = mmlim g (x) a.e. in [a, b]

or mmlim g f = 0 a.e. in [a, b] … (iii)

Also, gm (x) = m 1

1 kk 1k 1

g g g

|gm| |g1| + m 1

kk 1k 1

g g

|g1| + kk 1k 1

g g = g,

|gm| g, m N

mmlim g (x) g

(iii) |f| g.

Again, |gm – f| |gm| + |f| g + g = 2g.

|gm – f| 2 g.

Thus there exists a function g Lp [a, b] s.t.

|gm – f| 2g, m

and mmlim g f = 0 a.e. in [a, b] … (iv)

Applying Lebesgue dominated convergence theorem,

bp

mma

lim g f dx = b b

pmm

a a

lim g f dx 0 dx 0 [Using (iv)]



Notes 1b p

pmm

a

lim g f dx = 0

m pmlim g f = 0

nm pmlim f f = 0 as gm = nm

f

nm pf f < .

Also m nm pf f < .

fm – f p = m n nm m pf f f f

m n nm mp pf f f f

< ( + ) = .

Hence m pmlim f f = 0

or mmlim f = f Lp [a, b].

This proves the theorem.

Alternative Statement of this Theorem

A convergent sequence <fn> in Lp-spaces has a limit in Lp-space.

Or

Every Cauchy sequence <fn> in the Lp-space converges to a function in Lp-space.

Theorem: Prove that a normed linear space is complete iff every absolutely summable sequenceis summable.

Proof: Necessary part

Let X be a complete normed linear space with norm . and <fn> be an absolutely summablesequence of elements of X

nn 1

f = M < ,

For arbitrary > 0, however small, no N

s.t. nn no

f < , … (i)

Now, if Sn = n

ii 1

f , then n m no, we get



Notes Sn – Sm p =

n

ii m 1

f n

ii m 1

f

ii no

f

Sequence <Sn> of partial sums is a Cauchy sequence

<Sn> converges.

Sequence <fn> is summable to some element S X.

But X is a complete space. Therefore <Sn> will converge to some element S X.

Sufficient part: Given that every absolutely summable sequence in the space X is summable.

To show that X is complete.

Let <fn> be a Cauchy sequence in X.

For each positive integer k, we can choose a number nk N such that

fn – fm < k

12

, n, m nk … (ii)

We can choose these nk’s such that nk+1 > nk.

Then nk 1f is a subsequence of <fn>.

Setting g1 = n1f and gk = n nk k 1

f f , (k > 1), we get a sequence <gk> s.t. its kth partial.

Sum = Sk = g1 + g2 + … + gk = n n n n n n1 2 1 k k 1 kf (f f ) (f f ) f .

Now, gk = n n k 1k k 1

1f f2

, [by (ii)], k > 1

k 1 k 1k 1 k 2

1g g2

= g1 + 1 (a finite quantity)

The sequence <gk> is absolutely summable and hence by the hypothesis, it is a summablesequence.

The sequence of partial sums of this sequence converges to some S X.

The sequence <Sk> converges and hence nkf converges to some f X.

Now, we shall show that the limit fn = f.

Again, since <fn> is a Cauchy sequence, we get that for each > 0, however small, n N s.t.n, m > n .

fn – fm < 2

.

Also since nkf f , n N such that k n ,



Notesnk

f f2 .

Choosing a number k, as large that k > n and nk > n , we get

n n n nk kf f f f f f ,

2 2

n > n , we obtain fn – f < , where is an arbitrary quantity.

fn f X and hence X is a complete space.

Theorem: Let {fn} be a sequence in Lp, 1 p < , such that fn f a.e. and that f Lp.

If n pnlim f = f p, then

n pnlim f f = 0.

Proof: Without any loss of generality, we may assume that each fn 0 a.e. so that f is also 0 a.e.since the result in general case follows by considering f = f+ – f–.

Now, let a and b be any pair of non-negative real numbers, we have

|a – b|p 2p (|a|p + |b|q),

1 p <

So, we get

2p (|fn|p + (|f|p) – |fn – f|p 0 a.e.

Thus, by Fatou’s Lemma and by the given hypothesis,

We get

p 1 p2 |f| = p ppp

n nnlim 2 f f f f

p pppn nn

lim inf 2 f f f f

= p ppp 1 pn nn n

2 lim f 2 f lim inf f f

= ppp 1nn

2 f limsup f f .

Since pf it follows that

pnn

lim sup f f 0.

Therefore p pn nn n

limsup f f liminf f f 0 ,

So that pnn

lim f f = 0



Notes1pp

nnlim f f dt = 0

n pnlim f f = 0.

Theorem: In a normed linear space, every convergent sequence is a Cauchy sequence.

Proof: Let the sequence <xn> in a normed linear space N, converges to a point xo N1. We shallshow that it is a Cauchy sequence.

Let > 0 be given. Since the sequence converges to xo a positive integer mo s.t.

n mo xn – xo < /2 … (1)

Hence for all m, n mo, we have

xm – xn = xm – xo + xo – xn

xm – xo + xo – xn

< 2 2

= by (1)

It follows that the convergent sequence <xn> is a Cauchy sequence.

Theorem: Prove that L [0, 1] is complete.

Proof: Let (fn) be any Cauchy sequence in L , and let

Ak = {x : |fk (x)| > fk },

Bm, n = {x : |fk (x) – fm (x)| > fk – fm }.

Then m (Ak) = 0 = m (Bm, n) (k, m, n = 1, 2, 3, …),

So that if E is the union of these sets, we have m (E) = 0.

Now, if x F = [0, 1] – E, then

|fk (x) fk

|fn (x) – fm (x) fn – fm 0 as n, m .

Hence the sequence (fn) converges uniformly to a bounded function on F.

Define f : [0, 1] R by

f (x) = n

nlim f (x) if x F

0, if x E

Then f L and fn – f 0 as n .

Thus L is

Hence proved.

7.2 Summary

A sequence <xn> in a normal linear space X with norm . is said to converge to an elementx X if for arbitrary > 0, however small, no N s.t. xn – x < , n > no. Then we write

nnlim x x .



Notes A complete normed linear space is also called Banach space.

The normed Lp-spaces are complete for (p 1).

A convergent sequence <fn> in Lp-spaces has a limit in Lp-space.

A normed linear space is complete iff every absolutely summable sequence is summable.

In a normed linear space, every convergent sequence is a Cauchy sequence.

7.3 Keywords

Banach Space: A complete normed linear space is also called Banach space.

Cauchy Sequence: A sequence <xn> in a normal linear space (X, . ) is said to be a Cauchysequence if for arbitrary > 0, n0 N s.t.

xn – xm < , n, m n0.

Complete Normed Linear Space: A normed linear space (X, . ) is said to be complete if everyCauchy sequence <xn> in it converges to an element x X.

Convergence almost Everywhere: Let <fn> be a sequence of measurable functions defined over ameasurable set E. Then <fn> is said to converge almost everywhere in E if there exists a subset Eo

of E s.t.

(i) fn (x) f (x), x E – Eo.

and (ii) m (Eo) = 0.

Convergent Sequence: A sequence <xn> in a normal linear space X with norm . is said toconverge to an element x X if for arbitrary > 0, however small, n0 N such that xn – x <

, n > n0.

Then we write nnlim x x .

Normed Linear Space: A linear space N together with a norm defined on it, i.e., the pair (N, )is called a normed linear space.

Summable Series: A series nn 1

u in N1 is said to be summable to a sum u if u N1 and nnlim S u ,

where

Sn = u1 + u2 + … + un


1. Prove that np is complete.

2. Prove that the vector space L equipped with L

is a complete vector space.

3. Suppose f L is supported on a set of finite measure.

Then f Lp for all p < , and

L Lf f as p .



NotesIf f Lp (p > 0), f 0 and fn = min (f, n) (n N), show that fn Lp and n pnlim f f = 0.


Books H.L. Royden, Real analysis.

Walter Rudin, Real and Complex Analysis, Third, McGraw-Hill Book Co., NewYork, 1987.

Online links www.public.iastate.edu

www.scribd.com/doc/49732162/103.Convergence-and-Completeness.



Notes Unit 8: Bounded Linear Functional on the Lp-spaces

CONTENTS

Objectives

Introduction

8.1 Bounded Linear Functionals on Lp-spaces

8.1.1 Linear Functional

8.1.2 Bounded Linear Functional

8.1.3 Bounded Linear Functional on Lp-spaces

8.1.4 Norm

8.1.5 Continuous Linear Functional

8.1.6 Theorems

8.2 Summary

8.3 Keywords



Objectives


Understand bounded linear functional on Lp-spaces

Understand related theorems.

Solve problems on bounded linear functionals.

Introduction

In this unit, we obtain the representation of bounded linear functionals on Lp-space. We shallalso study about linear functional, continuous linear functionals and norm of f *

p . Further, weshall prove important theorems on bounded linear functionals.

8.1 Bounded Linear Functionals on Lp-spaces

8.1.1 Linear Functional

Definition: Let N1 be a normed space over a field R (or C). A mapping f : N1 R (or C) is called alinear functional on N1 if f ( x + y) = f (x) + f (y), x, y N1 and , R (or C).


Definition: A linear functional f on a normed space N1 is said to be bounded if there is a constantk > 0 such that

|f (x)| k x , x N1 … (1)


Unit 8: Bounded Linear Functional on the Lp-spaces

NotesThe smallest constant k for which (1) holds is called the norm of f, written f .

Thus f = 1|f(x)|sup : x 0 and x N

x or equivalently

f = sup {|f (x)| : x X and x = 1}.

Also |f (x)| f x x N1.

Definition: Let p R, p > 0. We define Lp = Lp [0, 1] to be the set of all real-valued functions on[0, 1] such that

(i) f is measurable and (ii) f p =

11 p

p

0

|f| < .

Note L1 or simply L denotes the class of measurable function f (x) which are alsoL-integrable.

8.1.3 Bounded Linear Functional on Lp-spaces

If x p and f is bounded linear functional on p , then f has the unique representation of the

form as an infinite series

f (x) = k kk 1

x f(e )

8.1.4 Norm

The norm of f *p is given by

f =

1q

qk

k 1

|f(e )|

Likewise in finite dimensional case, the bounded linear functionals are characterised by thevalues they assume on the set ek, k = 1, 2, 3, … .

8.1.5 Continuous Linear Functional

A linear functional f is continuous if given > 0 there exists > 0 so that

|f (x) – f (y)| whenever x – y .

8.1.6 Theorems

Theorem 1: Suppose 1 p < , and 1 1 1p q

, then, with = Lp we have

* = qL ,



Notes in the following sense: For every bounded linear functional on Lp there is a unique g Lq sothat

(f) = p

x

f(x)g(x)d (x), for all f L

Moreover,*

= qLg .

This theorem justifies the terminology where by q is usually called the dual exponent of p.

The proof of the theorem is based on two ideas. The first, as already seen, is Hölder’s inequality;to which a converse is also needed. The second is the fact that a linear functional on Lp, 1 p< , leads naturally to a (signed) measure . Because of the continuity of the measure isabsolutely continuous with respect to the underlying measure , and our desired function g isthen the density function of in terms of .

We begin with:

Lemma: Suppose 1 p, q , are conjugate exponents.

(i) If g Lq, then qLf pL 1

g sup fg .

(ii) Suppose g is integrable on all sets of finite measure and

f pL 1f simple

sup fg = M <

Then g Lq, and qLg = M.

For the proof of the lemma, we recall the signum of a real number defined by

sign (x) = 1 if x 01 if x 00 if x 0

Proof: We start with (i). If g = 0, there is nothing to prove, so we may assume that g is not 0 a.e.,

and hence qLg 0 . By Hölder’s inequality, we have that

qLg

f pL 1

sup fg .

To prove the reverse inequality we consider several cases.

First, if q = 1 and p = , we may take f (x) = sign g (x). Then, we have Lf 1 and

clearly 1Lfg g .

If 1 < p, q < , then we set f (x) = |g (x)|q–1 sign q 1qL

g(x) g . We observe that

p p(q 1) p(q 1)p qL L

f g(x) d g 1 since p (q – 1) = q, and that qLfg g .



Notes Finally, if q = and p = 1, let > 0, and E a set of finite positive measure, where |g (x)|

Lg . (Such a set exists by the definition of

Lg and the fact that the measure is -

finite). Then, if we take f (x) = E (x) sign g (x)/ (E), where E denotes the characteristicfunction of the set E, we see that 1L

f 1 , and also

E

1fg |g| g(E)

.

This completes the proof of part (i).

To prove (ii) we recall that we can find a sequence {gn} of simple functions so that |gn (x) |g (x)|

while gn (x) g (x) for each x. When p > 1 (so q < ), we take fn (x) = |gn (x)|q–1 sign q 1qn L

g(x) g .

As before, pn Lf 1 . However

qn

qn q 1 n Lqn L

g (x)f g g

g,

and this does not exceed M. By Fatou’s Lemmas if follows that q q|g| M , so g Lq with

qLg M . The direction qL

g M is of course implied by Hölder’s inequality. When p = 1 theargument is parallel with the above but simpler. Here we take fn (x) = (sign g (x)) En (x), whereEn is an increasing sequence of sets of finite measure whose union is X. The details may be left tothe reader.

With the lemma established we turn to the proof of the theorem. It is simpler to consider first thecase when the underlying space has finite measure. In this case, with the given functional onLp, we can then define a set function by

(E) = ( E),

where E is any measurable set. This definition make sense because E is now automatically in Lp

since the space has finite measure. We observe that

| (E)| C ( (E))1/p … (1)

where C is the norm of the linear functional, taking into account the fact that 1/ppE L

(E) .

Now the linearity of clearly implies that is finitely-additive. Moreover, if {En} is a countable

collection of disjoint measurable sets, and we put *n 1 n N n N 1 nE E , E E , then obviously

N

*E nENn 1

E .

Then N

*N n

n 1(E) E (E ) . However *

NE 0 , as N because of (1) and the

assumption p < . This shows that is countably additive and moreover (1) also shows us that is absolutely continuous with respect to .

We can now invoke the key result about absolutely continuous measures, the Lebesgue-Radon

– Nykodin theorem. It guarantees the existence of an integrable function g so that (E) = E

g du



Notesfor every measurable set E. Thus we have ( E) Eg d . The representation (f) fg d

then extends immediately to simple function f, and by a passage to the limit, to all f Lp since the

simple functions are dense in Lp, 1 p < . Also by lemma, we see that qLg .

To pass from the situation where the measure of X is finite to the general case, we use an

increasing sequence {En} of sets of finite measure that exhaust X, that is, X = n 1 nE . According

to what we have just proved, for each n there is an integrable function gn on En (which we can set

to be zero in cnE ) so that

(f) = nfg d … (2)

whenever f is supported in En and f Lp. Moreover by conclusion (ii) of the lemma pn Lg .

Now it is easy to see because of (2) that gn = gm a.e. on mE , whenever n m. Thus limn gn (x)

= g (x) exists for almost every x, and by Fatou’s lemma, g Lq . As a result we have that

(f) f g du for each f Lp supported in En, and then by a simple limiting argument, for all f

Lp supported in En. The fact that g Lq is already contained in Hölder’s inequality andtherefore the proof of the theorem is complete.

Theorem 2: Let f be a linear functional defined on a normed linear space N, then f is bounded f is continuous.

Proof: Let us first show that continuity of f boundedness of f.

If possible let f is continuous but not bounded. Therefore, for any natural number n, howeverlarge, there is some point xn such that

|f (xn)| n xn … (1)

Consider the vector nn

n

xyn x

so that

yn = 1n

.

yn 0 as n .

yn 0 in the norm.

Since any continuous functional maps zero vector into zero, and f is continuous f (yn) f (0) = 0.

But |f (yn)| = nn

1 f(x )n x … (2)

It now follows from (1) and (2) that

|f (yn)|> 1, a contradiction to the fact that

f (yn) 0 as n .

Thus if f is bounded then f is continuous.



NotesConversely, let f is bounded. Then for any sequence (xn), we have

|f (xn)| k xn n = 1, 2, … and k 0.

Let xn 0 as n then

f (xn) 0

f is continuous at the origin and consequently it is continuous everywhere.

This completes the proof of the theorem.

Theorem 3: If L is a linear space of all n-tuples, then

(i) n np q*

(ii) n n1 *

(iii) n n1*

Proof: Let (e1, e2, …, en) be a standard basis for L so that any x = (x 1, x2, …, xn) L can be writtenas x = x1e1 + x2e2 + … + xnen.

If f is a scalar valued linear function defined on L, then we get

f (x) = x1 f (e1) + x2f (e2) + … + xnf (en) … (1)

f determines and is determined by n scalars y i = f (ei).

Then the mapping

y = (y1, y2, … yn) f where f (x) = n

i ii 1

x y is an isomorphism of L onto the linear space L of all

function f. We shall establish (i) – (iii) by using above given facts.

(i) If we consider the space L = np (1 p < ) with the pth norm, then f is continuous and L

represents the set of all continuous linear functionals on np so that L =

*np .

Now for y f as an isometric isomorphism we try to find the norm of y’s.

For 1 < p < , we show that

np * = n

p .

For x np , we have defined,

x =

1n p

pi

i 1

|x |

Now |f (x)| = n n

i i i ii 1 i 1

x y |x ||y |

By using Holder’s inequality, we get

n

i ii 1

|x y | 1 1

n np qp q

i ii 1 i 1

|x | |y | so that



Notes

|f (x)|

1 1n nq p

q pi i

i 1 i 1

|y | |x |

Using the definition of norm for f, we get

f

1n q

qi

i 1

|y | … (2)

Consider the vector, defined by

xi = q

ii

i

|y | , y 0y

and xi = 0 if yi = 0 … (3)

Then, x =

11pn n pqp

p ii

ii 1 i 1

|y ||x ||y |

… (4)

Since q = p (q – 1) we have from (4),

x =

1n q

qi

i 1

|y | … (5)

Now f (x) = n n q

ii i i

ii 1 i 1

|y |x y yy … (4)

= n

qi

i 1

|y | (By (3))

So that

nq

ii 1

|y | = |f (x) f x … (6)

From (5) and (6) we get,

11n pq

ii 1

|y | f

1n q

qi

i 1

|y | f … (7)

Also from (2) and (7) we have

f =

1n q

qi

i 1

|y | , so that



Notesy f is an isometric isomorphism.

Hence np * = n

q .

(ii) Let L = n1 with the norm defined by

x = n

ii 1

|x |

Now f defined in (1), above is continuous as in (i) and L here represents the set of continuous

linear functional on n1 so that

L = n1 * .

We now determine the norm of y’s which makes y f an isometric isomorphism.

Now, f (x) = n

i ii 1

x y

n

i ii 1

|x ||y |

Butn

i ii 1

|x ||y | n

i ii 1

max. |y | |x | so that

f (x) n

i ii 1

max. |y | |x |.

From the definition of the norm for f, we have

f = max. {|yi| : i = 1, 2, …, n} … (8)

Now consider the vector defined as follows:

If |yi| = i1 i nmax |y | , let us consider the vector x as

xi = ii i i1 i n

i

|y |when|y | max |y | and x 0y

… (9)

otherwise

From the definition, xk = 0 k i, so that we have

(x) = iyy = 1

Further | f (x)| = n

i ii 1

(x y ) = |yi|.

Hence |yi| = |f (x)| f x



Notes |yi| f or max {|yi|} [ x = 1]

f … (10)

From (8) and (10), we obtain

f = max. {|yi|} so that

y f is an isometric isomorphism of L to n1 * .

Hence n n1 * .

(iii) Let L = n with the norm

x = max {|xi| : 1, 2, 3, …, n}.

Now f defined in (1) above is continuous as in (1). Let L represents the set of all continuous

linear functionals on n so that

L = n * .

Now we determine the norm of y’s which makes y f as isometric isomorphism.

|f (x)| = n n

i i i ii 1 i 1

x y |x ||y |.

But n

i ii 1

|x ||y | n

i ii 1

max (|x |) |y |

Hence we have

|f (x)|n n

i ii 1 i 1

|y | ( x ) so that f |y | … (11)

Consider the vector x defined by

xi = i

i

|y |y

when yi 0 and xi = 0 otherwise. … (12)

Hence x = i

i

|y |max 1|y |

.

and |f (x)| = n n

i i ii 1 i 1

|x y | |y |

Thereforen

ii 1

|y | = |f (x)| f x = f .

n

ii 1

|y | f … (13)



NotesIt follows now from (11) and (13) that

f = n

ii 1

|y | so that y f is an isometric isomorphism.

Hence, n n1* .


Note We need the signum function for finding the conjugate spaces of some infinitedimensional space which we define as follows:

If is a complex number, then

sgn if 0| |0 if 0

(i) |sgn | = 0 if = 0 and | sgn | = 1 if 0

(ii) sgn = 0 if = 0 and sgn = | | =| |, if 0.

Theorem 4: The conjugate space of p is q , where

1 1p q

= 1 and 1 < p < .

or *p q .

Proof: Let x = (xn) p so that pn

n 1

|x | … (1)

Let en = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the n th place.

en p for n = 1, 2, 3, … .

We shall first determine the form of f and then establish the isometric isomorphism of *p onto

q .

By using (en), we can write any sequence

(x1, x2, … xn, 0, 0, 0, …) in the form n

k kk 1

x e and

n

k k n 1 n 2k 1

x x e (0, 0, 0, , x , x , ).



Notes

Now

1n p

pk k k

k 1 k n 1

x x e |x | … (2)

The R.H.S. of (2) gives the remainder after n terms of a convergent series (1).

Hence1p

pk

k n 1

|x | 0 as n . … (3)

From (2) and (3) if follows that

x = k kk 1

x e … (4)

Let f p* and Sn = n

k kk 1

x e then

Sn x as n (Using (4))

Since f is linear, we have

f (Sn) = n

k kk 1

x f(e ) .

Also f is continuous and Sn x, we have

f (Sn) f (x) as n .

f (x) = n

k kk 1

x f (e ) … (5)

which gives the form of the functional on p .

Now we establish the isomeric isomorphism of p* onto q , for which proceed as follows:

Let f (ek) = k and show that the mapping

T : p* q given by

T (f) = ( 1, 2, …, k, …) is an isomeric isomorphism of p* onto q .

First, we show that T is well defined.

For let x p , where x = ( 1, 2, …, n, 0, 0, …) where

k = g 1

kk 1 k n| | sgn ,n k0

| k| = | k|q–1 for 1 k n.

| k|p = p(q 1)

k| | = | k|q. 1 1 q p(q 1) qp q



NotesNow q 1 q 1k kk k k k k| | sgn | | sgn

k k = | k|q = | k|p … (7)

(Using property of sgn function)

x =

1n p

pk

k 1

| |

x =

1n p

pk

k 1

| |

=

1n q

qk

k 1

| | … (8)

Since we can write

x = n

k kk 1

e , we get

f (x) = n n

k k k kk 1 k 1

f (e )

f (x) = n

qk

k 1

| | (Using (7)) … (9)

We know that for every x p

| f (x) | f x ,

which upon using (8) and (9), gives

|f (x)| 1

n n pq q

k kk 1 k 1

| | f | |

which yields after simplification.

1n p

qk

k 1

| | f … (10)

Since the sequence of partial sum on the L.H.S. of (10) is bounded; monotonic increasing, itconverges. Hence

1n p

qk

k 1

| | f … (11)

So the sequence ( k) which is the image of f under T belongs to q and hence T is well defined.



Notes We next show that T is onto q .

Let ( k) q , we shall show that is a g *p such that T maps g into ( k).

Let x p so that

x = n

k kk 1

x e

We shall show that

g (x) = n

k kk 1

x is the required g.

Since the representation for x is unique, g is well defined and moreover it is linear on p . To

prove it is bounded, consider

|g (x)| = n n

k k k kk 1 k 1

x x

1 1n np q

qpk k

k 1 k 1

x (Using Hölder’s inequality)

|g (x)|

1n q

qk

k 1

x | |

g is bounded linear functional on p .

Since ek p for k = 1, 2, …, we get

g (ek) = k for any k so that

Tg = ( k) and T is on p* onto q .

We next show that

Tf = f so that T is an isometry.

Since Tf q , we have from (6) and (10) that

1q

qk

k 1

= Tf f

Also, x p x = k kk 1

x e . Hence

f (x) = k k k kk 1 k 1

x (e ) x



Notes|f (x)| k k

k 1

|x || |

1 1n q p

pqk k

k 1 k 1

| | x (Using Hölder’s inequality)

or |f (x)|

1n q

qk p

k 1

| | x x .

Hence we have

x 0

|f(x)|supx

1q

qk

k 1

| | = Tf (Using (6))

which upon using definition of norm yields

f Tf … (13)

Thus f = Tf (Using (12) and (13))

From the definition of T, it is linear. Also since it is an isometry, it is one-to-one and onto(already shown). Hence T is an isometric isomorphism of p* onto q , i.e.,

p q* .


Theorem 5: Let p > 1 with 1 1 1p q

and let g Lp (X). Then the function defined by

F (f) = p

X

fg d for f L (X)

is a bounded linear functional on Lp (X) and

F = g q … (1)

Proof: We first note that

F is linear on Lp (X). For if f1, f2 Lp (X), then we get

F (f1 + f2) = 1 2 1 2

X X X

(f f )g d f g d f g d

= F (f1) + F (f2)

So that

F (f1 + f2) = F (f1) + F (f2)

and F ( F) = X

fg d F(f) .

Now |F (f)| = X X

fg d |fg|d … (2)



Notes Making use of Hölder’s inequality, we get

X

|fg|d

1 1p q

p q

X X

|f| d |g| d

= f p g q … (3)

From (2) and (3) it follows that

|F (f)| f p g q.

Hence q

p

F|f|sup : f Lp(X) and f 0 g

f

F g q (Using definition of the norm) … (4)

Further, let f = |g|q–1 sgn g … (5)

Since sgn g = 1, we get

|f|p = |g|p(q–1) = |g|q ( p (q – 1) = q)

Thus, f Lp (X) and

1p

p

X

|f| d =

1q

q

X

|g| d … (6)

But

1p

q p

X X

|g| d |g| d = q /p

qg

which implies on using (6) that

f p = q /p

qg … (7)

Now F (f) = q 1

X X

fg d |g| g sgng d

= qqq

X

|g| d g

Hence q

qg g = F (f) F f p.

and this on using (7) yields that

q

qg = F (f) F

q /p

qg

q q /p

qg = g q F … (8)

( g 0)



NotesFrom (4) and (8) it finally follows that

F = g q.


Approximation by Continuous Function

Theorem 6: If f is a bounded measurable function defined on [a, b], then for given > 0, acontinuous function g on [a, b], such that

f – g 2 <

Proof: Let F (x) = x

a

f(t) dt where x [a, b].

Then |F (x + h) – F (x)| = x h x

a a

f(t) dt f(t) dt

= x h x h

x x

f(t) dt f(t) dt

Mh, where |f (x)| M, x [a, b].

Taking h < , and Mh < 1, we get

|x + h – x| < | F (x + h) – F (x) | < 1

F (x) is continuous on [a, b].

Let Gn (x) = x h

x

n f(t) dt : x [a, b] and n N;

then Gn (x) = 1n F x F(x)n

( F (x) is continuous on [a, b]

Gn (x) is continuous on [a, b] n)

Again, since F (x) = x

a

f (t) dt, x [a, b] .

F (x) = f (x) a.e. in [a, b].

Now, nnLim G (x) =

n

F(x (1/n) F(x)Lim1/n

= h 0

F(x h) F(x) 1Lim , hh n

= F (x) = f (x) a.e. in [a, b]

and hence2

nnLim G (x) f(x) = 0.



Notes

Also |Gn (x)| = x (1 /n) x (1 /n)

x x

f (t) dt n |f(t)|dt M .

Hence |Gn (x)| M, n N and x [a, b].

[Gn(x) – f (x)]2 (M + M)2 = 4M2, x [a, b].

On applying Lebesgue bounded convergence theorem, we get

b

2nn

a

Lim (G f) = b

2n

a

Lim (G f) 0

2n 2n

Lim G f = 0

n 2nLim G f

or nnLim f G = 0

for given > 0, no N, such that n no

f – Gn 2 <

Particularly for n = no.

no 2f G <

f – g 2 < (Taking noG = g)

Thus there exists a continuous function noG (x) g(x)

= x (1 /n )o

o

x

n f(t) dt, x [a, b] ,

which satisfies the given condition.

8.2 Summary

A linear functional f on a normed space N1 is said to be bounded if there is a constantk > 0 such that

|f (x)| k x , x N1

If x p and f is bounded linear functional on p , then f has the unique representation of

the form as an infinite series.

f (x) = k kk 1

x f (e )

The norm of f *p is given by

f =

1q

qk

k 1

|f(e )|



Notes8.3 Keywords

Bounded Linear Functional on Lp-spaces: If x p and f is bounded linear functional on p , then

f has the unique representation of the form as an infinite series

f (x) = k kk 1

x f(e )

Bounded Linear Functional: A linear functional f on a normed space N1 is said to be bounded ifthere is a constant k > 0 such that

|f (x)| k x , x N1

Continuous Linear Functional: A linear functional f is continuous if given > 0 there exists >0 so that

|f (x) – f (y)| whenever x – y .

Linear Functional: Let N1 be a normed space over a field R (or C). A mapping f : N1 R (or C)is called a linear functional on N1 if f ( x + y) = f (x) + f (y), x, y N1 and , R (or C).

Norm: The norm of f *p is given by

f =

1q

qk

k 1

|f(e )|


1. Account for bounded linear functionals on Lp-space.

2. State and prove different continuous linear functional theorems.

3. Describe approximation by continuous function.

4. How will you explain norms of bounded linear functional on Lp-space?

5. What is Isometric Isomorphism?


Books Rudin, Walter (1991), Functional Analysis, Mc-Graw-Hill Science/Engineering/Math

Kreyszig, Erwin, Introductory Functional Analysis with Applications, WILEY 1989.

T.H. Hilderbrandt, Transactions of the American Mathematical Society . Vol. 36,No. = 4, 1934.

Online links www.math.psu.edu/yzheng/m597k/m597kLIII4.pdf

www.public.iastate.edu/…/Royden_Real_Analysis_Solutions.pdf



Notes Unit 9: Measure Spaces

CONTENTS

Objectives

Introduction

9.1 Measure Space

9.1.1 Null Set in a Measure Space

9.1.2 Complete Measure Space

9.1.3 Measurable Mapping

9.2 Summary

9.3 Keywords



Objectives


Define measure space.

Define null set in a measure space.

Understand theorems based on measure spaces.

Solve problems on measure spaces.

Introduction

A measurable space is a set S, together with a non-empty collection, S, of subsets of S, satisfyingthe following two conditions:

1. For any A, B in the collection S, the set1 A – B is also in S.

2. For any A1, A2, … S, Ai S.

The elements of S are called measurable sets. These two conditions are summarised by sayingthat the measurable sets are closed under taking finite differences and countable unions.

9.1 Measure Space

Measurable Space: Let be a -algebra of subsets of set X. The pair (X, ) is called a measurablespace. A subset E of X is said to be -measurable if E .

(a) If is a measure on a -algebra of subsets of a set X, we call the triple (X, , u) a measurespace.

(b) A measure on a -algebra of subsets of a set X is called a finite measure if m (X) < . Inthis case (X, , ) is called a finite measure space.


Unit 9: Measure Spaces

Notes(c) A measure on a -algebra of subsets of a set X is called a -finite measure if there existsa sequence (En : n ) in such that n En = X and (En) < for every n . In this case(X, , ) is called a -finite measure space.

(d) A set D in an arbitrary measure space (X, , ) is called a -finite set if there exists asequence (Dn : n ) in such that Un Dn = D and (Dn) < for every n .

Lemma 1: (a) Let (X, , ) be a measure space. If D is a -finite set, then there exists anincreasing sequence (Fn : n ) in such that nn

lim F D and (Fn) < for every n and there

exists a disjoint sequence (Gn : n ) in such that n Gn = D and (Gn) < for every n .

(b) If (X, , ) is a -finite measure space then every D is a -finite set.

Proof 1: Let (X, , ) be a measure space. Suppose D is a -finite set. Then there exists asequence (Dn : n ) in such that Un Dn = D and (Dn) < for every n . For each n ,

let nn k 1 kF U D . Then (Fn : n ) is an increasing sequence in such that n nnn

lim F U F

nnU D D and

n n

n k kk 1k 1

(F ) D (D ) for every n .

Let G1 = F1 and Gn = Fn\n 1k 1 kU F for n 2. Then (Gn: n ) is a disjoint sequence in such that

n nn nU G U F D as in the proof of Lemma “let (En : n ) be an arbitrary sequence in analgebra of subsets of a set X. Then there exists a disjoint sequence (Fn: n ) in such that

(1)N N

n nn 1 n 1

E F for every N ,

and

(2) n nn n

E F

”.

(G1) = (F1) < and n 1

n n k nk 1

(G ) F F (F ) for n 2. This proves (a).

2. Let (X, , ) be a -finite measure space. Then there exists a sequence (En : n ) in suchthat Un En = X and (En) < for every n . Let D . For each n , let Dn = D En.Then (Dn : n ) is a sequence in such that Un Dn = D and m (Dn) (En) < for everyn . Thus D is a -finite set. This proves (b).

9.1.1 Null Set in a Measure Space

Definition: Given a measure on a -algebra of subsets of a set X . A subset E of X is called a nullset with respect to the measure if E and (E) = 0. In this case we say also that E is a null setin the measure space (X, , ). (Note that is a null set in any measure space but a null set in ameasure space need not be .)

Theorem 1: A countable union of null sets in a measure space is a null set of the measure space.

Proof: Let (En : n ) be a sequence of null sets in a measure space (X, , ). Let E = Un En. Since is closed under countable unions,

we have E .



Notes By the countable subadditivity of on ,

we have (E) n (En) = 0.

Thus (E) = 0.

This shows that E is a null set in (X, , ).

9.1.2 Complete Measure Space

Definition: Given a measure on a -algebra of subsets of a set X. We say that the -algebra is complete with respect to the measure if an arbitrary subset E0 of a null set E with respect to

is a member of (and consequently has (E0) = 0 by the Monotonicity of ). When iscomplete with respect to , we say that (X, , ) is a complete measure space.

Example: Let X = {a, b, c}. Then = { , {a}, {b, c}, X} is a -algebra of subsets of X. If wedefine a set function on by setting ( ) = 0, ({a}) = 1, ({b, c}) = 0, and (X) = 1, then isa measure on . The set {b, c} is a null set in the measure space (X, , ), but its subset {b} is nota member of . Therefore, (X, , ) is not a complete measure space.

9.1.3 Measurable Mapping

Let f be a mapping of a subset D of a set X into a set Y. We write D (f) and R (f) for the domain ofdefinition and the range of f respectively. Thus

D (f) = D X,

R (f) = {y Y : y = f (x) for some x D (f)} Y.

For the image of D (f) by f, we have f (D (f)) = R (f). For an arbitrary subset E of y we define thepreimage of E under the mapping f by

F–1 (E) : = {x X : f (x) E} = {x D (f) : f (x) E}.

Notes

1. E is an arbitrary subset of Y and need not be a subset of R (f). Indeed E may be disjointfrom (f), in which case f–1 (E) = . In general, we have f (f–1 (E)) E.

2. For an arbitrary collection C of subsets of Y, we let f–1 (C) : = {f–1 (E) : E C}.

Theorem 2: Given sets X and Y. Let f be a mapping with D (f) X and (f) Y. Let E and E bearbitrary subsets of Y. Then

1. f–1 (Y) = D (f),

2. f–1 (EC) = f–1 (Y\E) = f–1 (Y)\f–1 (E) = D (f) \ f–1 (E),

3. f–1 (U E ) = U f–1 (E ),

4. f–1 ( E ) = f–1 (E ).

Theorem 3: Given sets X and Y. Let f be a mapping with D (f) X and R (f) Y. If is a -algebraof subsets of Y then f–1 () is a -algebra of subsets of the set D (f). In particular, if D (f) = X thenf–1 () is a -algebra of subsets of the set X.



NotesProof: Let be a -algebra of subsets of the set Y. To show that f–1 () is a -algebra of subsets ofthe set D (f) we show that D (f) f–1 (); if A f–1 () then D (f)\A f–1 (); and for any sequence(An : n ) in f–1 () we have Un An f–1 (B).

1. By (1) of above theorem, we have D (f) = f–1 (Y) f–1 (B) since Y .

2. Let A f–1 (). Then A = f–1 () for some B . Since BC we have f–1 (BC) f–1 (). On theother hand by (2) of above theorem, we have f–1 (BC) = D (f)\f–1 (B) = D (f)\A. Thus D (f)\A

f–1 ().

3. Let (An : n ) be a sequence in f–1 (). Then An = f–1 (Bn) for some Bn for each n .Then by (3) of above theorem, we have

1 1 1n n n

n n n

A f (B ) f B f ( )

,

since nn

B ( )

.

Measurable Mapping

Definition: Given two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and (f) Y. We say that f is a / measurable mapping if f–1 (B) for every B , that is, f–1 ()

.

Theorem 4: Given two measurable spaces (X, ) and (Y, ). Let f be a /-measurable mapping.

(a) If , is a -algebra of subsets of X such that , , then f is 1/-measurable.

(b) If 0 is a -algebra of subsets of Y such that 0 , then f is /0-measurable.

Proof: (a) Follows from f–1 () 1 and (b) from f–1 (B0) f–1 () .

Composition of two measurable mappings is a measurable mapping provided that the twomeasurable mappings from a chain.

Theorem 5: Given two measurable spaces (X, ) and (Y, ), where = () and is arbitrarycollection of subsets of Y. Let f be a mapping with D (f) and (f) Y. Then f is a /-measurable mapping of D (f) into Y if and only if f–1 () .

Proof: If f is a /-measurable mapping of D (f) into Y, then f–1 () so that f–1 () .Conversely if f–1 () , then (f–1 () () = . Now by theorem,

“Let f be a mapping of a set X into a set Y. Then for an arbitrary collection C of subsets of Y, wehave (f–1 ()) = f–1 ( ().”

(f–1 () = f–1 ( ()) = f–1 (). Thus f–1 () and f is a /- measurable mapping of D (f).

Theorem 6: If X0 is a thick subset of a measure space (X, S, ), if S0 = S X0, and if, for E in S, 0 (E X0) = (E), then (X0, S0, 0) is a measure space.

Proof: If two sets, E1 and E2, in S are such that E1 X0 = E2 X0, then (E1 E2) Xo = 0, so that (E1 E2) = 0 and therefore (E1) = (E2). In other words 0 is indeed unambiguously defined onS0.

Suppose next that {Fn} is a disjoint sequence of sets in S0, and let En be a set in S such that

Fn = En X0, n = 1, 2, … .



NotesIf n n iE E {E : 1 i n}, n 1, 2, , then

n n 0E E X = n i nF F : 1 i n F

= Fn Fn = 0,

So that n nE E = 0, and therefore

n 1 0 n(F ) = n 1 0 n n 1 0 n n 1 n(E ) (E ) E

= n 0 nn 1 n 1

E F

In other word 0 is indeed a measure, and the proof of the theorem is complete.

9.2 Summary

Let be a -algebra of subsets of a set X. The pair (X, ) is called a measurable space. Asubset E of X is said to be -measurable if E .

If is a measure on a -algebra of subsets of a set X, we call the triple (X, , ) a measurespace.

A subset E of X is called a null set with respect to the measure if E and (E) = 0.

Two measurable spaces (X, ) and (Y, ). Let f be a mapping with D (f) X and (f) Y.We say that f is a /-measurable mapping if f–1 (B) for every B , that is f–1 () .

9.3 Keywords

Complete Measure Space: Given a measure on a -algebra of subsets of a set X. We say thatthe -algebra is complete with respect to the measure if an arbitrary subset E0 of a null setE with respect to is a member of (and consequently has (E0) = 0 by the Monotonicity of ).When is complete with respect to , we say that (X, , ) is a complete measure space.

Measurable Mapping: Given two measurable spaces (X, ) and (Y, ). Let f be a mapping with D(f) X and (f) Y. We say that f is a / measurable mapping if f–1 (B) for every B , thatis, f–1 () .

Measurable Space: A measurable space is a set S, together with a non-empty collection, S, ofsubsets of S.

Null Set in a Measure Space: A subset E of X is called a null set with respect to the measure ifE and (E) = 0. In this case we say also that E is a null set in the measure space (X, , ).

Sigma Algebra: is sigma algebra which establishes following relations:

(i) Ak for all k implies kk 1

A

(ii) A implies AC

(iii)




1. Let be a -algebra of subsets of a set X and let Y be an arbitrary subset of X. Let = {A Y : A }. Show that is a -algebra of subsets of Y.

2. Let (X, , ) be a measure space. Show that for any E1, E2 we have the equality: (E1 E2) + (E1 E2) = (E1) + (E2).


Books Paul Halmos, (1950). Measure Theory. Van Nostrand and Co.

Bogachev, V.I. (2007), Measure Theory, Berlin : Springer

Online links planetmath.org/measurable space.html

mathworld.wolfram.com > Calculus and Analysis > Measure Theory



Notes Unit 10: Measurable Functions

CONTENTS

Objectives

Introduction

10.1 Measurable Functions

10.1.1 Lebesgue Measurable Function/Measurable Function

10.1.2 Almost Everywhere (a.e.)

10.1.3 Equivalent Functions

10.1.4 Non-Negative Functions

10.1.5 Characteristic Function

10.1.6 Simple Function

10.1.7 Step Function

10.1.8 Convergent Sequence of Measurable Function

10.1.9 Egoroff's Theorem

10.1.10 Riesz Theorem

10.2 Summary

10.3 Keywords



Objectives


Understand measurable functions.

Define equivalent functions and characteristic function.

Describe Egoroff's theorem and Riesz theorem.

Define simple function and step function.

Introduction

In this unit, we shall see that a real valued function may be Lebesgue integrable even if thefunction is not continuous. In fact, for the existence of a Lebesgue integral, a much less restrictivecondition than continuity is needed to ensure integrability of f on [a, b]. This requirement givesrise to a new class of functions, known as measurable functions. The class of measurable functionsplays an important role in Lebesgue theory of integration.


Unit 10: Measurable Functions

Notes10.1 Measurable Functions

10.1.1 Lebesgue Measurable Function/Measurable Function

Definition: Let E be a measurable set and R* be a set of extended real numbers. A functionf : E R* is said to be a Lebesgue measurable function on E or a measurable function on E iff theset

E (f > ) = {x E : f (x) > } = f–1 { , )} is a measurable subset of E R.

Notes

1. The definition states that f is a measurable function if for every real number , theinverse image of ( , ) under f is a measurable set.

2. The measure of the set E (f > ) may be finite or infinite.

3. A function whose values are in the set of extended real numbers is called an extendedreal valued function.

4. If E = R, then the set E (f > ) becomes an open set.

Example: A constant function with measurable domain is measurable.

Sol: Let f be a constant function defined over a measurable set E so that f (x) = x E.

Then for any real number ,

E (f > ) = E, if c, if c

The sets E and are measurable and hence E (f > ) is measurable i.e. the function f is measurable.

Theorem 1: Let f and g be measurable real valued functions on E, and c is a constant. Then each ofthe following functions is measurable on E.

(a) f c (b) c f

(c) f + g (d) f – g

(e) |f| (f) f2

(g) fg (h) f/g (g vanishes no where on E)

Proof: Let be an arbitrary real number.

(a) Since f is measurable and

E (f ± c > ) = E (f > c),

the function f ± c is measurable.

(b) To prove c f is measurable over E.

If c = 0, then cf is constant and hence measurable because a constant function is measurable.



Notes Consider the case in which c 0, then

E (cf > ) = E f if c 0

c

E f if c 0c

Both the sets on R.H.S. are measurable.

Hence E (cf > ) is measurable and so cf is measurable c R.

(c) Before proving f + g is measurable, we first prove that if f and g are measurable over E thenthe set E (f > g) is also measurable.

Now f > g a rational number r such that

f (x) > r > g (x)

Thus E (f > g) = r Q

[(E (f r) (E(g r))]= an enumerable union of measurable sets.

= measurable set, since Q is an enumerable set.

Now, we shall prove that f + g is measurable over E. Let a be any real number.

Now E (f + g > a) = E (f > a – g) … (1)

Again, g is measurable

cg is measurable, c is constant.

( We know that if f is a measurable function and c is constant then cf is measurable)

a + cg is measurable a, c R

a – g is measurable by taking c = –1,

since f and a – g are measurable

E (f > a – g) is measurable.

E (f + g > a) is a measurable set.

f + g is a measurable function.

(d) To prove that f – g is measurable. Before proving f – g is measurable, we first prove that iff and g are measurable over E then the set E (f > g) is also measurable.

Now f > g a rational number r, such that f (x) > r > g (x).

Thus E (f > g) = r Q

[(E (f r) (E(g r))]= an enumerable union of measurable sets.

= measurable sets, since Q is an enumerable set.

Now we shall prove that f – g is measurable over E.

Let a be any real number.



NotesNow E (f – g > a) = E (f > a + g)

since g is measurable.

cg is measurable, c is constant.

a + cg is measurable a, c R

a + g is measurable by taking c = 1,

since f and a + g are measurable

E (f > a + g) is measurable.

E (f – g > a) is a measurable set.

f – g is a measurable function.

(e) To prove |f| is measurable.

We have

E (|f| > ) = E if 0[E (f )] [E(f )] if 0

[because we know that |x| > a x > a or x < –a]

since f is measurable therefore E (f > ) and E (f < – ) are measurable by definition.

Also we know that finite union of two measurable sets is measurable.

E (f > ) E (f < – ) is measurable.

E (|f| > ) is measurable.

|f| is measurable.

(f) To prove f2 is measurable.

We have E (f2 > ) = E if 0E (|f| )] if 0

But E (|f| > ) = [E(f )] [E(f )], if 0 ( |x| > a x > a or x < – a)

E (f2 > ) = E if 0[E (f )] [E(f )] if 0

But f is measurable over E.

E (f ) and E (f ) are measurable sets.

[E (f )] [E(f )] is measurable.

( union of two measurable sets is measurable)

E (f2 > ) is measurable because both the sets on RHS are measurable.

f2 is measurable over E.

(g) To prove fg is measurable.

Clearly, f + g and f – g are measurable functions over E.



Notes (f + g)2, (f – g)2 are measurable functions over E.

(f + g)2 – (f – g)2 is a measurable function over E.

2 21 (f g) (f g)4

is a measurable function over E.

fg is a measurable function over E.

(h) To prove f/g is measurable.

Let g vanish nowhere on E, so that

g (x) 0 x E.

1g

exists.

Now we shall show that 1g

is measurable.

We have

E(g 0) if 01 1E [E(g 0)] E g if 0g

1[E(g 0)] E(g 0) E g

Also finite union and intersection of measurable sets are measurable. Hence 1Eg

is

measurable in every case.

Since f and 1g

are measurable.

1(f)g

is measurable over E.

fg

is measurable over E.

10.1.2 Almost Everywhere (a.e.)

Definition: A property is said to hold almost everywhere (a.e.) if the set of points where it fails tohold is a set of measure zero.

Example: Let f be a function defined on R by

f (x) = 0, if x is irrational1, if x is rational

Then f (x) = 0 a.e.



Notes10.1.3 Equivalent Functions

Definition: Two functions f and g defined on the same domain E are said to be equivalent on E,written as f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1 E with m (E1)= 0.

Theorem 2: If f, g : E R (E M) such that g (E).

Proof: Let be any real number and let E1 = E (f > ) and E2 = E (g > )

Then E1 E2 = (E1 – E2) (E2 – E1)

= {x E : f (x) g (x)}

so that by given hypothesis we have

m (E1 E2) = 0.

This together with the fact that E1 is measurable

E2 is measurable.

Hence g (E).

10.1.4 Non-negative Functions

Definition: Let f be a function, then its positive part, written f+ and its negative part, written f–1, aredefined to be the non-negative functions given by

f+ = max (f, 0) and f–1 = max (–f, 0) respectively.

Note f = f+ – f–1

and |f| = f+ + f–1

Theorem 3: A function is measurable iff its positive and negative parts are measurable.

Proof: For every extended real valued function f, we may write

f+ = 12

[f + |f|]

and f–1 = 12

[|f| – f]

But f is measurable then |f| is measurable and hence positive and negative parts of f i.e. f+ andf– are measurable.

Conversely, let f+ and f–1 be measurable.

Since f = f+ – f–1

Since we know that if f and g are measurable functions defined on a measurable set E then f – gis measurable on E.

Here f+ – f–1 is measurable.

and hence f is measurable.



Notes Theorem 4: If f is a measurable function and f = g a.e. then g is measurable.

Proof: Let E = {x : f (x) g (x)}.

Then m (E) = 0

Let be a real number.

{x : g (x) > } = {x : f (x) > } {x E : g (x) > } – {x E : g (x) }

since f is measurable, the first set on the right is measurable i.e. {x : f (x) > } is measurable.

The last two sets on the right are measurable since they are subsets of E and m (E) = 0.

Thus, {x : g (x) > } is measurable.

So, g is measurable.

Example: Give an example of function for which f is not measurable but |f| is measurable.

Sol: Let k be a non-measurable subset of E = [0, 1).

Define a function f : E R by

f (x) = 1 if x k1 if x k

The function f is not measurable, since E (f > 0) (=k) is a non-measurable set. But |f| is measurableas the set

E (|f| > ) = E if 1

if 1 is measurable

10.1.5 Characteristic Function

Definition: Let A be subset of real numbers. We define the characteristic function A of the set A asfollows:

A (x) = 1 if x A0 if x A

Note The characteristic function A of the set A is also called the indicator function of A.

Theorem 5: Show that the characteristic function A is measurable iff A is measurable.

Proof: Let A be measurable.

Since A = {x : A (x) > 0} is measurable.

But A is measurable, therefore the set {x : A (x) > 0} is measurable.

A is measurable.

Conversely, let A be measurable and be any real number.

then E ( A > ) = if 1

A if 0 1E if 0



NotesEvery set on R.H.S. is measurable.

Therefore E ( A > ) is measurable.

Hence A is measurable.

Note The above theorem asserts that the characteristic function of non-measurable setsare non-measurable even though the domain set is measurable.

10.1.6 Simple Function

A real valued function is called simple if it is measurable and assumes only a finite number ofvalues.

If is simple and has the values 1, 2, … n, then

= n

i Aii 1

where Ai = {x : (x) = i}

and Ai Aj is a null set.

Thus we can always express a simple function as a linear combination of characteristic function.

Notes

(i) is simple A is are measurable.

(ii) sum, product and difference of simple functions are simple.

(iii) the representation of as given above is not unique.

But if is simple and { 1, 2, ……, n} is the set of non-zero values of f, then

= n

i Aii 1


This representation of is called the Canonical representation. Here A is are disjointand is are distinct and non-zero.

(iv) Simple function is always measurable.

10.1.7 Step Function

A real valued function S defined on an interval [a, b] is said to be a step function if these is apartition a = xo < x1 … < xn = b such that the function assumes one and only one value in eachinterval.



Notes

Notes

(i) Step function also assumes finite number of values like simple functions but the sets{x : S (x) = Ci} are intervals for each i.

(ii) Every step function is also a simple function but the converse is not true.

e.g. f : R R such that f (x) = 1, x is rational0, x is irrational

is a simple function but not step as the sets of rational and irrational are not intervals.

Theorem 6: If f and g are two simple functions then f + g is also a simple function.

Proof: Since f and g are simple functions and we know that every simple function can beexpressed as the linear combination of characteristic function.

f and g can be expressed as the linear combination of characteristic function.

f = m

i Aii 1

and g = m

j Bjj 1

where A is and B js are disjoint.

Ai = {x : f (x) = i}

Bj = {x : g (x) = j}

The set Ek obtained by taking all intersections Ai Bj from a finite disjoint collection of measurablesets and we may write

f = n

k Ekk 1

a

and g = n

k Ekk 1

b

where n = mm .

f + g = n n

k E k Ek kk 1 k 1

a b

= n

k b Ek kk 1

( a )

which is a linear combination of characteristic functions, therefore it is simple.



NotesSimilarly fg =

n

k k Ekk 1

a b

which is again a linear combination of characteristic function, therefore fg is simple.

Theorem 7: Let E be a measurable set with m (E) < and {fn} a sequence of measurable functionsconverging a.e. to a real valued function defined on E. Then, given > 0 and > 0, there is a setA E with m (A) < and an integer N such that |fn (x) – f (x)| < for all x E – A and all n N.

Proof: Let F be the set of points of E for which fn f. Then m (F) = 0 and fn (x) f (x) for all x E – F = E1 (say). Then by the previous theorem for the set E1, we get A1 E1 with m (A1) < andan integer N such that

|fn (x) – f (x) | < for all n N and x E1 – A1.

We get the required result by taking

A = A1 F since m (F) = 0 and E – A = E1 – A1

Note Before proving this theorem first prove the previous theorem.

10.1.8 Convergent Sequence of Measurable Function

Definition: A sequence {fn} of measurable functions is said to converge almost uniformly to ameasurable function f defined on a measurable set E if for each > 0 there exists a measurableset A E with m (A) < such that {fn} converges to f uniformly an E – A.

10.1.9 Egoroff's Theorem

Statement: Let E be a measurable set with m (E) < and {fn} a sequence of measurable functionswhich converge to f a.e. on E. Then, given > 0 there is a set A E with m (A) < with that thesequence {fn} converges to f uniformly on E – A.

Proof: Applying the theorem, “Let E be a measurable set with m (E) < and {fn} a sequence ofmeasurable function converging a.e. to real valued function f defined on E. Then given > 0 and

> 0 there is a set A E with m (A) < and an integer N such that

|fn (x) – f (x) | < for all x E – A and all n N”

with = 1, = /2, we get a measurable set

A1 E with m (A1) < /2 and a positive integer N1, such that

|fn (x) – f (x)|< 1 for all x N1

and x E1, where E1 = E – A1.

Again, taking = 1/2 and = n/22,

we get a measurable set A2 E1 with m (A2) < /22, and a positive integer N2 such that

|fn (x) – f (x)| < 12

n N2 and x E2 where E2 = E1 – A2, and so on.



Notes At the pth stage, we get a measurable set

Ap Ep–1 with m (Ap) < p

n2

and a positive integer Np such that

|fn (x) – f (x)| < 1p n Np and x Ep where

Ep = Ep – 1 – Ap.

Let A = pp 1

A ,

then m (A) pp 1

m(A )

But m (Ap) < n

2p

m (A) < pp 1

n2

But pp 1

12

is a G.P. series so S =

1 1a 2 2

1 11 r 12 2

= 1

m (A) <

Also, E – A = E – pp

A

= pp

(E A )

= p 1 pp

(E A )

= pp

ELet x E – A. Then x Ep p and so

|fn (x) – f (x)| < 1p

, n Np.

Let us choose p such that 1p

< , we get

|fn (x) – f (x)| < x E – A and n Np = N



Notes

Note Egoroff’s theorem can be stated as: almost every where convergence implies almostuniform convergence.

10.1.10 Riesz Theorem

Let {fn} be a sequence of measurable functions which converges in measure to f. Then there is a

subsequence nkf which converges in measure to f a.e.

Proof: Let { n} and { n} be two sequences of positive real numbers such that n 0 as n and

nn 1

.

Let us now choose an increasing sequence {nk} of positive integers as follows.

Let n be a positive integer such that

nm x E f (x) f(x)

Since fn f in measure for a given 1 > 0 and 1 > 0, a positive integer n1 such that

n 1 1 11m x E, f (x) f(x) , n n

Similarly, let n2 be a positive number such that n 2 2 2 12m x E, f (x) f(x) , n n and so

on.

In general let nk be a positive number such that

n k k k k 1km x : x E, f (x) f (x) and that n n .

We shall now prove that the subsequence nkf converges to f a.e.

Let Ak = n i kii k k 1

x : x E, f (x) f (x) ,k N and A A .

Clearly, {Ak} is a decreasing sequence of measurable sets and m (A1) < .

Therefore, we have

m (A) = kklimm(A )

But m (Ak) ii k

0 as k .

Hence m (A) = 0.

Now it remains to show that {fn} converges to f on E – A. Let xo E – A.



Notes Then xo koA for some positive integer ko.

or xo {x E : |fn (x) – f (x)| k}, k ko

which gives |fn (xo) – f (xo)| < k, k ko

But k 0 as k .

Hence n o okklim f (x ) f(x ) .

10.2 Summary

Let E be a measurable set and R* be a set of extended real numbers. A function f : E R* issaid to be a Lebesgue measurable function on E or a measurable function on E iff the set E(f > ) = {x E : f (x) > } = f–1 {( , )} is a measurable subset of E R.

A property is said to hold almost everywhere (a.e.) if the set of points where it fails to holdis a set of measure zero.

Two functions f and g defined on the same domain E are said to be equivalent on E, writtenas f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1 E with m (E1) = 0.

f+ = max (f, 0) and f–1 = max (–f, 0)

|f| = f+ + f–1

Let A be subset of real numbers. We define the characteristic function A of the set A asfollows:

A (x) = 1, if x A0, if x A

A real valued function is called simple if it is measurable and assumes only a finitenumber of values.

10.3 Keywords

Almost Everywhere (a.e.): A property is said to hold almost everywhere (a.e.) if the set of pointswhere it fails to hold is a set of measure zero.

Characteristic Function: Let A be subset of real numbers. We define the characteristic functionA of the set A as follows:


Egoroff’s Theorem: Let E be a measurable set with m (E) < and {fn} a sequence of measurablefunctions which converge to f a.e. on E. Then given n > 0 there is a set A E with m (A) < n suchthat the sequence {fn} converges to f uniformly on E – A.

Equivalent Functions: Two functions f and g defined on the same domain E are said to beequivalent on E, written as f ~ g on E, if f = g a.e. on E, i.e. f (x) = g (x) for all x E – E1, where E1

E with m (E1) = 0.



NotesHaracteristic Function: Let A be subset of real numbers. We define the characteristic function A

of the set A as follows:


Lebesgue Measurable Function: A function f : E R* is said to be a Lebesgue measurable functionon E or a measurable function on E iff the set

E (f > ) = {x E : f (x) > } = f–1 { , )} is a measurable subset of E R.

Measurable Set: A set E is said to be measurable if for each set T, we have

m* (T) = m* (T E) + m* {T Ec)

Non-negative Functions: Let f be a function, then its positive part, written f+ and its negativepart, written f–1, are defined to be the non-negative functions given by

f+ = max (f, 0) and f–1 = max (–f, 0) respectively.

Riesz Theorem: Let {fn} be a sequence of measurable functions which converges in measure to f.

Then there is a subsequence nkf which converges to f a.e.

Simple Function: A real valued function is called simple if it is measurable and assumes onlya finite number of values.

If is simple and has the values 1, 2, … n, then

= n

i Aii 1


and Ai Aj is a null set.

Step Function: A real valued function S defined on an interval [a, b] is said to be a step functionif these is a partition a = xo < x1 … < xn = b such that the function assumes one and only one valuein each interval.

Subsequence: If (xn) is a given sequence in X and (nk) is an strictly increasing sequence of positive

integers, then nkx is called a subsequence of (xn).


1. If f is a measurable function and c is a real number, then is it true to say that cf is measurable?

2. A non-zero constant function is measurable if and only if X is measurable comment.

3. Let Q be the set of rational number and let f be an extended real-valued function such that{x : f (x) > } is measurable for each Q. Then show that f is measurable.

4. Show that if f is measurable then the set {x : f (x) = } is measurable for each extended realnumber .

5. If f is a continuous function and g is a measurable function, then prove that the compositefunction fog is measurable.



Notes 6. Show that

(i) A B A B

(ii) A B A B A B

(iii) c AA1


Books Dudley, R.M. (2002). Real Analysis and Probability (2 ed.). Cambridge UniversityPress

Strichartz, Robert (2000). The Way of Analysis. Jones and Bortlett.

Online links mathworld.wolfram.com>calculus and Analysis > Measure theory

planetmath.org/Measurable functions.html

zeta.math.utsa.edu/~ mqr 328/class/real2/mfunct.pdf


Unit 11: Integration

NotesUnit 11: Integration

CONTENTS

Objectives

Introduction

11.1 Integration

11.1.1 The Riemann Integral

11.1.2 Lebesgue Integral of a Bounded Function over a Set of Finite Measure

11.1.3 The Lebesgue Integral of a Non-negative Function

11.1.4 The General Lebesgue Integral

11.2 Summary

11.3 Keywords



Objectives


Define the Riemann integral and Lebesgue integral of bounded function over a set offinite measure.

Understand the Lebesgue integral of a non-negative function.

Solve problems on integration.

Introduction

We now come to the main use of measure theory: to define a general theory of integration. Theparticular case of the integral with respect to the Lebesgue measure is not, in any way, simplerthe general case, which will give us a tool of much wider applicability.

11.1 Integration

11.1.1 The Riemann Integral

Let f be a bounded real valued function defined on the interval [a, b] and let a = x0 < x1 < … < xn= bbe a sub-division of [a, b].

Then for each sub-division we can define the sums

S = n

i i 1 ii 1

(x x ) M

and s = n

i i 1 ii 1

(x x ) m



Notes where Mi = x x xi 1 i

sup f(x) ,

mi = x x xi 1 i

inf f (x)

We then define the upper Riemann integral of f by

b

a

R f(x) dx inf S

where the infimum is taken over all possible sub-divisions of [a, b].

Similarly, we define the lower Riemann integral

b

a

R f(x) dx sup S

The upper integral is always at least as large as the lower integral, and if the two are equal, wesay that f is Riemann integrable and we call this common value the Riemann integral of f.

It will be denoted by

b

a

R f(x) dx

Note By a step function we mean function s.t.

(x) = i x [xi–1, xi]

for some sub-division of [a, b] and some set of constant i then

b

a

(x) dx = x x1 n

x xo n 1

(x) dx (x) dx

= x x x1 2 n

1 2 n

x x xo 1 n 1

dx dx dx

= 1 (x1 – x0) + 2 (x2 – x1) + … + n (xn–1 – xn)

= n

i i i 1i 1

(x x ) … (1)

with this in mind, we see that

b

a

R f (x) dx = pinf (f)



Notes=

n

i i i 1i 1

inf M (x x )

= b

a

inf (x) dx for all step functions

(x) f (x)

Similarlyb

a

R f (x) dx = sup Lp (f)

= n

i i i 1i 1

sup m (x x )

= b

a

sup (x) dx for all step function

(x) f (x).

11.1.2 Lebesgue Integral of a Bounded Function over a Set of FiniteMeasure

Characteristic Function

The function E defined by

E (x) = 1 if x E0 if x E

is called the characteristic function of E.

Simple Function

A linear combination (x) = n

i Eii 1

(x) is called a simple function if the sets Ei are measurable.

This representation of is not unique.

However, a function is simple if and only if it is measurable and assumes only a finite numberof values.

Canonical Representation

If is simple function and { 1, 2, …, n} the set of non-zero values of , then

= n

i Eii 1

,

where Ei = {x : (x) = i}.



Notes This representation of is called the canonical representation. Here Ei’s are disjoint and i’s arefinite in number, distinct and non-zero.

Elementary Integral

Definition: If vanishes outside a set of finite measure, we define the elementary integral of byn

i ii 1

(x) dx mE when has the canonical representation.

= n

i Eii 1

.

We sometimes abbreviate the expression for this integral . If E is any measurable set, we

define the elementary integral of over E by E

E

.

If E = [a, b], then the integral [a , b]

will be denoted by b

a

.

Theorem 1: Let and be simple functions which vanish outside a set of finite measure, then

(a b ) = a b and if a.e., then

Proof: Since and are simple functions.

Therefore these can be written in the canonical form

=m

i Ai

and = m

j Bjj 1

B

where {Ai} and {Bj} are disjoint sequences of measurable sets and

Ai = {x : (x) = i}

and Bj = {x : (x) = j}

The set Ek obtained by taking all intersections Ai Bj form a finite disjoint collection of measurablesets. We may write

= N

k Ekk 1

a and

= N

k Ekk 1

b (where N = mm )



NotesNow a + b =

N N

k E k Ek kk 1 k 1

a a b b

= N

k k Ekk 1

(aa bb )

which is again a simple function.

Since = i ii

a m E

(a b ) = N

k k kk 1

(aa bb ) m E (by definition)

= N N

k k k kk 1 k 1

a a m E b b m E

= a b

Now since a.e.

– 0 a.e.

We have proved that

(a b ) = a b

Put a = 1, b = –1 in the first part, we get

( ) =

Since – 0 a.e. is a simple function, by the definition of the elementary integral, we have

( ) 0

0

Theorem 2: Riemann integrable is Lebesgue integrable.

Proof: Since f is Riemann integrable over [a, b], we have

b b b

1 1f1 fa a a

inf (x) dx sup (x) dx R f(x) dx

where 1 and 1 vary over all step functions defined on [a, b].



Notes Since we know that every step function is a simple function,

b

1f1 a

sup (x) dx b

fa

sup (x) dx

andb

1f1a

inf (x) dx b

fa

inf (x) dx

where and vary over all the simple functions defined on [a, b]. Thus from the above relation,we have

b b b b

ffa a a a

R f(x) dx sup (x) dx inf (x) dx R f(x) dx

b b

ffa a

sup (x) dx inf (x) dx

b b

a a

f (x) dx R f(x) dx

Note The converse of this theorem is not true i.e.

A Lebesgue integrable function may not be Riemann integrable

e.g. Let f be a function defined on the interval [0, 1] as follows:

f (x) = 1, if x is rational0, if x is irrational

Let us consider a partition p of an interval [0, 1].

U (p, f) = N

i ii 1

M x

= 1 x1 + 1 x2 + …… + 1 xn = 1 – 0

= 1.

1

0

f dx = inf U (p, f) = 1 – 0 = 1.

1

0

f dx = sup L (p, f)

= sup {0 x1 + 0 x2 + …… + 0 xn}

= 0



NotesThus f dx f dx

The function is not Riemann integrable.

Now for Lebesgue Integrability

Let A1 be the set of all irrational numbers and A2 be the set of all rational numbers in [0, 1].

The partition P = {A1, A2} is a measurable partition of [0, 1] and mA1 = 0, mA2 = 1.

L (p, f) = 1 2A A1 2inf f(x) mA inf f(x) mA

= 1 20 mA 1 mA = 1.

U (p, f) = 1 2A A1 2

sup f(x) mA sup f(x) mA

= 1 20 mA 1 mA = 1.

psup L (p, f) =

p1 inf U(p, f)

f is Lebesgue integrable over [0, 1].

Theorem 3: If f and g are bounded measurable functions defined on the set E of finite measure,then

(1)E E E

(af bg) a f b g

(2) If f = g a.e., then E E

f g

(3) If f g a.e., then E E

f g

Hence f |f|

(4) If A and B are disjoint measurable set of finite measure, then

A B A B

f f f

Proof of 1: Result is true if a = 0

Let a 0.

If is a simple function then so is a and conversely.

Hence for a > 0

E

af = a af

E

inf a



Notes=

fE

inf a ( a > 0)

= f

E

inf a

= f

E

a inf

= E

a f

Again if a < 0,

E

af = a afE

inf a

= f

E

sup a ( a < 0)

= f

E

sup a

= f

E

a sup

= E

a f

Therefore in each case

E

af = E

a f … (i)

Now we prove that

E

(f g) = E E

f g

Let 1 and 2 be two simple functions such that 1 > f and 2 g, then 1 + 2 is a simplefunction and 1 + 2 f + g.

or f + g = 1 + 2

E

(f g) 1 2

E

( )

But 1 2

E

= 1 2

E E



Notes

E

(f g) 1 2

E E

Since 1f 1E

inf = E

f

and 2g 2E

inf = E

g

E

(f g) E E

f g … (2)

On the other hand if 1 and 2 are two simple functions such that 1 < f and 2 g. Then 1 + 2 issimple function and

1 + 2 f + g,

or f + g 1 + 2

E

(f g) 1 2

E

( )

But 1 2

E

( ) = 1 2

E E

E

(f g) 1 2

E E

Since 1f 1 E

sup = E

f

and 2f 2 E

sup = E

g

E

(f g)E E

f g … (3)

From (2) and (3), we get

E

(f g) = E E

f g

E

(af bg) = E E

af bg

= E E

a f b g from (i)

Proof of 2: Since f = g a.e.

f – g = 0 a.e.



Notes Let F = {x : f (x) g (x)}

Then by definition of a.e., we have mF = 0 and F E.

E

(f g) = F (E F) F E F

(f g) (f g) (f g)

= (f – g) mF + (f – g) m (E – F)

= (f – g) . 0 + 0 . m (E – F) [ mF = 0 and f – g = 0 over E – F]

= 0

E

(f g) = 0

E E

f g = 0 E E

f g

Note Converse need not be true

e.g. Let the functions f : [–1, 1] R and g : [–1, 1] R be defined by

f (x) = 2 if x 00 if x 0

and g (x) = 1 x.

Then1

1

f(x) dx = 2 = 1

1

g(x) dx

But f g a.e.

In other words, they are not equal even for a single point in [–1, 1].

Proof of 3: f g a.e.

f – g 0 a.e.

Let be simple function,

= f – g

0 [ f – g = 0 a.e.]

E

0

E

(f g) 0

E E

f g 0



Notes

E

f E

g

Since f |f|

E

f E

|f| … (1)

Again – f |f|

E

fE

|f|

orE

|f|E

f … (2)

From (1) and (2) we get

E

|f|E

fE

|f|

E

fE

|f|

Proof of 4: It follows from (3) and the fact that E

1 mE .

Proof of 5:A B

f = A Bf

Now A B = A B A B

where A and B are disjoint measurable sets i.e.

A B =

A B

f = A B A Bf ( ) f

= A Bf f 0 [ A B = and m ( ) = 0]

= A B

f f

11.1.3 The Lebesgue Integral of a Non-negative Function

Definition: If f is a non-negative measurable function defined on a measurable set E, we define

E

f = h f

E

sup h ,



Notes where h is a bounded measurable function such that

m {x : h (x) 0} <

Theorem 4: Let f be a non-negative measurable function. Show that f 0 implies f = 0 a.e.

Proof: Let be any measurable simple function such that

f.

Since f = 0 a.e. on E

0 a.e.

E

(x) dx 0

Taking supremum over all those measurable simple functions f, we get

E

f dx 0 … (1)

Similarly let be any measurable simple function such that f

Since f = 0 a.e.

0 a.e.

E

(x) dx 0

Taking infimum over all those measurable simple functions f, we get

E

f dx 0 … (2)


E

f dx = 0.

Conversely, letE

f dx = 0

If En = 1x : f (x)n , then

E

f dx > EnE

1 (x) dxn

But EnE

1 (x) dxn

= n1 m En

(By definition)



Notes

E

f dx > n1 m En

Or n1 m En

< E

f dx

ButE

f dx < 0

n1 m En

< 0

m En < 0

But m En 0 is always true

m En = 0

But {x : f (x) > 0} = nn 1

E

and m En = 0

nn 1

m E = 0

m {x : f (x) > 0} = 0

f = 0 a.e. on E

Theorem 5: Let f and g be two non-negative measurable functions. If f is integrable over E andg (x) < f (x) on E, then g is also integrable over E, and

E

(f g) = E E

f g .

Proof: Since we know that if f and g are non-negative measurable functions defined on a set E,then

E

(f g) = E E

f g

Since f = (f – g) + g,

therefore we have

E

f = E E E

(f g g) (f g) g … (1)

Since the functions f – g and g are non-negative and measurable. Further, f being integrable over

E, E

f < (by definition)



Notes Therefore, each integral on the right of (1) is finite.

In particular, E

g < ,

which shows that g is an integrable function over E.

SinceE

f = E E

(f g) g

E E

f g = E

(f g) .

11.1.4 The General Lebesgue Integral

For the positive part f+ of a function f, we define

f+ = max (f, 0)

and negative part f– by f–

f– = max (–f, 0)

and that f is measurable if and only if both f+ and f– are measurable.

Note f = f+ – f–

and |f| = f+ + f–

Definition: A measurable function f is said to be Lebesgue integrable over E if f+ and f– are both

Lebesgue integrable over E. In this case, we define E E E

f f f .

Theorem 6: Let f and g be integrable over E, then

(a) The function of f is integrable over E, and E E

cf c f .

(b) Sum of two integrable functions is integrable i.e. the function f + g is integral over E, and

E

(f g) = E E

f g

(c) If f g a.e., then E E

f g .

(d) If A and B are disjoint measurable sets contained in E, then

A B

f = A B

f f .



NotesProof: (a) If c 0, then

(cf)+ = cf+

(cf)– = cf

and if c < 0, then

(cf)* = (–c) . f

(cf)– = (–c) . f+

Since f is integrable so f+ and f– are also integrable and conversely. Hence the resultfollows.

(b) In order to prove the required result first of all we show that if f1 and f2 are non-negativeintegrable functions such that f = f1 – f2, then

E

f = 1 2

E E

f f … (1)

Since f = f+ – f–,

Also f = f1 – f2,

then f+ – f– = f1 – f2

f+ + f2 = f1 + f– … (2)

Also we know that if f and g are non-negative measurable functions defined on a set E,then

E

(f g) = E E

f g

Then from (2), we get

2

E E

f f = 1

E E

f f

E E

f f = 1 2

E E

f f … (3)

But f is integrable so f+ and f– are integrable i.e.

f = f f

Therefore (3) becomes

HenceE

f = 1 2

E E

f f

which proves (1).

Now, if f and g are integrable functions over E, then

f+ + g+, f– + g– and f + g = (f+ + g+) – (f– + g–)

and also integrable functions over E.



Notes Furthermore f and g are integrable, it implies that |f| and |g| are integrable.

( A measurable function f is integrable over E if and only if |f| is integrable over E.)

Thus |f| + |g| is integrable over E.

( E E E

(f g) f g and by the definition of integrable)

Since |f + g| |+|f| + |g|

which shows that f + g is integrable.

Hence sum of two integrable functions is integrable.

ThusE

(f g) = E E

(f g ) (f g )

= E E E E

f g f g

= E E E E

f f g g

= E E

f g

(c) f g a.e.

f – g 0 a.e.

g – f 0 a.e.

E

(g f) 0

Since g = f + (g – f) and f, g – f are integrable over E.

Then by the given hypothesis (g – f)– = 0 a.e.

then E

(g f) = 0,

(Since we know that if f = 0 a.e. then E

f = 0)

E E E E E E

g f (g f) f (g f) (g f)

becomes

E

g = E E E E

f (g f) 0 f (g f) ( (g – f) 0)



Notes

E

f

E

g E E E

f f g

(d)A B

f = A Bf

= A Bf f

= A B

f f .

Example: Let f be a non-negative integrable function. Show that the function F definedby

F (x) = x

f (t) dt is continuous on R.

Solution: Since f is a non-negative integrable function, then given > 0 there is a > 0 such that forevery set A R with m (A) < , we have

A

f <

If xo R, then x R with |x – xo| < , we have

x

xo

f (t) dt <

x

xo

f (t) dt f (t) dt <

x

xo

f (t) dt f (t) dt <

xx o

f (t) dt f (t) dt <

|F (x) – F (xo)| <

Hence F is continuous at xo. Since xo R is arbitrary, F is continuous on R.



Notes 11.2 Summary

Let and be simple functions which vanish outside a set of finite measure, then

(a b ) a b and if a.e., then

A Lebesgue integrable function may not be Riemann integrable.

Let A1 be the set of all irrational numbers and A2 be the set of all rational numbers in [0, 1].

If f is a non-negative measurable function defined on a measurable set E, we define

h fE E

f sup h ,

where h is a bounded measurable function such that

m {x : h (x) 0} <

Let f and g be two non-negative measurable functions. If f is integrable over E and g (x) <f (x) on E, then g is also integrable over E, and

E E E

(f g) f g .

11.3 Keywords

Canonical Representation: If is simple function and { 1, 2, …, n} the set of non-zero values of, then

= n

i Eii 1

,

where Ei = {x : (x) = i}.

Characteristic Function: The function E defined by

E (x) = 1 if x E0 if x E

is called the characteristic function of E.

Elementary Integral: If vanishes outside a set of finite measure, we define the elementary

integral of by n

i ii 1

(x) dx mE when has the canonical representation.

= n

i Eii 1

.

Lebesgue Integrable: A measurable function f is said to be Lebesgue integrable over E if f+ and f–

are both Lebesgue integrable over E. In this case, we define E E E

f f f .



NotesSimple Function: A linear combination (x) =

n

i Eii 1

(x) is called a simple function if the sets

Ei are measurable.

Simple Function: A linear combination (x) = n

i Eii 1

(x) is called a simple function if the sets Ei

are measurable.

This representation of is not unique.

However, a function is simple if and only if it is measurable and assumes only a finite numberof values.

The Lebesgue Integral of a Non-negative Function: If f is a non-negative measurable functiondefined on a measurable set E, we define

E

f = h f

E

sup h ,

where h is a bounded measurable function such that

m {x : h (x) 0} <

The Riemann Integral: Let f be a bounded real valued function defined on the interval [a, b] andlet a = x0 < x1 < … < xn= b be a sub-division of [a, b].

Then for each sub-division we can define the sums

S = n

i i 1 ii 1

(x x ) M

and s = n

i i 1 ii 1

(x x ) m

where Mi = x x xi 1 i

sup f(x) ,

mi = x x xi 1 i

inf f (x)


1. Prove that E E

a f a f real number a.

2. If f is bounded real valued measurable function defined on a measurable set E of finite

measure such that a f (x) b, then show that amE E

f bmE.

3. If f and g are non-negative measurable functions defined on E then prove that

(a)E E

cf c f, c 0



Notes(b)

E E E

(f g) f g

(c) f 0 f 0 a.e.

(d) If f g a.e. then E E

f g

4. If f is integrable over E, then show that |f| is integrable over E, and E E

f |f|.

5. Show that if f is a non-negative measurable function then f = 0 a.e. on E iff E

f = 0.

6. If E

f = 0 and f (x) 0 on E, then f = 0 a.e.


Books Erwin Kreyszig, Introductory Functional Analysis with Applications, John Wiley &Sons Inc., New York, 1989

Walter Rudin, Real and Complex Analysis, Third McGraw Hill Book Co., New York,1987

R.G. Bartle, The Elements of Integration and Lebesgue Measure, Wiley Interscience,1995

Online links www.maths.manchester.ac.uk

www.uir.ac.za


Unit 12: General Convergence Theorems

NotesUnit 12: General Convergence Theorems

CONTENTS

Objectives

Introduction

12.1 General Convergence Theorems

12.1.1 Convergence almost Everywhere

12.1.2 Pointwise Convergence

12.1.3 Uniform Convergence, Almost Everywhere (a.e.)

12.1.4 Bounded Convergence Theorem

12.1.5 Fatou’s Lemma

12.1.6 Monotone Convergence Theorem

12.1.7 Lebesgue Dominated Convergence Theorem

12.2 Summary

12.3 Keywords



Objectives


Understand bounded convergence theorem.

State and prove monotone convergence theorem and Lebesgue dominated convergencetheorem.

Solve related problems on these theorems.

Introduction

Convergence of a sequence of functions can be defined in various ways and there are situationsin which each of these definitions is natural and useful. In this unit, we shall study aboutconvergence almost everywhere, pointwise and uniform convergence. We shall also provebounded convergence theorem and monotone convergence theorem which are so useful insolving problems on convergence. The dominated convergence theorem is one of the mostimportant results of Lebesgue’s integration theory. It gives a general sufficient condition for thevalidity of proceeding to the limit of a sequence of functions under the integral sign. It is aninvaluable tool to study functions defined by integrals.

12.1 General Convergence Theorems

12.1.1 Convergence almost Everywhere

Let <fn> be a sequence of measurable functions defined over a measurable set E. Then <fn> is saidto converge almost everywhere in E if there exists a subset E0 of E s.t.



Notes (i) fn(x) f (x), x E – E0,

and (ii) m (E0) = 0.

12.1.2 Pointwise Convergence

Let <fn> be a sequence of measurable functions on a measurable set E. Then <fn> is said toconverge “pointwise” in E, if a measurable function f on E such that

fn(x) f (x) x E or

nnlt f (x) = f (x)

12.1.3 Uniform Convergence, Almost Everywhere (a.e.)

Let <fn> be a sequence of measurable functions defined over a measurable set E. Then thesequence <fn> is said to converge uniformly a.e. to f, if a set E0 E s.t.

(i) m (E0) = 0 and

(ii) <fn> converges uniformly to f on the set E – E0.

12.1.4 Bounded Convergence Theorem

Theorem 1: State and Prove: Bounded Convergence Theorem

Statement: Let {fn} be a sequence of measurable functions defined on a set E of finite measure, andsuppose that there is a real number M such that |fn(x)| M n and x. If f (x) = nn

lim f (x) for each

x in E, then

E

f = nnE

lim f

Proof: Since f (x) = nnE

lim f (x) and fn is measurable on E

f is also measurable on E

Let > 0 be given

Then measurable set A E with mA < 4M

and a positive integer N such that

|fn(x) – f (x)| < 2mE

n N and x E – A

n

E E

f f = n

E

(f f)

n

E

(f f)



Notes= n n

E A A

(f f) (f f) as (E – A) A =

n

E A A

1 f f2mE

A

m(E A) 2M 12mE

mE 2M mA2mE

as m (E – A) mE

< 2M2 4M

= 2 2

=

Thus n

E E

f f <

But was arbitrary

nnE

lim f = E

f

12.1.5 Fatou’s Lemma

If {fn} is a sequence of non-negative measurable functions and fn(x) f (x) almost everywhere ona set E, then

E

f nn

E

lim f

i.e.E

f nnE

lim inf f

Proof: Since integrals over sets of measure zero are zero.

Without loss of generality, we may assume that the convergence is everywhere. Let h bea bounded measurable function with h f and h (x) = 0 outside a set E E of finitemeasure.

Define a function hn by

hn(x) = Min. {h(x), fn(x)}

then hn(x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as x E – E h(x) = 0 hn(x) = 0 because



Notes Since hn = h or hn = fn

hn is measurable function on E

If hn = h, then hn h

If hn = fn < h f

then fn h as fn f

hn h

Thus hn h

Since hn (x) h (x) for each x E and {hn} is a sequence of bounded measurable functionson E

By Bounded Convergence Theorem

nnE E E E E E

h h h h lim h

as E = (E – E ) E & (E – E ) E =

= nn

E

lim h

n n nn

E

lim f as h f

nn

E

lim f as E E

nn

E E

h lim f

Taking supremum over all h f, we get

n fE

sup h nn

E E

h lim f

E

f nn

E E

f lim f

Remarks:

(1) If in Fatou’s Lemma, we take

fn(x) = 1, n x n 10, otherwise

with E = R

then nn

E E

f lim f

Thus in Fatou’s Lemma, strict inequality is possible.



Notes(2) fn 0 x E is essential for Fatou’s Lemma

However, if we take

fn(x) = 1 2n, xn n

0, otherwise

with E = [0, 2]

Then nn

E E

f lim f .

12.1.6 Monotone Convergence Theorem

Statement: Let {fn} be an increasing sequence of non-negative measurable functions and let

f = nnlim f . Then

nnf lim f

Proof: Let h be a bounded measurable function with h f and h (x) = 0 outside a set E E of finitemeasure


hn(x) = Min. {h(x), fn(x)}

then hn(x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as

x E – E h(x) = 0 hn(x) = 0 because fn(x) 0

Since hn = h or hn = fn



then fn h as fn f

hn h

Thus hn h

Since hn(x) h(x) for each x E and {hn} is a sequence of measurable functions on E


nnE E E E E E

h h h h lim h

as E = (E – E ) E & (E – E ) E = .

= nn

E

lim h



Notes= n

nE

lim f

nn

E

lim f as E E

E

h nn

E

lim f


E

sup h nn

E

lim f

E

f nn

E

lim f … (1)

Since {fn} is monotonically increasing sequence and fn f

fn f

nf f

nnlim f f … (2)

From (1) and (2), we have

f n nnnlim f lim f f

f nnlim f

Theorem 2: Let {un} be a sequence of non-negative measurable functions, and let f = nh 1

u .

Then nh 1

f u

Proof: Let fn = u1 + u2 + … + un = n

jj 1

u

then fn f

i.e. nnlim f f

Let h be a bounded measurable function with h f and h(x) = 0 outside a set E E of finitemeasure.


hn(x) = Min. {h (x), fn(x)}



Notesthen hn (x) h(x) and hn(x) fn(x)

hn is bounded by the boundedness of h and vanishes outside E as

x E – E h(x) = 0 hn(x) = 0 because fn(x) 0.

Since hn = h or hn = fn



If hn = fn < h < f

then fn h as fn f

hn h

Thus hn h

Since hn(x) h(x) for each x E and {hn} is a sequence of measurable function on E


nnE E E E E E

h h h h lim h

as E = (E – E ) E & (E – E ) E =

= nn

E

lim h

= nn

E

lim f

nn

E

lim f as E E

E

h nn

E

lim f


h fE

sup h nn

E

lim f … (1)

E

f nn

E

lim f

Since {fn} is monotonically increasing sequence and fn f

fn f

nf f

nnlim f f … (2)



Notes From (1) and (2), we have

f n nnnlim f lim f f

f nnlim f

= nnlim f

= n

jnj 1

lim u

= n

jnj 1

lim u

= n

jj 1

u

Hence f = nn 1

u

Theorem 3: Let f be a non-negative function which is integrable over a set E. Then given > 0there is a > 0 such that for every set A E with mA < , we have

A

f <

Proof: If f is bounded function on E

Then positive real number M such that

|f (x)| M x E

For given > 0 = M

such that for every set A E with mA < , we have

A A

f M M mA M. MM

i.e.A

f

Thus the result is true if f is a bounded function. So assume that f is not a bounded function on E.

Define a function fn on E by

nf(x) if f(x) nf (x) n otherwise



NotesThen each fn is bounded and

fn f at each point

Since {fn} is an increasing sequence of bounded functions such that fn f on E

By the monotone convergence theorem

nnE E

lim f f

For given > 0 a positive integer N such that

n

E E

f f for n N2

N

E E

f f2

N

E E

f f2 2

N

E

(f f )2

Choose 2N

If mA < , then we have

A

f = N N

A

(f f ) f

= N N

A A

(f f ) f

N N

E A

(f f ) N as f N

< NmA2

< N2

< N2 2N

= 2 2

=



Notes

A

f

12.1.7 Lebesgue Dominated Convergence Theorem

Theorem 4: State and prove Lebesgue dominated convergence theorem

Statement: Let g be an integrable function on E and let {fn} be a sequence of measurable functionssuch that |fn| g on E and nn

lim f = f a.e. on E. Then

nnE E

f lim f .

Proof: Since we know that if f is a measurable function over a set E and there is an integrablefunction g such that |f| g, then f is integrable over E. So clearly, each fn is integrable over E.

Also nnlim f = f a.e. on E.

and |fn| g a.e. on E

|f| g a.e. on E.

Hence f is integrable over E.

Let { n} be a sequence of functions defined by n = fn + g. Clearly, n is a non-negative and integrablefunction for each n.

Therefore, by Fatou’s Lemma, we have

E

(f g) nn

E

lim (f g)

E

f nn

E

lim f … (1)

Similarly, let { n} be a sequence of functions defined by n = g – fn. Clearly n is a non-negativeand integrable function for each n. So, given by Fatou’s Lemma, we have

E

(g f) nn

E

lim (g f )

E E

g f nn

E E

g lim f

E

f nn

E

lim f

E

f n

E

lim f … (2)

Hence from (1) & (2), we get

E

f = n n

E E

lim f lim f



NotesBut n

E

lim f = n n

E E

lim f lim f

HenceE

f = n

E

lim f .

Corollary: Let {un} be a sequence of integrable functions on E such that nn 1

u converges a.e. on

E. Let g be a function which is integrable on E and satisfy n

ii 1

u g a.e. on E for each n. Then

nn 1

u is integrable on E and n nn 1 n 1E E

u u .

Proof: Let n

i ni 1

u f .

Applying Lebesgue Dominated Convergence Theorem for the sequence {fn}, we get

n nn 1 n 1E E

u u

Corollary: If f is integrable over E and {Ei} is a sequence of disjoint measurable sets such that

ii 1

E E , then

i 1E Ei

f f

Proof: Since {Ei} is a sequence of disjoint measurable sets, we may write.

f = Eii 1

f

The function f. Ei is integrable over E since Eif |f| and |f| is integrable over E. Moreover

n

Eii 1

f |f|, n N

Thus the conditions of above corollary are satisfied and hence

E

f = Eii 1E

f



Notes= Ei

i 1 E

f

= i 1 Ei

f

Example: Show that the theorem of bounded convergence applies to fn(x) = 2 2

nx1 n x

, 0

x 1.

Sol: fn(x) = 2 2

nx1 n x

= 1

1 nxnx

= 21

1 nx 2nx

12

Thus a number 12

such that |fn(x)| 12

.

Hence it satisfies the conditions of bounded convergence theorem.

Now

1

nn0

lim f (x) dx = 1

2 2

0

nxlim dx1 n x

= 2 2

n

1lim log(1 n x ) form2n

= 2 2 2

n

[1/(1 n x )] 2nxlim2

[Using L’Hospital Rule]

= 2

2 2n

nxlim1 n x

= 2

n 22

1 xnlim 01 x

n

and1

nn0

lim f (x) dx = 1

2 2n0

nxlim dx1 n x



Notes=

1

0

(0) dx 0

1

nn0

lim f (x) dx = 1

nn0

lim f (x) dx

This verifies the result of bounded convergence theorem.

Example: Use Lebesgue dominated convergence theorem to evaluate 1

nn0

lim f (x) dx ,

where

fn(x) = 3/2

2 2

n x1 n x

, n = 1, 2, 3, … 0 x 1.

Solution: fn(x) = 3/2

2 2

n x1 n x

= 3/2 2

2 2

1 n xx 1 n x

1 g(x), (say)x

fn(x) g (x)

and g (x) L (0, 1],

Hence by Lebesgue Dominated Convergence Theorem.

1

nn0

lim f (x) dx = 1

nn0

lim f (x) dx

= 1 3/2

2 2n0

n xlim dx1 n x

= 1

n 20 2

1 xlim dx1n xn

= 1

0

0 dx = 0.

Example: If (fn) is a sequence of non-negative function s.t. fn f and fn f for each n, showthat

nf lim f



Notes Solution: From the given hypothesis it follows that

nlim f f … (1)

Also by Fatou’s Lemma, we have

f nlim f … (2)

Then from (1) and (2), we get

f n nlim f lim f f .

Hence f = n n nlim f lim f lim f .

Example: If > 0, prove that n n

1 x 1

no o

xLim 1 x dx e .x dxn

, where the integrals are

taken in the Lebesgue sense.

Solution: If fn(x) =n

1x1 .x 0n

, then fn(x) g(x), where g(x) = e–x.x –1 n

x

n

xrecall Lim 1 en

Also g(x) L[0, ], hence by Lebesgue dominated convergence theorem, we get

n

nno

Lim f (x) dx = nno

Lim f (x) dx

= n

1

no

xLim 1 .x dxn

= x 1

o

e .x dx

Example: Show that if > 1,

1

1

o

xsin x dx 0(n ) as n1 (nx)

.

Solution: Consider the sequence <fn(x)> s.t.

fn(x) = nx sin x1 (nx)

, n = 1, 2, ………

Obviously since > 1, and x [0, 1]



Notesnx sin x 11 (nx)

If (x) = 1, x, then | fn (x) | (x), x.

Hence by dominated convergence theorem, we get

1 1 1

n n0 0 0

nx sin x nx sin xLim dx Lim dx (0) dx 01 (nx) 1 (nx)

1

n0

x sin xLim n dx1 (nx)

= 0

1

0

x sin x dx1 (nx)

= 0 (n–1).

Example: Show that 2 22 n x

2na

n xeLim dx1 x

= 0, if a > 0, but not for a = 0.

Solution: If a > 0, putting nx uand du ndx , we get

2 2 22 n 2 u u

(na, )2 2 2 2 2

a na o

n xe x ue du uedx du1 x 1 u /n 1 u /n

Also 2u

2u(na, )2 2

u.e u.e L [0, ]1 (u /n )

and 2u

(ne , ) ( , )2 2n

u.eLim 0 as 01 u /n

.

Hence by Lebesgue dominated convergence theorem, we obtain

2 2 22 n x u

(na, )2 2 2n na a

n xe u.eLim dx Lim du1 x 1 u /n

= 2u

(na, ) 2 2na o

u.eLim du 0 dx 01 u /n .

Now when a = 0,

12 2 2 22 n x 2 n x

2 2

o 0

n xe n xedx dx1 x 1 x



Notes 12 22 n x

0

1 n x.e dx2

(putting 1 in place of x2)

= 12 2n x

0

1 1e4 4

12.2 Summary

Bounded Convergence Theorem: Let {fn} be a sequence of measurable functions defined ona set E of finite measure, and suppose that there is a real number M such that |fn(x)| < M

n and all x. If f (x) = nnlim f (x) for each x in E, then

nnE E

f lim f

Monotone Convergence Theorem: Let {fn} be an increasing sequence of non-negativemeasurable functions and let f = nn

lim f . Then

nnf lim f

Lebesgue Dominated Convergence Theorem: Let g be an integrable function on E and let {fn}be a sequence of measurable functions such that |fn| g on E and nn

lim f = f a.e. on E. Then

nnE E

f lim f

12.3 Keywords

Convergence almost Everywhere: Let <fn> be a sequence of measurable functions defined over ameasurable set E. Then <fn> is said to converge almost everywhere in E if there exists a subset E0

of E s.t.

(i) fn(x) f (x), x E – E0,

and (ii) m (E0) = 0.

Convergence: Refers to the notion that some functions and sequence approach a limit undercertain conditions.

Fatou’s Lemma: If {fn} is a sequence of non-negative measurable functions and fn(x) f (x) almosteverywhere on a set E, then

nnE E

f lim inf f

Pointwise Convergence: Let <fn> be a sequence of measurable functions on a measurable set E.Then <fn> is said to converge “pointwise” in E, if a measurable function f on E such that

fn(x) f (x) x E or

nnlt f (x) = f (x)



NotesUniform Convergence, Almost Everywhere (a.e.): Let <fn> be a sequence of measurable functionsdefined over a measurable set E. Then the sequence <fn> is said to converge uniformly a.e. to f,if a set E0 E s.t.

(i) m (E0) = 0 and

(ii) <fn> converges uniformly to f on the set E – E0.


1. Show that we may have strict inequality in Fatou’s Lemma.

2. Let <fn> be an increasing sequence of non-negative measurable functions, and let f = lim fn.

Show that nf lim f .

Deduce that nn 1

f u , if un is a sequence of non-negative measurable functions and

nn 1

f u .

3. State the Monotone Convergence theorem. Show that it need not hold for decreasingsequences of functions.

4. Let {gn} be a sequence of integrable functions which converge a.e. to an integrable functiong. Let {fn} be a sequence of measurable functions such that |fn| gn and {fn} converges to fa.e.

If nng lim g

then prove that nnf lim f .

5. State and prove monotone convergence theorem.


Books G.F. Simmons, Introduction to Topology and Modern Analysis, New York: McGrawHill, 1963.

H.L. Royden, Real analysis, Prentice Hall, 1988.

Online links dl.acm.org

math.stanford.edu

www.springerlink.com



Notes Unit 13: Signed Measures

CONTENTS

Objectives

Introduction

13.1 Signed Measures

13.1.1 Signed Measure: Definition

13.1.2 Positive Set, Negative Set and Null Set

13.1.3 Hahn Decomposition Theorem

13.1.4 Hahn Decomposition: Definition

13.2 Summary

13.3 Keywords



Objectives


Define signed measure.

Describe positive and negative and null sets.

Solve problems on signed measure.

Introduction

We have seen that a measure is a non-negative set function. Now we shall assume that it takesboth positive and negative values. Such assumption leads us to a new type of measure known assigned measure. In this unit, we shall start with definition of signed measure and we shall provesome important theorems on it.

13.1 Signed Measures

13.1.1 Signed Measure: Definition

Definition: Let the couple (X, ) be a measurable space, where represents a -algebra ofsubsets of X. An extended real valued set function

: [– , ]

defined on is called a signed measure, if it satisfies the following postulates:

(i) assumes at most one of the values – or + .

(ii) ( ) = 0.


Unit 13: Signed Measures

Notes(iii) If <An> is any sequence of disjoint measurable sets, then

n nn 1n 1

A (A ),

i.e., is countably additive.

From this definition, it follows that a measure is a special case of a signed measure. Thus, everymeasure on is a signed measure but the converse is not true in general, i.e. every signedmeasure is not a measure in general.

If – < (A) < , for very A , then we say that signed measure is finite.

13.1.2 Positive Set, Negative Set and Null Set

Definition

(a) Positive Set: Let (X, ) be a measurable space and let A be any subset of X. Then A X issaid to be a positive set relative to a signed measure defined on (X, ), if

(i) A , i.e. A is measurable.

(ii) (E) 0, E A s.t. E is measurable.

Obviously, it follows from the above definition that:

(i) every measurable subset of a positive set is a positive set,

(ii) is a positive set w.r.t. every signed measure.

Also for A to be positive. (A) 0 is the necessary condition, but not in general sufficientfor A to be positive.

(b) Negative Set: Let (X, ) be a measurable space. Then a subset A of X is said to be a negativeset relative to a signed measure defined on measurable space (X, ) if

(i) A i.e., A is measurable.


set A is negative w.r.t. , provided it is positive w.r.t. – .

(c) Null Set: A set A X is said to be a null set relative to a signed measure defined onmeasurable space (X, ) is, A is both positive and negative relative to .

Thus, measure of every null set is zero.

Now, we know that a measurable set is a set of measure zero, iff every measurable subset of ithas measure zero. Thus, if A X is a null set relative to then (E) = 0, measurable subsetsE A. In other words.

A is a null set (E) = 0, measurable subsets E A.

Theorem 1: Countable union of positive sets w.r.t. a signed measure is positive.

Proof: Let (X, ) be a measurable space and let be a signed measure defined on (X, ). Let <An>

be a sequence of positive subsets of X, let A = n

ii 1

A and let B be any measurable subset of A.

Set n = C Cn n 1 1B A A A , n N.



Notes where CnA (n 1,2,3 n 1) denotes complement of An(n = 1, 2, 3, … n – 1) with respect to X.

Now, we know that complement of a measurable set is also measurable so that eachCnA (n 1,2,3 n 1) is measurable relative to . Again, intersection of countable collection of

measurable sets is also measurable. Hence Bn is a measurable subset of the positive set An. Thus

(Bn) 0 (by the definition of positive set) … (i)

Obviously, the set Bn are disjoint and

if B = nn 1

B , we get … (ii)

(B) = nn 1

(B ) … (iii)

In view of (i).

(B) 0.

Thus, we have

(1) A is measurable for

An is a positive set An is a measurable set

countable union nn 1

A is measurable,

A = nn 1

A is measurable

(2) (B) 0, B A s.t. B is a measurable set.

Hence A is a positive set, by definition.

Theorem 2: Let (X, ) be a measurable space and let be a signed measure defined on (X, A). If Bis a measurable set with finite negative measure i.e., – < (B) < 0, then prove that B contains anegative set A B with the property (A) < 0.

Proof: If B is itself a negative set, then we may take A = B and theorem is done. Thereforeconsider the case when B is not a negative set. Then there must exist a measurable subset E 1 Band a smallest positive integer n1, s.t.

(E1) > 1

1n

B = (B – E1) E1 and (B – E1) E1 = ,

(B) = (B – E1) + (E1) … (i)

or (B – E1) = (B) – (E1) … (ii)

Since (B) is finite, (i) implies that (B – E1) and (E1) are finite. Again (B) < 0, (ii) implies that(B – E1) < 0.



NotesNow, the set B – E1 is either negative or contains a subset of positive measure. If the set B – E1 isa negative set, then we may take A = B – E1 and the theorem is done. So, suppose that B – E1 is nota negative set. Then there must exist a measurable subset E2 of B – E1 and a smallest positivenumber n2 with a property

(E2) > 2

1n .

Since B = (B – E1 E2) (E2 E2),

and (B – E1 E2) (E1 E2) = ,

we have (B) = (B – E1 E2) + (E2 E2)

or (B – E1 E2) = (B) – (E2 E2)

= (B) – (E1) – (E2).

As before, (B – E1 E2) > 0 [ (B) < 0, (Er) > 0 for r = 1, 2]

Thus, B – E1 E2 is a set of negative measure, which is either a negative set or contains a subsetof positive measure. If B – E1 E2 is a negative set, then the theorem is done by taking B = A – E1

E2. Otherwise we repeat the above process.

On repeating this process, at some stage we shall get either a negative subset A B s.t. (A) < 0or a sequence <Er> of disjoint measurable sets and a sequence <nr : r N> of positive integers s.t.

Er B – r 1

nn 1

E and r

1n < (Er) <

In first case, we have nothing to do. In the latter case, let

A = B – nn 1

E or B = A nn 1

E … (iii)

Then as before, it follows that

(B) = (A) + nn 1

(E ) .

> (A) + kk 1

1n … (iv)

[ change of suffix is in material]

Since (B) is finite and assumes at most one of the values – and , it follows from (iv) that

(A) is finite and the series kk 1

1n

is convergent.

Then (A) < (B) – kk 1

1n

= a finite negative number

( (B) is a finite negative number)



Notes or (A) < 0.

Again, we know that difference of two measurable sets is measurable and enumerable union ofmeasurable sets is measurable therefore it follows from (iii) that A is a measurable set.

Now we shall prove that A is a negative set. Let E A be an arbitrary measurable set.

Since A = B – nn 1

E ,

E = B – nn 1

E .

Since nk , we can choose k so large that

(E) k 1

1n

Letting nk , we obtain

(E) 0.

Thus we have

(1) A is measurable.

(2) (E) 0, E A s.t. E is measurable.

Hence A is a negative set.

13.1.3 Hahn Decomposition Theorem

Theorem 3: Let be a signed measure on a measurable space (X, ). Then there exists a positiveset P and a negative set Q s.t.

P Q = and P Q = X.

Proof: Let (X, ) be a measurable space and let be a signed measure defined on a measurablespace (X, ). Since, by definition, assumes at most one of the values + or collection of allnegative subsets of X w.r.t. and let be a collection of all negative subsets of X w.r.t. and let

k = inf { (E) : E )

(i) that there exists a sequence <En> in such that

nnLim (E ) = k.

Let Q = nn 1

E .

Since is a family of negative sets, < En> is a sequence of negative sets. Again, we know byremark of theorem 1 that countable union of negative sets is negative, it follows that Q is anegative subset of X so that

(Q) K.



NotesNow, Q – En is a subset of Q, it follows that

(Q – En) 0.

Since (Q – En) En =

and Q = (Q – En) En,

we have (Q) = (Q – En) + (En)

(Q) (En), n N and En B.

Therefore (Q) K. … (iii)

(ii) and (iii) (Q) = K – < k. … (iv)

Now we shall show that p = QC, the complement of Q w.r.t. is a positive subset of X. Supposenot, i.e. P is negative. Then E P s.t. E is measurable and (E) < 0. Now we know that if– < (E) < 0, we get a negative set A E s.t. (A) < 0.

A, Q are distinct negative subsets of X

A Q is negative set

(A Q) K [using (i)]

(A) + (Q) K,

(A) + K K, [using (iv)]

(A) 0,

a contradiction, for (A) < 0

P = QC is a positive subset of X

Q is a negative subset of X.

Thus X = P Q, P Q = .

13.1.4 Hahn Decomposition: Definition

A decomposition of a measurable space X into two subsets s.t. X = P Q, P Q = ,

where P and Q are positive and negative sets respectively relative to the signal measure , iscalled as Hahn decomposition for the signed measure . P and Q are respectively called positiveand negative components of X.

Notice that Hahn decomposition is not unique.

13.2 Summary

Let the couple (X, ) be a measurable space, where represents a -algebra of subsets ofX. An extended real-valued set function

: [– , ]



(ii) ( ) = 0.

(iii) If <An> is any sequence of disjoint measurable sets, then is countably additive.



Notes Let (X, ) be a measurable space and then A X is said to be a positive set relative to asigned measure defined on (X, ) if

(i) A is measurable


Let (X, ) be a measurable space. Then A X is said to be negative set relative to a signedmeasure if

(i) A is measurable


A X is said to be a null set relative to a signed measure defined on measurable space(X, ) is: A is both positive and negative relative to .

13.3 Keywords

Hahn Decomposition: Definition: A decomposition of a measurable space X into two subsets s.t.X = P Q, P Q = .

Negative Set: Let (X, ) be a measurable space. Then a subset A of X is said to be a negative setrelative to a signed measure defined on measurable space (X, ) if

(i) A i.e., A is measurable.


Null Set: A set A X is said to be a null set relative to a signed measure defined on measurablespace (X, ) is, A is both positive and negative relative to .

Positive Set: Let (X, ) be a measurable space and let A be any subset of X. Then A X is said tobe a positive set relative to a signed measure defined on (X, ), if

(i) A , i.e. A is measurable.


Signed Measure: Let the couple (X, ) be a measurable space, where represents a -algebra ofsubsets of X. An extended real valued set function

: [– , ]



(ii) ( ) = 0.


1. If (E) = 2x

E

xe dx, then find positive, negative and null sets w.r.t. . Also give a Hahn

decomposition of R w.r.t. .

2. State and prove Hahn decomposition theorem for signed measures.

3. If is a measure and 1, 2 are the signed measures given by 1 (E) = (A E), 2 (E) = (B E), where (A B) = 0, show that 1 2.



Notes4. Show that if 1 and 2 are two finite signed measures, then so is a 1 + b 2 where a, b are realnumbers.


Books Bartle, Robert G., The Elements of Integration, New York – London – Sydney: JohnWiley and Sons

Cohn, Donald L. (1997) [1980], Measure Theory (reprint ed.), Boston – Based –Stuttgart: Birkhauser Verlag

Online links www.maths.bris.ac.uk

www.planetmath.org/signedmeasure.html



Notes Unit 14: Radon-Nikodym Theorem

CONTENTS

Objectives

Introduction

14.1 Radon-Nikodym Theorem

14.1.1 Absolutely Continuous Measure Function

14.2 Summary

14.3 Keywords



Objectives


Define Absolutely continuous measure function

State Radon-Nikodym theorem

Understand the proof of Radon-Nikodym theorem

Solve problems on this theorem

Introduction

In mathematics, the Radon-Nikodym theorem is a result in measure theory that states that givena measurable space (X, ), if a -finite measure on (X, ) is absolutely continuous with respect toa -finite measure on (X, ), then there is a measurable function f on X and taking values in [0, ],such that for any measurable set A.

The theorem is named after Johann Radon, who proved the theorem for the special case wherethe underlying space is RN in 1913, and for Otto Nikodym who proved the general case in 1930.In 1936 Hans Freudenthal further generalised the Radon-Nikodym theorem by proving theFreudenthal spectral theorem, a result in Riesz space theory, which contains the Radon-Nikodymtheorem as a special case.

If Y is a Banach space and the generalisation of the Radon-Nikodym theorem also holds forfunctions with values in Y, then Y is said to have the Radon-Nikodym property. All Hibertspaces have the Radon-Nikodym property.

14.1 Radon-Nikodym Theorem

14.1.1 Absolutely Continuous Measure Function

Let (X, ) be a measurable space and let and be measure functions defined on the space (X, A).The measure is said to be absolutely continuous w.r.t. if

(A) = 0 or | | (A) = 0, A (A) = 0, and is denoted by .


Unit 14: Radon-Nikodym Theorem

Notes

Notes

If is -finite, the converse is also true.

If and are signed measures on (X, ), then if | | | |.

Radon-Nikodym Theorem

Let (X, , ) be a -finite measure space. If be a measure defined on A s.t. is absolutelycontinuous w.r.t. , then there exists a non-negative measurable function f on s.t.

(A) = A

f d , A .

The function f is unique in the sense that if g is any measurable function with the propertydefined as above, then f = g almost everywhere with respect to .

Proof: To establish the existence of the function f, we shall use the following two Lemmas:

Lemma 1: Let E be a countable set of real numbers. Let for each a E there is a set Fa s.t.Fa Fb, whenever b < a i.e. <Fn> is a monotonic decreasing sequence of subsets of correspondingto the sequence <an> of real numbers in E. Then a measurable extended real valued function fon X s.t.

f (x) a, x Fa,

and f (x) a, x (X – Fa).

Proof: Let f (x) = inf {a : x Fa} x X and let, conventionally

inf {empty collection of real numbers} =

Now, x Fa f(x) a

x Fa x Fa for every b < a

f (x) a

Now, f (x) < a x Fb for some b < a

or {x : f (x) < a} = bb a

[F ] .

Also x Fb f (x) b < a for some b < a.

Hence f is measurable.

Again, by definition of f, we observe that

f (x) a, x Fa,

and f (x) a, x Fa.

Thus f is the required function.

Lemma 2: Let E be a countable set of real numbers. Let corresponding to each a E, there is a setFa s.t.

(Fa – Fb) = 0 whenever b > a.

Then there exists a measurable function f with the property

x Fa f (x) a a.e.

and x (X – Fa) f (x) > a a.e.



NotesProof: Let P = a b

b a

{F F } .

Evidently (P) = 0.

Let aF = Fa P.

This a bF F = (Fa – Fb) – P = for a < b.

In view of Lemma 1, it follows that a measurable function f s.t.

f (x) a, x Fa

and f (x) a, x X – Fa

Thus we have

a

a

x F f(x) a a.e., except for x P.x X F f(x) a a.e.,

Proof of the main theorem

At first, suppose that is finite.

( – a ) is a signed measure on for each rational number a.

Let (Pa, Qa) be a Hahn decomposition for the measure ( – a ).

Let PO = X and QO = .

By the definition of Hahn decomposition theorem,

Pa Qa = X,

and Pb Qb = X.

Therefore, Qa – Qb = Qa – (X – Pb)

= Qa Pb.

Thus, ( – a ) (Qa – Qb) 0 … (i)

Similarly, we can prove that

( – b ) (Qa – Qb) 0 … (ii)

Let a < b, then from (i) and (ii), we have

(Qa – Qb) = 0.

Therefore, by Lemma (ii)

f (x) a, a.e. x Pa

and f (x) a, a.e. x Qa,

where f is measurable

Since QO = , it follows that f is non-negative

Again, let A be arbitrary.

Define Ar = A r 1 r

o o

Q Qn n



NotesA = A – r

o

Qn

.

Evidently, A = A rr o

A ,

where A is disjoint union of measurable sets.

(A) = (A ) + rr o

(A ) .

Obviously Ar r 1 r

o o

Q Qn n

ro o

r r 1f(x) , x An n

r ro oAr

r r 1(A ) fd (A )n n [by first mean value theorem]

Again r r ro o

r r 1(A ) (A ) (A )n n

, we have

r r r ro oAr

1 1(A ) (A ) fd (A ) (A )n n … (iii)

Now, if (A ) > 0, then g (A ) = 0, [ ( – a ) A is positive, a]

and (A ) = 0 if (A ) = 0 [ )]

In either case, (A ) = Ar

f d .

Adding the inequalities (iii) over r, we get

ro oA

1 1(A) (A) f d (A) (A ).n n

Since no is arbitrary and (A) is assumed to be finite, it follows that

A(A) f d A .

To show that the theorem is true for -finite measure , decompose X into a countable union ofXi of finite -measure. Applying the same argument as above for each X i, we get the requiredfunction.

To show the second part, let g be any measurable function satisfying the condition,

(A) = A

f d A .



Notes For each n N, define

A n = 1x X : f(x) g(x)n

and Bn = 1x X : g(x) f(x) .n

Since f (x) – g (x) n1 , x An

, we have by first mean value theorem

n

An

1(f g)d (A )n

n

A An n

1fd gd (A )n

(An) – (An) n n1 1(A ) or 0 (A )n n

(An) 0.

Since (An) is always greater than equal to zero, we have (An) = 0.

Similarly, we can show that

(Bn) 0.

If C = {x X : f (x) g (x)}

= n nn 1

(A B ),then (C) = 0 f = g a.e. on X w.r.t. .

Hence the theorem.

Theorem 1: If 1, 2 are -finite signed measures on (X, A) and 1 , 2 , then

1 2 1 2 1 1d( ) d d d d( )andd d d d d

Proof: Since 1, 2 are -finite and 1 , 2 , we have that 1 + 2 is also -finite and

1 + 2 .

Now for any A ,

( 1 + 2) (A) = 1 (A) + 2 (A)

= 1 2 1 2

A A A

d d d dd d dd d d d

1 2 1 2

A A

d d dd dd d d



Notes1 2 1 2d( ) d dd d d

Prove the other result yourself.

Theorem 2: If is a -finite signed measures and is a -finite measure s.t. , show that

d| | d .d d

Proof: Let = + – – with Hahn decomposition A, B.

Then on A, d dd d

and on B, d dd d

d d d d( ) d| |d d d d d .

Theorem 3: If be a -finite signed measure and , be -finite measures on (X, A) s.t. , : then show that

d d dd d d

Proof: Since we may write = + – – and

d d( ) d d( ),d d d d

.

we need to prove the above result for measures only.

If d fd

and d gd

, (f, g are non-negative functions as obtained in Radon-Nikodym Theorem),

then we need to prove that

(F) = F

fg d .

Let be a measurable simple function s.t.

= n

i Eii 1

a ,

then n

i ii 1F

d a (E F)

= n

i gi 1 E F Fi

a gd d .

Let < n> be a sequence of measurable simple function which converges to f, then

(F) = n

F F

fd lim d .



Notes= n g n

F F

lim gd f d as g fg

= d d dufgd d d .

14.2 Summary

Let (X, ) be a measurable space and let r and m be measure functions, defined on the space(X, A). The measure is said to be absolutely continuous w.r.t. if


Let (X, , ) be a -finite measure space. If Y be a measure defined on A s.t. is absolutelycontinuous w.r.t. , then there exists a non-negative measurable function f s.t.

(A) = A

fd , A


14.3 Keywords

Absolutely Continuous Measure Function: Let (X, ) be a measurable space and let and bemeasure functions defined on the space (X, A). The measure is said to be absolutely continuousw.r.t. if


Radon-Nikodym Theorem: Let (X, , ) be a -finite measure space. If be a measure defined onA s.t. is absolutely continuous w.r.t. , then there exists a non-negative measurable function fon s.t.

(A) = A

f d , A .



1. Show that 1

d d ,d d

where and are -finite signed measures and , .

2. If (E) = E

fd , , where E

fd exists, then find | | (E).

3. State and prove Radon-Nikodym theorem.



Notes14.5 Further Readings

Book G.E. Shilov and B.L. Gurevich, Integral, Measure and Derivative: A Unified Approach,Richard A. Silverman, trans. Dover Publications, 1978.

Online links www.math.ksu.edu/nnagy/real-an/4-04-rn.pdf

mathworld.woltram.com

www.csun.edu

pioneer.netserv.chula.ac.th



Notes Unit 15: Banach Space: Definition and Some Examples

CONTENTS

Objectives

Introduction

15.1 Banach Spaces

15.1.1 Normed Linear Space

15.1.2 Convergent Sequence in Normed Linear Space

15.1.3 Subspace of a normed Linear Space


15.1.5 Banach Space

15.2 Summary

15.3 Keywords



Objectives


Know about Banach spaces.

Define Banach spaces.

Solve problems on Banach spaces.

Introduction

Banach space is a linear space, which is also, in a special way, a complete metric space. Thiscombination of algebraic and metric structures opens up the possibility of studying lineartransformations of one Banach space into another which have the additional property of beingcontinuous. The concept of a Banach space is a generalization of Hilbert space. A Banach spaceassumes that there is a norm on the space relative to which the space is complete, but it is notassumed that the norm is defined in terms of an inner product. There are many examples ofBanach spaces that are not Hilbert spaces, so that the generalization is quite useful.

15.1 Banach Spaces

15.1.1 Normed Linear Space

Definition: Let N be a complex (or real) linear space. A real valued function n : N R is said todefine, a norm on N if for any x, y N and any scalar (complex number) , the followingconditions are satisfied by n:

(i) n (x) 0, n (x) = 0, x = 0;

(ii) n (x + y) n (x) + n (y); and


Unit 15: Banach Space: Definition and Some Examples

Notes(iii) n ( x) = | | n (x)

It is customary to denote n (x) by

n (x) = x (read as norm x)

With this notation the above conditions (i) – (iii) assume the following forms:

(i) x 0, x = 0 x = 0;

(ii) x + y x + y ; and

(iii) x = x .

A linear space N together with a norm defined on it, i.e., the pair (N, ) is called a normed linearspace and will simply be denoted by N for convenience.

Notes

1. The condition (ii) is called subadditivity and the condition (iii) is called absolutehomogeneity.

2. If we drop the condition viz. x = 0 x = 0, then is called a semi norm (or pseudonorm) or N and the space N is called a semi-normed linear space.

Theorem 1: If N is a normed linear space and if we define a real valued function d : N × N R byd (x, y) = x – y (x, y N), then d is a metric on N.

Proof: We shall verify the conditions of a metric

(i) d (x, y) 0, d (x, y) = 0 x – y = 0 x = y;

(ii) d (x, y) = x – y = (–1) (y – x) = |–1| y – x = y – x = d (y, x);

(iii) d (x, y) = x – y = x – z + z – y (z = N)

x – z + z – y = d (x, z) + d (z, y)

Hence, d defines a metric on N. Consequently, every normed linear space is automatically ametric space.


Notes

1. The above metric has the following additional properties:

(i) If x, y, z N and is a scalar, then

d (x + z, y + z) = (x + z) – (y + z) = x – y = d (x, y).

(ii) d ( x, y) = x – y = (x – y)

= | | x – y = | | d (x, y).

2. Since every normed linear space is a metric space, we can rephrase the definition ofconvergence of sequences by using this metric induced by the norm.



Notes 15.1.2 Convergent Sequence in Normed Linear Space

Definition: Let (N, ) be a normed linear space. A sequence (xn) is N is said to converge to anelement x in N if given > 0, there exists a positive integer no such that

xn – x < for all n no.

If xn converges to x, we write nnLim x x .

or xn x as n

It follows from the definition that

xn x xn – x 0 as n

Theorem 2: If N is a normed linear space, then

x y x – y for any x, y N

Proof: We have

x = (x – y) + y

x – y + y

x – y x – y … (1)

Using (1), we have

– ( x – y ) = y – x y – x

But y – x = (–1) (x – y) = |–1| x – y

Therefore

– ( x – y ) x – y so that

x – y – x – y … (2)


x y x – y


15.1.3 Subspace of a Normed Linear Space

Definition: A subspace M of a normed linear space is a subspace of N consider as a vector spacewith the norm obtain by restricting the norm of N to the subset M. This norm on M is said to beinduced by the norm on N. If M is closed in N, then M is called a closed subspace of N.

Theorem 3: Let N be a normed linear space and M is a subspace of N. Then the closure M of M isalso a subspace of N.

(Note that since M is closed, M is a closed subspace).

Proof: To prove that M is a subspace of N, we must show that any linear combination of

element in M is again in M. That is if x and y M , then x + y M for any scalars and .



NotesSince x, y M , there exist sequences (xn) and (yn) in M such that

xn x and yn y,

By joint continuity of addition and scalar multiplication in M.

xn + yn x + y for every scalars and .

Since xn + yn M, we conclude that

x + y M and consequently M is a subspace of N.


Notes

1. The scalars , can be assumed to be non-zero.

For if = 0 = , then

x + y = 0 M M

2. In a normed linear space, the smallest closed subspace containing a given set ofvectors S is just the closure of the subspace spanned by the set S. To see this, let S bethe subset of a normed linear space N and let M be the smallest closed subspace of N,

containing S. We show that M = [S] , where [S] is the subspace spanned by S.

By theorem, [S] is a closed subspace of N and it contains S.

Since M is the smallest closed subspace containing S, we have

M [S] .

But [S] M and M = M , we must have

[S] M = M so that [S] M.

Hence [S] = M.


Definition: A normed linear space N is said to be complete if every Cauchy sequence in Nconverges to an element of N. This means that if xm – xn 0 as m, n , then there exists x N such that

xn – x 0 as n .

15.1.5 Banach Space

Definition: A complete normed linear space is called a Banach space.

OR

A normed linear space which is complete as a metric space is called a Banach space.



Notes In the definition of a Banach space completeness means that if

xm – xn 0 as m, n , where (xn) N, then

a x N such that

xn – x 0 as n .

Note A subspace M of a Banach space B is a subspace of B considered as a normed linearspace. We do not require M to be complete.

Theorem 4: Every complete subspace M of a normed linear space N is closed.

Proof: Let x N be any limit point of M.

We have to show that x M.

Since x is a limit point of M, there exists a sequence (xn) in M and xn x as n .

But, since (xn) is a convergent sequence in M, it is Cauchy sequence in M.

Further M is complete (xn) converges to a point of M so that x M.

Hence M is closed.


Theorem 5: A subspace M of a Banach space B is complete iff the set M is closed in B.

Proof: Let M be a complete subspace of a Banach space M. They be above theorem, M is closed(prove it).

Conversely, let M be a closed subspace of Banach space B. We shall show that M is complete.

Let x = (xn) be a Cauchy sequence in M. Then

xn x in B as B is complete.

We show that x M.

Now x M x M ( M being closed M = M )

Thus every Cauchy sequence in M converges to an element of M. Hence the closed sequence Mof B is complete. This completes the proof of the theorem.

Example 1: The linear space R of real numbers or C of complex numbers are Banachspaces under the norm defined by

x = |x|, x R (or C)

Solution: We have

x = |x| > 0 and x = 0 |x| = x = 0

Further, let z1, z2 C and let 1z and 2z be their complex conjugates, then

|z1 + z2|2 = (z1 + z2) 1 2(z z )

= 1 2 1 21 1 2 2z z z z z z z z

2 221 1 2z 2 z z z



Notes= 2 2

1 1 2 2z 2 z z z 1 2 1 2 1 2z , z z , z z z

= (|z1| + |z2|)2

|z1 + z2| |z1| + |z2|

or z1 + z2 z1 + z2 ( x = |x|)

Also x = | x| = | | |x| = | | x

Hence all the conditions of normed linear space are satisfied. Thus both C or R are normed linearspace. And by Cauchy general principle of convergence, R and C are complete under the matricesinduced by the norm. So R and C are Banach spaces.

Example 2: Euclidean and Unitary spaces: The linear space Rn and Cn of all n-tuples (x1,x2 …, xn) of real and complex numbers are Banach spaces under the norm

x =

1/2n

2i

i 1

|x |

[Usually called Euclidean and unitary spaces respectively].

Solution: (i) Since each |xi| 0, we have

x 0

and x = 0 n

2i

i 1

|x | = 0 xi = 0, i = 1, 2, …, n

(x1, x2, … xn) = 0

x = 0

(ii) Let x = (x1, x2, …, xn)

and y = (y1, y2, … yn) be any two numbers of Cn (or Rn). Then

x + y 2 = (x1, x2, …, xn) + (y1, y2, … yn) 2

= (x1+ x1), (x2 + y2), …, (xn + yn) 2

= n

2i i

i 1

|x y |

n

i i i ii 1

|x y |(|x | |y |)

n n

i i i i i ii 1 i 1

|x y ||x | |x y ||y |

Usually Cauchy inequality for each sum, we get

x + y 2 =

1 1 12 2 2 2n n n n2 2 2

i i i i i ii 1 i 1 i 1 i 1

|x y | |x | |x y | |y |



Notes = x + y x + x + y y

= ( x + y ) ( x + y ).

If x + y = 0, then the above inequality is evidently true.

If x + y 0, we can divide both sides by it to obtain

x + y x + y .

(iii) x =

1 12 2n n

2 2i i

i 1 i 1

| x | | | |x |

= | | x .

This proves that Rn or Cn are normed linear spaces.

Now we show the completeness of Cn (or Rn).

Let < x1, x2, … xn > be a Cauchy sequence in Cn (or Rn). Since each xm is an n-tuple of complex (orreal) numbers, we shall write

xm = (m) (m) (m)1 2 nx , x , , x

So that (m)kx is the kth coordinate of xm.

Let > 0 be given, since <xm> is a Cauchy sequence, there exists a positive integer mo, such that

, m mo mx x

2 2mx x

n

(m) ( ) 2i i

i 1

x x … (1)

(m) ( ) 2i ix x (i = 1, 2, ……, n)

(m) ( )i ix x

Hence (m )i m 1

x is a Cauchy sequence of complex (or real) numbers for each fixed but

arbitrary i.

Since C (or R) is complete, each of these sequences converges to a point, say 2i in C (or R) so that

(m)im

Lim x = zi (i = 1, 2, …, n) … (2)

Now we show that the Cauchy sequence <xm> converges to the point z = (z1, z2, ……, zn) Cn (orRn).

To prove this let in (1). Then by (2) we have

n2(m) 2

i ii 1

x z



Notes xm – z 2 < 2

xm – z <

It follows that the Cauchy sequence <xm> converges to z Cn (or Rn).

Hence Cn or Rn are complete spaces and consequently they are Banach spaces.

15.2 Summary

A linear space N together with a norm defined on it, i.e. the pair (N, ) is called a normedlinear space.

Let (N, ) be a normed linear space. A sequence (xn) in N is said to converge to an elementx in N if given > 0, there exists a positive integer no such that

xn – x < for all n no.

If N is a normed linear space, then

x y x y for any x, y N.

A normed linear space N is said to be complete if every Cauchy sequence in N convergesto an element of N.

A complete normed linear space is called a Banach space.

15.3 Keywords

A Subspace M of a Normed Linear Space: A subspace M of a normed linear space is a subspace ofN consider as a vector space with the norm obtain by restricting the norm of N to the subset M.If norm on M is said to be induced by the norm on N. If M is closed in N, then M is called a closedsubspace of N.

Banach Space: A complete normed linear space is called a Banach space.

Complete Normed Linear Space: A normed linear space N is said to be complete if every Cauchysequence in N converges to an element of N. This means that if xm – xn 0 as m, n , thenthere exists x N such that

xn – x 0 as n .

Normed Linear: A linear space N together with a norm defined on it, i.e., the pair (N, ) is calleda normed linear space and will simply be denoted by N for convenience.


1. Let N be a non-zero normed linear space, prove that N is a Banach space {x : x = 1} iscomplete.

2. Let a Banach space B be the direct sum of the linear subspaces M and N, so that B = M N.If z = x + y is the unique expression of a vector z in B as the sum of vectors x and y in M andN, then a new norm can be defined on the linear space B by z = x + y .

Prove that this actually is a norm. If B symbolizes the linear space B equipped with thisnew norm, prove that B is a Banach space of M and N are closed in B.



Notes 15.5 Further Readings

Books Bourbaki, Nicolas (1987), Topological Vector Spaces, Elements of Mathematics, Berlin:Springer-Verlag.

Beauzamy, Bernard (1985), Introduction to Banach Spaces and their Geometry (Secondrevised ed.), North-Holland.

Online links mathword.wolfram.com?Calculus and Analysis>Functional Analysis

homepage.ntlworld-com/ivan.wilde/notes/fal/fal.pdf

www.math.ucdavis.edu/~ hunter/book.chs.pdf


Unit 16: Continuous Linear Transformations

NotesUnit 16: Continuous Linear Transformations

CONTENTS

Objectives

Introduction

16.1 Continuous Linear Transformation

16.1.1 Continuous Linear Functionals Definition


16.1.3 Norm of a Bounded Linear Functional

16.1.4 Equivalent Methods of Finding F

16.1.5 Representation Theorems for Functionals

16.2 Summary

16.3 Keywords



Objectives


Understand continuous linear transformation

Define bounded linear functional and norm of a bounded linear functional

Understand theorems on continuous linear transformations.

Introduction

In this unit, we obtain the representation of continuous linear functionals on some of Banachspaces.

16.1 Continuous Linear Transformation

16.1.1 Continuous Linear Functionals Definition

Let N be a normed linear space. Then we know the set R of real numbers and the set C ofcomplex numbers are Banach spaces with the norm of any x R or x C given by theabsolute value of x. Thus with our previous notations, (N, R) or (N, C) denote respectivelythe set of all continuous linear transformations from N into R or C.

We denote the Banach space (N, R) or (N, C) by N* and call it by the conjugate space (ordual space or adjoint space) of N.

The elements of N* will be referred to as continuous linear functionals or simply functionalson N.



Notes

Note The conjugate space (N*)* of N* is called the second conjugate space of N and shallbe denoted by N**. Also note that N** is complete too.

Theorem 1: The conjugate space N* is always a Banach space under the norm

f = sup f(x)

: x N, x 0x

… (i)

= sup f(x) : x 1

= inf k, k 0 and f(x) k x x

Proof: As we know that if N, N are normed linear spaces, (N, N ) is a normed linear space. IfN is a Banach space, (N, N ) is Banach space. Hence (N, R) or (N, C) is a Banach space becauseR and C are Banach spaces even if N is not complete.


Theorem 2: Let f be a linear functional on a normed linear space. If f is continuous at xo N, itmust be continuous at every point of N.

Proof: If f is continuous at x = xo, then

xn xo f (xn) f (x)

To show that f is continuous everywhere on N, we must show that for any y N,

yn y f (yn) f (y)

Let yn y as n

Now f (yn) = f (yn – y + xo + y – xo)

since f is linear.

f (yn) = f (yn – y + xo) + f (y) – f (xo) … (1)

As yn y yn – y + xo xo by hypothesis

Also f is continuous, f (yn – y + xo) f (xo) … (2)

From (1) and (2), it follows that

f (yn) f (y) as n .

f is continuous at y N and consequently as it is continuous everywhere on N.

Hence proved.


A linear functional on a normed linear space N is said to be bounded, if there exists a constant ksuch that

f (x) K x x N .



Notes

Note

We may find many K’s satisfying the above condition for a given bounded function. If itis satisfied for one K, it is satisfied for a K1 > K.

Theorem 3: Let f be a linear functional defined on a normed linear space N, then f is bounded f is continuous.

Proof: Let us first show that continuity of f boundedness of f.

If possible let f is continuous but not bounded. Therefore, for any natural number n, howeverlarge, there is some point xn such that

|f (xn)| n || xn|| … (1)

Consider the vector, yn = n

n

xn x so that

ny = 1n .

ny 0 as n

yn 0 in the norm.

Since any continuous functional maps zero vector into zero and f is continuous f (yn) f (0) = 0.

But |f (yn)| = n

1n x f (xn) … (2)

It now follows from (1) & (2) that |f (yn)| > 1, a contradiction to the fact that f (yn) 0 as n .

Thus if f is bounded, then f is continuous.

Conversely, let f is bounded. Then for any sequence (xn), we have

|f (xn)| K || xn|| n = 1, 2, …, and K 0.

Let xn 0 as n then

f (xn) 0 f is continuous at the origin and consequently it is continuous everywhere.


Note The set of all bounded linear function on N is a vector space denoted by N*. As in thecase of linear operators, we make it a normed linear space by suitably defining a norm ofa functional f.

16.1.3 Norm of a Bounded Linear Functional

If f is a bounded linear functional on a normed space N, then the norm of f is defined as:

|| f|| = x 0

f(x)sup

x … (1)



Notes We first note that the above norm is well defined. Since f is bounded, we have

|f (x)| M || x||, M 0.

Let M be the set of real numbers M satisfying this relation. Then the set f(x)

; x 0x

is

bounded above so that it must possess a supremum. Let it be f . So f is well defined and we

must have

f(x)x

f x 0.

or |f (x)| f x .

Let us check that defined by (1) is truly a norm on N*:

If f, g N*, then

f g = x 0

f(x) g(x)sup

x

x 0 x 0

f(x) g(x)sup sup

x x

f g f g .

Similarly, we can see that f f .

16.1.4 Equivalent Methods of Finding F

If f is a bounded linear functional on N, then

|f (x)| M x , M 0.

(I) f = inf {M : M M } where M is the set of all real numbers satisfying

|f (x)| M x ,

Since f M and M is the set of all non-negative real numbers, it is bounded below byzero so that it has an infimum. Hence

f inf {M : M M } … (2)

For x 0 and M M we have f(x) M.x

Since M is the only upper bound then from

definition (2), we have

M x 0

f (x)sup

x = f for any M M .



NotesSince M is bounded below by f , it has an infimum so that we have

M Minf M = inf {M : M M } f … (3)


f = inf {M : M M }

(II) f = x 0sup f (x)

Let us consider x 1. Then

f (x) f x f .

Therefore, we have

xsup f (x) f . … (4)

Now by definition,

f = x 0

f (x)sup

x

It follows from the property of the supremum that, given > 0, an x N such that

f (x )x > ( f ) … (5)

Define

xxx

. Then x is a unit vector.

Since x 1 x 1 , we have

x 1sup f(x)

1f(x) f (x ) ( f )x [by (2)]

Hence > 0 is arbitrary, we have

x 1sup f(x) > f … (6)


x 1sup f(x) = f .

(III) f = x 1

sup f(x) .

Consider x = 1, we have

f (x) f x f



Notes So that

x 1sup f(x) f … (7)

Now consider

f = x 0

f (x)sup

x

By supremum property, given > 0, x 0

Such that |f (x )| > (||f|| – ) ||x ||

Define xxx

.

Since f is continuous in ||x|| 1 and reaches its maximum on the boundary ||x|| = 1,

We get

x 1

1sup f (x) f (x) f (x ) fx

.

x 1sup f (x) f .

The arbitrary character of yields that

x 1sup f (x) f … (8)

Hence from (7) and (8), we get

f = x 1

sup f (x) .

Note If N is a finite dimensional normed linear space, all linear functions are boundedand hence continuous. For, let N be of dimension n so that any x N is of the form

n

i ii 1

x , where x1, x2, …, xn is a basis of N and 1, 2, …, n are scalars uniquely determined

by the basis.


f (x) = n

ii 1

f (xi) so that

| f (x)| n

ii 1

|f (xi)| … (1)



NotesWe have from (1) by using the notation of the Zeroth norm in a finite dimensional space,

| f (x)| n

i0i 1

x f(x ) … (2)

If n

ii 1

f(x ) = M, then from (2), we have

| f (x)| M 0x .

Hence f is bounded with respect to 0 .

Since any norm on N is equivalent to 0, f is bounded with respect to any norm on N.

Consequently, f is continuous on N.

16.1.5 Representation Theorems for Functionals

We shall prove, in this section, the representation theorems for functionals on some concreteBanach spaces.

Theorem 4: If L is a linear space of all n-tuples, then (i) n np q* .

Proof: Let (e1, e2, …, en) be a standard basis for L so that any x = (x 1, x2, …, xn) L can be writtenas

x = x1e1 + x2e2 + … + xnen.

If f is a scalar valued linear function defined on L, then we get

f (x) = x1 f (e1) + x2 f (e2) + … + xn f (en) … (1)

f determines and is determined by n scalars

yi = f (ei).

Then the mapping

y = (y1, y2, …, yn) f

where f (x) = n

i ii 1

x y is an isomorphism of L onto the linear space L of all function f. We shall

establish (i) – (iii) by using above given facts.

(i) If we consider the space

L = np (1 p < ) with the pth norm, then f is continuous and L represents the set of all

continuous linear functionals on np so that

L = np * .

Now for y f as an isometric isomorphism we try to find the norm for y’s.

For 1 < p < , we show that

np * = n

q .



Notes For x np , we have defined

x =

1n p

pi

i 1

x

Now |f (x)| = n n

i i i ii 1 i 1

x y x y

By using Hölder’s inequality, we get

n

i ii 1

x y

1 1n np q

qpi i

i 1 i 1

x y

so that

|f (x)|

1 1n nq p

q pi i

i 1 i 1

y x

Using the definition of norm for f, we get

f

1n q

qi

i 1

y … (2)

Consider the vector, defined by

xi = q

i

i

yy , yi 0 and xi = 0 if yi = 0 … (3)

Then

x =

11 p pqn npp i

iii 1 i 1

yx

y… (4)

Since q = p (q – 1) we have from (4),

x =

1n p

qi

i 1

y … (5)

Now

|f (x)| = qn n

ii i i

ii 1 i 1

yx y y

y

= n

qi

i 1

y , (By (3))



NotesSo that

nq

ii 1

y = |f(x)| f x … (6)


11n pq

ii 1

y f

1n q

qi

i 1

y f … (7)

Also from (2) and (7), we have

f =

1n q

qi

i 1

y , so that

y f is an isometric isomorphism.

Hence n np q* .

(ii) Let L = n1 with the norm defined by

n

ii 1

x x .

Now f defined in (1), above is continuous as in (i) and L here represents the set of continuous

linear functional on n1 so that

L = n1 * .

We now determine the norm of y’s which makes y f an isometric isomorphism.

Now,

|f (x)| = n

i ii 1

x y

n

i ii 1

x y

But n n

i i i ii 1 i 1

x y max. y x so that n

i ii 1

f (x) max. y x .

From the definition of norm for f, we have

f = imax. y : i 1,2, ,n … (8)

Now consider the vector defined as follows:

If |yi| = i1 i nmax y , let us consider vector x as



Notesxi = i

i

yy when |yi| = i1 i n

max y

and xi = 0 otherwise.

From the definition, xk = 0 k i. So that we have

x = iyy = 1

Further |f (x)| = n

i ii 1

(x y ) = |yi|

Hence |yi| = |f (x)| f x

|yi| f or max. {|yi|} [ ||x|| = 1]

||f|| … (10)


||f|| = max. {|yi|} so that

y f is an isometric isomorphism of L to n1 * .

Hence n n1 * .

(iii) Let L = n with the norm

x = max {|xi| : i = 1, 2, 3, …, n}.

Now f defined in (1) above is continuous as in (1).

Let L represents the set of all continuous linear functionals on n so that

L = n * .

Now we determine the norm of y’s which makes y f as isometric isomorphism

|f (x)| = n n

i i i ii 1 i 1

x y x y .

But n

i ii 1

x y n

i ii 1

max( x ) y

Hence we have

| f (x) | n

ii 1

y x so that

f n

ii 1

y … (11)



NotesConsider the vector x defined by

xi = i

i

yy when yi 0 and xi = 0 otherwise. … (12)

Hence i

i

yx max 1

y.

and |f (x)| = n n

i i ii 1 i 1

x y y .

Therefore n

ii 1

y f(x) f x f .

n

ii 1

y f … (13)

It follows now from (11) and (13) that n

ii 1

f y so that y f is an isometric

isomorphism.

Hence, n n1* .


Theorem 5: The conjugate space of p is q , where

1 1 1p q

and 1 < p < .

or *p q .

Proof: Let x = (xn) p so that pn

n 1

x . … (1)

Let n = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the mth place.

en p for n = 1, 2, 3, …

We shall first determine the form of f and then establish the isometric isomorphism of *p onto

q .

By using (en), we can write any sequence (x1, x2, …, xn, 0, 0, 0, …) in the form n

k kk 1

x e and

n

k kk 1

x x e = (0, 0, 0, …, xn+1, xn+2, …).



Notes

Now

1n p

pk k k

k 1 k n 1

x x e x … (2)

The R.H.S. of (2) gives the remainder after n terms of a convergent series (1).

Hence

1p

pk

k n 1

x 0 as n … (3)


x = k kk 1

x e . … (4)

Let *pf and

n

n k kk 1

s x e then

sn x as n . (Using (4))

since f is linear, we have

f (sn) = n

k kk 1

x f(e ) .

Also f is continuous and sn x, we have

f (sn) f (x) as n

f (x) = n

k kk 1

x f(e ) … (5)

which gives the form of the functional on p .

Now we establish the isometric isomorphism of *p onto q , for which we proceed as follows:

Let f (ek) = k and show that the mapping

T : *p q given by … (6)

T (f) = ( 1, 2, …, k, …) is an isometric isomorphism of *p onto q .

First, we show that T is well defined.

For let x p , where x = ( 1, 2, …, n, 0, 0, …)

where k = g 1

kksgn , 1 k n

n k0

| k| = | k|q – 1 for 1 k n.

| k|p = p(q 1)

k = | k|q.1 1 q p(q 1) qp q



NotesNow k k =

q 1 q 1k k k k k ksgn sgn

k k = | k|q = | k|p (Using property of sgn function) … (7)

x =

1n p

pk

k 1

=

1n q

qk

k 1

… (8)

Since we can write

x = n

k kk 1

e , we get

f (x) = n n

k k k kk 1 k 1

f (e )

f (x) = qn

kk 1

( Using (7)) … (9)

We know that for every x p

| f (x)| f x ,

which upon using (8) and (9), gives

|f (x)|

1n n p

q qk k

k 1 k 1

f

which yields after simplification.

1n p

qk

k 1

f … (10)

since the sequence of partial sums on the L.H.S. of (10) is bounded, monotonic increasing, itconverges. Hence

1n q

qk

k 1

f … (11)

so the sequence ( k) which is the image of f under T belongs to q and hence T is well defined.

We next show that T is onto q .

Let ( k) q , we shall show that there is a *pg such that T maps g into ( k).



Notes Let x p so that

x = k kk 1

x e .

We shall show that

g (x) = k kk 1

x is the required g.

Since the representation for x is unique, g is well defined and moreover it is linear on p . Toprove it is bounded, consider

|g (x) | = n

k k k kk 1 k 1

x x

1 1p q

p qk k

k 1 k 1


|g (x)|

1q

qk

k 1

x .

g is bounded linear functional on p .

since ek p for k = 1, 2, …, we get

g (ek) = k for any k so that

Tg = ( k) and T is on *p onto p .

We next show that

Tf f so that T is an isometry.

Since Tf q , we have from (6) and (10) that

1q

qk

k 1

= || Tf || || f || … (12)

Also, p k kk 1

x x x e . Hence

f (x) = k k k kk 1 k 1

x (e ) x .

|f (x)| k kk 1

x



Notes

1 1q p

q pk k

k 1 k 1


or |f (x)|

1q

qk p

k 1

x x .

Hence, we have

x 0

f (x)sup

x

1q

qk

k 1

= Tf (Using (6))

which upon using definition of norm yields.

f Tf … (13)

Thus f = Tf (Using (12) and (13))

From the definition of T, it is linear. Also since it is an isometry, it is one-to-one and onto.

Hence T is an isometric isomorphism of *p onto p , i.e.

*p q

Theorem 6: Let N and N be normed linear and let T be a linear transformation of N into N . Thenthe inverse T–1 exists and is continuous on its domain of definition if and only if there exists aconstant m > 0 such that

m x T (x) x N. … (1)

Proof: Let (1) holds. To show that T–1 exists and is continuous.

Now T–1 exists iff T is one-one.

Let x1, x2, N. Then

T(x1) = T(x2) T (x1) – T(x2) = 0

T (x1 – x2) = 0

x1 – x2 = 0 by (1)

x1 = x2

Hence T is one-one and so T–1 exists. Therefore to each y in the domain of T–1, there exists x in Nsuch that

T (x) = y T–1 (y) = x … (2)

Hence (1) is equivalent to

m T–1(y) y T–1(y) 1m y

T–1 is bounded T–1 is continuous converse.

Let T–1 exists and be continuous on its domain T(N). Let x be an arbitrary element in N. SinceT–1 exists, there is y T(N) such that T–1 (y) = x T(x) = y.



Notes Again since T–1 is continuous, it is bounded so that there exists a positive constant k such that

T–1 (y) k y x k T(x)

m x T(x) where m = 1k

> 0.


Theorem 7: Let T : N N be a linear transformation. Then T is bounded if and only if T mapsbounded sets in N onto bounded set in N .

Proof: Since T is a bounded linear transformation,

T (x) k x for all x N.

Let B be a bounded subset of N. Then

x k1k x B.

We now show that T(B) is bounded subset of N .

From above we see that

T (x) k1 x B.

T (B) is bounded in N .

Conversely, let T map bounded sets in N into bounded sets in N . To prove that T is a boundedlinear transformation, let us take the closed unit sphere S [0, 1] in N as a bounded set. Byhypothesis, its image T (S[1, 0]) must be bounded set in N .

Therefore there is a constant k1 such that

T (x) k1 for all x S [0, 1]

Let x be any non-zero vector in N. Then x S[0,1]x

and so we get

1xT kx

T (x) k1 x .

Since this is true for x = 0 also, T is a bounded linear transformation.


16.2 Summary

Let N be a normed linear space. Then we know the set R of real numbers and the set C ofcomplex numbers are Banach spaces with the norm of any x R or x C be the absolutevalue of X. (N, R) or (N, C) denote respectively the set of all continuous lineartransformations from N into R or C.

A linear functional on a normed linear space N is said to be bounded, if there exists aconstant k such that

|f (x)| k x x N.



Notes If f is a bounded linear functional on a normed space N, then the norm of f is defined as:

f = x 0

f (x)sup

x

16.3 Keywords

Bounded Linear Functional: A linear functional on a normed linear space N is said to be bounded,if there exists a constant k such that

f (x) K x x N .

Continuous Linear Transformations: Let N be a normed linear space. Then we know the set R ofreal numbers and the set C of complex numbers are Banach spaces with the norm of any x R orx C given by the absolute value of x. Thus with our previous notations, (N, R) or (N, C)denote respectively the set of all continuous linear transformations from N into R or C.

Norm of a Bounded Linear Functional: If f is a bounded linear functional on a normed space N,then the norm of f is defined as:

|| f|| = x 0

f(x)sup

x

Second Conjugate: The conjugate space (N*)* of N* is called the second conjugate space of N .


1. Prove that the conjugate space of 1 is ,

i.e. *1 .

2. Prove that the conjugate space of co is 1 .

or *o 1c

3. Let p > 1 with 1 1p q = 1 and let g Lq (X).

Then prove that the function defined by

F (f) = X

fg d for f Lp (X)

is a bounded linear functional on Lp (X) and

F = g q

4. Let N be any n dimensional normed linear space with a basis B = {x1, x2, ..., xn}. If(r1, r2, ..., rn) is any ordered set of scalars, then prove that, there exists a unique continuouslinear functional f on N such that

f (xi) = ri for i = 1, 2, …, n

5. If T is a continuous linear transformation of a normed linear space N into a normed linearspace N , and if M is its null space, then show that T induces a natural linear transformationT of N/M into N and that T = T .



Notes 16.5 Further Readings

Books JB Conway (1990), A Course in Functional Analysis.

E Hille (1957), Functional Analysis and Semigroups.

Online links pt.scribd.com/doc/86559155/14/Continuous-Linear-Transformations

www.math.psu.edu/bressan/PSPDF/fabook.pdf


Unit 17: The Hahn-Banach Theorem

NotesUnit 17: The Hahn-Banach Theorem

CONTENTS

Objectives

Introduction

17.1 The Hahn-Banach Theorem

17.1.1 Theorem: The Hahn-Banach Theorem – Proof


17.2 Summary

17.3 Keywords



Objectives


State the Hahn-Banach theorem

Understand the proof of the Hahn-Banach theorem

Solve problems related to it.

Introduction

The Hahn-Banach theorem is one of the most fundamental and important theorems in functionalanalysis. It is most fundamental in the sense that it asserts the existence of the linear, continuousand norm preserving extension of a functional defined on a linear subspace of a normed linearspace and guarantees the existence of non-trivial continuous linear functionals on normed linearspaces. Although there are many forms of Hahn-Banach theorem, however we are interested inBanach space theory, in which we shall first prove Hahn-Banach theorem for normed linearspaces and then prove the generalised form of this theorem. In the next unit, we shall discusssome important applications of this theorem.

17.1 The Hahn-Banach Theorem

17.1.1 Theorem: The Hahn-Banach Theorem – Proof

Let N be a normed linear space and M be a linear subspace of N. If f is a linear functional definedon M, then f can be extended to a functional fo defined on the whole space N such that

fo = f .

Proof: We first prove the following lemma which constitutes the most difficult part of thistheorem.



Notes Lemma: Let M be a linear subspace of a normed linear space N let f be a functional defined on M.If xo N such that xo M and if Mo = M + [xo] is the linear subspace of N spanned by M and xo, thenf can be extended to a functional fo defined on Mo s.t.

fo = f .

Proof: We first prove the following lemma which constitutes the most difficult part of thistheorem.

Lemma: Let M be a linear subspace of a normed linear space N let f be a functional defined on M.If xo N such that xo M and if Mo = M + [xo] is the linear subspace of N spanned by M and xo, thenf can be extended to a functional fo defined on Mo s.t.

fo = f .

Proof: The lemma is obvious if f = o. Let then f 0.

Case I: Let N be a real normed linear space.

Since xo M, each vector y in Mo is uniquely represented as

y = x + xo, x M and R.

This enables us to define

fo : Mo R by

fo (y) = fo (x + xo) = f (x) + ro,

where ro is any given real number … (1)

We show that for every choice of the real number ro, fo is not only linear on M but it also extendsf from M to Mo and

fo = f .

Let x1, y1 Mo. Then these exists x and y M and real scalars and such that

x1 = x + xo and y1 = y + xo,

Hence, fo (x1 + y1) = fo (x + xo + y + xo)

= fo (x + y + ( + ) xo)

= f (x + y) + ( + ) ro, ro is a real scalar … (2)

Since f is linear M, f (x + y) = f (x + y) … (3)

From (2) and (3) it follows after simplification that

fo (x1 + y1) = f (x) + ro + f (y) + ro

= fo (x + xo) + fo (y + xo)

= fo (x1) + fo (y1)

fo (x1 + y1) = fo (x1) + fo (y1) … (4)

Let k be any scalar. Then if y Mo, we have

fo (ky) = fo [k (x + xo)]

= fo (kx + k xo)

But fo (kx + k xo) = f (kx) + k ro

= k f (x) + k ro



NotesHence fo (ky) = k [f (x) + ro] = k fo (y) … (5)

From (4) and (5) it follows that fo is linear on Mo.

If y M, then = 0 in the representation for y so that

y = x.

Hence fo (x) = f (x) x M.

fo extends f from M to Mo.

Next we show that

fo = f .

If = 0 this is obvious. So we consider when 0. Since M is a subspace of Mo we then have

fo = sup {|fo (x)| : x Mo, x 1}

sup {|fo (x)| : x M, x 1}

= sup {|f (x)| : x M, x 1} ( fo = f on M)

= f .

Thus, fo f … (A)

So our problem now is to choose ro such that fo f .

Let x1, x2 M. Then we have

f (x2) – f (x1) = f (x2 – x1)

|f (x2 – x1)|

f (x2 + xo) – (x1 + xo)

f ( x2 + xo + – (x1 + xo) )

= f x2 + xo + f x1 + xo

Thus – f (x1) – f x1 + xo – f (x2) + f x2 + xo … (6)

Since this inequality holds for arbitrary x1, x2 M, we see that

oy Msup f(y) f y x oy M

int f (y) f y x

Choose ro to be any real number such that

oy Msup f(y) f y x ro

oy Minf f (y) f y x

From this, we get for all y M

sup {– f (y) – f ( y + xo )} ro

inf {– f (y) + f ( y + xo )}

Let us take y = x in the above inequality, we have



Noteso

y M

x xsup f f x ro

oy M

x xinf f f x … (7)

If > 0 the right hand side of (7) becomes

ro o1 1f (x) f x x which implies that

f (x) + ro = fo (x + xo) f x + xo

If z = x + xo Mo then we get from above

|fo (z)| f z … (8)

If < 0, then from L.H.S. of (7) we have

ox xf f x ro

o1 1f (x) f x x ro, since < 0,

1 1.

f (x) + ro f x + xo

fo (z) f z for every z Mo … (9)

Replacing z by –z in (9) we get

– fo (z) f z , since fo is linear on Mo … (10)

Hence we get from (9) and (10).

|fo (z)| f z … (11)

Since f is functional on M, f is bounded.

Thus ( ) shows that fo is a functional on Mo.

Since fo = sup {|fo (z)| : z Mo, z |}, it follows from ( ) that

fo f

We finally obtain from (A) and (B) that

fo = f

This power the lemma for real normed linear space.

Case II: Let N be a complex normed linear space.

Let N be a normed linear space over C and f be a complex valued functional on a subspace M ofN.

Let g = Re (f) and h = Im (f) so that we can write

f (x) = g (x) + i h (x). We show that g (x) and h (x) are real valued functionals.


f (x + y) = f (x) + f (y)



Notesg (x + y) + i h (x + y) = g (x) + i h (x) + g (y) + i h (y)

= g (x) + g (y) + i (h (x) + h (y))

Equating the real and imaginary parts, we get

g (x + y) = g (x) + g (y)

and h (x + y) = h (x) + h (y)

If R, then we have

f ( x) = g ( x) + i h ( x)

Since f is linear

f ( x) = f (x) = g (x) + i h (x)

f ( x) = g (x) and h ( x) = h (x)(equating real and imaginary parts)

g, h are real linear functions on M.

Further |g (x)| |f (x)| f x

If f is bounded on M, then g is also bounded on M.

Similarly h is also bounded on M.

Since a complex linear space can be regarded as a real linear space by restricting the scalars to bereal numbers, we consider M as a real linear space. Hence g and h are real functional on realspace M.

For all x in M we have

f (i x) = i f (x) = i {g (x) + i h (x)}

or g (i x) + i h (i x) = – h (x) + i g (x)

Equating real and imaginary parts, we get

g (i x) = – h (x) and h (i x) = g (x)

Therefore we can express f (x) either only by g or only h as follows:

f (x) = g (x) – i g (i x)

= h (i x) + i h (x).

Since g is a real functional on M, by case I, we extend g to a real functional go on the real space Mo

such that go = g . For x Mo, we define

fo (x) = go (x) – i go (i x)

First note that fo is linear on the complex linear space Mo. Such that fo = f on M.

Now fo (x + y) = go (x + y) – i go (i x + i y)

= go (x) + go (y) – i go (i x) – i go (i y)

= fo (x) + fo (y).

Now for a, b R, we have

fo ((a + i b) x) = go (ax + i bx) – i go (– bx + i ax)

= a go (x) + b go (i x) – i (–b) go (x) – i a go (i x)

= (a + ib) {go (x) – i go (i x)}



Notes So that

fo ((a + i b) x) = (a + i b) fo (x)

fo is linear on Mo and also go = g on M.

fo = f on M.

Now have to show that fo = f on Mo.

Let x Mo and fo (x) = rei

|fo (x) | = r = e–i rei = e–i fo (x) … (12)

Since fo (x) is linear,

e–i fo (x) = fo (e–i x) … (13)

So we get from (12) and (13) that

|fo (x) | = r = fo (r e–i x).

Thus the complex valued functional fo is real and so it has only real part so that

|fo (x) | = go (e–i x) |go (e–i x)|

But |go (e–i x)| go e–i x

= go x ,

We get |fo (x)| go x

Since go is the extension of g, we get

go x = g x f x

Therefore

|fo (x) f x so that from the definition of the norm of fo, we have

fo f

As in case I, it is obvious that f fo

Hence fo = f .



Theorem: The generalized Hahn-Banach Theorem for Complex Linear Space.

Let L be a complex linear space. Let p be a real valued function defined on L such that

p (x + y) p (x) + p (y)

and p ( x) = | | p (x) x L and scalar .

If f is a complex linear functional defined on the subspace M such that |f (x)| p (x) for x M,then f can be extended to a complex linear functional to be defined on L such that |fo (x)| p (x)for every x L.

Proof: We have from the given hypothesis that f is a complex linear functional on M such that

| f (x) p (x) x M.



NotesLet g = Re (f) then g (x) |f (x)| p (x).

So by the generalised Hahn-Banach Theorem for Real Linear space, can be extended to a linearfunctional go on L into R such that go = g on M and go (x) p (x) x L.

Define fo (x) = go (x) – i go (i x) for x L as in the Hahn-Banach Theorem, fo is linear functional onL such that fo = f on M.

To complete the proof we have to prove that

|fo (x)| p (x) x L.

Let x L and fo (x) = r ei , r > o and real. Then

|fo (x)| = r = e– rei = e–i fo (x)

= fo (e–i x).

Since r = fo (e–i x), fo is real so that we can take

|fo (x)| = r = fo (e–i x) = go (e–i x) … (1)

Since go (x) p (x), go (e–i x) p (e–i x) for x L.

But p (e–i x) = |e–i | p (x) so that go (e–i x) p (x) … (2)

It follows from (1) and (2) that

|fo (x)| p (x)


Corollary 1: Deduce the Hahn-Banach theorem for normed linear spaces from the generalisedHahn-Banach theorem.

Proof: Let p (x) = f x for x N.

We first note that p (x) 0 for all x N.

Then for any x, y N, we have

p (x + y) = f x + y

f ( x + y )

= f x + f y

= p (x) + p (y)

p (x + y) p (x) + p (y)

Also p ( x) = f x = | | f x = | | p (x).

Hence p satisfies all the conditions of the generalized Hahn-Banach Theorem forComplex Linear space. Therefore a functional fo defined on all of N such that fo = f on M and|fo (x)| p (x) = f x x N.

fo f … (3)

Since fo is the extension of f from a subspace M, we get

f fo … (4)


fo = f



Notes

Notes Let L be a linear space. A mapping p : L R is called a sub-linear functional on L ifit satisfies the following two properties namely,

(i) p (x + y) p (x) + p (y) x, y L (sub additivity)

(ii) p ( x) = p (x), 0 (positive homogeneity)

Thus p defined on L in the above theorems is a sub-linear functional on L.

Some Applications of the Hahn-Banach Theorem

Theorem: If N is a normed linear space and xo N, xo 0 then there exists a functional fo N* suchthat

fo (xo) = xo and fo = 1.

Proof: Let M denote the subspace of N spanned by xo, i.e.,

M = { xo : any scalar}.

Define f : M F (R or C) by

f ( xo) = xo .

We show that f is a functional on M with f = 1.

Let x1, x2 M so that

x1 = 1 xo and x2 = 2 xo. Then

f (x1 + x2) = f ( 1 xo + 2 xo)

= ( 1 + 2) xo

But ( 1 + 2) xo = 1 xo + 2 xo

= f (x1) + f (x2)

Hence f (x1 + x2) = f (x1) + f (x2) … (1)

Let k be a scalar (real or complex). Then if x M, then x = xo.

Now f (kx) = f (k xo) = k xo = k f (x) … (2)

If follows from (1) and (2) that f is linear.

Further, we note that since xo M with = 1, we get

f (xo) = xo .

For any x M, we get, | f (x)| = | | || xo|| = xo = x

|f (x) | = x

f is bounded and we have

|f(x)|supx

= 1 for x M and x 0.

So by definition of norm of a functional, we get

f = 1.



NotesHence by Hahn-Banach theorem, f can be extended to a functional fo N* such that fo (M) = f (M)and fo = f = 1, which in particular yields that

fo (xo) = f (xo) = xo and fo = 1.


Corollary 2: N* separates the vector (points) in N.

Proof: To prove the cor. it suffices to show that if x, y N with x y, then there exists a f N*such that f (x) f (y).

Since x y x – y 0.

So by above theorem there exists a functional f N* such that

f (x – y) = f (x) – f (y) 0

and hence f (x) f (y).

This shows that N* separates the point of N.

Corollary 3: If all functional vanish on a given vector, then the vector must be zero, i.e.

if f (x) = 0 f N* then x = 0.

Proof: Let x be the given vector such that f (x) = 0 f N*.

Suppose x 0. Then by above theorem, there exists a function f N* such that

f (x) = x > 0, which contradicts our supposition that

f (x) = 0 f N*. Hence we must have x = 0.

17.2 Summary

The Hahn-Banach Theorem: Let N be a normed linear space and M be a linear subspace ofN. If f is a linear functional defined on M, then f can be extended to a functional f o definedon the whole space N such that

fo = f

If f is a complex linear functional defined on the subspace M such that |f (x)| p (x) forx M, then f can be extended to a complex linear function fo defined on L such that| fo (x) | p (x) for every x L.

17.3 Keywords

Hahn-Banach theorem: The Hahn-Banach theorem is one of the most fundamental and importanttheorems in functional analysis. It is most fundamental in the sense that it asserts the existenceof the linear, continuous and norm preserving extension of a functional defined on a linearsubspace of a normed linear space and guarantees the existence of non-trivial continuous linearfunctionals on normed linear spaces.

Sub-linear Functional on L: Let L be a linear space. A mapping p : L R is called a sub-linearfunctional on L if it satisfies the following two properties namely,

(i) p (x + y) p (x) + p (y) x, y L (sub additivity)

(ii) p ( x) = p (x), 0 (positive homogeneity)



Notes Thus p defined on L in the above theorems is a sub-linear functional on L.

The Generalized Hahn-Banach Theorem for Complex Linear Space: Let L be a complex linearspace. Let p be a real valued function defined on L such that

p (x + y) p (x) + p (y)

and p ( x) = | | p (x) x L and scalar .


1. Let M be a closed linear subspace of a normed linear space N and xo is a vector not in M.Then there exists a functional fo N* such that

fo (M) = 0 and fo (xo) 0

2. Let M be a closed linear subspace of a normed linear space N, and let x o be a vector not inM. If d is the distance from xo to M, then these exists a functional fo N* such thatfo (M) = 0, fo (xo) = d, and fo = 1.

3. Let M is a closed linear subspace of a normed linear space N and x o N such that xo M.If d is the distance from xo to M, then there exists a functional fo N* such that fo (M) = 0, fo

(xo) = 1 and fo = 1d .

4. Let N be a normed linear space over R or C. Let M N be a linear subspace. Then

M = N f N* is such that f (x) = 0 for every x M, then f = 0.

5. A normed linear space is separable if its conjugate (or dual) space is separable.


Books Walter Rudin, Functional Analysis (2nd ed.). McGraw-Hill Science/Engineering/Math 1991.

Eberhard Zeidler, Applied Functional Analysis: Main Principles and their Applications,Springer, 1995.

Online links mat.iitm.ac.in

www.math.ksu.edu

mizar.uwb.edu.pl/JFW/Vol5/hahnban.html


Unit 18: The Natural Imbedding of N in N**

NotesUnit 18: The Natural Imbedding of N in N**

CONTENTS

Objectives

Introduction

18.1 The Natural Imbedding of N into N**

18.1.1 Definition: Natural Imbedding of N into N**

18.1.2 Definition: Reflexive Mapping

18.1.3 Properties of Natural Imbedding of N into N**


18.2 Summary

18.3 Keywords



Objectives


Define the natural imbedding of N into N**.

Define reflexive mapping.

Describe the properties of natural imbedding of N into N*.

Introduction

As we know that conjugate space N* of a normed linear space N is itself a normed linear space.So, we can find the conjugate space (N*)* of N*. We denote it by N** and call it the secondconjugate space of N. Likewise N*, N** is also a Banach space. The importance of the space N**lies in the fact that each vector x in N given rise to a functional Fx in N** and that there exists anisometric isomorphism of N into N**, called natural imbedding of N into N**.

18.1 The Natural Imbedding of N into N**

18.1.1 Definition: Natural Imbedding of N into N**

The map J : N N** defined by

J (x) = Fx x N,

is called the natural imbedding of N into N**.

Since J (N) N**, N can be considered as part of N** without changing its basic norm structure.We write N N** in the above sense.



Notes 18.1.2 Definition: Reflexive Mapping

If the map J : N N** defined by

J (x) = Fx x N,

is onto also, then N (or J) is said to be reflexive (or reflexive mapping). In this case we writeN = N**, i.e., if N = N**, then N is reflexive.

Note Equality in the above definition is in the sense of isometric isomorphism under thenatural imbedding. Since N** must always be a complete normed linear space, noincomplete space can be reflexive.

18.1.3 Properties of Natural Imbedding of N into N**

I. Let N be a normed linear space. If x N, then

x = sup {|f (x)|: f N* and f = 1}.

Using natural imbedding of N into N**, we have for every x N,

Fx (f) = f (x) and Fx = x .

Now, Fx = xf 1 f 1

sup {|F (f)|} sup {|f(x)|, f N*}

therefore, x = sup {|f (x)|: f N*, f = 1}.

II. Every normed linear space is a dense linear subspace of a Banach space.

Let N be a normed linear space. Let

J : N N** be the natural imbedding of N into N**.

The image of the mapping is linear subspace J (N) N**. Let J(N) be the closure of N(N)in N**.

Since N** is a Banach space, its closed subspace J(N) is also a Banach space. Hence if weidentity N with J(N), then J(N) is a dense subspace of a Banach space.


Theorem 1: Let N be an arbitrary normal linear space. Then each vector x in N induces a functionalFx on N* defined by

Fx(f) = f(x) for every f N* such that Fx = x .

Further, the mapping J : N N** : J (x) = Fx for every x N defines and isometric isomorphismsof N into N**.

Proof: To show that Fx is actually a function on N*, we must prove that Fx is linear and bounded(i.e. continuous).

We first show Fx is linear.



NotesLet f, g N* and , be scalars. Then

Fx ( f + g) = ( f + g) x = f (x) + g (x)

= Fx (f) + Fx (g)

Fx is linear

Fx is bounded.

For any f N*, we have

|Fx (f)| = |f (x)|

f x … (1)

Thus the constant x is bounded (in the sense of a bounded linear functional) for Fx. Hence Fx isa functional on N*.

We now prove Fx = x

We have Fx = sup {|Fx (f)| : f 1}

sup { F x : f 1 } (Using (1))

x

Hence Fx x … (2)

To prove the reverse inequality we consider the case when x = 0. In this case (2) gives Fo = 0 = 0.

But Fx = 0 always. Hence Fo = 0 i.e. Fx = x for x = 0.

Not let x 0 be a vector in N. Then by theorem (If N is a normal linear space and xo N, xo 0,then there exists a functional fo N* such that

fo (xo) = xo and fo = 1.)

a functional f N* such that

f (x) = x and f = 1.

But Fx = sup {|Fx (f)| : f 1}

= sup { f (x) : f = 1}

and since x = f (x) sup {|f (x)| : f = 1}

we conclude that Fx x … (3)

[Note that since f (x) = x 0 we have f (x) = |f (x)|]

From (2) and (3); we have

Fx = x … (4)

Finally, we show that J is an isometric isomorphism of N into N**. For any x, y N and scalar.

Fx+y (f) = f (x + y) = f (x) + f (y)

= Fx (f) + Fy (f)

Fx+y (f) = (Fx + Fy) f … (5)

Fx+y = Fx + Fy … (6)

Further, F x (f) = f ( x) = f (x) = ( Fx) (f)



Notes Hence F x = Fx … (7)

Using definition of J and equations (6) and (7) we get

J(x + y) = Fx+y = Fx + Fy = J (x) + J (y) … (8)

and J( x) = F x = Fx = J (x) … (9)

(8) and (9) J is linear and also (4) shows that J is norm preserving.

For any x and y in N, we have

J (x) – J (y) = Fx – Fy = Fx–y = x – y … (10)

Thus J preserve distances and it is an isometry. Also (10) shows that

J (x) – J (y) = 0 J (x – y) = 0 x – y = 0

i.e. J (x) = J (y) x = y so that J is one-one.

Hence J defines an isometric isomorphism of N into N**. This completes the proof of the theorem.

Example 1: The space np (1 p < ) are reflexive.

Solution: We know that if 1 p < , then

np

* = np .

But nq

* = np

Hence np

* * = np

Similarly we have n1

* * = n1 for p = 1

and n * * = n for p =

So that np spaces are reflexive for 1 p < .

Example 2: The space p for 1 < p < are reflexive.

Sol: We know that if p p* and q p*

q p* * .

p are reflexive for 1 < p < .

A similar result can be seen to hold for Lp (X).

Example 2: If N is a finite dimensional normed linear space of dimension m, then N* alsohas dimension m.

Solution: Since N is a finite dimensional normed linear space of dimension m then {x 1, x2, …, xm}is a basis for N, and if ( 1 2 … m) is any set of scalars, then there exists a functional f on N suchthat f (xi) = i, i = 1, 2, …m.



NotesTo show that N* is also of dimension m, we have to prove that there is a uniquely determinedbasis (f1, f2, …, fm) in N*, with fj (xi) = y.

By the above fact, for each i = 1, 2, …, m, a unique fj in N* exists such that fj (xi) = ij. We show nowthat {f1, f2, …, fn} is a basis in N* to complete our proof.

Let us consider 1 f1 + 2 f2 + …… m fm = 0 … (1)

For all x N, we have 1 f1 (x) + 2 f2 (x) + …… + m fm (x) = 0.

We have j j j j i j ij j

f (x ) 0 for i = 1, 2, …, m, when x = x i.

f1, f2, …… fm are linearly independent in N*.

Now let f (xi) = i.

Therefore if x = i id

x , we get

f (x) = 1f (x1) + 2f (x2) + …… + mf (xm) … (2)

Further fj (x) = 1fj (x1) + …… + ifj (xi) + …… + mfj (xm)

fj (x) = j


f (x) = 1 f1 (x) + 2 f2 (x) + …… + m fm (x)

= ( 1f + 2f2 + …… + m fm) (x)

(f1, f2, …… fm) spans the space.

N* is m-dimensional.

18.2 Summary

The map J : N N** defined by

J (x) = Fx x N,


If the map J : N N** defined by

J (x) = Fx x N,

is onto also, then N (or J) is said to be reflexive. In this case we write N = N**, i.e., if N = N**,then N is reflexive.

Let N be an arbitrary normal linear space. Then each vector x in N induces a functional F x

on N* defined by Fx (f) = f (x) for every f N* such that Fx = x .

18.3 Keywords

Natural Imbedding of N into N**: The map J : N N** defined by

J (x) = Fx x N,




Notes Reflexive Mapping: If the map J : N N** defined by

J (x) = Fx x N,

is onto also, then N (or J) is said to be reflexive (or reflexive mapping).


1. Let X be a compact Hausdorff space, and justify the assertion that C (X) is reflexive if X isfinite.

2. If N is a finite-dimensional normed linear space of dimension n, show that N* also hasdimension n. Use this to prove that N is reflexive.

3. If B is a Banach space, prove that B is reflexive B* is reflexive.

4. Prove that if B is a reflexive Banach space, then its closed unit sphere S is weakly compact.

5. Show that a linear subspace of a normed linear space is closed it is weakly closed.


Books G.F. Simmons, Introduction to topology and Modern Analysis. McGraw-Hill,Kogakusha Ltd.

J.B. Conway, A Course in Functional Analysis. Springer-Verlag.

Online links www.mathoverflow.net/…/natural-embedding

www.tandfonline.com


Unit 19: The Open Mapping Theorem

NotesUnit 19: The Open Mapping Theorem

CONTENTS

Objectives

Introduction

19.1 The Open Mapping Theorem

19.1.1 Lemma

19.1.2 Proof of the Open Mapping Theorem


19.2 Summary

19.3 Keywords



Objectives


State the open mapping theorem.

Understand the proof of the open mapping theorem.

Solve problems on the open mapping theorem.

Introduction

In this unit, we establish the open mapping theorem. It is concerned with complete normedlinear spaces. This theorem states that if T is a continuous linear transformation of a Banachspace B onto a Banach space B , then T is an open mapping. Before proving it, we shall prove alemma which is the key to this theorem.

19.1 The Open Mapping Theorem

19.1.1 Lemma

Lemma 1: If B and B are Banach spaces and T is a continuous linear transformation of B onto B ,then the image of each sphere centered on the origin in B contains an open sphere centered onthe origin in B .

Proof: Let Sr and Sr respectively denote the open sphere with radius r centered on the origin in Band B .

We one to show that T (Sr) contains same Sr .

However, since T (Sr) = T (r S1) = r T (S1), (by linearity of T).

It therefore suffices to show that T (S1) contains some Sr for then S , where = r2, will be contained

in T (Sr). We first claim that 1T(S ) (the closure of T (S1)) contains some rS .



Notes If x is any vector in B we can by the Archimedean property of real numbers find a positiveinteger n such that n > x , i.e., x Sn,

therefore

B = nn 1

S

and since t is onto, we have

B = T (B)

= T nn 1

S

= nn 1

T (S )

Now B being complete, Baire’s theorem implies that some n0T S possesses an interior point

Z0. This in turn yields a point y0 n0T S such that y0 is also an interior point of n0

T S .

Further, maps j : B B and g : B B

defined respectively by j (y) = y = y – y0 and g (y) = 2 n0 y

where no is a non-zero scalars, are homeomorphisms as shown below f is one-to-one and onto.To show f, f–1 are continuous, let yn B and yn y in B.

Then f (yn) = yn – y0 y – y0 = f (y)

and f–1 (yn) = yn + y0 y + y0 = f–1 (y)

Hence f and f–1 are both continuous so that is a homeomorphism.

Similarly g : B B : g (x) = 2n0y is a homeomorphism for, g is one-to-one, onto and bicontinuousfor n0 0.

Therefore we have

(i) f (y0) = 0 = origin in B is an interior point of nf T S .

(ii) n0f T S = n0

f T S

= n 00T S y

2n0T S 0 n0

y T S

(iii) 2n0T S = n 1 0 10

T 2 S 2n T S

= 1 1g (T(S )) g (T(S ))

= n 102 T(S )



NotesCombining (i) – (iii), it follows that origin is also an interior point of 1(T(S )) . Consequently,

there exists > 0 such that

S 1T(S )

This justifies our claims.

We conclude the proof of the lemma by showing that

/3S 1T(S ) , i.e., 3S T(S )

Let y B such that y < . Then y 1T(S ) and therefore there exists a vector x1 B such that

x1 < 1, y – y1 < /2 and y1 = T (x1)

We next observe that

/2S 1/2T(S ) and y – y1 /2S

Therefore there exists a vector x2 B such that

3 1 2 2

1x , y y y2 2 and y2 = T (x2)

Continuing this process, we obtain a sequence (xn) in B such that

n n n 1 2 nn 1 2

1x , y T(x ) and y (y y y )2 2

Let sn = x1 + x2 + … + xn, then

sn = x1 + x2 + … + xn

x1 + x2 + … + xn

< 2 n 1

1 1 112 2 2

n

12 12

< 2

Also for n > m, we have

sn – sm = sm+1 + … + xn

xm+1 + … + xn

< m n 1

1 12 2

= m n m

1 112 2

112

(summing the G.P.)

= m 1 n m 1

1 12 2

0 as m, n



Notes Thus (sn) is a Cauchy sequence in B and since B is complete, a vector x B such that

sn x and therefore

x = n nn nlim s lim s 2 3 ,

i.e., x S3.

It now follows by the continuity of T that

T (x) = nnT lim s

= nnlim T(s )

= 1 2 nnlim (y y y )

= y

Hence y T (S3)

Thus y S y T (S3). Accordingly

S T (S3)

This completes the proof of the lemma.

Note If B and B are Banach spaces, the symbol S (x; r) and S (x; r) will be used to denoteopen spheres with centre x and radius r in B and B respectively. Also Sr and rS will denotethese spheres when the centre is the origin. It is easy to see that

S (x; r) = x + Sr and Sr = r S1

For, we have

y S (x; r) y – x < r

z < r and y – x = z, z Sr

y = x + z and z < r

y x + Sr

Thus S (x; r) = x + Sr

and Sr = {x : x < r} = x

x : 1r

= {r . y y < 1}

= r S1

Thus Sr = r S1

Now we prove an important lemma which is key to the proof of the open mapping theorem.



Notes19.1.2 Proof of the Open Mapping Theorem

Statement: If T is a continuous linear transformation of a Banach space B onto a Banach space B ,then T is an open mapping.

Proof: Let G be an open set in B. We are to show that T (G) is an open set in B

i.e. if y is any point of T (G), then there exists an open sphere centered at y and contained in T (G).

y T (G) y = T(x) for some x G.

x G, G open in B there exists an open sphere S (x; r) with centre x and radius r such thatS (x; r) G.

But as remarked earlier we can write S (x; r) = x + Sr, where Sr is open sphere of radius r centeredat the origin in B.

Thus x + Sr G … (1)

By lemma (2) (prove it),

T (Sr) contains some r1S . Therefore

S (y; r1) = y + r1S

y + T (Sr)

= T (x) + T (Sr)

= T (x + Sr)

T (G), (Using (1))

since x + Sr = S (x; r) G.

Thus we have shown that to each y T (G), there exists an open sphere in B centered at y andcontained in T (G) and consequently T (G) is an open set.



Theorem 1: Let B and B be Banach spaces and let T be an one-one continuous linear transformationof B onto B . Then T is a homeomorphism.

In particular, T–1 is automatically continuous.

Proof: We know that a one-to-one continuous open map from B onto B is a homeomorphism.

By hypothesis T : B B is a continuous one-to-one onto mapping.

By the open mapping theorem, T is open. Hence T is a homeomorphism. Since T ishomeomorphism, T– exists and continuous from B to B so that T– is bounded and hence

T–1 (B , B).


Cor. 1: Let B and B be Banach spaces and let T (B, B ). If T : B B is one-to-one and onto, thereare positive numbers m and M such that

m x T (x) M x .



Notes Proof: By the theorem,

T : B B is a homeomorphism. So that T and T–1 are both continuous and hence bounded. Henceby theorem,

Let N and N be normed linear spaces. Then N and N are topologically isomorphic if and onlyif there exists a linear transformation T of N onto N and positive constants m and M such that

m x T (x) M x , for every x N.

constants m and M such that

m x T (x) M x .

Theorem 2: If a one-to-one linear transformation T of a Banach space B onto itself is continuous,then its inverse T–1 is continuous.

Proof: T is a homeomorphism (using theorem of B onto B. Hence T–1 is continuous.


Note The following examples will show that the completeness assumption in the openmapping theorem and theorem can neither be omitted in the domain of definition of T norin the range of T.

Example 1: Let C [0, 1] be the set of all continuous differentiable function on [0, 1]. Weknow that C [0, 1] is an incomplete space with the norm

f = sup {|f (x)| : 0 x 1}

But it is complete with respect to the norm

f = f + f .

Now let us choose B = [C [0, 1], ] and N = [C [0, 1], ].

Consider the identity mapping I : B N. The identity mapping is one-to-one onto and continuous.I–1 is not continuous. For, if it were continuous, then it is a homeomorphism. Mapping of acomplete space into an incomplete space which cannot be. Hence I does not map open sets intoopen sets.

Thus the open mapping theorem fails if the range of T is not a Banach space.

Example 2: Let B be an infinite dimensional Banach space with a basis { i : i I} with i = 1 for each i I. Let N be the set of all functions from I to C which vanish everywhere except

a finite member of points in I. Then N is a linear space under addition and scalar multiplication.We can define the norm on N as

f = |f (i)|, i I.

Then N is an incomplete normed linear space. Now consider the transformation

T : N B defined as follows.

For each f N, let T (f) = f (i) i.



NotesThen T is linear and

T (f) ii I

|f(i)|

= ii I

|f | = f for every f N.

Hence T is bounded transformation from N to B . It is also one-to-one and onto. But T does notmap open subsets of N onto B . For, if it maps, it is a linear homeomorphism from N onto Bwhich cannot be since N is incomplete.

Theorem 3: Let B be a Banach space and N be a normed linear space. If T is a continuous linearopen map on B onto N, then N is a Banach space.

Proof: Let (yn) be a Cauchy sequence in N. Then we can find a sequence of positive integer (nk)such that nk < nk + 1 and for each k

n n kk 1 k

1y y2

Hence by theorem: “Let N and N be normed linear spaces. A linear map T : N N is open andonto if and only if there is a M > 0 such that for any y N , there is a x N such that Tx = y and x M y .”

For n nk 1y y N , there is a nk B and a constant M such that

T (xk) = n nk 1y y and k n nk 1

x y y .

By on choice n nk 1 kk 1

y y is convergent so that nkk 1

x is convergent. Since B is a Banach

space there is a x B such that

x = n

nknk 1

Lim x

Since T is continuous n

kk 1

T(x ) T (x) as n

But k n nk 1 1k 1

T(x ) y y so that

n n n nk 1 1 k 1 1y y T(x) y y T(x)

Since (yn) is a Cauchy sequence such that every subsequences is convergent, (yn) itself converges

and yn n1y + T (x) in N.

Hence N is complete. Consequently, N is a Banach space.


Example: Let N be complete in two norms 1 and 2 respectively. If there is a numbera > 0 such that x 1 a x 2 for all x N, then show that the two norms are equivalent.

Solution: The identity map

i : (N, 2) (N, 1) is an one-one onto map.



Notes Also x 1 a x 2 i is bounded … (1)

i is continuous.

Hence by open-mapping theorem, i is open and so it is homeomorphism of (N, x 2) onto(N, x 1). Consequently i is bounded as a map from (N, x 1) (N, x 2)

Since i–1 (x) = x, a, b s.t. x 2 b 1 … (2)

(1) & (2) imply that the norms are equivalent.

19.2 Summary

If B and B are Banach spaces and T is a continuous linear transformation of B onto B , thenthe image of each sphere centered on the origin in B contains an open sphere centered onthe origin in B .

The open mapping theorem : If T is a continuous linear transformation of a Banach spaceB onto a Banach space B , then T is an open mapping.

19.3 Keywords

Banach Space: A normed space V is said to be Banach space if for every Cauchy sequence

n n 1 V then there exists an element V such that nn

lim .

Homeomorphism: A map f : (X, T) (Y, U) is said to be homeomorphism if

(i) f is one-one onto.

(ii) f and f–1 are continuous.

Open Sphere: Let xo X and r R+. Then set {x X : p (xo, x) < r} is defined as open sphere withcentre xo and radius r.


1. If X and Y are Banach spaces and A : X Y is a bounded linear transformation that isbijective, then prove that A–1 is bounded.

2. Let X be a vector space and suppose 1 , and 2 are two norms on X and that T1 and T2

are the corresponding topologies. Show that if X is complete in both norms and T1 T2,then T1 = T2.


Books Walter Rudin, Functional Analysis, McGraw-Hill, 1973.

Jean Diendonne, Treatise on Analysis, Volume II, Academic Press (1970).

Online links euclid.colorado.edu/ngwilkin/files/math 6320…/OMT_CGT.pdf

people.sissa.it/nbianchin/courses/…/lecture05.banachstein.pdf


Unit 20: The Closed Graph Theorem

NotesUnit 20: The Closed Graph Theorem

CONTENTS

Objectives

Introduction

20.1 The Closed Graph Theorem

20.1.1 Graph of Linear Transformation

20.1.2 Closed Linear Transformation

20.1.3 The Closed Graph Theorem - Proof

20.2 Summary

20.3 Keyword



Objectives


State the closed graph theorem.

Understand the proof of the closed graph theorem

Solve problems based on the closed graph theorem.

Introduction

Though many of the linear transformations in analysis are continuous and consequently bounded,there do exist linear transformation which are discontinuous. The study of such kind oftransformation is much facilitated by studying the graph of transformation and using the graphof the transformation as subset in the Cartesian product space to characterise the boundedness ofsuch transformations. The basic theorem in this regard is the closed graph theorem.

20.1 The Closed Graph Theorem

20.1.1 Graph of Linear Transformation

Definition: Let N and N be a normed linear space and let T : N N be a mapping with domainN and range N . The graph of T is defined to be a subset of N × N which consists of all orderedpairs (x, T (x)). It is generally denoted by GT.

Therefore the graph of T : N N is

GT = {(x, T (x) : x N}.

Notes GT is a linear subspace of the Cartesian product N × N with respect to coordinate-wise addition and scalar multiplications.



Notes Theorem 1: Let N and N be normed linear spaces. Then N × N is a normed linear space withcoordinate-wise linear operations and the norm.

(x, y) = 1

p p px x , where x N, y N

and | p < . Moreover, this norm induces the product topology on N × N , and N × N iscomplete iff both N and N are complete.

Proof:

(i) It needs to prove the triangle inequality since other conditions of a norm are immediate.

Let (x, y) and (x , y ) be two elements of N × N .

Then (x, y) + (x , y ) = (x + x , y + y )

1p p px x y y

= 1

p p px x y y

= 1 1

p p p pp px y x y

(By Minkowski’s inequality)

= (x, y) + (x , y ) .

This establishes the triangular inequality and therefore N × N is a normed linear space.

Furthermore (xn, yn) (x, y) xn = x and yn = y. Hence theorem on N × N induces theproduct topology.

(ii) Next we show that N × N is complete N, N are complete.

Let (xn, yn) be a Cauchy sequence in N × N . Given > 0, we can find a no such that

(xn, yn) – (xm, ym) < m, n no. … (1)

(xn – xm) < and yn – ym < m, n no

(xn) and (yn) are Cauchy sequences in N and N respectively.

Since N, N are complete, let

xn xo N and yn yo N in their norms,

i.e. (xn – xo) < and yn – ym < m, n no. … (2)

since xo N, yo N , (xo, yo) N × N .

Further (xn, yn) – (xo, yo) < n no (using (2))

(xn, yn) (xo, yo) in the norm of N × N and (xo, yo) N × N .

N × N is complete.

The converse follows by reversing the above steps.




Notes

Notes The following norms are equivalent to above norm

(i) (x, y) = max { x , y }

(ii) (x, y) = x + y (p = 1 in the above theorem)

20.1.2 Closed Linear Transformation

Definition: Let N and N be normed linear spaces and let M be a subspace of N. Then a lineartransformation

T : M N is said to be closed

iff xn M, xn x and T (xn) y imply x M and y = T (x).

Theorem 2: Let N and N be normed linear spaces and B be a subspace of N. Then a lineartransformation T : M N is closed its graph GT is closed.

Proof: Let T is closed linear transformation. We claim that its graph GT is closed i.e. GT containsall its limit point.

Let (x, y) be any limit point of GT. Then a sequence of points in GT, (xn, T (xn), xn M, convergingto (x, y). But

(xn, T (xn)) (x, y)

xn, T (xn) – (x, y) 0

(xn – x), T (xn) – y 0

xn – x + T (xn) – y 0

xn – x 0 and T (xn) – y 0

xn x and T (xn) y ( T is closed)

(x, y) GT. (By def. of graph)

Thus we have shown that every limit point of GT is in GT and hence GT is closed.

Conversely, let the graph of T, GT is closed.

To show that T is closed linear transformation.

Let xn M, xn x and T (xn) y.

Then it can be seen that (x, y) is an adherent point of GT so that

(x, y) TG . But TG = GT ( GT is closed)

Hence (x, y) GT and so by the definition of GT we have x M and y = T (x).

Consequently, T is a closed linear transformation. This completes the proof of the theorem.

20.1.3 The Closed Graph Theorem – Proof

If B and B are Banach spaces and if T is linear transformation of B into B , then T is continuous Graph of T (GT) is closed.

Proof: Necessary Part:

Let T be continuous and let GT denote the graph of T, i.e.

GT = {(x, T (x) : x B} B × B .



Notes We shall show that TG = GT.

Since GT TG always, it suffices to show that TG GT.

Let (x, y) TG . Then there exists a sequence (xn, T (xn)) in GT such that

(xn, T (xn)) (x, y)

xn x and T (xn) y.

But T is continuous T (xn) T (x) and so y = T (x)

(x, y) = (x, T (x)) GT

TG GT

Hence GT = TG i.e. GT is closed.

Sufficient Part:

Let GT is closed. Then we claim that T is continuous. Let B1 be the given linear space B renormedby 1 given by

x 1 = x + T (x) for x B.

Now T (x) x + T (x) = x 1.

T is bounded (continuous) as a mapping from B1 to B .

So if B and B1 have the same topology then T will be continuous from B to B . To this end, we haveto show that B and B1 are homeomorphic.

Consider the identity mapping

I : B1 B defined by

I (x) = x for every x B1.

Then I is always one-one and onto.

Further I (x) = x x + T (x) = x 1

I is bounded (continuous) as a mapping from B1 onto B.

Therefore if we show that B1 is complete with respect to 1, then B1 is a Banach space so bytheorem.

“Let B and B be Banach spaces and let T be one-one continuous linear transformation of B ontoB . Then T is a homeomorphism. In particular, T–1 is automatically continuous.”

I is homeomorphism. Therefore to complete the proof, we have to show that B 1 is completeunder the norm 1.

Let (xn) be a Cauchy sequence in B1. Then

xn – xm 1 = xn – xm + T (xn – xm) 0 as m, n

(xn) and (T (xn)) are Cauchy sequences in B and B respectively.

Since B and B are complete, we have

xn x in B and T (xn) T (x) in B … (1)



NotesSince GT is closed, we have

(x1 T (x)) GT and if we take

y = T(x); then (x, y) GT.

Now xn – x 1 = xn – x + T (xn – x)

= xn – x + T (xn) – T(x)

= xn – x + T (xn – y) 0 as n . (Using (1))

Hence, the sequence (xn) in B1 x B1 and consequently B1 is complete.


Theorem 3: Let B and B be Banach spaces and let T : B B be linear. If GT is closed in B × B andif T is one-one and onto, then T is a homeomorphism from B onto B .

Proof: By closed graph theorem, T is continuous.

Let T = T–1 : B B. Then T is linear.

Further (x, y) GT (y, x) TG .

TG is closed in B × B.

T is continuous (By closed graph theorem)

T is a homeomorphism on B onto B .


Theorem 4: Let a Banach space B be made into a Banach space B by a new norm. Then thetopologies generated by these two norms are the same if either is stronger than the other.

Proof: Let the new norm on B be . Let is stronger than . Then a constant k such that x k x for every x B.

Consider the identity map

I : B B .

We claim that G1 is closed.

Let xn x in B and xn y in B .

Then x k x x B, I (xn) = xn y in also.

Since a sequence cannot converge to two distinct points in , y = x. Consequently G1 is closed.

Hence closed graph theorem, I is continuous. Therefore a k s such that

x = I(x) k x for every x B. Hence is stronger than . Hence two topologies aresame.

20.2 Summary

Let N and N be a normal linear space and let T : N N be a mapping with domain N andrange N . The graph of T is defined to be a subset of N × N which consist of all orderedpairs (x, T (x)). It is generally denoted by GT.



Notes Let N and N be normed linear spaces and let M be a subspace of N. Then a lineartransformation T : M N is said to be closed iff xn M, xn x and T (xn) y implyx M and y = T (x).

If B and B are Banach spaces and if T is linear transformation of B into B , then T iscontinuous Graph of T (GT) is closed.

20.3 Keyword

Closed Linear Transformation: Let N and N be normed linear spaces and let M be a subspace ofN. Then a linear transformation

T : M N is said to be closed

iff xn M, xn x and T (xn) y imply x M and y = T (x).


1. If X and Y are normed spaces and A : X Y is a linear transformation, then prove thatgraph of A is closed if and only if whenever xn 0 and Axn y, it must be that y = 0.

2. If P is a projection on a Banach space B, and if M and N are its range and null space, thenprove that M and N are closed linear subspaces of B such that B = M N.


Books Folland, Gerald B, Real Analysis: Modern Techniques and their Applications (1st ed.),John Wiley & Sons, (1984).

Rudin, Walter, Functional Analysis, Tata McGraw-Hill (1973).

Online links euclid.colorado.edu/ngwilkin/files/math6320.../OMT_CGT.pdf

mathworld.wolfram.com


Unit 21: The Conjugate of an Operator

NotesUnit 21: The Conjugate of an Operator

CONTENTS

Objectives

Introduction

21.1 The Conjugate of an Operator

21.1.1 The Linear Function

21.1.2 The Conjugate of T

21.2 Summary

21.3 Keywords



Objectives


Understand the definition of conjugate of an operator.

Understand theorems on it.

Solve problems relate to conjugate of an operator.

Introduction

We shall see in this unit that each operator T on a normed linear space N induces a correspondingoperator, denoted by T* and called the conjugate of T, on the conjugate space N*. Our first task isto define T* and our second is to investigate the properties of the mapping T T*.

21.1 The Conjugate of an Operator

21.1.1 The Linear Function

Let N* be the linear space of all scalar-valued linear functions defined on N. Clearly the conjugatespace N* is a subspace of N*. Let T be a linear transformation T of N* into itself as follows:

If f N+, then T (f) is defined as

[T (f)]x = f (T (x))

Since f (T (x)) is well defined, T is a well-defined transformation on N+.

Theorem 1: Let T : N+ N+ be defined as

[T (j)] x = f (T (x)), f N+, then

(a) T (j) is a linear junction defined on N.

(b) T is a linear mapping of N+ into itself.



Notes (c) T (N*) N* T is continuous, where T is a linear transformation of N into itself which isnot necessarily continuous.

Proof:

(a) x, y N and , be any scalars. Then

[T (f)] ( x + y) = f (T ( x + y))

Since T and f are linear, we get

f (T ( x + y)) = f (T (x) + f (T (y))

= [T (f)] (x) + [T (f)]y

part (a).

(b) Let f, g N+ and , be any scalars. Then

[T ( f + g) (x)] = ( f + g) (T (x)) = [T (f)] (x) + (T (g)] (x)

T is linear on N+

part (b)

(c) Let S be a closed unit sphere in N. Then we know that T is continuous T (S) is bounded

f (T (S)) is bounded for each f N*.

By definition of T , f (T (S)) is bounded if and only if [T (f)] (S) is bounded for each f in

N* = T (f) is in N* for each f in N*.

T (N) N*

part (c)


Note: Part (c) of the above theorem enables us to restrict T to N* iff T is continuous. Henceby making T continuous we define an operation called the conjugate of T by restricting Tto N*. We see it below.

21.1.2 The Conjugate of T

Definition: Let N be normed linear space and let T be a continuous linear transformation of N intoitself (i.e. T is an operator). Define a linear transformation T* of N* into itself as follows:

If f N*, then, T* (f) is given by

[T* (f)] (x) = f (T (x))

We call T* the conjugate of T.

Theorem 2: If T is a continuous linear transformation on a normed linear space N, then itsconjugate T* defined by

T* : N* N* such that

T* (f) = f.T where

[T* (f)] (x) = f (T (x)) f N* and all x N

is a continuous linear transformation on N* and the mapping T T* given by

: (N) (N*) such that



Notes (T) = T* for every (N)

is an isometric isomorphism of b (N) into b (N*) reverses products and preserves the identifytransformation.

Proof: We first show that T* is linear

Let f, g N* and , be any scalars

then [T* ( j + g)] (x) = ( j + g) (T (x))

= ( j) T (x) + ( g) T (x)

= [j (T (x))] + [g (T (x))]

= [T* (j)] (x) + [T* (g)] (x)

= [ T* (j) + T* (g)] (x) x N

Hence T* ( f + g) = T* (f) + T* (g)

T* is linear on N*.

To show that T* is continuous, we have to show that it is bounded on the assumption that T isbounded.

T* = sup { T* (f) : f 1}

= sup { [T* (f)] (x) : f 1 and x }

= sup { f (T (x))|: f 1 and x }

= sup { f T x : x 1 and x } … (1)

T

T* is a bounded linear transformation on N* into N*. Hence by application of Hahn-Banach theorem, for each non-zero x in N, a functional f N* such that

f = 1 and f (T (x)) = T (x) … (2)

Hence T = T(x)

sup : x 0x

= f T(x)

sup : f 1, x 0x (by (2))

= T * (f) (x)

sup : f 1, x 0x

(by (1))

T * (f) x

sup : f 1, x 0x

= sup T * (f) : f 1 T * … (3)


T = T* . … (4)



Notes Now we show that

: (N) (N*) given by(T) T * … (5)

for every T (N) is an isometric isomorphism which reverses the product and preserve theidentity transformation.

The isometric character of follows by using (5) as seen below:

(T) = T* = T .

Next we show that is linear and one-to-one. Let T, T1 (N) and , be any scalars. Then

( T + T1) = ( T + T)* by (3)

But [( T + T1)* (f)] (x) = f ( T + T1) (x)

= f ( T (x) + T1 (x) )

Since f is linear, we get

[( T + T1)* (f)] = f (T (x)) + f (T1 (x))

= [T* (f) (x) + 1T * (f) (x)]

= 1[T * (f)] [T * (f)] (x)

x N. Hence we get

[( T + T1)* (f)] = [T* (f)] + 1[T * (f) ]

= 1T * T * (f)

Hence ( T + T1)* = T* + T1* … (6)

Therefore ( T + T1) = ( T + T1)* = T* + 1T * = (T) + (T1)

is linear.

To show is one-to-one, let (T) = (T1)

Then T* = 1T *

1T * T * = 0

Using (6) by choosing = 1, = – 1 we get

(T – T1)* = 0 T – T1 = 0 or T = T1.

is one-to-one.

Hence is an isometric isomorphism on (N) onto (N*).

Finally we show that reverses the product and preserves the identity transformation.

Now [(T T1)* (f)] (x) = f ((T T1) (x))

= f (T (T1 (x))

= [T* (f)] [T1 (x)], since T1 (x) N and T* (f) N*.



Notes= [ 1T * (T* (f))] (x)

= [( 1T * T*) (f)] (x)

Hence, we get

(T T1)* = 1T * T* so that

(T T1) = (T T1)* = 1T * T.

reverses the product.

Lastly if I is the identity operator on N, then

[I* (f)] (x) = f (I (x)) = f (x) = (I f) (x).

I* = I so that (I) = I* = I

preserves the identity transformation.


Theorem 3: Let T be an operator on a normal linear space N. If N N* in the natural imbedding,then T** is an extension of T. If N is reflexive, then T** = T.

Proof: By definition, we have

(T*)* = T**

Using theorem 2, we have T* = T .

Hence T** = T* = T .

By definition of conjugate of an operator

T : N N, T* : N* N*, T** : N** N**.

Let J ; x Fx be the natural imbedding of N onto N** so that

Fx (f) = f (x) and J (x) = Fx.

Further, since T** is the conjugate operator of T*, we get

T** (x ) x = x (T* (x )) where x N* and x N**

T** (x ) x = T** (J (x)) x .

Using the definition of conjugate, we get

T** (J (x)) x = J (x) (T* (x )).

By definition of canonical imbedding

J (x) (T* (x )) = T* (x ) x.

Again T* (x ) (x) = x (T (x)) (By definition of conjugate)

Now x (T (x)) = J (T (x))x (By natural imbedding)

Hence T** (J (x))x = J (T (x))x .

T** . J = J T

and so T** is the norm preserving extension of T. If N is reflexive, N = N** and so T** coincideswith T.



Notes This completes the proof of the theorem.

Theorem 4: Let T be an operator on a Banach space B. Then T has an inverse T–1 T* has aninverse (T*)–1, and

(T*)–1 = (T–1)*

Proof: T has inverse T–1 TT–1 = T–1 T = I

By theorem 2, the mapping : T T* reverse the product and preserves the identity

(TT–1)* = (T–1 T)* = I*

(T–1)*T = T* (T–1)* = I

(T*)–1 exists and it is given by (T*)–1 = (T–1)*. This completes the proof of the theorem.

21.2 Summary

Let N+ be the linear space of all scalar-valued linear functions defined on N. Clearly theconjugate space N* is a subspace of N+. Let T be a linear transformation T of N into itselfas follows:


T (f) x = f (T (x))

Let N be a normed linear space and let T be a continuous linear transformation of N intoitself. Define a linear transformation T* of N* into itself as follows:

If f N+, then T (f) is given by

T (f) x = f (T (x))


21.3 Keywords

The Conjugate of T: Let N be normed linear space and let T be a continuous linear transformationof N into itself (i.e. T is an operator). Define a linear transformation T* of N* into itself asfollows:

If f N*, then, T* (f) is given by

[T* (f)] (x) = f (T (x))


The Linear Function: Let N* be the linear space of all scalar-valued linear functions defined on N.Clearly the conjugate space N* is a subspace of N*. Let T be a linear transformation T of N* intoitself as follows:


[T (f)]x = f (T (x))

Since f (T (x)) is well defined, T is a well-defined transformation on N+.


1. Let B be a Banach space and N a normed linear space. If {Tn} is a sequence in B (B, N) suchthat T(x) = lim Tn (x) exists for each x in B, prove that T is a continuous transformation.



Notes2. Let T be an operator on a normed linear space N. If N is considered to be part of N** bymeans of the natural imbedding. Show that T** is an extension of T. Observe that if N isreflexive, then T** = T.

3. Let T be an operator on a Banach space B. Show that T has an inverse T–1 T* has an inverse(T*)–1, and that in this case (T*)–1 = (T–1)*.


Books James Wilson Daniel, The Conjugate Gradient Method for Linear and Non-linearOperator Equations.

G.O. Okikiolu, Special Integral Operators: Poisson Operators, Conjugate Operators andrelated Integrals. Vol. Okikiolu Scientific and Industrial Organization, 1981.

Online links epubs.siam-org/sinum/resource/1/sjnamm/v9/i|/p165_s|

www.ima.umm.edu



Notes Unit 22: The Uniform Boundedness Theorem

CONTENTS

Objectives

Introduction

22.1 The Uniform Boundedness Theorem

22.1.1 The Uniform Boundedness Theorem – Proof


22.2 Summary

22.3 Keywords



Objectives


State the uniform boundedness theorem.

Understand the proof of this theorem.

Solve problems related to uniform boundedness theorem.

Introduction

The uniform boundedness theorem, like the open mapping theorem and the closed graphtheorem, is one of the cornerstones of functional analysis with many applications. The openmapping theorem and the closed graph theorem lead to the boundedness of T–1 whereas theuniform boundedness operators deduced from the point-wise boundedness of such operators.In uniform boundedness theorem we require completeness only for the domain of the definitionof the bounded linear operators.

22.1 The Uniform Boundedness Theorem

22.1.1 The Uniform Boundedness Theorem – Proof

If (a) B is a Banach space and N a normed linear space,

(b) {Ti} is non-empty set of continuous linear transformation of B into N, and

(c) {Ti (x)} is a bounded subset of N for each x B, then { Ti } is a bounded set of numbers, i.e.{Ti} is bounded as a subset of (B, N)

Proof: For each positive integer n, let

Fn = {x B : Ti (x) n i}.

Then Fn is a closed subset of B. For if y is any limit point of Fn, then a sequence (xk) of points ofFn such that

xk y as k


Unit 22: The Uniform Boundedness Theorem

NotesTixk Tiy as k (By continuity of Ti)

Tixk Tiy as k (By continuity of norm)

Tiy = klt Ti xk

n i ( xk Fn)

y Fn.

Thus Fn contains all its limit point and is therefore closed. Further, if x is any element of B, thenby hypothesis (c) of the theorem a real number k 0 s.t.

Tix k i

Let n be a positive integer s.t. n > k. Then

Tix < n i

so that x Fn.

Consequently, we have B = nn 1

F .

Since B is complete, it therefore follows by Baire’s theorem that closure of some Fn, say no noF F ,

possesses an interior point xo. Thus we can find a closed sphere So with centre xo and radius ro

such that So noF .

Now if y is any vector in Ti (So), then

y = i soT

where so So noF .

y = i oTs no.

Thus norm of every vector in Ti (So) is less than or equal to no. We write this fact as Ti (So) no.

Let S = o o

o

S xr . Then S is a closed unit sphere centred at the origin in B and

Ti (S) = o oi

o

S xTr

= i o i oo

1 T (S ) T (x )r

i o i oo

1 T (S ) T (x )r

o

o

2n ir

Hence Ti o

o

2n ir




Notes 22.1.2 Theorems and Solved Examples

Theorem 1: If B is a Banach space and (fi (x)) is sequence of continuous linear functionals on B suchthat (|fi (x)|) is bounded for every x B, then the sequence ( fi ) is bounded.

Proof: Since the proof of the theorem is similar to the theorem (1), however we briefly give itsproof for the sake of convenience to the readers.

For every m, let Fm B be the set of all x such that |fn (x)| m n

|fn (x)| m n

Now Fm is the intersection of closed sets and hence it is closed.

As in previous theorem, we have

B = mm 1

F . Since B is complete. It is of second category. Hence by Baire’s theorem, there is a

xo Fm and a closed sphere S [xo, ro] such that |fn (x)| m n.

Let x be a vector with x ro.

Now fn (x) = fn (x + xo – xo)

= fn (x + xo) – fn (xo)

|fn (x)| |fn (x + xo)| + |fn (xo)| … (1)

Since x + xo – xo = x < ro, we have (x + ro) S [xo, ro]

|fn (x + xo) | m … (2)

Also we have |fn (xo)| k n … (3)

From (1), (2) and (3), we have for x S [xo, r].

|fn (x)| < (m + k) n.

Now for x B, consider the vector or xx .

Then on n

o o

x xr x|f (x)| f (m k)r x r

so that n

o

f (x) m kx r .

In other words,

f o

m kr

.


Example 1: Show that the completeness assumption in the domain of (Ti) in the uniformboundedness theorem cannot be dropped.

Solution: Consider N= space of all polynomial x

= x (t) = n nn 0

a t , an 0

for finitely many n’s.



NotesIt we define the norm on N as

x = max {|an|, n = 0, 1, 2, …}

then N is an incomplete normed linear space.

Now define fn (x) = n 1

kk 0

a , n = 1, 2, …

The functions {fn} are continuous linear functional on N.

If we take x = a0 + a1t + …… + am tm then

|fn (x)| (m + 1) max {|ak|} = (m + 1) x ,

so that (|fn (x)|) is point-wise bounded.

Now consider x = 1 + t + t2 + … + tn–1. Then x = 1 and from the definition of |fn (x)| = n.

Hence fn nf (x)x = n.

( fn ) is unbounded.

Thus if we drop the condition of completeness in the domain of (T i), the uniform boundednesstheorem is not true anymore.

Theorem 2: Let N be a normed linear space and B be a Banach space. If a sequence (Tn) (B, N)such that T (x) = lim Tn (x) exists for each x in B, then T is a continuous linear transformation.

Proof: T is linear.

T ( x + y) = lim Tn ( x + y)

= lim {Tn ( x) + Tn ( y)|}

= lim Tn (x) + lim Tn (y)

= T (x) + T (y) for x, y B and for any scalars and .

since lim Tn (x) exists, (Tn (x)) is a convergent sequence in N. Since convergent sequences arebounded, (Tn (x)) is point-wise bounded.

Hence by uniform bounded theorem, ( Tn ) is bounded so that a positive constant such that

Tn n.

Now Tn(x) Tn x x .

Since Tn (x) T (x), we have

T (x) x

T is bounded (continuous) linear transformation. This completes the proof of the theorem.

Corollary 1: If f is a sequence in B* such that f (x) = nnlim f (x) exists for each x B, then f is

continuous linear functional on B.

Example 2: Let (an) be a sequence of real or complex numbers such that for each x = (xn)

co, n nn 1

a x converges. Prove that nn 1

|a | .



NotesSolution: For every x co, let

n

n i ii 1

f a x . Since each n

i ii 1

a x is a finite sum of scalars, (fn) is

sequence of continuous linear functional on co. Let f (x) = n

n i in ni 1

lim f (x) lim a x . By cor. 1, f (x)

exists and bounded. f = nn 1

|a |. Since f is bounded, nn 1

|a | < .

Theorem 3: A non-empty subset X of a normed linear space N is bounded f (X) is a bounded setof numbers for each f in N*.

Proof: Let X be a bounded subset of N so that a positive constant 1 such that

x 1 x X … (1)

To show that f (X) is bounded for each f N*. Now f N* f is bounded.

2 > 0 such that |f (x)| 2 x x N … (2)

It follows from (1) & (2) that

|f (x)| 1 2 x X.

f (X) is a bounded set of real numbers for each f N*.

Conversely, let us assume that f (X) is a bounded set of real numbers for each f N*.

To show that X is bounded. For convenience, we exhibit the vectors in X by writing X = {x i}. Wenow consider the natural imbedding J from N to N** given by

J : xi xiF

From the definition of this natural imbedding, we have

Fx (f) = f (x) for each x N.

Hence our assumption f (X) = {f (xi)} is bounded for each f N* is equivalent to the assumption

that xiF (f) is bounded set for each f N*.

Since N* is complete xiF is bounded subset of N** by uniform boundedness theorem.

That is, xiF is a bounded set of numbers. Since the norms are preserved in natural imbedding,

we have xiF = xi for every xi X.

Therefore ( xi ) is a bounded set of numbers. Hence is bounded subset of Ni.


Theorem 4: Let N and N be normed linear space A linear transformation.

T : N N is continuous for each f N*, f o T N*.

Proof: We first note that f o T is linear. Also f o T is well defined, since T (x) N for every x Nand f is a functional on N so that f (T (x)) is well defined and f o T N*. Since T is continuous andf is continuous, f o T is continuous on N.



NotesConversely, let us assume that f o T is continuous for each f N*. To show that T is continuousit suffices to show that

T(B) = {Tx : x N, B = x 1} is bounded in N .

For each f N*, f o T is continuous and linear on N and so (f o T) B = f (T(B)) is bounded set forevery f N*, where we have considered the unit sphere B with centre at the origin and radius 1.Since any bounded set in N can be obtained from B, T (B) is bounded by a non-empty subset X ofa normed linear space N of bounded f (X) is a bounded set of number for each f in N*.

22.2 Summary

Uniform Boundedness Theorem: If (a) B is Banach space and N a normed linear space,(b) {Ti} is non-empty set of continuous linear transformation of B into N and (c) {Ti (x)} is abounded subset of N for each x B, then { Ti } is a bounded set of numbers, i.e. {Ti} isbounded as a subset of (B, N).

If B is a Banach space and (fi (x)) is sequence of continuous linear functionals on B such that(|fi (x)|) is bounded for every x B, then the sequence ( Ti ) is bounded.

22.3 Keywords

Imbedding: Imbedding is one instance of some mathematical structure contained within anotherinstance, such as a group that is a subgroup.

Uniform Boundedness Theorem: The uniform boundedness theorem, like the open mappingtheorem and the closed graph theorem, is one of the cornerstones of functional analysis withmany applications.


1. If X is a Banach space and A X*, then prove that A is a bounded set if and only if for everyx in X, Sup {|f (x)| : f A} < .

2. Let be a Hilbert space and let be an orthonormal basis for . Show that a sequence {hn}in satisfies <hn, h> 0 for every h in if and only if sup { hn : n 1} < and <hn, e>

0 for every e in .


Books Bourbaki, Nicolas, Topological vector spaces, Elements of mathematics, Springer (1987).

Diendonne, Jean, Treatise on Analysis, Volume 2, Academic Press, (1970).

Rudin, Walter, Real and Complex Analysis, McGraw-Hill, 1966.

Online links www.jstor.org/stable/2035429

www.sciencedirect.com/science/article/pii/S0168007211002004



Notes Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

CONTENTS

Objectives

Introduction

23.1 Hilbert Spaces

23.1.1 Inner Product Spaces

23.1.2 Hilbert Space and its Basic Properties

23.1.3 Hilbert Space: Definition

23.1.4 Examples of Hilbert Space

23.2 Summary

23.3 Keywords



Objectives


Define inner product spaces.

Define Hilbert space.

Understand basic properties of Hilbert space.

Solve problems on Hilbert space.

Introduction

Since an inner product is used to define a norm on a vector space, the inner product are specialnormed linear spaces. A complete inner product space is called a Hilbert space. We shall also seefrom the formal definition that a Hilbert space is a special type of Banach space, one whichpossesses additional structure enabling us to tell when two vectors are orthogonal. From theabove information, one can conclude that every Hilbert space is a Banach space but not converselyin general.

We shall first define Inner Product spaces and give some examples so as to understand theconcept of Hilbert spaces more conveniently.

23.1 Hilbert Spaces

23.1.1 Inner Product Spaces

Definition: Let X be a linear space over the field of complex numbers C. An inner product on X isa mapping from X × X C which satisfies the following conditions:


Unit 23: Hilbert Spaces: The Definition and Some Simple Properties

Notes(i) ( x + y, z) = (x, z) + (y, z), x, y, z X and , C. (Linearity property)

(ii) (x, y) = (y, x) (Conjugate symmetry)

(iii) (x, x) 0, (x, x) = 0 x = 0 (Non-negativity)

A complex inner product space X is a linear space over C with an inner product defined on it.

Notes

1. We can also define inner product by replacing C by R in the above definition. In thatcase, we get a real inner product space.

2. It should be noted that in the above definition (x, y) does not denote the ordered pairof the vectors x and y. But it denotes the inner product of the vectors x and y.

Theorem 1: If X is a complex inner product space then

(a) ( x – y, z) = (x, z) – (y, z)

(b) (x, y + z) = (x, y) + (x, z)

(c) (x, y – z) = (x, y) – (x, z)

(d) (x, 0) and (0, x) = 0 for every x X.

Proof: (a) ( x – y, z) = ( x + (– ) y, z)

= (x, z) + (– ) (y, z)

= (x, z) – (y, z).

(b) (x, y + z) = ( y z, x) ( y, x) ( z, x)

= (y, x) (z,x)

= (x, y) (x,z)

(c) (x, y – z) = (x, y + (– ) z) = (x, y) ( ) (x,z)

= (x, y) (x,z)

(d) (0, x) = (0 , x) = 0 ( , x) = 0, where is the zero

element of x and (x, 0) (0, x) 0 0 .

Further note that (x, y + z) = (x, |y + 1|z) = 1 (x, y) + 1 (x, z)

Hence (x, y + z) = (x, y) + (x, z).




Notes

Notes

1. Part (b) shows an inner product is conjugate linear in the second variable.

2. If (x, y) = 0 x X, then y = 0. If (x, y) = 0 x X, it should be true for x = y also,so that (y, y) = 0 y = 0.

Example 1: The space n2 is an inner product space.

Solution: Let x = (x1, x2, ……, xn), y = (y1, y2, ……, yn) n2 .

Define the inner product on n2 as follows:

(x, y) = n

i ii 1

x y

Now

(i) ( x + y, z) = n

ii ii 1

x y z

= n n

i ii ii 1 i 1

x z y z

= (x, z) + (y, z)

(ii) (x,y) = n

i ii 1

x y

= 1 2 n1 2 n(x y x y x y )

= 1 2 n1 2 n(x y x y x y )

= 1 2 n1 2 nx y x y x y

= (y, x)

(iii) (x, x) = n

iii 1

x x

= n

2i

i 1

x 0

Hence (x, x) 0 and (x, x) = 0 xi = 0 for each i, i.e. (x, x) = 0 x = 0.

(i) – (iii) n2 is a inner product space.

23.1.2 Hilbert Space and its Basic Properties

By using the inner product, on a linear space X we can define a norm on X, i.e. for each x X, we

define x = (x,x) . To prove it we require the following fundamental relation known asSchwarz inequality.



NotesTheorem 2: If x and y are any two vectors in an inner product space then

|(x, y)| x y … (1)

Proof: If y = 0, we get y = 0 and also theorem 1 implies that |(x, y)| = 0 so that (1) holds.

Now, let y 0, then for any scalar C we have

0 x – y 2 = (x – y,x – y)

But

(x – y, x – y) = (x, x) – (x, y) – ( y, x) + ( y, y)

= (x, x) – (x, y) – (y, x) + (y, y)

= x 2 – (y, x) – (x, y) + | |2 y 2

x 2 – (y, x) – (x, y) +| |2 y 2 = (x, y, x – y)

= x – y 2 0 … (2)

Choose = 2(x,y) , y 0, y 0

y.

We get from (2)

22 2

2 2 4

(x,y)(x,y) (x y) (x, y)x (x,y) y 0y y y

22 22

2 2 2

(x,y)|(x,y)| |(x,y)|x 0y y y

22

2|(x,y)|x 0

y

x 2 y 2 |(x, y)|2

or |(x, y)| x y .


Theorem 3: If X is an inner product space, then (x, x) has the properties of a norm, i.e.

x = (x, x) is a norm on X.

Proof: We shall show that satisfies the condition of a norm.

(i) x = (x, x) x 2 = (x, x) 0 and x = 0 x = 0.

(ii) Let x, y X, then

x + y 2 = (x + y, x + y)

= (x, x) + (x, y) + (y, x) + (y, y) … (1)

= x 2 + (x, y) + (x, y) + y 2

= x 2 + 2Re (x, y) + y 2 [ (x,y) (x, y) 2 Re(x,y)]

x 2 + 2|(x, y)| + y 2 [ Re (x, y) |(x, y)|]



Notes x 2 + 2 x y + y 2 [using Schwarz inequality]

= ( x + y )2

Therefore x + y x + y

(iii) x 2 = ( x, x)= (x, x)

= | |2 x 2

, x = x

(i)-(iii) imply that x = (x, x) is a norm on X. This completes the proof of the theorem.

Note Since we are able to define a norm on X with the help of the inner product, the innerproduct space X consequently becomes a normed linear space. Further if the inner productspace X is complete in the above norm, then X is called a Hilbert space.

23.1.3 Hilbert Space: Definition

A complete inner product space is called a Hilbert space.

Let H be a complex Banach space whose norm arises from an inner product which is a complexfunction denoted by (x, y) satisfying the following conditions:

H1 : ( x + y, 2) = (x, 2) + (y, 2),

H2 : (x, y) = (y, x), and

H3 : (x, x) = x 2,

for all x, y, z H and for all , C.

23.1.4 Examples of Hilbert Space

1. The space n2 is a Hilbert space.

We have already shown in earlier example that n2 is an inner product space. Also n

2 is a

Banach space. Consequently n2 is a Hilbert space. Moreover n

2 , being a finite dimensional,

hence n2 is a finite dimensional Hilbert space.

2. 2 is a Hilbert space.

Consider the Banach space 2 consisting of all infinite sequence x = (xn), n = 1, 2, … of

complex numbers such that 2n

n 1

x with norm of a vector x = (xn) defined by x =

2n

n 1

x .



NotesWe shall show that if the inner product of two vectors x = (xn) and y = (yn) is defined by

(x, y) = n nn 1

x y , then 2 is a Hilbert space.

We first show that inner product is well defined. For this we are to show that for all x, y in 2 the

infinite series n nn 1

x y is convergent and this defines a complex number.

By Cauchy inequality, we have

n

i ii 1

x y

1 1n n2 2

22i i

i 1 i 1

x y

1 12 2

22n n

n 1 n 1

x y .

Since 2n

n 1

x and 2n

n 1

y are convergent, the sequence of partial sum n

i ii 1

x y is a monotonic

increasing sequence bounded above. Therefore, the series i in 1

x y is convergent. Hence

n nn 1

x y is absolutely convergent having its sum as a complex number.

Therefore (x, y) = n nn 1

x y is convergent so that the inner product is well defined. The condition

of inner product can be easily verified as in earlier example.

Theorem 4: If x and y are any two vectors in a Hilbert space, then

(x + y) 2 + x – y 2 = 2 ( x 2 + y 2)

Proof: We have for any x and y

(x + y) 2 = (x + y, x + y) (By def. of Hilbert space)

= (x, x + y) + (y, x + y)

= (x, x) + (x, y) + (y, x) + (y, y)

= x 2 + (x, y) + (y, x) + y 2 … (1)

x – y 2 = (x – y, x – y)

= (x, x – y) – (y, x – y)

= (x, x) – (x, y) – (y, x) + y 2 … (2)

Adding (1) and (2), we get

x + y 2 + (x – y) 2 = 2 x 2 + 2 y 2 = 2 ( x 2 + y 2)




Notes Theorem 5: In a Hilbert space the inner product is jointly continuous i.e.,

xn x, yn y (xn, yn) (x, y)

Proof: We have

|(xn, yn) – (x, y)| = |(xn, yn) – (xn, y) + (xn, y) – (x,, y)|

= |(xn, yn – y) + (xn – x, y)|

(by linearity property of inner product)

|(xn, yn – y)| + |(xn – x, y)| [ | + | | | + | |]

xn + yn – y + xn – x y [By Schwarz inequality]

Since xn x and yn y as n .

Therefore yn – yn 0 and xn – x 0 as h . Also (xn) is a continues sequence, it is boundedso that xn M n.

Therefore

| (xn, yn) – (x, y) | 0 as n .

Hence (xn, yn) (x, y) as n .


Theorem 6: A closed convex set E in a Hilbert space H continuous a unique vector of smallestnorm.

Proof: Let = inf { e ; e E}

To prove the theorem it suffices to show that there exists a unique x E s.t. x = .

Definition of yields us a sequence (xn) in E such that

nnLim x = … (1)

Convexity of E implies that m nx x E2

. Consequently

m nx x2

xm + xn 2 … (2)

Using parallelogram law, we get

xm + xn 2 + xm – xn 2 = 2 xn 2 + 2 xn 2

or xm – xn 2 = 2 xm 2 + 2 xn 2 – xm – xn 2

2 xm 2 + 2 xn 2 – d 2 (Using (2))

0 as m, n (Using (1))

xm – xn 2 0 as m, n

(xn) is a CAUCHY sequence in E.

x E such that nn

Lim x x , since H is complete and E is a closed subset of H, therefore

E is also complete and consequently (xn) is in E is a convergent sequence in E.



NotesNow x = nn

Lim x

= nnLim x ( norm is continuous mapping)

= .

Uniqueness of x.

Let us suppose that y E, y x and y = .

Convexity of E x y2

E

x y2 … (3)

Also by parallelogram law, we have

2yx2 2

= 2 2 2x y x y

2 2 2

= 2 22 2

2x y x y2 2 2 2

< 2.

So that

2x y2 < , a result contrary to (3).

Hence we must have y = x.


Example: Give an example of a Banach space which is not an Hilbert space.

Solution: C [a, b] is a Banach space with supremum norm, i.e. if x C [a, b] then

x = Sup {|x(t)| : t [a, b]}.

Then this norm does not satisfy parallelogram law as shown below:

Let x(t) = 1 and y (t) = t ab a . Then x = 1, y = 1

Now x (t) + y (t) = 1 + t ab a so that x + y = 2

x (t) – y (t) = 1 – t ab a so that x – y = 1

Hence 2 ( x 2 – y 2) = 4, and x + y 2 + x – y 2 = 5

So that x + y 2 + x – y 2 2 x 2 + 2 y 2.

C [a, b] is not a Hilbert space.



Notes 23.2 Summary

Let X be a linear space over the field of complex numbers C. An inner product on X is amapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z) x, y, z X and , C.

(ii) (x, y) = (y, x)

(iii) (x, x) 0, (x, x) = 0 x = 0

A complete inner product space is called a Hilbert space.

23.3 Keywords

Hilbert Space: A complete inner product space is called a Hilbert space.

Inner Product Spaces: Let X be a linear space over the field of complex numbers C. An innerproduct on X is a mapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z), x, y, z X and , C. (Linearity property)

(ii) (x, y) = (y, x) (Conjugate symmetry)

(iii) (x, x) 0, (x, x) = 0 x = 0


1. For the special Hilbert space n2 , use Cauchy’s inequality to prove Schwarz’s inequality.

2. Show that the parallelogram law is not true in n2 (n > 1).

3. If x, y are any two vectors in a Hilbert space H, then prove that

4 (x, y) = x + y 2 – x – y 2 + i x + iy 2 – i x – iy 2.

4. If B is complex Banach space whose norm obeys the parallelogram law, and if an innerproduct is defined on B by

4 (x, y) = x + y 2 – x – y 2 + i x + iy 2 – i x – iy 2,

then prove that B is a Hilbert space.


Books Bourbaki, Nicolas (1987), Topological vector Spaces, Elements of Mathematics, BERLIN:Springer – Verlag.

Halmos, Paul (1982), A Hilbert space Problem Book, Springer – Verlag.

Online links www.math-sinica.edu.tw/www/file_upload/maliufc/liu_ch04.pdf

mathworld.wolfram.com>Calculus and Analysis > Functional Analysis


Unit 24: Orthogonal Complements

NotesUnit 24: Orthogonal Complements

CONTENTS

Objectives

Introduction

24.1 Orthogonal Complement

24.1.1 Orthogonal Vectors

24.1.2 Pythagorean Theorem

24.1.3 Orthogonal Sets

24.1.4 Orthogonal Compliment: Definition

24.1.5 The Orthogonal Decomposition Theorem or Projection Theorem

24.2 Summary

24.3 Keywords



Objectives


Define Orthogonal complement

Understand theorems on it

Understand the Orthogonal decomposition theorem

Solve problems related to Orthogonal complement.

Introduction

In this unit, we shall start with orthogonality. Then we shall move on to definition of orthogonalcomplement. Let M be a closed linear subspace of H. We know that M is also a closed linearsubspace, and that M and M are disjoint in the sense that they have only the zero vector incommon. Our aim in this unit is to prove that H = M M , and each of our theorems is a step inthis direction.

24.1 Orthogonal Complement

24.1.1 Orthogonal Vectors

Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y, written as x y, if(x, y) = 0.

By definition,

(a) The relation of orthogonality is symmetric, i.e.,

x y y x



Notes For, x y (x, y) = 0

(x,y) 0

(y, x) = 0

y x

(b) If x y then every scalar multiple of x is orthogonal to y i.e. x y x y for every scalarC.

For, let be any scalar, then

( x, y) = (x, y)

= . 0

= 0

x y x y.

(c) The zero vector is orthogonal to every vector. For every vector x in H, we have

(0, x) = 0

0 x for all x H.

(d) The zero vector is the only vector which is orthogonal to itself. For,

if x x (x, x) = 0 x 2 = 0 x = 0

Hence, if x x, then x must be a zero vector.

24.1.2 Pythagorean Theorem

Statement: If x and y are any two orthogonal vectors in a Hilbert space H, then

x + y 2 = x – y 2 = x 2 + y 2.

Proof: Given x y (x, y) = 0, then we must have

y x i.e. (y, x) = 0

Now x + y 2 = (x + y, x + y)

= (x, x) + (x, y) + (y, x) + (y, y)

= x 2 + 0 + 0 + y 2

= x 2 + y 2

Also, x – y 2 = (x – y, x – y)

= (x, x) – (x, y) – (y, x) + (y, y)

= x 2 – 0 – 0 – y 2

= x 2 + y 2

x + y 2 = x – y 2 = x 2 + y 2

24.1.3 Orthogonal Sets

Definition: A vector x is to be orthogonal to a non-empty subset S of a Hilbert space H, denotedby x S if x y for every y in S.



NotesTwo non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted byS1 S2, if x y for every x S1 and every y S2.

24.1.4 Orthogonal Compliment: Definition

Let S be a non-empty subset of a Hilbert space H. The orthogonal compliment of S, written as Sand is read as S perpendicular, is defined as

S = {x H : x y y S}

Thus, S is the set of all those vectors in H which are orthogonal to every vectors in H which areorthogonal to every vector in S.

Theorem 1: If S, S1, S2 are non-empty subsets of a Hilbert space H, then prove the following:

(a) {0} = H (b) H = {0} (c) S S {0}

(d) S1 S2 2 1S S (e) S S

Proof:

(a) Since the orthogonal complement is only a subset of H, {0} H.

It remains to show that H {0} .

Let x H. Since (x, 0) = 0, therefore x {0} .

Thus x H x {0} .

H {0} .

Hence {0} = H

(b) Let x H. Then by definition of H, we have

(x, y) = 0 y H

Taking y = x, we get

(x, x) = 0 x 2 = 0 x = 0

Thus x H x = 0

H = {0}

(c) x S S .

Then x S and x S

Since x S , therefore x is orthogonal to every vector in S. In particular, x is orthogonal tox because x S.

Now (x, x) = 0 x 2 = 0 x = 0.

0 is the only vector which can belong to both S and S .

S S {0}

If S is a subspace of H, then 0 S. Also S is a subspace of H. Therefore 0 S . Thus, if S isa subspace of H, then 0 S S . Therefore, in this case S S = {0}.



Notes (d) Let S1 S2, we have

x 2S x is orthogonal to every vector in S2

x is orthogonal to every vector in S1 because S1 S2.

x 1S

2 1S S

(e) Let x S. Then (x, y) = 0 y S .

by definition of (S ) , x (S ) .

Thus x S x S .

S S .


Theorem 2: If S is a non-empty subset of a Hilbert space H, then S is a closed linear subspace ofH and hence a Hilbert space.

Proof: We have

S = {x H : (x, y) = 0 y S} by definition. Since (0, y) = 0 y S, therefore at least 0 S andthus S is non-empty.

Now let x1, x2 S and , be scalars. Then (x1, y) = 0, (x2, y) = 0 for every y S.

For every y S, we have

( x1 + x2, y) = (x1, y) + (x2, y)

= (0) + (0)

= 0

x1 + x2 S

S is a subspace of H.

Next we shall show that S is a closed subset of H.

Let (xn) S and xn x in H.

Then we have to show that x S .

For this we have to prove (x, y) = 0 for every y S.

Since xn S , (xn, y) = 0 for every y S and for n = 1, 2, 3, …

Since the inner product is a continuous function, we get

(xn, y) (x, y) as n

Since (xn, y) = 0 n, (x, y) = 0

x S .

Hence S is a closed subset of H.



NotesNow S is a closed subspace of the Hilbert space H.

So, S is complete and hence a Hilbert space. This completes the proof of the theorem.

Theorem 3: If M is a linear subspace of a Hilbert space H, then M is closed

M = M .

Proof: Let us assume that M = M ,

M being a subspace of H.

by theorem (2), (M ) is closed subspace of H.

Therefore M = M is a closed subspace of H. Conversely, let M be a closed subspace of H. Weshall show that M = M .

We know that M M .

Now suppose that M M .

Now M is a proper closed subspace of Hilbert space M . a non-zero vector zo in M such thatzo M or zo M .

Now zo M and M gives zo M M … (1)

Since M is a subspace of H, we have

M M = {0} … (2)

(by theorem 1 (iii))

From (1) and (2) we conclude that z = 0, a contradiction to the fact that zo is a non-zero vector.

M M can be a proper inclusion.

Hence M = M .


Cor. If M is a non-empty subset of a Hilbert space H, then M = M .

Proof: By theorem (2), M is a closed subspace of H. So by theorem (3),

M = (M ) = M .

Theorem 4: If M and N are closed linear subspace of a Hilbert space H such that M N, then thelinear subspace M N is closed.

Proof: To prove: M + N is closed, we have to prove that it contains all its limit point.

Let z be a limit point of M + N,

a sequence (zn) in M + N such that zn z in H.

Since M N, M N = {0} and M + N is the direct sum of the subspace M and N, zn can be writtenuniquely as

zn = xn + yn where xn M and yn N.

Taking two points zm = xm + ym and zn = xn + yn, we have

zm – zn = (xm – xn) + (ym – yn).

Since xm – xn M and ym – yn N, we get

(xm – xn) (ym – yn)



Notes So, by Pythagorean theorem, we have

(xm – xn) + (ym – yn) 2 = xm – xn 2 + ym – yn 2.

But (xm – xn) + (ym – yn) = zm – zn so that

zm – zn 2 = xm – xn 2 + ym – yn 2 … (1)

Since (zn) is a convergent sequence in H, it is a Cauchy sequence in H.

Hence zm – zn 2 0 as m, n … (2)

Using (2) in (1), we see that

xm – xn 2 0 and ym – yn 2 0

So that (xn) and (yn) are Cauchy sequence in M and N.

Since H is complete and M and N are closed subspace of a complete space H, M and N arecomplete.

Hence, the Cauchy sequence (xn) in M converges to x in M and the Cauchy sequence (yn) in Nconverges to y in N.

Now z = lim zn = lim (xn + yn)

= lim xn + lim yn

But lim xn + lim yn = x + y M + N

Thus, z = x + y M + N

M + N is closed.

24.1.5 The Orthogonal Decomposition Theorem or Projection Theorem

Theorem 5: If M is a closed linear subspace of a Hilbert space H, then H = M M .

Proof: If M is a subspace of a Hilbert space H, then we know that M M = {0}.

Therefore in order to show that

H = M M , we need to verify that

H = M + M .

Since M and M are closed subspace of H, M + M is also a closed subspace of H by theorem 4.

Let us take N = M + M and show that N = H.

From the definition of N, we get M N and M N. Hence by theorem (1), we have

N M and N M .

Hence N M M = {0}.

N = {0}

N = {0} = H … (1)

Since N = M + M is a closed subspace of H, we have by theorem (3),

N = N … (2)

From (1) and (2), we have

N = M + M = H.



NotesSince M M = {0} and

H = M + M ,

we have from the definition of the direct sum of subspaces,

H = M M .


Theorem 6: Let M be a proper closed linear sub space of a Hilbert space H. Then there exists anon-zero vector zo in H such that zo M.

Proof: Since M is a proper subspace of H, there exists a vector x in H which is not in M.

Let d = d (x, M) = inf { x – y } : y M}.

Since x M, we have d > 0.

Also M is a proper closed subspace of H, then by theorem: “Let M be a closed linear subspace ofa Hilbert space H. Let x be a vector not in M and let d = d (x, m) (or d is the distance from x to M).Then there exists a unique vector yo in M such that x – yo = d.”

There exists a vector yo in M such that

x – yo = d.

Let zo = x – yo. We then here

zo = x – yo = d > 0.

zo is a non-zero vector.

Now we claim that Zo M.

Let y be an arbitrary vector in M. We shall show that zo y. For any scalar , we have

zo – y = x – yo – y = x – (yo + y).

since M is a subspace of H and yo, y M,

yo + M M.

Then by definition of d, we have

x – (yo + y) d

Now zo – y = x – (yo + y) d = zo

zo – y 2 zo 2

or (zo – y, zo – y) – (zo, zo) 0

or (zo, zo) – (zo, y) – (y, zo) + (y, y) – (zo, zo) 0

or o o(z ,y) (z ,y) (y,y) 0 … (1)

The above result is true for all scalars .

Let us take o(z ,y) .

Putting the value of , in (1), we get

22o o o o o o(z ,y) (z ,y) (z ,y) (z ,y) (z ,y) (z ,y) y 0

or –2 |(zo, y)|2 + 2 |(zo, y)|2 y 2 0



Notes or |(zo, y)|2 { y 2 – 2} 0 … (2)

The above result is true for all real suppose that (zo, y) 0. Then taking positive and so smallthat y 2 < 2, we see from (2) that |(zo, y)|2 { y 2 – 2} < 0.

This contradicts (2).

Hence we must have (zo, yo) = 0 zo y, y M.

zo M.


24.2 Summary

Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y, written as x y, if(x, y) = 0.

If x and y are any two orthogonal vectors in a Hilbert space H, then

x + y 2 = x – y 2 = x 2 + y 2.

Two non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted byS1 S2, if x y for every x S1 and every y S2.

Let S be a non-empty subsets of a Hilbert space H. The orthogonal compliment of S,written as S and is read as S perpendicular, is defined as

S = {x H : x y y S}

The orthogonal decomposition theorem: If M is a closed linear subspace of a Hilbert spaceH, then H = M M .

24.3 Keywords

Orthogonal Compliment: Let S be a non-empty subset of a Hilbert space H. The orthogonalcompliment of S, written as S and is read as S perpendicular, is defined as

S = {x H : x y y S}

Orthogonal Sets: A vector x is to be orthogonal to a non-empty subset S of a Hilbert space H,denoted by x S if x y for every y in S.

Two non-empty subsets S1 and S2 of a Hilbert space H are said to be orthogonal denoted by S1 S2, if x y for every x S1 and every y S2.

Orthogonal Vectors: Let H be a Hilbert space. If x, y H then x is said to be orthogonal to y,written as x y, if (x, y) = 0.

Pythagorean Theorem: If x and y are any two orthogonal vectors in a Hilbert space H, then

x + y 2 = x – y 2 = x 2 + y 2.


1. If S is a non-empty subset of a Hilbert space, show that S = S .

2. If M is a linear subspace of a Hilbert space, show that M is closed M = M .

3. If S is a non-empty subset of a Hilbert space H, show that the set of all linear combinationsof vectors in S is dense in H S = {0}.



Notes4. If S is a non-empty subset of a Hilbert space H, show that S is the closure of the set of alllinear combinations of vectors in S.

5. If M and N are closed linear subspace of a Hilbert space h such that M N, then the linearsubspace M + N is closed.


Books Halmos, Paul R. (1974), Finite-dimensional Vector Spaces, Berlin, New York

Paul Richard Halmos, A Hilbert Space Problem Book, 2nd Ed.

Online links Itcconline.net/green/courses/203/…/orthogonal complements.html

www.math.cornell.edu/~andreim/Lec33.pdf

www.amazon.co.uk



Notes Unit 25: Orthonormal Sets

CONTENTS

Objectives

Introduction

25.1 Orthonormal Sets

25.1.1 Unit Vector or Normal Vector

25.1.2 Orthonormal Sets, Definition

25.1.3 Examples of Orthonormal Sets

25.1.4 Theorems on Orthonormal Sets

25.2 Summary

25.3 Keywords



Objectives


Understand orthonormal sets

Define unit vector or normal vector

Understand the theorems on orthonormal sets.

Introduction

In linear algebra two vectors in an inner product space are orthonormal if they are orthogonaland both of unit length. A set of vectors from an orthonormal set if all vectors in the set aremutually orthogonal and all of unit length.

In this unit, we shall study about orthonormal sets and its examples.

25.1 Orthonormal Sets

25.1.1 Unit Vector or Normal Vector

Definition: Let H be a Hilbert space. If x H is such that x = 1, i.e. (x, x) = 1, then x is said to bea unit vector or normal vector.

25.1.2 Orthonormal Sets, Definition

A non-empty subset { ei } of a Hilbert space H is said to be an orthonormal set if

(i) i j ei ej, equivalently i j (ei, ej) = 0

(ii) ei = 1 or (ei, ej) = 1 for every i.


Unit 25: Orthonormal Sets

NotesThus a non-empty subset of a Hilbert space H is said to be an orthonormal set if it consists ofmutually orthogonal unit vectors.

Notes

1. An orthonormal set cannot contain zero vector because 0 = 0.

2. Every Hilbert space H which is not equal to zero space possesses an orthonormalset.

Since 0 x H. Then x 0. Let us normalise x by taking e = xx , so that

e = x 1 xx x

= 1.

e is a unit vector and the set {e} containing only one vector is necessarily anorthonormal set.

3. If {xi} is a non-empty set of mutually orthogonal vectors in H, then {ei} = i

i

xx

is an

orthonormal set.

25.1.3 Examples of Orthonormal Sets

1. In the Hilbert space n2 , the subset e1, e2, …, en where ei is the i-tuple with 1 in the ith place

and O’s elsewhere is an orthonormal set.

For (ei, ej) = 0 i j and (ei, ej) = 1 in the inner product n

i ii 1

x y of n2 .

2. In the Hilbert space 2 , the set {e1, e2, …, en, …} where en is a sequence with 1 in the nth placeand O’s elsewhere is an orthonormal set.

25.1.4 Theorems on Orthonormal Sets

Theorem 1: Let {e1, e2, …, en} be a finite orthonormal set in a Hilbert space H. If x is any vector inH, then

n2

ii 1

(x, e ) x 2 ; … (1)

further,n

i ii 1

x (x, e ) e ej for each j … (2)

Proof: Consider the vector

y = n

i ii 1

x (x, e ) e



Notes We have y 2 = (y, y)

= n n

i i i ii 1 i 1

x (x, e ) e , x (x, e ) e

= n n

i i j ji 1 j 1

(x,x) (x, e ) (e , x) (x, e ) (x, e )

n n

i i i ji 1 j 1

(x, e ) (x, e ) (e , e )

= n n n

2i i i j j i

i 1 j 1 i 1

x (x, e ) (x, e ) (x, e ) (x, e ) (x, e ) (x, e )

On summing with respect to j and remembering that (ei, ej) = 1, i = j and (ei, ej) = 0, i j

= n n n

2 2 2 2i i i

i 1 i 1 i 1

x x, e x, e (x, e )

= n

2 2i

i 1

x (x, e )

Now y 2 0, therefore x 2 – n

2i

i 1

(x, e ) 0

n2

ii 1

(x, e ) x 2

result (1).

Further to prove result (2), we have for each j (1 j n),

n

i i ji 1

x (x, e ) e , e = n

j i i ji 1

(x, e ) (x, e ) e , e

= n

j i i ji 1

(x, e ) (x, e ) (e , e )

= (x, ej) – (x, ej) [ (ei, ej) = 1, i j 0, i = j]

= 0

Hence n

i i ji 1

x (x, e ) e e for each j.


Note The result (1) is known as Bessel’s inequality for finite orthonormal sets.



NotesTheorem 2: If {ei} is an orthonormal set in a Hilbert space H and if x is any vector in H, then theset S = {ei : (x, ei) 0} is either empty or countable.

Proof: For each positive integer n, consider the set

Sn = 2

2i i

xe : (x, e )

n .

If the set Sn contains n or more than n’ vectors, then we must have

2i

e Si n

(x, e ) > n2x

xn

… (1)

By theorem (1), we have

2i

e Si n

(x, e ) x 2 … (2)

which contradicts (1).

Hence if (2) were to be valid, Sn should have at most (n – 1) elements. Hence for each positive n,the set Sn is finite.

Now let ei S. Then (x, ei) 0. However small may be the value of |(x, ei)|2, we can take n solarge that

|(x, ei)|2 > 2x

n.

Therefore if ei S, then ei must belong to some Sn. So, we can write S = nn 1

S .

S can be expressed as a countable union of finite sets.

S is itself a countable set.

If (x, ei) = 0 for each i, then S is empty. Otherwise S is either a finite set or countable set.


Theorem 3: Bessel’s Inequality: If {ei} is an orthonormal set in a Hilbert space H, then |(x, ei)|2 x 2 for every vector x in H.

Proof: Let S = {ei : (x, ei) 0}.

By theorem (2), S is either empty or countable.

If S is empty, then (x, ei) = 0 i.

So if we define |(x, ei)|2 = 0, then

|(x, ei)|2 = 0 x 2.

Now let S is not empty, then S is finite or it is countably infinite.

If S is finite, then we can write S = {e1, e2, …, en} for some positive integer n.

In this case, we have

|(x, ei)|2 = n

2i

i 1

|(x, e )| x 2 … (1)

which represents Bessel’s inequality in the finite case.



Notes If S is countable infinite, let S be arranged in the definite order such as {e1, e2, …, en, …}.

In this case we can write

|(x, ei)|2 = 2n

n 1

|(x, e )| … (2)

The series on the R.H.S. of (2) is absolutely convergent.

Hence every series obtained from this by rearranging the terms is also convergent and all suchseries have the same sum.

Therefore, we define the sum |(x, ei)|2 to be 2n

n 1

|(x, e )| .

Hence the sum of |(x, ei)|2 is an extended non-negative real number which depends only on Sand not on the rearrangement of vectors.

Now by Bessel’s inequality in the finite case, we have

n2

ii 1

|(x, e )| x 2 … (3)

For various values of n, the sum on the L.H.S. of (3) are non-negative. So they form a monotonicincreasing sequence. Since this sequence is bounded above by x 2, it converges. Since thesequence is the sequence of partial sums of the series on the R.H.S. of (2), it converges and wehave ei S,

|(x, ei)|2 = 2i

n 1

|(x, e )| x 2


Note: From the Bessel’s inequality, we note that the series 2n

n 1

|(x, e )| is convergent series.

Corollary: If en S, then (x, en) 0 as n .

Proof: By Bessel’s inequality, the series 2n

n 1

|(x, e )| is convergent.

Hence |(x, en)|2 0 as n .

(x, en) 0 as n .

Theorem 4: If {ei} is an orthonormal set in a Hilbert space H and x is an arbitrary vector in H, then

{x – (x, ei) ei} ej for each j.

Proof: Let S = {ei : (x, ei) 0}

Then S is empty or countable. [See theorem (2)]

If S is empty, then (x, ei) = 0 for every i.

In this case, we define (x, ei) ei to be a zero vector and so we get

x – (x, ei) ei = x – 0 = x.



NotesHence in this case, we have to show x ej for each j.

Since S is empty, (x, ej) = 0 for every j.

x ej for every j.

Now let S is not empty. Then S is either finite or countably infinite. If S is finite, let

S = {e1, e2, …, en} and we define

(x, ei) ei = n

i ii 1

(x, e ) e ,

and prove that i in 1

x (x, e ) e ej for each j. This result follows from (2) of theorem (1).

Finally let S be countably infinite and let

S = {e1, e2, …, en, …}

Let sn = i ii 1

(x, e ) e

For m > n, sm – sn 2 = 2

m

i ii n 1

(x, e )e

= m

2i

i n 1

(x, e )

By Bessel’s inequality, the series 2n

n 1

(x, e ) is convergent.

Hence 2i

i n 1

(x, e ) as m, n .

sm – sn 2 0 as m, n .

(sn) is a Cauchy sequence in H.

Since H is complete sn s H. Now s H can be written as

s = n nn 1

(x, e ) e

Now we can define (x, ei) ei = n nn 1

(x, e ) e .

Before, completing the proof, we shall show that the above sum is well-defined and does notdepend upon the rearrangement of vectors.

For this, let the vector in S be arranged in a different manner as

S = {f1, f2, f3, …, fn, …}



NotesLet ns =

n

i ii 1

(x, f ) f

As shown for the case above for (sn), let

ns s in H where we can take

s = n nn 1

(x, f ) f .

We prove that s = s . Given > 0 we can find n0 such that n n0.

2i

i n 1o

x, e < 2, sn – s < , ns s < … (1)

For some positive integer m0 > n0, we can find all the terms of sn in mos also.

Hence m no os s contains only finite number of terms of the type (x, ei) ei for i = n0 + 1, n0 + 2, …

Thus, we get 2 2m n io o

i n 1o

s s x, e so that we have

m no os s < … (2)

Now s – s = m m n no o o os s s s s s

m m n n0 0 0 0s s s s s s

< + + = 3 (Using (1) and (2))

Since > 0 is arbitrary, s – s = 0 or s = s .

Now consider

(x – (x, ei) ei, ej) = (x – s, ej)

But (x – s, ej) = (x, ej) – (s, ej)

= (x, ej) – (lim sn, ej) … (3)

By continuity of inner product, we get

(lim sn, ej) = lim (sn, ej) … (4)

Using (3) in (4), we obtain

(x – (x, ei) ei, ej) = (x, ej) – lim (sn, ej)

If ej S, then

(sn, ej) = n

i i ji 1

(x, e ) e , e = 0

n jnlim (s , e ) = 0



NotesHence (x – (x, ei) ei, ej) = (x, ej) = 0 since ej S.

If ej S, then (sn, ej) = n

i i ji 1

(x, e ) e , e … (5)

If n > j, we get n

i i ji 1

(x, e ) e , e = (x, ej) … (6)

From (5) & (6), we get

n jnlim (s , e ) = (x, ej).

So, in this case

(x – (x, ei) ei, ej) = (x, ej) – (x, ej) = 0

Thus (x – (x, ei) ei, ej) = 0 for each j.

Hence x – (x, ei) ei ej for each j.


Theorem 5: A Hilbert space H is separable every orthonormal set in H is countable.

Proof: Let H be separable with a countable dense subset D so that H = D .

Let B be an orthonormal basis for H.

We show that B is countable.

For x, y B, x y, we have

x – y 2 = x 2 + y 2 = 2

Hence the open sphere

1 1S x; z : z x2 2

= as x B are all disjoint.

Since D is dense, D must contain a point in each 1S x,2

.

Hence if B is uncountable, then B must also be uncountable and H cannot be separable contradictingthe hypothesis. Therefore B must be countable.

Conversely, let B be countable and let B = {x1, x2, …}. Then H is equal to the closure of all finite

linear combinations of element of B. That is H = L(B) . Let G be a non-empty open subset of H.

Then G contains an element of the form n

i ii 1

a x with ai C. We can take ai C. We can take ai

to be complex number with real and imaginary parts as rational numbers. Then the set

D = n

i i ii 1

a x , n 1, 2, , a rational

is a countable dense set in H and so H is separable.




Notes Theorem 6: A orthonormal set in a Hilbert space is linear independent.

Proof: Let S be an orthonormal set in a Hilbert space H.

To show that S is linearly independent, we have to show that every finite subset of S is linearlyindependent.

Let S1 = {e1, e2, …, en} be any finite subset of S.

Now let us consider

1e1 + 2e2 + … + nen = 0 … (1)

Taking the inner product with ek (1 k n),

n

i i ki 1

e , e = n

i i ki 1

(e , e ) … (2)

Using the fact that (ei, ek) = 0 for i k and (ek, ek) = 1, we get

n

i j ki 1

(e , e ) = k … (3)

It follows from (2) on using (1) and (3) that

(0, ek) = k

k = 0 k = 1, 2, …, n.

S1 is linearly independent.


Example: If {ei} is an orthonormal set in a Hilbert space H, and if x, y are arbitrary vectors

in H, then i i(x, e ) (y, e ) x y .

Solution: Let i i iS e : (x, e )(y, e ) 0

Then S is either empty or countable.

If S is empty, then we have

i i(x, e )(y, e ) = 0 i

and in this case we define

i i(x, e ) (y, e ) to be number 0 and we have 0 x 2 y 2.

If S is non-empty, then S is finite or it is countably infinite. If S is finite, then we can write

S = {e1, e2, …, en} for some positive integer n.

In this case we define

i i(x, e ) (y, e ) = n

i ii 1

(x, e ) (y, e )



Notes1 1n n2 2

2 2i i

i 1 i 1

(x, e ) (y, e ) (By Cauchy inequality)

x 2 y 2 (by Bessel’s inequality for finite case)

n

i ii 1

(x, e ) (y, e ) x y … (1)

Finally let S is countably infinite. Let the vectors in S be arranged in a definite order as

S = {e1, e2, …, en, …}.

Let us define

i i(x, e ) (y, e ) = n ni 1

(x, e ) (y, e ) .

But this sum will be well defined only if we can show that the series n nn 1

(x, e ) (y, e ) is

convergent and its sum does not change by rearranging its term i.e. by any arrangement of thevectors in the set S.

Since (1) is true for every positive integer n, therefore it must be true in the limit. So

n nn 1

(x, e ) (y, e ) x y … (2)

From (2), we see that the series n nn 1

(x, e ) (y, e ) is convergent. Since all the terms of the series

are positive, therefore it is absolutely convergent and so its sum will not change by anyrearrangement of its terms. So, we are justified in defining

i i(x, e ) (y, e ) = n nn 1

(x, e ) (y, e )

and from (2), we see that this sum is x y .

25.2 Summary

Two vectors in an inner product space are orthonormal if they are orthogonal and both ofunit length. A set of vectors from an orthonormal set if all vectors in the set are mutuallyorthogonal and all of unit length.

Examples of orthonormal sets are as follows:

(i) In the Hilbert space n2 , the subset e1, e2, …, en where ei is the i-tuple with 1 in the ith

place and O’s elsewhere is an orthonormal set.

For (ei, ej) = 0 i j and (ei, ej) = 1 in the inner product n

i ii 1

x y of n2 .



Notes (ii) In the Hilbert space 2 , the set {e1, e2, …, en, …} where en is a sequence with 1 in thenth place and O’s elsewhere is an orthonormal set.

25.3 Keywords

Orthonormal Sets: A non-empty subset { ei } of a Hilbert space H is said to be an orthonormal setif

(i) i j ei ej, equivalently i j (ei, ej) = 0

(ii) ei = 1 or (ei, ej) = 1 for every i.

Unit Vector or Normal Vector: Let H be a Hilbert space. If x H is such that x = 1, i.e. (x, x) =1, then x is said to be a unit vector or normal vector.


1. Let {e1, e2, …, en} be a finite orthonormal set in a Hilbert space H, and let x be a vector in H.

If 1, 2, …, n are arbitrary scalars, show that n

i ii 1

x e attains its minimum value

i = (x, ei) for each i.

2. Prove that a Hilbert space H is separable every orthonormal set in H is countable.


Book Sheldon Axler, Linear Algebra Done Right (2nd ed.), Berlin, New York (1997).

Online links www.mth.kcl.ac.uk/~jerdos/op/w3.pdf

mathworld.wolfram.com

www.utdallas.edu/dept/abp/PDF_files


Unit 26: The Conjugate Space H*

NotesUnit 26: The Conjugate Space H*

CONTENTS

Objectives

Introduction

26.1 The Conjugate Space H*

26.1.1 Definition


26.2 Summary

26.3 Keywords



Objectives


Define the conjugate space H*.

Understand theorems on it.

Solve problems related to conjugate space H*.

Introduction

Let H be a Hilbert space. A continuous linear transformation from H into C is called a continuouslinear functional or more briefly a functional on H. Thus if we say that f is a functional on H, thenf will be continuous linear functional on H. The set H,C of all continuous linear functional onH is denoted by H* and is called the conjugate space of H. The elements of H* are called continuouslinear functional or more briefly functionals. We shall see that the conjugate space of a Hilbertspace H is the conjugate space H* of H is in some sense is same as H itself. After establishing acorrespondence between H and H*, we shall establish the Riesz representation theorem forcontinuous linear functionals. Thereafter we shall prove that H * is itself a Hilbert space and H isreflexive, i.e. has a natural correspondence between H and H** and this natural correspondenceis an isometric isomorphism of H onto H**.

26.1 The Conjugate Space H*

26.1.1 Definition

Let H be a Hilbert space. If f is a functional on H, then f will be continuous linear functional onH. The set H,C of all continuous linear functional on H is denoted by H* and is called theconjugate space of H. The conjugate space of a Hilbert space H is the conjugate space H* of H is insome sense is same as H itself.



Notes 26.1.2 Theorems and Solved Examples

Theorem 1: Let y be a fixed vector in a Hilbert space H and let fy be a scalar valued function onH defined by

fy x x,y x H.

Then fy is a functional in H* i.e. fy is a continuous linear functional on H and y fy .

Proof: From the definition

fy : H C defined as fy x x,y x H.

We prove that fy is linear and continuous so that it is a functional.

Let 1 2x ,x H and , be any two scalars. Then for any fixed y H,

1 2 1 2fy x x x x ,y

1 2x ,y x ,y

1 2fy x fy x

fy is linear.

To show fy is continuous, for any x H

fy x x, y x . y ...(1)

(Schwarz inequality)

Let y M. Then for M > 0

fy x M x so that fy is bounded and hence fy is continuous.

Now let y = 0, y 0 and from the definition fy = 0 so that fy y .

Further let y 0. Then from (1) we have Sup fy x

y .x

Hence using the definition of the norm of a functional,

we get fy y ...(2)

Further fy sup fy x : x 1 ...(3)

Since yy 0,y is a unit vector.

From (3), we get

yfy fyy ...(4)



NotesBut y y 1fy , y y, y y

y y y...(5)

Using (5) in (4) we obtain

fy y


fy y


Theorem 2: (Riesz-representation Theorem for Continuous Linear Functional on a Hilbert Space):Let H be a Hilbert space and let f be an arbitrary functional on H*. Then there exists a uniquevector y in H such that

f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

Proof: We prove the following three steps to prove the theorem.

Step 1: Here we show that any *f H has the representation f = fy.

If f = 0 we take y = 0 so that result follows trivially.

So let us take f 0.

We note the following properties of y in representation if it exists. First of all y 0, sinceotherwise f = 0.

Further (x,y) = 0 x for which f(x) = 0. This means that if x belongs to the null space N(f) of f, then(x,y) = 0.

y N f .

So let us consider the null space N(f) of f. Since f is continuous, we know that N(f) is a properclosed subspace and since f 0,N f H and so N f 0 .

Hence by the orthogonal decomposition theorem, 0 ay 0 in N f . Let us define anyarbitrary x H.

0 0z f x y f y x

Now 0 0f z f x f y f y f x 0

z N f .

Since 0y N f , we get

0 0 0 00 z,y f x y f y x,y

0 0 0 0f x y , y f y x,y

Hence we get

0 0 0 0f x y ,y f y x,y 0 ...(3)



Notes Noting that 20 0 0y , y y 0, we get from (3),

02 0

0

f yf x x,y

y...(4)

We can write (4) as

02 0

0

f yf x x, y

y

Now taking oo

o

f y y as y,

y we have established that there exists a y such that f(x) = (x,y) for x H.

Step 2: In this step we know that

f y

If f =0, then y = 0 and f y hold good.

Hence let f 0. Then y 0.

From the relation f(x) = (x,y) and Schwarz inequality we have

f x x,y x y .

x 0

f xsup y .

x

Using definition of norm of f, we get from above

f y ...(5)

Now let us take x = y in f(x) = (x,y), we get

2y y,y f y f y

y f ...(6)

(5) and (6) implies that

f y .

Step 3: We establish the uniqueness of y in f(x) = (x,y). Let us assume that y is not unique inf(x) = (x,y).

Let for all 1 2x H, y ,y such that

f(x) = (x,y1) = (x,y2)

Then (x,y1) – (x,y2) = 0

(x,y1 – y2) = 0 x H.



NotesLet us choose x to be y1 – y2 so that

21 2 1 2 1 2y y ,y y y y 0

1 2y y 0

1 2y y

y is unique in the representation of f(x) = (x,y)


Note The above Riesz representation theorem does not hold in an inner product spacewhich is not complete as shown by the example given below. In other words thecompleteness assumption cannot be dropped in the above theorem.

Example: Let us consider the subspace M of l2 consisting of all finite sequences. This is theset of all scalar sequence whose terms are zero after a finite stage. It is an incomplete innerproduct space with inner product

n nn 1

x,y x y x,y M

Now let us define

nn

n 1

xf x as x x M.n

Linearity of f together with Hölder’s inequality yields

22 2n

n2n 1 n 1 n 1

x 1f x xn n

2 2

2x,x x ,6 6

since 2

2n 1

1 .n 6

f is a continuous linear functional on M.

We now prove that there is no y M such that

f x x,y x M.

Let us take x = en = (0,0,....,1,0,0,.....) where 1 is in n th place.



NotesUsing the definition of f we have f(x) = 1 .

n

Suppose ny y M satisfying the condition of the theorem, then

n n n nf x x,y x y y as x e .

Thus Riesz representation theorem is valid if and only if n1y 0n

for every n.

Hence ny y M.

no y M such that f(x,y) = (x,y) for every x H.

the completeness assumption cannot be left out from the Riesz-representation theorem.

Theorem 3: The mapping *: H H defined by *: H H defined by y fy where

fy(x) = (x,y) for every x H is an (i) additive, (ii) one-to-one, (iii) onto, (iv) symmetry, (v) notlinear.

Proof:

(i) Let us show that is additive, i.e.,

1 2 1 2 1 2y y y y for y ,y H.

Now from the definition 1 2 y y1 2y y f

Hence for every x H, we get

y y 1 2 1 21 2f x x,y y x,y x,y

y y1 2f x f x

y y 1 2 y y 1 21 2 1 2f y y f f y y

(i) 1 2 is one-to-one. Let y ,y H

Then 1 y 2 y1 2y =f and y =f . Then

1 2 y y1 2y = (y ) f = f

y y1 2f x = f x x H. ...(1)

y 1 y 21 2f x = x,y and f x x,y

from (1), we get

1 2 1 2x,y x,y x,y x,y 0

1 2x,y y 0 x H ...(2)

Choose x = y1 – y2 then from (2) if follows that (y1 – y2,y1 – y2) = 0



Notes21 2y y 0

1 2y y

is one-to-one.

(iii) * is onto: Let f H . Then y H such that

f(x) = (x,y)

since f(x) = (x,y) we get

f = fy so that yy = f = f.

Hence for *f H , a pre-image y H. Therefore is onto.

(iv) 1 2 is isometry; let y ,y H, then

1 2 y y1 2y y f f

y y1 2f f

But y y y y 1 21 2 1 2f f f y y (By theorem (1))

Hence 1 2 1 2y y y y .

(v) To show is not linear, let y H and be any scalar. Then ,y f y. Hence for any

x H, we get

y yf (x) (x, y) (x,y) f (x)

y yf f

y y

is not linear. Such a mapping is called conjugate linear.


Note: The above correspondence is referred to as natural correspondence between H andH*.

Theorem 4: If H is a Hilbert space, then H* is also an Hilbert space with the inner product definedby

(fx, fy) = (y,x) … (1)

Proof: We shall first verify that (1) satisfies the condition of an inner product.

Let x,y H and , be complex scalars.

(i) We know (see Theorem 3) that

y yf f

y y yf f f .

Now x y z x zyf f , f f f , f ... (2)



NotesBut x yf f , fz z, x, y (by (1))

Now z, x y z,x z,y

x z y zf , f f , f … (3)

From (2) and (3) it follows that x y z x z y zf f , f f , f f , f

x y x y(ii) f , f y,x x,y f , f .

22x x x x x x x(iii) f , f x,x x f so f , f 0 and f 0 f 0.

i iii implies that (1) represents an inner product. Now the Hilbert space H is a completenormed linear space. Hence its conjugate space H* is a Banach space with respect to the normdefined on H*. Since the norm on H* is induced by the inner product, H* is a Hilbert space withthe inner product (fx, fy) = (y,x)


Cor. The conjugate space H** of H* is a Hilbert space with the inner product defined as follows:

If *f gf ,g H ,let F and F be the corresponding elements of H** obtained by the Riesz representation

theorem.

Then (Ff,Fg) = (g,f) defines the inner product of H**.

Theorem 5: Every Hilbert space is reflexive.

Proof: We are to show that the natural imbedding on H and H** is an isometric isomorphism.

Let x be any fixed element of H. Let Fx be a scalar valued function defined on H* by Fx(f) = f(x) forevery *f H . We have already shown in the unit of Banach spaces that * *

xF H . Thus each vectorx H gives rise to a functional Fx in H**. Fx is called a functional on H* induced by the vector x.

Let **J : H H be defined by xJ x F for every x H.

We have also shown in chapter of Banach spaces that J is an isometric isomorphism of H into H **.We shall show that J maps H onto H**.

Let int o *1T : H H defined by

1 x xT x f ,f y y,x for every y H.

and int o* **2T : H H defined by

*2 x f f xx x

T f F ,F f f , f for f H .

Then T2.T1 is a composition of T2 and T1 from H to H**. By Theorem 3, T1,T2 are one-to-one andonto.

Hence T2.T1 is same as the natural imbedding J.

For this we show that J(x) = (T2.T1)x for every x H.

Now (T2.T1)x = T2(T1(x)) = T2(fx) = fxF .



NotesBy definition of J, J(x) = Fx. Hence to show T2.T1 = J, we have to prove that Fx = fxF .

For this let *f H . Then f = fy where f corresponds to y in the representation

f x y xxF (f) (f , f ) (f , f ) (x,y) .

But (x,y) = fy(x) = f(x) = Fx(f).

Thus we get f xxF (f) F (f) for every *f H .

Hence the mapping fxF and Fx are equal.

T2.T1 = J and J is a mapping of H onto H**, so that H is reflexive.


Notes

1. Since x fxF F x H (From above theorem)

x y f f y xy yF ,F F ,F f , f x,y by using def. of inner product on H** and by the

def. of inner product on H*.

2. Since an isometric isomorphism of the Hilbert space H onto Hilbert space H**,therefore we can say that Hilbert space H and H** are congruent i.e. they are equivalentmetrically as well as algebraically. We can identify the space H ** with the space H.

26.2 Summary

Let H be a Hilbert space. If f is a functional on H, then f will be continuous linear functionalon H. The set H,C of all continuous linear functional on H is denoted by H* and is calledconjugate space of H. Conjugate space of a Hilbert space H is the conjugate space H * of H.

Riesz-representation theorem for continuous linear functional on Hilbert space:

Let H be a Hilbert space and let f be an arbitrary functional on H*. Then there exists aunique vector y in H such that f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

26.3 Keywords

Continuous Linear Functionals: Let N be a normal linear space. Then we know that the set R ofreal numbers and the set C of complex numbers are Banach spaces with the norm of anyx R or x C given by the absolute value of x. We denote the BANACH space N,R or N,Cby N*.

The elements of N* will be referred to as continuous linear functionals on N.

Hilbert space: A complete inner product space is called a Hilbert space.

Let H be a complex Banach space whose norm arises from an inner product which is a complexfunction denoted by (x,y) satisfying the following conditions:



Notes H1: x y,z x,z y,z

H2: x,y y,x

H3: 2x,x x

for all x,y,z H and for all , C.

Inner Product: Let X be a linear space over the field of complex numbers C. An inner product onX is a mapping from X × X C which satisfies the following conditions:

(i) ( x + y, z) = (x, z) + (y, z) x, y, z X and , C.

(ii) (x, y) = (y, x)

(iii) (x, x) 0, (x, x) = 0 x = 0

Riesz-representation Theorem for Continuous Linear Functional on a Hilbert Space: Let H be aHilbert space and let f be an arbitrary functional on H*. Then there exists a unique vector y in Hsuch that

f = fy, i.e. f(x) = (x,y) for every vector x H and f y .

The Conjugate Space H*: Let H be a Hilbert space. If f is a functional on H, then f will becontinuous linear functional on H. The set H,C of all continuous linear functional on H isdenoted by H* and is called the conjugate space of H. The conjugate space of a Hilbert space H isthe conjugate space H* of H is in some sense is same as H itself.


1. Let H be a Hilbert space, and show that H* is also a Hilbert space with respect to the innerproduct defined by (fx, fy) = (y, x). In just the same way, the fact that H* is a Hilbert spaceimplies that H** is a Hilbert space whose inner product is given by (Ff, Fg) = (g, f).

2. Let H be a Hilbert space. We have two natural mappings of H onto H**, the second ofwhich is onto: the Banach space natural imbedding x Fx, where fx (y) = (y, x) and

f xxF (f) (F, f ). Show that these mappings are equal, and conclude that H is reflexive. Show

that (Fx, Fy) = (x, y).


Books Hausmann, Holm and Puppe, Algebraic and Geometric Topology, Vol. 5, (2005)

K. Yosida, Functional Analysis, Academic Press, 1965.

Online links www.spot.colorado.edu

www.arvix.org


Unit 27: The Adjoint of an Operator

NotesUnit 27: The Adjoint of an Operator

CONTENTS

Objectives

Introduction

27.1 Adjoint of an Operator

27.2 Summary

27.3 Keywords



Objectives


Define the adjoint of an operator.

Understand theorems on adjoint of an operator.

Solve problems on adjoint of an operator.

Introduction

We have already proved that T gives rise to an unique operator T* and H* such that (T*f) (x) =f(Tx) f H * and x H. The operator T* on H* is called the conjugate of the operator T on H.

In the definition of conjugate T* of T, we have never made use of the correspondence between Hand H*. Now we make use of this correspondence to define the operator T* on H called theadjoint of T. Though we are using the same symbol for the conjugate and adjoint operator on H,one should note that the conjugate operator is defined on H*, while the adjoint is defined on H.

27.1 Adjoint of an Operator

Let T be an operator on Hilbert space H. Then there exists a unique operator T* on H such that

(Tx,y)= (x,T*y) for all x, y H

The operator T* is called the adjoint of the operator T.

Theorem 1: Let T be an operator on Hilbert space H. Then there exists a unique operator T* on Hsuch that

(Tx,y)= (x,T*y) for all x, y H ...(1)


Proof: First we prove that if T is an operator on H, there exists a mapping T* on H onto itselfsatisfying

(Tx,y)= (x,T*y) for all x,y H. ...(2)



Notes Let y be a vector in H and fy its corresponding functional in H*.

Let us define

int oT* : H H * by

y zT* : f f ...(3)

Under the natural correspondence between H and H*, let z H corresponding to zf H * . Thus

starting with a vector y in H, we arrive at a vector z in H in the following manner:

y y zy f T * f f z, ...(4)

where y zT* : H* H * and y f and z f are on H to H*

under the natural correspondence. The product of the above three mappings exists and it isdenoted by T*.

Then T* is a mapping on H into H such that

T * y z.

We define this T* to be the adjoint of T. We note that if we identify H and H* by the natural

correspondence yy f , then the conjugate of T and the adjoint of T are one and the same.

After establishing, the existence of T*, we now show (1). For x H, by the definition of theconjugate T* on an operator T,

y yT * f x f Tx ...(5)

By Riesz representation theorem,

yy f so that

yf Tx Tx,y ...(6)

Since T* is defined on H*, we get

y zT * f x f x x,z ...(7)

But we have from our definition T*y = z ...(8)


yT * f x Tx,y ...(9)


yT * f x x,T * y ...(10)

From (9) and (10), we thus obtain

Tx,y x,T * y x,y H.




Notes

Note The relation Tx,y x,T * y can be equivalently written as

T * x,y x,Ty since

T * x,y y,T * x = Ty,x = x,Ty = x,Ty

T * x, y x,Ty .

Example: Find adjoint of T if T is defined on 2 as 1 2Tx 0,x ,x ,... for every n 2x x .

Let T* be the adjoint of T. Using inner product in 2 , we have

T * x,y = x,Ty

since 1 2Ty 0,y ,y ,... , we have

n+1 nn=1

T * x,y x,Ty = x y Sx,y ,

where 2 3S x x ,x ,...

Hence T * x,y Sx,y for every x in 2 .

Since T* is unique, T*=S so that we have

2 3 4T * x x ,x ,x ,... .

Theorem 2: Let H be the given Hilbert space and T* be adjoint of the operator T. Then T* is abounded linear transformation and T determine T* uniquely.

Proof: T* is linear.

Let 1 2y ,y H and , be scalars. Then for x H, we have

1 2 1 2x,T * y y Tx, y y

But 1 2 1 2Tx, y y Tx,y Tx,y

1 2Tx,y x,T * y

1 2x, T * y x, T * y .

Hence for any x H,

1 2 1 2x,T * y y x, T * y x, T * y



Notes 1 2x, T * y T * y .

T* is linear.

T* is bounded

for any y H, let us consider

2T * y T * y,T * y

TT * yy

TT * y y using Schwarz inequality

T T * y y

Hence 2T * y T T * y y 0 ...(1)

If T * y 0 then T * y T y because T y 0

Hence let T * y 0.

Then we get from (1)

T * y T y .

since T is bounded,

T M so that

T * y M y for every y H.

T* is bounded.

T* is continuous.

Uniqueness of T*.

Let if T* is not unique, let T’ be another mapping of H into H with property

Tx,y = x,T * y x,y H.

Then we have

Tx,y = x,T'y ...(2)

and Tx,y = x,T * y ...(3)


x,T'y = x,T * y x,y H

x, T'y T * y =0



Notesx, T' T * y =0 x H

T' T * y = 0 for every y H

Hence T'y = T * y for every y H.

T T * .


Notes

1. We note that the zero operator and the identity operator I are adjoint operators. For,

(i) x,0 * y 0x,y 0,y 0 x,0 = x,0y

so from uniqueness of adjoint 0* = 0.

(ii) (x,Iy) = (Ix,y) = (x,y) = (x,Iy)

so from uniqueness of adjoint I*=I.

2. If H is only an inner product space which is not complete, the existence of T*corresponding to T in the above theorem is not guaranteed as shown by the followingexample.

Example: Let M be a subspace of L2 consisting of all real sequences, each one containingonly finitely many non-zero terms. M is an incomplete inner product space with the same inner

product for 2 given by

n nn=1

x,y = x y ...(1)

For each x M, define

n

n=1

xT x ,0,0,......n ...(2)

Then from the definition, for x,y M,

n1

n=1

xT x,y y .n

Now let ne 0,0,...,1,0,... where 1 is in the nth place.

Then using (3) we obtain

ne 1n

e ( j) 1T ,e 1. 1. .j n



NotesNow we check whether there is T* which is adjoint of T. Now n 1 1 ne ,T * e T * e .e , where the

R.H.S. gives the component wise inner product. Since 1 1 nT * e M,T * e .e cannot be equal to

1 n 1,2,...n

there is no T* on M such that

n 1 n 1T e ,e e ,T * e

Hence completeness assumption cannot be ignored from the hypothesis.

Notes

1. The mapping T T * is called the adjoint operation on H .

2. From Theorem (2), we see that the adjoint operation is mapping T T * on Hinto itself.

Theorem 3: The adjoint operation T T* on (H) has the following properties:

(i) 1 2 1 2T T * T * T * (preserve addition)

(ii) 1 2 2 1T T * T * T * (reverses the product)

(iii) T * T * (conjugate linear)

(iv) T * T

(v) 2T * T T

Proof: (i) For every x, y H, we have

(x, (T1 + T2)*y = ((T1 + T2) x, y) (By def. of adjoint)

= (T1x + T2x, y)

= (T1x, y) + (T2x, y)

= (x, T1*y) + (x, T2*y)

= (x, T1*y + T2*y)

= (x, (T1* + T2*)y)

(T1 +T2)* = T1* + T2* by uniqueness of adjoint operator

(ii) For every x,y H, we have

1 2 1 2x T T * y T T x,y

1 2T T x ,y



Notes 2 1T x,T * y

* *2 1x,T , T y

* *2 1x, T T ,y

Therefore from the uniqueness of adjoint operator, we have * *1 2 2 1T T * T T .

(iii) For every x,y H, we have

x, T * y T x,y Tx ,y

Tx,y

x,T * y x, T * y

x T * y .

Therefore from the uniqueness of adjoint operator, we have

T * T * .

(iv) For every y H we have

2T * y T * y,T * y

TT * y,y

TT * y,y

2T * y TT * y,y is a real number 0

TT * y y By Schwarz inequality

T T * y y Tx T x

Thus 2T * y T T * y y y H

T * y T y y H. ...(1)

Now T * Sup T * y : y 1

from (1), we see that if y 1 then T * y T

T * T ...(2)

Now applying (2) from the operator T* in place of operator T, we get

T * * T *



Notes T * * T *

T T * T * * T ...(3)


T T * .

(v) We have T * T T * T

T T T * T

2T ...(4)

Further for every x H, we have

2Tx Tx,Tx

T * Tx,x

T * T x,x

T * T x x (By Schwarz inequality)

2T * T x

Then we have

2 2Tx T * T x x H ...(5)

Now T sup Tx : x 1

22T sup Tx : x 1

2sup Tx : x 1

From (5) we see that

if x 1 , then 2Tx T * T .

Therefore, 2Sup Tx : x 1 T * T

2T T * T . ...(6)


2T * T T .




NotesCor: If nT is a sequence of bounded linear operators on a Hilbert space and

n nT T, then T * T * .

We have

n nT * T * T T *

nT T (By properties of T*)

Since nT T as n

* *nT T as n .

Theorem 4: The adjoint operation on H is one-to-one and onto. If T is a non-singular operatoron H, then T* is also non-singular and

1 1T * T * .

Proof: Let : H H is defined by

T T * for every T H .

To show is one-to-one, let 1 2T ,T H . Then we shall show that 1 2 1 2T T T T .

Now 1 2T T

1 2T * T *

1 2T * * T * * (using Theorem 4. prop (iv))

1 2T T

is one-to-one.

is onto:

For T* H ,we have on using Theorem 4 (iv),

T * = T* * =T.

Thus for every T* H ,there is a T* H such that

T * T is onto.

Next let T be non-singular operator on H. Then its inverse 1T exists on H and

1 1TT T T I.



Notes Taking the adjoint on both sides of the above, we obtain

1 1TT * T T * I * .

By using Theorem 4 and note (2) under Theorem 2, we obtain

1 1T * T* T * T * I.

T * is invertible and hence non-singular.

Further from the above, we conclude

1 1T * T * .


Note From the properties of the adjoint operation T T * on H discussed in Theorems(3) and (4), we conclude that the adjoint operation T T* is one-to-one conjugate linearmapping on H into itself.

Example: Show that the adjoint operation is one-to-one onto as a mapping of (H) intoitself.

Solution: Let : (H) (H) be defined

(T) = T* T (H)

We show is one-to-one and onto.

is one-one:

Let T1, T2 (H). Then

(T1) = (T2) T1* = T2*

(T1*)* = (T2*)*

T1** = T2**

T1 = T2

is one-to-one.

is onto:

Let T be any arbitrary member of (H). Then T* (H) and we have (T*) = (T*)* = T** = T. Hence,the mapping is onto.

27.2 Summary

Let T be an operator on Hilbert Space H. Then there exists a unique operator T* on H suchthat Tx,y x,T * y for all x,y H.The operator T* is called the adjoint of the operator T.



Notes The adjoint operation T T * on H has the following properties:

(i) 1 2 1 2T T * T * T *

(ii) 1 2 2 1T T * T * T *

(iii) T * T *

(iv) T * T

(v) 2T * T T

27.3 Keywords

Adjoint of the Operator T: Let T be an operator on Hilbert space H. Then there exists a uniqueoperator T* on H such that

(Tx,y)= (x,T*y) for all x, y H


Conjugate of the Operator T on H: T gives rise to an unique operator T* and H* such that (T*f) (x)= f(Tx) f H * and x H. The operator T* on H* is called the conjugate of the operator T on H.


1. Show that the adjoint operation is one-to-one onto as a mapping of H into itself.

2. Show that 2TT * T .

3. Show that O*=O and I*=I. Use the latter to show that if T is non-singular, then T* is also

non-singular, and that in this case 1 1T * T * .


Books N.I. Akhiezer and I.M. Glazman, Theory of Linear Operators in Hilbert Space, Vol. II,Pitman, 1981.

K. Yosida, Functional Analysis, Academic Press, 1965.

Online links www.math.osu.edu/ gerlach.1/math.BVtypset/node 78.html.

sepwww.standford.edu/sep/prof/pvi/conj/paper_html/node10.html.



Notes Unit 28: Self Adjoint Operators

CONTENTS

Objectives

Introduction

28.1 Self Adjoint Operator

28.1.1 Definition: Self Adjoint

28.1.2 Definition: Positive operator

28.2 Summary

28.3 Keywords



Objectives


Define self adjoint operator.

Define positive operator.

Solve problems on self adjoint operator.

Introduction

The properties of complex number with conjugate mapping z z motivate for the introduction

of the self-adjoint operators. The mapping z z of complex plane into itself behaves like the

adjoint operation in H as defined earlier. The operation z z has all the properties of the

adjoint operation. We know that the complex number is real iff z z . Analogue to thischaracterization in H leads to the motion of self-adjoint operators in the Hilbert space.

28.1 Self Adjoint Operator

28.1.1 Definition: Self Adjoint

An operator T on a Hilbert space H is said to be self adjoint if T*=T.

We observe from the definition the following properties:

(i) O and I are self adjoint O* O and I* I

(ii) An operator T on H is self adjoint if

Tx,y x,Ty x,y H and conversely.


Unit 28: Self Adjoint Operators

NotesIf T* is an adjoint operator T on H then we know from the definition that

Tx,y x,T * y x,y H .

If T is self-adjoint, then T = T*.

Tx,y x,Ty x,y H.

Conversely, if Tx,y x,Ty x,y H then we show that T is self-adjoint.

If T* is adjoint of T then (Tx, y) = (x, T*y)

We have x,Ty x,T * y

x, T T * y 0 x, y H

But since x 0 T T * y 0 y H

T = T*

T is self adjoint.

(iii) For any T H ,T T * and T * T are self adjoint. By the property of self-adjoint operators,

we have

T T * * T * T * *

T * T

T T *

T T * * T T*,

and T * T * T * T * * T * T

T * T * * T * T.

Hence T T * and T * T are self adjoint.

Theorem 1: If nA is a sequence of self-adjoint operators on a Hilbert space H and if nAconverges to an operator A, then A is self adjoint.

Proof: Let nA be a sequence of self adjoint operators and let nA A.

nA is self adjoint n nA * A for n 1,2,...

We claim that A A *

Now n n n nA A* A A A A * A * A *

n n n nA A * A A A A * A A *

n n n nA A A A A A n nA * A

n nA A 0 A A



Notes = 2 An – A

0 as n

A A * 0 or A A* 0 A A *

A is self-adjoint operator.


Theorem 2: Let S be the set of all self-adjoint operators in (H). Then S is a closed linear subspaceof (H) and therefore S is a real Banach space containing the identity transformation.

Proof: Clearly S is a non-empty subset of (H), since O is self adjoint operator i.e. O S.

H , since O is self adjoint operator i.e. O S.

Let 1 2A ,A S, We prove that 1 2A A S.

1 2 1 1 2 2A ,A S A * A and A * A ...(1)

For , R , we have

1 2 1 2A A * A * A *

1 2A * A *

1 2A A , are real numbers, ,

1 2A A is also a self adjoint operator on H.

1 2 1 2A ,A S A A S.

S is a real linear subspace of H .

Now to show that S is a closed subset of the Banach space H . Let A be any limit point of S.

Then a sequence of operator An is such that nA A. We shall show that A S i.e. A A * .

Let us consider

n nA A * A A A A *

n nA A A A *

n n n nA A A A * A * A *

n n n nA A A A * A * A *

n n n nA A A A A A * n n nA S A * A

n nA A 0 A A



Notesn2 A A 0 0, T * T and T T

n0 as A A

A A * 0 A A* 0

A A* A is self adjoint

A S

S is closed.

Now since S is a closed linear subspace of the Banach space H , therefore S is a real Banachspace. ( S is a complete linear space)

Also I* I the identity operator I S.


Theorem 3: If 1 2A ,A are self-adjoint operators, then their product 1 2A ,A is self adjoint

1 2 2 1A ,A A ,A (i.e. they commute)

Proof: Let 1 2A ,A be two self adjoint operators in H.

Then 1 1 2 2A * A ,A * A .

Let 1 2A ,A commute, we claim that 1 2A ,A is self-adjoint.

1 2 2 1 2 1 1 2A ,A * A * A * A A A A

1 2 1 2A ,A * A A

1 2A A is self adjoint.

Conversely, let 1 2A A is self adjoint, then

1 2 1 2A A * A A

2 1 1 2A * A * A A

1 2A ,A commute


Theorem 4: If T is an operator on a Hilbert space H, then T = T 0 Tx,y 0 x,y H.

Proof: Let T = 0 (i.e. zero operator). Then for all x and y we have

T x,y Ox,y O,y O.

Conversely, Tx,y O x,y H

Tx,Tx O x,y H (taking y = Tx)

Tx O x,y H



Notes T O i.e. zero operator.


Theorem 5: If T is an operator on a Hilbert space H, then

Tx,x 0 x in H T O.

Proof: Let T = O. Then for all x in H, we have

Tx,x Ox,x 0,x 0.

Conversely, let Tx,x 0 x,y H. Then we show that T is the zero operator on H.

If , any two scalars and x,y are any vectors in H, then

T x y , x y Tx Ty, x y

T x , x y T y , x y

Tx,x Tx,y Ty,x Ty,x

22 Tx,x Tx,y Ty,x Ty,x

22T x y , x y Tx,x Ty,y Tx,y Ty,x ...(1)

But by hypothesis Tx,x 0 x H.

L.H.S. of (1) is zero, consequently the R.H.S. of (1) is also zero. Thus we have

Tx,y Ty,x 0 ...(2)

for all scalars , and x,y H.

Putting 1, 1 in (2) we get

Tx,y Ty,x 0 ...(3)

Again putting i, 1 in (2) we obtain

i Tx,y i Ty,x 0 ...(4)

Multiply (3) by (i) and adding to (4) we get

2i Tx,y 0 x,y H

Tx,y 0 x,y H

Tx,Tx 0 x,y H (Taking y = Tx)

Tx 0 x,y H



NotesT = 0 (zero operator)


Theorem 6: An operator T on a Hilbert space H is self-adjoint.

Tx,x is real for all x.

Proof: Let T* =T (i.e. T is self adjoint operator)

Then for every x H,we have

Tx,Tx x,T * x x,Tx Tx,x

Tx,x equals its own conjugate and is therefore real.

Conversely, let Tx,x is real x H. We claim that T is self adjoint i.e. T*=T.

since Tx,x is real x H,

Tx,x Tx,x x,T * x T * x,x

Tx,x T * x,x 0 x H

Tx T * x,x 0 x H

T T * x,x 0 x H

T – T* = 0 [ if (Tx, x) = 0 T = 0]

T = T*

T is self adjoint.


Cor. If H is real Hilbert space, then A is self adjoint

Ax,y Ay,x x,y H.

A is self adjoint for any x,y H.

Ax,y x,A * y A * y,x .

since H is real Hilbert space A * y,x A * y,x so that Ax,y Ay,x A* A

Theorem 7: The real Banach space of all self-adjoint operators on a Hilbert space H is a partiallyordered set whose linear and order structures are related by the following properties:

(a) If 1 2 1 2A A then A +A A +A for every A S;

(b) If 1 2 1 2A A and 0, then A A .

Proof: Let S represent the set of all self-adjoint operators on H. We define a relation on S asfollows:

If 1 2 1 2 1 2A A S, we write A A if A x,x A ,x x in H.

We shall show that ' ' is a partial order relation on S. ' ' is reflexive.



Notes Let A S. Then

Ax,x Ax,x x H

Ax,x Ax,x x H

By definition A A.

' ' on S is reflexive.

' ' is transitive.

Let 1 2 2 3A A and A A then

1 2A x,x A x,x x H.

and 2 3A x, x A x, x x H.

From these we get

1 3A x,x A x,x x H.

and 2 3A x, x A x, x x H.

From these we get

1 3A x,x A x,x x H.

Therefore by definition 1 3A A and so the relation is transitive.

' ' is anti-symmetric.

Let 1 2 2 1 1 2A A and A A then to show that A A .

We have 1 2 1 2A A A x,x A x,x x H.

Also 2 1 2 1A A A x,x A x,x x H.

From these we get

1 2A x,x A x,x x H.

1 2A x A x,x 0 x H.

1 2A A x,x 0 x H.

1 2A A 0

1 2A A

' ' on anti-symmetric.

Hence ' ' is a partial order relation on S.

Now we shall prove the next part of the theorem.

(a) We have 1 2 1 2A A A x,x A x,x x H.



Notes1 2A x,x Ax,x A x,x Ax,x x H.

1 2 1A A x,x A A x,x x H.

1 2 2A A A A, by def. of .

(b) We have 1 2 1 2A A A x,x A x,x x H

1 2A x,x A x,x x H 0

1 2A x,x A x,x x H

1 2A x,x A x,x x in H

1 2A A ,by def. of ' '.


28.1.2 Definition – Positive Operator

A self adjoint operator on H is said to be positive if A 0 in the order relation. That is

if Ax,x 0 x H.

We note the following properties from the above definition.

(i) Identity operator I and the zero operator O are positive operators.

Since I and O are self adjoint and

2Ix, x x, x x 0

also (Ox, x) = (0, x) = 0

I,O are positive operators.

(ii) For any arbitrary T on H, both TT* and T*T are positive operators. For, we have

TT * * T * * T* TT *

TT * is self adjoint

Also T * T * T * T * * T * T

T * T is self adjoint

Further we see that

2TT * x,x T * x,T * x T * x 0

and 2T * Tx,x Tx,T * *x (Tx,Tx) Tx 0

Therefore by definition both TT* and T*T are positive operators.

Theorem 8: If T is a positive operator on a Hilbert space H, then I+T is non-singular.

Proof: To show I+T is non-singular, we are to show that I+T is one-one and onto as a mapping ofH onto itself.

I+T is one-one.

First we show I T x 0 x 0



Notes We have I T x 0 Ix Tx 0 x Tx 0

Tx x2Tx,x x,x x

2x 0 Tx,x 0

2x 0

2x 0 2x is always 0

x 0

I T x 0 x 0.

Now I T x I T y I T x y 0

x y 0 x y

Hence I+T is one-one.

I+T is onto.

Let M = range of I+T. Then I+T will be onto if we prove that M=H.

We first show that M is closed.

For any x H, we have

2 2I T x x Tx

x Tx,x Tx

x,x x,Tx Tx,x Tx,Tx

2 2x Tx Tx,x Tx,x

2 2x Tx 2 Tx,x T is positive T is self-adjoint Tx,x real

2x T is positive Tx,x 0

Thus x I T x x H

Now let nI T x be a CAUCHY sequence in M. For any two positive integers m,n we have

m n m nx x I T x x

m nI T x I T x 0,

since I T x is a CAUCHY sequence.

m nx x 0

nx is a CAUCHY sequence in H. But H is complete. Therefore by CAUCHY sequence nx in

H converges to a vector, say x in H.



NotesNow n nLim I T x I T lim x I T is a continuous mapping

I T x M range of I T

Thus the CAUCHY sequence nI T x in M converges to a vector I T x in M.

every CAUCHY sequence in M is a convergent sequence in M.

M is complete subspace of a complete space is closed.

M is closed.

Now we show that M = H. Let if possible M H.

Then M is a proper closed subspace of H.

Therefore, a non-zero vector 0x in H s.t. 0x is orthogonal in M.

Since 0I T x M, therefore

0 0x M I T x ,x 0

0 0 0x Tx ,x 0

0 0 0 0x ,x Tx ,x 0

20 0x Tx ,x 0

20 0x Tx ,x

2x 0 0 0T positive Tx ,x 0

2x 0

x 02x 0

x 0

a contradiction to the fact that 0x 0.

Hence we must have M = H and consequently I+T is onto. Thus I+T is non-singular.


Cor. If T is an arbitrary operator on H, then the operator I+TT* and I+T*T are non-singular.

Proof: We know that for an arbitrary T on H, T*T and TT* are both positive operators.

Hence by Theorem (8) both the operators I+TT* and I+T*T are non-singular.

28.2 Summary

An operator T on a Hilbert space H is said to be self adjoint if T*=T.

A self adjoint operator on H is said to be positive if A 0 in the order relation. That is if

Ax,x 0 x H.



Notes 28.3 Keywords

Positive Operator: A self adjoint operator on H is said to be positive if A 0 in the orderrelation. That is

if Ax,x 0 x H.

Self Adjoint: An operator T on a Hilbert space H is said to be self adjoint if T*=T.


1. Define a new operation of “Multiplication” for self-adjoint operators by

1 2 2 11 2

A A A AA A ,

2 and note that 1 2A A is always self-adjoint and that it equals 1 2A A

whenever A1 and A2 commute. Show that this operation has the following properties:

1 2 2 1A A A A ,

1 2 3 1 2 1 3A A A A A A A ,

1 2 1 2 1 2A A A A A A ,

and A I I A A. Show that

1 2 3 1 2 3A A A A A A whenever A1 and A3 commute.

2. If T is any operator on H, it is clear that 2Tx,x Tx x T x ; so if H 0 ,we have

sup 2Tx,x / x : x 0 T . Prove that if T is self-adjoint, then equality holds here.


Books Akhiezer, N.I.; Glazman, I.M. (1981), Theory of Linear Operators in Hilbert Space

Yosida, K., Functional Analysis, Academic Press

Online links www.ams.org/bookstore/pspdf/smfams-14-prev.pdf-UnitedStatesmathworld.wolfram.com


Unit 29: Normal and Unitary Operators

NotesUnit 29: Normal and Unitary Operators

CONTENTS

Objectives

Introduction

29.1 Normal and Unitary Operators

29.1.1 Normal Operator

29.1.2 Unitary Operator

29.1.3 Isometric Operator

29.2 Summary

29.3 Keywords



Objectives


Understand the concept of Normal and Unitary operators.

Define the terms Normal, Unitary and Isometric operator.

Solve problems on normal and unitary operators.

Introduction

An operator T on H is said to be normal if it commutes with its adjoint, that is, if TT*=T*T. Weshall see that they are the most general operators on H for which a simple and revealingstructure theory is possible. Our purpose in this unit is to present a few of their more elementaryproperties which are necessary for our later work. In this unit, we shall also study about Unitaryoperator and Isometric operator.

29.1 Normal and Unitary Operators

29.1.1 Normal Operator

Definition: An operator T on a Hilbert space H is said to be normal if it commutes with itsadjoint i.e. if TT* = T*T

Conclusively every self-adjoint operator is normal. For if T is a self adjoint operator i.e. T*=Tthen TT* =T*T and so T is normal.

Note A normal operator need not be self adjoint as explained below by an example.



NotesExample: Let H be any Hilbert space and I : H H be the identity operator.

Define T = 2iI. Then T is normal operator, but not self-adjoint.

Solution: Since I is an adjoint operator and the adjoint operation is conjugate linear,

T* = –2iI* = –2iI so that

TT* =T*T =4I.

T is a normal operator on H.

But T = T* T is not self-adjoint.

Note If T H is normal, then T* is normal.

since if T* is the adjoint of T; then T**= T.

T is normal TT*=T*T

Hence T*T**=T*T=TT*=T**T* so that T*T**=T**T

T* is normal if T H .

Theorem 1: The limit T of any convergent sequence (Tk) of normal operators is normal.

Proof: Now k k kT * T * T T * T T

k kT * T * as k since T T as k .

Now we prove TT* T * T so that T is normal.

* * * *k k k k k k k k k k k k kTT * T * T TT * T T * T T * T * T T * T TT * TT T T T T T T

k kT * T TT *

*k k k kTT * T * T TT * T T * T * T TT ...(1)

k k k k kT is normal i.e. T T* T * T

since k kT T as T * T*, R.H.S. of (1) 0

TT * T * T 0

TT* T * T

T is normal.


Theorem 2: The set of all normal operators on a Hilbert space H is a closed subspace of Hwhich contains the set of all set-adjoint operators and is closed under scalar multiplication.

Proof: Let M be the set of all normal operators on a Hilbert space H. First we shall show that Mis closed subset of H .



NotesLet T be any limited point of M. Then to show that T M i.e. to show that T is a normal operatoron H.

Since T is a limited point of M, therefore n a sequence T of distinct point of M such that

nT T. We have

n n nT * T * T T * T T 0.

n nT * T * 0 T * T * .

Now, n n n n n n nTT * T * T TT * T T * T T * T * T TT * T T * T T * T * T

n n n n n n n

n n n n n n n n

TT * T T * T T * T * T T * T T * T

TT * T T * T T * T * T T * T T * T

n n n nTT * T T * 0 T * T T * T

n n n n n nT M T is a normal operator on H i.e. T T* T* T and 0 0

n n n n n nTT * T T * T * T T * T 0 since T T and T * T *

Thus, TT * T * T 0 TT * T * T 0

TT* T * T T is normal operator on H.

T M and so M is closed.

Now every self adjoint operator is normal. Therefore the set M contains the set of all self-adjointoperators on H.

Finally, we show that M is closed with respect to scalar multiplication i.e. T M

T M, is any scalar.

In other words, we are to show that if T is a normal operator on H and is any scalar, then Tis normal operator on H. Since T is normal, therefore TT* =T*T.

We have T * T * .

Now T T * T T * TT * .

Also T * T T T T * T ( ) (TT*)

T T * T * T

T is normal.


Theorem 3: If N1, N2 are normal operators on a Hilbert space H with the property that eithercommutes with the adjoint of the other, then N1 + N2 and N1N2 are also normal operators.

Proof: Since N1, N2 are normal operators, therefore

1 1 1 1 2 2 2 2N N* = N* N and N N* N* N ...(1)



Notes Also by hypothesis, we have 1 2 2 1 2 1 1 2N N* =N* N and N N* N* N … (2)

we claim that N1 + N2 is normal.

i.e. (N1 + N2)(N1 + N2)* = (N1 + N2)*(N1 + N2) ...(3)

since adjoint operation preserves addition, we have

(N1 + N2)(N1 + N2)* = (N1 + N2) 1 2N* +N*

1 1 1 2 2 1 2 2N N* N N N N* N N* ...(4)

1 1 2 1 1 2 2 2N N* N* N N* N N* N

= 1 2 1 2N* N* N N

= 1 2 1 2N N * N N (using (1) and (2))

1 2 1 2 1 2 1 2N N N N * N N * N N

1 2N N is normal.

Now we show that N1N2 is normal i.e.

1 2 1 2 1 2 1 2N N N N * N N * N N .

L.H.S.= 1 2 1 2 1 2 2 1N N N N * N N N * N *

1 2 2 1N N N * N *

1 2 2 1N N * N N *

1 2 2 1N N * N N *

2 1 1 2N * N N * N

2 1 1 2N * N N * N

2 1 1 2N * N * N N

1 2 1 2N N * N N

1 2 1 2 1 2 1 2N N N N * N N * N N

N1N2 is normal.


Theorem 4: An operator T on a Hilbert space H is normal T * x Tx for every x H.

Proof: We have T is normal TT* T * T

TT * T * T 0



NotesTT * T * T x,x 0 x

TT * x;x T * Tx,x x

T * x,T * x Tx,T * *x x

2 2T * x Tx x T * * T

T * x Tx x.


Theorem 5: If N is normal operator on a Hilbert space H, then 2 2N N .

Proof: We know that if T is a normal operator on H then

Tx T * x x ...(1)

Replacing T by N, and x by Nx we get

NNx N * Nx x

2N x N * Nx x ...(2)

Now 2 2N Sup N x : x 1

Sup N * Nx : x 1 (by (2))

N * N

2N


Theorem 6: Any arbitrary operator T on a Hilbert space H can be uniquely expressed as

1 2 1 2T T iT where T ,T are self-adjoint operators on H.

Proof: Let 1 2T T * 1T and T T T *

2 2i

Then 1 2T iT T ...(1)

Now 1

*1T * T T *2

1 *T T *2

1 T * T * *2

11 1T * T T T * T2 2



Notes1 1T * T

1T is self-adjoint.

Also 2

*1T * T T *2i

1 *T T *2i

1 T * T * *2i

21 1T * T T T * T2 2i

2 2T * T

2T is self-adjoint.

Thus T can be expressed in the form (1) where T1,T2 are self adjoint operators.

To show that (1) is unique.

Let T = U1 + iU2, U1,U2 are both self-adjoint

We have 1 2T* U iU *

1 2U * iU *

1 2U * iU *

1 1 22U * iU * U iU

1 2 1 2T T* U iU U iU 2U,

*1 1

1U T T T2

and 1 2 1 2 2T T* U iU U iU 2iU

2 21U T T * T2i

expression (1) for T is unique.


Note The above result is analogous to the result on complex numbers that every complexnumber z can be uniquely expressed in the form z = x + iy where x, y are real. In the abovetheorem T =T1 + T2, T1 is called real part of T and T2 is called the imaginary part of T.



NotesTheorem 7: If T is an operator on a Hilbert space H, then T is normal its real and imaginaryparts commute.

Proof: Let T1 and T2 be the real and imaginary parts of T. Then T1, T2 are self-adjoint operatorsand T = T1 + i T2.

We have T* = (T1 + iT2)* = T1* + (iT2)*

= Ti* + i T2*

= Ti* – iT2*

= T1 – iT2

Now TT* = (T1 + iT2) (Ti – iT2)

= T12 + T2

2 + i (T2T1 – T1T2) … (1)

and T*T = (Ti – iT2) (T1 – iT2)

= T12 + T2

2 + i (T1T2 – T2T1) … (2)

Since T is normal i.e. TT* = T*T.

Then from (1) and (2), we see that

T12 + T2

2 + i (T2T1 – T1T2) = T12 + T2

2 + i (T1T2 – T2T1)

T2T1 – T1T2 = T1T2 – T2T1

2T2T1 = 2T1T2

T2T1 = T1T2 T1, T2 commute.

Conversely, let T1, T2 commute

i.e. T1T2 = T2T1, then from (1) and (2)

We see that

TT* = T*T T is normal.

Example: If T is a normal operator on a Hilbert space H and is any scalar, then T – I isalso normal.

Solution: T is normal TT* = T*T

Also (T – I)* = T* – ( I)*

= T* – I*

= T* – I.

Now (T – I) (T – I)* = (T – I) (T* – I)

= TT* – I – T* + | |2I … (1)

Also (T – I)* (T – I) = (T* – I) (T – I)

= T*T – I* – T + | |2I … (2)



Notes Since TT* = T*T, therefore R.H.S. of (1) and (2) are equal.

Hence their L.H.S. are also equal.

(T – I) (T – I)* = (T – I)* (T – I)

T – I is normal.

29.1.2 Unitary Operator

An operator U on a Hilbert space H is said to be unitary if UU* =U*U =I.

Notes

(i) Every unitary operator is normal.

(ii) U* = U-1 i.e. an operator is unitary iff it is invertible and its inverse is precisely equalto its adjoint.

Theorem 8: If T is an operator on a Hilbert space H, then the following conditions are allequivalent to one another.

(i) T*T = I.

(ii) (Tx,Ty) = (x,y) for all x,y H.

(iii) Tx x x H.

Proof: (i) (ii)

(Tx,Ty) = (x,T*Ty) = (x,Iy) = (x,y) x and y.

(ii) (iii)

We are given that

Tx,Ty x, y x, y H.

Taking y = x, we get

(Tx,Tx) = (x,x) 2 2Tx x

Tx x x H.

(iii) (i)

Given Tx x x

2 2Tx x

Tx,Tx x,x

T * Tx,x x,x



NotesT * T I x,x O x H

T * T I O

T * T I


29.1.3 Isometric OperatorDefinition: An operator T on H is said to be isometric if Tx Ty x y x, y H.

Since T is linear, the condition is equivalent to Tx x for every x H.

For example: let 1 2 ne ,e ,...,e ,... be an orthonormal basis for a separable Hilbert space H and

T H be defined as 1 1 2 2 1 2 2 3 nT x e x e ... x e x e ... where x x .

Then 22 2

nn 1

Tx x x

T is an isometric operator.

The operator T defined is called the right shift operator given by Ten = en+1.

Theorem 9: If T is any arbitrary operator on a Hilbert space H then H is unitary it is anisometric isomorphism of H onto itself.

Proof: Let T is a unitary operator on H. Then T is invertible and therefore T is onto.

Further TT* = I.

Hence Tx x for every x H. [By Theorem (7)]

T preserves norms and so T is an isometric isomorphism of H onto itself.

Conversely, let T is an isometric isomorphism of H onto itself. Then T is one-one and onto.Therefore T–1 exists. Also T is an isometric isomorphism.

Tx x x

T*T = I [By Theorem (7)]

1 1T * T T IT

1 1T * TT T

1T * I T

TT* I T * T and so T is unitary.




Notes

Note If T is an operator on a Hilbert space H such that Tx x x H and T is definitelyan isometric isomorphism of H onto itself. But T need not be onto and so T need not beunitary. The following example will make the point more clear.

Example: Let T be an operator on l2 defined by 1 2 1 2T x ,x ,... 0,x ,x ,...

2Tx x x I .

T is an isometric isomorphism of l2 into itself.

However T is not onto. If 1 2y ,y ,... is a sequence in l2 such that 1y 0, then no sequence in l2

whose T-image is 1 2y ,y ,... . Therefore T is not onto and so T is not unitary.

29.2 Summary

An operator T on a Hilbert space H is said to be normal if it commutes with its adjoint i.e.if TT* = T*T. Conclusively every self adjoint operator is normal.

The set of all normal operators on a Hilbert space H is a closed subspace of H whichcontains the set of all set-adjoint operators and is closed under scalar multiplication.

An operator U on a Hilbert space H is said to be unitary if UU* = U*U =I.

An operator T on H is said to be isometric if Tx Ty x y x,y H , since T is linear,

the condition is equivalent to Tx x for every x H.

29.3 Keywords

Normal Operator: An operator T on a Hilbert space H is said to be normal if it commutes withits adjoint i.e. if TT* = T*T.

Unitary Operator: An operator U on a Hilbert space H is said to be unitary if UU* = U*U = I.

Isometric Operator: An operator T on H is said to be isometric if Tx Ty x y x, y H.

Since T is linear, the condition is equivalent to Tx x for every x H.


1. If T is an operator on a Hilbert space H, then T is normal its real and imaginary partcommute.

2. An operator T on H is normal T * x Tx for every x.

3. The set of all normal operators on H is a closed subset of H which contains the set of all

self-adjoint operators and is closed under scalar multiplication.

4. If H is finite-dimensional, show that every isometric isomorphism of H into itself isunitary.

5. Show that the unitary operators on H form a group.



Notes29.5 Further Readings

Books Arch W. Naylor, R Sell George, Linear Operator Theory in Engineering and Sciences,New York Springer, (1982).

Paul Garret, Operators in Hilbert Spaces, 2005.

Online link www.maths.leeds.ac.uk/nkisilv/



Notes Unit 30: Projections

CONTENTS

Objectives

Introduction

30.1 Projections

30.1.1 Perpendicular Projections

30.1.2 Invariance

30.1.3 Orthogonal Projections

30.2 Summary

30.3 Keywords



Objectives


Define perpendicular projections.

Define invariance and orthogonal projections.

Solve problems on projections.

Introduction

We have already defined projections both in Banach spaces and Hilbert spaces and explainedhow Hilbert spaces have plenty of projection as a consequence of orthogonal decompositiontheorem or projection theorem. Now, the context of our present work is the Hilbert space H, andnot a general Banach space, and the structure which H enjoys in addition to being a Banach spaceenables us to single out for special attention those projections whose range and null space areorthogonal. Our first theorem gives a convenient characterisation of these projections.

30.1 Projections

30.1.1 Perpendicular Projections

A projection P on a Hilbert space H is said to be a perpendicular projection on H if the range Mand null space N of P are orthogonal.

Theorem 1: If P is a projection on a Hilbert space H with range M and null space N then M N Pis self-adjoint and in this case N M .

Proof: Let M N and z be any vector in H. Then since H M N, we can write z uniquely as

z x y,x M,y N.


Unit 30: Projections

NotesThus Pz P x y

Px Py

Px x y N

Pz,z x,z Pz = P x+y x,P being projection on H

x,x y

x,x x,y

2x

and Pz*,z z,Pz

x y,x x,x x,y

2x .

Hence Pz,z Pz*,z z H

P P * z,z 0 z H

P P* 0 i.e. P P *

P is self adjoint.

Further, M N N M

If N M , then N is a proper closed linear subspace of the Hilbert space M and therefore avector 0 0z 0 M s.t. z N.

Now z0 M and z0 N and H = M N.

0 0z H z 0, a contradiction.

Hence N M

Conversely, let P* P,x,y be any vectors in M and N respectively. Then

x,y Px,y

x,P * y x,Py

x,0 0

M N.


Theorem 2: If P is the projection on the closed linear subspace M of H, then

x M Px x Px x .

Proof: We have, P is a projection on H with range M then, to show x M Px x.



Notes Let Px = x. Then X is in the range of P because Px is in the range of P.

Px x x M.

Conversely, let x M. Then to show Px = x.

Let Px = y. Then we must show that y = x.

We have

2Px y P Px Py P x Py

2Px Py P =P

P x y 0

x y is a in null space of P.

x y M .

x y z,z M .

x y z.

Now y Px y is in the range of P.

i.e. y is in M. Thus we have expressed

x y z,y M,z M .

But x is in M. So we can write x = x+0, x M,0 M

But H M M .

Therefore we must have y = x, z = 0

Hence x M Px x.

Now we shall show that Px = x Px x .

If Px = x then obviously Px x .

Conversely, suppose that Px x .

We claim that Px = x. We have

22x Px I P x ...(1)

Now Px is in M. Also P is the projection on M.

I P is the projection on M .

I P x in M .

Px and I P x are orthogonal vectors.

Then by Pythagorean theorem, we get

2 22Px I P x Px I P x ...(2)



NotesFrom (1) and (2), we have

22 2x Px I P x

2I P x 0 by hypothesis Px x

I P x 0

x Px 0

Px 0


Theorem 3: If P is a projection on a Hilbert space H, then

(i) P is a positive operator i.e. P 0

(ii) 0 P 1

(iii) Px x x H.

(iv) P 1.

Proof: P, projection on H 2 *P P,P P.

Let M = range of P.

(i) Let x H. Then

Px,x PPx,x

2*Px,P x Px,Px Px 0

Px,x 0 x H.

P is a positive operator i.e. P 0.

(ii) P is a projection on H I – P is also a projection on H.

I – P 0. (by (i))

P I

But P 0 , consequently 0 P 1.

(iii) Let x H. If M is the range of P, then M is the range of (I – P).

Now Px is in M and (I – P)x is in M .

Therefore Px and (I – P)x are orthogonal vector. So by Pythagorean theorem, we have

2 22Px I P x Px I P x

22 2x Px I P x Px+ I–P x x



Notes 2 2x Px

Px x .

(iv) We have P sup Px : x 1

But Px x x H (by(iii))

sup Px : x 1 1

Hence P 1


Example: If P and Q are the projections on closed linear subspaces M and N of H. Showthat PQ is a projection PQ = QP. In this case, show that PQ is the projection on M N.

Solution: Since P and Q are projections on H, therefore P2 = P, P* = P, Q2 = Q, Q* = Q. Also it is giventhat M is range of P and N is the range of Q.

Now suppose PQ is projection on H. Then to prove that PQ = QP.

Since PQ is a projection on H.

(PQ)* = PQ

Q* P* = PQ

QP = PQ ( Q* = Q, P* = P)

Conversely, let PQ = QP. We shall show that PQ is a projection on H.

We have (PQ)* = Q*P* = QP = PQ.

Also (PQ)2 = (PQ) (PQ) = (PQ) (QP)

= PQ2P = PQP

= QPP = QP2

= QP = PQ

Thus (PQ)* = PQ and (PQ)2 = PQ.

PQ is a projection on H.

Finally we are to show that PQ is the projection on M N, i.e. we are to show that range of PQis M N.

Let R (PQ) = range of PQ.

Let x M N x M, x N we have

(PQ) (x) = P (Qx) = Px [ N is range of Q and x N Qx = x]

= x [ M is range of P and x P]

(PQ)x = x

x R (PQ)

x M N x R (PQ)



NotesM N R (PQ)

Now let x R (PQ). Then (PQ)x = x

Now (PQ) x = x

P [(PQ) x] = Px

[P (PQ)] x = Px

(P2Q) x = Px

(PQ) x = Px

But (PQ) x = x.

We have Px = x x M i.e. the range of P.

Also PQ = QP

x R (PQ) (PQ) x = x

(QP)x = x Q [(QP)x] = Qx

(Q2P)x = Qx (QP)x = Qx

But (QP)x = x, Qx = x x N.

Thus x R (PQ) x M and x N

x M N

R(PQ) M N

Hence R(PQ) = M N.

Example: Show that an idempotent operator on a Hilbert space H is a projection on H it is normal.

Solution: P is an idempotent operator on H i.e. P2 = P.

Let P be a projection on H. Then P* = P. We have

PP* = P* P* [taking P* in place of P in L.H.S.]

= P* P [ P* = P]

P is normal.

Conversely, let PP* = P*P.

Then to prove that P* = P.

For every vector y H, we have

(Py, Py) = (y, P* Py) = (y, PP*y) [ P*P = PP*]

= (P*y, P*y) [ (P*)* = P]

From this we conclude that

Py = 0 P*y = 0.

Now let x be any vector in H.

Let y = x – Px. Then



Notes Py = P(x – Px) = Px – P2x = Px – Px = 0

0 = P*y = P*(x – Px) = P*x – P*Px

P*x = P*Px x H

P* = P*P

Now P = (P*)* = (P*P)* = P*P = P*

P is a self adjoint operator.

Also P2 = P.

Hence P is a projection on H.

30.1.2 Invariance

Definition: Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then Mis said to be invariant under T if T M M. If we do not take into account the action of T onvectors outside M, then T can be regarded as an operator on M itself. The operator T on H induceson operator TM on M such that TM(x) = T(x) for every x M. This operator TM is called therestriction of T on M.

Further, let T be an operator on Hilbert space H. If M is a closed subspace of H and if M and Mare both invariant under T, then T is said to be reduced by M. If T is reduced by M, we also saythat M reduces T.

Theorem 4: A closed linear subspace M of a Hilbert space H is invariant under the operation

T M is invariant under T*.

Proof: Let M is invariant under T, we show M is invariant under T*.

Let y be any arbitrary vector in M . Then to show that T*y is also in M i.e. T*y is orthogonal toevery vector in M.

Let x be any vector in M. Then Tx M because M is invariant under T.

Also y M y is orthogonal to every vector in M.

Therefore y is orthogonal to Tx i.e.

(Tx,y) = 0

(x,Ty) = 0

T*y is orthogonal to every vector x in M.

T * y is in M and so M is invariant under T*.

Conversely, let M is invariant under T*. Thus to show that M is invariant under T. Since M is

a closed linear subspace of H invariant under T*, therefore by first case M is invariantunder T.

But M M M and T * * T * * T.



NotesHence M is invariant under T.


Theorem 5: A closed linear subspace M of a Hilbert space H reduces on operator M is invariantunder both T and T*.

Proof: Let M reduces T, then by definition both M and M are invariant under T*. But by

theorem 4, if M is invariant under T then M i.e. M is invariant under T*. Thus M is invariantunder T and T*.

Conversely, let M is invariant under both T and T*. Since M is invariant under T*, therefore Mis invariant under T * * = T (by theorem 4). Thus both M and M are invariant under T.Therefore M reduces T.

Theorem 6: If P is the projection on a closed linear subspace M of a Hilbert space H, then M isinvariant under an operator T TP PTP.

Proof: Let M is invariant under T.

Let x H. Then Px is in the range of T, Px M TPx M.

Now P is projection and M is the range of P. Therefore TPx M TPx will remain unchangedunder P. So, we have

PTPx = TPx

PTP = TP (By equality of mappings)

Conversely, let PTP = TP. Let x M. Since P is a projection with range M and x M , therefore

Px = x

TPx = Tx

PTPx =Tx PTP TP

PTPx = TPx TPx Tx

But P is a projection with range M.

P TPx TPx TPx M Tx M

Since TPx = Tx.

Thus x M Tx M

M is invariant under T.

Theorem 7: If P is the projection on a closed linear subspace of M of a Hilbert space H, then Mreduces an operator TP PT.

Proof: M reduces T M is invariant under T and T*.

TP PTP and T * P PT * P

TP PTP and T * P * PT * P *

TP PTP and P * T * * P * T * *P *

TP PTP and PT PTP P is projection P* P. AlsoTT* T



Notes Thus M reduces T.

TP PTP and PT PTP ...(1)

Now suppose M reduces T. Then from (1), TP = PTP and PT = PTP. This gives TP = PT.

Conversely, let TP = PT

PTP =P2T (Multiplying both sides on left by P.)

or PTP = PT 2P P

similarly multiplying both sides of TP = PT on the right of P, we get

TP2 = PTP or TP = PTP. Thus

TP = PT TP = PTP and PT = PTP.

Therefore from (1), we conclude that M reduces T.

Theorem 8: If M and N are closed linear subspace of a Hilbert space H and P and Q are theprojections on M and N respectively, then

(i) M N PQ O. and

(ii) PQ O QP O.

Proof: Since P and Q are projections on a Hilbert space H, therefore P* = P, Q* = Q.

We first observe that

PQ O PQ * O * Q * P* O *

QP O.

Therefore to prove the theorem it suffices to prove that

M N PQ O.

First suppose M N. If y is any vector in N, then M N y is orthogonal to every vector in M.

so y M .Consequently N M .

Now, let z be any vector in H. Then Qz is the range of Q i.e. Qz is in N.

Consequently Qz is in M which is null space of P.

Therefore P(Qz) = O.

Thus PQz = O z H

PQ = O

Conversely, let PQ = O and x M and y N.

since M is the range of P, therefore Px = x. Also N is the range of Q. Therefore

Qy = y

We have (x,y) = (Px, Qy) = (x,P*Qy)



Notes = (x,PQy) P* P

= (x,Oy) PQ O

= (x,O) = O

M and N are orthogonal i.e. M N.

30.1.3 Orthogonal Projections

Definition: Two projections P and Q on a Hilbert space H are said to be orthogonal if PQ = O.

Note: By theorem 8, P and Q are orthogonal iff their ranges M and N are orthogonal.

Theorem 9: If P1, P2, ... Pn are projections on closed linear subspaces M1, M2, ... Mn of a Hilbertspace H, then P = P1 + P2 + ... + Pn is a projection i the P 's are pair-wise orthogonal (in the sense

that i jP P 0,i j).

Also then P is the projection on M = M1 + M2 + ... + Mn.

Proof: Given that P1, P2, ... Pn are projections on H.

Therefore 2 *i i iP P P for each i = 1, 2, ...,n.

Let P = P1 + P2+ ... + Pn. Then P* = (P1 + P2+ ... + Pn)* = * *1 nP ... P

= P1 + P2 + ... + Pn = P.

Sufficient Condition:

Let i jP P O,i j. Then to prove that

P is a projection on H. We have

P2 = PP = (P1 + P2+ ... + Pn) (P1 + P2+ ... + Pn)

= 2 2 21 2 nP P ... P i jP P 0,i j

= P1 + P2+ ... + Pn

= P

Thus, P* = P = P*.

Therefore P is a projection on H.

Necessary Condition:

Let P is a projection on H.

Then P2 = P = P*.

We are to prove that i jP P 0 if i j.

We first observe that if T is any projection on H and z is any vector in H, then

(Tz, z) = (T Tz, z) = (Tz, T*z)

= (Tz, Tz)

= 2Tz ...(1)



Notes Now let x belongs to the range of some Pi so that Pix = x. Then

22ix P x

n

2 2 2j 1 n

j 1

P x P x ... P x

n

jj 1

P x,x [Using (i)]

1 nP x,x ... P x,x

1 2 nP P ... P x

Px,x

2Px [by (1)]

2x ...(2)

Thus we conclude that sign of equality must hold throughout the above computation. Thereforewe have

n22

i jj 1

P x P x

2

jP x O if j i

jP x O, j i

jP x O, j i

x is in the null space of jP ,i j

jx M ,if j i

x is orthogonal to the range jM of every jP with j i.

Thus every vector x in the range Pi(i = 1,...,n) is orthogonal to the range of every P j with j i.

Therefore the range of Pi is orthogonal to the range of every P j with j i. Hence

i jP P O, i j [By theorem (8)]

Finally in order to show that P is the projection on 1 2 nM M M ... M

We are to show that R(P) = M where R(P) is the range of P.

Let 1 2 nx M. Then x x x ... x

where i ix M , 1 i n. Now from (2), we observe that 2 2x Px if x is the range of some Pi.

i ix M, i.e. the range of P .



Notes2 2i i i iP x x P x x

i iPx x

ix the range of P.

ix R P , for each i 1,2,...,n

1 2 nx x ... x R P .

x R P .

Then x M x R P

M R P ...(3)

Now suppose that x R P . Then

Px = x

1 2 nP P ... P x x

1 2 nP x P x ... P x x

But 1 1 2 2 n nP x M ,P x M ,...,P x M .

1 2 nx M M ... M and so R P M ...(4)

Hence from (3) and (4), we get

M = R(P)


30.2 Summary

A projection P on a Hilbert space H is said to be a perpendicular projection on H if therange M and null space N of P are orthogonal.

Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then M is saidto be invariant under T if T M M.

Let T be an operator on Hilbert space H, if M is closed subspace of H and if M and M areboth invariant under T, then T is said to be reduced by M.

Two projections P and Q on a Hilbert space H are said to be orthogonal if PQ = O.

30.3 Keywords

Invariance: Let T be an operator on a Hilbert space H and M be a closed subspace of H. Then Mis said to be invariant under T if T M M.

Orthogonal Projections: Two projections P and Q on a Hilbert space H are said to be orthogonalif PQ = O.

Perpendicular Projections: A projection P on a Hilbert space H is said to be a perpendicularprojection on H if the range M and null space N of P are orthogonal.




1. If P and Q are the projections on closed linear subspaces M and N of H, prove that PQ is aprojection PQ QP. In this case, show that PQ is the projection on M N.

2. If P and Q are the projections on closed linear subspaces M and N of H, prove that thefollowing statements are all equivalent to one another:

(a) P Q;

(b) Px Qx for every x;

(c) M N;

(d) PQ P;

(e) QP = P.

3. If P and Q are the projections on closed linear subspaces M and N of H, prove that Q P isa projection P Q. In this case, show that Q – P is the projection on N M .


Books Borbaki, Nicolas (1987), Topological Vector Spaces, Elements of mathematics, Berlin:Springer – Verlag

Rudin, Walter (1987), Real and Complex Analysis, McGraw-Hill.

Online links www.math.Isu.edu/~ sengupta/7330f02/7330f02proiops.pdf

Planetmath.org/....OrthogonalProjections OntoHilbertSubspaces.html.


Unit 31: Finite Dimensional Spectral Theory

NotesUnit 31: Finite Dimensional Spectral Theory

CONTENTS

Objectives

Introduction

31.1 Finite Dimensional Spectral Theory

31.1.1 Linear Operators and Matrices on a Finite Dimensional Hilbert Space

31.1.2 Similar Matrices

31.1.3 Determinant of an Operator

31.1.4 Spectral Analysis

31.1.5 Spectrum of an Operator

31.1.6 Spectral Theorem

31.1.7 Spectral Resolution of an Operator

31.1.8 Compact Operators

31.1.9 Properties of Compact Operators

31.2 Summary

31.3 Keywords



Objectives


Understand finite dimensional spectral theory.

Describe spectral analysis and spectral resolution of an operator.

Define compact operators and understand properties of compact operators.

Solve problems on spectral theory.

Introduction

The generalisation of the matrix eigenvalue theory leads to the spectral theory of operators ona Hilbert space. Since the linear operators on finite dimensional spaces are determined uniquelyby matrices, we shall study to some extent in detail the relationship between linear operators ina finite dimension Hilbert spaces and matrices as a preliminary step towards the study ofspectral theory of operators on finite dimensional Hilbert spaces.



Notes 31.1 Finite Dimensional Spectral Theory

31.1.1 Linear Operators and Matrices on a Finite Dimensional HilbertSpace

Let H be the given Hilbert space of dimension n with ordered basis B = {e1, e2, …, en} where theordered of the vector is taken into consideration. Let T (H) (the set of all bounded linearoperators). Since each vector in H is uniquely expressed as linear combination of the basis, we

can express Tej as Tej = n

ij ii 1

e , where the n-scalars 1j, 2j, … nj are uniquely determined by Tej.

Then vectors Te1, Te2, …, Tej determine uniquely the n2 scalars ij, i, j = 1, 2, … n. These n2 scalarsdetermine matrix with ( i1, i2, …, in) as the ith row and ( 1j, 2j, … nj) as its jth column. Wedenote this matrix by {T} and call this matrix as the matrix of the operator T with respect to theordered basis B.

Hence = [ ij] = 11 12 1n

21 22 2n

n1 n2 nn

We note that

(i) [0] = 0, which is the zero matrix.

(ii) [I] = I = [ ij], which is a unit matrix of order n. Here ij is the Kronecker delta.

Definition: The set of all n × n matrices denoted by An is complex algebra with respect toaddition, scalar multiplication and multiplication defined for matrices.

This algebra is called the total matrices algebra of order n.

Theorem 1: Let B be an ordered basis for a Hilbert space of dimension n. Let T (H) with (T) =[ ij], then T is singular [ ij] is non-singular and we have [ ij]–1 = [T–1].

Proof: T is non-singular iff there exists an operator T–1 on H such that

T–1 T = T T–1 = I … (1)

Since there is one-to-one correspondence between T and [T–1],

(1) is true [T–1 T] = [TT–1] = [I]

from (2) [T–1] [T] = [T] [T–1] = [I] = [ ij]

so that [T–1] [ ij] = [ ij] [T–1] = [ ij], [T] = [ ij].

[ ij] is a non-singular and [ ij]–1 = [T–1].


31.1.2 Similar Matrices

Let A, B are square matrix of order n over the field of complex number. Then B is said to besimilar to A if there exists a n × n non-singular matrix C over the field of complex numbers suchthat

B = C–1 AC.



NotesThis definition can be extended similarly for the case when A, B are operators on a Hilbert space.

Notes1. The matrices in An are similar iff they are the matrices of a single operator on H

relative to two different basis H.

2. Similar matrices have the same determinant.

31.1.3 Determinant of an Operator

Let T be an operator on an n-dimensional Hilbert space H. Then the determinant of the operatorT is the determinant of the matrix of T, namely [T] with respect to any ordered basis for H.

Following we given properties of a the determinant of an operator on a finite dimensionalHilbert space H.

(i) det (I) = 1, I being identity operator.

Since det (I) = det ([I]) = det ([ ij]) = 1.

(ii) det (T1 T2) = (det T1) (det T2)

(iii) det (T) 0 [T] is non-singular

det ([T]) 0.

Hence det (T) 0 [T] is non-singular.

31.1.4 Spectral Analysis

Definition: Eigenvalues

Let T be bounded linear operator on a Hilbert space H. Then a scalar is called an eigenvalue ofT if there exists a non-zero vector x in H such that Tx = x.

Eigenvalues are sometimes referred as characteristic values or proper values or spectral values.

Definition: Eigenvectors

If is an eigenvalue of T, then any non-zero vector x H such Tx = x, is called a eigenvector(characteristic vector or proper vector or spectral vector) of T.

Properties of Eigenvalues and Eigenvectors

Notes If the Hilbert space has no non-zero vectors then T cannot have any eigenvectorsand consequently the whole theory reduces to triviality. So we shall assume that H 0throughout this unit.

1. If x is an eigenvector of T corresponding to the eigenvalue and is a non-zeroscalar, then x is also an eigenvector of T corresponding to the same eigenvalue .

Contd...



Notes Since x is an eigenvector of T, corresponding to the eigenvalue 0 and Tx = x.

0 x 0

Hence T ( x) = T(x) = ( x)

Therefore corresponding to an eigenvalue there are more than one eigenvectors.

2. If x is an eigenvector of T, then x cannot correspond to more than one eigenvalueof T.

If possible let 1, 2 be two eigenvalues of T, ( 1 2) for eigenvector x. Then

Tx = 1x and Tx = 2x

1x = 2x

( 1 – 2)x = 0

1 – 2 = 0 ( x 0)

l = 2

3. Let be an eigenvalue of an operator T on H. If M is the set consisting of alleigenvectors of T corresponding to together with the vector 0, then M is a non-zero closed linear subspace of H invariant under T.

By definition x M Tx = x … (1)

By hypothesis 0 M and 0 vector satisfies (1).

M = {x H : Tx = x} = {x H : (T – I) x = 0}

Since T and I are continuous, M is the null space of continuous transformation T – I.Hence M is closed.

Next we show that if x M , then Tx M . If x M then Tx = x.

Since M is a linear subspace of H, x M x = Tx M .

M is invariant under T.

Definition: Eigenspace

The closed subspace M is called the eigenspace of T corresponding to the eigenvalue .

From property (3), we have proved that each eigenspace of T is a non-zero linear subspace of Hinvariant under T.

Note It is not necessary for an operator T on a Hilbert space H to possess an eigenvalue.

Example: Consider the Hilbert space 2 and the operator T on 2 defined by

T (x1, x2, …) = (0, x1, x2, …)

Let be a eigenvalue of T. Then a non zero vector

y = (y1, y2, …) in 2 such that Ty = y.



NotesNow Ty = y T (y1, y2, …) = (y1, y2, …)

(0, y1, y2, …) = ( y1, y2 …)

y1 = 0, y2 = y1, ……

Now y is a non-zero vector y1 0

y1 = 0 = 0.

Then y2 = y1 y1 = 0 and this contradicts the fact that y is a non-zero vector. Therefore T cannothave an eigenvalue.

31.1.5 Spectrum of an Operator

The set of all eigenvalues of an operator is called the spectrum of T and is denoted by (T).

Theorem 1: If T is an arbitrary operator on a finite dimensional Hilbert space H, then the spectrumof T namely (T) is a finite subset of the complex plane and the number of points in (T) doesnot exceed the dimension n of H.

Proof: First we shall show that an operator T on a finite dimensional Hilbert space h is singularif and only if there exists a non-zero vector x H such that Tx = 0.

Let a non-zero vector x H s.t. Tx = 0. We can write Tx = 0 as Tx = T0. Since x 0, the two distinctelements x, 0 H have the same image under T. Therefore T is not one-to-one. Hence T–1 doesnot exist. Hence it is singular.

Conversely, let T is singular. Let no non-zero vector such that Tx = 0. This means Tx = 0 x =0. Then T must be one-to-one. Since H is finite dimensional and T is one-to-one, T is onto, so thatT is a non-singular, contradicting the hypothesis that T is singular. Hence there must be non-zero vector x s.t. Tx = 0.

Now if T is an operator on a finite dimensional Hilbert space H of dimension n. Then A scalar(T), if there exists a non-zero vector x such that (T – I)x = 0.

Now (T – I)x = 0 (T – I) is a singular.

(T – I) is singular det (T – I) = 0. Thus (T) satisfies the equation det (T – I) = 0.

Let B be an ordered basis for H. Thus det (T – I)

= det ([T – I]B)

But det ([T – I]B) = det ([T]B – [I]B)

Thus det (T – I) = det ([T]B – [ ij]).

So det (T – I) = 0 det ([T]B – [ ij]) = 0 … (1)

If [T]B = [ ij] is a matrix of T then (1) gives

11 12 1n

22 22 2n

n1 nn

… (2)

The expression of determinant of (2) gives a polynomial equation of degree n in with complexcoefficients in the variable . This equation must have at least one root in the field of complexnumber (by fundamental theorem of algebra). Hence every operator T on H has eigenvalue so



Notes that (T) . Further, this equation in has exactly n roots in complex field. If the equation hasrepeated roots, then the number of distinct roots are less than n. So that T has an eigenvalue andthe number of distinct eigenvalue of T is less than or equal to n. Hence the number of elementsof (T) is less than or equal to n. This completes the proof of the theorem.

Example: For a two dimensional Hilbert space H, let B = {e1, e2} be a basis and T be anoperator on H given by the matrix

A = 11 12

21 22… (1)

(i) If T is given by Te1 = e2 and Te2 = – e2, find the spectrum T.

(ii) If T is an arbitrary operator on H with the same matrix representation, then

T2 ( 11 + 22) T + ( 11 22 – 12 21) I = 0

Sol:

(i) Using the matrix A of the operator T, we have

Te1 = 11 e1 + 21 e2 = e2 so that 11 = 0 and 21 = 1

Te2 = 12 e1 + 22 e2 = –e1 so that 12 = –1 and 22 = 0

Hence [T] = 11 12

21 22

0 11 0 .

For this matrix, the eigenvalue are given by the characteristic equation

11 = 0

2 + 1 = 0 = i so (T) = { i}.

(ii) Let us consider the eigenvalues of A, which are given by

11 12

21 22= 0

2 – ( 11 + 22) + ( 11 22 – 12 21) = 0 … (2)

Since (2) is true for , we can take

T = I … (3)


T2 – ( 11 + 22) T + ( 11 22 – 12 21) I = 0 … (4)

The operator T on H having as an eigenvalue satisfies equation (4).

Theorem 2: If T is an operator on a finite dimension Hilbert space, then the following statementsare true.

(a) T is singular 0 (T)

(b) If T is non-singular, then (T) –1 (T–1)

(c) If A is non-singular, then (ATA–1) = (T)

(d) If (T) and if P is polynomial then P ( ) (P (T)).



NotesProof:

(a) T is singular a non-zero vector x H such that Tx = 0 or Tx = 0. Hence T is singular0 is the eigenvalue of T i.e. 0 (T).

(b) Let T be non-singular and (T).

Hence 0 by (a) so that –1 exists. Since is an eigenvalue of T, a non-zero vectorx H s.t. Tx = x.

Premultiplying by T–1 we get

T–1 Tx = T–1 ( x)

T–1 (x) = 1 x for x 0

Hence –1 ( (T–1))

Conversely, if –1 is an eigenvalue of T–1 then ( –1)–1 = is an eigenvalue of (T–1)–1 = T.

Hence (T).

(c) Let S = ATA–1. Then we find S – I.

Now S – I = ATA–1 – A ( I) A–1

= A (T – I) A–1

det (S – I) = det (A(T – I) A–1)

= det (T – I)

is an eigenvalue of T det (T – I) = 0.

Hence det (T – I) = 0 det (S – I) = 0

S and T have the some eigenvalues so that

(ATA–1) = (T).

(d) If = (T), is an eigenvalue of T. Then a non-zero vector x such that Tx = x.

Hence T (Tx) = T ( x) = Tx = 2x.

Hence if is an eigenvalue of T, then 2 is an eigenvalue of T2. Repeating this we get thatif is an eigenvalue of T, then n is an eigenvalue of Tn for any positive integer n.

Let P (t) = a0 + a1t + … + amtm, a0, a1, ……, am are scalars.

Then [P (T)]x = (a0I + a2T + …… + amTm)x

= a0x + a1 ( x) + …… + am ( mx)

= [a0 + a1 ( ) + …… + amm]x

Hence P( ) = a0 + a1 + …… + amm is an eigenvalue of P (T).

This if (T), then P ( ) (P (T)).


31.1.6 Spectral Theorem

Statement: Let T be an operator on a finite dimensional Hilbert space H with 1, 2, …, m as theeigenvalues of T and with M1, M2, …, Mm be then corresponding eigenspaces. If P1, P2, …, Pm arethe projections on the spaces, then the following statements are equivalent.



Notes (a) The Mi’s are pairwise orthogonal and span H:

(b) The Pi’s are pairwise orthogonal and P1 + P2 + … + Pm = I and T = 1P1 + 2P2 + … + mPm.

(c) T is normal operator on H.

Proof: We shall show that

(i) (ii) (iii) (i)

(i) (ii)

Assume that Mi’s are pairwise orthogonal and span H. Hence every x H can be representeduniquely as

x = x1 + x2 + … + xm … (1)

where xi Mi for i = 1, 2, …, m

by hypothesis Mi’s are pairwise orthogonal. Since P i’s are projections in M i’s Pi’s are pairwiseorthogonal, i.e. i j PiPj = 0.

If x is any vector in H, then from (1) for each i,

Pi(x) = Pi (x1 + x2 + … + xm)

= Pix1 + Pix2 + … + Pixm … (2)

Since Pi is the range of Mi, Pi xi = xi.

For i j Mj Mi since xj Mj for each j we have

xj Mi for j i.

Hence xj iM (null space of Pi)

xj iM Pixj = xi

from (2) we get

Pix = xi … (3)

Since I is the identity mapping on H, we get

Ix = x1 + x2 + … + xm … (by (1))

= P1x + P2x + … + Pmx … (by (3))

= (P1 + P2 + … + Pm) x x H.

This show that I = P1 + P2 + …… + Pm.

For every x H, we have from (1)

T (x) = T (x1 + x2 + … + xm)

= Tx1 + Tx2 + … + Txm

Since xi Mi Txi = xi

Tx = 1x1 + 2x2 + …… + mxm … (4)

= 1P1x + 2P2x + …… + mPmx … (5)

T = 1P1 + 2P2 + …… + mPm

(ii) (iii)



NotesLet T = 1P1 + 2P2 + …… + mPm, where Pi’s are pairwise orthogonal projections and to show thatT is normal.

Since Pi’s are projection and 2i i i iP * P and P P … (6)

Further we have PiPj = 0 for i j

Since adjoint operation is conjugate linear, we get

T* = ( 1P1 + 2P2 + …… + mPm)*

= 1 2 m1 2 mP * P * P *

= 1 2 m1 2 mP P P .

Now TT* = 1 2 m1 1 2 2 m m 1 2 m( P P P ) ( P P P )

= 2 2 2 2 2 21 1 2 2 1 m| | P | | P | | P ( PiPj = 0, i j)

= | 1|2P1 + | 2|2P2 + …… + | m|2Pm … (by (6))

Similarly T*T can be found s.t.

T*T = | 1|2P1 + | 2|2P2 + …… + | m|2 Pm

Hence T*T = TT* T is normal.

(iii) (i)

Let T is normal operator on H and prove that Mi’s are pairwise orthogonal and M i’s span H.

We know that if

T is normal on H its eigenspaces Mi’s are pairwise orthogonal.

So it suffices to show that Mi’s span H.

Let M = M1 + M2 + …… + Mm

and P = P1 + P2 + …… + Pm

Since T is normal on H, each eigenspace Mi reduces T. Also Mi reduces T PiT = TPi for each Pi.

TP = T (P1 + P2 + …… + Pm)

= TP1 + TP2 + …… + TPm

= P1T + P2T + …… + PmT

= (P1 + P2 + …… + Pm) T

= PT

Since P is projection on M and TP = PT, M reduces T and so M is invariant under T. Let Ube the restriction of T to M . Then U is an operator on a finite dimensional Hilbert spaceM and Ux = Tx x M . If x is an eigenvector for U corresponding to eigenvalue thenx M and Ux = x.

Tx = x and so x is also eigenvector for T.

Hence each eigenvector of U is also an eigenvector for T. But T has no eigenvector in M . HenceM M = {0}. So U is an operator on a finite dimensional Hilbert space M and U has noeigenvector and so it has no eigenvalue.



Notes M = {0}.

For if M {0}, then every operator on a non-zero finite dimensional Hilbert space must have aneigenvalue.

Now M = {0} M = H.

Thus M = M1 + M2 + …… + Mm = H and so Mi’s span H.

This complete the proof of the theorem.

31.1.7 Spectral Resolution of an Operator

Let T be an operator on a Hilbert space H. If there exist distinct complex numbers 1, 2, …, m

and non-zero pairwise orthogonal projections p 1, p2, …, pm such that

T = 1p1 + 2p2 + … + mpm

and p = p1 + p2 + … + pm, then the expression

T = 1p1 + 2p2 + … + mpm for T is called the spectral resolution for T.

Note We note that the spectral theorem coincides with the spectral resolution for anormal operator on a finite dimensional Hilbert space.

Theorem: The spectral resolution of the normal operator on a finite dimensional non-zero Hilbertspace is unique.

Proof: Let T = 1p1 + 2p2 + … + mpm

be a spectral resolution of a normal operator on a non-zero finite dimensional Hilbert space H.Then 1, 2, …, m are distinct complex numbers and pi s are non-zero pairwise orthogonalprojections such that p1 + p2 + … + pm = 1. We establish that 1 + 2, …, m are precisely the distincteigenvalues of T.

To this end we show first that for each i, i is an eigenvalue of T. Since pi 0 is a projection, anon-zero x in the range of p1 such that pix = x

Let us consider

Tx = ( 1p1 + 2p2 + … + mpm)x

= ( 1p1pi + 2p2pi + … + ipi2 + … + mpmpi)x

So pi’s are pairwise orthogonal p ipj = 0 for i j and pi2 = pi, we have Tx = ipix = ix by pix = x.

i is an eigenvalue of T.

Next we show that each eigenvalue of T is an element of the set ( 1, 2, …, m). Since T is anoperator on a finite dimensional Hilbert space, T must have an eigenvalue.

If is an eigenvalue of T then Tx = x = Ix.

( 1 p1 + 2 p2 + … + mpm) x = (p1 + p2 + … + pm) x

{( 1 – ) p1 + ( 2 – ) p2 + … + ( m – ) pm} x = 0 … (2)

Since pi2 = pi and pi pj = 0 for i j operating with pi throughout (2), we get

( i – ) pix = 0 for i = 1, 2, …, m.



NotesIf i for each i, pix = 0 for each i.

Hence pix + p2x + … + pmx = 0

(p1 + p2 + … + pm) x = 0

Ix = 0

x = 0, a contradiction to the fact that x 0. Hence must be equal to i for some i. This in thespectral resolution (1) of T, the scalar i are the precisely the eigenvalue of T.

If the spectral resolution is not unique.

Let T = 1Q1 + 2Q2 + … + nQn … (3)

be another revolution of T. Then 1, 2, … n is the same set of eigenvalues of T written indifferent order. Hence writing the eigenvalues in the same order as in (1) and renaming theprojections, we can write (3) as

T = 1Q1 + 2Q2 + … + mQm … (4)

To prove uniqueness, we shall show that p i in (1) and Qi in (4) are some.

Using the fact pi2 = pi, pipj = 0 i j, we can have

T0 = I = p1 + p2 + … + pm

T1 = 1p1 + 2p2 + … + mpm

T2 = 12p1 + … + m

2 pm

and Tn = 1np1 + … + m

n pm for any positive integer n. … (5)

Now if g (t) is a polynomial with complex coefficient in the complex variable t, we can writeg (T) as

g (T) = g ( 1) p1 + g ( 2) p2 + … + g ( m) pm

= m

j jj 1

g ( ) p … (by 5)

Let i be a polynomial such that i ( i) = 1 and i ( j) = 0

if i j

Taking i in place of g, we get

i (T) = m m

i j j ij j ij 1 j 1

( ) p p p

Hence for each i, let pi = i (T) which is a polynomial in T. The proof is complete if we show theexistence of i over the field of complex number.

Now 1 i 1 i 1 mi

i 1 i i 1 i i 1 i m

(t ) (t ) (t )...(t )(t)( )....( ) ( ) ( )

satisfies our requirements i.e. i ( i) = 1 and i ( j) = 0 if i j

Repeating the above discussion for Qi’s we get in a similar manner Q i = i (T) for each i.

pi = Qi for each i.




Notes 31.1.8 Compact Operators

Definition: A subset A in a normed linear space N is said to be relatively compact if its closure Ais compact.

A linear transformation T of a normed linear space N into a normed linear space N is said to bea compact operator if it maps a bounded set of N into a relatively compact set in N , i.e.

T : N N is compact of every bounded set B N, T(B) is compact in N .

31.1.9 Properties of Compact Operators

1. Let T : N N be a compact operator. Then T is bounded (continuous) linear operator. For,

let B be a bounded set in N. Since T is compact, T(B) is compact in N . So T(B) is complete

and totally bounded in N . Since a totally bounded set is always bounded, T(B) is boundedand consequently T(B) is bounded, since a subset of a bounded set is bounded.

T is a bounded linear transformation and it is continuous.

2. Let T be a linear transformation on a finite dimensional space N. Then T is compactoperator. For, N is finite dimensional and T is linear T(N) is finite dimensional. Since anylinear transformation on a finite dimensional space is bounded. T(B) is bounded subset of

T(N) for every bounded set B N. Now if T(B) is bounded so is T(B) and is closed. T(N) is

finite dimensional, any closed and bounded subset of T(N) is compact, so that T(B) iscompact, being closed and bounded subset of T(N).

3. The operator O on any normed linear space N is compact.

4. If the dimension of N is infinite, then identity operator I : N N is not compact operator.

For consider a closed unit sphere.

S = {x N : x 1} then S is bounded.

Since N is a infinite dimensional.

I (S) = S = S is not necessarily compact.

Hence I : N N is not compact operator. But I is a bounded (continuous) operator.

Theorem: A set A in a normed linear space N is relatively compact every sequence of pointsin A contains a convergent sub sequence.

Proof: Let A is relatively compact.

Since A A , every sequence in A is also sequence in A . Since A is compact, such a sequence

in A contains a convergent subsequence. Hence every sequence in A has a convergentsubsequence.

Conversely, let every sequence in A has a convergent subsequence.

Let (yn) be a sequence of points in A . Since A is dense in A , a sequence (xn) of points of A s.t.

n nx y 1n

… (1)



Notesand nkx x A … (2)

we can find a nk(y ) of (yn) s.t.

nky x = n n nk k k

y x x x

n n nk k ky x x x

0 as n .

A is compact.


31.2 Summary

If T is an arbitrary operator on a finite dimensional Hilbert space H, then the spectrum ofT namely (T) is a finite subset of the complex plane and the number of points in (T) doesnot exceed the dimension n of H.

Let T be bounded linear operator on a Hilbert space H. Then a scalar is called an eigenvalueof T if there exists a non-zero vector x in H such that Tx = x.

The closed subspace M is called the eigenspace of T corresponding to the eigenvalue .

The set of all eigenvalues of an operator is called the spectrum of T. It is denoted by (T).

The spectral resolution of the normal operator on a finite dimensional non-zero Hilbertspace is unique.

A subset A in a normed linear space N is said to be relatively compact if its closure A iscompact.

31.3 Keywords

Eigenspace: The closed subspace M is called the eigenspace of T corresponding to the eigenvalue .

Eigenvalues: Let T be bounded linear operator on a Hilbert space H. Then a scalar is called aneigenvalue of T if there exists a non-zero vector x in H such that Tx = x.

Eigenvalues are sometimes referred as characteristic values or proper values or spectral values.

Eigenvectors: If is an eigenvalue of T, then any non-zero vector x H such Tx = x, is called aeigenvector.

Similar Matrices: Let A, B are square matrix of order n over the field of complex number. ThenB is said to be similar to A if there exists a n × n non-singular matrix C over the field of complexnumbers such that

B = C–1 AC.

Spectrum of an Operator: The set of all eigenvalues of an operator is called the spectrum of T andis denoted by (T).

Total Matrices Algebra: The set of all n × n matrices denoted by An is complex algebra withrespect to addition, scalar multiplication and multiplication defined for matrices.

This algebra is called the total matrices algebra of order n.




1. If T (H) is a self-adjoint operator, then (T) = {m, M} where m, M are spectral values.

2. If T is self-adjoint operator then (T) is the subset of the real line [– T , T ].

3. Let R (T) = (T – I)–1 for a T B (X, X). Prove that R T 0 as .

4. Prove that the projection of a Hilbert space H onto a finite dimensional subspace of H iscompact.


Books Walter Rudin, Real and complex analysis, Third, McGraw-Hill Book Co., New York,1987.

Erwin Kreyszig, Introductory functional analysis with applications, John Wiley &Sons Inc., New York, 1989.

Online links www.math.washington.edu

chicago.academia.edu

www.math.ethz.ch/~ Kowalski/spectral-theory.pdf

Jalandhar-Delhi G.T. Road (NH-1)Phagwara, Punjab (India)-144411For Enquiry: +91-1824-300360Fax.: +91-1824-506111Email: [email protected]

LOVELY PROFESSIONAL UNIVERSITY

Documents

Measure Theorey and Functional Analysisebooks.lpude.in/arts/ma_mathematics/year_2/DMTH505...Unit 9: Measure Spaces 100 Unit 10: Measurable Functions 106 Unit 11: Integration 121 Unit