Math 247 Advanced Calculus

8/13/2019 Math 247 Advanced Calculus

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mlbaker.org presents

MATH 247Advanced Calculus 3

Dr. Alexandru NicaFall 2010 (1109)University of Waterloo

Disclaimer: These notes are providedas-is, and may be incomplete or contain errors.

Contents

1 Euclidean n-space

2 Sequences

3 Topology

4 Continuous functions

5 Continuity and compactness

6 Integrable functions

7 Linearity of the integral

8 Integration of functions that are continuous modulo null sets

9 Fubinis theorem

10 Integration on more general domains 2

11 Partial derivatives 2

12C1 functions 213 Chain rule 2

14 Higher-order partial derivatives 3

15 Functions inC1(A, Rm) (INCOMPLETE) 3

16 Inverse function theorem 3

17 Implicit function theorem 3

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1 Euclideann-space

In this section, we will review the algebra of vectors and the structure of the Euclidean n-space,Rn.

Course overview

Sequences, limits, continuity in Rn

Sets and topology ofRn (open, closed, compact, etc.) Derivatives (requires a good grasp of linear algebra)

Multivariable integrals Connections between derivatives and integrals

Notation 1.1 (review of vectors). We write x= (x(1), x(2), . . . , x(n)) for a vector xRn, where each x(i) R, which is calltheith componentof the vector x.

Operations with vectors.

Letx= (x(1), . . . , x(n))and y= (y(1), . . . , y(n))be vectors in Rn, and R. Addition. Addition is defined by:

x + y= (x(1) + y(1), x(2) + y(2), . . . , x(n) + y(n)).

Scalar multiplication. Multiplication by a scalar is defined by:x= (x(1), x(2), . . . , x(n)).

Standard inner product. The standard inner product onRn is defined by:

x, y=n

i=1

x(i)y(i) =x(1)y(1) + x(2)y(2) + . . . + x(n)y(n).

Norm or length. The norm(or length) of a vector is defined by:

x=

x, x=

n

i=1x(i)y(i).

We note that this square root makes sense, because the sum of the squares of real numbers is always non-negative. Nothatx 0, holding with equality if and only ifxis the zero vector, that is, the vector whose coordinates are all zero.

Remark 1.2 (basic properties of the inner product).

Bilinearity. Let1, 2, 1, 2 R. Then1x1+ 2x2, y= 1x1, y + 2x2, yx, 1y1+ 2y2= 1x, y1 + 2x, y2.

Symmetry.x, y=y, x for all x, y Rn.

Positivity. x, x 0 with equality holding if and only ifx= 0.Proposition 1.3 (Cauchy-Schwarz inequality). We have

|x, y| x y for all x, y Rn.Proof. The proof is trivial wheny= 0because the above holds with equality. So assume this is not the case. Define the functif : R R by f(t) =x ty, x ty for all t R. We can also write this as

f(t) =x, x x, ty ty, x + ty,ty=x2 2tx, y + t2 y2 .

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Hence we see that f is a quadratic function f(t) = at2 + bt + c, where

a=y2b=2x, yc=x2

Observe that f(t) =x ty2 0 for all t R, from the positivity of the norm. Hence the discriminant = b2 4ac musatisfy0. We have

= (2x, y)2 4 y2 x2 = 4(x, y2 x2 y2).

Hence0 implies that x, y2 x2 y2 .Taking square roots, we obtain the desired inequality.

Exercise 1.4. Determine all cases when the Cauchy-Schwarz inequality holds with equality.

Corollary 1.5 (triangle inequality).x + y x + y for all x, y Rn.

Proof. Observe that

x + y2 =x + y, x + y=x, x + x, y + y, x + y, y=

x2

+ 2x, y

+

y2

x2 + 2 x y + y2 by Cauchy-Schwarz= (x + y)2.

Taking square roots yields the inequality.

Definition 1.6. Forx= (x(1), x(2), . . . , x(n))and y = (y(1), y(2), . . . , y(n))in Rn, we define the distance betweenx and y to b

d(x, y) =x y=

(x(1) y(1))2 + . . . + (x(n) y(n))2.

Corollary 1.7 (triangle inequality for distance).

d(x, z)d(x, y) + d(y, z) for all x, y, z Rn.

Proof. Observe that

d(x, z) =x z=(x y) + (y z) x y + y z by the triangle inequality=d(x, y) + d(y, z).

Definition 1.8. Fora Rn andr >0, we define the open ballof radius r centered ata byB(a; r) ={x Rn |d(x, a)< r}={x Rn | x a< r}.

Similarly, we define the closed ballof radius r centered ataby

B(a; r) ={

x R

n

|d(x, a)

r

}=

{x R

n

| x

a

r}

.

Exercise 1.9. Letx be a vector inRn. Prove that

(1)x(i) xfor all i (1in).

(2)x x(1) + x(2) + . . . + x(n).

Note. Letx Rn. We define: The 1-normofx as

x1 =n

i=1

x(i) .

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The-normofx asx= maxi

x(i) . Thep-norm ofx as

xp =

ni=1

x(i)p1/p

.

2 Sequences

In this section, we will cover the n-dimensional versions of the two important theorems from MATH 147 (Advanced Calcul1): the Cauchy criterion and the Bolzano-Weierstrass theorem. A sequence (xk)

k=1 inR

n determines, for every k N, a vectxk. Note that we may havexi = xj fori=j (repetition is permitted in the sequence). A subsequenceof a sequence (xk)k=1obtained by taking (infinitely many) terms corresponding to certain indices from the original sequence:

1k(1)< k(2)< k(3)< . . . < k(p)< . . .We obtain a new sequence, which we denote by (xk(p))

p=1.

Definition 2.1. Let (xk)k=1 be a sequence in R

n, and a Rn. We say that (xk)k=1 converges toa if for every > 0 theexists somek0 N such thatxka< for all kk0. When this is the case, we often write xk a.Note. In the definition above we could have said d(xk, a) < or xk B(a; ), rather thanxka < . These are simpdifferent notations for the same concept.

Definition 2.2. Let (xk)k=1 be a sequence in R

n. We say that (xk)k=1 is Cauchy(or is a Cauchy sequence) if for every >

there exists k0 N such thatxp xq< for all p, qk0.Definition 2.3. Let (xk)k=1 be a sequence in R

n. We say that(xk)k=1 is boundedwhen there exists r >0 such thatxk for all k N. In other words, xk B(0, r)for all k N.Definition 2.4. Let(xk)

k=1 be a sequence in R

n. Write explicitly

x1 = (x(1)1 , . . . , x

(n)1 )

x2 = (x(1)2 , . . . , x

(n)2 )

...

xk = (x(1)k , . . . , x(n)k )

and look at the verticals to obtain n sequences of real numbers

(x(1)k )

k=1, (x

(2)k )

k=1, . . . , (x

(n)k )

k=1.

These are called the component sequences of(xk)k=1. Conversely, note that if one has n sequences in R, they can be assembltogether in this way to create a sequence in Rn.

Proposition 2.5. Let (xk)k=1 be a sequence in R

n, and a Rn. Thenxk converges to a in Rn if and only if each of tcomponentsx

(i)k converge toa

(i) inR.

Proof. Assume first thatxk a in Rn. Fix i between 1 and n. Simply note that

0 |x(i)k a(i)|=|(xka)(i)| xka .

By squeeze theorem, we obtain that x(i)k a(i).

Assume next that for all i (1in) we have x(i)k a(i). Hence|x(i)k a(i)| 0, and so also

|x(1)k a(1)| + |x(2)k a(2)| + . . . + |x(n)k a(n)| 0.However, observe that

0 xka n

i=1

|x(i)k a(i)|.

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Then again by squeeze theorem we conclude that xk a as required.Proposition 2.6. Let (xk)

k=1 be a sequence in R

n. Then (xk)k=1 is Cauchy in R

n if and only if each of the compone

sequences(x(i)k )

k=1 are Cauchy in R.

Proof. Exercise.

Proposition 2.7. Let (xk)k=1 be a sequence in R

n. Then (xk)k=1 is bounded in R

n if and only if each of the compone

sequences(x(i)k )

k=1 are bounded in R.

Proof. Exercise.

Theorem 2.8 (Cauchy criterion in Rn). Let (xk)k=1 be a sequence inRn. Then (xk)

k=1 is convergent (to some limit aR

if and only if it is Cauchy.

Proof. Note that(xk)k=1is convergent in R

n if and only if each component sequence (x(i)k )

k=1is convergent in R (by Propositi

2.5), if and only if each component sequence (x(i)k )

k=1 is Cauchy in R (by the Cauchy criterion for R), if and only if(xk)

k=1

Cauchy inRn (by Proposition 2.6).

Theorem 2.9 (Bolzano-Weierstrass in Rn). Let (xk)k=1 be a bounded sequence in Rn. Then one can find indices 1k(1)

k(2)< . . . < k(p)< . . . such that the subsequence (xk(p))p=1 is convergent.

Notation 2.10. Let(xk)k=1 be a sequence in Rn, and leta Rn. LetMbe an infinite subset of the natural numbers (M N

Say thatlimk

kM

xk = a

if for every >0 there exists k0 N such thatxka< for all kM such that kk0.Remark 2.11. We writeM explicitly as M ={k(1), k(2), . . . , k(p), . . .}. Observe that Notation 2.10 is just a convenient wof stating that limp xk(p) = a, in the original sense of Definition 2.1. We now restate Theorem 2.9 as Theorem 2.9*.

Theorem 2.9*. Let (xk)k=1 be a bounded sequence in Rn. Then we can find an infinite subset M N and a Rn such thalimkkM

xk = a.

Proof. We will assume n = 4. It is left as an exercise to use induction on n to provide a fully rigorous proof. We ha

xk = (x(1)k , x

(2)k , x

(3)k , x

(4)k ). Look at (x

(1)k )

k=1. This is, by Proposition 2.7, a bounded sequence in R. Apply B-W-147 (t

Bolzano-Weierstrass theorem from MATH 147) to this sequence to obtain an infinite subset M1

N and some a(1)

R su

thatlimkkM1

x(1)k =a(1).

We now proceed to the second component, but only look at those elementsx(2)k wherekM1. Apply B-W-147 to this to obta

an infinite subset M2M1 N and some a(2) R such thatlimkkM2

x(2)k =a

(2).

At this point we note that since M2M1 andlimkkM1

x(1)k =a

(1)

we certainly havelimkkM2

x(1)k =a

(1).

Continuing this process, we obtain a hierarchy of subsets M4 M3 M2 M1 N and we observe that for i = 1, 2, 3, 4 whave

limkkMi

x(i)k =a

(i).

Hence by the previous note we see that for i = 1, 2, 3, 4 we have

limkkM4

x(i)k =a

(i).

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Definea= (a(1), a(2), a(3), a(4)). Finally, apply Proposition 2.5 to conclude

limkkM4

xk = a.

Hence, taking M=M4 in the statement of the theorem, the proof is complete.

3 Topology

In this section we study the topology of Euclidean n-space.

Definition 3.1. Let A be a subset ofRn, and letaA. We calla an interior point ofA when B(a; r) A for some r >The set of all interior points ofA is called the interior ofA and is denoted by int(A)or sometimes A.

Example 3.2. Let A R2 be given byA={a= (s, t)|t >0} {a= (s, 0)|s >0}.

What is int(A)?

Definition 3.3. Let ARn. If everyaA is an interior point ofA, then we say that A is open. Note that a set A is openand only ifint(A) = A.

Definition 3.4. LetC Rn. We say that C is closed whenRn \ Cis open.

Note. Contrary to intuition, openness and closedness are not mutually exclusive properties. Most sets are neither open nclosed.

Example 3.6. Let U={(s, t) R2 |s R, t > 0}. This is an open set. Hence L = R2 \ U is closed.Definition 3.7 (the cannot escape property). LetCRn be nonempty. We say that Chas the cannot escape propertywhenever(xk)k=1 is a sequence in Cconverging to some b Rn, we must have bC.Proposition 3.8. Let C Rn be nonempty. Then Cis closed if and only ifChas the cannot escape property.Proof. First, assumeCis closed, that is, assume Rn\Cis open. Let(xk)k=1 be a sequence inC, withxk b Rn. Assume, fthe sake of contradiction, that b /C. Then b Rn \C. Since Rn \Cis an open set, there existsr >0 such thatB(b; r) Rn \But then, sincexk b, we can find k0 Nsuch thatxkB(b; r)for all kk0. In particular,xkB(b; r) Rn \ C. Therefoxk Rn \ Cwhich contradicts xkC. So we must have bC.Now, assumeChas the cannot escape property. We will show Rn\Cis open. Let b Rn\C. We want to show that there exisr >0 such thatB (b; r)Rn \ C. Assume, for the sake of contradiction, that no such r exists. In particular, B (b; 1) Rn \So there existsx1B(b, 1)such thatx1C. Also,B(b; 12) Rn \ C. So there existsx2B(b, 12)such thatx2C. Continuithis process, we obtain for every k 1 that B(b; 1k ) Rn \ C, and hence there exists xk B (b; 1k ) with xk C. So we haconstructed a sequence(xk)

k=1 inCthat converges to

b /C! This is a contradiction with the cannot escape property. Henour assumption that b was not an interior point ofRn \ Chas lead to a contradiction, henceRn \ Cis open, and C is closed.Note. What about the case when C=? In this case, C is closed, since Rn \ = Rn is open. However,is also open, becauwe can not find an a that is not an interior point. Moreover, sinceRn is a connected space, and Rn are the only subsethat are both open and closed.

Definition 3.9. For A Rn, we define the closureofA to becl(A) ={b Rn |there exists a sequence (xk)k=1 inA converging to b}

That is, the closure ofA is the set of all its limit points.

Remark 3.10. It is always the case that Acl(A). This is because for everyaA, the sequence (xk)k=1 defined by lettixk =a for every k N trivially converges toa. Hence every point inA is a limit point ofA. The inclusion A cl(A) holwith equality if and only ifA is closed. This is due to the cannot escape property. Observe the analogy between interior aclosure:

int(A)A, with equality if and only ifA is open. Acl(A), with equality if and only ifA is closed.

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Exercise 3.11. Prove that

For every A Rn we have int(Rn \ A) = Rn \ cl(A). For every B Rn we have cl(Rn \ B) = Rn \ int(B).

Exercise 3.12. Prove that

For every A Rn, int(A)is an open set, and moreover it is the largest open set contained in A. For every B Rn, cl(B)is a closed set, and moreover it is the smallest closed set containing B .

Definition 3.13. A set B Rn

is said to bebounded

when there exists r >0 such thatx r for all xB, or alternativesuch thatB B(0; r).Definition 3.14. A set K Rn is said to be compactif it is both closed and bounded.Theorem 3.15. LetK Rn be a compact set and let (xk)k=1 be a sequence inK. Then there exists a subsequence of(xk)kwhich converges to a limit in K.

Proof. SinceKis bounded then the sequence (xk)k=1 is also bounded, and hence by the Bolzano-Weierstrass theorem, the

exists a convergent subsequence (xk(p))p=1 converging to some

b Rn. However,Kis closed, and hence has the cannot escapproperty. Therefore bK.Definition 3.16. For A Rn we define the boundary ofA to be

bd(A) = cl(A)

\int(A).

Remark 3.17. We can alternatively writebd(A) = cl(A) cl(Rn \ A).

Proof. Observe that

bd(A) = cl(A) \ int(A)= cl(A) (Rn \ int(A))= cl(A) cl(Rn \ A).

Note. From the above it is clear that bd(A) = bd(Rn \ A) in all cases. Also, bd(A) is always a closed set, because the abogives the boundary as an intersection of two closed sets. It is easy to show that an intersection of two closed sets is againclosed set (this is immediate from the cannot escape property).

4 Continuous functions

Definition 4.1. LetA Rn be nonempty, and let f :A Rm. LetaA. We say that f is continuousata if for every >there exists some >0 such thatf(x) f(a)< for all xA such thatx a< . Moreover, we say f is continuousA if it is continuous at every point aA.Definition 4.2. Let A Rn be nonempty, and let f : A Rm. Leta A. We say thatf respects sequences in A whiconverge toa if whenever (xk)k=1 is a sequence in A such thatxk a, it follows that f(xk)f(a)in Rm.Proposition 4.3. Let A Rn be nonempty, f : A Rm be a function, anda A. Thenfrespects sequences in A whiconverge toa if and only iff is continuous ata.

Proof. First, assumef respects sequences in A which converge toa. Fix >0. We want to show that there exists >0 su

thatf(x) f(a)< for all xA such thatx a< . Assume, for the sake of contradiction, that no such exists. Thif= 1 we obtain x1A withx1 a< 1, but such thatf(x1) f(a) . If= 12 , we obtain x2A withx2 a0 satisfying the definition of continuity. So f is continuous ata.

The other direction is left as an exercise.

Definition 4.4. LetA Rn be nonempty, and let f :A Rm be a function. For every aA, we write

f(a) =

f(1)(a), f(2)(a), . . . , f (m)(a) Rm.

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Then we obtain functions f(1), f(2), . . . , f (m) which map A intoR. These are called the component functions off.

Proposition 4.5. Let f :ARm and its components be as above. FixaA. Then f is continuous at a if and only if eaf(i) is continuous ata.

Proof. fis continuous ataif and only iff respects sequences inA converging toaif and only if each of the f(1), f(2), . . . , f (

respect sequences in A converging toa if and only if each of the f(1), f(2), . . . , f (m) are continuous ata.

Note. The second equivalence above holds because for a sequence (xk)k=1 inA such thatxk a, we can write

f(xk) = f(1)(xk), . . . , f (m)(xk) Rm

and apply Proposition 2.5 to conclude that f(xk)f(a) if and only iff(i)(xk)f(i)(a)for all i (1im).Example 4.6. Consider the function f : R2 R2 defined by

f((s, t)) =

sts2+t2 if(s, t)= (0, 0)0 if(s, t) = (0, 0)

Observe that ( 1k , 1k )(0, 0)as k , butf(( 1k , 1k )) = 12 for all k1, hence fdoes not respect sequences converging to (0,

and so f is not continuous at(0, 0).

Remark 4.7. Linear combinations of continuous functions are again continuous. This is easily verified using sequences.

Remark 4.8. To check continuity in concrete examples, use the above remark, and use the one-dimensional continuity froMATH 147.

Example 4.9. Let f : R3 R2 be defined by

f((p, q, r)) = (cos(p + q+ r), sin(pqr)).

We verify the continuity offat a given point a= (p0, q0, r0) R3. We have f= (f(1), f(2)), where

f(1)(p, q, r) = cos(p + q+ r)

f(2)(p, q, r) = sin(pqr)

It suffices to check the continuity at a of each of these component functions. We know these functions are continuous froMATH 147, so fis continuous everywhere.

5 Continuity and compactness

In this section we will discuss the connections between compactness and continuity.

Definition 5.1. Let A Rn be nonempty, and f :A Rm be a function. We say that f is uniformly continuous (onA) if fevery >0 there exists >0 such thatf(x) f(a)< for all x, aA such thatx a< .Remark 5.2. Uniform continuity says that

(1) f is continuous on A (continuous at everyaA)(2) The choice ofin the -definition of continuity does not depend on the pointa in question.

Make sure to carefully compare the definitions of continuity and uniform continuity.

Example 5.3. Let A = (0, 1) (0, 1)R2. Let f : AR be defined by f((s, t)) = st . Thenf is continuous on A, but nuniformly continuous on A. This is easily verified.

Proposition 5.4. LetA Rn be compact, and letf :A Rn be a continuous function. Thenfis uniformly continuous on Proof. Fix >0. We want to find > 0 such thatf(x) f(a)< for all x, aA withx a< . Assume, for the saof contradiction, that no such exists. Pickk Nand use= 1k . Observe that this yields xk, akA such thatxkak< but such thatf(xk) f(ak) . We have therefore constructed two sequences, (xk)k=1 and (ak)k=1 in A. However, Acompact, so we can find indices

1k(1)< k(2)< . . . < k(p)< . . .

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such that (xk(p))p=1 converges to a limit x0 A (see Theorem 3.15). We claim that for the same selection of indices, we g

ak(p) x0. To see this, note that

0ak(p) x0 ak(p) xk(p) + xk(p) x0< 1k(p)+ xk(p) x0 .

The rightmost expression above converges to 0 as k , and hence by squeeze theorem, ak(p)x0, which verifies our claimNow, f is continuous at x0 and so it respects the convergence ofxk(p) x0 andak(p) x0. So therefore f(xk(p))f(x0) anf(ak(p))f(x0), and so

f(xk(p)) f(ak(p))

f(xk(p)) f(x0)+

f(ak(p)) f(x0)which converges to 0. However, this is a contradiction since we previously had f(xk(p)) f(ak(p)) . Therefore o

assumption that no such >0 existed yielded a contradiction. So such a >0 does exist, and this completes the proof.

Theorem 5.5 (Extreme Value Theorem). LetA Rn be nonempty and compact, and let f :A R be a continuous functioThen fattains a minimum and maximumvalue on A, that is, there exists a1, a2 A such that f(a1) f(x) f(a2) for xA.Remark. The Extreme Value Theorem is proved by combining two observations:

Continuous images of compact sets are compact Every compact subset ofR contains its infimum and supremum

Theorem 5.6. LetA Rn be nonempty and compact. Letf :A Rm be a continuous function. Then f(A) Rm is compac

Proof. We must showf(A)is closed and bounded. To showf(A)is closed, we will show B = cl(B)whereB = f(A). It sufficto check Bcl(B), since B cl(B) always holds. So fixbcl(B). By the definition ofcl(B) there exists a sequence (yk)kwithyk b. For each k1, we have ykB, hence there exists xkA such thatf(xk) = yk. Note that (xk)k=1 is a sequenin A (which is a compact set) and so we can find a subsequence (xk(p))

p=1 converging to some x0 A. f is continuous an

hence respects convergent sequences inA, sof(xk(p))f(x0), so we haveyk(p)f(x0)whileyk b, which impliesyk(p)By the uniqueness of the limit, f(x0) = b. Sobf(A) =B , proving that B is closed. Next, we give a sketch of the proof thf(A) is bounded. Suppose otherwise. Then we can find (yk)

k=1 inB such thatyk> k for all k1. For everyk1 pick

such that f(xk) =yk. Select a convergent subsequence (xk(p))p=1 and reach a contradiction in the same way as above (left

exercise). Hence B is closed and bounded, so by definition it is compact.

Remark 5.7. LetKR be a nonempty compact set. Then Kis bounded, so in particular Khas a least upper bound aa greatest lower bound , which lie withinK.

Proof. Suppose for instance that /K. Then for every m N we can find an element tmKsuch that tm( 1m , ), fotherwise 1m would be an upper bound for K, contradicting the definition of. We obtain a sequence (tm)m=1 in K atm by squeeze theorem. Since K is closed, we get K(by the cannot escape property). This shows K, and tproof that Kis done in the same way.Proof (of Theorem 5.5). Denotef(A)by K. Proposition 5.6 implies thatK is compact. Denote = infKand= sup K. Wknow that , K by Remark 5.7. SoK= f(A) implies there exists a1 A such that f(a1) = . Also, K= f(Aimplies there existsa2A such that f(a2) = . Now for every xA we obtain f(x)Kand hence

f(a1)f(x)f(a2).

6 Integrable functions

Definition 6.1. A closed rectangle inRn is a set of the formP = [a1, b1] . . . [an, bn]. For such sets P, we define the volumofP byvol(P) = (b1 a1)(b2 a2) . . . (bn an). We also define the diameters, diam(P) =b awhere b= (b1, b2, . . . , bn)ana= (a1, a2, . . . , an), anddiam(P) =b a.Definition 6.2. Let Pbe a closed rectangle in Rn. A division ofPis a collection of closed rectangles ={P1, . . . , P k} suthat

ki=1 Pi= P andint(Pi) int(Pj) = for i=j . For such a division, we define the mesh

= max1ik

diam(Pi)

.

Hence is small means that diam(Pi) is small for every 1 i k. Furthermore, for two divisions ={P1, . . . , Pand ={Q1, . . . , Ql}, we say that refines and write


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such thatQjPi.Remark 6.3.

(1) Let be a refinement of as above. Then we can write ={P1, . . . , P k}, and

={Q1,1, . . . , Q1,m1 , . . . , . . . , . . . , Qk,1, Qk,mk}

whereQi,1 . . . Qi,mi =Pi for every i (1ik).(2) Let Pbe a closed rectangle in Rn, and let 1, 2 be divisions ofP. Then we can find a division such that < 1,

(that is, a common refinement of both 1 and 2). Indeed, let 1 ={P1, P2, . . . , P k} and 2 ={P1, P2, . . . , P l}. P ={Pi Pj |1ik, 1jl, Pi Pj =}. Note: IfPi Pj =, thenPi Pj is itself a closed rectangle - check!Claim. is also a division ofP. For instance, applying one of De Morgans Laws, we have

1ik1jl

Pi Pj

=

ki=1

Pi

l

j=1

Pj

= P P =P.

The other property of a division (interiors of distinct rectangles being pairwise disjoint) is easily proved.

Definition 6.4. Let A be a nonempty subset ofRn, and f : AR be a function. We say that f is boundedwhen there exisC >0 such that|f(x)| C for all xA.

Note. Iffis bounded on the set A, it makes sense to talk about the supremum and infimum offon the set A as follows:

sup{f(x)| xA}= supA

f

inf{f(x)| xA}= infA

f.

Definition 6.5. LetPbe a closed rectangle. Let f :P R be a bounded function. Let ={P1, P2, . . . , P k} be a division P. Then we introduce what are known respectively as the upper and lower Darboux sums (offwith respect to ):

U(f, ) =k

i=1

vol(Pi) supPi

f

L(f, ) =

k

i=1 vol(Pi) infPi f.

Note. We always have L(f, )U(f, )simply because the infimum is always less than or equal to the supremum.Lemma 6.6. Let Pbe a closed rectangle in Rn, and let f : P R be bounded. Let, be divisions ofP such that < Then

L(f, )L(f, )U(f, )U(f, ).Proof. First, write

={P1, P2, . . . , P k} ={Q1,1, . . . , Q1,m1 , . . . , . . . , . . . , Qk,1, . . . , Qk,mk}

whereQi,1

. . .

Qi,mi =Pi for all 1

i

k. Then

U(f, ) =k

i=1

mi

j=1

vol(Qi,j) supQi,j

f

.

Observe that for every 1ik, 1jmi, we have

supQi,j

f= sup{f(x)| xQi,j} sup{f(x)| xPi}= supPi

f

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simply becauseQi,jPi. Substituting this into the above expression for the upper Darboux sum, we obtain

U(f, )k

i=1

mi

j=1

vol(Qi,j) supPi

f

=k

i=1

sup

Pi

fmi

j=1

vol(Qi,j)

=k

i=1

vol(Pi)

supPi

f

=U(f, ).

The proofs of the other inequalities are similar.

Proposition 6.7. LetPbe a closed rectangle in Rn andf :P Rbe a bounded function. Let 1, 2 be divisions ofP. ThL(f, 1)U(f, 2).

Proof. Take such that < 1 and < 2. Then by Lemma 6.6 we have that

L(f, 1)L(f, )U(f, )U(f, 2).Remark 6.8 (Review from MATH 147). LetSandTbe nonempty subsets ofR such that s

t whenever s

S, t

T. Th

Smust be bounded above and Tis bounded below and hence Shas a supremum, Thas an infimum, and hence

sinfTsup St.

Definition 6.9. LetPbe a closed rectangle in Rn, and let f :P R be a bounded function. Look atS={L(f, )| a division ofP}T ={U(f, )| a division ofP}.

Denote the lower and upper integrals off onP, respectively, by

Pf:= sup S and Pf:= infT.

From Remark 6.8, we have

sup S=

P

fP

f= infT.

WhenPf=

Pf, we say that f is integrable (onP) and we define the integral offto be

P

f :=

P

f=

P

f.

Note that integrability is only defined iff is bounded.

Proposition 6.10. LetPbe a closed rectangle in Rn

, and letfbe a bounded function. Then fis integrable if and only if fevery >0 there exists a division ofP such that U(f, ) L(f, )< .Proof. First, assume f is integrable. Let >0. Then

sup{L(f, )| a division ofP}=P

f=

P

f= inf{U(f, )| a division ofP}

So certainly, by definition of the supremum, we can find a division 1 ofP such that

L(f, 1)>

P

f

2

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Similarly, we can find a division2 ofP such that

U(f, 2)0, we obtain

U(f, ) = U(f, )

L(f, ) = L(f, ).

In the case where 0, we can find a finite family Q1, . . . , Qm of closrectangles inRn such that

(1) Q1 . . . QmC.

(2) vol(Q1) + . . . + vol(Qm)< .

Null sets are also said to have no content.

Example 8.2. Consider C={(t, t) R2 |0t1}, that is, the diagonal of the unit square. We claim that Cis a null set.Verification. Let >0. Choosek N such that

1

k < .

Look at the squaresQ1, Q2, . . . , Qk, where

Qi=

i 1

k ,

i

k

i 1k

,i

k

.

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where the Pi are contained in Q1 . . . Qm and the Pi are contained in K. Moreover we can arrange, by further refinementhat

diam(Pj ) < , 1jr.Note that for every1jr, we have thatx, yPj impliesx y< and hence

|f(x) f(y)|< 0.This in turn implies

supPj

(f) infPj

(f)0.

Properties of this division

()qi=1vol(Pi )mi=1vol(Qi)< 0.This is due to the fact that each Pi is contained within Qi, and none of the P

i overlap due to the disjointness conditi

in the definition of division.

() For every 1jr we have: supPj

f

infPj

f

= sup{|f(x) f(y)| | x, yPj} 0.

This is simply due to the uniform continuity of f on each Pj , which comes from the fact that for each Pj we ha

diam(Pj) < which will imply|f(x) f(y)|< 0 by construction of.Claim. The we found satisfies

U(f, ) L(f, )< .Proof. We calculate

U(f, ) =

qi=1

vol(Pi ) supPi

f+r

j=1

vol(Pj ) supPj

f

L(f, ) =

qi=1

vol(Pi ) infPi

f+r

j=1

vol(Pj) infPj

f

Hence we obtain

U(f, ) L(f, ) = S1+ S2

= qi=1

vol(Pi ) supPi

f infPi f + r

j=1vol(Pj) supP

jf infPj f .

We now estimate S1. For every 1iqwe havesupPi

f

infPi

f

So that, using the property () of, we get

S1q

i=1

vol(Pi ) ( )

= (

)

q

i=1 vol(Pi )

( ) 0.Now estimateS2. We get, using the property ()of:

S2r

j=1

vol(Pj ) 0

=0r

j=1

vol(Pj )

0vol(P).

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So now we have, by the construction of0, that

U(f, ) L(f, ) = S1+ S2( )0+ 0vol(P)=0[( ) + vol(P)]< .

This completes the proof of Theorem 8.5.

Corollary 8.6. IfP

Rn is a closed rectangle, and f :P

R is a continuous function, then f is integrable.

Proof. Note thatf is certainly bounded, by the Extreme Value Theorem, since the set P is compact. TakeG = P, B =Then by Theorem 8.5 we have that f is integrable on P.

Corollary 8.7. Let A be a compact subset ofRn, such that bd(A)is a null set. Let Pbe a closed rectangle such that P Define

f(x) =

1 ifxA0 ifxP\ A

Thenfis integrable.

Proof. Take B = bd(A), and G = P\ bd(A). By the hypotheses, B is a null set. Let us check that f is continuous on Observe thatbd(A) = cl(A) \ int(A) = A \ int(A), sinceA is closed. For every aint(A), we can find r >0 such that f(x) =for all xB (a; r). This implies thatf is continuous ata. For everyv P\ A, we can find r > 0 such that f(x) = 0 for xB(v; r) P. This clearly shows thatf is continuous at v also. Thereforef is continuous on G, and we apply Theorem 8to conclude that it is integrable on A.

Remark 8.8. The function ffrom Corollary 8.7 is called the characteristic functionofA, and it is denoted A. Note that wcan useA to define volume ofA, by

vol(A) :=

P

A.

9 Fubinis theorem

Remark 9.1. We let n = p + qwith p, q N. For A Rp, B Rq, we define the Cartesian product

A B={(a,b)| aA,bB} Rn

Every closed rectangleP = [a1, b1] . . . [an, bn] Rn

can be rewritten as P=M Nwhere

M= [a1, b1] . . . [ap, bp] RpN= [ap+1, bp+1] . . . [an, bn] Rq

For P =M Nas above, f :P R a function. For every vM can define partial function

fv :N R

fv( w) = f(v, w)

We considern = 2, p = q= 1.

P= [a1, b1] [a2, b2]M= [a1, b1], N= [a2, b2]

Theorem 9.2 (Fubini). LetP=M N, and f :P R as above. Suppose that:(i) f is integrable on P.

(ii) For allvM, the function fv :N R is integrable on N.

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DefineF :M R by putting, for all vM,F(v) =

N

fv.

ThenF is integrable on M and M

F =

P

f.

Note. Theorem 9.2 has benefit that it reduces an n-dimensional integral to some integral in p dimensions, that is,M

F (sinM Rp), and one in qdimensions.Notation 9.3. Pfis also denoted f(x)dx, wherexP. Alternatively we can write

P

f(x(1), x(2), . . . , x(n))dx(1)dx(2) . . . d x(n)

or, instead, break the vectorx into its first p components and its last qcomponents and write, where x= (v, w), the followinP

f(v, w)dv d w.

With this notation we have

F(v) =

N

fv =

N

fv( w)d w=

N

f(v, w)d w

whereby applying Fubinis formula we obtainM

F(v)dv=

M

N

f(v, w)d w

dv=

P

f(x) dx.

Fubinis theorem justifies the calculation of iterated integrals.

Example 9.4. TakeP = [0, 1] [0, 1] R2 and consider the function f :P R defined byf(v, w) = v+ w for all v, w[0,It was computed in Assignment 4 that

P

f= 1

We now calculate this using the Theorem of Fubini.P =MN, andM=N= [0, 1]. For every v[0, 1]we definefv : [0, 1]byfv( w) = f(v, w) = v+ w. By Fubini, we have

P

f= [0,1][0,1]

(v+ w) dv d w= [0,1]

[0,1]

(v+ w)d w dv= [0,1]

10

(v+ w)d w dv=

10

v+

1

2

dv

=1

2+

1

2= 1.

Notation 9.5. P =M Nas before. Consider the divisions

={M1, . . . , M r} Rp ={N1, . . . , N s} Rq

Then get division for P (inRn, n = p + q):

={Mi Nj|1ir, 1js}

In the case where n = 2, is what we would call a grid.Lemma 9.6. Let P = MN as before. Then for every division of P, one can find divisions and , for M and respectively, such that refines .Proof. Exercise (geometry). The basic idea is that we extend the horizontal and vertical lines to establish a grid. Look at tcoordinates of the corners of the rectangles within .

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Lemma 9.7. Let P =M N, and letf :P R be a function, and define F(v) = Nfv for everyvM, as in the statemeof Theorem 9.2. Let AMbe a closed rectangle. Let ={N1, . . . , N s} be a division ofN. Then

supA

(F)s

j=1

vol(Nj) supANj

(f)

infA

(F)s

j=1

vol(Nj) infANj

(f).

Proof (of Lemma 9.7). Verifications of the two inequalities are similar. We will prove the infimum inequality. Define

C=s

j=1

vol(Nj) infANj

(f).

Our goal is to show that C F(v) for all vA. We note thatC is a lower bound for{F(v)|vA} hence we want to shoinfA F= inf{F(v)| vA} C. So fixv0A, for which we prove that CF(v0). We note that by definition

F(v0) =

N

fv0L(fv0 , ) =s

j=1

vol(Nj) infNj

(fv0) =s

j=1

vol(Nj) inf{fv0( w)| wNj}.

Now we realize this is equal to

sj=1

vol(Nj) inf{f(v0, w)|wNj} s

j=1

vol(Nj) infANj

(f) = C.

So we obtainF(v0)Cfor all v0A. HenceCinf

A(f).

Remark 9.8. The case A = M in Lemma 9.7 contains in particular a verification of the fact that the functionFis bounded

Lemma 9.9. Let ={M1, . . . , M r} be a division ofM, and let ={N1, . . . , N s} be a division ofN. Let = be tdivision ofP.

={Mi Nj|1ir, 1js}.Then we claim U(F, )U(f, )and L(F, )L(f, ).

Proof. The verification of the two inequalities is similar. Let us prove the inequality involving the lower sums.

L(F, ) =

ri=1

vol(Mi) infMi

(F)

r

i=1

vol(Mi) s

j=1

vol(Nj) infMiNj

(f)

=r

i=1

sj=1

vol(Mi) vol(Nj) infMiNj

(f)

=r

i=1s

j=1vol(Mi Nj) inf

MiNj(f)

=L(f, ).

This completes the proof of the lemma.

Proof (of Theorem 9.2). fis integrable, so we can find a sequence (k)k=1of divisions ofP, such thatU(f, k)L(f, k)

as k . Recall that every k can be refined to a division of the form k = k k, wherek is a division ofM andka division ofN. For every k1, Lemma 9.9 yields

U(F, k)U(f, k)U(f, k)L(F, k)L(f, k)L(f, k).

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Hence we getU(F, k) L(F, k)U(f, k) L(f, k)0.

So by squeeze we getU(F, k) L(F, k)0.

HenceF is integrable by the criterion presented in Proposition 6.12. By the same proposition we haveM

F = limk

U(F, k).

However, we have U(f, k)U(F, k)L(f, k)L(f, k).Now we observe that U(f, k)

Pfand alsoL(f, k)

Pf. Also, U(F, k)

MF, andL(F, k)

MF. So ask

we obtain by squeeze again that P

fM

FP

f.

This proves that P

f=

M

F.

Hence the proof of Fubinis theorem is complete.

Remark 9.10. Theorem 9.2 had two hypotheses. They are both important (they do not imply each other). Hence there exist examples where one hypothesis holds but the other does not. So we can also give examples where the expressions

M

N

f(v, w)d w

dv

N

M

f(v, w)dv

d w.

exist, but are not the same, and hence the theorem will not hold in these cases.

Note. The homework assignments relevant to the midterm will be assignments 1 to 5.

10 Integration on more general domains

Example 10.1. Consider the functionf((s, t)) = 1 (s2 + t2). What is the domain off? The natural domain seems to D={(s, t) R2 |s2 + t2 1}. How do we integrate f onD? We dont know, because D is not a rectangle.Lemma 10.2 Let P P Rn be a closed rectangle, and let f : P R be a bounded function, such that f(x) = 0 for xP\ P. Then f is integrable on Pif and only if it is integrable on P. In this case, we have

P

f=

P

f.

Proof. IncorporateP into a division ={P1, P2, . . . , P r}of the rectangle P, say by allowing P =P1. We know from Rema7.6 thatf is integrable on P if and only if it is integrable on P1, . . . , P r. Furthermore, if this happens then

P f=r

i=1 Pif.

But for everyi (2r), we have that fi= 0 on int(Pi)becauseint(Pi)P \ P wherefis given to be 0. This implies thfor every 2ir we have that f is integrable on Pi with

Pi

f= 0.

So then f is integrable on P1, P2, . . . , P r if and only iff is integrable on P1 and if this happens then

ri=1

Pi

f=

P1

f=

P

f.

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Putting together these two things we obtain the lemma.

Definition and Proposition 10.3. LetA be a nonempty, bounded subset ofRn, and f :A R be a bounded function. Piany closed rectangleP Rn such thatP A, and extend f to f :P R by

f(x) =

f(x) xA0 otherwise

Then it makes sense to declare f is integrable onA if and only iffis integrable on P, and iffis integrable on A, then it maksense to define

A f :=

P f .

Proof. We have to prove the definition doesnt depend on our choice of the rectangleP. So suppose QRn is another closrectangle such that QA, and define f :Q R, by putting

f(x) =

f(x) xA0 xQ \ A

We must verify that f is integrable on P if and only iff is integrable on Q, and that if this happens, thenP

f=

Q

f .

Consider the closed rectangle P Q. We surely haveA P Q. Observe that f is defined to be 0 for all x P\ (P Qbecause P\ (P Q)P\ A. Similarly, f is defined to be 0 for all xQ \ (P Q), because Q \ (P Q)Q \ A. So f andcoincide on P Q. So then f is integrable on P if and only iff is integrable on P Q by Lemma 10.2. But this occurs if aonly iffis integrable onPQ, which occurs if and only iffis integrable on Q by Lemma 10.2 again, and if these are true thwe have

P

f=

PQ

f=

PQ

f=

Q

f .

Proposition 10.4. Let A be a bounded, nonempty subset ofRn. Suppose bd(A) = cl(A) \ int(A) is a null set. Letf :Abe a bounded and continuous function. Then f is integrable onA.

Proof. TakeP Rn to be a closed rectangle, such that int(P) cl(A) (in particular have that P A). Let f : P R defined by

f(x) =

f(x) xA0 x

P

\A

DecomposeP= int(A) (cl(A) \ int(A)) (P\cl(A)) = B G, whereB = cl(A) \ int(A)and G = int(A) (P\cl(A)). We mathe observation that B is null by hypothesis. Observe that fis continuous at every point in the interior ofA. (This is becauthere exists r > 0 such that B (x; r)A and f = f on B(x; r) and fis given to be continuous at x.) Also, f is continuous every xP\ cl(A). (This is because for suchx there exists r > 0 such that f = 0 on B(x; r).) Hence fis continuous on tgood set G. Apply Theorem 8.5 to conclude that f is integrable onA.

Example 10.5. Look again at Example 10.1. The function f :D R whereD={(s, t)|s2 + t2 1}

defined by f((s, t)) =

1 (s2 + t2). We know thatf is continuous on D (this is a quick exercise with sequences) and thbd(D)is a null set (i.e., the unit circle). Hence we can apply Proposition 10.4 to obtain that f is integrable onD. How muchthe integral? 23? We enclose DP := [1, 1] [1, 1]. Extendf to f :P R. Then

f((s, t)) =

1 (s2 + t2) s2 + t2 1

0 s2 + t2 >1

Then D

f=

[1,1][1,1]

f

Note that in this example the function f is actually continuous on P. In particular, the theorem of Fubini applies. For evefixeds[1, 1] we have

f((s, t)) =

1 (s2 + t2) t2 1 s2

0 t2 >1 s2

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So for fixed s:

f((s, t)) =

(1 s2) t2 1 s2 t 1 s2

0 t 1 s2

[1,1][1,1]

f=

11

+1s21s2

(

1 (s2 + t2))dt

ds

We denote r =

1 s2. Have to find rrr2 t2 dt= 12 r2 = 2 (1 s2)

So we get D

f=

11

[1,1]

f((s, t)) dt

ds =

2

3

Definition 10.6. LetA be a nonempty subset ofRn. Consider the functionf :A R,f(x) = 1for allxA. Iffis integrabon A, then we say that A has volume, and define vol(A) :=

A

f=A1.

Corollary 10.7. Let A be a nonempty subset ofRn. Ifbd(A)is a null set in Rn, thenA has volume.

Proof. The function fwhich is constantly equal to 1 is surely continuous, and by hypothesis has null boundary, so it certainis integrable on A by Proposition 10.4.

Example 10.8. LetS:={(s,t,z)|s2

+ t2

1 and 0z 1 (s2 + t2)}. Note that this again half of the closed unit ball R3 with the usual Euclidean metric. Then Sis compact and bd(S)is a null set in R3. (This is somewhat tedious to show, so wwill just accept this). So then Shas volume by Corollary 10.7. What is the volume ofS? EncloseS[1, 1] [1, 1] [0, 1

vol(S) =

S

1

=

[1,1][1,1][0,1]

S

So by Fubini, we get

vol(S) =

[1,1][1,1]

[0,1]

S((s,t,z))dz

ds dt

Hence the volume of the closed unit ball in R3 is 43 , which is 2 vol(S).

11 Partial derivatives

Motivation

We had a function f : D R, where D ={(s, t) R2 | s2 +t2 1}, defined by f((s, t)) = 1 s2 t2. Calculated wiFubini:

Df=

23 . We note that this functions value is constant on concentric circles around the origin. So another approa

to evaluate this integral is by using polar coordinates, r and , wherer is the distance from the origin and is the angle whiis made between the line from the origin to the point and the horizontal axis. Sos = r cos , and t = r sin . We ask if we cwrite

D f(s, t)ds dt=

[0,1][0,2] f((r cos , r sin ))dr d=

10 20

1 r2 drbut sadly the answer is no. What we have here is a change of variable. We want to have

D

f(s, t) ds dt=

[0,1][0,2]

f(((r, ))) J(r, )dr d.

That is, we need the correction factor J(r, ). What isJ(r, )? has a matrix of partial derivatives which is called the Jacobiamatrix (of at the point (r, )). The quantity J(r, ) is the absolute value of the determinant of the Jacobian matrix.

Midterm tomorrow. DWE 3516, 4:30pm. Statements of definitions and theorems. Two proofs: one from Sections 1-5, aone from Sections 6-8. Also, three problems similar to homework problems from Assignments 1-5. Emphasis on calculatio

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with Darboux sums.

Definition 11.1. Let A Rn, and letaint(A). Also let f :A R. Pick a vector v Rn. If the limit

limt0

f(a + tv) f(a)t

exists, then say that f has directional derivativeata, in directionv. When this happens, the limit above is denoted

(vf)(a).

Remark 11.2. Make the same notations as in Definition 11.1.

(1) Ifv= 0, then (vf)(a) exists and is 0.

(2) Suppose v= 0 and hencev = 0. Since a is interior, there exists r > 0 such that B(a; r) A. We can define rv , + rv

R, by(t) = f(a + tv). We claim that this makes sense. Indeed|t|< rv implies(a + tv) a=tv

|t| v< r. This shows that (a + tv)B(a; r)A. Hencefis defined at the pointa + tv. The quantity(vf)(a) is thgiven by

limt0t=0

f(a + tv) f(a)t

= limt0t=0

(t) (0)t

= (0).

We note that (vf)(a) is the derivative of the partial function obtained by cutting in direction v.

Proposition 11.3. Let A

Rn, anda

int(A). Definef : A

R, v

Rn, v=0. Suppose that(vf)(a) exists. Then f

every R have that(vf)(a) exists, and (vf)(a) = (vf)(a). (This is called homogeneity).Proof. If = 0 then the homogeneity formula is clear, comes to 0 = 0. So assume = 0. Denotev= w. We look at

limt0t=0

f(a + t w) f(a)t

= limt0t=0

f(a + (t)v) f(a)t

= lims0s=0

f(a + sv) f(a)s

= (vf)(a).

This completes the proof.

Definition 11.4. Let A Rn, andaint(A), and define f :A R. For every 1in consider the vector

ei = (0, . . . , 0, 1, 0, . . . , 0)

If(eif)(a)exists, then we call it the ith partial derivative off ata, denoted (if)(a).

Example 11.5. Define the function f : R2 R by

f((s, t)) =

s2ts2+t2 (s, t)= (0, 0)0 (s, t) = (0, 0)

Thisf is continuous at everya R2 includinga= (0, 0). Leta= (0, 0)and pickv= (1, 2). Does (vf)(a) exist? If so, whatit?

Note thata + tv= (0, 0) + t(1, 2) = (t, 2t). Since t= 0, f((t, 2t)) = t2(2t)t2+(2t)2 = 2t3

5t2 = 2t5. So

limt

0

t=0

f(a + tv) f(a)t

= limt

0

t=0

2t5 0

t =

2

5.

So we see that (vf)(a) exists and is equal to 25 . What about (1f)(a)and (2f)(a)? We get

limt0t=0

f(a + te1) f(a)t

= 0.

So (1f)(a)and (2f)(a)exist, and we have (1f)(a) = (2f)(a) = 0. Observe that(vf)(a) = 25= 0 = (1f)(a) + 2(2f)(a)

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12 C1 functionsDefinition 12.1. LetA be a nonempty, open subset ofRn, and letf :A Rbe a function. Let v Rn. Suppose that(vf)(exists for everyaA. Then we get a new function vf :AR, such that every aA is mapped to the value (vf)(a). Tfunctionvf is called the directional derivative off, in directionv. We also denote, in the case wherev = ei (1in), tith partial derivativeoff asif=eif.

Definition 12.2. LetA be a nonempty subset ofRn. A functionf :A Ris said to be aC1 functionwhen it has the followiproperties:

fis continuous

fhas partial derivatives at every vA The new functions if :A R are all continuous on A

The collection of allC1 functions from A to R is denoted asC1(A, R).Note. One has a hierarchy ofCk functions.C0(A, R) ={f : AR|f is continuous on A}. Will also haveC2(A, R), that the set of all functions that have continuous partial derivatives of orders 1 and 2.

Theorem 12.3. LetA be a nonempty, open subset ofRn. Letf C1(A, R). For everyv= (v(1), v(2), . . . , v(n)). The directionderivativevfexists, and moreover

vf=v(1)(1f) + . . . + v

(n)(nf).

Lemma 12.4. Let A be a nonempty, open subset ofRn. Let f C

1(A, R) and leta

A. Let r >0 be such thatB (a; r)

Fix an i (1in), and suppose x, yB(a; r) such that they only differ in the ith component. Then there existsbB(a;such that

f(y) f(x) = (y(i) x(i)) (if)(b)Proof. The case x(i) = y (i) is trivial, because we have x = y, and so we get 0 = 0. Assume x(i) = y(i), say for example thx(i) < y(i). Note that (x(1), . . . , x(i1), s , x(i+1), . . . , x(n)) B(a; r) for every s [x(i), y(i)]. (Exercise; due to the fact thopen balls are convex). Hence we can define : [x(i), y(i)]R by putting (s) = f(x(1), . . . , x(i1), s , x(i+1), . . . , x(n)) for s[x(i), y(i)]. Then is continuous on [x(i), y(i)], and differentiable on (x(i), y(i)), moreover, for every s (x(i), y(i)) we ha(s) = (if)(b)where b= (x(1), . . . , x(i1), s , x(i+1), . . . , x(n)). We apply the Mean Value Theorem to to conclude that theexists ans in (x(i), y(i))such that

(y(i)) (x(i))y(i) x(i) =

(s).

Verification. We will verify that the functionis differentiable on(x(i)

, y(i)

). Let s(x(i)

, y(i)

), and denote(x(1)

, . . . , x(i1)

, sb. Look at(s + h) (s)

h =

f(b + hei) f(b)h

(if)(b), as h0 (h= 0).Hence(s) exists and moreover

(s) = (if)(b)

whereb = (x(1), . . . , x(i1), s , x(i+1), . . . , x(n)). We can apply the Mean Value Theorem to from MATH 147. So there exiss0(x(i), y(i))such that

(y(i)) (x(i))y(i) x(i) =

(s0)

However(x(i)) = f(x)and (y(i)) = f(y)and (s0) = (if)(b), for bas calculated above.

Lemma 12.5. Let A be a nonempty, open subset ofRn, and let f C1(A, R). FixaA. Then have:

limxax=a

f(x) f(a) ni=1(x(i) a(i)) (if)(a)x a = 0.

Proof. Let >0. We want to find a >0 such thatB (a; )A and such that for everyxB(a; ) \ {a} we havef(x) f(a) n

i=1(x(i) a(i)) (if)(a)

x a < .

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So

limt0t=0

f(a + tv) f(a)t

must exist and be equal to L. Given >0, use the limit provided by Lemma 12.5 for

1 + vto get 0 > 0 such that B (a, 0)A and for everyx= a in B (a, 0) we have

|f(x) f(a) ni=1(x(i) a(i)) (if)(a)|x a < 1 + v . (1)Make also = 01+v . Claim. For every t= 0 with|t|< we havef(a + tv) f(a)t L

< .Verification of claim. Let t= 0be such that|t|< . Putx= a + tv. Then

x a=t v=|t| v< v= 01 + v v< 0.

Therefore x

B(a; 0

) hence (1) applies to x. So replacex = a+ tv in (1). We have

x

a

=|

t|

v

, and x(i)

a(i)

(a(i) + tv(i)) a(i) =tv(i), for all i (1in). Hencen

i=1

(x(i) a(i)) (if)(a) =n

i=1

tv(i) (if)(a) = tL.

So (1) becomes|f(a + tv) f(a) tL|

|t| v <

1 + vwhich implies

f(a + tv) f(a)

t L

< v1 + v < .

Definition 12.6. LetA be a nonempty, open subset ofRn

, and let f C1

(A, R). For everyaA, the gradient vector offa by(grad f)(a) = (f)(a) =

(1f)(a), . . . , (nf)(a)

.

Remark 12.7.

(1) Theorem 12.3 gives formula

(vf)(a) =n

i=1

v(i) (if)(a) =

v, (f)(a)

(2) So(f)(a) says something about how to do a linear approximation offarounda.Remark 12.8 (linear algebra/geometry). Supposep Rm. How do we give a formula for a hyperplane H Rm going throup? One possibility is that we start from p and pick several important vectors starting from p and going in the directions of t

hyperplane, that is:

H=

p +m1i=1

ivi|1 . . . m1 R and v1 . . . vm1 are linearly independent

You take the orthogonal complement of the space generated by v1, . . . , vm1. This complement will have dimension 1. In generathe orthogonal complement of a subspace of dimensionk in a space of dimension m has dimension m k.Another possibility: give the normal vector w and write

H=

y Rm |(y p) w

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where(y p) w means by definition thaty p, w= 0. Hence

H=

y Rm | y p, w= 0

.

Remark 12.9. LetA be a nonempty, open subset ofRn, and let fC1(A, R). Consider the graph off

=

(a, t)| aA, t= f(a) Rn+1.

Consider a point p = (a, f(a)) . Then has a tangent hyperplane at p, which can be calculated by using the gradien(f)(a). We look at the points at = a + tv for sufficiently small t. Also define pt = (at, f(at)). Look at

limt0t=0

1t

(pt p) .

Note first thatpt p= (tv, f(a + tv) f(a))

and hence

limt0t=0

1

t(tv, f(a + tv) f(a)) = (v, (vf)(a)).

What we can conclude so far is that for every vRn, the vector (v, (vf)(a))Rn+1 gives a tangent direction to at p. Wnow ask the question: what is the normal direction to at the point p? We want to find a vector w Rn+1 such that worthogonal to every vector (v, (vf)(a))for all v Rn. However we have

(v, (vf)(a)) =

v(1), . . . , v(n),n

i=1

v(i) (if)(a) .By inspection we see that

w= ((1f)(a), . . . , (nf)(a), 1)is such a vector. So we conclude that the normal direction for at p= (a, f(a))is given by

w=

(f)(a), 1

.

Example 12.10. Consider the surface

E=

(x,y,z) R3 | x

2

4 +

y2

91, z =

1 x

2

4 y

2

9 This set represents the surface of an ellipsoid. Fix p= (1, 2, 116 ). We want to find the normal direction to p. Definef :Awhere

A=

(x, y) R2 | x2

4 +

y2

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Letf , g C1(A, R). Then f+ g C1(A, R)and (v(f+ g)) = (vf) + (vg) for all v Rn. Letf C1(A, R) and R. Then f C1(A, R)and (vf) = (vf) for allv Rn. Letf , g C1(A, R). Then f g C1(A, R)and (vf g) = (vf) g+ f (vg) for all v Rn.

We say thatC1(A, R)is an algebra of functions.Proof. We will check the last equality. Fix v Rn, andaA for which we prove that

(vf g)(a) = (vf)(a) g(a) + f(a) (vg)(a).

Pick r >0 such that B (a; r)A (we can do this becauseA is open). Define : ( r1+v , r1+v ) R by

(t) = (fg)(a + tv) = f(a + tv) g(a + tv) = 1(t)2(t).

with 1, 2 : ( r1+v , r1+v ) R given by 1(t) = f(a+ tv) and 2(t) = g(a+ tv). Notice that 1(0) = (vf)(a) a2(0) = (vg)(a), and so

(0) = 1(0) 2(0) + 1(0) 2(0) = (vf)(a) g(a) + f(a) (vg)(a).

Note. Lemma 12.4 is a form of the Mean Value Theorem in the direction ofei involving (if). This can be extended togeneral directionv involving (vf).

Remark 12.12. If we have two vectors in Rn, how do I describe the line segment which connects the two vectors? We l

t[0, 1] and define it by the formula (1 t)a + tb. This is called a convex combination. We will talk about extending the MeValue Theorem to the situation where we have a domain that is convex.

Denote b a=v, and note that b= a + v= a + 1 v, anda= a + 0 = a + 0 v. The line segment fromato bis then the convcombination

Co(a,b) ={a + tv|t[0, 1]}={(1 t)a + tb|t[0, 1]}.Definition 12.13. Let A Rn. The setA is said to be convex if for everya,bA we have Co(a,b)A.Proposition 12.14(MVT forC1 functions). Appeared already (in a special case) in Lemma 12.4. LetA be a nonempty opsubset ofRn. Letf C1(A, R). Supposea,bA are such that Co(a,b)A. Then there existscCo(a,b) withc=a,bsuthat

f(b) f(a) =b a, (f)(c).Denotev= b

a. Applying Theorem 12.3, we can obtain another form of this mean value formula:

f(b) f(a) =b a v

, (f)(c)=n

i=1

v(i)(if)(c) = (vf)(c)

wherec is an intermediate point in Co(a,b).

Note. Ifv is proportional to ei for some i (1 i n), then Proposition 12.14 becomes Lemma 12.4 proved earlier in thsection.

Sketch of proof. The proof of the general case in Proposition 12.14 is similar to that of Lemma 12.4. Define the auxiliafunction : [0, 1]R by (t) =f((1 t)a +tb). Due to hypotheses, f is aC1 function, so we obtain that is continuous [0, 1]and differentiable on (0, 1). So apply the Mean Value Theorem from MATH 147 to obtain an intermediate point t0(0,such that

(1) (0)1 0 =

(t0).

See that(0) =f(a), (1) =f(b), and (t0)becomes (vf)(c)for somec= (1 t0)a + t0b, so that we obtain the theorem.

13 Chain rule

Definition 13.1. Let I R be an open interval (that is, I = (a, b) or I = (, b) or I = (a, ) or I = R). A functi: I Rn has n components (1), (2), . . . , (n) :I R, and

(t) = ((1)(t), . . . , (n)(t)) Rn

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If every (i) is continuous on I, then we say that is a path in Rn. If every (i) is differentiable on I, we say that isdifferentiable path inRn.

Definition 13.2. Let : I Rn be a differentiable path in Rn. For every tI, the vector(t) = (((1))(t), . . . , ((n))(t)) Rn

is called the velocity vector ofatt.

Proposition 13.3 (chain rule). Let A be a nonempty open subset ofRn, and let f C1(A, R). Suppose : I Rn isdifferentiable path so that (t)

A for all t

I. Consider the composed function : I

R defined by = f

, that

(t) = f((t))for every tI. Then is differentiable and we have

(t) =n

i=1

(if)((t)) ((i))(t).

Remark 13.4. Denotea= (t)and v= (t). Then the chain rule formula above says that

(t) =(f)(a), v.

That is,(f )(t) =(f)((t)), (t).

Note the analogy with the chain rule from MATH 147. In fact, the chain rule in Proposition 13.3 has the single-variable charule as a special case.

Lemma 13.5. Let f C1(A, R)as above. Let KA be convex and compact. Then there exists c >0 such that|f(x) f(y)| cx y.

Any suchc is called a Lipschitz constant forf.

Proof of Proposition 13.3. Fixt0I. Denotea= (t0)and v= (t0). Observe that the right-hand side of the chain rule(f)(a), v= (vf)(a).

by Theorem 12.3. So we must prove that

limh0h

=0

(t0+ h) (t0)h

= (vf)(a).

Pick r >0 such that B (a; r)A. Fixc >0 such that whenever x, y B(x; r2), we have|f(x) f(y)| cx y

Fixl >0 such that(t0 l, t0+ l) is in the domain of the path. Will work with h= 0such that|h|< l and moreover

|h|< r2(1 + v)

This givesa + hvB(a; r2). Indeed, we will have

a + hv a=|h| v< r v2(1 + v) 0 suthatB (a; r)A. Let : ( r

2, r

2) R be defined by

(t) = f(a + tei+ tej) f(a + tei) f(a + tej) + f(a)

Take a sequence (tn)n=1 in (0,

r2

) such thattn

0. Then we have

(1)

limn

(tn)

t 2n= (i(jf))(a)

(2)

limn

(tn)

t 2n= (j(if))(a)

Observe thata + tei+ tejB(a; r), because

(a + tei+ tej) a=t(ei+ ej)=|t| ei+ ej< r

2 2 = r.

Proof of Theorem 14.4. Ifi = j then the statement is clear. Otherwise, we simply apply Lemma 14.5 and get

(i(jf))(a) = limn

(tn)

t 2n= (j(if))(a)

Proof of Lemma 14.5. We will prove the following claim: Given t(0, r2

) we can find 0 < , < t such that

(t)

t2 = (i(jf))(a + ei+ ej)

The statement of the lemma will follow once we prove the claim, because we have tn

0+ in the lemma. For every m

1 t

claim gives0 < m, m< tm such that(tm)

t 2m= (i(jf))(a + mei+ mej)

Asn we have m, m0 by squeeze. Hence as n we have

a + mei+ mej a

and hence(i(jf))(a + mei+ mej)(i(jf))(a)

by the continuity of(i(jf)). So we have obtained that

(tm)

t 2m (i(jf))(a)

The limit (2) in Lemma 14.5 is obtained in exactly the same way, by reversing the roles of i and j. We need only prove texistence of such , now.

Invalid proof. Write(t)

t2 =

1

t

f(a + tei+ tej) f(a + tei)

t f(a + tej) f(a)

t

The MVT gives a (0< < t) such that

f(a + tej) f(a)t

= (jf)(a + ej)

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Repeat to getf(a + tei+ tej) f(a + tei)

t = (jf)(a + tei+ ej)

Hence(t)

t2 =

1

t [(jf)(a + tei+ ej) (jf)(a + ej)]

Apply again the MVT for jf, to get 0 < < t such that

(t)

t2 = (i(jf))(a + ej+ ei)

However the problem is that the two may not be the same. We will see the valid proof on Friday.

Real proof. Define another auxiliary function : (2, 2) Rby

(q) = 1

t2[f(a + tei+ qtej) f(a + tei) f(a + qtej) + f(a)] .

Note. (1)is (t)t2 . Also, (0) = 0.

We now verify that is differentiable and calculate . Calculate separately the two derivatives corresponding to the two term

d

dq(f(a + tei+ qtej)) =

d

dq(f((q)))

where(q) = a + tei + qtej . So(q) = (. . . , a(j) + q t , . . .), where all the other components do not depend onq. Hence(q) = tefor all q(2, 2). Now the chain rule says that

(f )(q) =n

l=1

(lf)(a + tei+ qtej) ()(l)(q) = (jf)(a + tei+ qtej) t

Similarly,d

dq(f(a + qtej)) = t (jf)(a + qtej)

So is differentiable with

(q) = 1

t2[t (jf)(a + tei+ qtej) t (jf)(a + qtej)] = (jf)(a + tei+ qtej) (jf)(a + qtej)

t

Apply MVT to . Obtain some q0 (0, 1) such that (t)t2 = (1)(0)10 = (q0) = (jf)(a+tei+qtej)(jf)(a+q0tej)t . In the laexpression apply MVT for (jf), get some 0 < < t such that

(jf)(a + tei+ q0tej) (jf)(a + q0tej)t

= (i(jf))(a + ei+ q0tej)

Letq0t= and we are done with the proof of Theorem 14.4.

Note. One writes ijf instead ofi(jf). For i = j , one writes 2i f rather than iif.

Definition 14.6. For a function f C2(A, R),aA, the matrix

(Hf)(a) = ( 21 f)(a) (12f)(a) . . . (1mf)(a)

...(n1f)(a) . . . . . . (

2nf)(a)

is called the Hessian matrix off ata. (Hf)(a)Mn(R). The ij entry of(Hf)(a) is (ijf)(a). Theorem 14.4 says that tHessian matrix ata is always a symmetric matrix.

Remark 14.7 (linear algebra review). Let H = Ht Mn(R). Then the eigenvalues 1, . . . , n of H are all in R. Wh1, . . . , n > 0 we say that H is positive definite. WhenHis positive definite then the corresponding linear transformatioTH : Rn Rn satisfies

TH(v), v> 0for allv Rn withv= 0.

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Remark 14.8 (review from MATH 147; second derivative test). Suppose : (a, b)R is a twice differentiable function. Lt(a, b) such that(t) = 0and such that (t)> 0. Thent is a local minimum.Definition 14.9. LetA Rn be nonempty and open, and let f C1(A, R). LetaA. If(f)(a) = 0, then we say thata isstationary point forf.

Remark 14.10. Let f C1(A, R) anda A. Ifa is a point of local minimum or a point of local maximum for f, thenaa stationary point. Conversely, given thata A is stationary, are there methods to verify if it is a local minimum or a locmaximum?

Theorem 14.11. Let A

Rn be nonempty and open. Let f

C2(A, R). Leta

A be a stationary point for f. If(Hf)(a)

positive definite, thenais a point of local minimum for f.

Lemma 14.12. Let f C2(A, R), and a A, such that (Hf)(a) is positive definite. Then there exists r > 0 such thB(a; r)A and such that (Hf)(x)is positive definite at all xB(a; r).Comment. We will not prove this lemma. Note that this is a fact about stability of eigenvaluesunder small perturbations ofsymmetric matrix. The entries of the Hessian matrix are the second-order partial derivatives at a. But these are continuous, if we move froma tox by a small amount, so the entries of the Hessian matrix at x will be close to the entries of the Hessimatrix ata. The eigenvalues depend continuously on the entries of the matrix.

Lemma 14.13. Let f C2(A, R), and a A be stationary. Let r > 0 be the radius such that B(a; r) A. Suppose thbB(a; r2), and denote b a=v. Then there exists t0(0, 1)and there existscCo(a,b) such that

f(b)

f(a) = t0

T(Hf)(c)(v), v

.

Proof. Define : (2, 2) Rby (t) = f(a + tb) for all t(2, 2). So (0) = a and (1) =f(a + v) = f(b). So by the charule we have

(t) =n

i=1

(if)(a + tv) v(i)

In particular, noting thatais stationary, we have

(0) =n

i=1

(if)(a) v(i) = 0.

For1in definei : (2, 2) R by i(t) = (if)(a + tv)for all t(2, 2). Becauseif is inC1(A, R)we can do a similcalculation fori to get a formula for

i(t). For every fixed value ofi,

i(t) =n

j=1

(jif)(a + tv) v(j)

by the chain rule. Hence

(t) =n

i=1

i(t) v(i) =n

i=1

n

j=1

(jif)(a + tv) v(j) v(i) =T(Hf)(a+tv)(v), v

So for every t(2, 2)we have(t) =T(Hf)(a+tv)(v), v

Finally, we calculate

f(b) f(a) = (1) (0)1 0 =(t0)

for somet0(0, 1)by the Mean Value Theorem on . Then

(t0) = (t0) (0) =t0 (t0) (0)

t0 0 =t0 (t1) = t0T(Hf)(c)(v), v

for somet1(0, t0) by the Mean Value Theorem on .Proof (of Theorem 14.11). We pick some r >0 as in Lemma 14.12. Then for every b= ain B (a; r2)we have

f(b) f(a) = t0T(Hf)(c)(v), v

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wherecCo(a,b)B(a; r2)so (Hf)(c) is positive definite. Hence f(b)> f(a), so we are done.Remark 14.14. Theorem 14.11 covered the situation of a local minimum (appearing when(f)(a) = 0, and the Hessian matris positive definite). We have an analogous statement giving thata is a local maximum when (f)(a) = 0 and the Hessimatrix is negative definite. That is,(Hf)(a) is positive definite.That is, the case of local maxima follows from the previous theorem by replacing fwithf.

15 Functions inC1(A,Rm) (INCOMPLETE)Definition 15.1. Let A Rn be open, and f : A Rm be a function where m 1. Write f = (f(1), . . . , f (m)), wif(i) :A R for 1im. If each of the f(i) is inC1(A, R)then we say that f is inC1(A, Rm).Remark 15.2. Similar definitions hold for every possible integer p = 1, 2, 3, . . . , , namely thatCp(A, Rm)is the set of functiowith components inCp(A, R).Definition 15.3. Let A Rn be open, and f C1(A, Rm). For everyaA, the matrix

(Jf)(a) =

(f(1))(a)(f(2))(a)

...(f(m))(a)

=

(1f

(1))(a) (2f(1))(a) . . . (nf(1))(a)

...(1f

(m))(a) (2f(m))(a) . . . (nf

(m))(a)

Mmn(R)

is called the Jacobian matrix off ata. This is the m n counterpart for the concept of derivative off ata.Proposition 15.4 (chain rule). LetARn and BRm be non-empty open sets. Letf C1(A, Rn) be such that f(x)for all xA. Suppose we also have g C1(B, Rp). Then we can form the composed function h= (g f) :ARp defined h(x) = g(f(x))for all xA. Then h C1(A, Rp)and we have

(Jh)(x) = (Jg)(f(x)) (Jf)(x)where therepresents matrix multiplication in this case.Remark 15.5. Equation above is usually written in terms of the Jacobian matrices, but can be rewritten as

(jh(k))(x) =

mi=1

(ig(k))(f(x)) (jf(i))(x)

Note also that for every 1kp we have thath(k)(x) = g(k)(f(x))by definition ofh.Remark 15.6 (review from MATH 147). Linear approximation in Calculus 1: for xa, we have

f(x) = f(a) + (x a)f(a) +higher order termsDenote

D(x) := f(x) f(a) (x a)f(a)Then

limxax=a

D(x)

x a = limxax=a

f(x) f(a)x a f

(a) = 0

We now move to Calculus 3. What is the analogue ofD(x) for f

C1(A, Rm)? So let A be a nonempty supset ofRn and l

f C1(A, Rm), and letaA. DefineD : A Rm byD(x) = f(x) f(a) T(Jf)(a)(x a)

where T(Jf)(a) :Rn Rm is the linear transformation associated to (Jf)(a). In general ifM = [ij ] (1 i m, 1 j thenTM : Rn Rm is defined by

TM(y) = M

y(1)

y(2)

y(3)

...y(n)

.

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Proposition 15.7. For D : A Rm defined as above, we have

limxax=a

1

x aD(x) =0

The above is the linear approximation formula forC1 functions with values inRm. This proposition was actually already provwhen we talked aboutC1(A, R). This was Lemma 12.5. To see that this implies Proposition 15.7, we write the componentsD(x):

D(x)(i) =f(i)(x) f(i)(a) (f(i))(a), x aThen apply Lemma 12.5 to g = f(i) and obtain

limxax=a

D(i)(x)

x a = 0

So the limits of all the components ofD(x)are all zero, hence D(x) 0 as required.Remark 15.8. ForA Rn andf C1(A, Rm), the role of the derivative is now played by the Jacobian, (Jf)(x) Mmn(R)Rmn. Taking the derivative mutliplies the dimension of the target space by n.

16 Inverse function theorem

We will look at open subsets A Rn, and f C1(A, Rm).Remark 16.1. Suppose X, Y are nonempty sets andf :XY is a function. We say f is one-to-one or injective if whenevx1=x2 are elements ofX, we have f(x1)=f(x2)in Y. Also, we say f is onto or surjectivewhen f(X) = Y where

f(X) ={yY| there exists xXwithf(x) = y}

Finally, we say f is bijective when it is both injective and surjective. If f is bijective, then we define the inverse functif1 :YXby setting f1(y) to be the unique xX with f(x) = y.Remark 16.2 (review from MATH 146). IfM Mnn(R), we say M is invertible if

The determinantdet M= 0

There exists N Mnn(R)such that M N=In= N M The transformation TM : Rn Rn is a bijective mapping

Theorem 16.3(inverse function theorem). LetA Rn be nonempty and open, and let f C1(A, Rn). LetaA be such th(Jf)(a) is invertible. Denote f(a) = b Rn. Then there exist open sets U, V Rn such that

(i) aU, bV f(A)(ii) f mapsU ontoVbijectively, that is, f(U) = V andf is one-to-one on U

(iii) The functiong : V Uwhich inverts f is aC1 function and moreover (Jg)(b) =

(Jf)(a)1

.

We will show two aspects:

Why does the relationship between the Jacobians in (iii) hold? Why is fone-to-one on a neighbourhood ofa?

Remark 16.4. Assume (i) and (ii) hold in Theorem 16.3, also that g C1(V, Rn). Then the relationship between the Jacobiain (iii) follows by an immediate application of the chain rule. Indeed, let h : U Rn be defined byh(x) = g(f(x))for all xThen the chain rule (Proposition 15.4) applies toh. Why? Well,h = g f0 wheref0 : U Rn is the restriction off toU. Thf0 C1(U, Rn), f0(U)V, also we have g C1(V, Rn). Hence h = g f0 makes sense and isC1. Chain rule (Proposition 15.gives

(Jh)(x) = (Jg)(f(x)) (Jf)(x) xU.

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But g is the inverse for f on U, so actuallyh(x) = x for all xU, that is, h = (h(1), . . . , h(m))where

h(i)(x) = x(i)

Hence(jh(i))(x)is 1 ifi = j and otherwise 0, and hence (Jh)(x) = In, the identity matrix ofMnn. So the chain rule formu

becomesIn= (Jg)(f(x)) (Jf)(x) xU

and hence

(Jg)(b) =

(Jf)(a)

1

where b= f(x) and a= x as needed.

Now, why isfone-to-one on a (sufficiently small) neighbourhood ofa? The point is that invertibility of an n nmatrix isstaunder small perturbations of the entries.

Lemma 16.5. Suppose M= [ij]Mnn(R)such that M is invertible. There exists some >0 with the following propertifN= [ij ] is|ij ij |< for all 1i, jn then N is invertible as well.Proof. Denote| det M| = . Note that > 0 because M is invertible. Recall from linear algebra that the determinant of n n matrix is a polynomial function Pn of the entries of the matrix

T = [tij ] = det(T) = Pn(t11, t12, . . . , tnn)

For example ifn = 2then

T =t11 t12

t21 t22

= P2(t11, t12, t21, t22) = t11t22 t12t21

Write the continuity ofPn at(11, 12, . . . , nn). We obtain that there exists some >0 such that if

(11, . . . , nn) (11, . . . , nn)<

then we obtain|Pn(11, . . . , nn) Pn(11, . . . , nn)|<

2.

We take = n . IfN= [ij] has|ij ij |< then we obtain

(11, . . . , nn)

(11, . . . , nn)

=

n

i,j=1(ij ij)2

n22 =

n= .

But then we get

|Pn(11, . . . , nn) Pn(11, . . . , nn)|< 2

= | det N det M|< 2

.

Finally,

=| det M| | det N| + | det M det N|< 2

+ | det N|Hence

| det N|> 2

>0

which shows Nis invertible.

Proposition 16.6. Let A

Rn be an open set, a

A, and let f C

1(A, Rn) such that (Jf)(a) is an invertible matrix. Ththere exists some r >0 such that B (a; r)A and such that f is injective onB (a; r)as well.Proof. Denote(Jf)(a) = M= [ij ]. Mis invertible, so by Lemma 16.5 we have >0 such that ifN= [ij ]with|ij ij| 0 such that B(a; rij) A and|jf(i)(x)jf(i)(a)| < for all 1 i, j n and for all x B(a; rij). If we lr = maxi,j{rij}, we claim this will work. Assume, for the sake of contradiction, that fis not injective on B(a; r). Then theexist two distinct points p, qB(a; r) such that f(p) =f(q). Since open balls are convex, the line segment Co(p, q) lies withthe ball. We apply the Mean Value Theorem (Proposition 12.14) to all the component functionsf(i) (1in) which are inC1(A, R) and obtain a point ciCo(p, q) such that

f(i)(q) f(i)(p) =(f(i))(ci), q p.

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Theorem 17.4 (implicit function theorem). LetARn be open, and let a A. Let f C 1(A, Rm) where m < n. Denod:= n m. Denotef(a) =: band consider the level set

L={xA|f(x) = b}

Supposea is a regular point forfwith respect to the directions 1, 2, 3, . . . , m. (This is to simplify the writing; the result wouhold for any combination of directions j1, j2, . . . , jm). Then there is a radius r > 0 such that LB(a; r) is a parametrizC1 manifold of dimension d in Rn with parametrization obtained by solving for the firstm components. More precisely, wria = (p, q) with p Rm and q Rd. Then there exists an open set V in Rd such that q V, and there exists a functih

C1(V, Rm) such that

(i) h(q) = p

(ii){(h(z), z)| zV}= L B(a; r)Hence : V Rn, (z) = (h(z), z) is a parametrization for L B(a; r).Example 17.5. n= 3, m = 2, d = 3 2 = 1. f : R3 R2 defined byf((x,y,z)) = (x2 + y2 + z2 1, x + y+ z 1). Take

L={(x,y,z)|f((x,y,z)) = (0, 0)}

Theorem 17.4 says that we can parametrize L by solving for x, y in terms ofz at any point a= (x,y,z) which is regular wirespect to directions 1,2. All points are regular with respect to directions 1,2 except a= (0, 0, 1)or a= (23 ,

23

, 13).

Example 17.6. Take

S={x= (x(1), x(2), . . . , x(9)) R9 |M is orthogonal}where

M=

x(1) x(2) x(3)x(4) x(5) x(6)

x(7) x(8) x(9)

ThenSis a manifold of dimension 3. (M Mt =I3 = Mt M). This comes from the fact that orthogonality ofMis given byequations. So Sis a level set for a function f C1(R9, R6). Then use implicit function theorem and 9 6 = 3.Final. Friday, 7:30-10pm, at PAC 11. Similar to midterm, but 6 questions instead of 5. Must know all definitions, examplstatements of lemmas, propositions, corollaries, theorems, etc. done in class. Two proofs from lectures 9-16.

Office hours. Wednesday 3-5pm. Friday 12-2pm.

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Math 247 Advanced Calculus