Introduction to Functional Analysis

Goetz Grammel

February 4, 2005

Contents

1 Introduction 2

2 The One-dimensional Laplace Equation 3

3 Basic Functional Analysis 4

3.1 Linear Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3.2 Linear Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3.3 Inner Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.4 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Convergence, Continuity and Completeness 13

4.1 Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

4.3 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5 Sobolev Spaces and Variational Equalities 19

5.1 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

5.2 Variational Equalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5.3 Application to Linear Second Order Boundary Value Problems . . . . . . . 23

6 Approximation of Solutions 27

6.1 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6.2 Splines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.3 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

6.4 Homogenization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Bibliography 33

1

1 Introduction

In computational mechanics one solves problems involving partial differential equations.

Frequently those differential equations cannot be solved analytically and one has to find

approximate solutions. The standard way to compute approximate solutions is the so-

called finite element method. It relies on characterizing the solutions of partial differential

equations by related minimization problems, which in turn can be described by variational

equalities. Then the approximate solutions are minimizers within a prescribed finite-

dimensional subspace. The purpose of this course is to develop the mathematical tools

and skills needed to understand the abstract framework related to the finite element

method.

2

2 The One-dimensional Laplace Equation

Given two real numbers a and b, find a twice continuously differentiable function u :

[0, 1] R with the boundary conditions u(0) = a, u(1) = b, such that the Laplaceequation

u(x) = 0

is fulfilled for all x (0, 1). Multiplying both sides of the Laplace equation with acontinuously differentiable function w : [0, 1] R and integrating leads to the variationalsystem 1

0

u(x)w(x)dx = 0.

Integration by parts yields

10

u(x)w(x)dx+ u(1)w(1) u(0)w(0) = 0.

If we only take those functions w, whose values at the boundary points vanish, we finally

obtain the variational system 10

u(x)w(x)dx = 0.

We claim that the solution to the Laplace equation minimizes the functional

F (u) =

10

u(x)2dx.

To this end we note that all function of the type u + w satisfy the boundary conditions.

We calculate 10

(u(x) + w(x))2dx = 10

u(x)2dx+ 2 10

u(x)w(x)dx+ 10

w(x)2dx 10

u(x)2dx

for all continuously differentiable functions w : [0, 1] R with w(0) = w(1) = 0.

3

3 Basic Functional Analysis

In this chapter we focus on the algebraic side of functional analysis and present the

algebraic structures needed to formulate finite element methods in a mathematical way.

3.1 Linear Vector Spaces

Definition 3.1 (Linear spaces) A real linear space (vector space, linear vector space)

is a set V together with two operations

(i) Addition + : V V V , (u, v) 7 u+ v,(ii) Scalar Multiplication : R V V , (, u) 7 u,with the following properties.

u+ v = v + u

u+ (v + w) = (u+ v) + w

(+ ) u = u+ u (u+ v) = u+ v

() u = ( u)1 u = u

for all u, v, w V and , R. Furthermore, there is an element 0 V with u+ 0 = ufor all u V , and for all u V there exists an element w V with u+w = 0. We writew = u.

Remark 3.2 In any real vector space V , the equality

0 u = 0

is valid for all u V . Indeed, we can calculate

0 u = (0 + 0) u = 0 u+ 0 u

and adding the expression (0 u) to both sides, we have 0 = 0 u. Consequently, we canwrite

u = (1) ufor all u V , since

(1) u+ u = (1) u+ 1 u = (1 + 1) u = 0 u = 0.

Example 3.3 The smallest linear space is given by V = {0}.

4

Example 3.4 For any natural number n, the set Rn with the addition

u+ v = (u1, u2, ..., un) + (v1, v2, ..., vn) = (u1 + v1, u2 + v2, ..., un + vn)

and the scalar multiplication

u = (u1, u2, ..., un) = (u1, u2, ..., un)

is a real linear vector space.

Example 3.5 Let Rn. We write A(;R) for the set of all functions from to R.We define the addition of two functions f, g A(;R) by

(f + g)(x) = f(x) + g(x)

for all x . A scalar multiplication is obtained by setting

( f)(x) = f(x)

for all x . Using the fact that the range R is a linear space, it is easy to show thatA(;R) is a linear space as well.

Lemma 3.6 Let V be a linear space and W V . Then W is a linear space (a subspaceof V ), if and only if the following two conditions are satisfied.

(i) w1 + w2 W for all w1, w2 W ,(ii) w W for all R, w W .

Example 3.7 The following sets are subspaces of A(;R).B(;R), the set of all bounded functions from to R.C(;R), the set of all continuous functions from to R.Ck(;R), the set of all k-times continuously differentiable functions from to R.L1(;R), the set of all integrable functions from to R.

Example 3.8 Let L2(;R) be the set of all square integrable functions, i.e. of all func-tions f : R with

f(x)2dx

Let f, g L2(;R). We show that f + g L2(;R). To this end we apply the inequalityabove to a = |f(x)|, b = |g(x)|. Then we obtain

(f(x) + g(x))2 dx

f(x)2dx+

g(x)2dx+ 2

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

2

f(x)2dx+ 2

g(x)2dx

Definition 3.13 (Dimension) Let V be a linear space and B be a basis of V . The

cardinality of B is called the dimension of V .

Example 3.14 The smallest possible linear space {0} has dimension dim({0}) = 0, sinceits basis is the empty set.

Example 3.15 The Euclidean space R3 has the basis

B = {(1, 0, 0), (0, 1, 0), (0, 0, 1)}.

Accordingly, dim(R3) = 3.

Example 3.16 The Euclidean space Rn has the basis

B = {(1, 0, . . . , 0), (0, 1, 0, . . . , 0), . . . , (0, . . . , 0, 1)}.

Accordingly, dim(Rn) = n.

Example 3.17 Let P be the linear space of polynomials (of arbitrary degree). Then

B = {1, x1, x2, . . .}

is a basis of P . Accordingly dim(P ) =.

3.2 Linear Forms

Definition 3.18 (Mapping) Given two sets X, Y , a mapping f : X Y assigns to anyx X an element f(x) Y . The set X is called the domain of definition, the set Y isthe range of f .

Definition 3.19 (Linear mapping) Let V , W be a real linear vector spaces. A map-

ping L : V W is called a linear if

L(u+ v) = L(u) + L(v)

L( u) = L(u)

for all u, v V and R.

Example 3.20 The simplest linear mapping L : V W is the zero mapping given byL(v) = 0 for all v V .

Example 3.21 For the particular case that V = W = R, any linear mapping L : V Wis of the form L(v) = v, where R.

7

Remark 3.22 L(0) = 0 for any linear mapping L on V . Indeed, we can calculate

L(0) = L(0 0) = 0 L(0) = 0.Remark 3.23 A linear mapping L : V W is completely determined by its values on abasis B of V , since we can write

L(v) = L

(ki=1

i bi)=

ki=1

L(i bi) =ki=1

i L(bi)

for any v V , where v =ki=1 i bi is the unique (w.r.t. a fixed basis) representation ofv.

Definition 3.24 (Kernel, image) Let L : V W be linear. The setKer(L) := {v V : L(v) = 0}

is called the kernel of L. The set

Im(L) := {w W : w = L(v) for a v V }is called the image of L.

Remark 3.25 The smaller the kernel, the larger the image. For instance, if V = Rn onecan show that

dim(Ker(L)) + dim(Im(L)) = n = dim(Rn)

for any linear mapping L : V W .Definition 3.26 (Linear forms) Let V be a real linear vector space. A linear mapping

L : V R is called a linear form.Example 3.27 L : C([0, 1],R) R defined by L(f) = f(0) is a linear form on V =C([0, 1],R).

Example 3.28 L : C([0, 1],R) R defined by

L(f) =

10

f(x)dx

is a linear form on C([0, 1],R).

Example 3.29 L : C1([0, 1],R) R defined by L(f) = f (12) is a linear form on

V = C1([0, 1],R).

Example 3.30 Let y = (y1, y2, ..., yn) Rn be fixed. Ly : Rn R defined by

Ly(x) =ni=0

xiyi

is a linear form on V = Rn.

8

3.3 Inner Products

Definition 3.31 (Bilinear forms) Let V and W be two linear spaces. A mapping a :

V W R is a bilinear form if the mappings

aw : V R, u 7 aw(u) := a(u,w),

and

av : W R, u 7 av(u) := a(v, u),are linear forms for all v V and w W .

Example 3.32 Let V = R3 and W = R2. Then

a(v, w) := v1 w1 + v2 w2

is a bilinear form.

Definition 3.33 (Inner products) Let V be a linear vector space. An inner product

on V is a bilinear form a : V V R such that

a(v, v) 0

for all v V (positivity), a(v, v) = 0 if and only if v = 0 (definiteness),

a(v, w) = a(w, v)

for all v, w V (symmetry).A linear space which is endowed with an inner product is called an inner product space.

Remark 3.34 For inner products on V we write a(v, w) = v, w for u, v V .

Example 3.35 On the linear vector space V = Rn, the bilinear form a(v, w) = v, w =ni=0 viwi is called the Euclidean inner product.

Example 3.36 On the linear vector space V = C([0, 1];R), the bilinear form

a(f, g) :=

10

f(x)g(x)dx

defines an inner product.

Remark 3.37 Let V be an inner product space. Then the equality v, 0 = 0 is valid forall v V . Indeed, we can calculate

v, 0 = v, 0 v = 0v, v = 0

9

Theorem 3.38 (Schwarz inequality) Let V be an inner product space. Then the in-

equality

u, v2 u, uv, vis valid for all u, v V .

Proof. For v = 0, the claim is trivial. Let v 6= 0. Then we can calculate

0 u u, vv, vv, u

u, vv, vv

= u, u

u,u, vv, vv

u, vv, vv, u

+u, v2v, vv, v2

= u, u 2u, vv, v u, v+u, v2v, v

= u, u u, v2

v, vand the claim easily follows. 2

3.4 Norms

Definition 3.39 (Norms) Let V be a linear vector space. A function

: V R, v 7 v

is called a norm on V if

v 0 for all v V (positivity),v = 0 if and only if v = 0 (definiteness), v = ||v for all R and v V (positive homogeneity),u+ v u+ v for all u, v V (triangle inequality).A linear space which is endowed with a norm is called a normed space.

Example 3.40 The absolute value function x 7 |x| defines a norm in R.

Example 3.41 On Rn a norm is given by

x|1 := |x1|+ + |xn|.

Example 3.42 On Rn a norm is given by

x|2 :=|x1|2 + + |xn|2.

This norm is called the Euclidean norm.

10

Example 3.43 On C([0, 1];R) we obtain a norm by setting

f := maxx[0,1]

|f(x)|.

This norm is called the maximum norm.

Example 3.44 On C1([0, 1];R) we obtain a norm by setting

fC1 := maxx[0,1]

|f(x)|+ maxx[0,1]

|f (x)|.

Theorem 3.45 (Natural norms) Let V be a linear vector space with an inner product

, . Then we obtain a norm on V by setting

v := v, v 12 .

This norm is called the natural norm on an inner product space.

Proof. As for the positivity we immediately obtain v = v, v 12 0, since v, v 0for all v V . The definiteness follows from the definiteness of the inner product. Thepositive homogeneity is obtained as follows.

v = v, v 12 = (2v, v) 12 = (2) 12 v, v 12 = || v.As for the triangle inequality, we calculate

u+ v2 = u+ v, u+ v= u, u+ u, v+ v, uv, v= u, u+ 2u, v+ v, v,

and using the Schwarz inequality we obtain

u+ v2 u2 + 2uv+ v2= (u+ v)2.

Hence u+ v u+ v. 2

Theorem 3.46 (Parallelogram law) Let V be an inner product space. Then the par-

allelogram law

u+ v2 + u v2 = 2u2 + 2v2

is valid for all u, v V .Conversely, If V is a normed space such that the parallelogram law is valid for all u, v V ,then the norm is a natural norm of an inner product, i. e. there is an inner product on

V such that u = (u, u) 12 for all u V .

11

Example 3.47 The Euclidean norm on Rn given by

x := n

i=1

x2i

is the natural norm corresponding to the Euclidean inner product on Rn.

Example 3.48 (Energy norms) We set V := {u C([0, 1];R) : u(0) = u(1) = 0}.Obviously V is a linear vector space. On V we define a bilinear form by

a : V V R, a(u, v) = 10

u(x)v(x)dx.

The bilinear form a is an inner product on V . Indeed, we immediately obtain the positivity

a(u, u) =

10

(u(x))2dx 0

for all u V . As for the definiteness we note that a(u, u) = 0 if and only if 10(u(x))2dx =

0. Since u C([0, 1];R), we conclude that u(x) = 0 for all x [0, 1]. Since the domain[0, 1] is an interval, there is a constant c R with u(x) = c for all x [0, 1]. By thedefinition of V , we have c = u(0) = 0, and hence u = 0. The symmetry is obvious. The

natural norm

ua :=( 1

0

u(x)u(x)dx) 1

2

induced by a is called the energy norm on V .

Example 3.49 L2([0, 1];R) is an inner product space with inner product

f, g := 10

f(x)g(x)dx.

The corresponding natural norm defined by

fL2 :=( 1

0

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

is called the L2-norm. Here, the bilinearity, positivity and symmetry are easy to show. In

contrast to the space of continuous functions, definiteness, in the usual sense, does not

hold, since there are many square integrable functions f with 10

f(x)2dx = 0.

However, there is an artificial mathematical procedure to obtain definiteness. We put all

those functions with square integral equal to zero into one class, the zero-class, and more-

over say that two square integrable functions f, g are equivalent, if their difference fg isan element of this zero-class. In this way we obtain equivalence classes of functions. So,

we do not talk about single functions any more, but about equivalence classes of functions.

A single function can be viewed as a representative of its equivalence class.

12

4 Convergence, Continuity and Completeness

In this chapter, we present the notions that are necessary in order to understand the

analytical aspects of functional analysis.

4.1 Convergence

Definition 4.1 (Convergence) Let V be a normed space and (vn)nN be a sequence in

V . We say that the sequence vn converges to a limit v V , and writevn v, as n,

if

vn v 0, as n.Hence, the norm allows us to reduce the convergence in a vector space to a convergence

of real numbers. However, note that the convergence of a given sequence in a linear space

depends on the norm we have chosen. For instance, since we have the inclusions

C([0, 1];R) L2([0, 1];R) L1([0, 1];R),we can equip the linear space of real valued continuous functions on [0, 1] with at least

three different norms. The following theorem tells us that the maximum norm is nicely

adapted to C([0, 1];R).

Theorem 4.2 Let (fn)nN be a sequence in C([0, 1];R) and f B([0, 1];R) withfn f 0, as n.

Then the function f is continuous as well.

Proof. Let x [0, 1] be arbitrary (but fixed) and > 0. We take an arbitrary sequence(xk)kN in [0, 1] with xk x, as k . We choose n N large enough such that

fn f < 3,

and, for this particular n N, we choose k0 N large enough such that|fn(xk) fn(x)| <

3

for all k k0. Then we obtain (using the triangle inequality) the estimation|f(xk) f(x)| |f(xk) fn(xk)|+ |fn(xk) fn(x)|+ |fn(x) f(x)|

0 such that

v2 Cv1

for all v V .

When the norm 1 is stronger than the norm 2, a sequence vn in V converges w.r.t.the norm 2 whenever it is convergent w.r.t. the norm 1.

Definition 4.5 (Equivalence of norms) Let 1 and 2 be two norms on a linearspace X. We say that both norms are equivalent, if there is a real constant C > 0 such

that we can estimate

x1 Cx2 and x2 Cx1for all x X.

14

Remark 4.6 On the n-dimensional space Rn all norms are equivalent.

We are not going to prove this result, but present a simple example.

Example 4.7 Consider the two-dimensional space R2 with the two normsx1 = |x1|+ |x2| and x2 =

x21 + x

22.

A short look at the corresponding unit spheres tells us that

x2 x1 and x1 2x2

for all x R2. Hence, we can chose C = 2 and see that both norms are equivalent.

4.2 Continuity

Definition 4.8 (Continuity) Let V,W be normed spaces and V . A function f : W is continuous at v , if f(vn) f(v), as n, for any sequence (vn)nN in with vn v, as n . A mapping f : W is continuous, if it is continuous atany v .

Proposition 4.9 (Continuity of linear mappings) Let V,W be normed linear spaces

and L : V W be a linear mapping. Then L is continuous if and only if it is continuousat 0 V .

Proof. The only if part is obviuos, we only show the if part. Let L be continuous

at 0 V and let v V . We take a sequence (vn)nN in V with vn v, as n. Thenwe can write

Lvn Lv = L(vn v) 0, as n,since (vn v) 0, as n. 2

Example 4.10 Let V = C1([0, 1];R) and W = C([0, 1];R). We define a mapping D :V W by f 7 Df = f . Clearly, D is a linear mapping by the rules of differentiation. Itturns out that the continuity of D strongly depends on the norms we choose. For instance,

D is not continuous, if we equip both spaces V,W with the maximum norm. This can be

seen as follows. Consider the sequence (fn)nN in V given by

fn(x) =sin(nx)

n

for all x [0, 1]. Then we immediately calculate that fn 0, as n , butDfn = max0x1|cos(nx)| = 1 for all n N.In contrast, D becomes continuous, if we equip only W with the maximum norm and take

for f V the stronger norm

fC1 = f + f .

15

Indeed, we easily show that D is continuous at 0 V . Let fn be a sequence in V withfnC1 0, as n infty. Then we have Dfn = f n fn + f n = fnC1for all n N, and hence Dfn 0 in W , as n.

Proposition 4.11 (Continuity of the inner product) Let V be an inner product space.

Then the mapping

P : V V R, (v1, v1) 7 P (v1, v2) := v1, v2

is continuous, where the product space V V is equipped with the norm

(v1, v2)VV =v12 + v22.

Proof. For v1, v2, w1, w2 V we can estimate

|P (v1, v2) P (w1, w2)| = |v1, v2 w1, w2| |v1, v2 w1, v2|+ |w1, v2 w1, w2|= |v1 w1, v2|+ |w1, v2 w2| v1 w1v2+ w1v2 w2 L (v1 w1+ v2 w2) 2L

v1 w12 + v2 w22

= 2L(v1, v2) (w1, w2)VV ,

where L = max (v2, w1). 2

Definition 4.12 (Bounded linear mappings) Let V,W be normed linear spaces with

norms V , W . We say that a linear mapping L : V W is bounded, if there existsa constant C 0 such that

L(v)W CvVfor all v V .

Proposition 4.13 Let V,W be normed linear spaces and L : V W be a linear mapping.Then L is continuous if and only if it is bounded.

Proof. Let L be bounded. We show that L is continuous at the origin. Let vn be a

sequence in V with vn 0, as n. Then we can estimate

L(vn)W CvnV

for all n N. Hence, L(vn) 0, as n, and L is continuous at 0 V .

16

As for the converse, let L be continuous. Assume that L is not bounded, then there is a

sequence vn in V \ {0} withL(vn)WvnV > n

for all n N. Hence, for wn := vnnvnV we obtain wn 0, as n, but

L(wn)W > 1

for all n N, a contradiction to the continuity of L. Thus, our assumption was wrongand L is bounded. 2

Proposition 4.14 Let V,W be normed linear spaces and a : V W R be a bilinearmapping. We equip the product space V W with the norm

(v, w)VW = vV + wW .

Then a is continuous if and only if there is a constant C 0 such that the inequality

|a(v, w)| CvV wW

it valid for all v V , w W .

4.3 Completeness

Definition 4.15 (Cauchy sequence) Let V be a normed space. A sequence (vn)nN is

called a Cauchy sequence, if for any real > 0 there is a number n0 N such that

vn vm <

for all n,m n0.

Proposition 4.16 Let V be a normed space and (vn)nN a convergent sequence in V with

limit v V . Then (vn)nN is a Cauchy sequence.

Proof. For > 0 there is an n0 N such that

vn v < 2

for all n n0. Hence we can estimate

vn vm vn v+ v vm < 2+

2=

for all n,m n0. 2

17

Remark 4.17 The converse of the Proposition 4.16 above is wrong, i.e. it may happen

that a Cauchy sequence does not converge towards an element of the normed space con-

sidered. Recall, for instance, Example 4.3. There, the sequence fn converges w.r.t. the

L2-norm to an element of L2((0, 2);R). According to Proposition 4.16, fn necessarily is aCauchy sequence w.r.t. the L2-norm. However, the sequence fn does not converge to an

element of the linear space C([0, 2];R), which has been originally considered.

Definition 4.18 (Completeness) Let V be a normed space. V is called complete, if

any Cauchy sequence in V is convergent in V , i.e. if any Cauchy sequence in V has a

limit in V .

According to our considerations above, the linear space C([0, 2];R) cannot be completew.r.t. the L2-norm. One could also say that it is not large enough to contain all limits of

Cauchy sequences.

Normally, one tries to verify the convergence of a sequence by showing that the distance

to the limit is tending to zero. However, this procedure requires that the limit is already

known. So, convergence within complete spaces is much easier to show, since one only has

to check whether the distance between the elements of the sequence is tending to zero,

withput referring to a limit. Fortunately, many important infinite-dimensional function

spaces are complete, as well as all finite-dimensional spaces.

Definition 4.19 (Banach space, Hilbert space) A complete normed space is called a

Banach space. An inner product space, which is complete w.r.t. the natural norm induced

by the inner product, is called a Hilbert space.

Example 4.20 The finite-dimensional Euclidean space Rn is a Hilbert space with theinner product x, y =ni=1 xiyi.Example 4.21 The linear space of continuous functions, C([a, b];R), endowed with themaximum norm is a Banach space.

Example 4.22 The linear space of continuously differentiable functions, C1([a, b];R),endowed with the C1-norm C1 is a Banach space.

Example 4.23 The linear space of integrable functions, L1([a, b];R), endowed with theL1-norm L1 is a Banach space.

Example 4.24 The linear space of square integrable functions, L2([a, b];R), endowed withthe L2-norm L2 is a Hilbert space.

18

5 Sobolev Spaces and Variational Equalities

5.1 Sobolev Spaces

The linear space C([a, b];R) is not complete w.r.t. the L2-norm. But there is a larger linearspace, which is a Hilbert space w.r.t. the L2-norm, namely the linear space L2([a, b];R).Actually, one can show that L2([a, b];R) is a minimal complete extension of C([a, b];R).However, the space L2([a, b];R) is not suitable for solving boundary value problems, sinceeven the variational equalities constructed in the second section require some kind of dif-

ferentiability, let alone the original differential equation. In this section, we present an ap-

propriate notion of differentiability which is applicable to certain elements of L2([a, b];R).First of all, we introduce the inner product, which is related to variational equalities.

Proposition 5.1 We consider the linear space V := C1([a, b];R). Then the mapping

, H1 : V V R

defined by

f1, f2H1 := ba

f1(x)f2(x)dx+

ba

f 1(x)f2(x)dx

is an inner product on V .

Proof. This is an easy exercise. Note that the first term represents the L2-inner product,

whereas the second term defines a positive bilinear form. 2

C1([a, b];R) is not a Hilbert space w.r.t. this inner product. In the sequel we construct aminimal complete extension of C1([a, b];R). To this end recall the rule of integration byparts. For continuously differentiable functions f it yields the variational equality b

a

f (x)v(x)dx+ ba

f(x)v(x)dx = 0

for all v C10([a, b];R), where we set

C10([a, b];R) := {v C1([a, b];R) : v(a) = v(b) = 0}.

The variational equality above is used to generalize the notion of differentiability.

Definition 5.2 Let f L2([a, b];R). We say that g L2([a, b];R) is the generalizedderivative of f , if b

a

g(x)v(x)dx+

ba

f(x)v(x)dx = 0

for all v C10([a, b];R). In this case we write g = f

19

Example 5.3 In L2([1, 1];R) we consider the function f given byf(x) := |x|.

Surely, f is not contained in C1([1, 1];R), since it is not differentiable at x = 0. How-ever, f does possess a generalized derivative, namely the function g L2([1, 1];R) givenby

g(x) =

{1 for 1 x 01 for 0 < x 1.

This can be seen as follows. We calculate 11g(x)v(x)dx

=

01g(x)v(x)dx+

10

g(x)v(x)dx

=

01v(x)dx+

10

v(x)dx

=

01f (x)v(x)dx+

10

f (x)v(x)dx

=

01f(x)v(x)dx+ f(0)v(0) f(1)v(1) +

10

f(x)v(x)dx+ f(1)v(1) f(0)v(0)

=

11f(x)v(x)dx

for all v C10([1, 1];R). Here, we make use of the fact that f is continuously differen-tiable on the intervals (1, 0) and (0, 1).Definition 5.4 (Sobolev space) The Sobolev space H1([a, b];R) is defined as the set ofall functions f L2([a, b];R), which possess a generalized derivative f L2([a, b];R).Since

C1([a, b];R) H1([a, b];R),we know that H1([a, b];R) 6= . On the other hand, the example above shows that

C1([a, b];R) 6= H1([a, b];R).Remark 5.5 The definition of the Sobolev space H1([a, b];R) given above easily can begeneralized to functions defined on finite dimensional domains Rn. Let us notethat for one-dimensional domains, i.e. for intervals [a, b], the Sobolev space H1([a, b];R)equivalently can be regarded as the space of all continuous functions f C([a, b];R) thatcan be written as

f(x) = f(a) +

xa

g(y)dy,

where g L2([a, b];R). In this case g is the generalized derivative of f . Hence, we canapply the fundemental theorem of calculus.

20

Theorem 5.6 The Sobolev space H1([a, b];R) is a Hilbert space with the inner product

f1, f2H1 := ba

f1(x)f2(x)dx+

ba

f 1(x)f2(x)dx.

Actually, one can show that the Sobolev space H1([a, b];R) is the smallest completeextension of C1([a, b];R) w.r.t. the inner product , H1 . Hence the Sobolev functionscan be approximated by continuously differentiable functions w.r.t. the inner product

, H1 . However, the Sobolev space H1([a, b];R) is still too large for finding the solutionsto boundary value problems, since we did not prescribe boundary values. For this reason

we finally define the subspace H10 ([a, b];R) H1([a, b];R), which consists of those Sobolevfunctions f : [a, b] R with f(a) = f(b) = 0.

Definition 5.7 The Sobolev space H10 ([a, b];R) consists of those functions f H1([a, b];R),that can be approximated w.r.t. the inner product , H1 by functions taken from C10([a, b];R).

5.2 Variational Equalities

The following result describes the strong relation between variational equalities and quadratic

minimization problems, which underlies the finite element method.

Theorem 5.8 Let V be a linear space, a : V V R be a symmetric, positive definitebilinear form and L : V R be a linear form. Then a vector u V minimizes thefunctional F : V R given by

F (v) :=1

2a(v, v) L(v).

if and only if u V solves the variational equality

a(u, v) = L(v) v V.

Proof. For t R and v V we calculate

F (u+ tv) = F (u) + t(a(u, v) L(v)) + 12t2a(v, v).

If u V minimizes the functional F , we obtain

0 = limt0

F (u+ tv) F (u)t

= a(u, v) L(v)

for all v V . As for the converse, let u V satisfy the variational equality

a(u, v) = L(v)

for all v V . Then setting t = 1, we obtain

F (u+ v) = F (u) +1

2a(v, v) > F (u)

21

For all v V \ {0} and the proof is finished. 2

Note that the theorem above does not say anything about the existence of a solution

to the problems involved. Indeed, one needs more structure in order to guarantee the

solvability of the variational equality (and hence of the minimization problem). The

following fundamental theorem is the corner stone for obtaining various existence results

for boundary value problems.

Theorem 5.9 (Riesz representation theorem) Let V be a Hilbert space and L : V R be a continuous linear form. Then there exists a unique vector u V such that

u, v = L(v)

for all v V .

Many versions of this basic result running under the title Lax-Milgram theorem can be

found in the literature. Now we are nearly in a position to solve the variational equalities

related to linear second order boundary value problems. We only need one more property

of the bilinear form involved, that makes sure that the norm induced by the bilinear form

and by the original inner product are equivalent.

Definition 5.10 Let V be an inner product space. A bilinear form a : V V R iscalled coercive, if there is a constant c > 0 such that

|a(v, v)| cv, v

for all v V .

Theorem 5.11 Let V be a Hilbert space and L : V R be a continuous linear form.Moreover, let a : V V R be a coercive, positive definite, symmetric, continuousbilinear form. Then there exists a unique vector u V solving the variational equality

a(u, v) = L(v) v V.

Moreover, we can estimate

u 2Cc,

where c > 0 is the coercivity constant of the bilinear form a and C 0 is the boundednessconstant of the linear form L.

Proof. The bilinear form a defines a new inner product on the linear space V . Hence,

according to the representation theorem of Riesz, we only have to show that V is complete

and L is continuous w.r.t. the norm

va :=a(v, v).

22

To this end, we calculate

va =a(v, v)

Cvv =

Cv

and

v =v, v

ca(v, v) =

cva

for all v V . Hence the norms and a are equivalent. As for the estimation, weobserve that u minimizes the corresponding functional F (v) = 1

2a(v, v) L(v), hence we

have

F (u) F (0) = 0and

F (u) 12cu2 Cu,

which gives us1

2cu2 Cu 0

and the proof is finished. 2

5.3 Application to Linear Second Order Boundary Value Prob-

lems

In this part, we finally apply the abstract framework developped to linear second order

boundary value problems. For a given real number R and f L2([0, 1];R) considerthe following differential equation

u(x) + f(x) = u(x),

where x [0, 1]. We are interested in the unique existence of a solution u H10 ([0, 1];R)for prescribed boundary values

u(0) = u(1) = 0.

Multiplying both sides of the differential equation by test functions v H10 ([0, 1];R) andintegrating by parts leads to the variational equation 1

0

u(x)v(x)dx+ 10

u(x)v(x)dx =

10

f(x)v(x)dx v H10 ([0, 1];R).

In order to apply our abstact existence result, Theorem 5.11, we have to check three points.

(I) The bilinear form given by

a(w, v) :=

10

w(x)v(x)dx+ 10

w(x)v(x)dx

23

should be positive definite, coercive and continuous on H10 ([0, 1];R).

(II) The linear form given by

L(v) :=

10

f(x)v(x)dx

should be continuous on H10 ([0, 1];R).

(III) The space H10 ([0, 1];R) should be a Hilbert space w.r.t. the inner product , H1 .

As for (I), we calculate using the fundamental theorem of calculus and Hoelders inequality 10

v(x)2dx =

1/20

( x0

v(y)dy)2

dx+

11/2

( 1x

v(y)dy)2

dx

1/20

x

x0

v(y)2dydx+ 11/2

(1 x) 1x

v(y)2dydx

1/20

x

1/20

v(y)2dydx+ 11/2

(1 x) 11/2

v(y)2dydx

=1

8

1/20

v(y)2dy +1

8

11/2

v(y)2dy

=1

8

10

v(y)2dy

for all v C10([0, 1];R), and hence for all v H10 ([0, 1];R). This estimation is of impor-tance in order to obtain coercivity. For 8 < 1, it yields

a(v, v) =

10

v(x)2dx+ 10

v(x)2dx

=

(8 +

9

) 10

v(x)2dx+(1 8 +

9

) 10

v(x)2dx+ 10

v(x)2dx

(8 +

9

) 10

v(x)2dx+ 8(1 8 +

9

) 10

v(x)2dx+

10

v(x)2dx

=

(8 +

9

) 10

v(x)2dx+(8 +

9

) 10

v(x)2dx

=

(8 +

9

)v, vH1

for all v H10 ([0, 1];R). For 1, we immediately obtain

a(v, v) =

10

v(x)2dx+ 10

v(x)2dx 10

v(x)2dx+ 10

v(x)2dx = v, vH1

for all v H10 ([0, 1];R). The continuity of the bilinear form a is obvious, since we canwrite

|a(v, w)| (1 + ||) 1

0

v(x)w(x)dx+ 10

v(x)w(x)dx

(1 + ||)vH1wH124

for all v, w H10 ([0, 1];R).

As for (II), we calculate

|L(v)| fL2vL2 fL2vH1

for all v H10 ([0, 1];R).

As for (III), let vn be a Cauchy sequence in H10 ([0, 1];R). Since H1([0, 1];R) is a Hilbert

space, there is a v H1([0, 1];R) with vn v, as n . It remains to show thatv H10 ([0, 1];R). To this end let wn C10([0, 1];R) with vnwnH1 1/n for all n N.Then we can estimate

v wnH1 v vnH1 + vn wnH1 ,

where the right-hand side tends to zero, as n. Hence, v H10 ([0, 1];R).

Overall, we can conclude from the existence theorem, Theorem 5.11, that the variational

equation has a unique solution u H10 ([0, 1];R), at least for parameters > 8. Forparameters 8, we cannot prove the coercivity of the bilinear form involved. Actuallyit is not surprising that negative parameters < 0 might cause difficulties. Consider, for

instance, the homogeneous differential equation

u(x) = u(x),

where x [0, 1] and < 0. It possesses a two-dimensional space of solutions, which canbe represented by

u(x) = sin(||x) + cos(

||x),

where (, ) R2. In order to fulfil the boundary condition u(0) = 0, we have to set = 0. Accordingly, for

|| = kpi, k N, the boundary conditions are fulfiled by thesolution

u(x) = sin(||x).

Hence, for = k2pi2, k N, the boundary problem has more than one solution, i.e.uniqueness is destroyed. This happens, in particular, for = pi2, which is slightlysmaller than our deduced lower bound for the parameters .

Remark 5.12 The inequality 10

v(x)2dx 18

10

v(x)2dx

25

derived above is called Poincare-inequailty and is valid in a more general setting. Namely,

one can show that for any bounded domain Rn there is a constant C = C() suchthat we can estimate

v(x)2dx C

gradv(x), gradv(x)dx

for all v C10(;R).

26

6 Approximation of Solutions

Definition 6.1 Let (V, ) be a normed space and X V be a subset. We say that Xis dense in V , if for any v V and any > 0 there is an x X with

x v .Density of a subset means that any element of the larger space can be approximated by

elements of the smaller subset.

Example 6.2 The set of rational numbers Q is dense in the linear space of real numbers(R, | |).In connection with finite element methods, one is interested in the approximation of

functions.

Theorem 6.3 (Weierstrass approximation theorem) The space of real polynomials

P := {p C([a, b];R) : p(x) = a0 + a1x+ . . .+ anxn, n N, a0, . . . , an R} is dense inthe space of continuous functions (C([a, b];R), ).Theorem 6.4 The space of continuous functions C([a, b];R) is dense in the space ofsquare integrable functions (L2([a, b];R), L2).Theorem 6.5 The space of continuously differentiable functions C1([a, b];R) is dense inthe space of Sobolev functions (H1([a, b];R), H1).All these statements are quite interesting from a theoretical point of view. But for prac-

tical considerations one needs more information on the rate of approximation.

6.1 Taylor Series

Theorem 6.6 (Taylor) Let f CN+1([a, b];R). Then the N-th Taylor polynomial of fat a

TNf (x) := f(a) +Df(a)(x a) +D2f(a)(x a)2

2!+ +DN(a)(x a)

N

N !

fulfils the estimation

f TNf DN+1f(b a)N+1(N + 1)!

.

For f C([a, b];R) the Taylor series is given by

Tf (x) :=k=0

Dkf(a)(x a)k

k!.

Note that there are functions for which the Taylor series does not converge at any x [a, b]or for which the Taylor series does converge for all x [a, b] but its limit has nothing incommon with the original function f .

27

6.2 Splines

We consider linear splines on the interval [0, 1]. To this end we introduce a decomposition

[0, 1] =Nk=1

[xk1, xk],

where

xk :=k

N(k = 0, . . . , N).

For k = 0, . . . , N we define the continuous mapping bk : [0, 1] R by

bk(x) :=

1 + (x xk)N for xk 1N x xk1 (x xk)N for xk x xk + 1N

0 for |x xk| 1N .The subspace given by the basis BN = {b0, . . . , bN} C([0, 1];R) is called the spaceof linear splines and denoted by S1N([0, 1];R). In other words, S1N([0, 1];R) consists ofcontinous functions in C([0, 1];R) which are piecewise affine linear. Naturally we havedim(S1N([0, 1];R)) = N + 1.

Theorem 6.7 (Approximation with linear splines) For any N N and any f C2([0, 1];R) the linear spline s S1N([0, 1];R) given by

s(x) =Nk=0

f(xk)bk(x)

fulfills the estimation

f s 1N2

f 2

.

This approximation theorem relies on the Taylor approximation and often is used for

numerical integration.

Theorem 6.8 (Trapezoid rule for integration) For any N N and any f C2([0, 1];R)we can estimate 1

0

f(x)dx INf 1N2 f 2 ,

where

INf :=f(0) + 2f(x1) + + 2f(xN1) + f(1)

N.

Proof. We can estimate 10

f(x)dx 10

s(x)dx

10

f sdx 1N2

f 2

.

On the other hand, it is obvious that

INf =

10

s(x)dx

and the proof is finished. 2

28

6.3 Fourier Series

Here we consider the Hilbert space L2([0, 2pi];R) together with particular functions of theform

x 7 12pi

, x 7 cos(kx)pi

, x 7 sin(kx)pi

(k = 1, 2, . . .).

Those functions form a so-called orthogonal system, since they are pairwise orthogonal

w.r.t. the standard inner product on L2([0, 2pi];R). Furthermore those functions are unitvectors w.r.t. the L2-norm. For a given function f L2([0, 2pi];R) we define the Fouriercoefficients by

a0 :=

2pi0

f(x)2pi

dx, ak :=

2pi0

f(x)cos(kx)

pidx, bk :=

2pi0

f(x)sin(kx)

pidx (k = 1, 2, . . .).

Theorem 6.9 The trigonometric polynomial of order N

FNf := a012pi

+

(a1cos(x)

pi+ b1

sin(x)pi

)+ +

((aN

cos(Nx)pi

+ bNsin(Nx)

pi

)fulfils the estimation

f FNf 2L2 =

k=N+1

(a2k + b2k).

The Fourier series is given by

Ff (x) := a012pi

+k=1

(akcos(kx)

pi+ bk

sin(kx)pi

).

Note that f2L2 = a20 +

k=1(a2k + b

2k) < . Hence, the Fourier series does converge to

f w.r.t. the L2-norm.

6.4 Homogenization

Sometimes it is advisable to approximate the solution of differential equation by a solution

of a simpler differential equation before starting the numerical calculation. For instance, if

the material considered consitsts of different thin layers that are ordered in a regular, say

periodic way. A direct numerical calculation has to take into account the fine structure of

the material and hence leads to fine discretization grids. For this reason one simplifies the

equation before starting discretizations, a procedure which is often called homogenization.

We consider the linear second order differential equation

d

dx

(a(x)

d

dxu(x)

)= f(x) (1)

on the interval x (0, 1) with boundary conditions

29

u(0) = u(1) = 0.

Here, we set

a(x) := a(x

),

where the mapping a : R R is bounded and periodic. In particular, we assume thatthere are constants 0 < 0 such that

0 < a(y) 0.

Proof. We set

(x) := a(x)d

dxu(x).

Then the differential equation (1) is equivalent to

d

dx(x) = f(x),

d

dxu(x) =

(x)

a(x).

30

Hence, is absolutely continuous and we can write

(x) = (0) +

x0

f(w)dw, u(x) =

x0

(0) + z0f(w)dw

a(z)dz

for all > 0. Similarly, we set

0(x) := a0d

dxu0(x).

Then the homogenized differential equation (2) is equivalent to

d

dx0(x) = f(x),

d

dxu0(x) =

0(x)

a0.

Hence, 0 is absolutely continuouswe can write

0(x) = 0(0) +

x0

f(w)dw, u0(x) =

x0

0(0) + z0f(w)dw

a0dz.

We conclude that

(x) 0(x) = (0) 0(0)and

u(x) u0(x)=

x0

(0) + F (z)

a(z)dz

x0

0(0) + F (z)

a0dz

=

x0

(0) 0(0)a(z)

dz +

x0

0(0)

(1

a(z) 1a0

)dz +

x0

F (z)

(1

a(z) 1a0

)dz

=

x0

(0) 0(0)a(z)

dz + 0(0)A(x

) x0

f(z)A(z

)dz + F (x)A

(x

),

where

F (z) :=

z0

f(w)dw, A(y) :=

y0

(1

a(q) 1a0

)dq.

The boundary condition 0 = u(1) = u0(1) implies that 10

(0) 0(0)a(z)

dz = 0(0)A(1

)+

10

f(z)A(z

)dz F (1)A

(1

)and that accordingly

|(0) 0(0)| (0(0)A(1

)+ 10

f(z)A(z

)dz

+ F (1)A(1)) .

Overall, we obtain

|u(x) u0(x)| (

+ 1

)(0(0)A(1)+ 1

0

f(z)A(z

)dz

+ F (1)A(1))

(

+ 1

) (|0(0)|+ 2fL2)max

yR|A(y)|.

31

By the periodicity of a : R R and A(0) = A(Y ) = 0 we easily obtain the estimation

maxyR

|A(y)| = max0yY

|A(y)| Y(1

1

)Furthermore, the boundary condition u0(1) = 0 gives us

|0(0)| fL2

and the proof is finished. 2

32

References

[1] T.J.R. Hughes, The finite element method. Linear static and dynamic finite element

analysis. Englewood Cliffs, New Jersey: Prentice-Hall International, Inc. (1987).

[2] J.L. Nowinski, Applications of functional analysis in engineering. Mathematical Con-

cepts and Methods in Science and Engineering, Vol. 22. New York and London:

Plenum Press (1981).

[3] M. Pedersen, Functional analysis in applied mathematics and engineering. Studies in

Advanced Mathematics. Boca Raton, FL: Chapman & Hall/CRC (2000).

[4] B. Daya Reddy, Introductory functional analysis. With applications to boundary

value problems and finite elements. Texts in Applied Mathematics. 27. New York,

NY: Springer (1998).

[5] W. Rudin, Principles of mathematical analysis. 3rd ed. International Series in Pure

and Applied Mathematics. Dsseldorf etc.: McGraw-Hill Book Company (1976).

[6] W. Rudin, Real and complex analysis. 3rd ed. New York, NY: McGraw-Hill (1987).

[7] K. Yosida, Functional analysis. Repr. of the 6th ed. Berlin: Springer (1994).

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