## ABSTRACT

Compact operators are linear operators on Banach spaces that maps bounded set to relatively

compact sets. In the case of Hilbert space H it is an extension of the concept of matrix acting on

a nite dimensional vector space. In Hilbert space, compact operators are the closure of the nite

rank operators in the topology induced by the operator norm. In general, operators on innite

dimensional spaces feature properties that do not appear in the nite dimension case; i.e matrices.

The compact operators are notable in that they share as much similarity with matrices as one can

expect from a general operator. Spectral decomposition of compact operators on Banach spaces

takes the form that is very similar to the Jordan canonical form of matrices. In the context of

Hilbert spaces, the spectral properties of compact operators resembles those of square matrices.

** **

## TABLE OF CONTENTS

Certication i

1 Linear Operators and Boundedness 3

1.1 Denitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2 Examples of Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3.1 Examples of linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Bounded linear operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5 Examples of bounded operators on innite dimensional spaces . . . . . . . . . . . . 10

1.6 Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.7 Some properties of Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.7.1 Examples of Hilbert spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2 Compact linear Operators on Banach spaces 18

2.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.2 Compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3 Spectral Decomposition of Compact operators on Hilbert spaces 28

3.1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.2 Spectral theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.3 Classication of 2 (T) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.3.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.4 Spectral decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.5.1 CONCLUSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

## CHAPTER ONE

LINEAR OPERATORS AND

BOUNDEDNESS

In this chapter, some well-known results which will be needed in sequel are provided.

1.1 Denitions

Denition 1.1. (Norm): A non-negative function jj:jj on a vector space X over R is called a norm

on X if and only if the following are satised.

(N1) jjxjj > 0 8 x 2 X (positivity).

(N2) jjxjj = 0 if and only if x = 0 (Nondegeneracy).

(N3) jjxjj = jjjjxjj 8 x 2 X, for all 2 R (Homogeneity).

(N4) jjx + yjj jjxjj + jjyjj for all x; y 2 X (Sub-additivity).

A vector space X endowed with a norm jj:jj denoted by (X, jj:jj) is called a normed linear space

(or just a normed space).

Denition 1.2. A sequence (xn)n1 is said to be Cauchy if given ” > 0 there exists N0 2 N such

that jjxn ? xmjj < ” for all m; n N0.

Denition 1.3. A space (X; d), where d is a metric is said to be complete if every Cauchy sequence

in X converges to a point in it.

3

Remark

Completeness is a metric space concept. In a normed space, the metric is d(x; y) = jjx?yjj where

it satises the following special properties:

(a) The underlying space is a vector space

(b) Homogenity: d(x; y) = jjd(x; y)

(c) Translation invariance d(x + z; y + z) = d(x; y)

Conversely, every metric satisfying those three conditions denes a norm: jjxjj = d(x; 0)

Denition 1.4. A complete normed vector space is called a Banach space.

Denition 1.5. Space C([a; b];R)

The space C([a; b];R) denotes the set of all real valued continous functions on [a; b] into R:

1.2 Examples of Banach spaces

1. The space C([a; b];R) endowed with the sup-norm is Banach.

Proof. Let (fn)n1 be a Cauchy sequence in C[a; b]. This implies for every x 2 [a; b] and for all

” > 0 there exists an N 2 N such that

jjfn ? fmjjC[a; b] = sup jfn(x) ? fm(x)j ”

for all x 2 [a; b] and for all m; n N:

This implies jfn(x) ? fm(x)j ” for all x 2 [a; b] and m; n N,

thus (fn(x))n1 is a Cauchy sequence in R and since R is complete, it implies

fn(x) ?! f(x) 2 R as n ?! 1

this implies jfn(x) ? f(x)j ” for all x 2 [a; b] and for all n N we have sup jfn(x) ? f(x)j ”

for all x 2 [a; b] and for all n N

thus

jjfn ? fjjC[a;b] ” for all n N;

this implies f 2 C[a; b]

Hence C[a; b] endowed with the sup-norm is Banach.

2.The space Rn with jjxjjRn = (

P1

n=1 jxij2)1=2 is Banach.

4

1.3 Linear operators

Denition 1.6. Let T be an operator from a vector space X to a vector space Y, then the domain

D(T) is given by D(T) = f x 2 X : Tx exists in Y g

and the range R(T) is given by R(T) = fy 2 Y : 9 x 2 X such that Tx = yg.

Denition 1.7. (Null space)

Let T be an operator from a vector space X to a vector space Y, then the null space N(T) is given

by N(T) = f x 2 X : Tx = 0 g.

Denition 1.8. (Injectivity)

An operator T from X to a vector space Y is said to be injective if 8 x1; x2 2 D(T), such that

Tx1 = Tx2 implies x1 = x2.

Remark: If T is injective, then there exists an operator

T?1 : R(T) Y ?! D(T) X such that T?1(y0) = x0 =) Tx0 = y0.

Denition 1.9. (Continuity)

An operator T from a vector space X to a vector space Y said to be continous at a point x0 2 X

if given any > 0 9 > 0 such that

jjx ? x0jj =) jjTx ? Tx0jj

Denition 1.10. (Linear Operators)

Let X and Y be vector spaces. Let T : X ?! Y . Then T is said said to be linear if:

i. The domain D(T) is a vector space and the range R(T) lies in a vector space over the same

eld.

ii. 8 x; y 2 D(T) and scalars ,

T(x + y) = Tx + Ty (1.1)

T(x) = Tx (1.2)

1.3.1 Examples of linear operators

1. Dierential operator: Let X be the vector space of all polynomials on [a; b]. We dene a linear

operator T on X by setting Tx(t) = x0(t) 8 x 2 X, where the prime denotes dierentiation

with respect to t. This operator maps X into itself.

2. Integral operator: A linear operator T from C[a; b] into itself can be dened by Tx(t) =

R t

a x(s)ds; t 2 [a; b].

5

3. Multiplication by t: This is linear operator from C[a; b] into itself dened by: Tx(t) = tx(t).

Theorem 1.11. Let T : X ?! Y be a linear operator space, then

(a) The range R(T) is a vector space.

(b) If dimX = n < 1, then dimR(T) n.

(c) The null space N(T) is a vector space.

Proof. (a). Let y1; y2 2 R(T), we show that y1 + y2 2 R(T) for any scalars , . Since

y1; y2 2 R(T), we have y1 = Tx1; y2 = Tx2 for some x1; x2 2 D(T), and x1+x2 2 D(T) because

D(T) is a vector space. The linearity of T yields

T(x1 + x2) = Tx1 + Tx2 = y1 + y2:

Hence y1 + y2 2 R(T). Since y1; y2 2 R(T) were arbitrary and so were the scalars, this prove

that R(T) is a vector space.

(b). We choose n + 1 elements y1; y2; :::; yn+1 in R(T) arbitrary. Then we have

y1 = Tx1; :::; yn+1 = Txn+1 for some x1; x2; :::; xn+1 in X. Since dimX = n, the set fx1; :::; xn+1g

must be linearly dependent. Hence

1×1 + + n+1xn+1 = 0 (1.3)

for some scalars 1; ; n+1 not all zero. Since T is linear then T(0) = 0. Applying T to both

sides of (1.3) gives T(1×1 + + n+1xn+1) = 1y1 + ::: + n+1yn+1 = 0. This shows that

fy1; :::; yn+1g is linearly independent set because the {‘s are not all zero.

Remembering that this subset of R(T) was chosen arbitrary, we conclude that R(T) has no linearly

independent subsets of n+1 or more elements, this implies dimR(T) n.

(c). Let x1; x2 2 N(T), then Tx1 = Tx2 = 0. Since T is linear then for any ; we have

T(x1 + x2) = Tx1 + Tx2 = 0:

It implies x1 + x2 2 N(T). Hence N(T) is a vector space.

Theorem 1.12. ( Inverse of a linear operator)

Let X and Y be vector spaces over R. Let T : X ?! Y be linear operator then:

(a) The inverse T?1 : R(T) ?! X exists if and only if Tx = 0 =) x = 0 (T is injective).

(b) If T?1 exists, then it is a linear operator.

(c) If dimX = n < 1 and T?1 exists, then dimR(T) = dimX

6

Proof. (a).Suppose Tx = 0 =) x = 0. Let Tx1 = Tx2. Since T is linear,

T(x1 ? x2) = Tx1 ? Tx2 = 0;

so that x1 ? x2 = 0 by hypothesis. Hence Tx1 = Tx2 =) x1 = x2 and T?1 exist by remark on

Denition 1.4. Conversely T?1 exists then remark on denition 1.4 holds.

From Denition 1.4 with x2 = 0, we obtain Tx1 = T0 = 0 =) x1 = 0.

(b). We assume T?1 exists and show that it is linear. The domain of T?1 is R(T) and it is a

vector space, then by Theorem 1.7a, we consider any x1; x2 2 D(T) and their images

y1 = Tx1 and y2 = Tx2, then x1 = T?1y1 and x2 = T?1y2. T is linear so that for any scalar

and , we have

y1 + y2 = Tx1 + Tx2 = T(x1 + x2):

It implies T?1(y1 + y2) = x1 + x2 = T?1y1 + T?1y2: It implies T?1 is linear.

(c). We have dimR(T) dimX by Theorem 1.7b and dimX dimR(T) by the same theorem

applied to T?1. Hence, dim X = dimR(T).

Lemma 1.13. (Inverse of product)

Let T : X ?! Y and S : Y ?! Z be bijective linear operator, where X; Y;Z are vector spaces.Then

the inverse (ST)?1 : Z ?! X of the product (the composite) ST exists and (ST)?1 = T?1S?1.

Proof. The operator ST : X ?! Z is bijective, so (ST)?1 exists. We have

(ST)(ST)?1 = IZ;

where IZ is the identity operator on Z. Applying S?1 and using S?1S = IY (the identity operator

on Y ), we obtain

S?1(ST)(ST)?1 = T(ST)?1 = S?1IZ = S?1:

Applying T?1 and using T?1T = IX, we obtain the desired result

T?1T(ST)?1 = (ST)?1 = T?1S?1:

Implies (ST)?1 = T?1S?1.

Theorem 1.14. Every linear operator on a nite dimensional vector space can be represented by

means of matrix.

Proof. Let X and Y be nite dimensional vector spaces over thesame eld. Let T : X ?! Y

be a linear operator, let dimX = n and dimY = r, then there exists a basis fe1; e2; :::eng for X and

a basis fb1; b2; :::brg for Y.

Let x 2 X ) x =

Pn

i=1 iei where 0i

s are scalars . Since T is linear

y = T(x) =

Xn

i=1

iT(ei):

7

So T is uniquely determined if the images Tei 1 i n are prescribed. Since y and Tei are in

Y so y =

Pr

j=1 jbj and Tei =

Pr

j=1 jibj where j and ji are scalars, thus

y =

Xr

j=1

jbj =

Xn

i=1

iT(ei) =

Xn

i=1

i

Xr

j=1

jibj =

Xr

j=1

(

Xn

i=1

jii)bj :

Hence

j =

Xn

i=1

jii: 1 j r:

1.4 Bounded linear operators

Denition 1.15. (Bounded linear operator): Let X and Y be normed spaces and T:X ?! Y be

linear operator. The operator T is said to be bounded if there exist a real number c >0 such that

jjTxjj cjjxjj for all x 2 D(T).

Theorem 1.16. Let T : X ?! Y be a bounded linear operator. Then

jjTjj := sup

x2X;jjxjj=1

jjTxjj = sup

x2X;x6=0

jjTxjj

jjxjj

:

Proof. Let jjxjj = a, set y = ( 1

a )x, where x 6= 0. Then jjyjj = jjxjj

a = 1. Since T is linear, then

sup

x2X;x6=0

jjTxjj

jjxjj

= sup

jjTxjj

a

= sup jjT(

1

a

)x)jj = sup

y2X;jjyjj=1

jjTyjj := jjTjj:

Remark: jj:jj denes a norm on X.

Theorem 1.17. (Finite dimension): If a normed space X is nite dimensional, then every linear

operator on X is bounded.

Proof. Let dim X = n and fe1; e2; : : : ; eng be a basis for X, then for all x 2 X;

x =

Xn

i=1

iei

i scalars. Since T is linear,

jjTxjj = jj

Xn

i=1

iTeijj

Xn

i=1

jijjjTeijj max

i

jjTeijj

Xn

i=1

jij = jjxjj1 where ( = max jjTeijj)

= cjjxjj (where c= k by equivalence of norms on nite dimensional vector space)

Theorem 1.18. (Continuity and boundedness): Let T : X ?! Y be a linear operator, where X

and Y are normed spaces. Then:

(a) T is continuous if and only if T is bounded.

8

(b) If T is continuous at the origin, then T is continuous.

Proof. (a) For T = 0, the statement is trivial. Let T 6= 0,then jjTjj 6= 0. We assume T is

bounded and consider x0 2 X such that jjx ? x0jj < where = =jjTjj, we obtain

jjTx ? Tx0jj = jjT(x ? x0)jj jjTjjjjx ? x0jj < jjTjj = :

Since x0 2 X was arbitrary, this shows that T is continuous.

Conversely, assume that T is continuous at an arbitrary x0 2 X, then given any > 0, there exist

> 0 such that

jjTx ? Tx0jj

for all x 2 X satisfying

jjx ? x0jj :

We now take y 6= 0 2 X and set x = x0 +

jjyjjy. Then x ? x0 =

jjyjjy. Hence jjx ? x0jj = . Since

T is linear we have

jjTx ? Tx0jj = jjT(x ? x0)jj = jjT(

jjyjj

y)jj =

jjyjj

jjTyjj :

Thus jjTyjj

jjyjj; jjTyjj cjjyjj =) T is bounded, where c =

(b) Suppose T is continuous at a point x0 = 0, then it suces to show that T is bounded

(continuous). T is continuous at x0 = 0, take = 1, there exist > 0. such that

jjxjj =) jjTxjj 1:

Let z 2 D(T) z 6= 0, then jj z

jjzjj

2 jj =

2 < =) jjT( z

jjzjj

2 )jj < 1

(By linearity of T) =) jjTzjj 2

jjzjj 8z 2 D(T) =) T is continuous.

Corollary 1.19. (Continuity and null space)

Let T : X ?! Y be a bounded linear operator. Then

(a) xn ?! x implies Txn ?! Tx:

(b)The null space N(T) is closed.

Proof. (a) Suppose xn ?! x in X i.e jjxn ? xjj ?! 0. Since T is linear and bounded, then

jjTxn ? Txjj = jjT(xn ? x)jj kjTjjjjxn ? xjj ?! 0:

It implies Txn ?! Tx as n ?! 1

(b) Let x 2 N(T) it implies there exist (xn)n1 N(T) such that xn ?! x. Since T is bounded

by corollary 1.19a Txn ?! Tx, but xn 2 N(T): It implies Txn = 0 8n 1, thus T(x) = 0:

It implies x 2 N(T). Hence N(T) is closed.

9

1.5 Examples of bounded operators on innite dimensional spaces

1. Let K : [0; 1] [0; 1] ?! R be continuous. Let T : C([0; 1];R) ?! C([0; 1];R) be dened by

T(f)(x) =

Z 1

0

K(x; y)f(y)dy:

Then T 2 L(C[0; 1]) for f 2 C[0; 1] and bounded.

Proof. Clearly T is linear. We next show boundedness.

jT(f)(x)j

Z 1

0

jK(x; y)jjf(y)jdy supjf(y)j

Z 1

0

jK(x; y)jdy jjfjj

Z 1

0

jK(x; y)jdy:

It implies

jjT(f)(x)jj1 cjjfjj1

where

R 1

0 jK(x; y)j c since K is continuous. It implies T is bounded.

2. Let p 1, we dene

lp = f(xn)n1 R :

1X

n=1

jxnjp < 1g

Let T : lp ?! lp be dened by

T((xn)n1) = (xn+1)n1 (The left shift operator) is bounded

where (xn)n1 = (x1; x2; x3; :::) and T((xn)n1) = (x2; x3; :::)

Proof.

jjT((xn)n1)jj = (

1X

n=2

jxnjp)1=p (

1X

n=1

jxnjp)1=p = jj(xn)n1jj:

Thus T is bounded with jjTjj 1

3. Let T : L2([0; 1];R) ?! L2([0; 1];R) be dened by

(Tf)(t) = tf (t) for a:e t 2 [0; 1]:

Then T is bounded.

Proof. jjTfjj2

L2[0;1] =

R 1

0 j(Tf)(t)j2dt =

R 1

0 jtj2jf(t)j2dt

R 1

0 jf(t)j2dt = jjfjj2

L2[0;1]:

It implies jjTfjjL2[0;1] jjfjjL2[0;1]: Hence T is bounded with jjTjj 1

10

1.6 Hilbert spaces

Denition 1.20. Let E be a real vector space. An inner product on E is a function,

h:; :i : E E ?! R such that

(a) jjxjj2 hx; xi 0 with equality jjxjj2 = 0 i x = 0

(b) hx; yi = hy; xi

(c)hax + by; zi = ahx; zi + bhy; zi i.e x ?! hx; zi is linear.

A real vector space E endowed with the inner product i.e (E; h:; :i) is called an inner product space.

Lemma 1.21. (Cauchy-Schwartz Inequality) Let E be an inner product space. Then for arbitrary

x; y 2 E,

jhx; yij jjxjjjjyjj

Lemma 1.22. (The Parallelogram Law) Let E be a real inner product space. Then for arbitary

vector x; y 2 E,

jjx + yjj2 + jjx ? yjj2 = 2(jjxjj2 + jjyjj2):

Proof. Expanding the LHS jjx+yjj2+jjx?yjj2 = hx; xi+2hx; yi+hy; yi+hx; xi?2hx; yi+hy; yi

= 2(hx; xi + hy; yi) = 2(jjxjj2 + jjyjj2) = RHS.

Denition 1.23. A complete inner product space is called a Hilbert space.

Denition 1.24. Let x, y be vectors in a Hilbert space H, then we say that x and y are orthog-

onal,written x ? y, if hx; yi = 0. We say that subsets A and B are orthogonal, written A ? B, if

x ? y for every x 2 A and y 2 B. The orthogonal complement A? of a subset of A is the set of

vectors orthogonal to A,

A? = fx 2 H : x ? y for all y 2 Ag:

Denition 1.25. Let M and N be closed linear subspaces of a Hilbert space H, we dene the

orthogonal direct sum or simply the direct sum M

L

N of M and N by

M

M

N = fy + z : y 2M and z 2 Ng:

Denition 1.26. A subset U of nonzero vectors in a Hilbert space H is orthogonal if any two

distinct elements in U are othorgonal. A set of vectors U is orthonormal if it is orthogonal and

jjujj = 1 for all u 2 U.

1.7 Some properties of Hilbert spaces

Theorem 1.27. The orthogonal complement of a subset of a Hilbert space is a closed linear

subspace.

11

Proof. Let H be a Hilbert space and A a subset of H. if y; z 2 A? and ; 2 R:Then the

linearity of the inner product implies that

hx; y + zi = hx; yi + hx; zi = 0

for all x 2 A:

Therefore, y + z 2 A?, so A? is a linear subspace.

To show that A? is closed, we show that if (yn)n1 is a convergent sequence in A?, then the limit

y also belongs to A?. Let x 2 A then by continuity of inner product we have

hx; yi = hx; lim

n?!1

yni = lim

n?!1

hx; yni = 0:

Since hx; yni = 0 for every x 2 A and yn 2 A?. Hence y 2 A?

Theorem 1.28. Let M be a closed linear subspace of a Hilbert space H

(a) For every x 2 H there is a unique closest point y 2M such that

jjx ? yjj = minjjx ? zjj; z 2M

(b)The point y 2Mclosest to x 2 H is the unique element ofMwith the property that (x?y) ?M

Proof. (a). Let d be the distance of x from M i.e

d = inffjjx ? zjj : z 2 Mg:

First, we prove that there is a closest point y 2 M at which this inmum is attained, meaning

that jjx ? yjj = d: From the denition of d, there is a sequence of elements yn 2M such that

lim

n?!1

jjx ? ynjj = d:

Thus, for any ” > 0; there is an N such that

jjx ? ynjj d + ” when n N:

We show that the sequence (yn)n1 is Cauchy. From the parallelogram law, we have

jjym ? ynjj2 + jj2x ? ym ? ynjj2 = 2jjx ? ymjj2 + 2jjx ? ynjj2:

Since (ym + yn)=2 2M; it implies that jjx ? (ym + yn)=2jj d. Thus for all m; n N

jjym ? ynjj2 = 2jjx ? ymjj2 + 2jjx ? ynjj2 ? jj2x ? ym ? ynjj2 4(d + “)2 ? 4d2 = 4″(2d + “):

Therefore,(yn)n1 is Cauchy.Since a Hilbert space is complete, there is a y such that yn ?! y and

since M is closed, we have y 2M: By continuity of norm we have

jjx ? yjj = lim

n?!1

jjx ? ynjj = d

12

We prove the uniqueness of the vector y 2Mthat minimizes jjx?yjj. Suppose that y and y0 both

minimize the distance to x, meaning that jjx ? yjj = d, jjx ? y0jj = d:

Then the parallelogram law implies that

2jjx ? yjj2 + 2jjx ? y0jj2 = jj2x ? y ? y0jj2 + jjy ? y0jj2:

Since (y + y0)=2 2M,

jjy ? y0jj2 = 4d2 ? 4jjx ? (y + y0)=2jj2 0:

Therefore, jjy ? y0jj = 0 so that y = y0: (b) We show that the unique y 2M found above satises

the condition that the vector x ? y is orthogonal to M. Since y minimizes the distance to x, we

have for every 2 C and z 2M that

jjx ? yjj2 jjx ? y + zjj2:

Expanding the right-hand side of this equation, we obtain that

2Rehx ? y; zi jj2jjzjj2:

Suppose that hx ? y; zi = jhx ? y; zijei’: Choosing = “e?i’; where ” > 0 and dividing by “, we

get

2jhx ? y; zij “jjzjj2:

Taking the limit as ” ?! 0+, we get hx ? y; zi = 0 so (x ? y) ?M.

Finally, we show that y is the only element in M such that (x ? y) ? M: Suppose that y0 is

another such element in M: Then y ? y0 2M; and for any z 2M; we have

hz; y ? y0i = hz; x ? y0i ? hz; x ? yi = 0:

In particular, we may take z = y ? y0 and therefore we have y = y0

Denition 1.29. Let

be an open set in Rn, then

Lp(

) = ff :

?! R; measurable :

R

jfjpdx < 1g, 1 p < 1.

Denition 1.30. Let

be an open set in Rn and n 2 N, the Sobolev space Hm(

) is dened by

Hm(

) = ff 2 L2(

);Df 2 L2(

); 2 Nn; jj mg where Df = @jj

@x

1

1 :::@xn

n

f, jj = 1 + ::: +

n and jjujjHm(

) = jjujjL2 +

P

:jjm jDujL2 ; u 2 Hm(

)

Denition 1.31. Let ‘ :

?! R be continous, then the support of ‘ is dened by

Supp(‘) = fx 2

: ‘(x) 6= 0g

Denition 1.32. D(

) the space of test functions is dened by

D(

) = ff 2 C1 : Supp(f) is compact in

g

13

Denition 1.33. A distribution is a continuous linear map T : D(

) ?! R such that

lim

n?!1

T(‘n) = T(‘)

for any sequence

‘n

D(

)

?! ‘:

The space of distribution on

is denoted by D0(

).

Denition 1.34. A sequence of distribution Tn 2 D0(

) is said to converge in the sense of

distribution to T 2 D0(

), if for every test function ‘ 2 D(

) one has

limn?!1hTn; ‘i = hT; ‘i

1.7.1 Examples of Hilbert spaces

(a). L2(

) equipped with the norm jjfjjL2(

) = (

R

jfj2dx)1=2 is Hilbert.

(b). H1(

) equipped with the norm jjujj2

H1(

) =

R

u2dx +

R

jruj2dx is Hilbert, where

H1(

) = ff 2 L2(

) :

@f

@xi

2 L2(

)g

Proof. (a).Let (fn)n1 be a Cauchy sequence in L2(

), then we can nd a subsequence (fnk )k1

such that

jjfnk ? fnk+1jj <

1

2k ; k = 1; 2; 3; :::

Choose a function g 2 L2(

). By the Schwartz inequality,

Z

jg(fnk ? fnk+1jd

jjgjj

2k :

Hence

1X

k=1

Z

jg(fnk ? fnk+1jd jjgjj:

Thus

jg(x)j

1X

k=1

jfnk (x) ? fnk+1(x)j < +1

almost everywhere on X.

It implies

1X

k=1

jfnk (x) ? fnk+1(x)j < +1

almost everywhere on X.

Since the kth partial sum of the series

P1

k=1(fnk (x)?fnk+1(x)) which converges almost everywhere

on X is fnk (x) ? fnk+1(x):

It implies

f(x) = lim

k?!1

fnk (x):

14

Let ” > 0 be given, there exists N0 2 N such that

jjf ? fnk jj lim inf

j?!1

jjfnj ? fnk jj “:

Thus f ? fnk 2 L2(

), and since f = (f ? fnk ) + fnk , we see that f 2 L2(

).

Also, since ” is arbitrary,

lim

k?!1

jjf ? fnk jj = 0:

Finally, the inequality

jjf ? fnjj jjf ? fnk jj + jjfnk ? fnjj

shows that (fn) converges to f in L2(

).

(b). Let (un)n1 be a Cauchy sequence in H1(

) then given ” > 0 there exist n0 2 N such that

8m; n > n0

jjun ? umjjH1(

) < “:

which implies that

Z

jun ? umj2dx +

Z

jrun ? rumj2dx

1=2

< ” ():

Thus (un)n1 be a Cauchy sequence in L2(

) and (run)n1 is also a Cauchy sequence in L2(

).

Since L2(

) is complete,

un ?! u 2 L2(

) and run ?! wi 2 L2(

):

We need to show that ru = wi.

But

un ?! u 2 L2(

) ) un ?! u 2 D0(

);

thus

run ?! ru 2 D0(

):

By uniqueness of limit in D0(

), we have ru = wi

From (**), let n be xed and let m ?! 1 we have

Z

jun ? uj2dx +

Z

jrun ? ruj2dx

1=2

< “;

thus un ?! u in H1(

). Hence H1(

) is Hilbert.

15

Theorem 1.35. (Riesz Theorem) Let H be a Hilbert space over R or C. If T is a bounded linear

functional on H i.e T is a bounded operator from H to the eld R or C, then there exists some

g 2 H such that for every f 2 H we have

T(f) = hf; gi . Moreover, jjTjj = jjgjj:

Proof. We can choose an orthonormal basis j ; j 1 for H. Let T be bounded linear

functional and set aj = T(j ). Choose f 2 H, let cj = hf; ji and dene

fn =

Xn

j=1

cjj :

Since j forms a basis we know that jjfn ? fjj ?! 0 as n ?! 1.

Since T is linear we have

T(fn) =

Xn

j=1

ajcj (1)

Since T is bounded, say with norm jjTjj < 1 we have

jjT(fn) ? T(f)jj jjTjjjjfn ? fjj (2)

Because jjfn ? fjj ?! 0 as n ?! 1, we conclude from equations (1) and (2) that

T(f) = lim

n?!1)

T(fn) =

1X

j=1

ajcj (3)

Infact, the sequence aj must itself be square-summable. To see this, rst note that since jT(f)j

jjTjjjjfjj we have

j

1X

j=1

ajcj j jjTjj(

1X

j=1

c2j

)1=2 (4)

Equation (4) must hold for any square-summable sequence cj (since any cj corresponds to some

elements in H).Fix a positive integer N and dene a sequence cj = aj for j N, cj = 0 for j N.

Clearly such a sequence is square-summable and equation (4) then yields

j

XN

j=1

a2j

j jjTjj(

XN

j=1

a2j

)1=2

or

(

XN

j=1

a2j

)1=2 jjTjj (5)

Thus aj is square-summable the function g =

P

j ajj is well dened as an element of H and

T(f) =

X

j

ajcj = hf; gi:

16

Finally, equation (5) makes it clear that jjgjj jjTjj. But from Cauchy-Schwartz we also have

jT(f)j = jhf; gij jjfjjjjgjj implying jjTjj jjgjj, so jjTjj = jjgjj.

17

**GET THE COMPLETE PROJECT»**

Do you need help? Talk to us right now: (+234) 8111770269, 08111770269 (Call/WhatsApp). Email: [email protected]

**IF YOU CAN’T FIND YOUR TOPIC, CLICK HERE TO HIRE A WRITER»**

Disclaimer: This PDF Material Content is Developed by the copyright owner to Serve as a RESEARCH GUIDE for Students to Conduct Academic Research. You are allowed to use the original PDF Research Material Guide you will receive in the following ways: 1. As a source for additional understanding of the project topic. 2. As a source for ideas for you own academic research work (if properly referenced). 3. For PROPER paraphrasing ( see your school definition of plagiarism and acceptable paraphrase). 4. Direct citing ( if referenced properly). Thank you so much for your respect for the authors copyright. Do you need help? Talk to us right now: (+234) 8111770269, 08111770269 (Call/WhatsApp). Email: [email protected]

*Related Current Research Articles*

Purchase Detail

Hello, we’re glad you stopped by, you can download the complete project materials to this project with Abstract, Chapters 1 – 5, References and Appendix (Questionaire, Charts, etc) for ~~N~~4000 ($15) only, To pay with **Paypal**, **Bitcoin** or **Ethereum**; please click here to chat us up via Whatsapp.

You can also call **08111770269** or **+2348059541956** to place an order or use the whatsapp button below to chat us up.

Bank details are stated below.

**Bank:** UBA

**Account No:** 1021412898

**Account Name:** Starnet Innovations Limited

### The Blazingprojects Mobile App

**Tags:**current project topics in mathematics education m.sc maths project topics in differential equations m.sc maths project topics in graph theory pdf m.sc maths project topics in linear algebra m.sc maths project topics in topology pdf m.sc project topics in maths maths project topics in algebra msc maths project topics in algebra msc maths project topics in algebra pdf msc maths project topics in complex analysis msc maths project topics in differential equations msc maths project topics in graph theory msc maths project topics in graph theory pdf msc maths project topics in linear algebra msc maths project topics in number theory msc maths project topics in numerical analysis msc maths project topics in operations research msc maths project topics in operations research pdf msc maths project topics in real analysis msc maths project topics in topology msc maths project topics in topology pdf project ideas on maths project topic in mathematics project topic in mathematics education project topics for applied mathematics project topics for bsc maths students project topics for mathematics students project topics for maths project topics for maths and statistics department project topics for maths for class 8 project topics for msc maths project topics in discrete mathematics project topics in financial mathematics project topics in fuzzy mathematics project topics in industrial mathematics project topics in mathematics and statistics project topics in mathematics department project topics in mathematics education for undergraduate project topics in mathematics education pdf project topics in mathematics for degree students project topics in mathematics for pg project topics in mathematics pdf project topics in maths education project topics in pure mathematics project topics on maths project topics on maths education project topics related mathematics undergraduate project topics in mathematics

## Recent Comments