~~
~~
## Page 26

FUNCTIONAL ANALYSIS

1){ei}is complete.

2)x?{ei}=)x=0.

3)Ifxis an arbitrary vector in Hthen x=Pe i.

4)Ifxis an arbitrary vector in Hthen ||x||2=P||2.

Proof. a) To prove 1) implies 2)

Let us assume that 2) is not true. Then there exists a vector x6=0

such that x?{ei}. Deﬁne e=x

||x||then, {ei,e}is an orthonormal set

which properly contains {ei}. This is a contradiction as it is given that

{ei}is complete. Hence our assumption was wrong.

Therefore x?{ei}=)x=0 .

b) To prove 2) implies 3)

We know that xPe iis orthogonal to {ei}(by Theorem

2.9.4) So by assumption 2) we get xPe i=0 .

Thus x=Pe i. Hence proved.

c) To prove 3) implies 4)

||x||2= =e i,Pe i>(by assumption

3)) =PP

=P(as{ei}is orthonormal )

=P||2.

d) To prove 4) implies 1)

Suppose that {ei}is not complete.

i.e. it is a proper subset of an orthnormal set say {ei,e}

Since eis orthogonal to all e0

isso we get =0 . . . ( I ) .

Now using the assumption 4) we get ||e||2=P||2. . . (II)

From (I) and (II) we get ||e||2=0 . T h u s e=0,w h i c hi sc o n t r a d i c t i o n ,

aseis a unit vector. Hence our assumption is wrong.

Thus {ei}is complete.

e) To prove 2) implies 1)

Let us suppose that {ei}is not complete so {ei}is contained in an

orthonormal set say {e, ei}.Now eis orthogonal to every element in

{ei}. So by assumption in 2) we get e=0 ,w h i c hi sac o n t r a d i c t i o n .

Hence our assumption was wrong. Therefore {ei}is complete.

The equation, ||x||2=P||2is called Parseval’s Identity.

2.11 Separable Hilbert space

Deﬁnition 2.11. Dense set:

A subset Aof a metric space Xis said to be dense in Xif¯A=X

Deﬁnition 2.12. Separable space:

A metric space Xis said to be seprable if it has a countable subset

which is dense in X.

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CHAPTER 2. HILBERT SPACES

Example 5. The space of real numbers is seprable

Solution: The set Qof all rational numbers is countable and is dense

inR.

Theorem 2.11.1. IfHis separable, every orthonormal set in His

countable.

Proof. Given that His seperable.

LetAbe any dense set in HandNbe any orthonormal set in H.

To show that Nis countable.

Letx, ybe two distinct element in Nso =0. . . ( 1 )

Now ||xy||2== - - +

= + (from (1)).

=||x||2+||y||2= 2 . Hence ||xy||2=2 .T h u s ||xy||=p

2.

LetBxandBybe the neighbourhood of x, ywith radiusp

2/3.

Since Ais dense in H, soBx\A6=andBy\A6=...( 2 )

Also Bx\By=. . . (3)

IfNis uncountable we would have uncountably many such pairwise

disjoint neighbourhood hence Awould become uncountable. Thus we

will not be able to get dense set which is countable in H, which is

contradiction to the seperability of H. Therefore our assumption that

Nis uncountable is wrong. Thus Nis countable. Hence proved.

Theorem 2.11.2. LetYbe a subspace of a separble Hilbert space H

then Yis also seperable.

2.12 LET US SUM UP

•A space which is complete with respect to metric deﬁned by inner

product is called Hilbert space.

•The equation ||x+y||2+||xy||2=2 ( ||x||2+||y||2)i sc a l l e d

parallelogram equality.

•If a norm does not satisfy Parallelogram law then that norm can-

not be obtained from inner product. So all normed spaces are not

Hilbert spaces.

•Forx, ybe any two vectors in a Hilbert space

| |||x|| ||y||is called Schwarz iequality.

•Forx, ybe any two vectors in a Hilbert space

||x+y||||x||+||y||is called triangle inequality.

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FUNCTIONAL ANALYSIS

•IfYbe a ﬁnite dimensional subspace of a Hilbert space Hthen

Y is complete.

•IfYbe a subspace of a Hilbert space H, then Yis complete if

and only if Yis closed in H.

•There exists a unique vector of smallest norm in a closed convex

subset Cof a Hilbert Space H.

•A?is closed linear subspace.

•IfMis a closed linear subspace of a Hilbert space H,xis a vector

not in Msuch that dbe the distance from xtoM. Then there

exists a unique vector y0inMsuch that ||xy0||=d.

•IfMis a proper closed linear subspace of a Hilbert space H, then

there exists a non-zero vector z02Hsuch that z0?M.

•IfAis closed linear subspace of a Hilbert space H, then H=

AA?.

•If{ei}is an orthonormal set in a Hilbert space Hand if xis any

vector in Hthen the set S={ei, < x, e i>6=0}is either empty or

countable.

•If{ei}is an orthonormal set in a Hilbert space HthenP|<

x, ei>|2||x||28every vector x2H.. This is called Bessel’s

inequality.

•An orthonormal set {ei}inHis said to be complete if it is max-

imal in its partially ordered set.

•Every non-zero Hilbert space Hcontains a complete orthonormal

set.

•Hbe a Hilbert space, {ei}be an orthonormal set in Handxis

an arbitrary vector in Hthen x=Pe i. This is called

Parseval’s Identity.

•A metric space Xis said to be seprable if it has a countable subset

which is dense in X.

2.13 Chapter End Exercise

1. Any two norms on ﬁnite dimensional space are equivalent.

2. Prove Theorem 2.7.2

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CHAPTER 2. HILBERT SPACES

3. Prove Theorem 2.8.1

4. Prove Theorem 2.8.3

5. State and Prove Schwarz inequality.

6. Prove Theorem 2.11.2

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FUNCTIONAL ANALYSIS

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Chapter 3

Normed Spaces

Unit Structure :

3.1I n t r o d u c t i o n

3.2O b j e c t i v e

3.3 Few deﬁnitions and examples

3.4 Convergent Sequence and Cauchy Sequence in a Normed Space

3.5 LET US SUM UP

3.6C h a p t e rE n dE x e r c i s e

3.1 Introduction

In this chapter, you will be introduced to the notion of a norm on

a vector space. The concept of norm of a vector is a generalization of

the notion of length. The deﬁnition of a normed space (a vector space

equipped with a norm on it) was given (independently) by S. Banach,

H. Hahn and N. Wiener in 1922. In one section of this chapter, you

will study the concept of normed spaces which is fundamental to the

development of the theory of Banach spaces. You will come to know

the relation between a normed space and a metric space. In another

section of this chapter, you will learn about convergent sequences and

Cauchy sequences in a normed space.

3.2 Objectives

The main objective of this chapter is to learn the normed spaces and

Cauchy sequences in it.

After going through this chapter you will be able to:

•Deﬁne a norm on a vector space.

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FUNCTIONAL ANALYSIS

•Deﬁne a normed space.

•Learn how to check a vector space is a normed space under the given

norm.

•Learn H¨ older’s inequality and Minkowski’s inequality (for ﬁnite sums

and for integrals).

•Prove that every normed space is a metric space.

•Show that a metric space need not be a normed space.

•Deﬁne a convergent sequence and Cauchy sequence in a normed space.

•Prove that the quotient space is a normed space under the given norm.

•Prove that every convergent sequence in a normed space is a Cauchy

sequence.

3.3 Few deﬁnitions and examples

You have learnt the deﬁnition of a norm (and properties of a norm)

on an inner product space Vin your B.Sc. So you can guess the deﬁ-

nition of norm on a vector space and get convinced with the following

deﬁnition.

Deﬁnition 3.1. LetVbe a vector space over the ﬁeld F(=RorC). A

norm kkonVis a real valued function (i.e. kk:V!R), satisfying

the following 4 properties/axioms:

(N 1) kxk>0 8x2V

(N 2) kxk=0 i fa n do n l yi f x=0 v8x2V

(N 3) kx+yk6kxk+kyk8x, y2V

(N 4) k↵xk=|↵|kxk8 x2Vand 8↵2F

Now, you will come to know when a vector space is called, a normed

space.

Deﬁnition 3.2. Anormed space Vis a vector space over the ﬁeld

F(=RorC)w i t han o r m kkdeﬁned on it. In such a case, we say, ( V,

kk)i san o r m e ds p a c eo v e r F. Here, if F=Rthen Vis called a real

normed space and if F=Cthen Vis called a complex normed space .

Now, you will notice that examples of normed spaces are in abun-

dance.

Example 6. Show that the vector space R={x|x2R}over Ris a

normed space under the norm kxk=|x|=a b s o l u t ev a l u eo f x2R.

Solution: Using the properties of absolute value of a real number, we

have

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CHAPTER 3. NORMED SPACES

(1)8x2R,kxk=|x|>0a n dh e n c ep r o p e r t y( N1 )o fn o r mi s

satisﬁed.

(2)8x2R,kxk=0 ()|x|=0 ()x=0a n dh e n c ep r o p e r t y

(N 2) of norm is satisﬁed.

(3)8x, y2R,kx+yk=|x+y|6|x|+|y|=kxk+kykand hence

property (N 3) of norm is satisﬁed.

(4)8↵2Randx2R,k↵xk=|↵x|=|↵||x|=|↵|kxkand hence

property (N 4) of norm is satisﬁed.

Thus, Ris a normed space under deﬁned norm.

Can you recall the name of the following inequality?

nX

i=1|xiyi|6✓nX

i=1|xi|2◆1/2✓nX

i=1|yi|2◆1/2

for any complex (or real) numbers x1,···,xn;y1,···,yn

It’s the Cauchy-Schwarz inequality in an inner product space which

you have learnt in your B.Sc. You will revisit this inequality in further

example(s).

Now, recall the vector space Cn={(x1,···,xn)|xi2C}over C,

where the vector addition and scalar multiplication is deﬁned as follows,

respectively: for every scalar ↵2Cand ( x1,···,xn),(y1,···,yn)2

Cn

(x1,···,xn)+(y1,···,yn)=( x1+y1,···,xn+yn)

↵(x1,···,xn)=( ↵x1,···,↵xn)

Also, ( x1,···,xn)=(y1,···,yn)()xi=yi8i=1,···,n

Example 7. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a normed space under the norm kxk=✓nX

i=1|xi|2◆1/2

.

Solution: Using the properties of modulus of a complex number, we

have

(1)8x=(x1,···,xn)2Cn, each |xi|>0, and hence✓nX

i=1|xi|2◆1/2

=

kxk>0. So property (N 1) of norm is satisﬁed.

(2)8x=(x1,···,xn)2Cn,

kxk=0()kxk2=0()nX

i=1|xi|2=0()|xi|=0 ( 1 6i6n)

()xi=0( 1 6i6n)()x=(x1,···,xn)=( 0 ,···,0) = 0

and hence property (N 2) of norm is satisﬁed.

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FUNCTIONAL ANALYSIS

(3)8x=(x1,···,xn),y=(y1,···,yn)2Cn,

consider

kx+yk2=nX

i=1|xi+yi|2

=nX

i=1|xi+yi||xi+yi|

6nX

i=1|xi+yi|(|xi|+|yi|) by triangle inequality

=nX

i=1|xi+yi||xi|+nX

i=1|xi+yi||yi|

=nX

i=1|(xi+yi)xi|+nX

i=1|(xi+yi)yi|

6kx+ykkxk+kx+ykkyk

by Cauchy Schwarz inequality

=kx+yk✓

kxk+kyk◆

Ifkx+yk6=0t h e nd i v i d i n gb o t hs i d e sb y kx+yk, we get,

kx+yk6kxk+kyk. Ifkx+yk=0t h e nt h ei n e q u a l i t y kx+yk6

kxk+kykis trivial, since both sides reduce to zero and hence in

any case, property (N 3) of norm is satisﬁed.

(4)8↵2Candx=(x1,···,xn)2Cn,

k↵xk2=nX

i=1|↵xi|2=nX

i=1|↵|2|xi|2=|↵|2✓nX

i=1|xi|2◆

=|↵|2kxk2

Taking positive square root on both sides, we get, k↵xk=|↵|kxk

and hence property (N 4) of norm is satisﬁed.

Thus, Cnis a normed space under deﬁned norm.

Note: This norm is referred as Euclidean norm inCnand is de-

noted as kxk2. Ifn= 1 then kxk=p

x2

1+x2

2where x=x1+ix2.

Clearly, the notion of norm is actually a generalisation of the concept

of (Euclidean) length.

Recall the inequality between arithmetic mean and geometric mean.

It states that 8↵,2R+,p

↵61

2(↵+)

The generalisation of inequality between arithmetic mean and geo-

metric mean is given in the following lemma.

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CHAPTER 3. NORMED SPACES

Lemma 3.3.1. Let0<<1. Then ↵16↵+( 1 )holds

good for every pair of non-negative real numbers ↵and.

Proof. If either ↵=0o r =0t h e nw ea r ed o n e . S o ,a s s u m et h a t

↵>0a n d >0. For every non-negative real number t, deﬁne a

function as(t)=( 1 )+tt. For extreme value of , we must

have, 0(t)=0w h i c hi m p l i e s (1t1)=0a n dh e n c e t1=1a s

6=0 . T h u s t=1 . S o ,a t t=1 ,w eh a v ee x t r e m ev a l u eo f . It is easy

to see that

0(t)(

<0i f t<1

>0i f t>1

This implies that attains minimum at t=1 . T h u s , (t)>(1) which

gives (1 )+t>t. In this last inequality, put t=↵

where ↵and

are non-negative real numbers. Then (1 )+✓↵

◆

>✓↵

◆

and

on multiplying by throughout, we are done.

Deﬁnition 3.3. Letpandqbe non-negative extended real numbers.

Forp>1,qis said to be conjugate ofpif

1

p+1

q=1,f o r 1

q=1,f o rp =1

q=1,f o r p =1

Let 1 6p<1. To show that Cnis a normed space under the

norm kxk=✓nX

i=1|xi|p◆1/p

, we need a special inequality called as

Minkowiski’s inequality for ﬁnite sums. To prove this Minkowiski’s

inequality, we need another special inequality called as H¨ older’s in-

equality for ﬁnite sums. You will see these inqualities as following two

Lemma’s.

Lemma 3.3.2. H¨older’s inequality for ﬁnite sums:

Let1

p+1

q=1. For any complex (or real) numbers

x1,···,xn;y1,···,yn

nX

i=1|xiyi|6✓nX

i=1|xi|p◆1/p✓nX

i=1|yi|q◆1/q

Proof. Ifxiandyiare all zero then the result is obvious. So let atleast

onexi6=0a n da t l e a s to n e yi6=0 .

By Lemma 3.3.1, for ↵>0a n d >0, we have ↵16↵+(1).

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## Page 36

FUNCTIONAL ANALYSIS

In this inequality, take =1

p,↵=✓|xi|

(Pn

i=1|xi|p)1/p◆p

and =

✓|yi|

(Pn

i=1|yi|q)1/q◆q

. Then we get, 1 =1

qand

|xi|

(Pn

i=1|xi|p)1/p|yi|

(Pn

i=1|yi|q)1/q61

p|xi|p

(Pn

i=1|xi|p)+1

q|yi|q

(Pn

i=1|yi|q)

8i=1,···,n.

Adding all these inequalities, we get

Pn

i=1|xi||yi|

(Pn

i=1|xi|p)1/p(Pn

i=1|yi|q)1/q61

pPn

i=1|xi|p

(Pn

i=1|xi|p)+1

qPn

i=1|yi|q

(Pn

i=1|yi|q)

Thus,nX

i=1|xiyi|6✓nX

i=1|xi|p◆1/p✓nX

i=1|yi|q◆1/q

.

Note: If we take p=2i nH ¨ older’s inequality then q=2a n dw e

get the Cauchy-Schwarz inequality. So, Cauchy-Schwarz inequality is a

special case of H¨ older’s inequality.

Lemma 3.3.3. Minkowski’s inequality for ﬁnite sums: Let 16

p<1. For any complex (or real) numbers x1,···,xn;y1,···,yn

✓nX

i=1|xi+yi|p◆1/p

6✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

Proof. Ifp=1t h e n✓nX

i=1|xi+yi|p◆1/p

=nX

i=1|xi+yi|6nX

i=1(|xi|+|yi|)

=nX

i=1|xi|+nX

i=1|yi|=✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

Therefore the inequality holds for p=1 . S ol e t p>1a n d1

q=11

p

so that q>1. Then p=(p1)qandpp

q=1

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CHAPTER 3. NORMED SPACES

Now by H¨ older’s inequality in Lemma 3.3.2 (for ﬁnite sums), we have

nX

i=1|xi||xi+yi|p16✓nX

i=1|xi|p◆1/pnX

i=1✓

|xi+yi|p1◆q1/q

=✓nX

i=1|xi|p◆1/pnX

i=1|xi+yi|q(p1)1/q

=✓nX

i=1|xi|p◆1/pnX

i=1|xi+yi|p1/q

=✓nX

i=1|xi|p◆1/p✓nX

i=1|xi+yi|p◆1/pp/q

!(⇤)

NownX

i=1|xi+yi|p=nX

i=1|xi+yi|·|xi+yi|p1

6nX

i=1(|xi|+|yi|)|xi+yi|p1

=nX

i=1|xi||xi+yi|p1+nX

i=1|yi||xi+yi|p1

On using ( ⇤) to the two summation terms on RHS, we get,

nX

i=1|xi+yi|p6✓nX

i=1|xi|p◆1/p✓nX

i=1|xi+yi|p◆1/pp/q

+✓nX

i=1|yi|p◆1/p✓nX

i=1|xi+yi|p◆1/pp/q

)nX

i=1|xi+yi|p6✓nX

i=1|xi+yi|p◆1/q✓nX

i=1|xi|p◆1/p

+

✓nX

i=1|yi|p◆1/p

Hence,

✓nX

i=1|xi+yi|p◆1/p

6✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

.

Example 8. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a normed space under the norm kxk=✓nX

i=1|xi|p◆1/p

where

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FUNCTIONAL ANALYSIS

16p<1.

Solution: Using the properties of modulus of a complex number, we

have

(1)8x=(x1,···,xn)2Cn, each |xi|>0, and hence✓nX

i=1|xi|p◆1/p

=kxk>0. So property (N 1) of norm is satisﬁed.

(2)8x=(x1,···,xn)2Cn,

kxk=0()✓nX

i=1|xi|p◆1/p

=0

()nX

i=1|xi|p=0

()|xi|=0 8i=1,···,n

()xi=0 8i=1,···,n

()x=(x1,···,xn)=( 0 ,···,0) = 0

and hence property (N 2) of norm is satisﬁed.

(3)8x=(x1,···,xn),y=(y1,···,yn)2Cn, by Minkowski’s in-

equality in Lemma 3.3.3 (for ﬁnite sums), we have,✓nX

i=1|xi+yi|p◆1/p

6✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

=)kx+yk6kxk+kykand hence property (N 3) of norm is

satisﬁed.

(4)8↵2Candx=(x1,···,xn)2Cn,

k↵xk=✓nX

i=1|↵xi|p◆1/p

=✓nX

i=1(|↵||xi|)p◆1/p

=✓

|↵|pnX

i=1|xi|p◆1/p

=|↵|✓nX

i=1|xi|p◆1/p

=|↵|kxk

and hence property (N 4) of norm is satisﬁed.

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CHAPTER 3. NORMED SPACES

Thus, Cnis a normed space under deﬁned norm.

Example 9. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a normed space under the norm kxk=max{|x1|,···,|xn|}.

(This norm is referred as kxk1onCn).

Solution: Using the properties of of modulus of complex number, we

have

(1)8x=(x1,···,xn)2Cn, each |xi|>0, and hence max{|x1|,

···,|xn|}=kxk>0. So property (N 1) of norm is satisﬁed.

(2)8x=(x1,···,xn)2Cn,

kxk=0()max{|x1|,|x2|,···,|xn|}=0

()|xi|=0 8i=1,···,n

()xi=0 8i=1,···,n

()x=(x1,···,xn)=( 0 ,···,0) = 0

and hence property (N 2) of norm is satisﬁed.

(3)8x=(x1,···,xn),y=(y1,···,yn)2Cn,

kx+yk=max{|x1+y1|,···,|xn+yn|}

6max{|x1|+|y1|,···,|xn|+|yn|}

6max{|x1|,···,|xn|}+max{|y1|,···,|yn|}

=kxk+kyk

and hence property (N 3) of norm is satisﬁed.

(4)8↵2Candx=(x1,···,xn)2Cn,

k↵xk=max{|↵x1|,···,|↵xn|}

=max{|↵||x1|,···,|↵||xn|}

=|↵|max{|x1|,···,|xn|}

=|↵|kxk

and hence property (N 4) of norm is satisﬁed.

Thus, Cnis a normed space under deﬁned norm.

Remark 3.3.1. In view of norms deﬁned on Cnin examples 8 and 9,

we have

kxkp=8

><

>:✓nX

i=1|xi|p◆1/p

if 1 6p<1

max{|x1|,···,|xn|}ifp=1

These norms are referred as p-norms on Cn.

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FUNCTIONAL ANALYSIS

Remark 3.3.2. From examples 7, 8 and 9, it is clear that on a vector

space V, we can deﬁne more than one norm and accordingly di ↵erent

normed spaces are obtained from same vector space V.

Recall a sequence space which is a vector space whose elements are

inﬁnite sequences of real or complex numbers where the vector addition

and scalar multiplication, respectively are deﬁned as follows:

{x1,···,xn,···}+{y1,···,yn,···}={x1+y1,···,xn+yn,···}

↵{x1,···,xn,···}={↵x1,···,↵xn,···}

for every scalar ↵2CorR

Also, {x1,···,xn,···}={y1,···,yn,···}if and only if xi=yi8i=

1,···, n,···.

Example 10. Let 1 6p<1. Show that the sequence space lp=⇢

{x1,···,xn,···}1X

i=1|xi|p<1and x i2C

over Cis a normed

space under the norm kxk=✓1X

i=1|xi|p◆1/p

.

Solution: Using the properties of modulus of a complex number, we

have

(1)8x={x1,···,xn,···}2lp, each |xi|>0, and hence✓1X

i=1|xi|p◆1/p

=kxk>0. So property (N 1) of norm is satisﬁed.

(2)8x={x1,···,xn,···}2lp,

kxk=0()✓1X

i=1|xi|p◆1/p

=0

()1X

i=1|xi|p=0

()|xi|=0 8i=1,···, n,···

()xi=0 8i=1,···, n,···

()x={x1,···,xn,···}={0,···,0,···}=0

and hence property (N 2) of norm is satisﬁed.

(3) Let x={x1,···,xn,···},y={y1,···,yn,···}2lp.

By Minkowski’s inequality in Lemma 3.3.3 (for ﬁnite sums), we

have,

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CHAPTER 3. NORMED SPACES

✓nX

i=1|xi+yi|p◆1/p

6✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

6✓1X

i=1|xi|p◆1/p

+✓1X

i=1|yi|p◆1/p

!(⇤)

This is true for all n2N. As the 2 series on RHS of ( ⇤)c o n v e r g e ,

it is clear that the series on LHS of ( ⇤)m u s tc o n v e r g e . T h e r e f o r e ,

✓1X

i=1|xi+yi|p◆1/p

6✓1X

i=1|xi|p◆1/p

+✓1X

i=1|yi|p◆1/p

=)kx+yk6kxk+kykand hence property (N 3) of norm is

satisﬁed.

(4)8↵2Candx={x1,···,xn,···}2lp,

k↵xk=✓1X

i=1|↵xi|p◆1/p

=✓1X

i=1(|↵||xi|)p◆1/p

=✓

|↵|p1X

i=1|xi|p◆1/p

=|↵|✓1X

i=1|xi|p◆1/p

=|↵|kxk

and hence property (N 4) of norm is satisﬁed.

Thus, lpis a normed space under deﬁned norm.

Example 11. Show that the sequence space lp=⇢

{x1,···,xn,···}sup

{|x1|,···,|xn|,···} <1and x i2C

over Cis a normed space

under the norm kxk=sup{|x1|,···,|xn|,···}.

(This norm is denoted as kxk1onlp).

Solution: Using the properties of modulues of a complex number, we

have

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FUNCTIONAL ANALYSIS

(1)8x={x1,···,xn,···} 2lp, each |xi|>0, and hence sup{|x1|

,···,|xn|···} =kxk>0. So property (N 1) of norm is satisﬁed.

(2)8x={x1,···,xn,···}2lp,

kxk=0()sup{|x1|,|x2|,···,|xn|,···}=0

()|xi|=0 8i=1,···, n,···

()xi=0 8i=1,···, n,···

()x={x1,···,xn···}={0,···,0,···}=0

and hence property (N 2) of norm is satisﬁed.

(3)8x={x1,···,xn,···},y={y1,···,yn,···}2lp,

kx+yk=sup{|x1+y1|,···,|xn+yn|,···}

6sup{|x1|+|y1|,···,|xn|+|yn|,···}

6sup{|x1|,···,|xn|,···}+sup{|y1|,···,|yn|,···}

=kxk+kyk

and hence property (N 3) of norm is satisﬁed.

(4)8↵2Candx={x1,···,xn,···}2lp,

k↵xk=sup{|↵x1|,···,|↵xn|,···}

=sup{|↵||x1|,···,|↵||xn|,···}

=|↵|sup{|x1|,···,|xn|,···}

=|↵|kxk

and hence property (N 4) of norm is satisﬁed.

Thus, lpis a normed space under deﬁned norm.

Remark 3.3.3. In view of norms deﬁned on lpin examples 10 and 11,

we have

kxkp=8

><

>:✓1X

i=1|xi|p◆1/p

if 1 6p<1

sup{|x1|,···,|xn|,···} ifp=1

These norms are referred as p-norms on lp.

Recall the measure theory and Lebesgue integration that you learnt.

Consider a measure space ( E,S,µ ), where Eis a measurable set, Sis

a-algebra and µis a measure on S.F o r1 6p<1, let Lp(E,µ)=⇢

f:E!Rfis a measurable function on E&Z

E|f(x)|pdx < 1

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CHAPTER 3. NORMED SPACES

For convenience, (1) we denoteZ

E|f(x)|pdx < 1asZ

E|f|p<1.

(2) we take E=[a, b]a n d µas a Lebesgue measure. We write this

space as Lp(E).

It is easy to verify that Lp(E)i sav e c t o rs p a c eo v e r R. i.e.,

f+g2Lp(E)a n d ↵f2Lp(E)8f,g2Lp(E),8↵2R. Note

that the elements of Lp(E)a r ee q u i v a l e n c ec l a s s e so ft h o s ef u n c t i o n s ,

where fis equivalent to gifZ

E|fg|p=0 . i . e . t h ee l e m e n t so f Lp(E)

are equivalence classes of measurable functions which are equal almost

everywhere (a.e.).

To show that Lp(E)i san o r m e ds p a c eu n d e rt h en o r m kfk=✓Z

E|f|p◆1/p

where 1 6p<1, we need a special inequality called

as Minkowiski’s inequality for integrals. To prove this Minkowiski’s in-

equality, we need another special inequality called as H¨ older’s inequality

for integrals. You will see these inqualities as following two Lemma’s.

Lemma 3.3.4. (H¨older’s inequality for integrals) :

Let1

p+1

q=1. For f2Lp(E)and g2Lq(E)

(where Eis bounded closed interval in R),Z

E|fg|6✓Z

E|f|p◆1/p✓Z

E|g|q◆1/q

Proof. The inequality is trivial if either f=0a . e . o r g=0a . e . S ol e t

f6=0a . e . a n d g6=0a . e . T h e nZ

E|f|p>0a n dZ

E|g|q>0. By

Lemma 3.3.1, for ↵>0a n d >0, we have ↵16↵+( 1 ).

In this inequality, take =1

p,↵=✓|f|

(R

E|f|p)1/p◆p

and =

✓|g|

(R

E|g|q)1/q◆q

. Then we get, 1 =1

qand

|f|

(R

E|f|p)1/p|g|

(R

E|g|q)1/q61

p|f|p

(R

E|f|p)+1

q|g|q

(R

E|g|q)

On integrating, we get

R

E|f||g|

(R

E|f|p)1/p(R

E|g|q)1/q61

pR

E|f|p

(R

E|f|p)+1

qR

E|g|q

(R

E|g|q)

Thus,Z

E|fg|6✓Z

E|f|p◆1/p✓Z

E|g|q◆1/q

Note: Recall that if fandgare Lebesgue measurable over Ethen

the product fgis also Lebesgue measurable over E.I fR

E|f|p<1

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FUNCTIONAL ANALYSIS

andR

E|g|q<1then by H¨ older’s inequality,R

E|fg|<1.

i.e. if f2Lpandg2Lqthen fg2L

Lemma 3.3.5. (Minkowski’s inequality for integrals) :

Let16p<1.I f f2Lp(E)and g2Lp(E)(where Eis bounded

closed interval in R), then✓Z

E|f+g|p◆1/p

6✓Z

E|f|p◆1/p

+✓Z

E|g|p◆1/p

Proof. It is easy to see that if f and g are Lebesgue measurable over E

then f+gis also Lebesgue measurable over E. Also, ifZ

E|f|p<1

andZ

E|g|p<1thenZ

E|f+g|p<1.

Ifp=1t h e na s |f+g|6|f|+|g|, we are done. So let p>1a n d

1

q=11

pso that q>1. Then p=(p1)qandpp

q=1 .

Clearly,Z

E✓

|f+g|p1◆q

=Z

E|f+g|p<1.

Also,✓Z

E(|f+g|p1)q◆1/q

=✓Z

E|f+g|p◆1/q

. !(1)

Now by H¨ older’s inequality in Lemma 3.3.4 (for integrals),Z

E|f||f+g|p16✓Z

E|f|p◆1/p✓Z

E(|f+g|p1)q◆1/q

!(2)

andZ

E|g||f+g|p16✓Z

E|g|p◆1/p✓Z

E(|f+g|p1)q◆1/q

!(3)

Now, considerZ

E|f+g|p=Z

E|f+g|p1|f+g|6Z

E|f+g|p1(|f|+|g|)

=)Z

E|f+g|p6Z

E|f||f+g|p1+Z

E|g||f+g|p1

Using (1) ,(2),(3), we get,Z

E|f+g|p6⇢✓Z

E|f|p◆1/p

+✓Z

E|g|p◆1/p✓Z

E|f+g|p◆1/q

IfZ

E|f+g|p6=0t h e no nd i v i d i n gt h r o u g h o u tb yZ

E|f+g|pand

using 1 1

q=1

p, we get,

✓Z

E|f+g|p◆1/p

6✓Z

E|f|p◆1/p

+✓Z

E|g|p◆1/p

.

IfZ

E|f+g|p=0t h e nt h e r ei sn o t h i n gt op r o v ea n dw ea r ed o n e .

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CHAPTER 3. NORMED SPACES

Example 12. Let 1 6p<1&Ebe a (bounded closed interval in R)

measurable set. Show that the vector space Lp(E)={f:E!R|f

is Lebesgue measurable function on Eand|f|pis Lebesgue integrable

over E}over Ris a normed space under norm kfkp=✓Z

E|f|p◆1/p

.

(This norm is referred as p-norms on Lp(E)).

Solution: Using the properties of Lebesgue measurable and Lebesgue

integrable functions, we have

(1)8f2Lp(E),kfkp=✓Z

E|f|p◆1/p

>0. So property (N 1) of

norm is satisﬁed.

(2) Let f2Lp(E). If f=0a . e . t h e n kfkp=✓R

E|0|p◆1/p

=0 .

Conversely,

kfkp=0= )✓Z

E|f|p◆1/p

=0

=)|f|p=0 a.e.

=)|f|=0 a.e.

=)f=0 a.e.

(note that the condition kfk=0 ()f= 0 is not satisﬁed)

If we do not distinguish between equivalent functions then the

property (N 2) of norm is satisﬁed.

(3) Let f, g2Lp(E).

By Minkowski’s inequality 3.3.5 (for integrals), we have,

✓Z

E|f+g|p◆1/p

6✓Z

E|f|p◆1/p

+✓Z

E|g|p◆1/p

Therefore, kf+gkp6kfkp+kgkpand hence property (N 3) of

norm is satisﬁed.

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FUNCTIONAL ANALYSIS

(4)8↵2Randf2Lp(E),

k↵fkp=✓Z

E|↵f|p◆1/p

=✓Z

E(|↵||f|)p◆1/p

=✓

|↵|pZ

E|f|p◆1/p

=|↵|✓Z

E|f|p◆1/p

=|↵|kfkp

and hence property (N 4) of norm is satisﬁed.

Thus, Lp(E)i san o r m e ds p a c eu n d e rd e ﬁ n e dn o r m .

Deﬁnition 3.4. LetEbe a (bounded closed interval in R)m e a s u r a b l e

set. A measurable function f:E!Ris said to be essentially bounded

onEif there exists a ﬁnite real number m> 0s u c ht h a t |f(x)|6m

a.e. on E. Here mis called essential (upper) bound for f. (i.e. fis

bounded except possibly on a set of measure zero)

Iffhas an essential upper bound then least upper bound exists.

The least such bound is denoted by ess sup |f|. Iffdoes not have any

essential bound, then its essential supremum is deﬁned to be 1.

We deﬁne L1(E)=t h ec l a s so fa l lt h o s em e a s u r a b l ef u n c t i o n sd e -

ﬁned on Ewhich are essentially bounded on E. The elements of L1(E)

are equivalence classes of f. It is easy to verify that L1(E)i sav e c t o r

space over R.

Example 13. LetEbe a (bounded closed interval in R)m e a s u r a b l e

set. Show that the vector space L1(E)={f:E!R|fi sam e a -

surbale function on Eandess sup |f|<1}over Ris a normed space

under norm kfk1=esssup

E|f(x)|=inf{m> 0||f(x)|6ma.e. on

E}.

Hint.(Check)

(1)8f2L1(E),kfkp>0.

(2)8f2L1(E),kfk1=0i fa n do n l yi f f=0a . e .

(3) Let f, g2L1(E). Clearly, |f|6kfk1a.e. and |g|6kgk1a.e.

As|f+g|6|f|+|g|6kfk1+kgk1a.e., it follows that,

kf+gk16kfk1+kgk1

(4)8↵2Randf2L1(E),k↵fk1=|↵|kfk1

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CHAPTER 3. NORMED SPACES

Remark 3.3.4. In view of norms deﬁned on Lp(E)i ne x a m p l e s1 2a n d

13, we have

kfkp=8

><

>:✓Z

E|f|p◆1/p

if 1 6p<1

ess sup |f| ifp=1

Recall that the set C[0,1] = {f:[ 0,1]!R(orC)|fis a con-

tinuous function }is a vector space over RorCunder the opera-

tions ( f+g)(x)=f(x)+g(x) 8x2[0,1]

(↵f)(x)=↵f(x) 8x2[0,1]

Example 14. Show that the vector space C[0,1] = {f:[ 0,1]!R|

fis a continuous function }over Ris a normed space under norm

kfk=Z1

0|f(t)|dt.

Solution: Using the properties of Riemann integration and absolute

value of a real number , we have

(1) Let f2C[0,1]. As |f(t)|>08t2[0,1], we haveZ1

0|f(t)|dt>0

and hence kfk>0. So property (N 1) of norm is satisﬁed.

(2) Let f2C[0,1].

kfk=0()Z1

0|f(t)|dt=0

()|f(t)|=0 8t2[0,1]

()f(t)=0 8t2[0,1]

()f=0( zero function )

and hence property (N 2) of norm is satisﬁed.

(3) Let f, g2C[0,1].

kf+gk=Z1

0|(f+g)(t)|dt

=Z1

0|f(t)+g(t)|dt

6Z1

0|f(t)|dt+Z1

0|g(t)|dt

=kfk+kgk

Therefore, kf+gk6kfk+kgkand hence property (N 3) of

norm is satisﬁed.

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FUNCTIONAL ANALYSIS

(4)8↵2Randf2C[0,1],

k↵fk=Z1

0|(↵f)(t)|dt

=Z1

0|↵f(t)|dt

=Z1

0|↵||f(t)|dt

=|↵|Z1

0|f(t)|dt

=|↵|kfk

and hence property (N 4) of norm is satisﬁed.

Thus, C[0,1] is a normed space under deﬁned norm.

Example 15. Show that the vector space C[0,1] = {f:[ 0,1]!

R(orC)|fis a continuous function }over R(orC)i san o r m e ds p a c e

under norm kfk=s u p

x2[0,1]{|f(x)|}.

(This norm is referred as kfk1or sup norm on C[0,1]).

Solution: Using the properties of supremum and absolute value (or

modulus in C)o far e a ln u m b e r,w eh a v e

(1) Let f2C[0,1]. As |f(x)|>08x2[0,1], we have sup

x2[0,1]{|f(x)|}>

0a n dh e n c e kfk>0. So property (N 1) of norm is satisﬁed.

(2) Let f2C[0,1].

kfk=0()sup

x2[0,1]{|f(x)|}=0

()|f(x)|=0 8x2[0,1]

()f(x)=0 8x2[0,1]

()f=0( zero function )

and hence property (N 2) of norm is satisﬁed.

(3) Let f, g2C[0,1].

kf+gk=s u p

x2[0,1]{|(f+g)(x)|}

=s u p

x2[0,1]{|(f(x)+g(x)|}

6sup

x2[0,1]{|(f(x)|+|g(x)|}

=s u p

x2[0,1]{|f(x)|+s u p

x2[0,1]{|g(x)|}

=kfk+kgk

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CHAPTER 3. NORMED SPACES

Therefore, kf+gk6kfk+kgkand hence property (N 3) of

norm is satisﬁed.

(4)8↵2R(orC)a n d f2C[0,1],

k↵fk=s u p

x2[0,1]{|(↵f)(x)|}

=s u p

x2[0,1]{|↵f(x)|}

=s u p

x2[0,1]{|↵||f(x)|}

=|↵|sup

x2[0,1]{|f(x)|}

=|↵|kfk

and hence property (N 4) of norm is satisﬁed.

Thus, C[0,1] is a normed space under deﬁned norm.

Remark 3.3.5. 1. You can mimic Examples 14 and 15 for the vec-

tor space C[a, b] over RorC.

2. In view of norms on C[a, b] deﬁned in examples 14 and 15, we have

kfkp=8

>><

>>:✓Zb

a|f(t)|pdt◆1/p

if 1 6p<1

sup

t2[a,b]{|f(t)|} ifp=1

3. Note that the vector space C[a, b] is a particular case of the vector

space C(X)w h e r e Xis a compact space. So you can have similar

versions of Examples 14 and 15 for the vector space C(X)o v e r R

orC.

Deﬁnition 3.5. A normed space is called ﬁnite dimensional if the un-

derlying vector space is ﬁnite dimensional, otherwise it is called inﬁnite

dimensional .

Remark 3.3.6. The normed spaces in examples 6, 7, 8, 9 are ﬁnite

dimensional and the normed spaces in examples 10, 11, 12, 14 are

inﬁnite dimensional .

In a given normed space ( V,kk), for x, y2V,xy2V. (Why?).

This suggests us to give following deﬁnition.

Deﬁnition 3.6. In a given normed space ( V,kk), a function d:V⇥

V!R+deﬁned as d(x, y)=kxykis called the distance from xto

y, where x, y2V. Here dis referred as distance function on V.

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FUNCTIONAL ANALYSIS

Clearly, kxkis the distance from the zero vector to vector x2V.

Recall the deﬁnition of a metric space. In the next results, you will see

the relation between normed space and metric space.

Theorem 3.3.1. Every normed space is a metric space with respect to

the distance function

Proof. Let ( V,kk)b ean o r m e ds p a c ea n d dbe distance function on

V. So for x, y2V,d(x, y)=kxyk. Let x, y, z 2V

(1) By property (N 1) of norm, kxyk>0a n dh e n c e d(x, y)>0.

(2) By property (N 2) of norm,

kxyk=0()xy=0

()x=y

Thus, d (x, y)=0 ()x=y

(3) By property (N 4) of norm, kxyk=k(yx)k=|1|kyxk=

kyxk. Thus, d(x, y)=d(y,x).

(4) By property (N 3) of norm,

kxyk=k(xz)+(zy)k6kxzk+kzykand thus

d(x, y)6d(x, z)+d(z,y)

Therefore, all the conditions of the metric are satisﬁed by d.S o dis a

metric on Vand hence ( V,d)i sam e t r i cs p a c e .

The metric deﬁned in this way is called as the natural metric induced

by the norm.

Remark 3.3.7. You will come to know from the following example

that the converse of Theorem 3.3.1 need not be true.

Example 16. Let 0

x={x1,···,xn,···}xi2Rand1X

i=1|xi|p<1

over R.F o r

x={x1,···,xn,···}andy={y1,···,yn,···}inlp, deﬁne a function

d:lp⇥lp!R+asd(x, y)=1X

k=1|xkyk|p. Then ( lp,d)i sam e t r i c

space (Check!) but not a normed space as property (N 4) of norm is

not satisﬁed as shown below:

forz={0,1,0,···}2lpand↵=22R, we have,

k↵zk=k↵z0k=d(↵z,0) = |00|p+|↵0|p+|00|p+···=|↵|p=↵p

&|↵|kzk=↵kz0k=↵d(z,0) = ↵(|00|p+|10|p+|00|p+···)=↵

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CHAPTER 3. NORMED SPACES

3.4 Convergent Sequence and Cauchy

Sequence in a Normed Space

You are already familiar with the deﬁnition of convergence of se-

quences in a set of points and its related results. Now, you will learn

the concepts of convergent sequences and Cauchy sequences in a normed

space to obtain analogous results, which you learnt in your B.Sc.

Deﬁnition 3.7. Let ( V,kk)b ean o r m e ds p a c e . As e q u e n c e {xn}of

vectors in Visconvergent tox2Vif8✏>09n02Nsuch that

8n>nowe have kxnxk<✏. (or equivalently, if kxnxkconverges

to 0 as n! 1). Here, we say, xis the limit of the sequence {xn}

or{xn}is a convergent sequence, converging to x. In this case, we

write, lim

n!1xn=xor (xn!xasn!1)o r l i m

n!1kxnxk=0o r

(kxnxk!0a s n!1)

In the following result, you will notice that, in a normed space, a

sequence can have at most one limit.

Theorem 3.4.1. The limit of a convergent sequence in a normed space

is unique.

Proof. Consider a convergent sequence {xn}in normed space ( V,kk).

Assume that {xn}converges to x2Vand{xn}converges to y2V.

Then as n!1, we have, kxnxk!0a n d kxnyk!0. Clearly,

kxnxk=kxxnk.

Consider

kxyk=k(xxn)+(xny)k

6kxxnk+kxnykby property (N3)of norm

60as n !1

Thus, kxyk=0 by property (N1)of norm

)xy=0 by property (N2)of norm

)x=yand we are done.

Example 17. In a normed space ( V,kk), show thatkxkkyk6

kxyk8x, y2V.

Solution: Using property (N 3) of norm, we have, kxk=k(xy)+yk6

kxyk+kyk.S o , kxkkyk6kxyk ! (1)

Similarly, kykkxk6kxykaskyxk=kxyk

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FUNCTIONAL ANALYSIS

So,(kxkkyk)6kxyk ! (2)

From (1) and (2), we get,kxkkyk6kxyk

Now, you will learn the deﬁnition of a continuous function on normed

spaces and then prove that norm is a continuous function on R.

Deﬁnition 3.8. Consider the normed spaces ( V,kkV)a n d( W,kkW).

The function f:V!Wis continuous at x02Vif8✏>09>0

such that kxx0kV<=)kf(x)f(x0)kW<✏

Equivalently, we write, as n!1,xn!xo=)f(xn)!f(xo)

i.e. for every sequence {xn}inVconverging to x02V, the sequence

{f(xn)}inWconverges to f(x0)2W.

Here, the notations kkVandkkWmean the norms in normed spaces

VandW, respectively.

Theorem 3.4.2. Let(V,kk)be a normed space. Deﬁne a function

f:V!Rasf(x)=kxk. Then the norm kkonVis a real valued

continuous function.

Proof. Letx02V. Consider a sequence {xn}inVsuch that as n!

1,xn!xo. i.e. as n!1,kxnx0k!0. Then using Example

17, we get,f(xn)f(x0)=kxnkkx0k6kxnx0k !0a s

n!1and thus as n!1,w eh a v e f(xn)!f(x0). Therefore, for

every sequence {xn}inVconverging to x02V, the sequence {f(xn)}

inRconverges to f(x0)2R. Hence, the norm kkonVis a real valued

continuous function.

Deﬁnition 3.9. Asubspace Mof a normed space Vis a subspace of

Vconsidered as a vector space, with the norm obtained by restricting

the norm on Vto the subset M. This norm on Mis said to be induced

by the norm on V.

Note that, if Mis closed in a normed space V, then Mis called a

closed subspace of V.

Recall the quotient space (or factor space). Let Mbe a subspace of

av e c t o rs p a c e V. The coset of an element x2Vwith respect to Mis

x+M={x+m|m2M}. Under the following algebraic operations

(x+M)+( x0+M)=( x+x0)+Mand↵(x+M)=↵x+M,t h e

quotient space of VbyM, denoted by V/ M is a vector space. Note

that, x+M=Mif and only if x2M.

Given a normed space, you will learn how to form a new normed

space.

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Theorem 3.4.3. LetMbe a closed subspace of a normed space (V,kk).

For each coset x+Min quotient space V/ M, deﬁne kx+Mk=inf{kx+

mk|m2M}. Then V/ M is a normed space under the norm kx+Mk.

Proof. Using the properties of inﬁmum and properties in normed space

(V,kk), we have

(1) As kx+mkis a non-negative real number and every set of non-

negative real numbers is bounded below, it follows that inf{kx+

mk|m2M}exists and is non-negative. i.e. kx+Mk>0

8x+M2V/ M. So property (N 1) of norm is satisﬁed.

(2) Let x+M=M(zero element of V/ M). Then x2M.S o

kx+Mk=inf{kx+mk|m2M}

=inf{kx+mk|m2M,x 2M}

=inf{kyk|y2M}where y =x+m

=0

Thus, x+M=M(zero element of V/ M)=)kx+Mk=0 .

Conversely, let kx+Mk=0

=)inf{kx+mk|m2M}=0

=)9as e q u e n c e {mk}inMsuch that kx+mkk!0a sk!1

=)lim

k!1mk=x

=)x2M asMis closed

=)x2M asMis subspace

=)x+M=M

Thus, kx+Mk=0 = )x+M=M(zero element of V/ M)

)kx+Mk=0 ()x+M=M(zero element of V/ M)a n d

hence property (N 2) of norm is satisﬁed.

(3) Let x+M, y +M2V/ M.

k(x+M)+(y+M)k=k(x+y)+Mk

=inf{kx+y+mk|m2M}

=inf{kx+y+m1+m2k|m1,m22M}

=inf{k(x+m1)+(y+m2)k|m1,m22M}

6inf{kx+m1k+ky+m2k|m1,m22M}

=inf{kx+m1k|m12M}

+inf{kx+m2k|m22M}

=kx+Mk+ky+Mk

Therefore, k(x+M)+( y+M)k6kx+Mk+ky+Mkand

hence property (N 3) of norm is satisﬁed.

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FUNCTIONAL ANALYSIS

(4) If ↵=0t h e no b v i o u s l y k↵(x+M)k=|↵|kx+Mk.

So, let ↵6=0 . T h e n

k↵(x+M)k=k↵x+Mk

=inf{k↵x+mk|m2M}

=inf{k↵(x+m0)k|m/↵=m02M}

=inf{|↵|kx+m0k|m02M}

=|↵|inf{kx+m0k|m02M}

=|↵|kx+Mk

and hence property (N 4) of norm is satisﬁed.

Thus, V/ M is a normed space under deﬁned norm.

As with metric spaces, you can understand the concept of norms

from a geometrical point of view. In a normed space ( V,kk), the open

ball, centered at awith radius ✏is deﬁned by the set B(a;✏)={x2

V|kxak<✏}and the open unit ball is given by B(0; 1) = {x2

V|kxk<1}. Also, the closed ball, centered at awith radius ✏is de-

ﬁned by the set B[a;✏]={x2V|kxak6✏}and the closed unit ball

is given by B[0; 1] = {x2V|kxk61}.

Recall that M=set of all limit points of M. For each x2Mand

for each ✏>0, the open ball {y|kyxk<✏}must contain a point of

M. Hence, for each element to be in M, it su ces to show that for any

✏>0,9some element of Mwhich is within ✏distance from it.

Theorem 3.4.4. IfMis a subspace of a normed space (V,kk)then

Mis a closed subspace of V.

Proof. Initially, we show that, Mis a subspace of V. Let ✏>0. Con-

sider x, y 2Mand non-zero scalars ↵,. Then 9x1,y12Msuch

that kxx1k<✏

2|↵|andkyy1k<✏

2||.

Using the properties (N 3) and (N 4) of norm, we get,

k(↵x+y)(↵x1+y1)k=k↵(xx1)+(yy1)k

6k↵(xx1)k+k(yy1)k

=|↵|kxx1k+||kyy1k

<|↵|✏

2|↵|+||✏

2||

=✏

2+✏

2

=✏

So9(↵x1+y1)2M(being subspace) such that k(↵x+y)(↵x1+

y1)k<✏. Thus ( ↵x+y)2Mifx, y 2Mand↵,are non-zero

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CHAPTER 3. NORMED SPACES

scalars.(if ↵=0= then ↵x+y=02M). Hence, Mis a subspace

ofV. Further, as Mis closed, Mis a closed subspace of V.

You have learnt the deﬁnition of a Cauchy sequence in Rnat your

B.Sc. So you can guess the deﬁnition of a Cauchy sequence in a normed

space and get convinced with the following deﬁnition.

Deﬁnition 3.10. Let ( V,kk)b ean o r m e ds p a c e . As e q u e n c e {xn}

of vectors in Vis said to be a Cauchy sequence if8✏>09n02N

such that 8m, n >nowe have kxnxmk<✏. (or equivalently, if

kxnxmk!0a s m, n !1).

You will come to know that the relation between convergent se-

quences and Cauchy sequences in normed spaces.

Theorem 3.4.5. Every convergent sequence in a normed space is a

Cauchy sequence.

Proof. In normed space ( V,kk), consider a convergent sequence {xn},

converging to x2V. Then as n! 1, we have, kxnxk !0.

Clearly, kxmxk=kxxmk.

Consider

kxnxmk=k(xnx)+(xxm)k

6kxnxk+kxmxkby property (N3)of norm

!0as m, n !1

Hence, {xn}is a Cauchy sequence.

Remark 3.4.1. The converse of Theorem 3.4.5 need not be true. i.e. A

Cauchy sequence in a normed space need not be a convergent sequence.

You will see its counter example in next chapter. In the next chapter,

you will come to know that Cauchy sequences play a vital role in the

theory of normed spaces.

Recall the concept of subsequences in Rthat you learnt at your

B.Sc. and think of it in a normed space ( V,kk). Can you guess, what

happens to a Cauchy sequence, having a convergent subsequence?. The

next result is related to it.

Theorem 3.4.6. In a normed space, every Cauchy sequence having a

convergent subsequence is convergent.

Proof. In normed space ( V,kk), consider a Cauchy sequence {xn},

having a convergent subsequence {xnk}. Let {xnk}converge to x2V.

Then as nk! 1, we have, kxnkxk !0. Also, since {xn}is a

Cauchy sequence, as n, n k!1, we have, kxnkxnk!0.

Asn!1, it follows that,

kxnxk=kxnxnk+xnkxk6kxnxnkk+kxnkxk!0.

Hence, the Cauchy sequence {xn}is convergent.

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FUNCTIONAL ANALYSIS

3.5 LET US SUM UP

1. Let Vbe a vector space over the ﬁeld F(=RorC). A norm kk

onVis a real valued function (i.e. kk:V!R), satisfying the

following 4 properties/axioms:

(N 1) kxk>0 8x2V

(N 2) kxk= 0 if and only if x=0 v8x2V

(N 3) kx+yk6kxk+kyk8x, y2V

(N 4) k↵xk=|↵|kxk8 x2Vand 8↵2F

2. A normed space Vis a vector space over the ﬁeld F(=RorC)

with a norm kkdeﬁned on it. In such a case, we say, ( V,kk)

is a normed space over F. Here, if F=Rthen Vis called a real

normed space and if F=Cthen Vis called a complex normed

space .

3. A normed space is called ﬁnite dimensional if the underlying vec-

tor space is ﬁnite dimensional, otherwise a normed space is called

inﬁnite dimensional .

4. The vector space Cn={(x1,···,xn)|xi2C}over Cis a normed

space under the Euclidean norm kxk=✓nX

i=1|xi|2◆1/2

.

5. Let pandqbe non-negative extended real numbers. For p>1,q

is said to be conjugate of pif

1

p+1

q=1,f o r 1

q=1,f o rp =1

q=1,f o r p =1

6. H¨older’s inequality for ﬁnite sums :

Let 1

p+1

q=1 . F o ra n yc o m p l e x( o rr e a l )

numbers x1,···,xn;y1,···,yn

nX

i=1|xiyi|6✓nX

i=1|xi|p◆1/p✓nX

i=1|yi|q◆1/q

(Cauchy-Schwarz inequality is a special case of H¨ older’s inequality

forp=2= q)

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7. Minkowski’s inequality for ﬁnite sums : Let 1 6p<1.F o r a n y

complex (or real) numbers x1,···,xn;y1,···,yn

✓nX

i=1|xi+yi|p◆1/p

6✓nX

i=1|xi|p◆1/p

+✓nX

i=1|yi|p◆1/p

8. The vector space Cn={(x1,···,xn)|xi2C}over Cis a normed

space under the norms

kxkp=8

><

>:✓nX

i=1|xi|p◆1/p

if 1 6p<1

max{|x1|,···,|xn|}ifp=1

These norms are referred as p-norms on Cn.

9. On a vector space V, one can deﬁne more than one norm and ac-

cordingly di ↵erent normed spaces are obtained from same vector

space V.

10. The sequence space lp=⇢

{x1,···,xn,···}1X

i=1|xi|p<1and

xi2C

over Cis a normed space under the norms

kxkp=8

><

>:✓1X

i=1|xi|p◆1/p

if 1 6p<1

sup{|x1|,···,|xn|,···} ifp=1

These norms are referred as p-norms on lp.

11. H¨ older’s inequality for integrals :

Let 1

p+1

q=1 . F o r f2Lp(E)a n d g2Lq(E)

(where Eis bounded closed interval in R),

Z

E|fg|6✓Z

E|f|p◆1/p✓Z

E|g|q◆1/q

12. Minkowski’s inequality for integrals :

Let 1 6p<1. Iff2Lp(E)a n d g2Lp(E)( w h e r e Eis bounded

closed interval in R), then

✓Z

E|f+g|p◆1/p

6✓Z

E|f|p◆1/p

+✓Z

E|g|p◆1/p

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FUNCTIONAL ANALYSIS

13. The vector space Lp(E)={f:E!R|fis Lebesgue measur-

able function on Eand|f|pis Lebesgue integrable over E}over

Ris a normed space under norms

kfkp=8

><

>:✓Z

E|f|p◆1/p

if 1 6p<1

ess sup |f| ifp=1

where esssup

E|f(x)|=inf{m> 0||f(x)|6ma.e. on E}and

L1(E)=t h ec l a s so fa l lt h o s em e a s u r a b l ef u n c t i o n s fdeﬁned on

Ewhich are essentially bounded on Ewith ess sup |f|<1.

14. The vector space C(X)= {f:X!R(orC)|fis bounded

continuous function on X}over R(orC)i san o r m e ds p a c eu n d e r

norm kfk=s u p

x2X{|f(x)|}.

15. The vector space C[0,1] = {f:[ 0,1]!R|fis a continuous

function }overRis a normed space under norm kfk=Z1

0|f(t)|dt.

Note that C[0,1] is a particular case of C(X).

16. In a given normed space ( V,kk), a function d:V⇥V!R+

deﬁned as d(x, y)=kxykis called the distance from xtoy,

where x, y2V. Here dis referred as distance function on V.

17. Every normed space is a metric space (with respect to the distance

function).

18. Every metric space need not be a normed space as shown in fol-

lowing example:-

Let 0

x={x1,···,xn,

···}xi2Rand1X

k=1|xi|p<1

overR.F o r x={x1,···,xn,···}

andy={y1,···,yn,···}inlp, deﬁne a function d:lp⇥lp!R+

asd(x, y)=1X

k=1|xkyk|p. Then ( lp,d)i sam e t r i cs p a c eb u tn o t

an o r m e ds p a c e .

19. Let ( V,kk)b ean o r m e ds p a c e .As e q u e n c e {xn}of vectors in V

is convergent tox2Vif8✏>09n02Nsuch that 8n>nowe

have kxnxk<✏. (or equivalently, if kxnxkconverges to 0 as

n! 1). Here, we say, xis the limit of the sequence {xn}or

{xn}is a convergent sequence, converging to x. In this case, we

write, lim

n!1xn=xor (xn!xasn!1)o r l i m

n!1kxnxk=

0o r( kxnxk!0a s n!1).

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CHAPTER 3. NORMED SPACES

20. The limit of a convergent sequence in a normed space is unique.

21. In a normed space ( V,kk),kxkkyk6kxyk8x, y2V.

22. Consider the normed spaces ( V,kkV)a n d( W,kkW). The func-

tionf:V!Wis continuous atx02Vif8✏>09>0s u c h

that kxx0kV<=)kf(x)f(x0)kW<✏

Equivalently, we write, as n! 1,xn!xo=)f(xn)!

f(xo). i.e. for every sequence {xn}inVconverging to x02V,

the sequence {f(xn)}inWconverges to f(x0)2W.

23. Let ( V,kk)b ean o r m e ds p a c e .D e ﬁ n eaf u n c t i o n f:V!Ras

f(x)=kxk. Then the norm kkonVis a real valued continuous

function.

24. Let Mbe a closed subspace of a normed space ( N,kk). For

each coset x+Min quotient space N/M , deﬁne kx+Mk=

inf{kx+mk|m2M}. Then N/M is a normed space under the

norm kx+Mk.

25. If Mis a subspace of a normed space ( V,kk)t h e n Mis a closed

subspace of V.

26. Let ( V,kk)b ean o r m e ds p a c e . As e q u e n c e {xn}of vectors in

Vis said to be a Cauchy sequence if8✏>09n02Nsuch

that 8m, n >nowe have kxnxmk<✏. (or equivalently, if

kxnxmk!0a s m, n !1).

27. Every convergent sequence in a normed space is a Cauchy se-

quence.

3.6 Chapter End Exercise

1. Prove that every inner product space Vis a normed space with

respect to the norm kxk=p 8x2Vwhere

denotes the inner product of vector xwith itself.

2. Show that the vector space C={z|z2C}over Cis a normed

space under the norm kzk=|z|=absolute value of z2C.

3. Show that the vector space Rn={(x1,···,xn)|xi2R}over Ris

a normed space under the (Euclidean) norm kxk=✓nX

i=1|xi|2◆1/2

.

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FUNCTIONAL ANALYSIS

4. Show that the vector space Rn={(x1,···,xn)|xi2R}over R

is a normed space under the norm kxk=✓nX

i=1|xi|p◆1/p

where

16p<1.

5. Show that the vector space Rn={(x1,···,xn)|xi2R}over R

is a normed space under the norm kxk=max{|x1|,···,|xn|}.

6. Show that the vector space Lp([a, b]) = {f:[a, b]!R|fis

Lebesgue measurable function on [ a, b]a n d |f|pis Lebesgue

integrable over [ a, b]}over Ris a normed space under the norm

kfk=✓Zb

a|f|2◆1/2

7. Show that the vector space C([a, b]) = {f:[a, b]!R|fis

continuous function }over Ris a normed space under the norm

kfk=✓Zb

a|f|2◆1/2

8. Show that the vector space R2={(x1,x2)|xi2R}over Ris a

normed space under the norm kxk=|x1|+|x2|.

9. If kk1and kk2are 2 norms on a vector space Vthen check

whether the function kk:V!Rdeﬁned as kxk=kxk1+kxk2

is a norm on V.

10. Let ( X,kkX)a n d( Y,kkY)b en o r m e ds p a c e s . T h e np r o v et h a t

X⇥Yis a normed space under the norm k(x, y)k=kxkX+kykY.

11. Show that a metric d induced by a norm on a normed space

(V,kk) is translation invariant. (Hint: to show (a) d(x+u, y+

u)=d(x, y)a n d( b ) d(ax, ay )=|a|d(x, y)8x, y2V, for every

scalar awhere uis a ﬁxed vector in V).

12. Let {xn}and {yn}be sequences in a normed space ( V,kk).

If lim

n!1xn=x2Vand lim

n!1yn=y2Vthen prove that

lim

n!1xn+yn=x+y

13. Let {xn}be a sequence in a normed space ( V,kk)a n d {n}be

as e q u e n c eo fr e a ln u m b e r s . I f l i m

n!1xn=x2Vand lim

n!1n=

2Rthen prove that lim

n!1nxn=x

14. Prove that in a normed space, if {xn}is a Cauchy sequence then

{kxnk}is a Cauchy sequence of real numbers.

15. Prove that in a normed space, every Cauchy sequence is bounded.

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CHAPTER 3. NORMED SPACES

16. Prove that in a normed space, a Cauchy sequence is convergent

if and only if it has a convergent subsequence.

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Chapter 4

Banach Space

Unit Structure :

4.1I n t r o d u c t i o n

4.2O b j e c t i v e

4.3 Few deﬁnitions and examples

4.4 Equivalent Norms and Finite-Dimensional Spaces

4.6 Arzela-Ascoli theorem

4.6 LET US SUM UP

4.7C h a p t e rE n dE x e r c i s e

4.1 Introduction

In this chapter, you will be introduced to the notion of a Banach

Space. The concept of Banach space was introduced by the Polish

mathematician Stefan Banach in 1922. Banach spaces are fundamental

parts of functional analysis. Banach thought of, when a norm is de-

ﬁned on a vector space, how to deal with Cauchy sequences and hence

about completeness. This chapter has 3 sections, of which in the ﬁrst

section, you will ﬁnd several examples on Banach spaces, along with a

characterization of Banach Spaces. In an attempt to obtain a criterion

for determining when a Cauchy sequence with respect to one norm will

also be a Cauchy sequence with respect to other norm, you will be in-

troduced to the notion of equivalent norms in the second section and

interesting results on ﬁnite dimensional normed spaces are obtained.

Further, through Riesz lemma , the concept of compactness is linked

to subspaces of ﬁnite dimensional normed spaces. In the last section

of this chapter, you will be introduced to Ascoli-Arzela theorem and

the purpose of this theorem is to show a sequence of continuous func-

tions on campact space has a uniformly convergent subsequence, under

certain conditions.

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FUNCTIONAL ANALYSIS

4.2 Objectives

The main objective of this chapter is to learn the Banach spaces and

interesting results of ﬁnite dimensional normed spaces. After

going through this chapter you will be able to:

•Deﬁne a Banach space.

•Learn how to check a normed space is a Banach space under the given

norm.

•Prove that Lpspaces are Banach spaces (Riesz-Fisher theorem).

•Deﬁne equivalent norms on a normed space.

•Prove that on a ﬁnite dimensional normaed space, any two norms are

equivalent.

•Learn Riesz Lemma and results related to it.

•Prove Ascoli-Arzela theorem.

4.3 Few deﬁnitions and examples

In previous chapter, you have learnt the concept of a normed space

and at the end, it was mentioned that “A Cauchy sequence in a normed

space need not be a convergent sequence”. Here is a counter example

for it.

Example 18. In the normed space C[0,1] under the norm kfk=Z1

0|f(t)|dt, consider the sequence {fn(t)}where each function fn:

[0,1]!Ris deﬁned as follows:

fn(t)=8

>>>>><

>>>>>:0i f 0 6t61

2

2ntn if1

26t6✓1

2+1

2n◆

1i f✓1

2+1

2n◆

6t61

Show that the sequence {fn(t)}is a Cauchy sequence in C[0,1]. Is this

sequence convergent? Justify.

Solution: Clearly, fn(t)2C[0,1]8n2N.

Also, 8t21

2,✓1

2+1

2n◆

, it is easy to see that |fn(t)|61. So,

|fm(t)fn(t)|6|fm(t)|+|fn(t)|61+1=2 . !(⇤)

Now, with m>n , we have,

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CHAPTER 4. BANACH SPACE

kfmfnk=Z1

0|fm(t)fn(t)|dt

=Z1

2

0|fm(t)fn(t)|dt+Z1

2+1

2n

1

2|fm(t)fn(t)|dt+Z1

1

2+1

2n|fm(t)fn(t)|dt

=0+Z1

2+1

2n

1

2|fm(t)fn(t)|dt+0 b yd e ﬁ n i t i o no f fn(t)

62Z1

2+1

2n

1

2dt=2✓1

2n◆

using inequality ( ⇤)

Thus, kfmfnk61

n!0a s m, n!1

Hence, the sequence {fn(t)}is a Cauchy sequence in C[0,1].

Assume that this Cauchy sequence {fn}is convergent in C[0,1].

Then 9af u n c t i o n g2C[0,1] such that lim

n!1fn=g. It is easy to see

that lim

n!1Z1

2+1

2n

0|fn(t)g(t)|dt=0a n d l i m

n!1Z1

1

2+1

2n|1g(t)|dt=0 .

Thus,

g(t)=8

><

>:0i f 0 6t<1

2

1i f1

2It is clear that the function gis discontinuous at t=1

2and hence

g/2C[0,1], which contadicts g2C[0,1]. So, our assumption that the

Cauchy sequence {fn}is convergent in C[0,1] must be wrong. Hence,

the Cauchy sequence {fn}is not convergent in normed space C[0,1]

under norm kfk=Z1

0|f(t)|dt.

Now, you will come to know when a normed space is called, a Banach

space.

Deﬁnition 4.1. ABanach space is a normed space in which every

Cauchy sequence is convergent.

You know that, a metric space is called a complete space if every

Cauchy sequence of points in it converges to a point in the space. In

view of this deﬁnition, the normed space ( V,k.k)i ss a i dt ob e complete

ifVis complete as a metric space with the metric d(u, v)= ku

vk8u, v2V. Hence, you can redeﬁne the deﬁnition 4.1 as a Banach

space is a complete normed space or a Banach space is a normed space

which is a complete metric space.

Now, you will notice that examples of Banach spaces are in abun-

dance. Initially, you will ﬁnd an example of a normed space which is

not a Banach space.

Example 19. Show that the vector space C[0,1] = {f:[ 0,1]!R|

fis a continuous function }over Ris not a Banach space under norm

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FUNCTIONAL ANALYSIS

kfk=Z1

0|f(t)|dt.

Solution: Refer Example 14 to show that C[0,1] is a normed space

under deﬁned norm. Refer Example 18 to show that a Cauchy sequence

inC[0,1] is not convergent on it.

Thus, C[0,1] is not a Banach space under deﬁned norm.

You can generalize above Example 19 to get the following result.

Theorem 4.3.1. Prove that the vector space C[a, b]={f:[a, b]!

R(orC)|fis a continuous function }over R(orC)is not a Banach

space under norm kfk=✓Zb

a|f(t)|pdt◆1/p

where 16p<1

Proof. Left to the reader.

But you will see that C[0,1] is a Banach space with respect to sup

norm.

Example 20. Show that the vector space C[0,1] = {f: [0,1]!

R(orC)|fis a continuous function }over R(orC)i saB a n a c hs p a c e

under norm kfk=s u p

x2[0,1]{|f(x)|}.

Solution: Refer Example 15 to show that C[0,1] is a normed space

under deﬁned norm. Now, consider a Cauchy sequence {fn}inC[0,1].

Let✏>0. Then {fn}being a Cauchy sequence, 9m02Nsuch

that 8l,nm0and 8x2[0,1], we have, kflfnk<✏

3i.e.

sup

x2[0,1]{|fl(x)fn(x)|}<✏

3and hence 8l,nm0and 8x2[0,1]; we

have, |fl(x)fn(x)|<✏

3!(⇤)

This shows that for ﬁxed but arbitrary x2[0,1],{fn(x)}is a Cauchy

sequence in R(orC). As every Cauchy sequence in R(orC)i sc o n v e r -

gent, 9af u n c t i o n f:[ 0,1]!R(orC)s u c ht h a t l i m

n!1fn(x)=f(x),

for each ﬁxed x2[0,1]. So, making l!1in inequality ( ⇤), we get,

8nm0and for each ﬁxed x2[0,1];|f(x)fn(x)|<✏

3.T a k i n g

supremum over x2[0,1], (as m0is independent of xandxis arbi-

trary), we get, kffnk<✏

38nm0 !(⇤⇤)

Thus, fn!funiformly, i.e. kfnfk!0a s n!1.

Claim :f2C[0,1].

Consider a sequence {xn}in [0,1] such that xn!xwhere x2[0,1].

From inequality ( ⇤⇤), in particular n=m0gives, kffm0k<✏

3. And

therefore, for nsuciently large, we have,

|f(xn)f(x)||f(xn)fm0(xn)|+|fm0(xn)fm0(x)|+|fm0(x)f(x)|

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which implies |f(xn)f(x)|<✏

3+|fm0(xn)fm0(x)|+✏

3

and hence |f(xn)f(x)|<✏

3+✏

3+✏

3=✏asfm02C[0,1] and xn!x

So,f(xn)!f(x)i fxn!xwhich implies fis continuous and we

are done. Thus, the Cauchy sequence {fn}inC[0,1] is convergent to

f2C[0,1]. Hence, C[0,1] is a Banach space under deﬁned norm.

Remark 4.3.1. From above Example 20, it is clear that in the space

C[0,1] under norm kfk=s u p

x2[0,1]{|f(x)|},

fn!f

()fn(x)!f(x)u n i f o r m l yo n[ 0 , 1 ]

()8✏>0,9m02N(independent of x) such that 8x2[0,1] and

8n>m0, we have, |fn(x)f(x)|<✏

You can generalize above Example 20 to get the following results.

Theorem 4.3.2. Prove that the vector space C[a, b]={f:[a, b]!

R(orC)|fis a continuous function }over R(orC)is a Banach space

under norm kfk=s u p

x2[a,b]{|f(x)|}.

Proof. Left to the reader.

Theorem 4.3.3. Prove that the vector space C(X)over R(orC)is a

Banach space under norm kfk=s u p

x2X{|f(x)|}where Xis a compact

space.

Proof. Left to the reader.

Example 21. Show that the vector space R={x|x2R}over Ris a

Banach space under the norm kxk=|x|=absolute value of x2R.

Solution: Refer Example 6 of previous chapter to show that Ris a

normed space under deﬁned norm.

Claim : Every Cauchy sequence in Ris convergent in R.

Let{xn}be a Cauchy sequence. Then for any ✏>09n02Nsuch

that 8n, mn0we have |xnxm|<✏

2 (1)

{xn}being Cauchy, is bounded and hence {xn}has a convergent sub-

sequence {xnk} (using Bolzano Weirstrass theorem)

So suppose {xnk}converges to l2R. Then 9k02Nsuch that 8kk0,

we have, |xnkl|<✏

2 (2)

Note that nkkk0

Choose p=max {k0,n0}. Then pk0andpn08n, mpusing (1),

we have, |xnxm|<✏

2 (3)

8kpusing (2), we have, |xnkl|<✏

2

So in particular, for k=p, we have, |xnpl|<✏

2

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Now, in particular, for m= npp,8npby (3), we get, |xnxnp|<✏

2

Consider |xnl|=|(xnxnp)+(xnpl)||xnxnp|+|xnpl|

<✏

2+✏

28np

)8✏>0t h e r ee x i s t s p2Nsuch that 8np, we have, |xnl|<✏

Thus, the Cauchy sequence {xn}inRis convergent in Rand hence R

is a Banach space under deﬁned norm.

Example 22. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a Banach space under the norm kxk=✓nX

i=1|xi|p◆1/p

where

16p<1.

Solution: Refer Example 8 to show that Cnis a normed space under

deﬁned norm. Now, consider a Cauchy sequence {xm}inCn. As xm2

Cnis an n-tuple, denote xm=(x(m)

1,···,x(m)

n). Let ✏>0. Then

{xm}being a Cauchy sequence, 9m02Nsuch that 8l,mm0,w e

have, kxmxlk<✏which implies k(x(m)

1x(l)

1,···,x(m)

nx(l)

n)k=✓nX

i=1|x(m)

ix(l)

i|p◆1/p

<✏ !(⇤)

Eq.(⇤)i m p l i e s t h a t |x(m)

ix(l)

i|<✏8l,mm0and 8i=1,···,n

This shows that for ﬁxed but arbitrary i,{x(r)

i}is a Cauchy sequence in

C. As every Cauchy sequence in Cis convergent, {x(r)

i}must converge,

say to, zi2C. Thus lim

r!1x(r)

i=zi2C8i=1,···,n !(⇤⇤)

Making l!1in Eq.( ⇤)a n dt h e nu s i n gE q . ( ⇤⇤), we get, 8mm0,✓nX

i=1|x(m)

izi|p◆1/p

<✏which implies kxmzk<✏where z=

(z1,···,zn)2Cn. It follows that the Cauchy sequence {xm}inCn

is convergent to z2Cn. Hence, Cnis a Banach space under deﬁned

norm.

Example 23. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a Banach space under the norm kxk=max{|x1|,···,|xn|}.

(This norm is referred as kxk1onCn).

Solution: Refer Example 9 to show that Cnis a normed space under

deﬁned norm. Now, consider a Cauchy sequence {xm}inCn. As xm2

Cnis an n-tuple, denote xm=(x(m)

1,···,x(m)

n). Let ✏>0. Then

{xm}being a Cauchy sequence, 9m02Nsuch that 8l,mm0,

we have, kxmxlk<✏which implies k(x(m)

1x(l)

1,···,x(m)

nx(l)

n)k=

max{|x(m)

1x(l)

1|,···,|x(m)

nx(l)

n|}<✏ !(⇤)

Eq.(⇤)i m p l i e s t h a t |x(m)

ix(l)

i|<✏8l,mm0and 8i=1,···,n

This shows that for ﬁxed but arbitrary i,{x(r)

i}is a Cauchy sequence in

C. As every Cauchy sequence in Cis convergent, {x(r)

i}must converge,

say to, zi2C. Thus lim

r!1x(r)

i=zi2C8i=1,···,n !(⇤⇤)

Making l!1in Eq.( ⇤)a n dt h e nu s i n gE q . ( ⇤⇤), we get, 8mm0,

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max{|x(m)

1z1|,···,|x(m)

nzn|}<✏which implies kxmzk<✏where

z=(z1,···,zn)2Cn. It follows that the Cauchy sequence {xm}inCn

is convergent to z2Cn. Hence, Cnis a Banach space under deﬁned

norm.

Example 24. Show that the vector space Cn={(x1,···,xn)|xi2C}

over Cis a Banach space under the norm kxk=✓nX

i=1|xi|2◆1/2

.

Solution: Left to the reader.

Example 25. Let 1 6p<1. Show that the sequence space lp=⇢

{x1,···,xn,···}1X

i=1|xi|p<1and x i2C

over Cis a Banach

space under the norm kxk=✓1X

i=1|xi|p◆1/p

.

Solution: Refer Example 10 to show that lpis a normed space under

deﬁned norm. Now, consider a Cauchy sequence {xm}inlp. As xm2lp,

denote xm={x(m)

1,···,x(m)

n,···}where1X

i=1|x(m)

i|p<1. Let ✏>0.

Then as {xm}is a Cauchy sequence, 9m02Nsuch that 8l,mm0,w e

have, kxmxlk<✏which implies k{x(m)

1x(l)

1,···,x(m)

nx(l)

n,···}k=✓1X

i=1|x(m)

ix(l)

i|p◆1/p

<✏. !(⇤)

Eq.(⇤)i m p l i e s t h a t |x(m)

ix(l)

i|<✏ 8l,mm0and 8i

This shows that for ﬁxed but arbitrary i,{x(r)

i}is a Cauchy sequence in

C. As every Cauchy sequence in Cis convergent, {x(r)

i}must converge,

say to, zi2C. Thus lim

r!1x(r)

i=zi2C 8i !(⇤⇤)

From Eq.( ⇤), it is clear that, 8k2N,kX

i=1|x(m)

ix(l)

i|p<✏p8l,mm0.

Making l!1and then using Eq.( ⇤⇤), we get,✓kX

i=1|x(m)

izi|p◆

<✏p

8mm0. Further, making k!1,w eg e t ,✓1X

i=1|x(m)

izi|p◆

<✏p

8mm0. This implies that ( zxm)2lpwhere z={z1,···,zn,···}.

So,z=( (zxm)+xm)2lp.

Now, kxmzk=✓1X

i=1|x(m)

izi|p◆1/p

<✏8mm0. It follows that

the Cauchy sequence {xm}inlpis convergent to z2lp. Hence, lpis a

Banach space under deﬁned norm.

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Example 26. Show that the sequence space lp=⇢

{x1,···,xn,···}sup

{|x1|,···,|xn|,···} <1and x i2C

over Cis a Banach space

under the norm kxk=sup{|x1|,···,|xn|,···}.

(This norm is denoted as kxk1onlp).

Solution: Left to the reader.

To show that the normed space Lp(E)i saB a n a c hs p a c e ,w eﬁ r s t

prove characterization of a Banach space in the forem of lemma, for

which you are introduced with following terms.

Deﬁnition 4.2. A sequence {xk}in a normed space ( V,kk)i ss a i dt o

besummable to the sum sif the sequence {sn}of the partial sums of

the series1X

k=1xkconverges to s2V. i.e. ksnsk!0a s n!1.

In this case, we write, s=1X

k=1xk.

Deﬁnition 4.3. The sequence {xk}in a normed space ( V,kk)i ss a i d

to be absolutely summable if1X

k=1kxkk<1.

Lemma 4.3.1. A normed space (V,kk)is a Banach space if and only

if every absolutely summable sequence in Vis summable.

Proof. Let the normed space ( V,kk)b eaB a n a c hs p a c e . C o n s i d e ra n

absolutely summable sequence {xk}inV. Then1X

k=1kxkk=M< 1

where M> 0. Thus, 8✏>0,9r2Nsuch that1X

k=rkxkk<✏.S o , i f

sn=nX

k=1xkisnthpartial sum of the series1X

k=1xkthen 8n>m >r,

we have, ksnsmk=nX

k=m+1xk6nX

k=m+1kxkk61X

k=rkxkk<✏. It

follows that, the sequence {sn}of partial sums of the series1X

k=1xkis a

Cauchy sequence in Banach space Vand hence must converge to some

element; say s2V. Thus, {xk}is summable in Vand we are done.

Conversely, in a normed space ( V,kk), assume that every absolutely

summable sequence in Vis summable. Consider a Cauchy sequence

{xk}inV. Then, for each k2N,9nk2Nsuch that 8n, m>nk,w e

have, kxnxmk<1

2k!(⇤)

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Choose nksuch that nk+1>n k. Then {xnk}is a subsequence of Cauchy

sequence {xn}.

Deﬁne y1=xn1

y2=xn2xn1...

yk=xnkxnk1...

Then as nk>n k1,b y( ⇤), we have, kykk<1

2k1where k>1.

Consider the series1X

k=1yk. Its kthpartial sum is sk=kX

i=1yi=xnk. As

1X

k=21

2k1is a geometric series with ﬁrst term 0.5 and common ratio 0.5,

we have,1X

k=21

2k1=0.5

10.5=1 .

Now1X

k=1kykk=ky1k+1X

k=2kykk6ky1k+1X

k=21

2k1=ky1k+1 = M< 1

which implies that the sequence {yk}is absolutely summable and hence

summable (by hypothesis). So, the sequence {sk}converges to some

s2Vand hence the subsequence {xnk}is convergent. By Theorem

3.4.6, it follows that, the Cauchy sequence {xn}inVis convergent.

Therefore, the normed space ( V,kk)i saB a n a c hs p a c e .

Recall the following results which you studied in measure theory.

Fatou’s Lemma : Let {fn}be a sequence of non-negative measurable

functions and lim

n!1fn=fa.e. on E. ThenZ

Ef6lim

n!1Z

Efn

Lesbesgue Dominated Convergence Theorem : Let gbe an (Les-

besgue) integrable function on E. Let {fn}be a sequence of measurable

functions such that |fn|6gon E and lim

n!1fn=fa.e. on E. ThenZ

Ef=l i m

n!1Z

Efn

Example 27. Let 1 6p<1&Ebe a (bounded closed interval in R)

measurable set. Show that the vector space Lp(E)={f:E!R|f

is Lebesgue measurable function on Eand|f|pis Lebesgue integrable

over E}over Ris a Banach space under norm kfkp=✓Z

E|f|p◆1/p

.

(This norm is referred as p-norms on Lp(E)).

Solution: Refer Example 12 to show that Lp(E)i san o r m e ds p a c e

under deﬁned norm.

Consider an absolutely summable sequence {fk}inLp(E). Then

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1X

k=1kfkkp=M< 1where M> 0.

1. Deﬁne a sequence {gn}of functions where gn(x)=nX

k=1|fk|.

Clearly, for each x, {gn(x)}is an increasing sequence of (extended)

real numbers and 9some (extended) real number g(x)s u c ht h a t

gn(x)!g(x)8x2E

2. The function gis measurable, since the functions gn(x)a r em e a -

surable.

3. By Minkowski’s inequality (for ﬁnite sums) in Lemma 3.3.3, we

have,

kgnkp=nX

k=1|fk|

p6nX

k=1kfkkpE|gn|p6Mp.

4. As gn>0a n d l i m

n!1gp

n=gp, by Fatou’s lemma, we have,Z

Egp6lim

n!1Z

Egp

n6Mp

It follows that gpis integrable and hence g(x)i sﬁ n i t ea . e .o n E.

5. Thus, we ﬁnd that, for each x, for which g(x)i sﬁ n i t e ,t h es e -

quence {fn(x)}is an absolutely summable sequence of real num-

bers and therefore, must be summable to a real number, say s(x).

6. Deﬁne s(x)=0f o rt h o s e xwhere g(x)=1. Then, the function

sso deﬁned is the limit a.e. of partial sums sn(x)=nX

k=1fk(x).

i.e.sn(x)!s(x) a.e. Hence, sis a measurable function.

(note that |sn(x)s(x)|p!0a . e . 8x)

7. Clearly, |sn(x)|6nX

k=1|fk(x)|=gn(x)6g(x). Then as g2Lp(E),

we have, s2Lp(E). since, if h2Lp&|f|6|h|then f2Lp.

8. It is easy to see that |sn(x)s(x)|p62p(g(x))p.

9. As 2pgpis an integrable function and |sn(x)s(x)|p!0a . e .

8x, by Lesbesgue Dominated Convergence Theorem, we have,R

E|sns|p!0a n dh e n c e ksnskp!0

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10. So, the sequence {sn}of partial sums of series1X

k=1fkconverges to

s2Lp(E). i.e. the absolutely summable sequence {fk}inLp(E)

is summable in Lp(E).

Thus, by Lemma 4.3.1, Lp(E)i saB a n a c hs p a c eu n d e rd e ﬁ n e dn o r m .

Example 28. LetEbe a (bounded closed interval in R)m e a s u r a b l e

set. Show that the vector space L1(E)={f:E!R|fi sam e a -

surbale function on Eandess sup |f|<1}over Ris a Banach space

under norm kfk1=ess sup E|f(x)|=inf{m>0||f(x)|6ma.e. on

E}.

Solution: Refer Example 13 to show that L1(E)i san o r m e ds p a c e

under deﬁned norm.

Consider a Cauchy sequence {fn}inL1(E). Then |fn(x)fm(x)|6

kfnfmk1except on a set An,m⇢[a, b]=Ewith m(An,m)=0 .

IfA=[n,mAn,mthen m(A)=0a n d |fn(x)fm(x)|6kfnfmk1

8n, m and8x2(EA).

Therefore, it follows that, {fn}converges uniformly to a bounded

limit foutside Aand the result is proved by observing the fact that

the convergence in L1(E)i se q u i v a l e n tt ou n i f o r mc o n v e r g e n c eo u t s i d e

as e to fm e a s u r ez e r o .

Thus, as {fn}!foutside A, we have, L1(E)i saB a n a c hs p a c e

under deﬁned norm.

The Examples 27 and 28 are well known by following theorem.

Theorem 4.3.4. Riesz-Fischer theorem :

For16p61,Lpspaces are Banach spaces.

Proof. Combine the answers to the Examples 27 and 28.

Now, you will see next two results related to quotient space.

Theorem 4.3.5. LetMbe a closed subspace of a Banach space (V,kk).

For each coset x+Min quotient space V/ M, deﬁne kx+Mk=inf{kx+

mk|m2M}. Then V/ M is a Banach space under the norm kx+Mk.

Proof. Refer Theorem 3 .4.3t os h o wt h a t V/ M is a normed space under

deﬁned norm. Now, let {sn+M}be a Cauchy sequence in V/ M. Then

sn2V. We know that, a Cauchy sequence is convergent if and only if

it has a convergent subsequence. So in order to show that {sn+M}is

convergent, it is su cient to show that it has a convergent subsequence.

We construct a subsequence in the following manner:

As{sn+M}is Cauchy, given ✏=1

2>0,9n12Nsuch that

8n, m>n1, we have, k(sn+M)(sm+M)k<✏=1

2Setsn1=x12V.

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Similarly, ✏=1

22>0,9n22Nwith n2>n 1such that 8n, m >n1,

we have, k(sn+M)(sm+M)k<✏=1

22

Setsn2=x22V.

In general, having chosen x1,···,xkandn1,···,nk, let nk+1>n kbe

such that 8n, m>nk, we have, k(sn+M)(sm+M)k<✏=1

2k+1

Setsnk+1=xk+12V.

Thus, we have obtained a subsequence {xk+M}of{sn+M}such that

k(xk+M)(xk+1+M)k<1

2kfork=1,2,···

Claim : This subsequence converges to an element of V/ M.

Lety12x1+M. Choose y22x2+Msuch that ky1y2k<1

2.

Next, choose y32x3+Msuch that ky2y3k<1

22. Continuing this

process, we obtain a sequence {yn}inVsuch that kynyn+1k<1

2n.

Note that given ✏>0, we can choose m02Nso large that1

2m01<✏.

Then for n>m >m0, we have,

kymynk=k(ymym+1)+(ym+1ym+2)+···+(yn1yn)k

6k(ymym+1)k+k(ym+1ym+2)k+···+k(yn1yn)k

i=m1

2i

<1X

i=m1

2i=1

2m1,being a geometric series

<1

2m01

<✏

){yn}is a Cauchy sequence in Banach space Vand hence 9y2V

such that kynyk!0a s n!1.

Consider k(xn+M)(y+M)k=k(xny)+Mk

=inf{k(xny)+mk|m2M}

6k(xny)+mk8m2M

Asyn=xn+mnfor some mn2M, we conclude that,

k(xn+M)(y+M)k6kynyk!0a s n!1.

=)xn+M!y+M2V/ M

=)The Cauchy sequence {sn+M}has a subsequence {xn+M}

which is convergent in V/ M.

=)The Cauchy sequence {sn+M}is convergent in V/ M.

=)The normed space V/ M is complete and we are done.

Theorem 4.3.6. LetMbe a closed subspace of a normed space (V,k.k).

IfMandV/ M are Banach spaces then Vis a Banach space

Proof. Consider a Cauchy sequence in V. Let ✏>0b eg i v e n . T h e n 9

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n02Nsuch that 8m, n>n0, we have, kxnxmk<✏. Also, {xn+M}

is a sequence in V/ M.

Consider k(xn+M)(xm+M)k=k(xnxm)+Mk

=inf{k(xnxm)+yk|y2M}

6k(xnxm)+yk8y2M

Buty=02M.S o , k(xn+M)(xm+M)k6k(xnxm)k<✏

8m, n >n0. This implies that {xn+M}is a Cauchy sequence in

Banach space V/ M and so it must converge to some z+M2V/ M for

some z2V. Hence k(xn+M)(z+M)k=k(xnz)+Mk!0a s

n!1

Now, for each n2N,9yn2Msuch that

kxnz+Mk=inf{k(xnz)+ynk|yn2M}6k(xnz)+ynkand

thus kxnz+Mkn.

Consider kynymk=k(xnz+yn)(xmz+ym)(xnxm)k

)kynymk6k(xnz+yn)k+k(xmz+ym)k+k(xnxm)kand

thus kynymk6kxnxmk<✏8m, n>n0. This implies that {yn}

is a Cauchy sequence in Banach space Mand so it must converge to

some y2M. Hence yn!yinMasn!1.

Now, since as n! 1,w eh a v e kxnz+ynk !0a n dh e n c e

xn!(zyn)i nV.S o , l i m

n!1xn=l i m

n!1(zyn)=zy=x2V.

)the Cauchy sequence {xn}inVis convergent to x2V. Hence Vis

complete and we are done.

This section is concluded with following result.

Theorem 4.3.7. Every complete subspace of a normed space is closed.

Proof. Consider a normed space ( V,k.k). Let Mbe a complete subspace

ofV. Let zbe any limit point of M. Then 8n2N, the open ball

B✓

z,1

n◆

=⇢

xkxzk<1

n

must contain atleast one point ynother

than z.S o , 9y1,y2,···,yn,···inMsuch that

ky1zk<1

ky2zk<1

2

...

kynzk<1

n

...

Claim : the sequence {yn}of points in Mconverges to z.

Let✏>0. Choose m2Nsuch that1

m<✏. Then

8n>m, we have, kynzk<1

n61

m<✏

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Thus, 8✏>0,9m2Nsuch that kynzk<✏8n>m

=)the sequence {yn}converges to z. Hence the claim.

As every convergent sequence in a metric space is a Cauchy sequence,

the sequence {yn}is a Cauchy sequence in M. But, since Mis complete,

we have, every Cauchy sequence in Mconverges to a point in M.S o ,

yn!zimplies z2M.

Thus, we have shown that, every limit point of Mbelongs to M

and consequently, Mis closed.

4.4 Equivalent Norms and

Finite-Dimensional Spaces

You have seen that there are many norms on the same ﬁnite di-

mensional vector space X. It is interesting to see that all these norms

onXlead to same topology for X, that is, the open subsets of Xare

the same, regardless of the particular choice of a norm on X. In this

section, you will see the notion of equivalent norms and basic results

related to it.

Deﬁnition 4.4. Two norms kk1andkk2on a normed space Xare

said to be equivalent and written as kk1⇠kk2, if9positive real

numbers aandb(independent of x2X)s u c ht h a t

akxk16kxk26bkxk18x2X

With little e ↵ort you can show that equivalent of norms is an equiv-

alence relation on the set of all norms over a given space.

Two norms kk1andkk2on a normed space Xare equivalent if

and only if any (Cauchy) sequence in Xconverges with respect to kk1

converges with respect to kk2and conversely.

From following result, you will come to know that although one can

deﬁne many di ↵erent norms on ﬁnite dimensional linear spaces, there

is only one topology derived from these norms.

Theorem 4.4.1. On a ﬁnite dimensional normed space X, any two

norms are equivalent.

Proof. LetXbe a ﬁnite dimensional vector space over Fwith dim( X)=n.

Then X⇠=Fn, since if B={v1,···,vn}is a basis for Xthen each

x2Xcan be uniquely represented as x=nX

j=1ajvjfor some unique

scalars aj2Fwhich gives an element ¯ x=(a1,···,an)2Fn.

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Now, by Euclidean norm in Fn, we have, k¯xk2=✓nX

j=1|aj|2◆1/2

and for each x=nX

j=1ajvj2X, deﬁne kxk2=✓nX

j=1|aj|2◆1/2

. Then

k¯xk2=kxk2.

Suppose k.kis a norm on X. Then, for each x2X,

kxk=nX

j=1ajvj6nX

j=1|aj|kvjk6✓nX

j=1kvjk2◆1/2✓nX

j=1|aj|2◆1/2

IfM=✓nX

j=1kvjk2◆1/2

then M> 0s u c ht h a t kxk6Mkxk28x2X

which gives the one half of the desired inequality.

Now, to establish the other inequlaity, deﬁne S={¯x=(a1,···,an)2

Fn|k¯xk2=1}. Then Sis closed and bounded, and hence is compact

(by Heine-Borel theorem) with respect to the Euclidean norm.

Deﬁne f:S!Rasf(¯x)=kxk. As Bis a linearly independent

set and since ¯ x2S, i.e. ¯ x6=0 ,a l l ajcannot vanish simultaneously on

Sso that f(¯x)>0o n S.

Clearly, |f(¯x)f(¯y)|=kxkkyk6kxyk6Mkxyk2.

It follows that fis continuous on S. Thus, fis a positive valued

continuous function on the compact set Sand therefore, fattains its

minimum m> 0a ts o m ep o i n to nt h ec o m p a c ts e t S. Consequently,

whenever ¯ x2S, we have, f(¯x)=kxk>m.

Thus, for each 0 6=¯u=(c1,···,cn)2Fn,

kuk=nX

j=1cjvj=k¯uk2f✓¯u

k¯uk2◆

>mk¯uk2=mkuk2

Therefore, 9positive real numbers mandMsuch that

mkuk26kuk6Mkuk28u2X

This implies that, any given norm k.kis equivalent to the 2-norm

k.k2. Since, equivalence of norms is an equivalence relation, it follows

that any two norms on Xare equivalent.

Now, you will see some immediate consequences of this theorem in

the form of following corollaries.

Corollary 4.4.1. IfVis a ﬁnite dimensional normed space, then Vis

complete.

Proof. LetVbe a ﬁnite dimensional vector space over Fwhere F=

RorCwith dim( V)=n> 0a n d {e1,···,en}be basis for V. Then

each x2Vcan be uniquely represented as x=nX

j=1ajejfor some

unique scalars aj2F. It can be easily proved that kxk0=m a x

i|ai|is a

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norm on V. This norm is called the zeroth norm.

Since, by Theorem 4.4.1, on a ﬁnite dimensional normed space V,

any two norms are equivalent, to prove Vis complete, it su ces to

prove the completeness of Vwith respect to this zeroth norm.

Let{yn}be any Cauchy sequence in V. Consider the ithterm of

this sequence {yn}which is yi=nX

k=1a(i)

kekfor some uniquely determined

scalars a(i)

1,···,a(i)

ninF.

Since, {yn}is Cauchy, we have, kynymk0!0a s m, n!1.

=)nX

k=1✓

a(n)

ka(m)

k◆

ek

0!0a s m, n!1.

=)max

k|a(n)

ka(m)

k|!0a s m, n!1.

=)|a(n)

ka(m)

k|!0a s m, n!1.

=){a(m)

k}is Cauchy sequence in Ffork=1,···,n,

AsFis complete, 9scalars a1,···,aninFsuch that

a(m)

k!akasm!1 ! (⇤)

)the Cauchy sequence {a(m)

k}1

m=1converges for some ak(k=1,···,n).

Lety=nX

k=1akekthen y2V. To show that ym!yasm!1,

let✏>0b eg i v e n .T h e nu s i n g( ⇤), we get,

kymyk0=nX

k=1✓

a(m)

kak◆

ek

0=m a x

k|a(m)

kak|!0a sm!1

which implies ym!yasm!1

)the Cauchy sequence {ym}inVconverges to y2Vand hence Vis

complete.

Corollary 4.4.2. IfMis any ﬁnite dimensional subspace of a normed

space V, then Mis closed.

Proof. LetMbe a ﬁnite dimensional subspace of a normed space V. As

by Corollary 4 .4.1, every ﬁnite dimensional normed space is complete,

we have, Mis complete subspace of V. Further, as by Theorem 4 .3.7,

every complete subspace of a normed space is closed, we have, Mis

closed.

Now, you will see another interesting result (about closed subspaces)

in the form of lemma, which is due to the famous Hungarian mathe-

matician Riesz. This result/lemma is to prove very important theorem

that relates ﬁnite dimensional normed spaces with compactness of its

bounded subset.

Lemma 4.4.1. (Riesz Lemma ): (F.Riesz, 1918 )

LetMbe a closed proper subspace of a normed space Vand let a2R

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be such that 0andkxxak>a8x2M

Proof. LetMbe a closed proper subspace of a normed space V. Then

we have M=Mand9x12(VM). i.e. x1/2M=M.

Deﬁne h=i n f

x2Mkxx1k=d(x1,M). Then h6kxx1k8x2M.

Clearly, h>0b e c a u s e h=0i m p l i e s x12M=M, a contradiction.

Leta2Rbe such that 0 a>h.

By deﬁnition of inﬁmum, 9x02Msuch that kx0x1k6h

a. Clearly,

x06=x1. Further, as kx0x1k>h> 0, we have,1

kx0x1k>0.

Deﬁne xa=x1x0

kx0x1k. Then xa2Vsuch that kxak=1 .

Letx2Mbe arbitrary. Then ( kx1x0kx+x0)2Masx, x 02M

and hence k(kx1x0kx+x0)x1k>h.

Now,

kxxak=xx1

kx1x0k+x0

kx1x0k

=1

kx1x0kk(kx1x0kx+x0)x1k

>h

kx1x0k

>a

Thus, 9xa2Vsuch that kxak=1a n d kxxak>a8x2M.

The Riesz Lemma 4.4.1 states that for any closed proper subspace

Mof a normed space V,9points in the unit sphere S(0,1) = {x2

V|kxk=1}whose distance from Mis as near as we please to 1(but

not 1). There may not be a point, though, whose distance is exactly 1.

We conclude this section with the following required result, proved

using Reisz lemma.

Theorem 4.4.2. In a normed space (V,k.k), if the set S={x2

V|kxk=1}is compact then Vis ﬁnite dimensional.

Proof. We know that, in a metric space, a subset is compact if and only

if every sequence has a convergent subsequence,

Let the set S={x2V|kxk=1}in normed space ( V,k.k)b ec o m -

pact. Then every sequence in Smust have a convergent subsequence.

Suppose, if possible, Vis not ﬁnite dimensional. Choose x12S.

Then kx1k=1 .

LetV1be the subspace spanned by x1. Then V1is a proper subspace

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ofVandV1is ﬁnite dimensional. It follows that V1is closed, since by

Corollary 4.4.2 if Mis any ﬁnite dimensional subspace of a normed

space Vthen Mis closed.

Applying Riesz Lemma 4.4.1 to this closed proper subspace V1of

V, we get, 9x22Vsuch that kx2k=1a n d kx2xk>1

28x2V1.

=)9x22Sandkx2x1k>1

2asx12V1.

LetV2be the subspace spanned by x1,x2. Then V2is a proper sub-

space of VandV2is ﬁnite dimensional. So, by Corollary 4.4.2, V2is

closed. Applying Riesz Lemma 4.4.1 to this closed proper subspace V2

ofV, we get, 9x32Vsuch that kx3k=1a n d kx3xk>1

28x2V2.

=)9x32Sandkx3x2k>1

2asx22V2.

Continuing this argument, we obtain an inﬁnite sequence {xn}of

vectors in Ssuch that kxnxmk>1

2.

=)the sequence {xn}can have no convergent subsequence, which con-

tradicts the hypothesis that, Sis compact.

Hence, the assumption that Vis not ﬁnite dimensional must be

wrong and we msut have Vis ﬁnite dimensional.

4.5 Arzela-Ascoli theorem

You are already familiar with Bolzano-Weierstrass theorem which

states that every bounded sequence of real/complex numbers contains a

convergent subsequence. In this section, you will see something similar

is true for sequence of functions, but in connection with the additional

concept of equicontinuity.

Recall that the set C(X)i st h es e to fa l l K-valued continuous func-

tions on a compact metric space Xwhere K=RorC.F o r f,g2C(X),

letd1(f,g)=s u px2X{|f(x)g(x)|}. It is easy to see that d1is a met-

ric on C(X)a n di sc a l l e da ss u pm e t r i co n C(X).

In the metric space ( X,d), for x2Xandr>0, recall, the set

B(x, r)={y2X|d(x, y)dius r.

Also, recall that a subset Eof a metric space Xis said to be totally

bounded if8✏>0,9x1,···,xn2Esuch that E⇢[n

j=1B(xj,✏).

Further, recall that for functions, fn,f2C(X), we say that the se-

quence {fn}converges uniformly to fonXif8✏>0,9n02N(depends

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only on ✏)s u c ht h a t 8x2X,8n>n0, we have, |fn(x)f(x)|<✏.

Now, you will see two kinds of boundedness.

Deﬁnition 4.5. LetSbe subset of C(X). We say that Sispointwise

bounded onXifSis bounded at each x2X, that is, if 9aﬁ n i t e -

valued function deﬁned on Xsuch that |f(x)|<(x)8f2S.

Deﬁnition 4.6. LetSbe subset of C(X). We say that Sisuniformly

bounded onXif9an u m b e r Msuch that |f(x)|8x2X.

It is easy to see that a uniformly bounded subset SofC(X)i s

always pointwise bounded on X. The converse holds (as seen in Ascoli’s

theorem , the following result) under a certain condition which we now

introduce.

Deﬁnition 4.7. A subset SofC(X)i ss a i dt ob e equicontinuous at

x2Xif for every ✏>0,9>0s u c ht h a tf o re v e r y y2Xwith

d(x, y)<and 8f2S, we have, |f(x)f(y)|<✏, where may

depend on x,b u tn o to n f2S.

Here ddenotes the metric of X.

A subset SofC(X)i ss a i dt ob ee q u i c o n t i n u o u so n XifSis equicon-

tinuous at every x2X.

You can verify that the functions belonging to the equicontinuous

collection are uniformly continuous.

Example 29. Deﬁne a sequence {fn}onRbyfn(x)= sin(x+n).

Show that the family {fn|n2N}is equicontinuous on R.

Solution: Note that |cos✓|61a n df o rs m a l l ✓,|sin✓|6|✓|.

Also, sin(n+x)sin(n+x0)=2 cos✓2n+x+x0

2◆

sin✓xx0

2◆

Let✏>0b eg i v e na n d x2R. With x02R,

Conisder

|fn(x)fn(x0)|=2cos✓2n+x+x0

2◆sin✓xx0

2◆

62✓xx0

2◆

=|xx0|

If|xx0|<=✏then |fn(x)fn(x0)|<✏8n2N.

Thus, the family {fn|n2N}is equicontinuous at x2Rand hence is

equicontinuous on R.

A condition su cient to ensure that a sequence of continuous func-

tions on compact space Xhas a uniformly convergent subsequence will

come out of the following result (which is known as Arzela’s theorem ).

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Theorem 4.5.1. Let S be subset of C(X)where Xis a compact metric

space. Suppose that S is pointwise bounded on Xand is equicontinuous

at every x2X. Then

(a) (Ascoli, 1883) S is uniformly bounded on X. In fact, S is

totally bounded in the sup metric on C(X).

(b) (Arzela, 1889) Every sequence in S contains a uniformly

convergent subsequence.

Proof. (a)L e t ✏>0. Since ( X,d)i sac o m p a c tm e t r i cs p a c e ,w e

have, 9x1,···,xn2Xand positive numbers 1,···,nsuch that

X=[{B(xi,i)|i=1,···,n}. So, to every x2Xthere corresponds

at least one xiwith x2B(xi,i). Also, since S is equicontinuous at

every x2X,8f2Sa n d 8x2X, we have, d(x, x i)<i(16i6n)

which implies |f(x)f(xi)|<✏. Further, since S is pointwise bounded

onX, S is bounded at each x2X.S o , 9Mi<1(16i6n)s u c h

that |f(xi)|6Mi8f2S. Deﬁne M=max{M1,···,Mn}+✏. Then

it follows that |f(x)|6M8f2Sa n d 8x2X. Thus, S is uniformly

bounded on X.

Now, let EM={k2K||k|6M}where K2RorCand for f2S,

deﬁne e(f)=( f(x1),···,f(xn))2En

M. It is easy to see that En

Mis

totally bounded. Hence we can cover it by a ﬁnite union of open balls

of radius ✏, say V1,···,Vm. Ifj=1,···,mandVj\{e(f)|f2S}6=;,

choose fj2Ssuch that e(fj)2Vj.

Claim : S is union of open balls of radius 5 ✏about these f1,···,fm.

Letf2S. Then e(f)2Vjfor some j=1,···,m. Since, e(fj)2Vjand

the radius of V0

jsis✏, we see that |f(xi)fj(xi)|<2✏8i=1,···,n.

Now, each x2Xbelongs to some B(xi,i),i=1,···,n, which implies

|fj(x)fj(xi)|<✏and|f(x)f(xi)|<✏.

Consider

|f(x)fj(x)|=|f(x)f(xi)+f(xi)fj(xi)+fj(xi)fj(x)|

6|f(x)f(xi)|+|f(xi)fj(xi)|+|fj(xi)fj(x)|

<✏+2✏+✏

so that d1(f,fj)=s u px2X{|f(x)fj(x)|}64✏<5✏.

This proves that S is totally bounded in the sup metric d1.

(b)W ep r o v e( b)u s i n gt h ef o l l o w i n gr e s u l t so fm e t r i cs p a c e .

(i) The subset Aof metric space Xis totally bounded if and only if

every sequence of points of Acontains a Cauchy subsequence.

(ii) The sequence {fn}inC(X)c o n v e r g e st o f2C(X)( w i t hr e s p e c t

to sup metric on C(X)) if and only if {fn}converges uniformly to

fonX.

Consider a sequence {fn|fn2C(X),n=1,2,···} in S. Then by

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Ascoli theorem, {fn}✓C(X) is totally bounded. Hence, by result

(i),{fn}has a Cauchy subsequence {fnk}(with respect to sup metric

onC(X)). Since, by Theorem 4.3.3, C(X)i sc o m p l e t e,t h es e q u e n c e

{fnk}is convergent to some f2C(X). By result (ii), this implies that

{fnk}converges uniformly to fonXand we are done.

(b) (Alternative proof of ( b) without using Ascoli theorem)

Consider a sequence {fn|fn2C(X),n=1.2,···} in S. Let Abe a

countable dense subset of X. Then S is pointwise bounded on A.

Claim :{fn}has a subsequence {fnk}such that {fnk(x)}converges for

every x2A.

Let{xi},i=1,2,3,···be points of A, arranged in a sequence. Since,

{fn(x1)}is bounded, 9as u b s e q u e n c e ,w h i c hw es h a l ld e n o t eb y {f1,k},

such that {f1,k(x1)}converges as k!1.

Now, consider sequences S1,S2,S3,···, which we represent by the

array

S1:f1,1 f1,2f1,3 f1,4···

S2:f2,1 f2,2f2,3 f2,4···

S3:f3,1 f3,2f3,3 f3,4···

··· ··· ··· ··· ··· ···

and which have the following properties:

(1)Snis a subsequence of Sn1, for n=2,3,4,···

(2){fn,k(xn)}converges, as k!1

(3) Order in which the functions appear is the same in each sequence;

i.e., if one function precedes another in S1, they are in same relation

in every Sn, until one or the other is deleted. Hence, when going

from one row in the above array to the next below, functions may

move to the left but never to the right.

We now go down the diagonal of the array; i.e., we consider the

sequence P: f1,1f2,2f3,3f4,4···

By (3), the sequence P(except possibly its ﬁrst n-1 terms) is a subse-

quence of Sn, for n=1,2,3,···. Hence, (2) implies that {fn,n(xi)}

converges, as n!1, for every xi2A.

For convenience, put fni=gi. We shall prove that {gi}converges

uniformly on X.

Let✏>0 be given. As S is equicontinuous, choose >0s u c ht h a t

d(x, y)<implies |fn(x)fn(y)|<✏8n. Since Ais dense in X

andXis compact, 9ﬁnitely many points x1,···,xminAsuch that

X=[{B(xi,)|i=1,···,m}.

Since {gi(x)}converges for every x2A,9n02Nsuch that |gi(xs)

gj(xs)|<✏whenever i>n0,j>n0,16s6m.

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Also, if x2Xthen clearly, x2B(xs,)f o rs o m e s, so that |gi(x)

gi(xs)|<✏for every i. Ifi>n0andj>n0then it follows that

|gi(x)gj(x)|=|gi(x)gi(xs)+gi(xs)gj(xs)+gj(xs)gj(x)|

6|gi(x)gi(xs)|+|gi(xs)gj(xs)|+|gj(xs)gj(x)|

<✏+✏+✏

Thus, the subsequence {gi}of{fn}converges uniformly on X.

4.6 LET US SUM UP

1. A Banach space is a normed space in which every Cauchy se-

quence is convergent. In other words, a Banach space is a com-

plete normed space.

2. Every Banach space is a normed space but a normed space need

not be a Banach space.

3. The normed space C[0,1] = {f:[ 0,1]!R|fis a continu-

ous function }over Ris not a Banach space under norm kfk=Z1

0|f(t)|dt.

4. Prove that the vector space C[a, b]={f:[a, b]!R(orC)|fis

ac o n t i n u o u sf u n c t i o n }over R(orC)i sn o taB a n a c hs p a c eu n d e r

norm kfk=✓Zb

a|f(t)|pdt◆1/p

where 1 6p<1

5. The vector space C(X)= {f:X!R(orC)|fis bounded

continuous function on X}over R(orC)i saB a n a c hs p a c eu n d e r

norm kfk=s u p

x2X{|f(x)|}.

6. The vector space R={x|x2R}over Ris a Banach space under

the norm kxk=|x|=absolute value of x2R.

7. The vector space Cn={(x1,···,xn)|xi2C}over Cis a Banach

space under the norms

kxkp=8

><

>:✓nX

i=1|xi|p◆1/p

if 1 6p<1

max{|x1|,···,|xn|}ifp=1

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8. The sequence space lp=⇢

{x1,···,xn,···}1X

i=1|xi|p<1and

xi2C

over Cis a Banach space under the norms

kxkp=8

><

>:✓1X

i=1|xi|p◆1/p

if 1 6p<1

sup{|x1|,···,|xn|,···} ifp=1

9. A sequence {xk}in a normed space ( V,kk)i ss a i dt ob e summable

to the sum sif the sequence {sn}of the partial sums of the series

1X

k=1xkconverges to s2V. i.e. ksnsk!0a s n!1. In

this case, we write, s=1X

k=1xk.

10. The sequence {xk}in a normed space ( V,kk)i ss a i dt ob e abso-

lutely summable if1X

k=1kxkk<1.

11. A normed space ( V,kk)i saB a n a c hs p a c ei fa n do n l yi fe v e r y

absolutely summable sequence in Vis summable.

12. The vector space Lp(E)={f:E!R|fis Lebesgue measur-

able function on Eand|f|pis Lebesgue integrable over E}over

Ris a Banach space under norms

kfkp=8

><

>:✓Z

E|f|p◆1/p

if 1 6p<1

ess sup |f| ifp=1

where ess sup E|f(x)|=inf{m> 0||f(x)|6ma.e. on E}and

L1(E)=t h ec l a s so fa l lt h o s em e a s u r a b l ef u n c t i o n s fdeﬁned on

Ewhich are essentially bounded on Ewith ess sup |f|<1.

13.(Riesz-Fischer theorem) :Lpspaces are Banach spaces where

16p61.

14. Let Mbe a closed subspace of a Banach space ( V,kk). For

each coset x+Min quotient space V/ M, deﬁne kx+Mk=

inf{kx+mk|m2M}. Then V/ M is a Banach space under the

norm kx+Mk.

15. Let Mbe a closed subspace of a normed space ( V,k.k). If Mand

V/ M are Banach spaces then Vis a Banach space.

16. Every complete subspace of a normed space is closed.

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17. Two norms kk1andkk2on a normed space Xare said to be

equivalent and written as kk1⇠kk2,i f9positive real numbers

aandb(independent of x2X)s u c ht h a t

akxk16kxk26bkxk18x2X.

18. On a ﬁnite dimensional normed space X,a n yt w on o r m sa r e

equivalent.

19. If Vis a ﬁnite dimensional normed space, then Vis complete.

20.(Riesz Lemma) : Let Mbe a closed proper subspace of a normed

space Vand let a2Rbe such that 0 xa2Vsuch that kxak=1a n d kxxak>a8x2M.

21. In a normed space ( V,k.k), if the set S={x2V|kxk=1}is

compact then Vis ﬁnite dimensional.

22. Let Sbe subset of C(X). We say that Sispointwise bounded

onXifSis bounded at each x2X, that is, if 9aﬁ n i t e -v a l u e d

function deﬁned on Xsuch that |f(x)|<(x)8f2S.

23. Let Sbe subset of C(X). We say that Sisuniformly bounded on

Xif9an u m b e r Msuch that |f(x)|24. A subset SofC(X)i ss a i dt ob e equicontinuous atx2Xif for

every ✏>0,9>0s u c ht h a tf o re v e r y y2Xwith d(x, y)<

and8f2S, we have, |f(x)f(y)|<✏, where may depend on

x,b u tn o to n f2S. Here ddenotes the metric of X.

25. Let S be subset of C(X)w h e r e Xis a compact metric space.

Suppose that S is pointwise bounded on Xand is equicontinuous

at every x2X. Then

(a) (Ascoli, 1883) S is uniformly bounded on X. In fact, S

is totally bounded in the sup metric on C(X).

(b) (Arzela, 1889) Every sequence in S contains a uniformly

convergent subsequence.

4.7 Chapter End Exercise

1. Show that the vector space C={z|z2C}over Cis a Banach

space under the norm kzk=|z|=absolute value of z2C.

2. Show that the vector space Rn={(x1,···,xn)|xi2R}over Ris

aB a n a c hs p a c eu n d e rt h e( E u c l i d e a n )n o r m kxk=✓nX

i=1|xi|2◆1/2

.

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3. Show that the vector space Rn={(x1,···,xn)|xi2R}over R

is a Banach space under the norm kxk=✓nX

i=1|xi|p◆1/p

where

16p<1.

4. Show that the vector space Rn={(x1,···,xn)|xi2R}over R

is a Banach space under the norm kxk=max{|x1|,···,|xn|}.

5. Show that the normed space C([a, b]) = {f:[a, b]!R|fis

continuous function }over Runder the norm kfk=✓Zb

a|f|2◆1/2

is not a Banach space.

6. Let ( X,kkX)a n d( Y,kkY)b en o r m e ds p a c e s . T h e np r o v et h a t

X⇥Yis a Banach space under the norm k(x, y)k=kxkX+kykY.

7. Let Vbe a non-zero normed space and let S={x2V|kxk61}.

Prove that Vis complete if and only if Sis complete.

8. Prove that every ﬁnite dimensional normed space is Banach and

hence deduce each ﬁnite dimensional subspace of a normed space

is closed.

9. Give a counter example to show that any two norms on an inﬁnite

dimensional normed space are not equivalent.

10. Show that the p-norms on Rnare equivalent where 1 6p61.

11. Let Mbe a closed proper subspace of the normed space V. Then

prove that for every real number a>0,9an element y2Vwith

kyk=1s u c ht h a t kxyk>1a8x2M.

12. Let Mbe a closed proper subspace of a normed space V. Then

prove that for each a2(0,1),9ap o i n t xainVbut not in M

(not necessarily unique) such that kxak=1a n dd i s t ( xa,M)=

inf

y2Mkxayk>a.

13. Let Vbe a ﬁnite dimensional normed space and r>0. Then

prove that the closed ball B[0;r]={x2V|kxk6r}is compact.

14. Let Vbe a normed space such that the closed ball B[x0;r]={x2

V|kxx0k6r}is compact for some x02Vandr>0. Then

prove that Vis ﬁnite dimensional.

15. Prove that in a ﬁnite dimensional normed space V, any proper

subset MofVis compact if and only if Mis closed and bounded.

16. Let Vbe a normed space. Prove that the closed unit ball in Vis

compact if and only if Vis ﬁnite dimensional.

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17. Prove that a normed space Vis ﬁnite dimensional if and only if

every bounded closed subset in Vis compact.

18. If Mis a compact subset of a Banach space Vthen show that M

is also compact.

19. If Mis any ﬁnite dimensional subspace of a normed space V, then

Mis closed.

20. Let fn2C(X)f o r n=1,2,3,···where Xis a compact metric

space. If {fn}converges uniformly on Xthen prove that {fn}is

equicontinuous on X.

21. If {fn}is a pointwise bounded sequence of complex functions on

ac o u n t a b l es e tE ,t h e np r o v et h a t {fn}has a subsequence {fnk}

such that {fnk(x)}converges for every x2E.

22. Let Xbe a compact metric space. Prove that a closed subspace

ofC(X) is compact if and only if it is uniformly bounded and

equicontinuous.

23. Show that the family {sin nx }1

n=1is not an equicontinuous subset

ofC[0,⇡].

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Chapter 5

BOUNDED LINEAR

TRANSFORMATIONS

AND DUAL SPACES

Unit Structure :

5.1I n t r o d u c t i o n

5.2O b j e c t i v e

5.3D e ﬁ n i t i o n s ,n o t a t i o n s ,t h e o r e m s

5.4S e p a r a b l es p a c e s

5.5 LET US SUM UP

5.6C h a p t e rE n dE x e r c i s e

5.1 Introduction

The bounded linear transformations on normed linear spaces are

important operators , that satisfy many properties as a function be-

tween two metric spaces like continuity and their collections B(X,Y)

can be made into a normed linear space under pointwise addition and

scalar multiplication.Completeness of the normed space B(X,Y)i si n -

herited via the completeness of the space Y.The Dual space of Xis a

complete metric space even if Xis complete or not and hence it satisﬁes

the properties of being a complete space.The signiﬁcance of dual spaces

oflp,Lp,Rnis that it is useful to know the general form of bounded

linear functionals on spaces of practical importance.For Hilbert spaces,

Riesz’s theorem elucidates the form of such bounded linear functionals

in simple manner.The separable spaces are somewhat simpler than the

non separable spaces.The separability of the dual space X0implies that

the space Xis separable.

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5.2 Objectives

After going through this Chapter, you will be able to

•Deﬁne a bounded linear transformations between two normed linear

spaces.

•Characterise the bounded linear transformations as continuous func-

tions

•Identify the algebraic structure of the bounded linear transformations

as normed linear space

•Deﬁne the dual space of Xand describe its properties like complete-

ness and seperability

5.3 Deﬁnitions, notations, theorems

Deﬁnition 1: A metric space is a pair ( X,d), where X is a set and

dis a metric on X, that is dis a distance function deﬁned on X⇥X

such that for all x, y, z ✏Xwe have

(1)dis real valued , ﬁnite and nonnegative.

(2)d(x, y)=0i fa n do n l yi f x=y

(3)d(x, y)=d(y,x)

(4)d(x, y)d(x, z)+d(z,y)

Example 1 : Euclidean space Rnwith metric ddeﬁned as

d(x, y)=p

(x1y1)2+...+(xnyn)2

Example 2 : Sequence space l1with metric ddeﬁned as

d(x, y)=sup{|xiyi|:i1}

Example 3 :lpspace with metric ddeﬁned as d(x, y)=(P1

i=1|xiyi|p)1

p

Deﬁnition 2 : Let X, Y be normed linear spaces and and T be a

linear transformation of X into Y then T is said to be bounded linear

transformation, if there exists a real number K0s u c ht h a t

||T(x)||K||x||for every x ✏X. The set of all continuous or bounded

linear transformations from X into Y is denoted by B(X,Y)

Example 1 : The identity operator I:X!Xdeﬁned by I(x)=x

on a nonzero normed space Xis bounded linear operator with ||I(x)||=

||x||K||x||with K=1

Example 2 : The 0 operator 0 : X!Ydeﬁned by 0( x)=0 Yis

bounded linear operator with ||0(x)||=||0Y||=00||x||with K=0

Now we shall see equivalent characterisations of a bounded linear trans-

formation Texisting between two normed spaces.

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Theorem 5.3.1. Let X, Y be normed linear spaces and T is a linear

transformation of X into Y . Then the following statements are equiv-

alent

(1)T is continuous

(2)T is continuous at the origin, in other words T( xn)!0 as xn!

0

(3)There exists a real number K0such that ||T(x)||K||x||for

every

x✏X

(4) T carries the closed unit sphere in X to a bounded set in Y

Proof: (1))(2)

Suppose that T is continuous . Since we have T(0)=0 , therefore T is

continuous at the point x=0. If T is continuous at the origin then xn

!xif and only if xnx!0. This implies that T(xnx)!0.This

is true if and only if T(xn)!T(x). So T is continuous.

(2))(3)

Assume that for each positive integer n, we can ﬁnd a point xnin X

such that ||T(xn)||n||xn||. This implies that ||T(xn/nx n)||>1. Put

yn=xn/n||xn||. Then observe that yn!0a s n!1butT(yn)90.

Therefore, T is not continuous at the origin.This is a contradiction to

our hypothesis. Therefore There exists a real number K0s u c ht h a t

||T(x)||K||x||for every

x✏X

(3))(4)

Letxbe any point belonging to the closed unit sphere in X. This implies

that ||x||1.Hence,by hypothesis ||T(x)||K||x||K. Therefore, by

deﬁnition, T carries the closed unit sphere in X to a bounded set in Y.

(4))(1)

We ﬁrst show that (4) )(3). If x=0t h e n T(x)=0, beacuse T is Linear

transformation. Hence ||T(x)||K||x||. Ifx6=0 then x/||x||=1 . P u t

y=x/||x||, then y belongs to the closed unit sphere and therefore by

hypothesis, ||T(y)||Kf o rs o m er e a ln u m b e rK 0. This implies that

||T(x/||x||)K. Thus, ||T(x)||K||x||. Therefore, ||T(x)||K||x||for

every x✏X. Now, if xnis any convergent sequence in X such that xn

!x, then ||T(xnx)||K||xnx||implies that T(xn)T(x)!0

asxnx!0. Hence T is continuous.

Note:(1) T is continuous i ↵Ti sb o u n d e d

(2) If T is continuous then T carries the closed unit sphere to a bounded

set in Y , in this case, we denote the norm of T by ||T||and it is deﬁned

as

||T||=s u p {| |T(x)||:||x||1}.

We also have ||T||=i n f {K:K 0a n d ||T(x)||K||x||. From this, we

conclude that ||T(x)||||T|| ||x||for all x

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Now we shall see that B(X,Y) forms a normed linear space and it

is complete , when Yis complete space.

Theorem 5.3.2. If X, Y are normed linear spaces, then B(X,Y) is

a normed linear space with respect to pointwise addition and scalar

multiplication and the norm deﬁned as ||T||= sup {| |T(x)||:||x||1

}.Further, if Y is a Banach space then B(X,Y) is also a Banach space.

Proof: To show that B(X,Y) is a normed linear space, let T, U be any

two linear transformations belonging to B(X,Y), then T+U is deﬁned

as (T+U)(x):=T(x)+U(x)a n df o ra n ys c a l a r ↵in F , we have ( ↵

T)(x):=↵T(x).Therefore we have (U+T)( xn)=U( xn)+T( xn)a n d( ↵

T)(xn):=↵T(xn).

Therefore ( T+U)(xn)!0a n d( ↵T)(xn)!0a s xn!0

Hence, T+U and ↵Tb o t ha r ec o n t i n u o u sa tt h eo r i g i n . T h i si m p l i e s

that T+U ✏B(X,Y) and ↵T✏B(X,Y)

We have (T+U)( x)=T( x)+U( x)=U( x)+T( x)=(U+T)( x), therefore vec-

tor addition is commutative.

For any S,T,U ✏B(X,Y), we have [(S+T)+U]( x)=(S+T)( x)+U(x)=

S(x)+T( x)+U( x)=[S+(T+U)]( x)f o re v e r y x✏X.Therefore, (S+T)+U=S+(T+U)

There exist 0 linear transformation in B(X,Y) deﬁned as 0( x)=0 Yfor

every x✏X.

For every T ✏B(X,Y), there exist an additive inverse -T ✏B(X,Y) such

that T+(-T)=0.

we have (-T)( x)=-T( x)f o re v e r y x✏X.

Scalar multiplication is associative and distributive.For all ↵,✏Fa n d

T, U ✏B(X,Y), [ ↵(U)](x)=↵(U(x))=↵U(x)= (↵)U(x).This im-

plies that ↵(U)=( ↵)U. [( ↵+)U](x)=(↵+)U(x)=↵U(x)+U(x)=(↵U+U)(x).

Hence, ( ↵+)U=↵U+U. Now [ ↵(T+U)]( x)=↵(T+U)( x)=↵T(x)+↵U(x)=

(↵T+↵U)(x). Therefore, we have ↵(T+U)= ↵T+↵U.

Further, we have (1.U)( x)=1.U( x)=U( x).Hence 1.U=U.

Now we shall show that for T ✏B(X,Y), ||T||=s u p {| |T(x)||:||x||1

}is norm on the linear space B(X,Y).

(a) For every T ✏B(X,Y), we have sup {| |T(x)||:||x||1}0, there-

fore||T||0

(b)||T||=0 if and only if sup {| |T(x)||:||x||1}=0. This is possible

i↵||T(x)||=0. This is true i ↵T(x)=0 and therefore T=0.

(c)||↵T||=sup {| |(↵T)(x) ||:||x||1}=|↵|sup{| |T(x)||:||x||1}=|

↵|| |T||for every ↵✏F and For every T ✏B(X,Y)

(d)||T+U||=s u p {| |(T+U)(x) ||:||x||1}=sup {| |T(x)+U( x)||:||x||

1}

sup{| |T(x)||+||U(x)||:||x||1}sup{| |T(x)||:||x||1}+sup {

||U(x)||:||x||1}=||T||+||U||

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Therefore, ||T+U||||T||+||U||for every T, U ✏B(X,Y)

Next, we show that B(X,Y) is complete, when Y is complete.

LetTnbe any Cauchy sequence of linear transformations in B(X,Y).

For any vector x✏X,w eh a v e ||Tn(x)-Tm(x)||=||(TnTm)(x)||||

Tn-Tm||—— x||.This implies that {Tn(x)}is a Cauchy sequence

in Y.Since Y is Complete therefore, there exist a vector T( x)i nY

such that Tn(x)!T(x). T h i sd e ﬁ n e saf u n c t i o nT :X !Yb y x

!T(x).This function is a linear transformation from X into Y , for

ifx1,x2✏X, we have Tn(x1+x2)=Tn(x1)+Tn(x2)a n d Tn(↵x)=↵

Tn(x).Hence T( x1+x2)=T( x1)+T( x2) and T( ↵x)=↵T(x).Now we show

that T is continuous and Tn!Ta s n!1

It is enough to show that T is bounded linear transformation.

Consider ||T(x)||=||lim n!1Tn(x)||=lim n!1||Tn(x)||sup( ||Tn||

||x||)= (sup ||Tn||)||x||. Since the norms of the terms of the Cauchy

Sequence in a normed linear space is a bounded set, therefore there

exists K=sup ||Tn||0s u c ht h a t

||T(x)||K||x||.Hence, T is a bounded linear transformation.Now we

show that ||TnT||!0.Let ✏>0 be any number , let N be a positive

integer such that n, mN)||TnTm||<✏. Now, if ||x||1a n d m, n

N, then

||Tn(x)-Tm(x)||=||(Tn-Tm)(x)||||TnTm|| || x||||TnTm||<✏

Now hold n ﬁxed and letting m!1,w eo b t a i n

||Tn(x)-Tm(x)||!||Tn(x)-T(x)||.This implies that ||Tn(x)-T(x)||<

✏for all nN and every xsuch that ||x||1.Hence, ||TnT||<✏for

every nN.Therefore, we have ||TnT||!0a s n!1.

Deﬁnition 3 : If X is any normed linear space then then the set of

all continuous linear transformations from X into RorCis denoted

by B(X, R) or B(X, C), according to X is real or complex vector space.

Denote it by X0, it is called as the dual space of X.The elements of X0

are called as continuous linear functionals.

Note: (1) functional deﬁned on normed linear space X is a scalar- val-

ued continuous linear functional deﬁned on X.

(2)X0is a normed linear space with norm deﬁned by

||f||:=sup {|f(x)|

||x||:x✏X,x6=0}=sup {|f(x)|:||x||=1}

(3) Since RandCare complete normed linear spaces therefore X0is a

Banach space.

Deﬁnition 4 : A bijective linear operator from a normed space X onto

the normed space Y is called as an isomorphism if it preserves the

norm, that is, for every x✏X,||T(x)||=||x||. In this case, X is said to

be isomorphic to Y and X,Y are called isomorphic normed spaces.

Theorem 5.3.3. The dual space of RnisRn

Proof: Rnis a normed linear space with the norm deﬁned as follows:

For every x=(x1,x2,...,xn)✏Rn,||x||=p

x2

1+x2

2+...+x2

n. We recall

the theorem , which states that if the dimension of a normed linear space

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is ﬁnite, then every linear operator on X is bounded. Therefore, we have

Rn0=Rn⇤.Given any f✏Rn,i f{e1,e2,. . . ,e n}is a standard basis for

Rn, then f(x)=f(x1e1+x2e2+...+xnen)=Pn

i=1xif(ei)=Pn

i=1xii,

where i=f(ei)f o re v e r y i.Therefore, by the Cauchy-Schwarz inequality

we have |f(x)|Pn

i=1|xii|(Pn

i=1x2

i)1/2(Pn

i=12

i)1/2=||x||(Pn

i=12

i)1/2

Hence|f|

||x||(Pn

i=12

i)1/2.Now taking supremum over all xof norm

1, we get, ||f||(Pn

i=12

i)1/2, because of the equality obtained for

x=(1,2,. . . ,n)i nt h ea b o v ei n e q u a l i t y ,w em u s th a v e ||f||=(Pn

i=12

i)1/2.This

implies that the norm of fis the Euclidean norm on Rn

Hence the mapping :Rn0!Rn, deﬁned as (f)=(1,2,...,n)i s

norm preserving bijective linear map, hence it is an isomorphism.

Here, we shall try to identify dual spaces of some of the normed linear

spaces.

Theorem 5.3.4. The dual space of l1isl1.

Proof: Consider a Schauder basis for l1, namely ( ek), where ek=(kj),

where kj=1 if j=kandkj=0, if j6=k.Then every x✏l1can be

uniquely

represented as x=P1

k=1xkek. Let f✏l10be any linear functional.

Since fis linear and bounded , therefore, we have f(x)=P1

k=1xkk,

where k=f(ek).Here the k=f(ek)a r eu n i q u e l yd e t e r m i n e db y f.

Also, ||ek||=1 and |k|=|f(ek)||f|| || ek||=||f||.Taking supremum

on both th sides, we get, sup {|k|:k1}||f||...(1)

Hence, ( k)✏l1. Further, if d=(k)✏l1then deﬁne gonl1as follows:

g(x)=P1

k=1xkk, where x=(xk)✏l1. Observe that gis linear as well

as bounded map, because we have

|g(x)|P1

k=1|xk||k|sup{|k|:k1}P1

k=1|xk|=||x||sup{|k|:

k1}.Hence g✏l1.Finally, we prove that the norm of fis the norm

on the space l1.Since we have, f(x)=P1

k=1xkk, where k=f(ek),

therefore,

|f(x)|=|P1

k=1xkk|sup{|k|:k1}P1

k=1|xk|=||x||sup{|k|:k

1}.Thus, we get

|f(x)|

||x||sup{|k|:k1}.Taking supremum over all xof norm 1, we

get,||f||sup{|k|:k1}...(2). From (1) and (2), we conclude that

||f||=sup{|k|:k1}, which is the norm on l1.This shows that the

bijective linear mapping of l10onto l1deﬁned by f!(k)i sa n

isomorphism.

Theorem 5.3.5. The dual space of lpislqfor1

p+1

q=1

Proof: A Schauder basis for lpis (ek), where ek=(kj). Then every x

✏lphas a unique representation x=P1

k=1xkek. Let f✏lp0, since fis

linear and bounded, therefore f(x)=P1

k=1xkk, where k=f(ek)...(*)

Letqbe the conjugate of pand deﬁne yn=((n)

k)w i t h (n)

k=|k|q

k,i fk

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nandk6=0 and (n)

k=0, if k>n ork=0 . B ys u b s t i t u t i n gt h i si n

(*) we get,

f(yn)=P1

k=1(n)

kk=Pn

k=1|k|q

Using the deﬁnition of n

kand ( q1)p=q,w eo b t a i n ,

f(yn)||f|| ||yn||=||f||(P1

k=1|(n)

k|p)1/p=||f||(P1

k=1|k|(q1)p)1/p=

||f||(Pn

k=1|k|q)1/p

Hence, f(yn)=Pn

k=1|k|q||f||(Pn

k=1|k|q)1/p.Therefore,

(Pn

k=1|k|q)11/p=(Pn

k=1|k|q)1/q||f||.Since nwas arbitrary, there-

fore letting n!1,w eo b t a i n(P1

k=1|k|q)1/q||f||...(**)

Hence ( k)✏lq.

Conversely, for any ( ⌘k)✏lq, we can deﬁne corresponding bounded

linear functional gonlpas follows:

g(x)=g( k)=P1

k=1 k⌘k, where x=( k)✏lp. Then by the H¨ older in-

equality, we have

|g(x)|=|P1

k=1 k⌘k|(P1

k=1| k|p)1/p(P1

k=1|⌘k|q)1/q.Thus gis linear

and bounded. Hence, g✏lp0.Now, we prove that the norm of fis the

norm on the space lq. therefore, |f(x)|=|P1

k=1xkk|(P1

k=1|xk|p)1/p

(P1

k=1|k|q)1/q=||x||(P1

k=1|k|q)1/q.

Therefore, we have|f(x)|

||x||(P1

k=1|k|q)1/q.Now taking supremum over

allxof norm 1, we obtain , ||f||(P1

k=1|k|q)1/q. From (**), we have

||f||=(P1

k=1|k|q)1/q

Therefore, the mapping of lp0onto lqdeﬁned by f!(k)i sl i n e a r ,

bijective and norm preserving. Hence it is an isomorphism.

Its practically important to know the general form of bounded linear

fucntionals on various spaces.For general Banach spaces, such formulas

and their derivation can sometimes be complicated. But, for Hilbert

space the situation is simple as described by the following result:

Representation of functionals on Hilbert spaces:

Riesz’s representation of bounded linear functionals on Hilbert spaces:

Theorem 5.3.6. Every bounded linear functional fon a Hilbert space

H can be represented in terms of an inner product as follows:

f(x)= , where zis uniquely determined by fand||z||=||f||.

Proof: We prove the following claims

(a)fhas representation as f(x)=

(b)zis uniquely determined by f

(c)||z||=||f||

(a) If f=0 then f(x)=0= and||z||=||f||=0 for z=0. There-

fore, assume that f6=0. In this case, z6=0 since, otherwise f=0. Now

=0f o re v e r y xin the nullspace of f, denoted by N(f).So, con-

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sider N(f)a n di t so r t h o g o n a lc o m p l e m e n t N(f)?. Since fis bounded

functional, therefore N(f) is a closed vector subspace of H.Further f

6=0 implies that N(f)6=H. So that N(f)?6=0 by the following theorem:

Let Y be any closed subspace of a Hilbert space H .Then H=Y Y?

Hence N(f)?contains some z06=0.Put v=f(x)z0f(z0)x; where, x

✏H is arbitrary.Applying f,w eo b t a i n f(v)=f(x)f(z0)f(z0)f(x)=0.

This shows that v✏N(f)

Since z0?N(f), we have 0= =

=f(x)f(z0)=f(x)||z0||2f(z0)

Since ||z0||26=0, we can solve for f(x).The solution is given by

f(x)=f(z0)

. Therefore

z=f(z0)

z0. Since xwas arbitrary , therefore f(x)= .

(b) We prove that zis uniquely determined in (a). Suppose that for

allx✏H,f(x)==. Then =0f o ra l l

x.Choose x=z1z2,w eh a v e ==||z1

z2||2=0.

Hence z1z2=0 ,t h i si m p l i e st h a t z1=z2

(c) Now we show that ||z||=||f||. Assume that f6=0 then z6=0

Putx=zinf(x)= . This implies that ||z||2= =f(z)

||f|| || z||

Dividing by ||z||6=0, we get ||z||||f||. It remains to show that

||f||||z||, from Schwarz inequality, we have |f(x)|=| |

||x||||z||

This implies that ||f||=sup{| |:||x||=1}||z||.Hence

||f||=||z||.

Deﬁnition 5 : Let ( X,A,µ)b em e a s u r es p a c ea n d1 p<1.The

space Lp(X)c o n s i s t so fe q u i v a l e n c ec l a s s e so fm e a s u r a b l ef u n c t i o n s

f:X!Rsuch thatR

|f|pdµ < 1, where two measurable func-

tions are equivalent, if they are equal µa.e.The Lpnorm of f✏Lp(X)i s

deﬁned by ||f||Lp=(R

|f|pdµ)1/p

We say that fn!finLpif||ffn||Lp!0.For example, the

characteristic function Qof the rationals on Ris equivalent to 0 in

Lp(R)

Example 1: If Nis equipped with counting measure, then Lp(N)c o n -

sists of all sequences {xn✏R:n✏N}such thatP1

n=1|xn|p<1with

norm

||(xn)||Lp=(P1

n=1|xn|p)1/p

Deﬁnition 6 : Let ( X,A,µ)b eam e a s u r es p a c e . T h es p a c e L1(X)

consists of pointwise a.e.- equivalence classes of essentially bounded

measurable functions f:X!Rwith norm ||f||L1=ess sup |f|,

where for any measurable function f:X!R, the essential supre-

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mum of fonXis

ess sup f=inf{a✏R:µ{x✏X:f(x)>a}=0}, equivalently

ess sup f=inf{supg:g=fpointwise a.e }

Now we shall establish the isomorphism between Lq(X)a n dt h ed u a l

space of Lp(X)f o r1

Proposition 5.3.1. Suppose that (X,A,µ)be a measure space and

1

fgdµ deﬁnes a bounded linear

functional F:Lp(X)!Rwith ||F||Lp0=||f||Lq.IfXis-ﬁnite then

the same results hold for p=1.

Proof: Using H¨ older’s inequality, we have for 1 p1that

|F(g)|||f||Lq||g||Lp.This implies that Fis a bounded linear func-

tional on Lpwith ||F||Lp0||f||Lq.In proving the reverse inequality, we

may assume that f6=0, otherwise the result is trivial.

First assume that 1

||f||Lq)q/p, then, g✏Lp,

since f✏Lq,a n d ||g||Lp=1.Also, sinceq

p=q1,F(g)=R

(sgnf)f(|f|

||f||Lq)q1

dµ=||f||Lq

Since ||g||Lp=1 ,w eh a v e ||F||Lp0|F(g)|so that ||F||Lp0||f||Lq.

Ifp=1, we get the same conclusion by taking g=sgn(f)✏L1.Therefore,

in these cases the supremum deﬁning ||F||Lp0is actually attained for

suitable function g.

Now, suppose that p=1a n d Xis-ﬁnite. For ✏>0,

letA={x✏X:|f(x)|>||f||L1✏}. Then 0 <µ(A)1. Moreover,

since X is -ﬁnite, there is an increasing sequence of sets Anof ﬁnite

measure whose union is Asuch that µ(An)!µ(A), so we can ﬁnd a

subset B ⇢A such that 0 <µ(B)<1.

Letg=sgn(f)(B)

µ(B). Then g✏L1(X)a n d ||g||L1=1 and

F(g)=1

µ(B)R

B|f|dµ||f||L1✏. This implies that , ||F||L10

||f||L1✏and therefore we have ||F||L10||f||L1, this is because ✏

was arbitrary.

This result shows that the map F:Lq(X)! Lp(X)0deﬁned by F(g)=R

fgdµ

is an isometry from LqintoLp0

Fis onto, when 1

onLparises in this way from an Lqfunction.

Theorem 5.3.7. Let(X,A,µ)be a measure space.If 1

F:Lq(X)! Lp(X)0deﬁned by F(g)=R

fgdµ is an isometric isomor-

phism of Lq(X)onto the dual space of Lp(X).

Proof: Suppose that Xhas a ﬁnite measure and let F:Lp!R

be a bounded linear functional on Lp(X). If A✏A, then A✏Lp(X).

Since, Xhas ﬁnite measure, deﬁne ⌫:A!Rby⌫(A)=F(A). If

A=S1

i=1Aiis a disjoint union of measurable sets , then A=P1

i=1Ai

and the dominated convergence theorem implies that

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||AP1

i=1Ai||Lp!0a s n! 1.Hence, since Fis continuous

linear functional on Lp,

⌫(A)= F(A)= F(P1

i=1Ai)=P1

i=1F(Ai)=P1

i=1⌫(Ai). This im-

plies that ⌫is a signed measure on ( X,A).

Ifµ(A)=0 ,t h e n Ais equivalent to 0 in Lpand therefore ⌫(A)=0 ,

by the linearity of F.Thus ⌫is absolutely continuous with respect to µ.

By the Radon-Nikodym theorem,that is stated as follows:

Let⌫be a ﬁnite signed measure and µbe a ﬁnite measure on

am e a s u r a b l es p a c e( X,A).Then there exists unique ﬁnite signed

measures ⌫a,⌫ssuch that ⌫=⌫a+⌫s, where ⌫a<< µ and⌫s?µ.

Further, there exists a measurable function f:X!Runiquely de-

ﬁned upto µa.e. equivalence, such that ⌫a(A)=R

Afdµfor every A✏A,

where the integral is well deﬁned as an extended real number.

The decomposition ⌫=⌫a+⌫sis called the Lebesgue decomposition

of⌫and the representation of an absolutely continuous signed measure

⌫<< µ asd⌫=fdµis the Radon-Nikodym theorem.We call the func-

tion fhere as the Radon-Nikodym derivative of ⌫with respect to µ

and denote it by f=d⌫

dµ.

Thus there is a function f:X!Rsuch that d⌫=fdµ and

F(A)=R

fAdµfor every A✏A.Hence, by the linearity and bound-

edness of F,F()=R

fdµfor all simple functions ,a n d |R

fdµ|

M||||Lp, where M=||F||Lp0.

Taking =sgnf, which is a simple function, we observe that f✏

L1(X).We may then extend the integral of fagainst bounded func-

tions by continuity. If g✏L1(X), then from the following theorem:[

Suppose that ( X,A,µ)i sam e a s u r es p a c ea n d1 p1. Then the

simple functions that belong to Lp(X)a r ed e n s ei n Lp(X).], there exist

as e q u e n c eo fs i m p l ef u c n t i o n s nwith |n||g|such that n!g

inL1, and therefore, also in Lp. Since |fn|||g||L1|f|✏L1(X), the

dominated convergence theorem and the continuity of Fimplies that

F(g)=l i m n!1F(n)=lim n!1R

fndµ=R

fgdµ and that |R

fgdµ |

M||g||Lpfor every g✏L1(X)

Now we prove that f✏Lq(X). Let {n}be a sequence of simple func-

tions such that n!fpointwise a.e. as n!1and|n||f|.

Deﬁne gn=(sgnf)(|n|

||n||Lq)q/p.Then gn✏L1(X)a n d ||gn||Lp=1 . F u r t h e r ,

fgn=|fgn|

andR

|ngn|dµ=||n||Lq.Therefore, by Fatou’s lemma and inequality

|n||f|, we have, ||f||Lqlim n!1inf||n||Lqlim n!1infR

|ngn|dµ

lim n!1R

|fgn|dµM.

Thus f✏Lq.Since the simple functions are dense in Lpandgis a contin-

uous functional on Lp,i ff✏Lq, it follows that F(g)=R

fgdµ for every

g✏Lp(X).By the previous proposition, this implies that ||F||Lp0=||f||Lq.

This proves the result, when Xhas a ﬁnite measure.

IfXisfinite then there is an increasing sequence {An}of sets

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with ﬁnite measure such that1S

n=1An=X.By the previous result, there

is a unique function

fn✏Lq(An)s u c ht h a t F(g)=R

Anfngdµfor all g✏Lp(An). If m

nthen the functions fm,fnare equal pointwise a.e. on Anand the

dominated convergence theorem implies that f=l i m n!1fn✏Lq(X)

is the required function.

Finally , if Xis not finite then for each finite subset A⇢X, let

fA✏Lq(A) be the function such that F(g)=R

AfAgdµfor every g✏Lp(A).

Deﬁne M=sup{||fA||Lq(A):A⇢X is finite }||F||Lp(X)0,a n d

choose an increasing sequence of sets Ansuch that ||fAn||Lq(An)!M

asn!1

.Deﬁne B=1S

n=1An, verify that fBis the required function.

5.4 Separable spaces

Deﬁnition 7: Let ( X,d)b eam e t r i cs p a c et h e n Xis said to be

separable if there exist a countable subset AofXsuch that A=X

Examples:

1] the real line Ris separable because Qis a countable subset of Rsuch

that Q=R.

Now , we shall see some of the examples of separable and nonseparable

spaces through the following results.

Theorem 5.4.1. The space l1is not separable.

Proof: Lety=(⌘1,⌘2,. . .) be a sequence of 0,1.Then clearly, y✏

l1.Corresponding to y, we associate the real number ˜ y, whose binary

representation is⌘1

2+⌘2

22+⌘3

23+...

Observe that every ˜ y✏[0,1] has a unique binary representation.Hence,

there are uncountably many sequences made up of 0, 1.The metric on

l1shows that if x6=ythen d(x, y)=1 . I fw ec h o o s eas m a l lb a l lo f

radius1

3centred at these sequences then these balls do not intersect

and there are uncountably many such balls. If Ais any dense set in l1

then every such non-intersecting balls contain an element of A. This

shows that Acant be countable set.Since Ais an arbitrary dense set,

we conclude that l1cant have countable dense set.Therefore, l1is not

separable.

Theorem 5.4.2. The space lpis separable for 1p<1.

Proof: Let M be the set of all sequences y of the form y=(⌘1,⌘2,. . . ,⌘n,0,0,. . .)

where nis any positive integer and ⌘i’s are rational.Observe that M

is countable.We claim that Mis dense in lp.Let x=(⇣i)✏lpbe any

element.Then for every ✏>0t h e r ei sa n ndepending on ✏such that

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P1

i=n+1|⇣i|p<✏p/2.This is because LHS is the remainder of a con-

vergent series. Since, the rationals are dense in R, for each ⇣ithere

is rational number ⌘iclose to it.Hence, we can ﬁnd y✏MsatisfyingPn

i=1|⇣i⌘i|p<✏p/2.This implies that

[d(x, y)]p=Pn

i=1|⇣i⌘i|p+P1

i=n+1|⇣i|p<✏p.Thus, we have d(x, y)<

✏.Therefore, M is dense in lp.

Theorem 5.4.3. If the dual space X0of a normed space is separable ,

then X itself is separable.

Proof: We assume that X0is separable.Then the unit sphere

U={f✏X0:||f||=1}⇢X0also contains a countable dense subset

say ( fn). S i n c e fn✏U,w eh a v e ||fn||=sup{|fn(x)|:||x||=1}=1 .

Therefore, by deﬁnition of supremum, we can ﬁnd points xn✏Xof norm

1s u c ht h a t |fn(xn)|1

2. Let Y=span(xn).Then Y is separable,

because Y has a countable dense subset span(xn), which consists of all

linear combinations of x0

nswith coe cients, whose real and imaginary

parts are rational numbers.We claim that Y=X.Suppose that Y6=

X.Since, Yis closed in X, therefore, by the following lemma:

Lemma 5.4.1. (Existence of functional). Let Ybe proper closed sub-

space of a normed space X.Letx0✏XYbe arbitrary and =inf{||˜y

x0||:˜y✏Y}, the distance from x0to Y. Then there exists an ˜f✏X0such

that ||˜f||=1,˜f(y)=0 for all y✏Y,˜f(x0)=

We have ˜f✏X0with ||˜f||=1a n d ˜f(y)=0f o ra l l y✏Y

Since xn✏Y,w eh a v e ˜f(xn) = 0 and for all n,

1

2|fn(xn)|=|fn(xn)˜f(xn)|=|(fn˜f)(xn)|||fn˜f|| ||xn||, where

||xn||= 1.Hence ||fn˜f||1

2,but this contradicts the assumption that

(fn)i sd e n s ei n U, because we have ˜f✏U

5.5 Let us Sum Up

(1) If Tis a linear transformation existing between two normed

spaces XandYthen T is bounded i ↵Ti sc o n t i n u o u si ↵Ti sc o n t i n u -

ous at the origin i ↵Tm a p sc l o s e du n i ts p h e r et oab o u n d e ds e t .

(2)B(X,Y)i sn o r m e dl i n e a rs p a c ea n di ti sc o m p l e t e ,i f Yis complete.

(3) The dual space of RnisRn.

(4) The dual space of l1isl1.

(5) For 1

(6) Every bounded linear functional on Hilbert space can be represented

in terms of an inner product.

(7) For 1

then L1(X)0=L1(X).

(8) The space l1is not separable.

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(9) For 1 p<1, the space lpis separable.

(10)If the dual space X0of a normed space is separable then Xis sep-

arable.

5.5 List of References:

(1) Introductory Functional Analysis with applications by Erwin Kreyszig,

Wiley India

(2) Educational Resources , the University of California , Davis

5.6 CHAPTER END EXERCISES

(1) Show that C([a, b]) is separable.

(Hint: This follows from the Weierstrass approximation theorem that

states that P([a, b])={f✏C([a, b]) :fis a polynomial with real coef-

ﬁcients }is dense in C([a, b]).Further, Q([a, b]) = {f✏C([a, b]) :fis

ap o l y n o m i a lw i t hr a t i o n a lc o e cients }, we can show that Q([a, b]i s

countable and it is dense in P([a, b])

(2)C([a, b],R)w i t hs u p r e m u mn o r mi sB a n a c hs p a c e .

Proof: Let {fn}be any Cauchy sequence in C([a, b],R). This means

that given ✏>0t h e r ee x i s t sa ni n t e g e r N> 0s u c ht h a t ||fnfm||1<

✏

2, whenever, n, mN.That is , given any ✏>0t h e r ee x i s t sa ni n t e g e r

N> 0s u c ht h a t |fnfm|<✏

2for all n, mNand every x✏[a, b].Thus

{fk(x)}is a Cauchy sequence of real numbers for every x✏[a, b].Since

Ris complete.Therefore {fk(x)}converges to some real number for

each x; we will denote this value by f(x).This deﬁnes a new function

fsuch that fn!fpointwise. We prove that fn!funiformly

on [a, b]. Since fnis a sequence of continuous function therefore its

uniform limit fis also continuous.Let ✏>0b ea n yn u m b e r . T h e nt h e r e

exists an Nsuch that |fn(x)fm(x)|<✏

2for every choice of x✏[a, b]

andn, mN. IfmNandx✏[a, b] then fn(x)✏(fm(x)✏

2,fm(x)+✏

2),

for all nN.Therefore f(x)✏[fm(x)✏

2,fm(x)+✏

2], and hence |f(x)

fm(x)|✏

2<✏.Since xwas arbitrary, therefore we are done.

(3) Let Xbe a normed space of all polynomials on I=[ 0,1] with

norm given by ||x||=max|x(t)|, for t✏I.A d i ↵erentiation operator T

is deﬁned on XbyT(x(t)) = x0(t).Show that Tis linear operator and

it is not bounded.

(Hint: Consider xn(t)=tn, where n✏N. Then ||xn||=1a n d ||T(xn)||=

n)

(4) Let T be a bounded linear operator from a normed space Xonto a

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normed space Y.If there is positive bsuch that ||T(x)||b||x||for all

x✏Xthen show that T1exists and it is bounded.

(5) Show that the dual space of c0isl1

(6) Show that every bounded linear functional fonl2can be repre-

sented in the form f(x)=P1

j=1⇠j⇣jforz=(⇣j)✏l2

(Hint: Use Riesz representation theorem to express any bounded linear

functional fon Hibert space l2asf(x)= )

(7) Show that any Hilbert Space is isomorphic with its second dual

space H00=(H0)0(This property is called reﬂexivity of H)

Proof: We shall prove that the canonical mapping C:H!H00de-

ﬁned by C(x)=gxis onto, where gxis a functional on X0deﬁned for

ﬁxed x✏Xasgx(f)=f(x)f o r f✏X0, by showing that for every g✏H00

there exist an x✏Hsuch that g=C(x)

Deﬁne A:H0!HbyA(f)=z, where zis determined by the Riesz

representation theorem f(x)= ,w ek n o wt h a t Ais bijective

and isometric. A(↵f1+f2)=↵Af1+Af2implies that Ais conju-

gate linear.Observe that H0is complete and its an Hilbert space with

inner product deﬁned by =< Af 2, Af 1>.For all functionals

f1,f2,f3and scalars ↵we have

=< Af 3,A(f1+f2)>==

< Af 1, Af 3>+< Af 2, Af 3>=< Af 3, Af 1>+< Af 3, Af 2>

=+

<↵f1,f2>=< Af 2,A(↵f1)>=< Af 2,↵Af1>=↵

=< Af 2, Af 1>=< Af 1, Af 2>=

=< Af 1, Af 1>= 0a n d =0i m p l i e st h a t

z=0 ,h e n c e f1=0

Letg✏H”b ea r b i t r a r y . L e ti t sR i e s zr e p r e s e n t a t i o nb e g(f)==<

Af0, Af > .We know that f(x)= , where z=Af.W r i t i n g

Af0=x, we therefore have < Af 0, Af > = =f(x).Together

with g(f)=f(x)i m p l i e st h a t g=C(x), by the deﬁnition of C.Since

g✏H” was arbitrary, C is onto, so that His reﬂexive.

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Chapter 6

Four Pillars of Functional

Analysis

Unit Structure :

6.1I n t r o d u c t i o n

6.2O b j e c t i v e

6.3 Few Deﬁnitions and Notations

6.4 Hahn-Banach Theorem

6.5 Uniform Boundedness Principle

6.6O p e nM a p p i n gT h e o r e m

6.7C l o s e dG r a p hT h e o r e m

6.8 Applications of Hahn-Banach theorem

6.9C h a p t e rE n dE x e r c i s e

6.1 Introduction

In this chapter we shall see four important theorems, which are

also called sometimes called as four pillars of Functional Analysis. The

Hahn-Banach theorem, the Open Mapping Theorem, Closed Graph

Theorem and Uniform Boundedness Principle.

Hahn-Banach Theorems : It is so much important because it pro-

vides us with the linear functionals to work on various spaces as Func-

tional Analysis is all about the study of functionals.

Open Mapping Theorem : It provides us with the open sets in

the topology of the range of the mapping.

Uniform Boundedness Principle : An application of Baire Cat-

egory theorem. It is further used many times as the uniformity is an

important property.

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Closed Graph Theorem : Closeness of the graph of a map is

enough to prove its boundedness or continuity. This fact is further

used many times.

6.2 Objectives

After going through this chapter you will be able to:

•State and prove Hahn-Banach theorems

•State and prove Open Mapping theorem

•State and prove Closed Graph theorem

•State and prove Uniform Boundedness theorem

6.3 Few Deﬁnitions and Notations

LetXbe a normed space over the ﬁeld K(RorC).

Deﬁnition 6.1. LetXbe a normed space over K.A m a p p i n g f:

X!Kis said to be linear functional onXif:

f(↵x+y)=↵f(x)+f(y),8x, y2X,↵,2K

Deﬁnition 6.2. A linear functional f, as deﬁned in Deﬁnition 6.1 ,

is said to be bounded if there exists M> 0i nRsuch that

|f(x)|Mkxk,8x2X

.

Note : The branch of analysis of functionals as deﬁned above was ba-

sically called as functional analysis initially! The bounded linear func-

tional is a special case of bounded liner operator and hence all proper-

ties related to bounded linear operators holds true for bounded linear

functionals also. Here is a small activity for you to recollect these

properties.

Activity 1 : Let fbe a linear functional on a normed space X.

1.fis continuous i ↵ker (f)i sc l o s e di n X.

.....................................................................

.....................................................................

.....................................................................

.....................................................................

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.....................................................................

.....................................................................

2.Consider Rnwith usual norm and let a=(a1,a2,...,a n)b ea

vector in Rn. Deﬁne f:Rn!Rsuch that f(x)=x.a, where x=

(x1,x2,...,x n)a n d x.adenote the scalar product of xwith a.S h o w

that fis a bounded linear functional and kfk=kak.

.....................................................................

.....................................................................

.....................................................................

.....................................................................

.....................................................................

.....................................................................

3.fis continuous i ↵fis bounded.

.....................................................................

.....................................................................

.....................................................................

.....................................................................

4.Iffis bounded linear functional then |f(x)|kfkkxk,8x2X.

.....................................................................

.....................................................................

.....................................................................

.....................................................................

Remark: Iffis a bounded linear complex functional then fneed not

be linear.

Deﬁnition 6.3. LetXbe a linear space over R. A functional pis said

to be sublinear functional onXif it satisﬁes the following properties:

(i)p(x+y)p(x)+p(y),8x, y2X

(ii)p(↵x)=↵p(x),8↵0i nR,x2X

Deﬁnition 6.4. LetXbe a linear space over KandZbe it’s subspace.

Letf:Z!Kbe a linear functional. Then ˜f:X!Kis said to be

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an extension of fiff(x)= ˜f(x),8x2Z. Infact fis also called as

restriction of ˜fonZand is denoted as ˜f|Z=f.

6.4 Hahn-Banach Theorem

Theorem 6.4.1. (Hahn-Banach Lemma ) Let Xbe a real vector

space and pbe a sublinear functional on X. Let Zbe a subspace of X

andfbe linear functional deﬁned on Zsuch that

f(x)p(x),8x2Z

Then, there exists a linear functional ˜fonXsuch that ˜f|Z=fand

˜f(x)p(x),8x2X

Proof: Consider a set Lof all linear extensions ( Z↵,g↵)o f( Z, f)s u c h

that,

g↵(x)p(x),8x2Z↵

Since ( M,f)2L, clearly L6=.

Now, we deﬁne a relation ” ”o n Lsuch that:

(Z↵,g↵)(Z,g),Z↵⇢Zand

gis a extension of g↵i.e.g|Z↵=g↵

Activity 2: Check that the relation ” ” deﬁned above is reﬂexive,

antisymmetric and transitive.

.....................................................................

.....................................................................

.....................................................................

.....................................................................

From the above activity we can conclude that ( L,)i sap a r t i a lo r d e r e d

set. Let Qbe any totally ordered subset of Land let

Z0=[{Z:(Z,g)2Q}

We see that Z0is a subspace. Deﬁne g0:Z0!Rby

g0(x)=g(x),8x2Z

Clearly, g0is a linear functional on Z0andg0|Z=f.

Claim: ( Z0,g0)i sa nu p p e rb o u n do f Q(Try it yourself)

By Zorn’s lemma, Qhas a maximal element, say ( Z0,g0).

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To prove that, ( Z0,g0)=( X,˜f). It is enough to prove that Z0=X.

Suppose Z06=X. Then there exists x02XZ0. Consider the linear

space spanned by Z0andx0,Z0=Z0+[x0].

Each element z2Z0can be uniquely expressed as z=x+↵x0, where

x2Z0and↵2R.

Deﬁne g1:Z1!Rasg1(x+↵x0)=g0(x)+↵K, where Kis a real

constant.

We can see that g1is a linear functional on Z1 (Verify this !) And

g1|Z=g0

Thus, Z1is a linear subspace of Xcontaining Zandg1|Z=f

The constant Kcan be chosen appropriately so that g1(y)p(y),8y2

Z1.

This means, 9 Z1,g1)2Qand ( Z0,g0)(Z1,g1),Z 06=Z1. This

contradicts the maximality of ( Z0,g0). Hence our assumption that Z06=

Xis not true. Hence the proof.

Theorem 6.4.2. (Hahn-Banach ) Let Xbe a normed space over a

ﬁeld KandZbe a subspace of X.Then, for every bounded linear func-

tional fonZ, there exists a bounded linear functional ˜fonXsuch

that, ˜f|Z=fandk˜fk=kfk

6.5 Uniform Boundedness Principle

The uniform boundedness principle (or uniform boundedness theo-

rem) by S. Banach and H. Steinhaus (1927) is one of the fundamental

results in functional analysis. Together with the Hahn-Banach theo-

rem, the open mapping theorem and the closed graph theorem, it is

considered as one of the cornerstones of the ﬁeld.

The uniform boundedness principle answers the question of whether

a” p o i n t - w i s eb o u n d e d ”s e q u e n c eo fb o u n d e dl i n e a ro p e r a t o r sm u s ta l s o

be ”uniformly bounded”. As the proof of the Uniform Boundedness

Principle is an application of Baire’s Category Theorem. So, we shall

prove the Baire’s category theorem ﬁrst. Following are the basic con-

cepts needed for Baire’s theorem:

Deﬁnition 6.5. (Nowhere Dense or Rare) A subset Mof a metric

space Xis said to be Nowhere dense in Xif its closure Mhas no

interior points. That is, int(M)6=;,Mcontains no open ball.

Example 30. The set of all integers Zis nowhere dense set in R

Example 31. Let ( R,d)b et h eu s u a lm e t r i cs p a c e ,t h e ne v e r ys i n g l e t o n

is nowhere dense in Rsince {a}={a}for every a2R. And int(a)=;

since it contains no open interval.

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Deﬁnition 6.6. (Meager or of First Category) A subset Min metric

space Xis said to be of First Category if Mis the union of countably

many sets which are all nowhere dense in X.

Example 32. Since Qis countable and for every a2Q,{a}being

nowhere dense,

Q=[

a2Q{a}

is of ﬁrst category.

Deﬁnition 6.7. (Nonmeager or of Second Category) A subset Min

metric space Xis said to be of Second Category if Mis not of ﬁrst

category in X.

Theorem 6.5.1. (Baire’s Category Theorem) If a metric space X6=;

is complete, it is of second category. Hence, if X6=;is complete and

X=1[

k=1Ak(Akclosed )

then atleast one Akcontains a nonempty open subset.

Proof. Suppose the metric space X6=;is of ﬁrst category in itself.

Then

X=1[

k=1Mk,

where each Mkis rare in X.S i n c e M1is rare in X, so by deﬁnition, M1

does not contain a nonempty open set. But ( X,d) is complete, so it

will contain a nonempty set. So, M16=X. Therefore, M1C=XM1

which is nonempty and open. At the point p12M1C, we can get an

open ball

B1=B(p1,✏1)⇢M1C,

where ✏<1

2.

Further, M2is rare in X, so that M2does not contain a nonempty open

set. Hence, it does not contain open ball B(p1,✏1

2). This implies that

M2CTB(p1,✏1

2) is not empty and open. Now, we may choose an open

ball in this set, say,

B2=B(p2,✏2)⇢M2C\

B(p1,✏1

2),✏2<✏1

2.

Continuing this process, we obtain a sequence of balls by induction,

Bk=B(pk,✏k),✏k<2k

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such that BkTMk=;andBk+1⇢B(pk,✏k

2)⇢Bk,k=1,2,. . ..

As✏k<2kand the space Xis complete, the sequence ( pkof the

centers is Cauchy and converges, say, pk!p2X. Also, for every m

andn>m we have Bn⇢B(pm,✏m

2), so that

d(pm,p)d(pm,pn)+d(pn,p)

<✏m

2+d(pn,p)!✏m

2

as n approaches to 1. Hence, pbelongs to Bmfor every m. Since

Bm⇢MmC, we get p62Mmfor every m, so that p62SMm=X. This

contradicts p2X. Hence proved.

Theorem 6.5.2. Let{Tn}be a sequence of bounded linear operators

Tn:X!Yfrom a Banach space Xinto a normed space Y. If for

every x2X,{Tn(x)}is bounded, say,

kTn(x)kcx,n =1,2,3,. . .(pointwise boundedness ),

where cxis a real number, then the sequence kTnkis also bounded, that

is, there is a positive real csuch that

kTnkc, n =1,2,3,. . .(uniform boundedness ),

Proof. For every x2X,{Tn(x)}is bounded sequence in Y, that is,

kTn(x)kcxwhere c x0.

Suppose that Ak={x2X:kTn(x)kk}

Step 1: We claim that Akis nonempty, closed and X=S1

k=1Ak. Since

02XandTnis linear so Tn(0) = 0 kfor each k. This implies that

02Akfor each kwhich implies that Akis nonempty.

Let{xn}be a convergent sequence in Akwith xm!xasn!1.

This means that for every ﬁxed n,w eh a v e kTn(xm)kkand obtain

kTn(x)kkby applying limits for m!1because Tnis continuous

and so is the norm. Hence, x2Akand therefore Akis closed.

Also, Ak✓X(for all k))S1

k=1Ak✓X. On the other hand, let

x2X,kTn(x)kis a real number. Using Archimedean property, there

is a positive integer n0such that Tn(x)kcxn0.S o ,

x2An0✓S1

k=1Ak)X✓S1

k=1Ak

Hence,

X=1[

k=1Ak=1[

k=1Ak,(Since A k=Ak)

In view of Baire’s Category Theorem, every complete space is of second

category. That is, at least one of Akis not nowhere dense. Hence, some

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Akcontains an open ball, say, B0=B(x0,r)✓Ak0.

Step 2: Letx2Xbe arbitrary, not zero. We set

z=x0+x,

where =r

2kxk. Then

z=x0+r

2kxkx

zx0=r

2kxkx

kzx0k=r

2kxkkxk=r

2That is, z2B(x0,r)✓Ak0.

As,z,x 02Ak0)kTn(z)kk0,kTn(x0)kk0. We have,

x=2kxk

r(zx0).

Therefore,

kTn(x)k=2kxk

rkTn(zx0)k

=2kxk

rkTn(z)Tn(x0)k

2kxk

r[kTn(z)k+kTn(x0)k]

2kxk

r(2k0)

=4k0

rkxk

Hence,

kTn(x)k4k0

rkxk)kTn(x)k

kxk4k0

r=c.

which implies

sup

x2X,x6=0kTn(x)k

kxkc)kTnkc.

6.6 Open Mapping Theorem

Open mapping theorem is one of the basic theorems for the develop-

ment of the general theory of normed linear spaces. The theorem gives

conditions under which a linear mapping is open. In this theorem, we

begin to appreciate the importance of the completeness condition for

normed linear spaces.

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Deﬁnition 6.8. (Open Mapping) Let XandYbe metric spaces, a

mapping T:D(T)!Ywith domain D(T)⇢Xis called an open

mapping if for every open set in D(T)t h ei m a g ei sa no p e ns e ti n Y.

Deﬁnition 6.9. (Continuous Mapping) A continuous mapping T:

X!Yhas the property that for every open set in Ythe inverse

image is an open set in X.

Remark 6.6.1. Continuous mappings are not necessarily open map-

pings.

Example 33. Consider a mapping T:R!Rdeﬁned by T(x)=

Sinx. As Sine function is continuous, Tis a continuous mapping but

T[(0,2⇡)] = [ 1,1] that is, Tmaps an open set (0 ,2⇡)o n t o[ 1,1]

which is not open.

Example 34. Deﬁne T:R!RbyT(x)=x2. Then Tis continuous

mapping but not an open mapping because it maps an open set ( 1,1)

onto [0 ,1) which is not open.

Lemma 6.6.1. (Open Unit Ball) Let Tbe a bounded linear operator

from a Banach space Xonto a Banach space Y. Then the image of the

open ball B0=B1(0)⇢X, that is, T(B0), contains an open ball with

center 0inY.

Proof. The proof has three steps. We will prove

(a) The closure of the image of the open ball B1=B(0;1

2)c o n t a i n sa n

open ball B⇤.

(b)T(Bn)c o n t a i n sa no p e nb a l l Vnabout 0 2Y, where Bn=B(0; 2n)⇢

X.

(c)T(B0)c o n t a i n sa no p e nb a l la b o u t0 2Y.

(a) Clearly, we have

1[

k=1kB1⇢X

For any x2X, there is k(k>2||x||)s u c ht h a t x2kB1.S o ,

X⇢1[

k=1kB1.

Thus, we have

X=1[

k=1kB1.

Since Tis surjective,

Y=T(X)=T(1[

k=1kB1)=1[

k=1T(kB1).

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Due to the linearity of T,w eh a v e

Y=1[

k=1T(kB1)=1[

k=1kT(B1)=1[

k=1kT(B1).

Since Yis complete, by the Baire’s Category theorem, we conclude that

akT(B1)c o n t a i n sa no p e nb a l l .T h i si m p l i e st h a t T(B1)a l s oc o n t a i n s

an open ball, namely, there is B⇤=B(y0;✏)s u c ht h a t B⇤⇢T(B1). It

follows that B⇤y0=B(0;✏)⇢T(B1)y0.

(b) We will ﬁrst prove that B(0;✏)=B⇤y0⇢T(B0). Since B⇤⇢

T(B1)b y( a ) ,w eh a v e B⇤y0⇢T(B1)y0. It su ces to prove

T(B1)y0⇢T(B0). Let y2T(B1)y0. Then y+y02T(B1).

Notice that y02T(B1)s i n c e B⇤⇢T(B1). Then there are sequences

un=Twn2T(B1)a n d vn=Tzn2T(B1)s u c ht h a t un!y+y0,

vn!y0, where wn,zn2B1. Observing that

||wnzn||||wn||+||zn||<1

2+1

2=1.

So,wnzn2B0. Also, T(wnzn)=TwnTzn=unvn!y.

Hence, y2T(B0). This proves that B(0;✏)=B⇤y0⇢T(B0). Let

Bn=B(0;1

2n). Since Tis linear, we have T(Bn)=2nT(B0). Let

Vn=B(0;✏

2n). Then Vn=1

2nB(0;✏)⇢1

2nT(B0)=T(Bn). This proves

(b).

(c) We ﬁnally prove that V1=B(0;1

2✏)⇢T(B0)b ys h o w i n gt h a te v e r y

y2V1is in T(B0). So, let y2V1. Since V1⇢T(B1), there is x12B1

such that kyTx1k<✏

4. Then we have yTx12V2. Since V2⇢T(B2),

there is x22B2such that kyTx1Tx2k<✏

8. Continuing in this

manner, we have, for each n, there are xn2Bnsuch that

kynX

k=1Txkk<✏

2n+1.

Letzn=x1+x2+...+xn. The above inequality becomes

kyTznk<✏

2n+1,8n.

Namely, Tzn!y. Since xk2Bk,w eh a v e kxkk<1

2k. So, for n>m ,

kznzmknX

k=m+1kxkk<1X

k=m+11

2k!0, as m !1.

Thus, the sequence {zn}is Cauchy. Since Xis complete, there is x2X

such that zn!xandx=x1+x2+.... Notice that

kxk1X

k=1kxkk<1

2+1X

k=2kxkk1

2+1

2=1.

So,x2B0. Since Tis continuous, we have Tzn!Tx. Hence, y=Tx.

That is, y2T(B0).

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Theorem 6.6.1. (Open Mapping Theorem, Bounded Inverse Theorem)

LetXandYbe Banach spaces. Then any bounded linear operator T

from Xonto Yis an open mapping. Consequently, if Tis bijective,

then T1is continuous and hence bounded.

Proof. LetA⇢Xbe an arbitrary open subset of X. We will show that

the image T(A)i so p e ni n Y. That is, for any y=Tx2T(A), the set

T(A)c o n t a i n sa no p e nb a l lc e n t e r e da t y. Let y2T(A). Then y=Tx

with x2A. Since Ais open, there is r>0s u c ht h a t Br(x)⇢A. Thus

B1(0)⇢1

r(Ax).

By Lemma 6.6.1, the image T(1

r(Ax)) contains an open ball with

center 0. That is, there is ✏>0, such that

B(0;✏)⇢T(1

r(Ax))

Since Tis linear, we have

B(0;✏)⇢1

r(T(A)Tx).

Since y=Tx, the above relation implies B(y;r✏)⇢T(A). Hence, T(A)

contains an open ball with center y.

6.7 Closed Graph Theorem

Deﬁnition 6.10. (Cartesian product of two normed spaces)

Let ( X,k.k1)a n d( X,k.k2)b et w on o r m e ds p a c e s ,t h e n X⇥Yis also a

normed space where the two algebraic operations of a vector space and

the norm on X⇥Yare deﬁned as usual, that is

(x1,y1)+(x2,y2)=( x1+x2,y1+y2)

↵(x, y)=( ↵x,↵y)(↵a scalar )

k(x, y)k=kxk1+kyk2

Theorem 6.7.1. For any two Banach spaces XandY,X⇥Yis also

a Banach space.

Proof. We show that for any two Banach spaces XandY,X⇥Y=

{(x, y):x2X,y2Y}is also a Banach spaces. Let {zn}be a Cauchy

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sequence in X⇥Y, where zn=(xn,yn). Then for every ✏>0, there

exists a positive integer n0such that for every nn0,w eh a v e

kznzmk=k(xn,yn)(xm,ym)k

=k(xnxm,ynym)k

=kxnxmk+kynymk<✏

This implies that kxnxmk<✏,kynymk<✏thereby proving that

{xn}is a Cauchy sequence in Xand{yn}is a Cauchy sequence in Y.

Letxn!x2X,yn!y2Y, then

kznzk=k(xn,yn)(x, y)k=k(xnx, y ny)k=kxnxk+kynyk

Applying n!1,

kxnxk!0 and kynyk!0. This implies that lim n!1kznzk=0 .

Hence, zn!z=(x, y)2X⇥Y. Thus, X⇥Yis a Banach space.

Deﬁnition 6.11. (Closed Linear Operator) Let ( X,k.k1k)a n d( Y,k.k2k)

be normed spaces and let T:D(T)!Ya linear operator with domain

D(T)⇢X. Then Tis called a closed linear operator if its graph

G(T)={(x, y):x2D(T),y=Tx}✓X⇥Y

Theorem 6.7.2. LetXandYbe Banach spaces and T:D(T)!Y

be a closed linear operator, where D(T)⇢X.I fD(T)is closed in X,

then the operator Tis bounded.

Proof. First recall that for any two Banach spaces XandY,X⇥Yis

also a Banach space. By assumption, G(T)i sc l o s e di n X⇥Y. Hence,

G(T)i sc o m p l e t ea n da l s o D(T)i sc l o s e di n X. Therefore, D(T)i s

also complete space. We now consider the mapping P:G(T)!D(T),

deﬁned by P(x, Tx )= x.W e p r o v e t h a t Pis bijective, linear and

bounded. To show that Pis linear, we consider

P((x1,Tx 1)+(x2,Tx 2)) = P(x1+x2,Tx 1+Tx2)

=x1+x2

=P(x1,Tx 1)+P(x2,Tx 2)

And for ↵2K,w eh a v e

P(↵(x, Tx )) = P(↵x,↵Tx)=↵x=↵P(x, Tx )

Now, Pis bounded, because

kP(x, Tx )k=kxkkxk+kTxk=k(x, Tx )k

This implies that kP(x, Tx )k1.k(x, Tx )k. P is onto, since for all

x2D(T)9(x, Tx )2G(T)s u c ht h a t P(x, Tx )=x.Pis one to one,

since

P(x1,Tx 1)=P(x2,Tx 2))x1=x2)(x1,Tx 1)=( x2,Tx 2).

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Hence, Pis a bounded linear operator from Banach space G(T)t o

Banach space D(T)a n da l s oi ti sb i j e c t i v eh e n c e P1:D(T)!G(T)

given by P1(x)=( x, Tx )i sab o u n d e dl i n e a ro p e r a t o rb yB o u n d e d

inverse theorem. That is, there exists b>0s u c ht h a tf o re v e r y x2

D(T), we have

kP1(x)kbkxk)k(x, Tx )kbkxk)kxk+kTxkbkxk

Therefore, kTxkkxk+kTxkbkxk)kTxkbkxk. This implies

that Tis bounded.

6.8 Applications of Hahn-Banach theorem

1. Given a normed linear space Xover a ﬁeld Kand a non zero

member x02X, there is a bounded linear functional Fover Xsuch

that F(x0)=kx0kandkFk=1 .

Proof. Let,M=[x0]= the subspace spanned by {x0}={↵x0:↵is real }.

Deﬁne f:M!Rbyf(↵x0)=↵kx0k. Clearly, fis linear. Now, for

x2M,x=↵x0for some ↵. Now, |fx|=|f(↵x0)|=|↵kx0k|=

|↵|kx0k=k↵x0k=kxk. Clearly, kfk=1 ,t h a ti s , fis a bounded

linear functional on M. So, there exists an extension F(a bounded lin-

ear functional) of fover Xsuch that kfk=kFk. But f(x)=F(x),

8x2M. Now, x02M.S o , f(x0)=F(x0). But f(x0)=kx0k.S o ,

F(x0)=kx0kwith kFk=1 .

2. Let, Xbe a normed linear space over a ﬁeld Kand x06=✓be

an arbitrary member of Xand let M0be an arbitrary positive real.

Then 9ab o u n d e dl i n e a rf u n c t i o n a l fonX, such that kfk=M0and

f(x0)=kfkkx0k.

Proof. LetG=[x0]={tx0:t is real }. Clearly, Gis a subspace of X.

Deﬁne :G!Rby(tx0)=tM0kx0k. Clearly, is linear. Now, for

x2G,x=tx0for some t. Now,

|(x)|=|(tx0)|

=|tM0|kx0k

=M0|t|kx0k

=M0ktx0k

=M0kxk.

So,|(x)|=M0kxk. Clearly, kk=M0.S o , is a bounded linear

functional on G. So, 9an extension fofover Xsuch that kfk=

kk=M0. Hence, kfk=M0. But (x0)=M0kxk. Now, x02G,

f(x0)= (x0)= M0kx0k=kfkkx0k.S o , f(x0)= kfkkx0k. This

completes the proof.

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3. For every x2X,kxk=s u pf6=⇥f(x)

kfk.[⇥is zero functional on X].

Proof. We ﬁnd a non zero bounded linear functional f02X0such that

f0(x)=kxkandkf0k= 1. Now,

sup

f6=⇥|f(x)|

kfk|f0(x)

kf0k

Again, 8f6=⇥2X0,|f(x)|kfkkxk.S o ,

|f(x)|

kfkkxk.

That is,

sup

f6=⇥2X0|f(x)|

kfkkxk.

Hence,

kxk=s u p

f6=⇥2X0|f(x)|

kfk.

4. Let Mbe a closed subspace of normed linear space Xsuch that

M6=X. Let u2X\Mand let

d=dist(u, M)= i n f

m2MkuMk

Then there is a bounded linear functional f2X0such that

(i)f(x)=0 8x2M

(ii)f(u)=1a n d( i i i ) kfk=1

d

Proof. Clearly, d>0. Let N=[MS{u}]. Clearly, Nis a subspace of

X. So, every member of Nis of the form m+tu, where m2M,t2R.

Deﬁne g:N!Rbyg(m+tu)=t. Clearly, gis linear. Now, g(m)=0 ,

for some m2M,g(u)=1 . F o r t6=0 ,

|g(m+tu)|=|t|

=|t|km+tuk

km+tuk

=km+tuk

ku(m/t)k1

dkm+tuk,

as (m/t)2M. Let x=m+tu2N.S o , |g(x)|1

dkxkimplying that

gis bounded and kgk1

d.S o , g2N0. Again, d=i n f m2Mkumk.

So, there exists a sequence {mn}2Msuch that kumnk!das

n!1. Now,

|g(umn)|kgkkumnk)1kgkkumnk,

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since g(umn)=g(u)g(mn)=1 0=1 . L e t t i n g n!1,1

dkgk,

we get kgk=1

d.S o , 9ab o u n d e dl i n e a rf u n c t i o n a l f2X0which is an

extension of gonNsuch that f(x)=g(x),8x2Nandkfk=kgk.

Thus, we have

(i)f(x)=0 , 8x2N

(ii)f(u)=1

(iii)kfk=1

d.

5. Let Mbe a subspace of a normed linear space XandM6=X. If

u2XM such that d=dist(u, M)>0. Then there is a bounded

linear functional F2X0such that

(i)F(x)=0 8x2M

(ii)F(u)=d

(iii)kFk=1 .

Proof. Let,N=[MS{u}] = a subspace of Xspanned by Mand{u}.

Deﬁne f:N!Rbyf(x)=td, where x=m+tufor some m2M

and for some t. That is, f(m+tu)=td. Clearly, fis linear. Now,

f(m)=0f o r m2M,f(u)=d. Now, for t6=0

km+tuk=kt✓m

tu◆

k

=|t|km

tuk|t|d,

asm

t2Mandd=dist(u, M). So, |f(m+tu)|=|t|dkm+tuk.

Letx=m+tu2N.S o ,

|f(x)|kxk)kfk1. (6.1)

Clearly, fis bounded with kfk1. So, f2N0.W e h a v e d=

dist(u, M)=i n f m2Mkumk. Let ✏>0. By inﬁmum property there

exists an element m2Nsuch that kumkPut

v=mu

kmuk=m

kmuk1

kmuku2N,

where t=1

kmukandm

kmuk2M.S o , kvk=1 . B u t

|f(v)|=d

kmuk>d

d+✏kvk,kvk=1.

As,✏>0i sa r b i t r a r y ,

|f(v)|kvk)kfk1. (6.2)

So, by (6 .1) and (6 .2),kfk= 1. So, there exists a bounded linear

functional Fover Xwhich is an extension of fover Nsuch that f(x)=

F(X),8x2Nandkfk=kFk.S o ,( i ) F(x)=0 , 8x2M,

(ii)F(u)=dand (iii) kFk=1 .

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## Page 118

FUNCTIONAL ANALYSIS

6.9 Chapter End Exercises

1. Let fbe an additive functional on a normed space X. Prove that

iffis continuous then fis linear.

2. Prove that f:R2!Rdeﬁned by f(x, y)=f(x)+f(y), x, y 2R,

is a bounded linear functional on ( R,k.k2).

3. If Xbe a non trivial real normed linear space, that is, X6={0}.

Then its ﬁrst conjugate space X0is also nontrivial.

4. Let Xis a Banach space and A⇢Xad e n s es e t . C a nw eﬁ n d

af u n c t i o n f:X!Rsuch that, for every x2A,w eh a v e

lim t!x|f(t)|=1?

5. Let XandYbe Banach spaces and T2B(X,Y). Suppose Tis

bijective. Show that there exist real numbers a, b > 0s u c ht h a t

akxkkTxkbkxk,8x2X.

6. Let X,YandZbe Banach spaces. Suppose that T:X!Yis

linear, that J:Y!Zis linear, bounded and injective, and that

JT⌘JT:X!Zis bounded. Show that Tis also bounded.

7. Let ( X,k.k1)a n d( X,k.k2)b eB a n a c hs p a c e s .S u p p o s et h a t

9C0:kxk2Ckxk1,8x2X.

Show that the two norms k.k1andk.k2are equivalent.

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