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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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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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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 1q=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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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/pnX
i=1✓
|xi+yi|p1◆q1/q
=✓nX
i=1|xi|p◆1/pnX
i=1|xi+yi|q(p1)1/q
=✓nX
i=1|xi|p◆1/pnX
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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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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(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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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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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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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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(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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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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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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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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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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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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(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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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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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 1q=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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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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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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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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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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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, 8t21
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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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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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||=00||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 lpislqfor1p+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
1fgdµ 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 p1that
|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 1F: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 p1. 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||Lqlim n!1inf||n||Lqlim n!1infR
|ngn|dµ
lim n!1R
|fgn|dµM.
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 1p<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)kcx,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
kTnkc, n =1,2,3,. . .(uniform boundedness ),
Proof. For every x2X,{Tn(x)}is bounded sequence in Y, that is,
kTn(x)kcxwhere c x0.
Suppose that Ak={x2X:kTn(x)kk}
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)kkand obtain
kTn(x)kkby 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)kcxn0.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)kk0,kTn(x0)kk0. 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)k4k0
rkxk)kTn(x)k
kxk4k0
r=c.
which implies
sup
x2X,x6=0kTn(x)k
kxkc)kTnkc.
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 ,
kznzmknX
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
kxk1X
k=1kxkk<1
2+1X
k=2kxkk1
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=kxkkxk+kTxk=k(x, Tx )k
This implies that kP(x, Tx )k1.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)kbkxk)k(x, Tx )kbkxk)kxk+kTxkbkxk
Therefore, kTxkkxk+kTxkbkxk)kTxkbkxk. 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)|
kfkkxk.
That is,
sup
f6=⇥2X0|f(x)|
kfkkxk.
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)k1
dkm+tuk,
as (m/t)2M. Let x=m+tu2N.S o , |g(x)|1
dkxkimplying that
gis bounded and kgk1
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)1kgkkumnk,
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since g(umn)=g(u)g(mn)=1 0=1 . L e t t i n g n!1,1
dkgk,
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|dkm+tuk.
Letx=m+tu2N.S o ,
|f(x)|kxk)kfk1. (6.1)
Clearly, fis bounded with kfk1. 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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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
akxkkTxkbkxk,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:kxk2Ckxk1,8x2X.
Show that the two norms k.k1andk.k2are equivalent.
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