FUNCTIONAL ANALYSIS NOTES

(2011)
Mr. Andrew Pinchuck
Department of Mathematics (Pure & Applied)
Rhodes University
Contents
Introduction 1
1 Linear Spaces 2
1.1 Introducton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Subsets of a linear space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Subspaces and Convex Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Quotient Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.5 Direct Sums and Projections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.6 The H older and Minkowski Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 Normed Linear Spaces 13
2.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.2 Quotient Norm and Quotient Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.3 Completeness of Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.4 Series in Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5 Bounded, Totally Bounded, and Compact Subsets of a Normed Linear Space . . . . . . . 26
2.6 Finite Dimensional Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 28
2.7 Separable Spaces and Schauder Bases . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3 Hilbert Spaces 36
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
3.2 Completeness of Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.3 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.4 Best Approximation in Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.5 Orthonormal Sets and Orthonormal Bases . . . . . . . . . . . . . . . . . . . . . . . . . . 49
4 Bounded Linear Operators and Functionals 62
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
4.2 Examples of Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
4.3 The Dual Space of a Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5 The Hahn-Banach Theorem and its Consequences 81
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.2 Consequences of the Hahn-Banach Extension Theorem . . . . . . . . . . . . . . . . . . . 85
5.3 Bidual of a normed linear space and Reexivity . . . . . . . . . . . . . . . . . . . . . . . 88
5.4 The Adjoint Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90
5.5 Weak Topologies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91
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6 Baires Category Theorem and its Applications 99
6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
6.2 Uniform Boundedness Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101
6.3 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
6.4 Closed Graph Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
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2011 FUNCTIONAL ANALYSIS ALP
Introduction
These course notes are adapted from the original course notes written by Prof. Sizwe Mabizela when
he last gave this course in 2006 to whom I am indebted. I thus make no claims of originality but have made
several changes throughout. In particular, I have attempted to motivate these results in terms of applications
in science and in other important branches of mathematics.
Functional analysis is the branch of mathematics, specically of analysis, concerned with the study of
vector spaces and operators acting on them. It is essentially where linear algebra meets analysis. That is,
an important part of functional analysis is the study of vector spaces endowed with topological structure.
Functional analysis arose in the study of tansformations of functions, such as the Fourier transform, and in
the study of differential and integral equations. The founding and early development of functional analysis
is largely due to a group of Polish mathematicians around Stefan Banach in the rst half of the 20th century
but continues to be an area of intensive research to this day. Functional analysis has its main applications in
differential equations, probability theory, quantum mechanics and measure theory amongst other areas and
can best be viewed as a powerful collection of tools that have far reaching consequences.
As a prerequisite for this course, the reader must be familiar with linear algebra up to the level of a
standard second year university course and be familiar with real analysis. The aim of this course is to
introduce the student to the key ideas of functional analysis. It should be remembered however that we
only scratch the surface of this vast area in this course. We examine normed linear spaces, Hilbert spaces,
bounded linear operators, dual spaces and the most famous and important results in functional analysis
such as the Hahn-Banach theorem, Baires category theorem, the uniform boundedness principle, the open
mapping theorem and the closed graph theorem. We attempt to give justications and motivations for the
ideas developed as we go along.
Throughout the notes, you will notice that there are exercises and it is up to the student to work through
these. In certain cases, there are statements made without justication and once again it is up to the student
to rigourously verify these results. For further reading on these topics the reader is referred to the following
texts:
v G. BACHMAN, L. NARICI, Functional Analysis, Academic Press, N.Y. 1966.
v E. KREYSZIG, IntroductoryFunctional Analysis, John Wiley &sons, NewYork-Chichester-Brisbane-
Toronto, 1978.
v G. F. SIMMONS, Introduction to topology and modern analysis, McGraw-Hill Book Company, Sin-
gapore, 1963.
v A. E. TAYLOR, Introduction to Functional Analysis, John Wiley & Sons, N. Y. 1958.
I have also found Wikipedia to be quite useful as a general reference.
1
Chapter 1
Linear Spaces
1.1 Introducton
In this rst chapter we review the important notions associated with vector spaces. We also state and prove
some well known inequalities that will have important consequences in the following chapter.
Unless otherwise stated, we shall denote by R the eld of real numbers and by C the eld of complex
numbers. Let F denote either R or C.
1.1.1 Denition
A linear space over a eld F is a nonempty set X with two operations
: X X X (called addition). and
: F X X (called multiplication)
satisfying the following properties:
[1] x y X whenever x. y X;
[2] x y = y x for all x. y X;
[3] There exists a unique element in X, denoted by 0, such that x 0 = 0 x = x for all x X;
[4] Associated with each x X is a unique element in X, denoted by x, such that x (x) =
x x = 0;
[5] (x y) z = x (y z) for all x. y. z X;
[6] x X for all x X and for all F;
[7] (x y) = x y for all x. y X and all F;
[8] ( ) x = x x for all x X and all . F;
[9] () x = ( x) for all x X and all . F;
[10] 1 x = x for all x X.
We emphasize that a linear space is a quadruple (X. F. . ) where X is the underlying set, F a eld,
addition, and multiplication. When no confusion can arise we shall identify the linear space (X. F. . )
with the underlying set X. To show that X is a linear space, it sufces to show that it is closed under
addition and scalar multiplication operations. Once this has been shown, it is easy to show that all the other
axioms hold.
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2011 FUNCTIONAL ANALYSIS ALP
1.1.2 Denition
A real (resp. complex) linear space is a linear space over the real (resp. complex) eld.
A linear space is also called a vector space and its elements are called vectors.
1.1.3 Examples
[1] For a xed positive integer n, let X = F
n
= {x = (x
1
. x
2
. . . . . x
n
) : x
i
F. i =
1. 2. . . . . n] the set of all n-tuples of real or complex numbers. Dene the operations
of addition and scalar multiplication pointwise as follows: For all x = (x
1
. x
2
. . . . . x
n
).
y = (y
1
. y
2
. . . . . y
n
) in F
n
and F,
x y = (x
1
y
1
. x
2
y
2
. . . . . x
n
y
n
)
x = (x
1
. x
2
. . . . . x
n
).
Then F
n
is a linear space over F.
[2] Let X = Ca. b| = { x : a. b| F [ x is continuous ]. Dene the operations of addition and
scalar multiplication pointwise: For all x. y X and all R, dene
(x y)(t ) = x(t ) y(t ) and
( x)(t ) = x(t )
_
for all t a. b|.
Then Ca. b| is a real vector space.
Sequence Spaces: Informally, a sequence in X is a list of numbers indexed by N. Equivalently,
a sequence in X is a function x : N X given by n x(n) = x
n
. We shall denote a
sequence x
1
. x
2
. . . . by
x = (x
1
. x
2
. . . .) = (x
n
)
1
1
.
[3] The sequence space s. Let s denote the set of all sequences x = (x
n
)
1
1
of real or complex
numbers. Dene the operations of addition and scalar multiplication pointwise: For all x =
(x
1
. x
2
. . . .), y = (y
1
. y
2
. . . .) s and all F, dene
x y = (x
1
y
1
. x
2
y
2
. . . .)
x = (x
1
. x
2
. . . .).
Then s is a linear space over F.
[4] The sequence space
1
. Let
1
=
1
(N) denote the set of all bounded sequences of real or
complex numbers. That is, all sequences x = (x
n
)
1
1
such that
sup
i2N
[x
i
[ < o.
Dene the operations of addition and scalar multiplication pointwise as in example (3). Then

1
is a linear space over F.
[5] The sequence space
p
=
p
(N). 1 _ p < o. Let
p
denote the set of all sequences
x = (x
n
)
1
1
of real or complex numbers satisfying the condition
1

iD1
[x
i
[
p
< o.
Dene the operations of addition and scalar multiplication pointwise: For all x = (x
n
), y =
(y
n
) in
p
and all F, dene
x y = (x
1
y
1
. x
2
y
2
. . . .)
x = (x
1
. x
2
. . . .).
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2011 FUNCTIONAL ANALYSIS ALP
Then
p
is a linear space over F.
Proof. Let x = (x
1
. x
2
. . . .), y = (y
1
. y
2
. . . .)
p
. We must show that x y
p
. Since,
for each i N,
[x
i
y
i
[
p
_ 2 max{[x
i
[. [y
i
[]|
p
_ 2
p
max{[x
i
[
p
. [y
i
[
p
] _ 2
p
([x
i
[
p
[y
i
[
p
) .
it follows that
1

iD1
[x
i
y
i
[
p
_ 2
p
_
1

iD1
[x
i
[
p

iD1
[y
i
[
p
_
< o.
Thus, x y
p
. Also, if x = (x
n
)
p
and F, then
1

iD1
[x
i
[
p
= [[
p
1

iD1
[x
i
[
p
< o.
That is, x
p
.
[6] The sequence space c = c(N). Let c denote the set of all convergent sequences x = (x
n
)
1
1
of
real or complex numbers. That is, c is the set of all sequences x = (x
n
)
1
1
such that lim
n!1
x
n
exists. Dene the operations of addition and scalar multiplication pointwise as in example
(3). Then c is a linear space over F.
[7] The sequence space c
0
= c
0
(N). Let c
0
denote the set of all sequences x = (x
n
)
1
1
of real
or complex numbers which converge to zero. That is, c
0
is the space of all sequences
x = (x
n
)
1
1
such that lim
n!1
x
n
= 0. Dene the operations of addition and scalar multiplication
pointwise as in example (3). Then c
0
is a linear space over F.
[8] The sequence space
0
=
0
(N). Let
0
denote the set of all sequences x = (x
n
)
1
1
of real or
complex numbers such that x
i
= 0 for all but nitely many indices i . Dene the operations
of addition and scalar multiplication pointwise as in example (3). Then
0
is a linear space
over F.
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2011 FUNCTIONAL ANALYSIS ALP
1.2 Subsets of a linear space
Let X be a linear space over F. x X and A and B subsets of X and z F. We shall denote by
x A := {x a : a A].
A B := {a b : a A. b B].
zA := {za : a A].
1.3 Subspaces and Convex Sets
1.3.1 Denition
A subset M of a linear space X is called a linear subspace of X if
(a) x y M for all x. y M, and
(b) zx M for all x M and for all z F.
Clearly, a subset M of a linear space X is a linear subspace if and only if MM M and zM M
for all z F.
1.3.2 Examples
[1] Every linear space X has at least two distinguished subspaces: M = {0] and M = X.
These are called the improper subspaces of X. All other subspaces of X are called the
proper subspaces.
[2] Let X = R
2
. Then the nontrivial linear subspaces of X are straight lines through the origin.
[3] M = {x = (0. x
2
. x
3
. . . . . x
n
) : x
i
R. i = 2. 3. . . . . n] is a subspace of R
n
.
[4] M = {x : 1. 1| R. x continuous and x(0) = 0] is a subspace of C1. 1|.
[5] M = {x : 1. 1| R. x continuous and x(0) = 1 ] is not a subspace of C1. 1|.
[6] Show that c
0
is a subspace of c.
1.3.3 Denition
Let K be a subset of a linear space X. The linear hull of K, denoted by lin(K) or span(K), is the
intersection of all linear subspaces of X that contain K.
The linear hull of K is also called the linear subspace of X spanned (or generated) by K.
It is easy to check that the intersection of a collection of linear subspaces of X is a linear subspace of
X. It therefore follows that the linear hull of a subset K of a linear space X is again a linear subspace of X.
In fact, the linear hull of a subset K of a linear space X is the smallest linear subspace of X which contains
K.
1.3.4 Proposition
Let K be a subset of a linear space X. Then the linear hull of K is the set of all nite linear combinations
of elements of K. That is,
lin(K) =
_
_
_
n

jD1
z
j
x
j
[ x
1
. x
2
. . . . . x
n
K. z
1
. z
2
. . . . . z
n
F. n N
_
_
_
.
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2011 FUNCTIONAL ANALYSIS ALP
Proof. Exercise.
1.3.5 Denition
[1] A subset {x
1
. x
2
. . . . . x
n
] of a linear space X is said to be linearly independent if the equation

1
x
1

2
x
2

n
x
n
= 0
only has the trivial solution
1
=
2
= =
n
= 0. Otherwise, the set {x
1
. x
2
. . . . . x
n
] is
linearly dependent.
[2] Asubset K of a linear space X is said to be linearly independent if every nite subset {x
1
. x
2
. . . . . x
n
]
of K is linearly independent.
1.3.6 Denition
If {x
1
. x
2
. . . . . x
n
] is a linearly independent subset of X and
X = lin{x
1
. x
2
. . . . . x
n
], then X is said to have dimension n. In this case we say that {x
1
. x
2
. . . . . x
n
]
is a basis for the linear space X. If a linear space X does not have a nite basis, we say that it is innite-
dimensional.
1.3.7 Examples
[1] The space R
n
has dimension n. Its standard basis is {e
1
. e
2
. . . . . e
n
], where, for each
j = 1. 2. . . . . n, e
j
is an n-tuple of real numbers with 1 in the j-th position and zeroes
elsewhere; i.e.,
e
j
= (0. 0. . . . . 1. 0. . . . . 0). where 1 occurs in the j-th position.
[2] The space P
n
of polynomials of degree at most n has dimension n 1. Its standard basis is
{1. t. t
2
. . . . . t
n
].
[3] The function space Ca. b| is innite-dimensional.
[4] The spaces
p
, with 1 _ p _ o, are innite-dimensional.
1.3.8 Denition
Let K be a subset of a linear space X. We say that
(a) K is convex if zx (1 z)y K whenever x. y K and z 0. 1|;
(b) K is balanced if zx K whenever x K and [z[ _ 1;
(c) K is absolutely convex if K is convex and balanced.
1.3.9 Remark
[1] It is easy to verify that K is absolutely convex if and only if zx jy K whenever x. y K
and [z[ [j[ _ 1.
[2] Every linear subspace is absolutely convex.
1.3.10 Denition
Let S be a subset of the linear space X. The convex hull of S, denoted co(S), is the intersection
of all convex sets in X which contain S.
Since the intersection of convex sets is convex, it follows that co(S) is the smallest convex set which
contains S. The following result is an alternate characterization of co(S).
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2011 FUNCTIONAL ANALYSIS ALP
1.3.11 Proposition
Let S be a nonempty subset of a linear space X. Then co(S) is the set of all convex combinations of
elements of S. That is,
co(S) =
_
_
_
n

jD1
z
j
x
j
[ x
1
. x
2
. . . . . x
n
S. z
j
_ 0 V j = 1. 2. . . . . n.
n

jD1
z
j
= 1. n N
_
_
_
.
Proof. Let C denote the set of all convex combinations of elements of S. That is,
C =
_
_
_
n

jD1
z
j
x
j
[ x
1
. x
2
. . . . . x
n
S. z
j
_ 0 V j = 1. 2. . . . . n.
n

jD1
z
j
= 1. n N
_
_
_
.
Let x. y C and 0 _ z _ 1. Then x =
n

1
z
i
x
i
. y =
m

1
j
i
y
i
, where z
i
. j
i
_ 0,
n

1
z
i
= 1,
m

1
j
i
= 1, and x
i
. y
i
S. Thus
zx (1 z)y =
n

1
zz
i
x
i

m

1
(1 z)j
i
y
i
is a linear combination of elements of S, with nonnegative coefcients, such that
n

1
zz
i

m

1
(1 z)j
i
= z
n

1
z
i
(1 z)
m

1
j
i
= z (1 z) = 1.
That is, zx (1 z)y C and C is convex. Clearly S C. Hence co(S) C.
We now prove the inclusion C co(S). Note that, by denition, S co(S). Let x
1
. x
2
S,
z
1
_ 0. z
2
_ 0 and z
1
z
2
= 1. Then, by convexity of co(S), z
1
x
1
z
2
x
2
co(S). Assume that
n1

1
z
i
x
i
co(S) whenever x
1
. x
2
. . . . . x
n1
S, z
j
_ 0, j = 1. 2. . . . . n 1 and
n1

jD1
z
j
= 1. Let
x
1
. x
2
. . . . . x
n
S and z
1
. z
2
. . . . . z
n
be such that z
j
_ 0, j = 1. 2. . . . . n and
n

jD1
z
j
= 1. If
n1

jD1
z
j
= 0, then z
n
= 1. Hence
n

1
z
j
x
j
= z
n
x
n
co(S). Assume that =
n1

jD1
z
j
> 0. Then
z
j

_ 0
for all j = 1. 2. . . . . n 1 and
n1

jD1
z
j

= 1. By the induction assumption,

n1

jD1
z
j

x
j
co(S). Hence
n

jD1
z
j
x
j
=
_
_
n1

jD1
z
j

x
j
_
_
z
n
x
n
co(S).
Thus C co(S).
1.4 Quotient Space
Let M be a linear subspace of a linear space X over F. For all x. y X, dene
x y(mod M) x y M.
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2011 FUNCTIONAL ANALYSIS ALP
It is easy to verify that denes an equivalence relation on X.
For x X, denote by
x| := {y X : x y(mod M)] = {y X : x y M] = x M.
the coset of x with respect to M. The quotient space X,M consists of all the equivalence classes x|,
x X. The quotient space is also called a factor space.
1.4.1 Proposition
Let M be a linear subspace of a linear space X over F. For x. y X and z F, dene the operations
x| y| = x y| and z x| = z x|.
Then X,M is a linear space with respect to these operations.
Proof. Exercise.
Note that the linear operations on X,M are equivalently given by: For all x. y X and z F,
(x M) (y M) = x y M and z(x M) = zx M.
1.4.2 Denition
Let M be a linear subspace of a linear space X over F. The codimension of M in X is dened as the
dimension of the quotient space X,M. It is denoted by codim(M) = dim(X,M).
Clearly, if X = M, then X,M = {0] and so codim(X) = 0.
1.5 Direct Sums and Projections
1.5.1 Denition
Let M and N be linear subspaces of a linear space X over F. We say that X is a direct sum of M and N
if
X = M N and M N = {0].
If X is a direct sum of M and N, we write X = M N. In this case, we say that M (resp. N) is an
algebraic complement of N (resp. M).
1.5.2 Proposition
Let M and N be linear subspaces of a linear space X over F. If X = M N, then each x X has a
unique representation of the form x = mn for some m M and n N.
Proof. Exercise.
Let M and N be linear subspaces of a linear space X over F such that X = M N. Then
codim(M) =dim(N). Also, since X = M N, dim(X) =dim(M)dim(N). Hence
dim(X) = dim(M) codim(M).
It follows that if dim(X) < o, then codim(M) =dim(X)dim(M).
The operator P : X X is called an algebraic projection if P is linear (i.e., P(xy) = PxPy
for all x. y X and F) and P
2
= P, i.e., P is idempotent.
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2011 FUNCTIONAL ANALYSIS ALP
1.5.3 Proposition
Let M and N be linear subspaces of a linear space X over F such that X = M N. Dene P : X X
by P(x) = m, where x = mn, with m M and n N. Then P is an algebraic projection of X onto
M along N. Moreover M = P(X) and N = (I P)(X) = ker(P).
Conversely, if P : X X is an algebraic projection, then X = M N, where M = P(X) and
N = (I P)(X) = ker(P).
Proof. Linearity of P: Let x = m
1
n
1
and y = m
2
n
2
, where m
1
. m
2
M and n
1
. n
2
N. For
F,
P(x y) = P((m
1
m
2
) (n
1
n
2
)) = m
1
m
2
= Px Py.
Idempotency of P: Since m = m 0, with m M and 0 N, we have that Pm = m and hence
P
2
x = Pm = m = Px. That is, P
2
= P.
Finally, n = x m = (I P)x. Hence N = (I P)(X). Also, Px = 0 if and only if x N, i.e.,
ker(P) = N.
Conversely, let x X and set m = Px and n = (I P)x. Then x = m n, where m M and
n N. We show that this representation is unique. Indeed, if x = m
1
n
1
where m
1
M and n
1
N,
then m
1
= Pu and n
1
= (I P): for some u. : X. Since P
2
= P, it follows that Pm
1
= m
1
and
Pn
1
= 0. Hence m = Px = Pm
1
Pn
1
= Pm
1
= m
1
. Similarly n = n
1
.
1.6 The H older and Minkowski Inequalities
We now turn our attention to three important inequalities. The rst two are required mainly to prove the
third which is required for our discussion about normed linear spaces in the subsequent chapter.
1.6.1 Denition
Let p and q be positive real numbers. If 1 < p < oand
1
p

1
q
= 1, or if p = 1 and q = o, or if p = o
and q = 1, then we say that p and q are conjugate exponents.
1.6.2 Lemma
(Youngs Inequality). Let p and q be conjugate exponents, with 1 < p. q < o and . _ 0. Then
_

p
p

q
q
.
Proof. If p = 2 = q, then the inequality follows from the fact that ( )
2
_ 0. Notice also, that if = 0
or = 0, then the inequality follows trivially.
If p ,= 2, then consider the function : 0. o) R given by
() =

p
p


q
q
. for xed > 0.
Then,
0
() =
p1
= 0 when
p1
= . That is, when =
1
p1
=
q
p
> 0. We now apply
the second derivative test to the critical point =
q
p
.

00
() = (p 1)
p2
> 0. for all (0. o).
9
2011 FUNCTIONAL ANALYSIS ALP
Thus, we have a global minimum at =
q
p
. It is easily veried that
0 = (
q
p
) _ () =

p
p


q
q
= _

p
p


q
q
.
for each 0. o).

1.6.3 Theorem
(H olders Inequality for sequences). Let (x
n
)
p
and (y
n
)
q
, where p > 1 and 1,p 1,q = 1.
Then
1

kD1
[x
k
y
k
[ _
_
1

kD1
[x
k
[
p
_1
p
_
1

kD1
[y
k
[
q
_1
q
.
Proof. If
1

kD1
[x
k
[
p
= 0 or
1

kD1
[y
k
[
q
= 0, then the inequality holds. Assume that
1

kD1
[x
k
[
p
,= 0 and
1

kD1
[y
k
[
q
,= 0. Then for k = 1. 2. . . ., we have, by Lemma 1.6.2, that
[x
k
[
_
1
kD1
[x
k
[
p
_ 1
p

[y
k
[
_
1
kD1
[y
k
[
q
_ 1
q
_
1
p
[x
k
[
p

1
kD1
[x
k
[
p

1
q
[y
k
[
q

1
kD1
[y
k
[
q
.
Hence,

1
kD1
[x
k
y
k
[
_
1
kD1
[x
k
[
p
_ 1
p
_
1
kD1
[y
k
[
q
_ 1
q
_
1
p

1
q
= 1.
That is,
1

kD1
[x
k
y
k
[ _
_
1

kD1
[x
k
[
p
_1
p
_
1

kD1
[y
k
[
q
_1
q
.
1.6.4 Theorem
(Minkowskis Inequality for sequences). Let p > 1 and (x
n
) and (y
n
) sequences in
p
. Then
_
1

kD1
[x
k
y
k
[
p
_1
p
_
_
1

kD1
[x
k
[
p
_1
p

_
1

kD1
[y
k
[
p
_1
p
.
10
2011 FUNCTIONAL ANALYSIS ALP
Proof. Let q =
p
p 1
. If
1

kD1
[x
k
y
k
[
p
= 0, then the inequality holds. We therefore assume that
1

kD1
[x
k
y
k
[
p
,= 0. Then
1

kD1
[x
k
y
k
[
p
=
1

kD1
[x
k
y
k
[
p1
[x
k
y
k
[
_
1

kD1
[x
k
y
k
[
p1
[x
k
[
1

kD1
[x
k
y
k
[
p1
[y
k
[
_
_
1

kD1
[x
k
y
k
[
(p1)q
_1
q
_
_
_
1

kD1
[x
k
[
p
_1
p

_
1

kD1
[y
k
[
p
_1
p
_
_
=
_
1

kD1
[x
k
y
k
[
p
_1
q
_
_
_
1

kD1
[x
k
[
p
_1
p

_
1

kD1
[y
k
[
p
_1
p
_
_
.
Dividing both sides by
_
1

kD1
[x
k
y
k
[
p
_1
q
, we have
_
1

kD1
[x
k
y
k
[
p
_1
p
=
_
1

kD1
[x
k
y
k
[
p
_
1
1
q
_
_
1

kD1
[x
k
[
p
_1
p

_
1

kD1
[y
k
[
p
_1
p
.
1.6.5 Exercise
[1] Show that the set of all n m real matrices is a real linear space.
[2] Show that a subset M of a linear space X is a linear subspace if and only if x y M
for all x. y M and all . F.
[3] Prove Proposition 1.3.4
[4] Prove Proposition 1.4.1.
[5] Prove Proposition 1.5.2.
[6] Show that c
0
is a linear subspace of the linear space
1
.
[7] Which of the following subsets are linear subspaces of the linear space C1. 1|?
(a) M
1
= {x C1. 1| : x(1) = x(1)].
(b) M
2
= {x C1. 1| :
1
_
1
x(t )dt = 1].
(c) M
3
= {x C1. 1| : [x(t
2
) x(t
1
)[ _ [t
2
t
1
[ for all t
1
. t
2
1. 1|].
11
2011 FUNCTIONAL ANALYSIS ALP
[8] Show that if {M
z
] is a family of linear subspaces of a linear space X, then M =
z
M
z
is a
linear subspace of X.
If M and N are linear subspaces of a linear space X, under what condition(s) is M L N a
linear subspace of X?
12
Chapter 2
Normed Linear Spaces
2.1 Preliminaries
For us to have a meaningful notion of convergence it is necessary for the Linear space to have a notion
distance and therefore a topology dened on it. This leads us to the denition of a norm which induces a
metric topology in a natural way.
2.1.1 Denition
A norm on a linear space X is a real-valued function [ [ : X R which satises the following properties:
For all x. y X and z F,
N1. [x[ _ 0;
N2. [x[ = 0 x = 0;
N3. [zx[ = [z[[x[;
N4. [x y[ _ [x[ [y[ (Triangle Inequality).
A normed linear space is a pair (X. [ [), where X is a linear space and [ [ a norm on X. The number
[x[ is called the norm or length of x.
Unless there is some danger of confusion, we shall identify the normed linear space (X. [ [) with the
underlying linear space X.
2.1.2 Examples
(Examples of normed linear spaces.)
[1] Let X = F. For each x X, dene [x[ = [x[. Then (X. [ [) is a normed linear space.
We give the proof for X = C. Properties N1 -N3 are easy to verify. We only verify N4. Let
x. y C. Then
[x y[
2
= [x y[
2
= (x y)(x y) = (x y)(x y) = xx yx xy yy
= [x[
2
xy xy [y[
2
= [x[
2
2Re(xy) [y[
2
_ [x[
2
2[xy[ [y[
2
= [x[
2
2[x[[y[ [y[
2
= [x[
2
2[x[[y[ [y[
2
= ([x[ [y[)
2
= ([x[ [y[)
2
.
Taking the positive square root both sides yields N4.
13
2011 FUNCTIONAL ANALYSIS ALP
[2] Let n be a natural number and X = F
n
. For each x = (x
1
. x
2
. . . . . x
n
) X, dene
[x[
p
=
_
n

iD1
[x
i
[
p
_1
p
. for 1 _ p < o. and
[x[
1
= max
1in
[x
i
[.
Then (X. [ [
p
) and (X. [ [
1
) are normed linear spaces. We give a detailed proof that
(X. [ [
p
) is a normed linear space for 1 _ p < o.
N1. For each 1 _ i _ n,
[x
i
[ _ 0 =
n

iD1
[x
i
[
p
_ 0 =
_
n

iD1
[x
i
[
p
_1
p
_ 0 = [x[
p
_ 0.
N2. For any x X,
[x[
p
= 0
_
n

iD1
[x
i
[
p
_1
p
= 0
[x
i
[
p
= 0 for all i = 1. 2. 3. . . . . n
x
i
= 0 for all i = 1. 2. 3. . . . . n x = 0.
N3. For any x X and any z F,
[zx[
p
=
_
n

iD1
[zx
i
[
p
_1
p
=
_
[z[
p
n

iD1
[x
i
[
p
_1
p
= [z[
_
n

iD1
[x
i
[
p
_1
p
= [z[[x[
p
.
N4. For any x. y X,
[x y[
p
=
_
n

iD1
[x
i
y
i
[
p
_1
p
_
_
n

iD1
[x
i
[
p
_1
p

_
n

iD1
[y
i
[
p
_1
p
(by Minkowski
0
s Inequality)
= [x[
p
[y[
p
.
[3] Let X = Ba. b| be the set of all bounded real-valued functions on a. b|. For each x X,
dene
[x[
1
= sup
at b
[x(t )[.
Then (X. [ [
1
) is a normed linear space. We prove the triangle inequality: For any t a. b|
and any x. y X,
[x(t ) y(t )[ _ [x(t )[ [y(t )[ _ sup
at b
[x(t )[ sup
at b
[y(t )[ = [x[
1
[y[
1
.
14
2011 FUNCTIONAL ANALYSIS ALP
Since this is true for all t a. b|, we have that
[x y[
1
= sup
at b
[x(t ) y(t )[ _ [x[
1
[y[
1
.
[4] Let X = Ca. b|. For each x X, dene
[x[
1
= sup
at b
[x(t )[
[x[
2
=
_
_
b
_
a
[x(t )[
2
dt
_
_
1
2
.
Then (X. [ [
1
) and (X. [ [
2
) are normed linear spaces.
[5] Let X =
p
. 1 _ p < o. For each x = (x
i
)
1
1
X, dene
[x[
p
=
_

i2N
[x
i
[
p
_1
p
.
Then (X. [ [
p
) is a normed linear space.
[6] Let X =
1
. c or c
0
. For each x = (x
i
)
1
1
X, dene
[x[ = [x[
1
= sup
i2N
[x
i
[.
Then X is a normed linear space.
[7] Let X = L(C
n
) be the linear space of all n n complex matrices. For A L(C
n
), let
t(A) =
n

iD1
(A)
ii
be the trace of A. For A L(C
n
), dene
[A[
2
=
_
t(A

A) =

_
n

iD1
n

kD1
(A)
ki
(A)
ki
=

_
n

iD1
n

kD1
[(A)
ki
[
2
.
where A

is the conjugate transpose of the matrix A.

Notation
Let a be an element of a normed linear space(X. [ [) and r > 0.
B(a. r ) = {x X [ [x a[ < r ] (Open ball with centre a and radius r ):
Ba. r | = {x X [ [x a[ _ r ] (Closed ball with centre a and radius r ):
S(a. r ) = {x X [ [x a[ = r ] (Sphere with centre a and radius r ).
15
2011 FUNCTIONAL ANALYSIS ALP
1
1
y
x
1 1
[(x. y)[
1
= 1
1
1
y
x
1 1
[(x. y)[
2
< 1
1
1
y
x
1 1
[(x. y)[
1
_ 1
Equivalent Norms
2.1.3 Denition
Let [ [ and [ [
0
be two different norms dened on the same linear space X. We say that [ [ is equivalent
to [ [
0
if there are positive numbers and such that
[x[ _ [x[
0
_ [x[. for all x X.
2.1.4 Example
Let X = F
n
. For each x = (x
1
. x
2
. . . . . x
n
) X, let
[x[
1
=
n

iD1
[x
i
[. [x[
2
=
_
n

iD1
[x
i
[
2
_1
2
. and [x[
1
= max
1in
[x
i
[.
We have seen that [ [
1
. [ [
2
and [ [
1
are norms on X. We show that these norms are
equivalent.
Equivalence of [ [
1
and [ [
1
: Let x = (x
1
. x
2
. . . . . x
n
) X. For each k = 1. 2. . . . . n,
[x
k
[ _
n

iD1
[x
i
[ = max
1kn
[x
k
[ _
n

iD1
[x
i
[ [x[
1
_ [x[
1
.
Also, for k = 1. 2. . . . . n,
[x
k
[ _ max
1kn
[x
k
[ = [x[
1
=
n

iD1
[x
i
[ _
n

iD1
[x[
1
= n[x[
1
[x[
1
_ n[x[
1
.
Hence, [x[
1
_ [x[
1
_ n[x[
1
.
We now show that [ [
2
is equivalent to [ [
1
. Let x = (x
1
. x
2
. . . . . x
n
) X. For each
k = 1. 2. . . . . n,
[x
k
[ _ [x[
1
= [x
k
[
2
_ ([x[
1
)
2
=
n

iD1
[x
i
[
2
_
n

iD1
([x[
1
)
2
= n([x[
1
)
2
[x[
2
_
_
n[x[
1
.
16
2011 FUNCTIONAL ANALYSIS ALP
Also, for each k = 1. 2. . . . . n,
[x
k
[ _
_
n

iD1
[x
i
[
2
_
1{2
= [x[
2
= max
1kn
[x
k
[ _ [x[
2
[x[
1
_ [x[
2
.
Consequently, [x[
1
_ [x[
2
_
_
n[x[
1
, which proves equivalence of the norms [ [
2
and [ [
1
.
It is, of course, obvious now that all the three norms are equivalent to each other. We shall
see later that all norms on a nite-dimensional normed linear space are equivalent.
2.1.5 Exercise
Let N(X) denote the set of norms on a linear space X. For [ [ and [ [
0
in N(X), dene a relation .by
[ [ . [ [
0
if and only if [ [ is equivalent to [ [
0
.
Show that .is an equivalence relation on N(X), i.e., .is reexive, symmetric, and transitive.
Open and Closed Sets
2.1.6 Denition
A subset S of a normed linear space (X. [ [) is open if for each s S there is an c > 0 such that
B(s. c) S.
A subset F of a normed linear space (X. [ [) is closed if its complement X \ F is open.
2.1.7 Denition
Let S be a subset of a normed linear space (X. [ [). We dene the closure of S, denoted by S, to be the
intersection of all closed sets containing S.
It is easy to show that S is closed if and only if S = S.
Recall that a metric on a set X is a real-valued function d : X X R which satises the following
properties: For all x. y. z X,
M1. d(x. y) _ 0;
M2. d(x. y) = 0 x = y;
M3. d(x. y) = d(y. x);
M4. d(x. z) _ d(x. y) d(y. z).
2.1.1 Theorem
(a) If (X. [ [) is a normed linear space, then
d(x. y) = [x y[
denes a metric on X. Such a metric d is said to be induced or generated by the norm [ [. Thus,
every normed linear space is a metric space, and unless otherwise specied, we shall henceforth
regard any normed linear space as a metric space with respect to the metric induced by its norm.
(b) If d is a metric on a linear space X satisfying the properties: For all x. y. z X and for all z F,
(i) d(x. y) = d(x z. y z) (Translation Invariance)
(ii) d(zx. zy) = [z[d(x. y) (Absolute Homogeneity).
then
[x[ = d(x. 0)
denes a norm on X.
17
2011 FUNCTIONAL ANALYSIS ALP
Proof. (a) We show that d(x. y) = [x y[ denes a metric on X. To that end, let x. y. z X.
M1. d(x. y) = [x y[ _ 0 by N1.
M2.
d(x. y) = 0 [x y[ = 0 x y = 0 by N2
x = y.
M3.
d(x. y) = [x y[ = [(1)(y x)[ = [ 1[[y x[ by N3
= [y x[ = d(y. x).
M4.
d(x. z) = [x z[ = [(x y) (y z)[ _ [x y[ [y z[ by N4
= d(x. y) d(y. z).
(b) Exercise.
It is clear from Theorem 2.1.1, that a metric d on a linear space X is induced by a norm on X if and
only if d is translation-invariant and positive homogeneous.
2.2 Quotient Norm and Quotient Map
We now want to introduce a norm on a quotient space. Let M be a closed linear subspace of a normed
linear space X over F. For x X, dene
[x|[ := inf
y2xj
[y[.
If y x|, then y x M and hence y = x m for some m M. Hence
[x|[ = inf
y2xj
[y[ = inf
m2M
[x m[ = inf
m2M
[x m[ = d(x. M).
2.2.1 Proposition
Let M be a closed linear subspace of a normed linear space X over F. The quotient space X,M is a
normed linear space with respect to the norm
[x|[ := inf
y2xj
[y[. where x| X,M.
Proof.
N1. It is clear that for any x X, [x|[ = d(x. M) _ 0.
N2. For any x X,
[x|[ = 0 d(x. M) = 0 x M = M x M = M = 0|.
N3. For any x. y X and z F \ {0],
[zx|[ = [zx|[ = d(zx. M) = inf
y2M
[zx y[ = inf
y2M
_
_
_z
_
x
y
z
__
_
_
= [z[ inf
z2M
[x z[ = [z[d(x. M) = [z[[x|[.
18
2011 FUNCTIONAL ANALYSIS ALP
N4. Let x. y X. Then
[x| y|[ = [x y|[ = d(x y. M) = inf
z2M
[x y z[
= inf
z
1
,z
2
2M
[x y (z
1
z
2
)[
= inf
z
1
,z
2
2M
[(x z
1
) (y z
2
)[
_ inf
z
1
,z
2
2M
[x z
1
[ [y z
2
[
= inf
z
1
2M
[x z
1
[ inf
z
2
2M
[y z
2
[
= d(x. M) d(y. M) = [x|[ [y|[.
The norm on X,M as dened in Proposition 2.2.1 is called the quotient norm on X,M.
Let M be a closed subspace of the normed linear space X. The mapping Q
M
fromX X,M dened
by
Q
M
(x) = x M. x X.
is called the quotient map (or natural embedding) of X onto X,M.
2.3 Completeness of Normed Linear Spaces
Now that we have established that every normed linear space is a metric space, we can deploy on a normed
linear space all the machinery that exists for metric spaces.
2.3.1 Denition
Let (x
n
)
1
nD1
be a sequence in a normed linear space (X. [ [).
(a) (x
n
)
1
nD1
is said to converge to x if given c > 0 there exists a natural number N = N(c) such that
[x
n
x[ < c for all n _ N.
Equivalently, (x
n
)
1
nD1
converges to x if
lim
n!1
[x
n
x[ = 0.
If this is the case, we shall write
x
n
x or lim
n!1
x
n
= x.
Convergence in the norm is called norm convergence or strong convergence.
(b) (x
n
)
1
nD1
is called a Cauchy sequence if given c > 0 there exists a natural number N = N(c) such
that
[x
n
x
m
[ < c for all n. m _ N.
Equivalently, (x
n
) is Cauchy if
lim
n,m!1
[x
n
x
m
[ = 0.
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2011 FUNCTIONAL ANALYSIS ALP
In the following lemma we collect some elementary but fundamental facts about normed linear spaces.
In particular, it implies that the operations of addition and scalar multiplication, as well as the norm and
distance functions, are continuous.
2.3.2 Lemma
Let C be a closed set in a normed linear space (X. [ [) over F, and let (x
n
) be a sequence contained in C
such that lim
n!1
x
n
= x X. Then x C.
Proof. Exercise.
2.3.3 Lemma
Let X be a normed linear space and A a nonempty subset of X.
[1] [d(x. A) d(y. A)[ _ [x y[ for all x. y X;
[2] [ [x[ [y[ [ _ [x y[ for all x. y X;
[3] If x
n
x, then [x
n
[ [x[;
[4] If x
n
x and y
n
y, then x
n
y
n
x y;
[5] If x
n
x and
n
, then
n
x
n
x;
[6] The closure of a linear subspace in X is again a linear subspace;
[7] Every Cauchy sequence is bounded;
[8] Every convergent sequence is a Cauchy sequence.
Proof. (1). For any a A,
d(x. A) _ [x a[ _ [x y[ [y a[.
so d(x. A) _ [x y[ d(y. A) or d(x. A) d(y. A) _ [x y[. Interchanging the roles of x and y gives
the desired result.
(2) follows from (1) by taking A = {0].
(3) is an obvious consequence of (2).
(4), (5) and (8) follow from the triangle inequality and, in the case of (5), the absolute homogeneity.
(6) follows from (4) and (5).
(7). Let (x
n
) be a Cauchy sequence in X. Choose n
1
so that [x
n
x
n
1
[ _ 1 for all n _ n
1
. By (2),
[x
n
[ _ 1 [x
n
1
[ for all n _ n
1
. Thus
[x
n
[ _ max{ [x
1
[. [x
2
[. [x
3
[. . . . . [x
n
1
1
[. 1 [x
n
1
[]
for all n.
(8) Let (x
n
) be a sequence in X which converges to x X and let c > 0. Then there is a natural
number N such that [x
n
x[ <
e
2
for all n _ N. For all n. m _ N,
[x
n
x
m
[ _ [x
n
x[ [x x
m
[ <
c
2

c
2
= c.
Thus, (x
n
) is a Cauchy sequence in X.
2.3.4 Proposition
Let (X. [[) be a normed linear space over F. ACauchy sequence in X which has a convergent subsequence
is convergent.
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2011 FUNCTIONAL ANALYSIS ALP
Proof. Let (x
n
) be a Cauchy sequence in X and (x
n
k
) its subsequence which converges to x X. Then,
for any c > 0, there are positive integers N
1
and N
2
such that
[x
n
x
m
[ <
c
2
for all n. m _ N
1
and
[x
n
k
x[ <
c
2
for all k _ N
2
.
Let N = max{N
1
. N
2
]. If k _ N, then since n
k
_ k,
[x
k
x[ _ [x
k
x
n
k
[ [x
n
k
x[ <
c
2

c
2
= c.
Hence x
n
x as n o.
2.3.5 Denition
A metric space (X. d) is said to be complete if every Cauchy sequence in X converges in X.
2.3.6 Denition
A normed linear space that is complete with respect to the metric induced by the norm is called a Banach
space.
2.3.1 Theorem
Let (X. [ [) be a Banach space and let M be a linear subspace of X. Then M is complete if and only if
the M is closed in X.
Proof. Assume that M is complete. We show that M is closed. To that end, let x M. Then there
is a sequence (x
n
) in M such that [x
n
x[ 0 as n o. Since (x
n
) converges, it is Cauchy.
Completeness of M guarantees the existence of an element y M such that [x
n
y[ 0 as n o.
By uniqueness of limits, x = y. Hence x M and, consequently, M is closed.
Assume that M is closed. We show that M is complete. Let (x
n
) be a Cauchy sequence in M. Then
(x
n
) is a Cauchy sequence in X. Since X is complete, there is an element x X such that [x
n
x[ 0
as n o. But then x M since M is closed. Hence M is complete.
2.3.7 Examples
[1] Let 1 _ p < o. Then for each positive integer n, (F
n
. [ [
p
) is a Banach space.
[2] For each positive integer n, (F
n
. [ [
1
) is a Banach space.
[3] Let 1 _ p < o. The sequence space
p
is a Banach space. Because of the importance of
this space, we give a detailed proof of its completeness.
The classical sequence space
p
is complete.
Proof. Let (x
n
)
1
1
be a Cauchy sequence in
p
. We shall denote each member of this
sequence by
x
n
= (x
n
(1). x
n
(2). . . .).
Then, given c > 0, there exists an N(c) = N N such that
[x
n
x
m
[
p
=
_
1

iD1
[x
n
(i ) x
m
(i )[
p
_1
p
< c for all n. m _ N.
For each xed index i , we have
[x
n
(i ) x
m
(i )[ < c for all n. m _ N.
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2011 FUNCTIONAL ANALYSIS ALP
That is, for each xed index i , (x
n
(i ))
1
1
is a Cauchy sequence in F. Since F is complete,
there exists x(i ) F such that
x
n
(i ) x(i ) as n o.
Dene x = (x(1). x(2). . . .). We show that x
p
, and x
n
x. To that end, for each k N,
_
k

iD1
[x
n
(i ) x
m
(i )[
p
_
1
p
_ [x
n
x
m
[
p
=
_
1

iD1
[x
n
(i ) x
m
(i )[
p
_1
p
< c.
That is,
k

iD1
[x
n
(i ) x
m
(i )[
p
< c
p
. for all k = 1. 2. 3. . . . .
Keep k and n _ N xed and let m o. Since we are dealing with a nite sum,
k

iD1
[x
n
(i ) x(i )[
p
_ c
p
.
Now letting k o, then for all n _ N,
1

iD1
[x
n
(i ) x(i )[
p
_ c
p
. (2.3.7.1)
which means that x
n
x
p
. Since x
n

p
, we have that x = (x x
n
) x
n

p
. It also
follows from (2.3.7.1) that x
n
x as n o.
[4] The space
0
of all sequences (x
i
)
1
1
with only a nite number of nonzero terms is an in-
complete normed linear space. It sufces to show that
0
is not closed in
2
(and hence not
complete). To that end, consider the sequence (x
i
)
1
1
with terms
x
1
= (1. 0. 0. 0. . . .)
x
2
= (1.
1
2
. 0. 0. 0. . . .)
x
3
= (1.
1
2
.
1
2
2
. 0. 0. 0. . . .)
.
.
.
x
n
= (1.
1
2
.
1
2
2
. . . . .
1
2
n1
. 0. 0. 0. . . .)
.
.
.
This sequence (x
i
)
1
1
converges to
x = (1.
1
2
.
1
2
2
. . . . .
1
2
n1
.
1
2
n
.
1
2
nC1
. . . .).
Indeed, since x x
n
= (0. 0. 0. . . . . 0.
1
2
n
.
1
2
nC1
. . . .), it follows that
[x
n
x[
2
=
1

kDn
1
2
2k
0 as n o.
That is, x
n
x as n o, but x ,
0
.
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2011 FUNCTIONAL ANALYSIS ALP
[5] The space C
2
1. 1| of continuous real-valued functions on 1. 1| with the norm
[x[
2
=
_
_
1
_
1
x
2
(t ) dt
_
_
1{2
is an incomplete normed linear space.
To see this, it sufces to show that there is a Cauchy sequence in C
2
1. 1| which converges
to an element which does not belong to C
2
1. 1|. Consider the sequence (x
n
)
1
1
C
2
1. 1|
dened by
x
n
(t ) =
_

_
0 if 1 _ t _ 0
nt if 0 _ t _
1
n
1 if
1
n
_ t _ 1.
1
y
t 0
1
x
n
(t )
1
n
1
We show that (x
n
)
1
1
is a Cauchy sequence in C
2
1. 1|. To that end, for positive integers m
and n such that m > n,
[x
n
x
m
[
2
2
=
1
_
1
x
n
(t ) x
m
(t )|
2
dt
=
1{m
_
0
nt mt |
2
dt
1{n
_
1{m
1 nt |
2
dt
=
1{m
_
0
m
2
t
2
2mnt
2
n
2
t
2
| dt
1{n
_
1{m
1 2nt n
2
t
2
| dt
= (m
2
2mn n
2
)
t
3
3

1{m
0

_
t nt
2
n
2
t
3
3
_

1{n
1{m
=
m
2
2mn n
2
3m
2
n
=
(m n)
2
3m
2
n
0 as n. mo.
Dene
x(t ) =
_
0 if 1 _ t _ 0
1 if 0 < t _ 1.
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2011 FUNCTIONAL ANALYSIS ALP
Then x , C
2
1. 1|, and
[x
n
x[
2
2
=
1
_
1
x
n
(t ) x(t )|
2
dt =
1
n _
0
nt 1|
2
dt =
1
3n
0 as n o.
That is, x
n
x as n o.
2.4 Series in Normed Linear Spaces
Let (x
n
) be a sequence in a normed linear space (X. [ [). To this sequence we associate another sequence
(s
n
) of partial sums, where s
n
=
n

kD1
x
k
.
2.4.1 Denition
Let (x
n
) be a sequence in a normed linear space (X. [ [). If the sequence (s
n
) of partial sums converges to
s, then we say that the series
1

kD1
x
k
converges and that its sum is s. In this case we write
1

kD1
x
k
= s.
The series
1

kD1
x
k
is said to be absolutely convergent if
1

kD1
[x
k
[ < o.
We now give a series characterization of completeness in normed linear spaces.
2.4.1 Theorem
A normed linear space (X. [ [) is a Banach space if and only if every absolutely convergent series in X is
convergent.
Proof. Let X be a Banach space and suppose that
1

jD1
[x
j
[ < o. We showthat the series
1

jD1
x
j
converges.
To that end, let c > 0 and for each n N, let s
n
=
n

jD1
x
j
. Let K be a positive integer such that
1

jDKC1
[x
j
[ < c. Then, for all m > n > K, we have
[s
m
s
n
[ =
_
_
_
_
_
m

1
x
j

n

1
x
j
_
_
_
_
_
=
_
_
_
_
_
m

nC1
x
j
_
_
_
_
_
_
m

nC1
[x
j
[ _
1

nC1
[x
j
[ _
1

KC1
[x
j
[ < c.
Hence the sequence (s
n
) of partial sums forms a Cauchy sequence in X. Since X is complete, the sequence
(s
n
) converges to some element s X. That is, the series
1

jD1
x
j
converges.
Conversely, assume that (X. [ [) is a normed linear space in which every absolutely convergent series
converges. We show that X is complete. Let (x
n
) be a Cauchy sequence in X. Then there is an n
1
N
such that [x
n
1
x
m
[ <
1
2
whenever m > n
1
. Similarly, there is an n
2
N with n
2
> n
1
such that
[x
n
2
x
m
[ <
1
2
2
whenever m > n
2
. Continuing in this way, we get natural numbers n
1
< n
2
< such
24
2011 FUNCTIONAL ANALYSIS ALP
that [x
n
k
x
m
[ <
1
2
k
whenever m > n
k
. In particular, we have that for each k N, [x
n
kC1
x
n
k
[ < 2
k
.
For each k N, let y
k
= x
n
kC1
x
n
k
. Then
n

kD1
[y
k
[ =
n

kD1
[x
n
kC1
x
n
k
[ <
n

kD1
1
2
k
.
Hence,
1

kD1
[y
k
[ < o. That is, the series
1

kD1
y
k
is absolutely convergent, and hence, by our assumption,
the series
1

kD1
y
k
is convergent in X. That is, there is an s X such that s
j
=
j

kD1
y
k
s as j o. It
follows that
s
j
=
j

kD1
y
k
=
j

kD1
x
n
kC1
x
n
k
| = x
n
jC1
x
n
1
j!1
s.
Hence x
n
jC1
j!1
s x
n
1
. Thus, the subsequence
_
x
n
k
_
of (x
n
) converges in X. But if a Cauchy
sequence has a convergent subsequence, then the sequence itself also converges (to the same limit as the
subsequence). It thus follows that the sequence (x
n
) also converges in X. Hence X is complete.
We now apply Theorem 2.4.1 to show that if M is a closed linear subspace of a Banach space X, then
the quotient space X,M, with the quotient norm, is also a Banach space.
2.4.2 Theorem
Let M be a closed linear subspace of a Banach space X. Then the quotient space X,M is a Banach space
when equipped with the quotient norm.
Proof. Let (x
n
|) be a sequence in X,M such that
1

jD1
[x
j
|[ < o. For each j N, choose an element
y
j
M such that
[x
j
y
j
[ _ [x
j
|[ 2
j
.
It now follows that
1

jD1
[x
j
y
j
[ < o, i.e., the series
1

jD1
(x
j
y
j
) is absolutely convergent in X. Since
X is complete, the series
1

jD1
(x
j
y
j
) converges to some element z X. We show that the series
1

jD1
x
j
|
converges to z|. Indeed, for each n N,
_
_
_
_
_
_
n

jD1
x
j
| z|
_
_
_
_
_
_
=
_
_
_
_
_
_
_
_
n

jD1
x
j
_
_
z|
_
_
_
_
_
_
=
_
_
_
_
_
_
_
_
n

jD1
x
j
z
_
_
_
_
_
_
_
_
= inf
m2M
_
_
_
_
_
_
n

jD1
x
j
z m
_
_
_
_
_
_
_
_
_
_
_
_
_
n

jD1
x
j
z
n

jD1
y
j
_
_
_
_
_
_
=
_
_
_
_
_
_
n

jD1
(x
j
y
j
) z
_
_
_
_
_
_
0 as n o.
25
2011 FUNCTIONAL ANALYSIS ALP
Hence, every absolutely convergent series in X,M is convergent, and so X,M is complete.
2.5 Bounded, Totally Bounded, and Compact Subsets of a Normed
Linear Space
2.5.1 Denition
A subset A of a normed linear space (X. [ [) is bounded if A Bx. r | for some x X and r > 0.
It is clear that A is bounded if and only if there is a C > 0 such that [a[ _ C for all a A.
2.5.2 Denition
Let A be a subset of a normed linear space (X. [ [) and c > 0. A subset A
e
X is called an c-net for A
if for each x A there is an element y A
e
such that [x y[ < c. Simply put, A
e
X is an c-net for A
if each element of A is within an c distance to some element of A
e
.
A subset A of a normed linear space (X. [ [) is totally bounded (or precompact) if for any c > 0 there
is a nite c-net F
e
X for A. That is, there is a nite set F
e
X such that
A
_
x2F
B(x. c).
The following proposition shows that total boundedness is a stronger property than boundedness.
2.5.3 Proposition
Every totally bounded subset of a normed linear space (X. [ [) is bounded.
Proof. This follows from the fact that a nite union of bounded sets is also bounded.
The following example shows that boundedness does not, in general, imply total boundedness.
2.5.4 Example
Let X =
2
and consider B = B(X) = {x X [ [x[ _ 1], the closed unit ball in X. Clearly, B
is bounded. We show that B is not totally bounded. Consider the elements of B of the form: for
j N, e
j
= (0. 0. . . . . 0. 1. 0. . . .), where 1 occurs in the j-th position. Note that [e
i
e
j
[
2
=
_
2
for all i = j. Assume that an c-net B
e
X existed for 0 < c <
p
2
2
. Then for each j N,
there is an element y
j
B
e
such that [e
j
y
j
[ < c. This says that for each j N, there is an
element y
j
B
e
such that y
j
B(e
j
. c). But the balls B(e
j
. c) are disjoint. Indeed, if i ,= j, and
z B(e
i
. c) B(e
j
. c), then by the triangle inequality
_
2 = [e
i
e
j
[
2
_ [e
i
z[ [z e
j
[ < 2c <
_
2.
which is absurd. Since the balls B(e
j
. c) are (at least) countably innite, there can be no nite
c-net for B.
In our denition of total boundedness of a subset A X, we required that the nite c-net be a subset
of X. The following proposition suggests that the nite c-net may actually be assumed to be a subset of A
itself.
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2011 FUNCTIONAL ANALYSIS ALP
2.5.5 Proposition
A subset A of a normed linear space (X. [ [) is totally bounded if and only if for any c > 0 there is a nite
set F
e
A such that
A
_
x2F
B(x. c).
Proof. Exercise.
We now give a characterization of total boundedness.
2.5.1 Theorem
A subset K of a normed linear space (X. [ [) is totally bounded if and only if every sequence in K has a
Cauchy subsequence.
Proof. Assume that K is totally bounded and let (x
n
) be an innite sequence in K. There is a nite set of
points {y
11
. y
12
. . . . . y
1r
] in K such that
K
r
_
jD1
B(y
1j
.
1
2
).
At least one of the balls B(y
1j
.
1
2
). j = 1. 2. . . . . r , contains an innite subsequence (x
n1
) of (x
n
). Again,
there is a nite set {y
21
. y
22
. . . . . y
2s
] in K such that
K
s
_
jD1
B(y
2j
.
1
2
2
).
At least one of the balls B(y
2j
.
1
2
2
). j = 1. 2. . . . . s, contains an innite subsequence (x
n2
) of (x
n1
).
Continuing in this way, at the m-th step, we obtain a subsequence (x
nm
) of (x
n(m1)
) which is contained in
a ball of the form B
_
y
mj
.
1
2
m
_
.
Claim: The diagonal subsequence (x
nn
) of (x
n
) is Cauchy. Indeed, if m > n, then both x
nn
and x
mm
are
in the ball of radius 2
n
. Hence, by the triangle inequality,
[x
nn
x
mm
[ < 2
1n
0 as n o.
Conversely, assume that every sequence in K has a Cauchy subsequence and that K is not totally
bounded. Then, for some c > 0, no nite c-net exists for K. Hence, if x
1
K, then there is an x
2
K
such that [x
1
x
2
[ _ c. (Otherwise, [x
1
y[ < c for all y K and consequently {x
1
] is a nite c-net
for K, a contradiction.) Similarly, there is an x
3
K such that
[x
1
x
3
[ _ c and [x
2
x
3
[ _ c.
Continuing in this way, we obtain a sequence (x
n
) in K such that [x
n
x
m
[ _ c for all m = n. Therefore
(x
n
) cannot have a Cauchy subsequence, a contradiction.
2.5.6 Denition
A normed linear space (X. [ [) is sequentially compact if every sequence in X has a convergent subse-
quence.
2.5.7 Remark
It can be shown that on a metric space, compactness and sequential compactness are equivalent. Thus, it
follows, that on a normed linear space, we can use these terms interchangeably.
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2011 FUNCTIONAL ANALYSIS ALP
2.5.2 Theorem
A subset of a normed linear space is sequentially compact if and only if it is totally bounded and complete.
Proof. Let K be a sequentially compact subset of a normed linear space (X. [ [). We show that K is
totally bounded. To that end, let (x
n
) be a sequence in K. By sequential compactness of K, (x
n
) has a
subsequence (x
n
k
) which converges in K. Since every convergent sequence is Cauchy, the subsequence
(x
n
k
) of (x
n
) is Cauchy. Therefore, by Theorem 2.5.1, K is totally bounded.
Next, we show that K is complete. Let (x
n
) be a Cauchy sequence in K. By sequential compactness
of K, (x
n
) has a subsequence (x
n
k
) which converges in K. But if a subsequence of a Cauchy sequence
converges, so does the full sequence. Hence (x
n
) converges in K and so K is complete.
Conversely, assume that K is a totally bounded and complete subset of a normed linear space (X. [ [).
We showthat K is sequentially compact. Let (x
n
) be a sequence in K. By Theorem 2.5.1, (x
n
) has a Cauchy
subsequence (x
n
k
). Since K is complete, (x
n
k
) converges in K. Hence K is sequentially compact.
2.5.8 Corollary
A subset of a Banach space is sequentially compact if and only if it is totally bounded and closed.
Proof. Exercise.
2.5.9 Corollary
A sequentially compact subset of a normed linear space is closed and bounded.
Proof. Exercise.
We shall see that in nite-dimensional spaces the converse of Corollary 2.5.9 also holds.
2.5.10 Corollary
A closed subset F of a sequentially compact normed linear space (X. [ [) is sequentially compact.
Proof. Exercise.
2.6 Finite Dimensional Normed Linear Spaces
The theory for nite-dimensional normed linear spaces turns out to be much simpler than that of their
innite-dimensional counterparts. In this section we highlight some of the special aspects of nite-dimensional
normed linear spaces.
The following Lemma is crucial in the analysis of nite-dimensional normed linear spaces.
2.6.1 Lemma
Let (X. [ [) be a nite-dimensional normed linear space with basis {x
1
. x
2
. . . . . x
n
]. Then there is a
constant m > 0 such that for every choice of scalars
1
.
2
. . . . .
n
, we have
m
n

jD1
[
j
[ _
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
.
28
2011 FUNCTIONAL ANALYSIS ALP
Proof. If
n

jD1
[
j
[ = 0, then
j
= 0 for all j = 1. 2. . . . . n and the inequality holds for any m > 0.
Assume that
n

jD1
[
j
[ = 0. We shall prove the result for a set of scalars {
1
.
2
. . . . .
n
] that satisfy the
condition
n

jD1
[
j
[ = 1. Let
A = {(
1
.
2
. . . . .
n
) F
n
[
n

jD1
[
j
[ = 1].
Since A is a closed and bounded subset of F
n
, it is compact. Dene : A R by
(
1
.
2
. . . . .
n
) =
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
.
Since for any (
1
.
2
. . . . .
n
) and (
1
.
2
. . . . .
n
) in A
[(
1
.
2
. . . . .
n
) (
1
.
2
. . . . .
n
)[ =

_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_

_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_

_
_
_
_
_
_
_
n

jD1

j
x
j

n

jD1

j
x
j
_
_
_
_
_
_
=
_
_
_
_
_
_
n

jD1
(
j

j
)x
j
_
_
_
_
_
_
_
n

jD1
[
j

j
[[x
j
[
_ max
1jn
[x
j
[
n

jD1
[
j

j
[.
is continuous on A. Since is a continuous function on a compact set A, it attains its minimum on A,
i.e., there is an element (j
1
. j
2
. . . . . j
n
) A such that
(j
1
. j
2
. . . . . j
n
) = inf{(
1
.
2
. . . . .
n
) [ (
1
.
2
. . . . .
n
) A].
Let m = (j
1
. j
2
. . . . . j
n
). Since _ 0, it follows that m _ 0. If m = 0, then
_
_
_
_
_
_
n

jD1
j
j
x
j
_
_
_
_
_
_
= 0 =
n

jD1
j
j
x
j
= 0.
Since the set {x
1
. x
2
. . . . . x
n
] is linearly independent, j
j
= 0 for all j = 1. 2. . . . . n. This is a
contradiction since (j
1
. j
2
. . . . . j
n
) A. Hence m > 0 and consequently for all (
1
.
2
. . . . .
n
) A,
0 < m _ (
1
.
2
. . . . .
n
) m
n

jD1
[
j
[ _
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
.
Now, let {
1
.
2
. . . . .
n
] be any collection of scalars and set =
n

jD1
[
j
[. If = 0, then the
29
2011 FUNCTIONAL ANALYSIS ALP
inequality holds vacuously. If > 0, then
_

.

2

. . . . .

n

_
A and consequently
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
=
_
_
_
_
_
_
n

jD1

x
j
_
_
_
_
_
_
=
_

.

2

. . . . .

n

_
_ m = m
n

jD1
[
j
[.
That is, m
n

jD1
[
j
[ _
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
.
2.6.1 Theorem
Let X be a nite-dimensional normed linear space over F. Then all norms on X are equivalent.
Proof. Let {x
1
. x
2
. . . . . x
n
] be a basis for X and [ [
0
and [ [ be any two norms on X. For any x X
there is a set of scalars {
1
.
2
. . . . .
n
] such that x =
n

jD1

j
x
j
. By Lemma 2.6.1, there is an m > 0 such
that
m
n

jD1
[
j
[ _
_
_
_
_
_
_
n

jD1

j
x
j
_
_
_
_
_
_
= [x[.
By the triangle inequality
[x[
0
_
n

jD1
[
j
[[x
j
[
0
_ M
n

jD1
[
j
[.
where M = max
1jn
[x
j
[
0
. Hence
[x[
0
_ M
_
1
m
[x[
_
=
m
M
[x[
0
_ [x[ [x[
0
_ [x[ where =
m
M
.
Interchanging the roles of the norms [ [
0
and [ [, we similarly get a constant such that [x[ _ [x[
0
.
Hence, [x[
0
_ [x[ _ [x[
0
for some constants and .
2.6.2 Theorem
Every nite-dimensional normed linear space (X. [ [) is complete.
Proof. Let {x
1
. x
2
. . . . . x
n
] be a basis for X and let (z
k
) be a Cauchy sequence in X. Then, given any
c > 0, there is a natural number N such that
[z
k
z
I
[ < c for all k. > N.
Also, for each k N, z
k
=
n

jD1

kj
x
j
. By Lemma 2.6.1, there is an m > 0 such that
m
n

jD1
[
kj

Ij
[ _ [z
k
z
I
[.
Hence, for all k. > N and all j = 1. 2. . . . . n,
[
kj

Ij
[ _
1
m
[z
k
z
I
[ <
c
m
.
30
2011 FUNCTIONAL ANALYSIS ALP
That is, for each j = 1. 2. . . . . n, (
kj
)
k
is a Cauchy sequence of numbers. Since F is complete,
kj

j
as k ofor each j = 1. 2. . . . . n. Dene z =
n

jD1

j
x
j
. Then z X and
[z
k
z[ =
_
_
_
_
_
_
n

jD1

kj
x
j

n

jD1

j
x
j
_
_
_
_
_
_
=
_
_
_
_
_
_
n

jD1
(
kj

j
)x
j
_
_
_
_
_
_
_
n

jD1
[
kj

j
[[x
j
[ 0
as k o. That is, the sequence (z
k
) converges to z X. hence X is complete.
2.6.2 Corollary
Every nite-dimensional normed linear space X is closed.
Proof. Exercise.
2.6.3 Theorem
In a nite-dimensional normed linear space (X. [ [), a subset K X is sequentially compact if and only
if it is closed and bounded.
Proof. We have seen (Corollary 2.5.9), that a compact subset of a normed linear space is closed and
bounded.
Conversely, assume that a subset K X is closed and bounded. We show that K is compact. Let
{x
1
. x
2
. . . . . x
n
] be a basis for X and let (z
k
) be any sequence in K. Then for each k N, z
k
=
n

jD1

kj
x
j
.
Since K is bounded, there is a positive constant M such that [z
k
[ _ M for all k N. By Lemma 2.6.1,
there is an m > 0 such that
m
n

jD1
[
kj
[ _
_
_
_
_
_
_
n

jD1

kj
x
j
_
_
_
_
_
_
= [z
k
[ _ M.
It now follows that [
kj
[ _
M
m
for each j = 1. 2. . . . . n, and for all k N. That is, for each xed j =
1. 2. . . . . n, the sequence (
kj
)
k
of numbers is bounded. Hence the sequence (
kj
)
k
has a subsequence
(
kr j
) which converges to
j
for j = 1. 2. . . . . n. Setting z =
n

jD1

j
x
j
, we have that
[z
kr
z[ =
_
_
_
_
_
_
n

jD1

kr j
x
j

n

jD1

j
x
j
_
_
_
_
_
_
_
n

jD1
[
kr j

j
[[x
j
[ 0 as r o.
That is, z
kr
z as r o. Since K is closed, z K. Hence K is compact.
2.6.3 Lemma
(Rieszs Lemma). Let M be a closed proper linear subspace of a normed linear space (X. [ [). Then for
each 0 < c < 1, there is an element z X such that [z[ = 1 and
[y z[ > 1 c for all y M.
Proof. Choose x X \ M and dene
d = d(x. M) = inf
m2M
[x m[.
31
2011 FUNCTIONAL ANALYSIS ALP
Since M is closed, d > 0. By denition of inmum, there is a m M such that
d _ [x m[ < d cd = d(1 c).
Take z =
_
mx
[mx[
_
. Then [z[ = 1 and for any y M,
[y z[ =
_
_
_
_
y
_
m x
[m x[
__
_
_
_
=
[y([mx[) m x[
[mx[
_
d
[m x[
>
d
d(1 c)
=
1
1 c
= 1
c
1 c
> 1 c.
We now give a topological characterization of the algebraic concept of nite dimensionality.
2.6.4 Theorem
A normed linear space (X. [ [) is nite-dimensional if and only its closed unit ball B(X) = {x
X [ [x[ _ 1] is compact.
Proof. Assume that (X. [ [) is nite-dimensional normed linear space. Since the ball B(X) is closed and
bounded, it is compact.
Assume that the closed unit ball B(X) = {x X [ [x[ _ 1] is compact. Then B(X) is totally
bounded. Hence there is a nite
1
2
-net {x
1
. x
2
. . . . . x
n
] in B(X). Let M =lin{x
1
. x
2
. . . . . x
n
]. Then
M is a nite-dimensional linear subspace of X and hence closed.
Claim: M = X. If M is a proper subspace of X, then, by Rieszs Lemma there is an element x
0
B(X)
such that d(x
0
. M) >
1
2
. In particular, [x
0
x
k
[ >
1
2
for all k = 1. 2. . . . . n. However this contradicts
the fact that {x
1
. x
2
. . . . . x
n
] is a
1
2
-net in B(X). Hence M = X and, consequently, X is nite-
dimensional.
We nowgive another argument to showthat boundedness does not imply total boundedness. Let X =
2
and B(X) = {x X [ [x[
2
_ 1]. It is obvious that B(X) is bounded. We show that B(X) is not totally
bounded. Since X is complete and B(X) is a closed subset of X, B(X) is complete. If B(X) were totally
bounded, then B(X) would, according to Theorem 2.26, be compact. By Theorem 2.6.4, X would be
nite-dimensional. But this is false since X is innite-dimensional.
2.7 Separable Spaces and Schauder Bases
2.7.1 Denition
(a) A subset S of a normed linear space (X. [ [) is said to be dense in X if S = X; i.e., for each x X
and c > 0, there is a y S such that [x y[ < c.
(b) A normed linear space (X. [ [) is said to be separable if it contains a countable dense subset.
2.7.2 Examples
[1] The real line R is separable since the set Q of rational numbers is a countable dense subset
of R.
[2] The complex plane C is separable since the set of all complex numbers with rational real
and imaginary parts is a countable dense subset of C.
[3] The sequence space
p
, where 1 _ p < o, is separable. Take M to be the set of all
sequences with rational entries such that all but a nite number of the entries are zero. (If
32
2011 FUNCTIONAL ANALYSIS ALP
the entries are complex, take for M the set of nitely nonzero sequences with rational real
and imaginary parts.) It is clear that M is countable. We show that M is dense in
p
. Let
c > 0 and x = (x
n
)
p
. Then there is an N such that
1

kDNC1
[x
k
[
p
<
c
2
.
Now, for each 1 _ k _ N, there is a rational number q
k
such that [x
k
q
k
[
p
<
e
2N
. Set
q = (q
1
. q
2
. . . . . q
N
. 0. 0. . . .). Then q M and
[x q[
p
p
=
N

kD1
[x
k
q
k
[
p

kDNC1
[x
k
[
p
< c.
Hence M is dense in
p
.
[4] The sequence space
1
, with the supremum norm, is not separable. To see this, consider
the set M of elements x = (x
n
), in which x
n
is either 0 or 1. This set is uncountable since
we may consider each element of M as a binary representation of a number in the interval
0. 1|. Hence there are uncountably many sequences of zeroes and ones. For any two
distinct elements x. y M, [x y[
1
= 1. Let each of the elements of M be a centre of
a ball of radius
1
4
. Then we get uncountably many nonintersecting balls. If A is any dense
subset of
1
, then each of these balls contains a point of A. Hence A cannot be countable
and, consequently,
1
is not separable.
2.7.1 Theorem
A normed linear space (X. [ [) is separable if and only if it contains a countable set B such that lin(B) =
X.
Proof. Assume that X is separable and let A be a countable dense subset of X. Since the linear hull of A,
lin(A), contains A and A is dense in X, we have that lin(A) is dense in X, that is, lin(A) = X.
Conversely, assume that X contains a countable set B such that lin(B) = X. Let B = {x
n
[ n N].
Assume rst that = R, and put
C =
_
_
_
n

jD1
z
j
x
j
[ z
j
Q. j = 1. 2. . . . . n. n N
_
_
_
.
We rst show that C is a countable subset of X. The set Q B is countable and consequently, the family
F of all nite subsets of Q B is also countable. The mapping
{(z
1
. x
1
). (z
2
. x
2
). . . . . (z
n
. x
n
)]
n

jD1
z
j
x
j
maps F onto C. Hence C is countable.
Next, we show that C is dense in X. Let x X and c > 0. Since lin(B) = X, we can nd an n N,
points x
1
. x
2
. . . . . x
n
B and z
1
. z
2
. . . . . z
n
F such that
_
_
_
_
_
_
x
n

jD1
z
j
x
j
_
_
_
_
_
_
<
c
2
.
33
2011 FUNCTIONAL ANALYSIS ALP
Since Q is dense in R, for each z
i
R, we can nd a j
i
Q such that
[z
i
j
i
[ <
c
2n(1 [x
i
[)
for all i = 1. 2. . . . . n.
Hence,
_
_
_
_
_
_
x
n

jD1
j
j
x
j
_
_
_
_
_
_
_
_
_
_
_
_
_
x
n

jD1
z
j
x
j
_
_
_
_
_
_

_
_
_
_
_
_
n

jD1
z
j
x
j

n

jD1
j
j
x
j
_
_
_
_
_
_
<
c
2

jD1
[z
j
j
j
[[x
j
[
<
c
2

n

jD1
c[x
j
[
2n(1 [x
j
[)
<
c
2

c
2
= c.
This shows that C is dense in X.
If F = C, the set C is that of nite linear combinations with coefcients being those complex numbers with
rational real and imaginary parts.
We now give another argument based on Theorem 2.7.1 to show that the sequence space
p
, where
1 _ p < o, is separable. Let e
n
= (
nm
)
m2N
, where

nm
=
_
1 if n = m
0 otherwise.
Clearly, e
n

p
. Let c > 0 and x = (x
n
)
p
. Then there is a natural number N such that
1

kDnC1
[x
k
[
p
< c
p
for all n _ N.
Now, if n _ N, then
_
_
_
_
_
_
x
n

jD1
x
j
e
j
_
_
_
_
_
_
p
=
_
_
1

kDnC1
[x
k
[
p
_
_
1{p
< c.
Hence lin({e
n
[ n N]) =
p
. Of course, the set {e
n
[ n N] is countable.
2.7.3 Denition
A sequence (b
n
) in a Banach space (X. [ [) is called a Schauder basis if for any x X, there is a unique
sequence (
n
) of scalars such that
lim
n!1
_
_
_
_
_
_
x
n

jD1

j
b
j
_
_
_
_
_
_
= 0.
In this case we write x =
1

jD1

j
b
j
.
2.7.4 Remark
It is clear from Denition 2.7.3 that (b
n
) is a Schauder basis if and only if X = lin{b
n
[ n N] and
every x X has a unique expansion x =
1

jD1

j
b
j
.
Uniqueness of this expansion clearly implies that the set {b
n
[ n N] is linearly independent.
34
2011 FUNCTIONAL ANALYSIS ALP
2.7.5 Examples
[1] For 1 _ p < o, the sequence (e
n
), where e
n
= (
nm
)
m2N
, is a Schauder basis for
p
.
[2] (e
n
) is a Schauder basis for c
0
.
[3] (e
n
) L {e], where e = (1. 1. 1. . . .) (the constant 1 sequence), is a Schauder basis for c.
[4]
1
has no Schauder basis.
2.7.6 Proposition
If a Banach space (X. [ [) has a Schauder basis, then it is separable.
Proof. Let (b
n
) be a Schauder basis for X. Then {b
n
[ n N] is countable and
lin({b
n
[ n N]) = X.
Schauder bases have been constructed for most of the well-known Banach spaces. Schauder conjectured
that every separable Banach space has a Schauder basis. This conjecture, known as the Basis Problem,
remained unresolved for a long time until Per Eno in 1973 answered it in the negative. He constructed a
separable reexive Banach space with no basis.
2.7.7 Exercise
[1] Let X be a normed linear space over F. Show that X is nite-dimensional if and only if every
bounded sequence in X has a convergent subsequence.
[2] Complete the proof of Theorem 2.1.1.
[3] Prove Lemma 2.3.2.
[4] Prove the claims made in [1] and [2] of Example 2.3.7.
[5] Prove Theorem 2.5.5.
[6] Prove Corollary 2.5.8.
[7] Prove Corollary 2.5.9.
[8] Prove Corollary 2.5.10.
[9] Prove Corollary 2.6.2.
[10] Is (Ca. b|. [ [
1
) complete? What about (Ca. b|. [ [
1
)? Fully justify both answers.
35
Chapter 3
Hilbert Spaces
3.1 Introduction
In this chapter we introduce an inner product which is an abstract version of the dot product in elementary
vector algebra. Recall that if x = (x
1
. x
2
. x
3
) and y = (y
1
. y
2
. y
3
) are any two vectors in R
3
, then the
dot product of x and y is x y = x
1
y
1
x
2
y
2
x
3
y
3
. Also, the length of the vector x is [x[ =
_
x
2
1
x
2
2
x
2
3
=
_
x x.
It turns out that Hilbert spaces are a natural generalization of nite-dimensional Euclidean spaces.
Hilbert spaces arise naturally and frequently in mathematics, physics, and engineering, typically as innite-
dimensional function spaces.
3.1.1 Denition
Let X be a linear space over a eld F. An inner product on X is a scalar-valued function (. ) : XX F
such that for all x. y. z X and for all . F, we have
IP1. (x. x) _ 0;
IP2. (x. x) = 0 x = 0;
IP3. (x. y) = (y. x) (The bar denotes complex conjugation.);
IP4. (x. y) = (x. y);
IP5. (x y. z) = (x. z) (y. z).
An inner product space (X. (. )) is a linear space X together with an inner (. ) product dened on it. An
inner product space is also called pre-Hilbert space.
3.1.2 Examples
Examples of inner product spaces.
[1] Fix a positive integer n. Let X = F
n
. For x = (x
1
. x
2
. . . . . x
n
) and y = (y
1
. y
2
. . . . . y
n
) in X,
dene
(x. y) =
n

iD1
x
i
y
i
.
Since this is a nite sum, (. ) is well-dened. It is easy to show that (X. (. )) is an inner
product space. The space R
n
(resp. C
n
) with this inner product is called the Euclidean
n-space (resp. unitary n-space) and will be denoted by
2
(n).
36
2011 FUNCTIONAL ANALYSIS ALP
[2] Let X =
0
, the linear space of nitely non-zero sequences of real or complex numbers. For
x = (x
1
. x
2
. . . .) and y = (y
1
. y
2
. . . .) in X, dene
(x. y) =
1

iD1
x
i
y
i
.
Since this is essentially a nite sum, (. ) is well-dened. It is easy to show that (X. (. )) is
an inner product space.
[3] Let X =
2
, the space of all sequences x = (x
1
. x
2
. . . .) of real or complex numbers with
1

1
[x
i
[
2
< o. For x = (x
1
. x
2
. . . .) and y = (y
1
. y
2
. . . .) in X, dene
(x. y) =
1

iD1
x
i
y
i
.
In order to show that (. ) is well-dened we rst observe that if a and b are real numbers,
then
0 _ (a b)
2
. whence ab _
1
2
(a
2
b
2
).
Using this fact, we have that
[x
i
y
i
[ = [x
i
[[y
i
[ _
1
2
_
[x
i
[
2
[y
i
[
2
_
=
1

iD1
[x
i
y
i
[ _
1
2
_
1

iD1
[x
i
[
2

iD1
[y
i
[
2
_
< o.
Hence, (. ) is well-dened (i.e., the series converges).
[4] Let X = Ca. b|, the space of all continuous complex-valued functions on a. b|. For x. y X,
dene
(x. y) =
b
_
a
x(t )y(t ) dt.
We shall denote by C
2
a. b| the linear space Ca. b| equipped with this inner product.
[5] Let X = L(C
n
) be the linear space of all n n complex matrices. For A L(C
n
), let
t(A) =
n

iD1
(A)
ii
be the trace of A. For A. B L(C
n
), dene
(A. B) = t(B

A). where B

denotes conjugate transpose of matrix B.

Show that (L(C
n
). (. )) is an inner product space.
It should be mentioned that we could consider real Hilbert spaces but there are powerful methods that
can be applied by using the more general complex Hilbert spaces.
3.1.1 Theorem
(Cauchy-Bunyakowsky-Schwarz Inequality). Let (X. (. )) be an inner product space over a eld F.
Then for all x. y X,
[(x. y)[ _
_
(x. x)
_
(y. y).
Moreover, given any x. y X, the equality
[(x. y)[ =
_
(x. x)
_
(y. y)
holds if and only if x and y are linearly dependent.
37
2011 FUNCTIONAL ANALYSIS ALP
Proof. If x = 0 or y = 0, then the result holds vacuously. Assume that x ,= 0 and y ,= 0. For any F,
we have
0 _ (x y. x y) = (x. x) (y. x) (x. y) (y. y)
= (x. x) (x. y) (y. x) (y. y)|.
Now choosing =
(x. y)
(y. y)
, we have
0 _ (x. x)
(y. x)(x. y)
(y. y)
= (x. x)
[(x. y)[
2
(y. y)
.
whence
[(x. y)[ _
_
(x. x)
_
(y. y).
Assume that [(x. y)[ =
_
(x. x)
_
(y. y). We show that x and y are linearly dependent. If x = 0 or
y = 0, then x and y are obviously linearly dependent. We therefore assume that x ,= 0 and y ,= 0. Then
(y. y) ,= 0. With =
(x. y)
(y. y)
, we have that
(x y. x y) = (x. x)
[(x. y)[
2
(y. y)
= 0.
That is,
(x y. x y) = 0. = x = y.
That is, x and y are linearly dependent.
Conversely, assume that x and y are linearly dependent. Without loss of generality, x = zy for some
z F. Then
[(x. y)[ = [(zy. y)[ = [z[[(y. y)[ = [z[(y. y)
= [z[
_
(y. y)
_
(y. y) =
_
[z[
2
(y. y)
_
(y. y) =
_
(zyzy)
_
(y. y)
=
_
(x. x)
_
(y. y).
3.1.2 Theorem
Let (X. (. )) be an inner product space over a eld F. For each x X, dene
[x[ :=
_
(x. x). (3.1.2.1)
Then [ [ denes a norm on X. That is, (X. [ [) is a normed linear space over F.
Proof. Let x. y X and z F. Then
N1. [x[ =
_
(x. x) _ 0;
N2. [x[ = 0
_
(x. x) = 0 (x. x) = 0 x = 0. by IP2.
N3. [zx[ =
_
(zxzx) =
_
[z[
2
(x. x) = [z[
_
(x. x) = [z[[x[.
N4.
38
2011 FUNCTIONAL ANALYSIS ALP
[x y[
2
= (x y. x y) = (x. x) (x. y) (y. x) (y. y)
= (x. x) (x. y) (x. y) (y. y)
= [x[
2
2Re(x. y) [y[
2
_ [x[
2
2[(x. y)[ [y[
2
_ [x[
2
2
_
(x. x)
_
(y. y) [y[
2
(by Theorem 3.1.1)
= [x[
2
2[x[[y[ [y[
2
= ([x[ [y[)
2
.
Taking the positive square root both sides yields
[x y[ _ [x[ [y[.
In view of (3.1.2.1), the Cauchy-Bunyakowsky-Schwarz Inequality now becomes
[(x. y)[ _ [x[[y[.
Any inner product space can thus be made into a normed linear space in a natural way: by dening the norm
as in (3.1.2.1). The norm advertised in (3.1.2.1) is called the inner product norm or a norm induced or
generated by the inner product.
A natural question arises: Is every normed linear space an inner product space? If the answer is NO,
howthen does one recognise among all normed linear spaces those that are inner product spaces in disguise,
i.e., those whose norms are induced by an inner product?
These questions will be examined later.
3.1.3 Theorem
(Polarization Identity). Let (X. (. )) be an inner product space over a eld F. Then for all x. y X,
(x. y) =
[x y[
2
4

[x y[
2
4
if F = R. and
(x. y) =
[x y[
2
4

[x y[
2
4
i
_
kxCyik
2
4

[x yi [
2
4
_
if F = C.
Proof. Assume that F = R. Then
[x y[
2
[x y[
2
= (x y. x y) (x y. x y)
= (x. x) (x. y) (y. x) (y. y) (x. x) (x. y) (y. x) (y. y)
= 4(x. y). since (x. y) = (y. x).
The case when F = C is proved analogously and is left as an exercise.
3.1.4 Theorem
(Parallelogram Identity). Let (X. (. )) be an inner product space over a eld F. Then for all x. y X,
[x y[
2
[x y[
2
= 2[x[
2
2[y[
2
. (3.1.4.1)
Proof.
[x y[
2
[x y[
2
= (x y. x y) (x y. x y)
= (x. x) (x. y) (y. x) (y. y) (x. x) (x. y) (y. x) (y. y)
= 2[x[
2
2[y[
2
.
The geometric interpretation of the Parallelogram Identity is evident: the sum of the squares of the
lengths of the diagonals of a parallelogram is equal to the sum of the squares of the lengths of the four
39
2011 FUNCTIONAL ANALYSIS ALP
sides.
x
y
The following theorem asserts that the Parallelogram Identity (Theorem 3.1.4) distinguishes inner prod-
uct spaces among all normed linear spaces. It also answers the question posed after Theorem 3.1.2. That is,
a normed linear space is an inner product space if and only if its norm satises the Parallelogram Identity.
3.1.5 Theorem
A normed linear space X over a eld F is an inner product space if and only if the Parallelogram Identity
[x y[
2
[x y[
2
= 2[x[
2
2[y[
2
(PI)
holds for all x. y X.
Proof. =. We have already shown (Theorem 3.1.4) that if X is an inner product space, then the parallel-
ogram identity (PI) holds in X.
=. Let X be a normed linear space in which the parallelogram identity (PI) holds. We shall only
consider the case F = R. The polarization identity (Theorem 3.1.3) gives us a hint as to how we should
dene an inner product: For all x. y X, dene
(x. y) =
_
_
_
_
x y
2
_
_
_
_
2

_
_
_
x y
2
_
_
_
2
.
We claim that (. ) is an inner product on X.
IP1. (x. x) =
_
_
_
_
x x
2
_
_
_
_
2

_
_
_
x x
2
_
_
_
2
= [x[
2
_ 0.
IP2. (x. x) = 0 [x[
2
= 0 x = 0.
IP3. (x. y) =
_
_
_
_
x y
2
_
_
_
_
2

_
_
_
x y
2
_
_
_
2
=
_
_
_
_
y x
2
_
_
_
_
2

_
_
_
y x
2
_
_
_
2
= (y. x) = (y. x) since F = R.
IP5. Replace x by u : and y by n : in the parallelogram identity:
[u n 2:[
2
[u n[
2
= 2[u :[
2
2[n :[
2
. (3.1.5.1)
Replace x by u : and y by n : in the parallelogram identity:
[u n 2:[
2
[u n[
2
= 2[u :[
2
2[n :[
2
. (3.1.5.2)
Subtract (3.1.5.2) from (3.1.5.1):
[u n 2:[
2
[u n 2:[
2
= 2
_
[u :[
2
[u :[
2
[: n[
2
[: n[
2
_
.
40
2011 FUNCTIONAL ANALYSIS ALP
Use the denition of (. ),
4(u n. 2:) = 8(u. :) (n. :)| =
1
2
(u n. 2:) = (u. :) (n. :). (3.1.5.3)
Take n = 0:
1
2
(u. 2:) = (u. :). (3.1.5.4)
Now replace u by x y and : by z in (3.1.5.4) and use (3.1.5.3) to get
(x y. z) =
1
2
(x y. 2z) = (x. z) (y. z).
IP4. We show that (zx. y) = z(x. y) for all z R and all x. y X. If z = n is a nonzero integer, then
using IP5,
(nx. y) = n(x. y) = n
_
x
n
. y
_
=
_
nx
n
. y
_
= (x. y).
That is,
_
x
n
. y
_
=
1
n
(x. y).
If z is a rational number, z =
p
q
, say. Then
_
p
q
x. y
_
= p
_
x
q
. y
_
=
p
q
(x. y).
If z R, then there is a sequence (r
k
) of rational numbers such that r
k
z as k o. Using
continuity of the norm, we have that
(zx. y) = ( lim
k!1
r
k
x. y) =
1
4
_
_
_
_
lim
k!1
r
k
x y
_
_
_
_
2

1
4
_
_
_
_
lim
k!1
r
k
x y
_
_
_
_
2
=
1
4
lim
k!1
[r
k
x y[
2

1
4
lim
k!1
[r
k
x y[
2
= lim
k!1
_
_
_
_
_
r
k
x y
2
_
_
_
_
2

_
_
_
r
k
x y
2
_
_
_
2
_
= lim
k!1
(r
k
x. y)
= lim
k!1
r
k
(x. y) = z(x. y).
Thus, (zx. y) = z(x. y) for all z R and all x. y X.
3.1.3 Corollary
Let (X. [ [) be a normed linear space over a eld F. If every two-dimensional linear subspace of X is an
inner product space over F, then X is an inner product space.
3.1.4 Examples
[1] Let X =
p
. for p ,= 2. Then X is not an inner product space. We show that the norm
on
p
. p ,= 2 does not satisfy the parallelogram identity. Take x = (1. 1. 0. 0. . . .) and
y = (1. 1. 0. 0. . . .) in
p
. Then
[x[ = 2
1
p
= [y[ and [x y[ = 2 = [x y[.
Thus,
[x y[
2
[x y[
2
= 8 ,= 2[x[
2
2[y[
2
= 4 2
2
p
.
41
2011 FUNCTIONAL ANALYSIS ALP
[2] The normed linear space X = Ca. b|, with the supremum norm [ [
1
is not an inner product
space. We show that the norm
[x[
1
= max
at b
[x(t )[
does not satisfy the parallelogram identity. To that end, take
x(t ) = 1 and y(t ) =
t a
b a
.
Since
x(t ) y(t ) = 1
t a
b a
and x(t ) y(t ) = 1
t a
b a
.
we have that
[x[ = 1 = [y[. and [x y[ = 2. [x y[ = 1.
Thus,
[x y[
2
[x y[
2
= 5 ,= 2[x[
2
2[y[
2
= 4.
3.2 Completeness of Inner Product Spaces
The mathematical concept of a Hilbert space, named after David Hilbert, generalizes the notion of Euclidean
space. Hilbert spaces, as the following denition states, are inner product spaces which in addition are
required to be complete, a property that stipulates the existence of enough limits in the space to allow the
techniques of calculus to be used.
The earliest Hilbert spaces were studied from this more abstract point of view in the rst decade of
the 20th century by David Hilbert, Erhard Schmidt, and Frigyes Riesz. They are indispensable tools in
the theories of partial differential equations, quantum mechanics, Fourier analysis which includes applica-
tions to signal processing, and ergodic theory which forms the mathematical underpinning of the study of
thermodynamics.
3.2.1 Denition
Let (X. (. )) be an inner product space. If X is complete with respect to the norm induced by the inner
product (. ), then we say that X is a Hilbert space.
3.2.2 Examples
[1] The classical space
2
is a Hilbert space.
[2]
0
is an incomplete inner product space.
[3] The space C1. 1| is an incomplete inner product space.
3.3 Orthogonality
3.3.1 Denition
Two elements x and y in an inner product space (X. (. )) are said to be orthogonal, denoted by x J y, if
(x. y) = 0.
The set M X is called orthogonal if it consists of non-zero pairwise orthogonal elements.
If M is a subset of X such that (x. m) = 0 for all m M, then we say that x is orthogonal to M and write
x J M. We shall denote by
M
?
= {x X : (x. m) = 0 V m M]
42
2011 FUNCTIONAL ANALYSIS ALP
the set of all elements in X that are orthogonal to M. The set M
?
is called the orthogonal complement
of M.
3.3.2 Proposition
Let M and N be subsets of an inner product space (X. (. )). Then
[1] {0]
?
= X and X
?
= {0];
[2] M
?
is a closed linear subspace of X;
[3] M (M
?
)
?
= M
??
;
[4] If M is a linear subspace, then M M
?
= {0];
[5] If M N, then N
?
M
?
;
[6] M
?
= (linM)
?
= (linM)
?
.
Proof.
[1] Exercise.
[2] Let x. y M
?
, and . F. Then for each z M,
(x y. z) = (x. z) (y. z) = 0.
Hence, x y M
?
. That is, M
?
is a subspace of X. To show that M
?
is closed, let x M
?
.
Then there exists a sequence (x
n
) in M
?
such that x
n
x as n o. Thus, for all y M,
(x. y) = lim
n
(x
n
. y) = 0.
whence x M
?
.
[3] Exercise.
[4] Exercise.
[5] Let x N
?
. Then (x. y) = 0 for all y N. In particular, (x. y) = 0 for all y M since M N.
Thus, x M
?
.
[6] Since M linM linM, we have, by [5], that (linM)
?
(linM)
?
M
?
. It remains to show
that M
?
(linM)
?
. To that end, let x M
?
. Then (x. y) = 0 for all y M, and consequently
(x. y) = 0 for all y linM. If z linM, then there exists a sequence (z
n
) in linM such that z
n
z
as n o. Thus,
(x. z) = lim
n
(x. z
n
) = 0.
whence x (linM)
?
.
3.3.3 Examples
Let X = R
3
. The vectors (3. 0. 2) and (4. 1. 6) are orthogonal since
((3. 0. 2). (4. 1. 6)) = (3)(4) 0(1) (2)(6) = 0.
43
2011 FUNCTIONAL ANALYSIS ALP
If M =
0
, the linear subspace of
2
consisting of all scalar sequences (x
i
)
1
1
with only a nite
number of nonzero terms, then M
?
= {0]. Indeed, suppose that y = (y
i
)
1
iD1
M
?
. Let

ij
=
_
_
_
1 if i = j
0 if i ,= j.
and e
n
= (
nj
)
1
n,jD1
. Then e
n
M for each n N, and hence,
0 = (y. e
i
) =
1

jD1
y
j

ij
= y
i
for all i = 1. 2. . . . .
That is, y = 0, whence M
?
= {0].
3.3.1 Theorem
(Pythagoras). Let (X. (. )) be an inner product space over a eld F and let x. y X.
[1] If F = R, then x J y if and only if
[x y[
2
= [x[
2
[y[
2
.
[2] If F = C, then x J y if and only if
[x y[
2
= [x[
2
[y[
2
and [x iy[
2
= [x[
2
[y[
2
.
Proof. [1] =. If x J y, then
[x y[
2
= (x y. x y) = (x. x) 2(x. y) (y. y) = [x[
2
[y[
2
.
=. Suppose that [x y[
2
= [x[
2
[y[
2
. Then
(x y. x y) = (x. x) (y. y)
= (x. x) 2(x. y) (y. y) = (x. x) (y. y)
= 2(x. y) = 0 = (x. y) = 0.
[2] =. Assume that x J y. Then
[x yi [
2
= (x yi . x yi ) = (x. x) (x. yi ) (yi . x) (yi . yi )
= (x. x) i (x. y) i (y. x) (y. y) = [x[
2
[y[
2
.
=. Assume that [x y[
2
= [x[
2
[y[
2
and [x iy[
2
= [x[
2
[y[
2
. Then
(x y. x y) = (x. x) (y. y)
= (x. x) (x. y) (y. x) (y. y) = (x. x) (y. y)
= (x. y) (y. x) = 0 = 2Re(x. y) = 0 = Re(x. y) = 0.
Also,
(x yi . x yi ) = (x. x) (y. y)
= (x. x) i (x. y) i (y. x) (y. y) = (x. x) (y. y)
= i (x. y) i (y. x) = 0
= i (x. y) (y. x)| = 0
= i
_
(x. y) (x. y)
_
= 0
= i 2i Im(x. y)| = 0 = Im(x. y) = 0.
Since Re(x. y) = 0 = Im(x. y), we have that (x. y) = 0.
44
2011 FUNCTIONAL ANALYSIS ALP
3.3.4 Corollary
If M = {x
1
. x
2
. . . . . x
n
] is an orthogonal set in an inner product space (X. (. )) then
_
_
_
_
_
n

iD1
x
i
_
_
_
_
_
2
=
n

iD1
[x
i
[
2
.
Proof. Exercise.
3.4 Best Approximation in Hilbert Spaces
3.4.1 Denition
Let K be a closed subset of an inner product space (X. (. )). For a given x X \K, a best approximation
or nearest point to x from K is any element y
0
K such that
[x y
0
[ _ [x y[ for all y K.
Equivalently, y
0
K is a best approximation to x from K if
[x y
0
[ = inf
y2K
[x y[ = d(x. K).
The (possibly empty) set of all best approximations to x from K is denoted by P
K
(x). That is,
P
K
(x) = {y K : [x y[ = d(x. K)].
The (generally set-valued) map P
K
which associates each x in X with its best approximations in K is
called the metric projection or the nearest point map. The set K is called
[1] proximinal if each x X has a best approximation in K; i.e., P
K
(x) ,= 0 for each x X;
[2] Chebyshev if each x X has a unique best approximation in K; i.e., the set P
K
(x) consists of a
single point.
The following important result asserts that if K is a complete convex subset of an inner product space
(X. (. )), then each x X has one and only one element of best approximation in K.
3.4.1 Theorem
Every nonempty complete convex subset K of an inner product space (X. (. )) is a Chebyshev set.
Proof. Existence: Without loss of generality, x X \ K. Let
= inf
y2K
[x y[.
By denition of the inmum, there exists a sequence (y
n
)
1
1
in K such that
[x y
n
[ as n o.
We show that (y
n
)
1
1
is a Cauchy sequence. By the Parallelogram Identity (Theorem 3.1.3),
[y
m
y
n
[
2
= [(x y
n
) (x y
m
)[
2
= 2[x y
n
[
2
2[x y
m
[
2
[2x (y
n
y
m
)[
2
= 2[x y
n
[
2
2[x y
m
[
2
4
_
_
_
_
x
_
y
n
y
m
2
__
_
_
_
2
_ 2[x y
n
[
2
2[x y
m
[
2
4
2
.
45
2011 FUNCTIONAL ANALYSIS ALP
since
y
n
y
m
2
K by convexity of K. Thus,
[y
m
y
n
[
2
_ 2[x y
n
[
2
2[x y
m
[
2
4
2
0 as n. m o.
That is, (y
n
)
1
1
is a Cauchy sequence in K. Since K is complete, there exists y K such that y
n

y as n o. Since the norm is continuous,
[x y[ = [x lim
n!1
y
n
[ = [ lim
n!1
(x y
n
)[ = lim
n!1
[x y
n
[ = .
Thus,
[x y[ = = d(x. K).
Uniqueness: Assume that y. y
0
K are two best approximations to x from K. That is,
[x y
0
[ = [x y[ = = d(x. K).
By the Parallelogram Identity,
0 _ [y y
0
[
2
= [(y x) (x y
0
)[
2
= 2[x y[
2
2[x y
0
[
2
[2x (y y
0
)[
2
= 2
2
2
2
4
_
_
_
_
x
_
y y
0
2
__
_
_
_
2
_ 4
2
4
2
= 0.
Thus, y
0
= y.
3.4.2 Corollary
Every nonempty closed convex subset of a Hilbert space is Chebyshev.
The following theorem characterizes best approximations from a closed convex subset of a Hilbert
space.
3.4.2 Theorem
Let K be a nonempty closed convex subset of a Hilbert space (H. (. )), x H\ K and y
0
K. Then y
0
is the best approximation to x from K if and only if
Re(x y
0
. y y
0
) _ 0 for all y K.
Proof. The existence and uniqueness of the best approximation to x in K are guaranteed by Theorem 3.4.1.
Let y
0
be the best approximation to x in K. Then, for any y K and any 0 < z < 1, zy (1 z)y
0
K
since K is convex. Thus,
[x y
0
[
2
_ [x zy (1 z)y
0
|[
2
= [(x y
0
) z(y y
0
)[
2
= (x y
0
) z(y y
0
). x y
0
) z(y y
0
))
= (x y
0
. x y
0
) z(x y
0
. y y
0
) (y y
0
. x y
0
)|
z
2
(y y
0
. y y
0
)
= [x y
0
[
2
2zRe((x y
0
. y y
0
)) z
2
[y y
0
[
2
= 2zRe((x y
0
. y y
0
)) _ z
2
[y y
0
[
2
= Re((x y
0
. y y
0
)) _
z
2
[y y
0
[
2
.
46
2011 FUNCTIONAL ANALYSIS ALP
As z 0,
z
2
[y y
0
[
2
0, and consequently Re(x y
0
. y y
0
) _ 0.
Conversely, assume that for each y K, Re(x y
0
. y y
0
) _ 0. Then, for any y K,
[x y[
2
= [(x y
0
) (y y
0
)[
2
= ((x y
0
) (y y
0
). (x y
0
) (y y
0
))
= (x y
0
. x y
0
) (x y
0
. y y
0
) (y y
0
. x y
0
) (y y
0
. y y
0
)
= (x y
0
. x y
0
) (x y
0
. y y
0
) (y y
0
. x y
0
)| (y y
0
. y y
0
)
= (x y
0
. x y
0
) (x y
0
. y y
0
) (x y
0
. y y
0
)| (y y
0
. y y
0
)
= [x y
0
[
2
2Re(x y
0
. y y
0
) [y y
0
[
2
_ [x y
0
[
2
.
Taking the positive square root both sides, we have that [x y
0
[ _ [x y[ for all y K.
As a corollary to Theorem 3.4.2, one gets the following characterization of best approximations from a
closed linear subspace of a Hilbert space.
3.4.3 Corollary
(Characterization of Best Approximations from closed subspaces). Let M be a closed subspace of a
Hilbert space H and let x H\ M. Then an element y
0
M is the best approximation to x from M if
and only if (x y
0
. y) = 0 for all y M (i.e., x y
0
M
?
).
Corollary 3.4.3 says that if M is a closed linear subspace of a Hilbert space H, then y
0
= P
M
(x)
(i.e., y
0
is the best approximation to x from M) if and only if x P
M
(x) J M. That is, the unique best
approximation is obtained by dropping the perpendicular from x onto M. It is for this reason that the
map P
M
: x P
M
(x) is also called the orthogonal projection of H onto M.
x
M
P
M
x
3.4.4 Example
Let X = C
2
1. 1|, M = P
2
= lin{1. t. t
2
], and x(t ) = t
3
. Find P
M
(x).
Solution. Note that C
2
1. 1| is an incomplete inner product space. Since M is nite-dimensional, it
is complete, and consequently proximinal in C
2
1. 1|. Uniqueness of best approximations follows
from the Parallelogram Identity.
47
2011 FUNCTIONAL ANALYSIS ALP
Let y
0
=
2

iD0

i
t
i
M. By Corollary 3.4.3,
y
0
= P
M
(x) x y
0
M
?
(x y
0
. t
j
) = 0 for all j = 0. 1. 2

_
t
3

iD0

i
t
i
. t
j
_
= 0 for all j = 0. 1. 2

iD0

i
(t
i
. t
j
) = (t
3
. t
j
. ) for all j = 0. 1. 2

iD0

i
1
_
1
t
i
t
j
dt =
1
_
1
t
3
t
j
dt for all j = 0. 1. 2

iD0

i
1
_
1
t
iCj
dt =
1
_
1
t
3Cj
dt for all j = 0. 1. 2

iD0

i
t
iCjC1
i j 1
_1
1
=
t
jC4
j 4
_1
1
for all j = 0. 1. 2

iD0

i
1
i j 1
_
1 (1)
iCjC1
_
=
1
j 4
_
1 (1)
jC4
_
for all j = 0. 1. 2

_
_
_
2
0
0
1

2
3

2
= 0
0
0

2
3

1
0
2
=
2
5
2
3

0
0
1

2
5

2
= 0

0
= 0.
1
=
3
5
.
2
= 0.
Thus, P
M
(x) = y
0
=
3
5
t .
3.4.3 Theorem
(Projection Theorem). Let H be a Hilbert space, M a closed subspace of H. Then
[1] H = M M
?
. That is, each x H can be uniquely decomposed in the form
x = y z with y M and z M
?
.
[2] M = M
??
.
Proof.
[1] If x M, then x = x 0, and we are done. Assume that x , M. Let y = P
M
(x) be the unique
best approximation to x from M as advertised in Theorem 3.4.1. Then z = x P
M
(x) M
?
, and
x = P
M
(x) (x P
M
(x)) = y z
is the unique representation of x as a sum of an element of M and an element of M
?
.
48
2011 FUNCTIONAL ANALYSIS ALP
[2] Since the containment M M
??
is clear, we only show that M
??
M. To that end, let x
M
??
. Then by [1] above
x = y z. where y M and z M
?
.
Since M M
??
and M
??
is a subspace, z = x y M
??
. But z M
?
implies that
z M
?
M
??
which, in turn, implies that z = 0. Thus, x = y M.

3.4.5 Corollary
If M is a closed subspace of a Hilbert space H, and if M ,= H, then there exists z H \ {0] such that
z J M.
Proof. Let x H\ M. Then by the Projection Theorem,
x = y z. where y M and z M
?
.
Hence z ,= 0 and z J M.
3.4.6 Proposition
Let S be a nonempty subset of a Hilbert space H. Then
[1] S
??
= linS.
[2] S
?
= {0] if and only if linS = H.
Proof.
[1] Since S
?
= (linS)
?
by Proposition 3.3.2, we have, by the Projection Theorem, that
linS = (linS)
??
= S
??
.
[2] If S
?
= {0], then by [1]
linS = S
??
= {0]
?
= H.
On the other hand, if H = linS, then H = S
??
by [1], and so
S
?
= S
???
= H
?
= {0].
3.5 Orthonormal Sets and Orthonormal Bases
In this section we extend to Hilbert spaces the nite-dimensional concept of an orthonormal basis.
3.5.1 Denition
Let (X. (. )) be an inner product space over F. A set S = {x

: ] of elements of X is called an
orthonormal set if
(a) (x

. x

) = 0 for all ,= (i.e., S is an orthogonal set), and

(b) [x

[ = 1 for all .
If S = {x

: ] is an orthonormal set and x X, then the numbers (x. x

) are called the Fourier

coefcients of x with respect to S and the formal series

2
(x. x

)x

the Fourier series of x.

49
2011 FUNCTIONAL ANALYSIS ALP
3.5.1 Theorem
An orthonormal set S in a separable inner product space (X. (. )) is at most countable.
Proof. If S is nite, then there is nothing to prove. Assume that S is innite. Observe that if x. y S,
then [x y[ =
_
2 (since x and y are orthonormal). Let D = {y
n
[ n N] be a countable dense subset
of X. Then to each x S corresponds an element y
n
D such that [x y
n
[ <
p
2
4
. This denes a map
: S N given by (x) = n, where n corresponds to the y
n
as indicated above. Now, if x and y are
distinct elements of S, then there are distinct elements y
n
and y
m
in D such that
[x y
n
[ <
_
2
4
and [y y
m
[ <
_
2
4
.
Hence,
_
2 = [x y[ _ [x y
n
[ [y
n
y
m
[ [y
m
y[ <
_
2
2
[y
n
y
m
[
_
2
2
< [y
n
y
m
[.
and so y
n
= y
m
. In particular, n = m. Thus, we have a one-to-one correspondence between the elements
of S and a subset of N.
3.5.2 Denition
An orthonormal set S in an inner product space (X. (. )) is said to be complete in X if S T and T is
an orthonormal set in X, then S = T.
Simply put, a complete orthonormal set S in an inner product space is an orthonormal set that is not
properly contained in any other orthonormal set in X; in other words, S is complete if it is a maximal
orthonormal set in X.
It is easy exercise to showthat a set S is complete in an inner product (X. (. )) if and only if S
?
= {0].
3.5.3 Examples
[1] In R
3
the set S = {(1. 0. 0). (0. 1. 0). (0. 0. 1)] is orthonormal.
[2] In
2
, let S = {e
n
: n N], where e
n
= (
1n
.
2n
. . . .) with

ij
=
_
1 if i = j
0 otherwise.
Then S is an orthonormal set. Furthermore, for each x = (x
i
)
1
iD1

2
, (x. e
n
) = x
n
for all n.
Thus
(x. e
n
) = 0 for all n x
n
= 0 for all n x = 0.
That is, S
?
= {0], hence, S is complete in
2
.
3.5.2 Theorem
Let (X. (. )) be a separable inner product space over F.
[1] (Best Fit). If {x
1
. x
2
. . . . . x
n
] is a nite orthonormal set in X and M = lin{x
1
. x
2
. . . . . x
n
], then for
each x X there exists y
0
M such that
[x y
0
[ = d(x. M).
In fact, y
0
=
n

kD1
(x. x
k
)x
k
.
50
2011 FUNCTIONAL ANALYSIS ALP
[2] (Bessels Inequality). Let (x
n
)
1
nD1
be an orthonormal sequence in X. Then for any x X,
1

kD1
[(x. x
k
)[
2
_ [x[
2
.
In particular, (x. x
k
) 0 as k o.
Proof.
[1] For any choice of scalars z
1
. z
2
. . . . . z
n
,
_
_
_
_
_
x
n

kD1
z
k
x
k
_
_
_
_
_
2
=
_
x
n

iD1
z
i
x
i
. x
n

jD1
z
j
x
j
_
= [x[
2

iD1
z
i
(x
i
. x)
n

jD1
z
j
(x. x
j
)
n

iD1
z
i
z
i
= [x[
2

iD1
z
i
(x. x
i
)
n

jD1
z
j
(x. x
j
)
n

iD1
z
i
z
i
= [x[
2

iD1
z
i
z
i
z
i
(x. x
i
) z
i
(x. x
i
) (x. x
i
)(x. x
i
) |

iD1
(x. x
i
)(x. x
i
)
= [x[
2

iD1
(z
i
(x. x
i
))(z
i
(x. x
i
)) |
n

iD1
[(x. x
i
)[
2
= [x[
2

iD1
(z
i
(x. x
i
))(z
i
(x. x
i
)) |
n

iD1
[(x. x
i
)[
2
= [x[
2

iD1
[(x. x
i
)[
2

iD1
[z
i
(x. x
i
)[
2
.
Therefore,
_
_
_
_
_
x
n

kD1
z
k
x
k
_
_
_
_
_
2
is minimal if and only if z
k
= (x. x
k
) for each k = 1. 2. . . . . n.
[2] For each positive integer n, and with z
k
= (x. x
k
), the above argument shows that
0 _
_
_
_
_
_
x
n

kD1
z
k
x
k
_
_
_
_
_
2
= [x[
2

iD1
[(x. x
i
)[
2
.
Thus,
n

kD1
[(x. x
k
)[
2
_ [x[
2
.
Taking the limit as n o, we get
1

kD1
[(x. x
k
)[
2
_ [x[
2
.
51
2011 FUNCTIONAL ANALYSIS ALP
3.5.3 Theorem
(Riesz-Fischer Theorem). Let (x
n
)
1
1
be an orthonormal sequence in a separable Hilbert space H and let
(c
n
)
1
1
be a sequence of scalars. Then the series
1

kD1
c
k
x
k
converges in Hif and only if c = (c
n
)
1
1

2
. In
this case,
_
_
_
_
_
1

kD1
c
k
x
k
_
_
_
_
_
=
_
1

kD1
[c
k
[
2
_1
2
.
Proof. Assume that the series
1

kD1
c
k
x
k
converges to x. Then for each j. n N,
_
n

kD1
c
k
x
k
. x
j
_
=
n

kD1
c
k
(x
k
. x
j
) = c
j
.
Using continuity of the inner product
(x. x
j
) =
_
1

kD1
c
k
x
k
. x
j
_
= lim
n!1
_
n

kD1
c
k
x
k
. x
j
_
= lim
n!1
c
j
= c
j
.
By Bessels Inequality, we have that
1

kD1
[c
k
[
2
=
1

kD1
[(x. x
k
)[
2
_ [x[
2
< o.
That is, c = (c
n
)
1
1

2
.
Conversely, assume that c = (c
n
)
1
1

2
. Set z
n
=
n

kD1
c
k
x
k
. Then for 1 _ n _ m,
[z
n
z
m
[
2
=
_
_
_
_
_
_
m

kDnC1
c
k
x
k
_
_
_
_
_
_
2
=
m

kDnC1
[c
k
[
2
0 as n o.
Hence, (z
n
)
1
1
is a Cauchy sequence in H. Since H is complete the sequence (z
n
)
1
1
converges to some
x H. Hence the series
1

kD1
c
k
x
k
converges to some element in H.
Also,
_
_
_
_
_
1

kD1
c
k
x
k
_
_
_
_
_
2
= lim
n!1
_
_
_
_
_
n

kD1
c
k
x
k
_
_
_
_
_
2
= lim
n!1
n

kD1
[c
k
[
2
.
whence,
_
_
_
_
_
1

kD1
c
k
x
k
_
_
_
_
_
=
_
1

kD1
[c
k
[
2
_1
2
.
Note that Bessels Inequality says that
1

kD1
[(x. x
k
)[
2
_ [x[
2
< o.
52
2011 FUNCTIONAL ANALYSIS ALP
That is, ((x. x
n
))
1
1

2
. Hence, by Theorem 3.5.3, the series
1

kD1
(x. x
k
)x
k
converges. There is however
no reason why this series should converge to x. In fact, the following example shows that this series may
not converge to x.
3.5.4 Example
Let (e
n
)
2
, where e
n
= (
1n
.
2n
. . . .) with

ij
=
_
1 if i = j
0 otherwise.
For each n N, let
n
= e
nC1
. Then (
n
)
1
nD1
is an orthonormal sequence in
2
. For any x =
(x
n
)
1
1

2
,
1

kD1
(x.
k
)
k
=
1

kD1
(x. e
kC1
)e
kC1
= (0. x
2
. x
3
. . . .) ,= (x
1
. x
2
. x
3
. . . .) = x.
3.5.5 Denition
Let (X. (. )) be an inner product space over F. An orthonormal set {x
n
] is called an orthonormal basis
for X if for each x X,
x =
1

kD1
(x. x
k
)x
k
.
That is, the sequence of partial sums (s
n
), where s
n
=
n

kD1
(x. x
k
)x
k
, converges to x.
3.5.4 Theorem
Let Hbe a separable innite-dimensional Hilbert space and assume that S = {x
n
] is an orthonormal set in
H. Then the following statements are equivalent:
[1] S is complete in H; i.e., S
?
= {0].
[2] linS = H; i.e., the linear span of S is norm-dense in H.
[3] (Fourier Series Expansion.) For any x H, we have
x =
1

iD1
(x. x
i
)x
i
.
That is, S is an orthonormal basis for H.
[4] (Parsevals Identity.) For all x. y H,
(x. y) =
1

kD1
(x. x
k
)(y. x
k
).
[5] For any x H,
[x[
2
=
1

kD1
[(x. x
k
)[
2
.
53
2011 FUNCTIONAL ANALYSIS ALP
Proof. [1] [2]. This equivalence was proved in Proposition 3.4.6[2].
[1] =[3]. Let x H and s
n
=
n

iD1
(x. x
i
)x
i
. Then for all n > m,
[s
n
s
m
[
2
=
_
_
_
_
_
n

iDmC1
(x. x
i
)x
i
_
_
_
_
_
2
=
n

mC1
[(x. x
i
)[
2
_ [x[
2
.
Thus, (s
n
) is a Cauchy sequence in H. Since H is complete, this sequence converges to some element
which we denote by
1

iD1
(x. x
i
)x
i
. We show that x =
1

iD1
(x. x
i
)x
i
. Indeed, for each xed j N,
_
x
1

iD1
(x. x
i
)x
i
. x
j
_
=
_
x lim
n!1
n

iD1
(x. x
i
)x
i
. x
j
_
= lim
n!1
_
x
n

iD1
(x. x
i
)x
i
. x
j
_
= lim
n!1
_
(x. x
j
)
n

iD1
(x. x
i
)(x
i
. x
j
)
_
= lim
n!1
((x. x
j
) (x. x
j
)) = 0.
Thus, by [1], x
1

iD1
(x. x
i
)x
i
= 0, whence
x =
1

iD1
(x. x
i
)x
i
.
[3] =[4]. Let x. y H. Then
(x. y) = lim
n!1
_
n

iD1
(x. x
i
)x
i
.
n

jD1
(y. x
j
)x
j
_
= lim
n!1
n

iD1
n

jD1
(x. x
i
)(y. x
j
)(x
i
. x
j
)
= lim
n!1
n

iD1
(x. x
i
)(y. x
i
) =
1

iD1
(x. x
i
)(y. x
i
).
[4] =[5]. Take x = y in [4].
[5] =[1]. Since [x[
2
=

k
(x. x
k
)(x. x
k
), if x J S then (x. x
k
) = 0 for all k. Thus, [x[
2
= 0,
whence x = 0. That is, S
?
= {0].
3.5.6 Examples
[1] In
2
, the set S = {e
n
: n N], where e
n
= (
1n
.
2n
. . . .) with

ij
=
_
1 if i = j
0 otherwise
is an orthonormal basis for
2
.
54
2011 FUNCTIONAL ANALYSIS ALP
[2] In L
2
. |, the set
_
1
_
2
e
int
: n Z
_
is an orthonormal basis for the complex L
2
. |.
[3] The set S =
_
1
_
2
.
cos nt
_

.
sin nt
_

_
1
nD1
is an orthonormal basis for the real L
2
. |.
Hence, if x L
2
. |, then by Theorem 3.5.4[3],
x(t ) =
_
x(t ).
1
_
2
_
1
_
2

nD1
_ _
x(t ).
cos nt
_

_
cos nt
_

_
x(t ).
sin nt
_

_
sin nt
_

_
=
1
2
(x(t ). 1)
1

nD1
_
1

(x(t ). cos nt ) cos nt

1

(x(t ). sin nt) sin nt

_
=
1
2
t
_
t
x(t ) dt

nD1
_
_
_
_
1

t
_
t
x(t ) cos nt dt
_
_
cos nt
_
_
1

t
_
t
x(t ) sin nt dt
_
_
sin nt
_
_
= a
0

nD1
(a
n
cos nt b
n
sin nt ).
where
a
0
=
1
p
2t
_
x(t ).
1
p
2t
_
=
1
2t
_
t
t
x(t ) dt.
a
n
=
1
p
t
_
x(t ).
cos nt
p
t
_
=
1
t
_
t
t
x(t ) cos nt dt. and
b
n
=
1
p
t
_
x(t ).
sin nt
p
t
_
=
1
t
_
t
t
x(t ) sin nt dt
_

_
n = 1. 2. . . . .
That is, the Fourier series expansion of x is
x(t ) = a
0

nD1
(a
n
cos nt b
n
sin nt ). (3.5.6.1)
It is clear from above that for all n = 1. 2. . . .,
2[a
0
[
2
=

_
x(t ).
1
_
2
_

2
. [a
n
[
2
=

_
x(t ).
cos nt
_

2
. and
[b
n
[
2
=

_
x(t ).
sin nt
_

2
.
55
2011 FUNCTIONAL ANALYSIS ALP
By Theorem 3.5.4 [5] we have that
t
_
t
[x(t )[
2
dt = [x[
2
=

_
x(t ).
1
_
2
_

nD1
_

_
x(t ).
cos nt
_

_
x(t ).
sin nt
_

2
_
= 2[a
0
[
2

nD1
([a
n
[
2
[b
n
[
2
)
=
_
2[a
0
[
2

nD1
([a
n
[
2
[b
n
[
2
)
_
.
We now apply the above results to a particular function: Let x(t ) = t . Then
a
0
=
1
2
t
_
t
t dt = 0 since x(t ) = t is an odd function.
For n = 1. 2. . . . . a
n
=
1

t
_
t
t cos nt dt = 0 since t cos nt is an odd function,
b
n
=
1

t
_
t
t sin nt dt =
2

t
_
0
t sin nt dt
=
2

_
_
t cos nt
n

t
0

1
n
t
_
0
cos nt dt
_
_
=
2

n
cos n
_
=
2(1)
nC1
n
.
Hence, by Theorem 3.5.4[3],
x(t ) =
1

nD1
2(1)
nC1
n
sin nt =
1

nD1
2(1)
nC1
_

n
sin nt
_

.
It now follows that
2(1)
nC1
_

n
=
_
t.
sin nt
_

_
.
Now,
[x[
2
2
=
t
_
t
t
2
dt = 2
t
_
0
t
2
dt =
2
3
t
3

t
0
=
2
3
3
.
Also, by Theorem 3.5.4[5],
[x[
2
2
=
1

nD1

_
t.
sin nt
_

2
=
1

nD1

2(1)
nC1
_

2
=
1

nD1
4
n
2
.
56
2011 FUNCTIONAL ANALYSIS ALP
Thus,
1

nD1
1
n
2
=

2
6
.
We can express the Fourier Series Expansion (3.5.6.1) of x L
2
. | in exponential form. Recall
that
e
i0
= cos 0 i sin 0 (Eulers Formula).
Therefore
cos 0 =
e
i0
e
i0
2
and sin 0 =
e
i0
e
i0
2i
.
Equation (3.5.6.1) now becomes
x(t ) = a
0

nD1
(a
n
cos nt b
n
sin nt )
= a
0

nD1
_
a
n
_
e
int
e
int
2
_
b
n
_
e
int
e
int
2i
__
= a
0

nD1
__
a
n
i b
n
2
_
e
int

_
a
n
i b
n
2
_
e
int
_
= a
0

nD1
_
a
n
i b
n
2
_
e
int

nD1
_
a
n
i b
n
2
_
e
int
. (3.5.6.2)
For each n = 1. 2. 3. . . ., let c
n
=
1
2
(a
n
i b
n
). Then c
n
=
1
2
(a
n
i b
n
) for each n = 1. 2. 3. . . ., and
so equation (3.5.6.2) becomes
x(t ) = a
0

nD1
c
n
e
int

nD1
c
n
e
int
. (3.5.6.3)
Re-index the rst sum in (3.5.6.3) by letting n = k. Then
x(t ) = a
0

kD1
c
k
e
ikt

nD1
c
n
e
int
. (3.5.6.4)
For n = 1. 2. 3. . . . . dene
c
n
= c
n
and let c
0
= a
0
. The we can rewrite equation (3.5.6.4) as
x(t ) =
1

1
c
n
e
int
. (3.5.6.5)
This is the complex exponential form of the Fourier Series of x L
2
. |.
Note that,
c
0
= a
0
=
1
2
t
_
t
x(t )dt
57
2011 FUNCTIONAL ANALYSIS ALP
and for n = 1. 2. 3. . . . ,
c
n
=
1
2
(a
n
i b
n
) =
1
2
_
_
1

t
_
t
x(t ) cos nt dt i
1

t
_
t
x(t ) sin nt dt
_
_
=
1
2
t
_
t
x(t ) (cos nt i sin nt ) dt
=
1
2
t
_
t
x(t )e
int
dt.
and
c
n
= c
n
=
1
2
t
_
t
x(t )e
int
dt =
1
2
t
_
t
x(t )e
int
dt =
1
2
t
_
t
x(t )e
int
dt.
Therefore, for all n Z,
c
n
=
1
2
t
_
t
x(t )e
int
dt.
Now, for n = 1. 2. 3. . . . ,
[c
n
[
2
= c
n
c
n
=
1
2
(a
n
i b
n
)
1
2
(a
n
i b
n
)
=
1
4
_
a
2
n
b
2
n
_
=
1
4
_
[a
n
[
2
[b
n
[
2
_
.
Therefore
1

nD1
[c
n
[
2
=
1
4
1

nD1
_
[a
n
[
2
[b
n
[
2
_
. (3.5.6.6)
Since for n = 1. 2. 3. . . . , c
n
= c
n
, it follows that
[c
n
[
2
= c
n
c
n
= c
n
c
n
= c
n
c
n
= [c
n
[
2
.
Hence, for n = 1. 2. 3. . . . ,
1

nD1
[c
n
[
2
=
1

nD1
[c
n
[
2
=
1
4
1

nD1
_
[a
n
[
2
[b
n
[
2
_
. (3.5.6.7)
From (3.5.6.6) and (3.5.6.7), we have that

n2Z
[c
n
[
2
=
1

nD1
[c
n
[
2
[c
0
[
2

nD1
[c
n
[
2
=
1
4
1

nD1
_
[a
n
[
2
[b
n
[
2
_
[a
0
[
2

1
4
1

nD1
_
[a
n
[
2
[b
n
[
2
_
= [a
0
[
2

1
2
1

nD1
_
[a
n
[
2
[b
n
[
2
_
=
1
2

_
2 [a
0
[
2

nD1
_
[a
n
[
2
[b
n
[
2
_
_
=
1
2
t
_
t
[x(t )[
2
dt.
58
2011 FUNCTIONAL ANALYSIS ALP
That is,

n2Z
[c
n
[
2
=
1
2
t
_
t
[x(t )[
2
dt. (3.5.6.8)
Let {x
1
. x
2
. . . . . x
n
] be a basis of an n-dimensional linear subspace M of an inner product space
(X. (. )). We have seen in Theorem 3.5.2 that if the set {x
1
. x
2
. . . . . x
n
] is orthonormal, then the orthog-
onal projection (=best approximation) of any x X onto M is given by
P
M
(x) =
n

kD1
(x. x
k
)x
k
.
It is clearly easy to compute orthogonal projections from a linear subspace that has an orthonormal basis:
the coefcients in the orthogonal projection of x X are just the Fourier coefcients of x. If the basis of
M is not orthogonal, it may be advantageous to nd an orthonormal basis for M and express the orthogonal
projection as a linear combination of the new orthonormal basis. The process of nding an orthonormal
basis from a given (non-orthonormal) basis is known as the Gram-Schmidt Orthonormalisation Procedure.
3.5.5 Theorem
(Gram-Schmidt Orthonormalisation Procedure). If {x
k
]
1
1
is a linearly independent set in an inner
product space (X. (. )) then there exists an orthonormal set {e
k
]
1
1
in X such that
lin{x
1
. x
2
. . . . . x
n
] = lin{e
1
. e
2
. . . . . e
n
] for all n.
Proof. Set e
1
=
x
1
[x
1
[
. Then lin{x
1
] = lin{e
1
]. Next, let y
2
= x
2
(x
2
. e
1
)e
1
. Then
(y
2
. e
1
) = (x
2
(x
2
. e
1
)e
1
. e
1
) = (x
2
. e
1
) (x
2
. e
1
)(e
1
. e
1
) = (x
2
. e
1
) (x
2
. e
1
) = 0.
That is, e
1
J y
2
. Set e
2
=
y
2
[y
2
[
. Then {e
1
. e
2
] is an orthonormal set with the property that lin{x
1
. x
2
] =
lin{e
1
. e
2
]. In general, for each k = 2. 3. . . ., we let
y
k
= x
k

k1

iD1
(x
k
. e
i
)e
i
.
Then for k = 2. 3. . . .
(y
k
. e
1
) = (y
k
. e
2
) = (y
k
. e
3
) = = (y
k
. e
k1
) = 0.
Set e
k
=
y
k
[y
k
[
. Then {e
1
. e
2
. . . . . e
k
] is an orthonormal set in X with the property that
lin{e
1
. e
2
. . . . . e
k
] = lin{x
1
. x
2
. . . . . x
k
].
We have made the point that
2
is a Hilbert space. In this nal part of this chapter we want to show that
every separable innite-dimensional Hilbert space looks like
2
in the sense dened below.
3.5.7 Denition
Two linear spaces X and Y over the same eld F are said to be isomorphic it there is a one-to-one map T
from X onto Y such that for all x
1
. x
2
X and all . F,
T(x
1
x
2
) = T(x
1
) T(x
2
). (3.5.7.1)
59
2011 FUNCTIONAL ANALYSIS ALP
3.5.8 Remark
Any map that satises condition (3.5.7.1) of Denition 3.5.7 is called a linear operator. Chapter 4
is devoted to the study of such maps. Clearly, the linear structures of the two linear spaces X and
Y are preserved under the map T .
3.5.9 Denition
Let (X. [ [) and (Y. [ [) be two normed linear spaces and T : X Y . Then T is called an isometry if
[Tx[ = [x[ for all x X. (3.5.9.1)
Simply put, an isometry is a map that preserves lengths.
3.5.10 Remark
It is implicit in the above denition that the norm on the left of equation (3.5.9.1) is in Y and that on
the right is in X. In order to avoid possible confusion, we should perhaps have labelled the norms
as [ [
X
and [ [
Y
for the norms in X and Y respectively. This notation is however cumbersome
and will therefore be avoided.
Normed linear spaces that are isometrically isomorphic are essentially identical.
3.5.11 Lemma
Let M = lin{x
n
] be a linear subspace of X. Then there exists a subsequence {x
n
k
] of {x
n
] which has the
following properties:
(i) lin{x
n
k
] = M;
(ii) {x
n
k
] is linearly independent.
Proof. We dene the subsequence inductively as follows: Let x
n
1
be the rst nonzero element of the
sequence {x
n
]. Therefore x
n
= 0 x
n
1
for all n < n
1
. If there is an F such that x
n
= x
n
1
for all
n > n
1
, then we are done. Otherwise, let x
n
2
be the rst element of the sequence {x
n
]
n>n
1
that is not a
multiple of x
n
1
. Thus there is an F such that x
n
= x
n
1
0x
n
2
for all n < n
2
. If x
n
= x
n
1
x
n
2
for some . F and all n > n
2
then we are done. Otherwise let x
n
3
be the rst element of the sequence
{x
n
] which is not a linear combination of x
n
1
and x
n
2
. Then x
n
= x
n
1
x
n
2
0x
n
3
for all n < n
3
. If
x
n
= x
n
1
x
n
2
;x
n
3
for all n > n
3
, then we are done. Otherwise let x
n
4
be the rst element of the
sequence {x
n
] that is not in lin{x
n
1
. x
n
2
. x
n
3
]. Continue in this fashion to obtain elements x
n
1
. x
n
2
. . . ..
If x lin{x
1
. x
2
. . . . . x
n
], then x lin{x
n
1
. x
n
2
. . . . . x
nr
] for r sufciently large. That is lin{x
n
k
] = M.
The subsequence {x
n
k
] is, by its construction, linearly independent.
3.5.6 Theorem
Every separable Hilbert space Hhas a countable orthonormal basis.
Proof. By Theorem 2.7.1 there is a set {x
n
[ n N] such that lin{x
n
[ n N] = H. Using Lemma 3.5.11
extract from {x
n
[ n N] a linearly independent subsequence {x
n
k
] such that lin{x
n
] =lin{x
n
k
]. Apply
the Gram-Schmidt Orthonormalisation Procedure to the subsequence {x
n
k
] to obtain an orthonormal basis
for H.
3.5.7 Theorem
Every separable innite-dimensional Hilbert space H is isometrically isomorphic to
2
.
Proof. Let {x
n
[ n N] be an orthonormal basis for H. Dene T : H
2
by
Tx = ((x. x
n
))
n2N
for each x H.
It follows from Bessels Inequality that the right hand side is in
2
. We must show that T is a surjective
linear isometry. (One-to-oneness follows from isometry.)
60
2011 FUNCTIONAL ANALYSIS ALP
(i) T is linear: Let x. y H and z F. Then
T(x y) = ((x y. x
n
))
n2N
= ((x. x
n
) (y. x
n
))
n2N
= ((x. x
n
))
n2N
((y. x
n
))
n2N
= T x Ty.
and
T(zx) = ((zx. x
n
))
n2N
= (z(x. x
n
))
n2N
= z ((x. x
n
))
n2N
.
(ii) T is surjective: Let (c
n
)
n2N

2
. By the Riesz-Fischer Theorem (Theorem 3.5.3), the series
1

kD1
c
k
x
k
converges to some x H. By continuity of the inner product, we have that for each
j N,
(x. x
j
) = lim
n!1
_
n

kD1
c
k
x
k
. x
j
_
= lim
n!1
c
j
= c
j
.
Hence, Tx = ((x. x
n
))
n2N
= (c
n
)
n2N
.
(iii) T is an isometry: For each x H,
[Tx[
2
2
=

n2N
[(x. x
n
)[
2
= [x[
2
.
where the second equality follows from the fact that {x
n
]
n2N
is an orthonormal basis and Theo-
rem 3.5.4[5].
61
Chapter 4
Bounded Linear Operators and
Functionals
4.1 Introduction
An essential part of functional analysis is the study of continuous linear operators acting on linear spaces.
This is perhaps not surprising since functional analysis arose due to the need to solve differential and
integral equations, and differentiation and integration are well known linear operators. It turns out that it is
advantageous to consider this type of operators in this more abstract way. It should also be mentioned that
in physics, operator means a linear operator from one Hilbert space to another.
4.1.1 Denition
Let X and Y be linear spaces over the same eld F. A linear operator from X into Y is a mapping
T : X Y such that
T(x
1
x
2
) = Tx
1
Tx
2
for all x
1
. x
2
X and all . F.
Simply put, a linear operator between linear spaces is a mapping that preserves the structure of the under-
lying linear space.
We shall denote by L(X. Y ) the set of all linear operators fromX into Y . We shall write L(X) for L(X. X).
4.1.2 Exercise
Let X and Y be linear spaces over the same eld F. Show that if T is a linear operator from X
into Y , then T(0) = 0.
The range of a linear operator T : X Y is the set
ran(T ) = {y Y [ y = T x for some x X] = TX.
and the null space or the kernel of T L(X. Y ) is the set
N(T ) = ker(T ) = {x X : Tx = 0] = T
1
(0).
If T L(X. Y ), then ker(T) is a linear subspace of X and ran(T ) is a linear subspace of Y .
An operator T L(X. Y ) is one-to-one (or injective) if ker(T) = {0] and onto (or surjective) if
ran(T ) = Y . If T L(X. Y ) is one-to-one, then there exists a map T
1
: ran(T) dom(T ) which
maps each y ran(T ) onto that x dom(T ) for which T x = y. In this case we write T
1
y = x and
the map T
1
is called the inverse of the operator T L(X. Y ). An operator T : X Y is invertible if it
has an inverse T
1
.
It is easy exercise to show that an invertible operator can have only one inverse.
62
2011 FUNCTIONAL ANALYSIS ALP
4.1.3 Proposition
Let X and Y be linear spaces over F. Suppose that T L(X. Y ) is invertible. Then
(a) T
1
is also invertible and (T
1
)
1
= T.
(b) T T
1
= I
Y
and T
1
T = I
X
.
(c) T
1
is a linear operator.
Proof. We shall leave (a) and (b) as an easy exercise.
(c) Linearity of T
1
: Let x. y Y and z F. Then
T
1
(x y) = T
1
(T T
1
x T T
1
y) = T
1
T(T
1
x T
1
y)|
= T
1
T(T
1
x T
1
y) = T
1
x T
1
y.
Let X and Y be linear spaces over F. For all T. S L(X. Y ) and F, dene the operations of
addition and scalar multiplication as follows:
(T S)(x) = Tx Sx and
(T)(x) = Tx for each x X.
Then L(X. Y ) is a linear space over F.
The most important class of linear operators is that of bounded linear operators.
4.1.4 Denition
Let X and Y be normed linear spaces over the same eld F. A linear operator T : X Y is said to be
bounded if there exists a constant M > 0 such that
[Tx[ _ M[x[ for all x X.
(It should be emphasised that the norm on the left side is in Y and that on the right side is in X.)
An operator T : X Y is said to be continuous at x
0
X if given any c > 0 there is a > 0 such that
[Tx T x
0
[ < c whenever [x x
0
[ < .
T is continuous on X if it is continuous at each point of X.
We shall denote by B(X. Y ) the set of all bounded linear operators from X into Y . We shall write B(X)
for B(X. X).
4.1.5 Denition
Let X and Y be normed linear spaces over the same eld F and let T B(X. Y ). The operator norm (or
simply norm) of T , denoted by [T[, is dened as
[T[ = inf{M : [Tx[ _ M[x[. for all x X].
Since T is bounded, [T[ < o. Furthermore,
[Tx[ _ [T[[x[ for all x X.
4.1.1 Theorem
Let X and Y be normed linear spaces over a eld F and let T B(X. Y ). Then
[T[ = sup
_
[Tx[
[x[
: x ,= 0
_
= sup{[Tx[ : [x[ = 1] = sup{[Tx[ : [x[ _ 1].
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2011 FUNCTIONAL ANALYSIS ALP
Proof. Let = sup
_
[Tx[
[x[
: x ,= 0
_
. = sup{[Tx[ : [x[ = 1], and ; = sup{[Tx[ : [x[ _
1]. We rst show that [T[ = . Now, for all x X \ {0] we have that
[Tx[
[x[
_ , and therefore
[Tx[ _ [x[. By denition of [T[ we have that [T[ _ . On the other hand, for all x X,
we have that [Tx[ _ [T[[x[. In particular, for all x X \ {0],
[Tx[
[x[
_ [T[, and therefore
= sup
_
[Tx[
[x[
: x ,= 0
_
_ [T[. Thus, = [T[.
Next, we show that = = ;. Now, for each x X
_
[Tx[
[x[
: x ,= 0
_
=
__
_
_
_
T
_
x
[x[
__
_
_
_
: x ,= 0
_
{[Tx[ : [x[ = 1] {[Tx[ : [x[ _ 1].
Thus,
= sup
_
[Tx[
[x[
: x ,= 0
_
_ = sup{[Tx[ : [x[ = 1] _ ; = sup{[Tx[ : [x[ _ 1].
But for all x ,= 0
[Tx[
[x[
_ = [Tx[ _ [x[ _ for all x such that [x[ _ 1.
Therefore,
; = sup{[Tx[ : [x[ _ 1] _ .
That is, _ _ ; _ . Hence, = = ;.
4.1.2 Theorem
Let X and Y be normed linear spaces over a eld F. Then the function [ [ dened above is a norm on
B(X. Y ).
Proof. Properties N1 and N2 of a norm are easy to verify. We prove N3 and N4. Let T B(X. Y ) and
F.
N3. [T[ = sup{[Tx[ : [x[ = 1] = [[ sup{[Tx[ : [x[ = 1] = [[[T[.
N4. Let T. S B(X. Y ). Then for each x X,
[(T S)(x)[ = [Tx Sx[ _ [Tx[ [Sx[ _ ([T[ [S[)[x[.
Thus, [T S[ _ [T[ [S[.
4.1.6 Examples
[1] Let X = F
n
with the uniform norm [ [
1
. For x = (x
1
. x
2
. . . . . x
n
) F
n
, dene
T : F
n
F
n
by
Tx = T(x
1
. x
2
. . . . . x
n
) =
_
_
n

jD1

1j
x
j
.
n

jD1

2j
x
j
. . . . .
n

jD1

nj
x
j
_
_
.
It is easy to show that T is a linear operator on X. We show that T is bounded.
[Tx[
1
= sup
1in

jD1

ij
x
j

_ sup
1in
n

jD1
[
ij
[[x
j
[
_ sup
1in
n

jD1
[
ij
[ sup
1jn
[x
j
[ = M[x[
1
.
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2011 FUNCTIONAL ANALYSIS ALP
where M = sup
1in
n

jD1
[
ij
[. Hence, [T[ _ M.
We claim that [T[ = M. We need to show that [Tx[
1
_ M[x[
1
. To that end, choose
an index k such that
n

jD1
[
kj
[ = M = sup
1in
n

jD1
[
ij
[ and let x be the unit vector whose j-th
component is

kj
[
kj
[
. Then
[Tx[
1
= sup
1in

jD1

ij
x
j

jD1

kj
x
j

=
n

jD1
[
kj
[ = M[x[
1
.
Thus [T[ = sup
1in
n

jD1
[
ij
[.
[2] Let {x
n
[ n N] be an orthonormal set in a Hilbert space H. For (z
i
)
1
iD1

1
, dene
T : H H by
Tx =
1

iD1
z
i
(x. x
i
)x
i
.
Then T is a bounded linear operator on H. Linearity is an immediate consequence of the
inner product.
Boundedness:
[Tx[
2
=
_
_
_
_
_
1

iD1
z
i
(x. x
i
)x
i
_
_
_
_
_
2
=
1

iD1
[z
i
[
2
[(x. x
i
)[
2
[x
i
[
2
_ M
2
1

iD1
[(x. x
i
)[
2
. where M = sup
i2N
[z
i
[
_ M
2
[x[
2
by Bessels Inequality.
Thus, [Tx[ _ M[x[, and consequently [T[ _ M.
We show that [T[ = sup
i2N
[z
i
[. Indeed, for any c > 0, there exists z
k
such that [z
k
[ > M c.
Hence,
[T[ _ [Tx
k
[ = [z
k
x
k
[ = [z
k
[ > M c.
Since c is arbitrary, we have that [T[ _ M.
[3] Dene an operator L :
2

2
by
Lx = L((x
1
. x
2
. x
3
. . . .)) = (x
2
. x
3
. . . .).
The L is a bounded linear operator on
2
.
Linearity: Easy.
Boundedness: For all x = (x
1
. x
2
. x
3
. . . .)
2
,
[Lx[
2
2
=
1

iD2
[x
i
[
2
_
1

iD1
[x
i
[
2
= [x[
2
2
.
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2011 FUNCTIONAL ANALYSIS ALP
That is, L is a bounded linear operator and [L[ _ 1. We show that [L[ = 1. To that end,
consider e
2
= (0. 1. 0. 0. . . .)
2
. Then
[e
2
[
2
= 1 and Le
2
= (1. 0. 0. . . .) which implies that [Le
2
[
2
= 1.
Thus, [L[ = 1. The operator L is called the left-shift operator.
[4] Let BC0. o) be the linear space of all bounded continuous functions on the interval 0. o)
with the uniform norm [ [
1
. Dene T : BC0. o) BC0. o) by
(Tx) (t ) =
1
t
t
_
0
x(t)dt.
Then T is a bounded linear operator on BC0. o).
Linearity: For all x. y BC0. o) and all . F,
(T(x y)) (t ) =
1
t
t
_
0
(x y)(t)dt =
_
_
1
t
t
_
0
x(t)dt
_
_

_
_
1
t
t
_
0
y(t)dt
_
_
.
Boundedness: For each x BC0. o),
[Tx[
1
= sup
t
[(T x)(t )[ = sup
t

1
t
t
_
0
x(t)dt

_ sup
t
1
t
t
_
0
[x(t)[dt _
_
_
sup
t
1
t
t
_
0
dt
_
_
[x[
1
= [x[
1
.
[5] Let M be a closed subspace of a normed linear space X and Q
M
: X X,M the qoutient
map. Then Q
M
is bounded and [Q
M
[ = 1. Indeed, since [Q
M
(x)[ = [x M[ _ [x[, Q
M
is bounded and [Q
M
[ _ 1. But since Q
M
maps the open unit ball in X onto the open unit
ball in X,M, it follows that [Q
M
[ = 1.
[6] Let X = mat hcalP0. 1| - the set of polynomials on the interval 0. 1| with the uniform norm
[x[
1
= max
0t 1
[x(t )[. For each x X, dene T : X X by
Tx = x
0
(t ) =
dx
dt
(differentiation with respect to t ).
Linearity: For x. y X and all . F,
T(x y) = (x y)
0
(t ) = x
0
(t ) y
0
(t ) = Tx Ty.
T is not bounded: Let x
n
(t ) = t
n
. n N. Then
[x
n
[ = 1. Tx
n
= x
0
n
(t ) = nt
n1
. and [Tx
n
[ =
[Tx
n
[
[x
n
[
= n.
Hence T is unbounded.
4.1.3 Theorem
Let X and Y be normed linear spaces over a eld F. Then T L(X. Y ) is bounded if and only if T maps
a bounded set into a bounded set.
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2011 FUNCTIONAL ANALYSIS ALP
Proof. Assume that T is bounded. That is, there exists a constant M > 0 such that [Tx[ _ M[x[ for all
x X. If [x[ _ k for some constant k, then [Tx[ _ M[x[ _ kM. That is, T maps a bounded set into
a bounded set.
Now assume that T maps a bounded set into a bounded set. Then T maps the unit ball B = {x X :
[x[ _ 1] into a bounded set. That is, there exists a constant M > 0 such that [Tx[ _ M for all x B.
Therefore, for any nonzero x X,
[Tx[
[x[
=
_
_
_
_
T
_
x
[x[
__
_
_
_
_ M.
Hence, [Tx[ _ M[x[. That is, T is bounded.
4.1.7 Exercise
Show that the inverse of a bounded linear operator is not necessarily bounded.
4.1.8 Proposition
Let T B(X. Y ). Then T
1
exists and is bounded if and only if there is a constant K > 0 such that
[Tx[ _ K[x[ for all x X.
Proof. Assume that there is a constant K > 0 such that [Tx[ _ K[x[ for all x X. If x = 0, then
Tx = 0 and so T is one-to-one and hence T
1
exists. Also, given y ran(T ), let y = Tx for some
x X. Then
[T
1
y[ = [T
1
(T x)[ = [x[ _
1
K
[Tx[ =
1
K
[y[.
i.e., [T
1
y[ _
1
K
[y[ for all y Y . Thus T
1
is bounded.
Assume that T
1
exists and is bounded. Then for each x X,
[x[ = [T
1
(T x)[ _ [T
1
[[Tx[
1
[T
1
[
[x[ _ [Tx[ K[x[ _ [Tx[.
where K =
1
kT
1
k
.
The following theorem asserts that continuity and boundedness are equivalent concepts for linear oper-
ators.
4.1.4 Theorem
Let X and Y be normed linear spaces over a eld F and T L(X. Y ). The following statements are
equivalent:
(1) T is continuous on X;
(2) T is continuous at some point in X;
(3) T is bounded on X.
Proof. The implication (1) =(2) is obvious.
(2) =(3): Assume that T is continuous at x X, but T is not bounded on X. Then there is a sequence
(x
n
) in X such that [Tx
n
[ > n[x
n
[ for each n N. For each n N, let
y
n
=
x
n
n[x
n
[
x.
Then
[y
n
x[ =
1
n
0as n o:
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2011 FUNCTIONAL ANALYSIS ALP
i.e., y
n
n!1
x, but
[Ty
n
Tx[ =
[Tx
n
[
n[x
n
[
>
n[x
n
[
n[x
n
[
= 1.
That is, Ty
n
,Tx as n o, contradicting (2).
(3) =(1): Assume that T is bounded on X. Let (x
n
) be a sequence in X which converges to x X.
Then
[Tx
n
T x[ = [T(x
n
x)[ _ [T[[x
n
x[ 0 as n o.
Thus, T is continuous on X.
4.1.5 Theorem
Let (X. [ [) and (Y. [ [) be normed linear spaces with dim(X) < o and T : X Y be a linear
operator. Then T is continuous. That is, every linear operator on a nite-dimensional normed linear space
is automatically continuous.
Proof. Dene a new norm [ [
0
on X by
[x[
0
= [x[ [Tx[ for all x X.
Since X is nte-dimensional, the norms [ [
0
and [ [ on X are equivalent. Hence there are constants
and such that
[x[
0
_ [x[ _ [x[
0
for all x X.
Hence,
[Tx[ _ [x[
0
_
1

[x[ = K[x[.
where K =
1

. Therefore T is bounded.
4.1.9 Denition
Let X and Y be normed linear spaces over a eld F.
(1) A sequence (T
n
)
1
1
in B(X. Y ) is said to be uniformly operator convergent to T if
lim
n!1
[T
n
T[ = 0.
This is also referred to as convergence in the uniform topology or convergence in the operator
norm topology of B(X. Y ). In this case T is called the uniform operator limit of the sequence
(T
n
)
1
1
.
(2) A sequence (T
n
)
1
1
in B(X. Y ) is said to be strongly operator convergent to T if
lim
n!1
[T
n
x T x[ = 0 for each x X.
In this case T is called the strong operator limit of the sequence (T
n
)
1
1
.
Of course, if T is the uniform operator limit of the sequence (T
n
)
1
1
B(X. Y ), then T B(X. Y ).
On the other hand, the strong operator limit T of a sequence (T
n
)
1
1
B(X. Y ) need not be bounded in
general.
The following proposition asserts that uniform convergence implies strong convergence.
4.1.10 Proposition
If the sequence (T
n
)
1
1
in B(X. Y ) is uniformly convergent to T B(X. Y ), then it is strongly convergent
to T.
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2011 FUNCTIONAL ANALYSIS ALP
Proof. Since, for each x X, [T
n
x Tx[ = [(T
n
T )(x)[ _ [T
n
T[[x[, if [T
n
T[ 0 as
n o, then
[(T
n
T)(x)[ 0 as n o.
The converse of Proposition 4.1.10 does not hold.
4.1.11 Example
Consider the sequence (T
n
) of operators, where for each n N,
T
n
:
2

2
is given by
T
n
(x
1
. x
2
. . . .) = (0. 0. . . . . 0. x
nC1
. x
nC2
. . . .).
Let c > 0 be given. Then for each x = (x
i
)
1
iD1

2
, there exists N such that
1

nC1
[x
i
[
2
< c
2
. for all n _ N.
Hence, for all n _ N,
[T
n
x[
2
2
=
1

nC1
[x
i
[
2
< c
2
.
That is, for each x
2
, T
n
x 0. Hence, T
n
0 strongly.
Now, since
[T
n
x[
2
2
=
1

nC1
[x
i
[
2
_
1

1
[x
i
[
2
= [x[
2
2
for n N and x = (x
i
)
1
iD1

2
, it follows that [T
n
[ _ 1 for each n N.
But [T
n
[ _ 1 for all n. To see this, take x = (0. 0. . . . . 0. x
nC1
. 0. . . .)
2
, where x
nC1
,= 0.
Then
T
n
x = x and hence [T
n
x[
2
= [x
nC1
[. and consequently. [T
n
[ _ 1.
That is, (T
n
) does not converge to zero in the uniform topology.
4.1.6 Theorem
Let X and Y be normed linear spaces over a eld F. Then B(X. Y ) is a Banach space if Y is a Banach
space.
Proof. We have shown that B(X. Y ) is a normed linear space. It remains to show that it is complete if Y
is complete. To that end, let (T
n
)
1
1
be a Cauchy sequence in B(X. Y ). Then given any c > 0 there exists a
positive integer N such that
[T
n
T
nCr
[ < c for all n > N.
whence,
[T
n
x T
nCr
x[ _ [T
n
T
nCr
[[x[ < c[x[ for all x X. (4.1.6.1)
Hence, (T
n
x)
1
1
is a Cauchy sequence in Y . Since Y is complete there exists y Y such that T
n
x y
as n o. Set Tx = y. We show that T B(X. Y ) and T
n
T. Let x
1
. x
2
X, and . F. Then
T(x
1
x
2
) = lim
n!1
T
n
(x
1
x
2
) = lim
n!1
T
n
x
1
T
n
x
2
|
= lim
n!1
T
n
x
1
lim
n!1
T
n
x
2
= Tx
1
Tx
2
.
That is, T L(X. Y ). Taking the limit as r oin (4.1.6.1) we get that
[(T
n
T )x[ = [T
n
x Tx[ _ c[x[ for all n > N. and all x X.
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2011 FUNCTIONAL ANALYSIS ALP
That is, T
n
T is a bounded operator for all n > N. Since B(X. Y ) is a linear space,
T = T
n
(T
n
T) B(X. Y ).
Finally,
[T
n
T [ = sup{[T
n
x T x[ : [x[ _ 1] _ sup{[x[c : [x[ _ 1] _ c for all n > N.
That is, T
n
T as n o.
4.1.12 Denition
Let T : X Y and S : Y Z. We dene the composition of T and S as the map ST : X Z
dened by
(ST )(x) = (S T )(x) = S(T x).
4.1.7 Theorem
Let X, Y and Z be normed linear spaces over a eld F and let T B(X. Y ) and S B(Y. Z). Then
ST B(X. Z) and [ST[ _ [S[[T[.
Proof. Since linearity is trivial, we only prove boundedness of ST . Let x X. Then
[(ST )(x)[ = [S(T x)[ _ [S[[Tx[ _ [S[[T[[x[.
Thus, [ST[ _ [S[[T[.
Let X be a normed linear space over F. For S. T
1
. T
2
B(X) it is easy to show that
(ST
1
)T
2
= S(T
1
T
2
)
S(T
1
T
2
) = ST
1
ST
2
(T
1
T
2
)S = T
1
S T
2
S.
The operator I dened by Ix = x for all x X belongs to B(X). [I[ = 1, and it has the property that
IT = TI = T for all T B(X). We call I the identity operator. The set B(X) is therefore an algebra
with an identity element. In fact, B(X) is a normed algebra with an identity element. If X is a Banach
space then B(X) is a Banach algebra.
We now turn our attention to a very special and important class of bounded linear operators, namely,
bounded linear functionals.
4.1.13 Denition
Let X be a linear space over F. A linear operator : X F is called a linear functional on X. Of course,
L(X. F) denotes the set of all linear functionals on X.
Since every linear functional is a linear operator, all of the foregoing discussion on linear operators
applies equally well to linear functionals. For example, if X is a normed linear space then we say that
L(X. F) is bounded if there exists a constant M > 0 such that [ (x)[ _ M[x[ for all x X. The
norm of is dened by
[ [ = sup{[ (x)[ : [x[ _ 1].
We shall denote by X

= B(X. F) the set of all bounded (i.e., continuous) linear functionals on X. We

call X

the dual of X. It follows from Theorem 4.1.6 that X

is always a Banach space under the above

norm.
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2011 FUNCTIONAL ANALYSIS ALP
4.1.14 Examples
[1] Let X = Ca. b|. For each x X, dene : X R by
(x) =
b
_
a
x(t )dt.
Then is a bounded linear functional on X.
Linearity: For any x. y X and any . R,
(x y) =
b
_
a
(x y)(t )dt =
b
_
a
x(t )dt
b
_
a
y(t )dt = (x) (y).
Boundedness:
[ (x)[ =

b
_
a
x(t )dt

_
b
_
a
[x(t )[dt _ max
at b
[x(t )[(b a) = [x[
1
(b a).
Hence is bounded and [ [ _ b a. We show that [ [ = b a. Take x = 1, the constant
function 1. Then
(1) =
b
_
a
dt = b a. i.e., [(1)[ = b a.
Hence
b a =
[(1)[
1
_ sup
_
[(x)[
[x[
: x ,= 0
_
= [ [ _ b a.
That is, [ [ = b a.
[2] Let X = Ca. b| and let t (a. b) be xed. For each x X, dene
t
: X R by

t
(x) = x(t ). (i.e.,
t
is a point evaluation at t ).
Then
t
is a bounded linear functional on X. Linearity of
t
is easy to verify.
Boundedness: For each x X,
[
t
(x)[ = [x(t )[ _ max
arb
[x(t)[ = [x[
1
.
That is,
t
is a bounded linear operator and [
t
[ _ 1. We show that [
t
[ = 1. Take x = 1,
the constant 1 function. Then
t
(1) = 1 and so
1 =
[
t
(1)[
1
_ sup {[
t
(x)[ : [x[ = 1] = [
t
[ _ 1.
That is, [
t
[ = 1.
[3] Let c
1
. c
2
. . . . . c
n
be real numbers and let X = Ca. b|. Dene : X R by
(x) =
n

iD1
c
i
x(t
i
). where t
1
. t
2
. . . . . t
n
are in a. b|.
Then is a bounded linear functional on X.
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2011 FUNCTIONAL ANALYSIS ALP
Linearity: Clear.
Boundedness: For any x X,
[ (x)[ =

iD1
c
i
x(t
i
)

_
n

iD1
[c
i
x(t
i
)[ =
n

iD1
[c
i
[[x(t
i
)[ _ [x[
1
n

iD1
[c
i
[.
Hence, is bounded and
[ [ _
n

iD1
[c
i
[.
[4] Let X be a linear space. The norm [ [ : X R is an example of a nonlinear functional on
X.
4.2 Examples of Dual Spaces
4.2.1 Denition
Let X and Y be normed linear spaces over the same eld F. Then X and Y are said to be isomorphic to
each other, denoted by X . Y , if there is a bijective linear operator T from X onto Y . If, in addition, T is
an isometry, i.e., [Tx[ = [x[ for each x X, then we say that T is an isometric isomorphism. In this
case, X and Y are are said to be isometrically isomorphic and we write X Y .
Two normed linear spaces which are isometrically isomorphic can be regarded as identical, the isometry
merely amounting to a relabelling of the elements.
4.2.2 Proposition
Let X and Y be normed linear spaces over the same eld F and T a linear operator from X onto Y . Then
T is an isometry if and only if
(i) T is one-to-one;
(ii) T is continuous on X;
(iii) T has a continuous inverse (in fact, [T
1
[ = [T[ = 1);
(iv) T is distance-preserving: For all x. y X, [Tx Ty[ = [x y[.
Proof. If T satises (iv), then, taking y = 0, we have that [Tx[ = [x[ for each x X; i.e., T is an
isometry.
Conversely, assume that T is an isometry. If x = y, then
[Tx Ty[ = [T(x y)[ = [x y[ > 0.
Hence, Tx = Ty. This shows that T is one-to-one and distance-preserving. Since [Tx[ = [x[ for each
x X, it follows that T is bounded and [T[ = 1. By Theorem 4.1.4, T is continuous on X.
Let y
1
. y
2
Y and
1
.
2
F. Then there exist x
1
. x
2
X such that T x
i
= y
i
for i = 1. 2. Therefore

1
y
1

2
y
2
=
1
Tx
1

2
T x
2
= T(
1
x
1

2
x
2
) or
T
1
(
1
y
1

2
y
2
) =
1
x
1

2
x
2
=
1
T
1
y
1

2
T
1
y
2
.
That is, T
1
is linear. Furthermore, for y Y , let x = T
1
y. Then,
[T
1
y[ = [x[ = [Tx[ = [y[.
Therefore T
1
is bounded and [T
1
[ = 1.
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2011 FUNCTIONAL ANALYSIS ALP
4.2.3 Remark
It is clear from Proposition 4.2.2 that two normed linear spaces X and Y are isometrically isomor-
phic if and only if there is a linear isometry from X onto Y .
[1] The dual of
1
is (isometrically isomorphic to)
1
; i.e.,

1

1
.
Proof. Let y = (y
n
)
1
and dene :
1

1
by
(y)(x) =
1

jD1
x
j
y
j
for x = (x
n
)
1
.
Claim 1: y

1
.
Linearity of y: Let x = (x
n
) . z = (z
n
)
1
and F. Then
(y)(x z) =
1

jD1
(x
j
z
j
)y
j
=
1

jD1
x
j
y
j

1

jD1
z
j
y
j
=
1

jD1
x
j
y
j

1

jD1
z
j
y
j
= (y)(x) (y)(z).
Boundedness of y: For any x = (x
n
)
1
,
[(y)(x)[ =

jD1
x
j
y
j

_
1

jD1
[x
j
y
j
[ _ [y[
1
1

jD1
[x
j
[ = [y[
1
[x[
1
.
That is, y

1
and
[y[ _ [y[
1
. (=)
Claim 2: is a surjective linear isometry.
(i) is a surjective: A basis for
1
is (e
n
), where e
n
= (
nm
) has 1 in the n-th position and zeroes
elsewhere. Let

1
and x = (x
n
)
1
. Then x =
1

nD1
x
n
e
n
and therefore
(x) =
1

nD1
x
n
(e
n
) =
1

nD1
x
n
z
n
.
where, for each n N, z
n
= (e
n
). We show that z = (z
n
)
1
. Indeed, for each n N
[z
n
[ = [ (e
n
)[ _ [ [[e
n
[ = [ [.
Hence, z = (z
n
)
1
. Also, for any x = (x
n
)
1
,
(z)(x) =
1

nD1
x
n
z
n
=
1

nD1
x
n
(e
n
) = (x).
That is, z = and so is surjective. Furthermore,
[z[
1
= sup
n2N
[z
n
[ = sup
n2N
[ (e
n
)[ _ [ [ = [z[. (==)
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2011 FUNCTIONAL ANALYSIS ALP
(ii) is linear: Let y = (y
n
). z = (z
n
)
1
and F. Then, for any x = (x
n
)
1
,
(y z)|(x) =
1

jD1
x
j
(y
j
z
j
) =
1

jD1
x
j
y
j

1

jD1
x
j
z
j
= (y)(x) (z)(x) = y z|(x).
Hence, (y z) = y , which proves linearity of .
(iii) is an isometry: This follows from (=) and (==).
[2] The dual of c
0
is (isometrically isomorphic to)
1
, i.e., c

0

1
.
Proof. Let y = (y
n
)
1
and dene :
1
c

0
by
(y)(x) =
1

jD1
x
j
y
j
for x = (x
n
) c
0
.
Proceeding as in Example 1 above, one shows that y is a bounded linear functional on c
0
and
[y[ _ [y[
1
. (=)
Claim: is a surjective linear isometry.
(i) is a surjective: A basis for c
0
is (e
n
), where e
n
= (
nm
) has 1 in the n-th position and zeroes
elsewhere. Let c

0
and x = (x
n
) c
0
. Then x =
1

nD1
x
n
e
n
and therefore
(x) =
1

nD1
x
n
(e
n
) =
1

nD1
x
n
n
n
.
where, for each n N, n
n
= (e
n
). For n. k N, let
z
nk
=
_

_
[n
k
[
n
k
if n
k
= 0 and k _ n
0 if n
k
= 0 or k > n.
and let
z
n
= (z
n1
. z
n2
. . . . . z
nn
. 0. 0. . . .).
Then z
n
c
0
and
[z
n
[
1
= sup
k2N
[z
nk
[ = 1.
Also,
(z
n
) =
1

kD1
z
nk
n
k
=
n

kD1
[n
k
[.
Hence, for each n N,
n

kD1
[n
k
[ = [ (z
n
)[ _ [ [[z
n
[ _ [ [.
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2011 FUNCTIONAL ANALYSIS ALP
Since the right hand side is independent of n, it follows that
1

kD1
[n
k
[ _ [ [ Hence, n =
(n
n
)
1
. Also, for any x = (x
n
) c
0
,
(n)(x) =
1

nD1
x
n
n
n
=
1

nD1
x
n
(e
n
) = (x).
That is, n = and so is surjective. Furthermore,
[n[
1
=
1

kD1
[n
k
[ _ [ [ = [n[. (==)
(ii) is linear: Let y = (y
n
). z = (z
n
)
1
and F. Then, for any x = (x
n
) c
0
,
(y z)|(x) =
1

jD1
x
j
(y
j
z
j
) =
1

jD1
x
j
y
j

1

jD1
x
j
z
j
= (y)(x) (z)(x) = y z|(x).
Hence, (y z) = y z, which proves linearity of .
(iii) is an isometry: This follows from (=) and (==).
[3] Let 1 < p < o.
1
p

1
q
= 1. Then the dual of
p
is (isometrically isomorphic to)
q
, i.e.,

p

q
.
Proof. Let y = (y
n
)
q
and dene :
q

p
by
(y)(x) =
1

jD1
x
j
y
j
for x = (x
n
)
p
.
It is straightforward to show that y is linear. We show that y is bounded. By H olders Inequality,
[(y)(x)[ =

jD1
x
j
y
j

_
1

jD1
[x
j
y
j
[ _
_
_
1

jD1
[x
j
[
p
_
_
1
p
_
_
1

jD1
[y
j
[
q
_
_
1
q
= [x[
p
[y[
q
.
That is, y

p
and
[y[ _ [y[
q
. (=)
Claim: is a surjective linear isometry.
(i) is a surjective: A basis for
p
is (e
n
), where e
n
= (
nm
) has 1 in the n-th position and zeroes
elsewhere. Let

p
and x = (x
n
)
p
. Then x =
1

nD1
x
n
e
n
and therefore
(x) =
1

nD1
x
n
(e
n
) =
1

nD1
x
n
n
n
.
where, for each n N, n
n
= (e
n
). For n. k N, let
z
nk
=
_

_
[n
k
[
q
n
k
if k _ n and n
k
= 0
0 if n
k
= 0 or k > n.
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2011 FUNCTIONAL ANALYSIS ALP
and let
z
n
= (z
n1
. z
n2
. . . . . z
nn
. 0. 0. . . .).
Then z
n

p
and
(z
n
) =
1

kD1
z
nk
n
k
=
n

kD1
[n
k
[
q
.
Hence, for each n N,
n

kD1
[n
k
[
q
= [(z
n
)[ _ [ [[z
n
[
p
.
Since
[z
n
[
p
=
_
1

kD1
[z
nk
[
p
_
1{p
=
_
n

kD1
[z
nk
[
p
_
1{p
=
_
n

kD1
[n
k
[
p(q1)
_
1{p
=
_
n

kD1
[n
k
[
q
_
1{p
.
it follows that , for each n N,
n

kD1
[n
k
[
q
_ [ [[z
n
[
p

n

kD1
[n
k
[
q
_ [ [
_
n

kD1
[n
k
[
q
_
1{p

_
n

kD1
[n
k
[
q
_
11{p
_ [ [

_
n

kD1
[n
k
[
q
_
1{q
_ [ [.
Since the right hand side is independent of n, it follows that
_
1

kD1
[n
k
[
q
_
1{q
_ [ [, and so
n = (n
n
)
q
. Also, for any x = (x
n
)
p
,
(n)(x) =
1

nD1
x
n
n
n
=
1

nD1
x
n
(e
n
) = (x).
That is, n = and so is surjective. Furthermore,
[n[
q
=
_
1

kD1
[n
k
[
q
_
1{q
_ [ [ = [n[. (==)
(ii) is linear: Let y = (y
n
). z = (z
n
)
q
and F. Then, for any x = (x
n
)
p
,
(y z)|(x) =
1

jD1
x
j
(y
j
z
j
) =
1

jD1
x
j
y
j

1

jD1
x
j
z
j
= (y)(x) (z)(x) = y z|(x).
Hence, (y z) = y z, which proves linearity of .
76
2011 FUNCTIONAL ANALYSIS ALP
(iii) is an isometry: This follows from (=) and (==).
The following result is an immediate consequence of Theorem 4.1.5.
4.2.1 Theorem
Every linear functional on a nite-dimensional normed linear space is continuous.
4.2.4 Proposition
Let X be a normed linear space over F. If X is nite-dimensional, then X

is also nite-dimensional and

dimX = dimX

.
Proof. Let {x
1
. x
2
. . x
n
] be a basis for X. For each j = 1. 2. . . . . n, let x

j
be dened by x

j
(x
k
) =

jk
for k = 1. 2. . . . . n. Then each x

j
is a bounded linear functional on X. We show that {x

j
[ j =
1. 2. . . . . n] is a basis for X

. Let x

be an element of X

and dene z
j
= x

(x
j
) for each j =
1. 2. . . . . n. Then for any k = 1. 2. . . . . n,
_
_
n

jD1
z
j
x

j
_
_
(x
k
) =
n

jD1
z
j

jk
= z
k
= x

(x
k
).
Hence x

=
n

jD1
z
j
x

j
; i.e., {x

j
[ j = 1. 2. . . . . n] spans X

. It remains to show that {x

j
[ j =
1. 2. . . . . n] is linearly independent. Suppose that
n

jD1

j
x

j
= 0. Then, for each k = 1. 2. . . . . n,
0 =
_
_
n

jD1

j
x

j
_
_
(x
k
) =
n

jD1

jk
=
k
.
Hence {x

j
[ j = 1. 2. . . . . n] is a linearly independent set.
4.3 The Dual Space of a Hilbert Space
If H is a Hilbert space then bounded linear functionals on H assume a particularly simple form.
Let (X. (. )) be an inner product space over a eld F. Choose and x y X \ {0]. Dene a map

y
: X F by
y
(x) = (x. y). We claim that
y
is a bounded (= continuous) linear functional on X.
Linearity: Let x
1
. x
2
X and . F. Then

y
(x
1
x
2
) = (x
1
x
2
. y) = (x
1
. y) (x
2
. y) =
y
(x
1
)
y
(x
2
).
Boundedness: For any x X,
[
y
(x)[ = [(x. y)[ _ [x[[y[ (by the CBS Inequality).
That is,
y
is bounded and [
y
[ _ [y[. Since

y
(y) = (y. y) = [y[
2
=
[
y
(y)[
[y[
= [y[.
we have that [
y
[ = [y[.
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2011 FUNCTIONAL ANALYSIS ALP
The above observation simply says that each element y in an inner product space (X. (. )) determines
a bounded linear functional on X.
The following theorem asserts that if H is a Hilbert space then the converse of this statement is true.
That is, every bounded linear functional on a Hilbert space H is, in fact, determined by some element
y H.
4.3.1 Theorem
(Riesz-Fr echet Theorem). Let Hbe a Hilbert space over F. If : H F is a bounded linear functional
on H (i.e., H

) then there exists one and only one y H such that

(x) = (x. y) for all x H.
Moreover, [ [ = [y[.
Proof. Existence: If = 0 then take y = 0. Assume that ,= 0. Let N = {x H [ (x) = 0], the
kernel of . Then N is a closed proper subspace of H. By Corollary 3.4.5 there exists z N
?
\ {0].
Without loss of generality, [z[ = 1. Put u = (x)z (z)x. Then
(u) = ( (x)z (z)x) = (x) (z) (z) (x) = 0. i.e., u N.
Thus,
0 = (u. z) = ((x)z (z)x. z) = (x)(z. z) (z)(x. z) = (x) (z)(x. z).
whence, (x) = (z)(x. z) = (x. (z)z). Take y = (z)z. Then (x) = (x. y).
Uniqueness: Assume that (x) = (x. y) = (x. y
0
) for each x H. Then
0 = (x. y) (x. y
0
) = (x. y y
0
). for all x H.
In particular, take x = y y
0
,
0 = (y y
0
. y y
0
) = [y y
0
[
2
= y y
0
= 0 = y = y
0
.
Finally, for any x H,
[ (x)[ = [(x. y)[ _ [x[[y[ (by the CBS Inequality).
That is, [ [ _ [y[. Since
(y) = (y. y) = [y[
2
=
[(y)[
[y[
= [y[.
we have that [ [ = [y[.
4.3.1 Remarks
(a) The element y H as advertised in Theorem 4.3.1 is called the representer of the functional
.
(b) The conclusion of Theorem 4.3.1 may fail if (X. (. )) is an incomplete inner product space.
4.3.2 Example
Let X be the linear space of polynomials over R with the inner product dened by
(x. y) =
1
_
0
x(t )y(t ) dt .
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2011 FUNCTIONAL ANALYSIS ALP
For each x X, let : X R be dened by
(x) = x(0). (i.e., is a point evaluation at 0).
Then is a bounded linear functional on X. We show that there does not exist an element y X
such that
(x) = (x. y) for all x X.
Assume that such an element exists. Then for each x X
(x) = x(0) =
1
_
0
x(t )y(t ) dt .
Since for any x X the functional maps the polynomial tx(t ) onto zero, we have that
1
_
0
tx(t )y(t ) dt = 0 for all x X.
In particular, with x(t ) = ty(t ) we have that
1
_
0
t
2
y(t )|
2
dt = 0:
whence y 0, i.e. y is the zero polynomial. Hence, for all x X,
(x) = (x. y) = (x. 0) = 0.
That is, is the zero functional, a contradiction since maps a polynomial with a nonzero con-
stant term to that constant term.
4.3.2 Theorem
Let H be a Hilbert space.
(a) If H is a real Hilbert space, then H H

.
(b) If H is a complex Hilbert space, then H is isometrically embedded onto H

.
Proof. For each y H, dene : H H

by
y =
y
. where
y
(x) = (x. y) for each x H.
Let y. z H. Then, for each x H,
y = z (x. y) = (x. z)
y
=
z
y = z.
Hence, is well dened and one-to-one. Furthermore, since
[y[ = [
y
[ = [y[
for each y H, is an isometry.
If H

, then by Riesz-Fr echet Theorem (Theorem 4.3.1), there is a unique y

f
H such that (x) =
(x. y
f
). Hence y
f
= , i.e., is onto.
The inverse
1
of is given by

1
= y. where (x) = (x. y) for all x H.
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2011 FUNCTIONAL ANALYSIS ALP
Since [
1
[ = [y[ = [ [ for each H

,
1
is bounded (in fact an isometry).
If H is real, then is linear. Indeed, for all x. y. z H and all R, then
((y z)) (x) =
.yCz/
(x) = (x. y z)
= (x. y) (x. z) = (x. y) (x. z)
= (y)(x) (z)(x) = (y z) (x).
Hence, (y z) = y z.
If H is complex, then is conjugate-linear; i.e., (y z) = y z.
4.3.3 Exercise
[1] Let X and Y be linear spaces over the same eld F and T L(X. Y ).
(a) Show that ran(T ) is a linear subspace of Y and ker(T ) is a linear subspace of X.
(b) T is one-to-one if and only if ker(T ) = {0].
[2] Let X and Y be normed linear spaces over the same eld F. Show that if T B(X. Y ) then
ker(T ) is a closed linear subspace of X.
[3] Show that the mapping R :
2

2
given by
Rx = R(x
1
. x
2
. x
3
. . . .) = (0. x
1
. x
2
. x
3
. . . .)
is a bounded linear operator on
2
and nd its norm. The operator R is called the right-shift
operator.
[4] Fix x C. |. Dene an operator M
x
: L
2
. | L
2
. | by
M
x
y = xy where (M
x
y)(t ) = x(t )y(t ) for all t . |.
Show that M
x
is a bounded linear operator on L
2
. |. The operator M
x
is called a
multiplicationoperator. The function x is the symbol of M
x
.
[5] Fix x = (x
1
. x
2
. . . .)
1
. Dene an operator M
x
:
2

2
by
M
x
y = M
x
(y
1
. y
2
. . . .) = (x
1
y
1
. x
2
y
2
. . . .).
Show that M
x
is a bounded linear operator on
2
and [M
x
[ = [x[
1
.
[6] Show that if S is a subset of a Hilbert space Hthat is dense in Hand T
1
and T
2
are operators
such that T
1
x = T
2
x for all x S, then T
1
= T
2
.
[7] Find the general form of a bounded linear functional on L
2
. |.
[8] Find the general form of a bounded linear functional on
2
.
[9] Dene :
2
C by
(x) =
1

nD1
x
n
n
2
. where x = (x
1
. x
2
. . . .)
2
.
Show that is a bounded linear functional on
2
and that [ [ =

2
3
_
10
.
80
Chapter 5
The Hahn-Banach Theorem and its
Consequences
The Hahn-Banach theorem is one of the most important results in functional analysis since it is required
for many other results and also because it encapsulates the spirit of analysis. The theorem was proved
independently by Hahn in 1927 and by Banach in 1929 although Helly proved a less general version much
earlier in 1912. Intersetingly, the complex version was proved only in 1938 by Bohnenblust and Sobczyk.
We prove the Hahn-Banach theorem using Zorns lemma which is equivalent to the axiom of choice. It
should be noted, however, that the Hahn-Banach is in fact strictly weaker than the axiom of choice. Since
the publication of the original result, there have been many versions published in different settings but that
is beyond the scope of this course.
5.1 Introduction
In this chapter the Hahn-Banach theorem is established along with a few of its many consequences. Before
doing that, we briey discuss Zorns Lemma.
5.1.1 Denition
A binary relation on a set P is a partial order if it satises the following properties: For all x. y. z P,
(i) is reexive: x x;
(ii) is antisymmetric: if x y and y x, then x = y;
(iii) is transitive: if x y and y z, then x z.
A partially ordered set is a pair (P. ), where P is a set is a partial order on P.
5.1.2 Examples
[1] Let P = R and take to be _, the usual less than or equal to relation on R.
[2] Let P = P(X) the power set of a set X and take to be _, the usual set inclusion relation.
[3] Let P = C0. 1|, the space of continuous real-valued functions on the interval 0. 1| and take
to be the relation _ given by _ g if and only if (x) _ g(x) for each x 0. 1|.
5.1.3 Denition
Let C be a subset of a partially ordered set (P. ).
(i) An element u P is an upper bound of C if x u for every x C;
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2011 FUNCTIONAL ANALYSIS ALP
(ii) An element m C is said to be maximal if for any element y C, the relation m y implies that
m = y.
5.1.4 Denition
Let (P. ) be a partially ordered set and x. y P. We say that x and y are comparable if either x y or
y x. Otherwise, x and y are incomparable.
A partial order is called a linear order (or a total order) if any two elements of P are comparable. In
this case we say that (P. ) is a linearly ordered (or totally ordered) set. A linearly ordered set is also called
a chain.
5.1.1 Theorem
(Zorns Lemma). Let (P. ) be a partially ordered set. If each linearly ordered subset of P has an upper
bound, then P has a maximal element.
5.1.5 Denition
Let M and N be linear subspaces of a linear space X with M N and let be a linear functional on M.
A linear functional F on N is called an extension of to N if F[
M
= ; i.e., F(x) = (x) for each
x M.
5.1.6 Denition
Let X be a linear space. A function p : X R is called a sublinear functional provided that:
(i) p(x y) _ p(x) p(y) for x. y X;
(ii) p(zx) = zp(x). z _ 0.
Observe that any linear functional or any norm on X is a sublinear functional. Also, every positive
scalar multiple of a sublinear functional is again a sublinear functional.
5.1.7 Lemma
Let M be a proper linear subspace of a real linear space X, x
0
X\M, and N = {mx
0
[ m M.
R]. Suppose that p : X R a sublinear functional dened on X, and a linear functional dened on M
such that (x) _ p(x) for all x M. Then can be extended to a linear functional F dened on N such
that F(x) _ p(x) for all x N.
Proof. Since x
0
, M, it is readily veried that N = M lin{x
0
]. Therefore each x N has a unique
representation of the form x = mzx
0
for some unique m M and z R. Dene a functional F on N
by
F(x) = (m) zc for some c R.
This functional F is well dened since each x N is uniquely determined. Furthermore F is linear and
F(y) = (y) for all y M. It remains to show that it is possible to choose a c R such that for each
x N,
F(x) _ p(x).
Let y
1
. y
2
M. Since (y) _ p(y) for all y M, we have that
(y
1
) (y
2
) = (y
1
y
2
) _ p(y
1
y
2
) = p(y
1
x
0
y
2
x
0
)
_ p(y
1
x
0
) p(y
2
x
0
)
(y
2
) p(y
2
x
0
) _ p(y
1
x
0
) (y
1
).
Therefore, for xed y
1
M, the set of real numbers {(y
2
) p(y
2
x
0
) [ y
2
M] is bounded above
and hence has the least upper bound. Let
a = sup{(y
2
) p(y
2
x
0
) [ y
2
M].
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2011 FUNCTIONAL ANALYSIS ALP
Similarly, for xed y
2
M, the set {p(y
1
x
0
) (y
1
) [ y
1
M] is bounded below. Let
b = inf{p(y
1
x
0
) (y
1
) [ y
1
M].
Of course, a _ b. Hence there is a real number c such that a _ c _ b. Therefore
(y) p(y x
0
) _ c _ p(y x
0
) (y)
for each y M.
Now, let x = y zx
0
N. If z = 0, then F(x) = (x) _ p(x). If z > 0, then
c _ p
_
y
z
x
0
_
(y,z) zc _ p(y zx
0
) (y)
(y) zc _ p(y zx
0
)
F(x) _ p(x).
Finally, if z < 0, then
(y,z) p(y,z x
0
) _ c
1
z
(y)
1
z
p(y zx
0
) _ c
(y) p(y zx
0
) _ zc
(y) zc _ p(y zx
0
)
F(x) _ p(x).
We now state our main result. What this theorem essentially states is that there are enough bounded
(continuous) linear functionals for a rich theory and as mentioned before it is used ubiquitously thoughout
functional analysis.
5.1.2 Theorem
(Hahn-Banach Extension Theorem for real linear spaces). Let p be a sublinear functional on a real
linear space X and let M be a subspace of X. If is a linear functional on M such that (x) _ p(x) for
all x M, then has an extension F to X such that F(x) _ p(x) for all x X.
Proof. Let F be the set of all pairs (M

), where M

is a subspace of X containing M,

(y) = (y)
for each y M, i.e.,

is an extension of , and

(x) _ p(x) for all x M

. Clearly, F ,= 0 since
(M. ) F. Dene a partial order on F by:
(M

) - (M

) M

and

[
M
=

.
Let T be a totally ordered subset of F and let
X
0
=
_
{M

[ (M

) T ].
Then X
0
is a linear subspace of X since T is totally ordered. Dene a functional
0
: X
0
R by
0
(x) =

(x) for all x M

. Then
0
is well-dened, since if x M

, then x M

and x M

.
Therefore
0
(x) =

(x) and
0
(x) =

(x). By total ordering of T , either

extends

or vice versa.
Hence

(x) =

(x). It is clear that

0
is a linear extension of . Furthermore
0
(x) _ p(x) for all
x X
0
and (M

) - (X
0
.
0
) for all (M

) T , i.e., (X
0
.
0
) is an upper bound for T . By Zorns
lemma, F has a maximal element (X
1
. F). To complete the proof, it sufces to show that X
1
= X. If
X
1
,= X, then choose y X \ X
1
. By Lemma 5.1.7, we can extend F to a linear functional

F dened
on

X = X
1
lin{y] and extending such that

F(x) _ p(x) for all x

X. Thus (

X.

F) F and
(X
1
. F) - (

X.

F), which contradicts the maximality of (X
1
. F).
5.1.8 Denition
A seminorm p on a (complex) linear space X is a function p : X R such that for all x. y X and
z C,
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2011 FUNCTIONAL ANALYSIS ALP
(i) p(x) _ 0 and p(0) = 0;
(ii) p(x y) _ p(x) p(y), and
(iii) p(zx) = [z[p(x).
5.1.3 Theorem
(Hahn-Banach Extension Theorem for (complex) linear spaces). Let X be a real or complex linear
space, p be a seminorm on X and a linear functional on a linear subspace M of X such that [ (x)[ _
p(x) for all x M. Then there is a linear functional F on X such that F[
M
= and [F(x)[ _ p(x) for
all x X.
Proof. Assume rst that X is a real linear space. Then, by Theorem 5.1.2, there is an extension F of
such that F(x) _ p(x) for all x X. Since
F(x) = F(x) _ p(x) = p(x) for all x X.
it follows that p(x) _ F(x) _ p(x), or [F(x)[ _ p(x) for all x X.
Nowassume that X is a complex linear space. Then we may regard X as a real linear space by restricting
the scalar eld to R. We denote the resulting real linear space by X
R
and the real linear subspace by M
R
.
Write as =
1
i
2
, where
1
and
2
are real linear functionals given by
1
(x) = Re(x)| and

2
(x) = Im(x)|. Then
1
is a real linear functional of M
R
and
1
(x) _ [ (x)[ _ p(x) for all x M
R
.
Hence, by Theorem 5.1.2,
1
has a real linear extension F
1
such that F
1
(x) _ p(x) for all x X
R
. Since,
(i x) = i(x)
1
(i x) i
2
(i x) = i
1
(x)
2
(x)

2
(x) =
1
(i x) and
2
(i x) =
1
(x).
we can write (x) =
1
(x) i
1
(i x). Set
F(x) = F
1
(x) iF
1
(i x) for all x X.
Then F is a real linear extension of and, for all x. y X,
F(x y) = F
1
(x y) iF
1
(i x iy) = F
1
(x) iF
1
(i x) F
1
(y) iF
1
(iy)
= F(x) F(y).
For all x X,
F(i x) = F
1
(i x) iF
1
(x) = F
1
(i x) iF
1
(x) = i (F
1
(x) iF
1
(i x)) = iF(x).
If = a bi for a. b R, and x X, then
F(x) = F((a bi )x) = F(ax bi x) = F(ax) F(bi x)
= aF(x) bF(i x) = aF(x) biF(x) = (a bi )F(x)
= F(x).
Hence, F is also complex linear. Finally, for x X, write F(x) = [F(x)[e
i0
. Then, since ReF = F
1
,
[F(x)[ = F(x)e
i0
= F(xe
i0
) = F
1
(xe
i0
) _ p(xe
i0
) = [e
i0
[p(x) = p(x).
Suppose that M is a subspace of a normed linear space X and is a bounded linear functional on M.
If F is any extension of to X, then the norm of F is at least as large as [ [ because
[F[ = sup{ [F(x)[ : x X. [x[ _ 1 ] _ sup{ [F(x)[ : x M. [x[ _ 1 ]
= sup{ [(x)[ : x M. [x[ _ 1 ] = [ [.
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2011 FUNCTIONAL ANALYSIS ALP
The following consequence of the Hahn-Banach theorem states that it is always possible to nd a
bounded extension of to the whole space which has the same, i.e., smallest possible, norm.
5.1.4 Theorem
(Hahn-Banach Extension Theorem for Normed linear spaces). Let M be a linear subspace of the
normed linear space (X. [ [) and let M

. Then there exists an extension x

of such that
[x

[ = [ [.
Proof. Dene p on X by p(x) = [ [[x[. Then p is a seminorm on X and [(x)[ _ p(x) for all
x M. By Theorem 5.1.3, has an extension F to X such that [F(x)[ _ p(x) for all x X. That is,
[F(x)[ _ [ [[x[. This shows that F is bounded and [F[ _ [ [. Since F must have norm at least as
large as [ [, [F[ = [ [ and the result follows
with x

= F.
5.2 Consequences of the Hahn-Banach Extension Theorem
5.2.1 Theorem
Let M be a linear subspace of a normed linear space (X. [ [) and x X such that
d = d(x. M) := inf
y2M
[x y[ > 0.
Then there is an x

such that
(i) [x

[ = 1
(ii) x

(x) = d
(iii) x

(m) = 0 for all m M.

Proof. Let Y = M lin{x] := {mx. m M. F]. Then each y in Y is uniquely expressible in
the form y = mx for some m M and some scalar . Indeed, if
y = m
1
x = m
2
x
for some m
1
. m
2
M and some scalars and , then ( )x = m
1
m
2
M.
Claim: = . If ,= , then since M is a subspace
x =
1

(m
1
m
2
) M.
a contradiction since x , M. Hence, = and consequently m
1
= m
2
.
Dene : Y F by
(y) = (mx) = d.
Since the scalar is uniquely determined, is well dened.
Claim: is a bounded linear functional on Y .
Linearity: Let y
1
= m
1

1
x and y
2
= m
2

2
x be any two elements of Y and z F. Then
(zy
1
y
2
) = ((zm
1
m
2
) (z
1

2
)x) = (z
1

2
)d = z
1
d
2
d = z(y
1
) (y
2
).
Boundedness: Let y = mx Y . Then
[y[ = [mx[ = [[
_
_
_
m

x
_
_
_ = [[
_
_
_x
_

__
_
_ _ [[d = [ (y)[.
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2011 FUNCTIONAL ANALYSIS ALP
i.e., [(y)[ _ [y[ for all y Y . Thus, is bounded and [ [ _ 1.
We show next that [ [ = 1. By denition of inmum, given any c > 0, there is an element m

M such
that [x m

[ < d c. Let z =
x m

[x m

[
. Then z Y. [z[ = 1 and
[ (z)[ =
d
[x m

[
>
d
d c
.
Since c is arbitrary, it follows that [(z)[ _ 1. Thus
1 _ [ (z)[ _ [ [[z[ = [ [.
Thus, [ [ = 1.
It is clear that and (m) = 0 for all m M and (x) = d. By Theorem 5.1.4, there is an x

such
that
x

(y) = (y) for all y Y and [x

[ = [ [.
Hence, [x

[ = 1 and x

(m) = 0 for all m M and x

(x) = d.
5.2.1 Corollary
Let (X. [ [) be a normed linear space and x
0
X \ {0]. Then there exists an x

, such that
x

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

[ = 1.
Proof. Consider M = {0]. Since x
0
X \{0], it follows that x
0
, M and so d = d(x
0
. M) = [x
0
[ > 0.
By Theorem 5.2.1, there is an x

such that x

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

[ = 1.
The following result asserts that X

is big enough to distinguish the points of X.

5.2.2 Corollary
Let (X. [ [) be a normed linear space and y. z X. If y ,= z, then there exists an x

, such that
x

(y) ,= x

(z).
Proof. Consider M = {0]. Since y = z, it follows that y z , M and consequently
d = d(y z. M) = [y z[ > 0.
By Theorem 5.2.1, there is an x

such that x

(y z) = d > 0. Hence x

(y) ,= x

(z).
5.2.3 Corollary
For each x in a normed linear space (X. [ [),
[x[ = sup{[x

(x)[ [ x

. [x

[ = 1].
Proof. If x = 0, then the result holds vacuously. Assume x X \ {0]. For any x

with [x

[ = 1,
[x

(x)[ _ [x

[[x[ = [x[.
Hence, sup{[x

(x)[ [ x

. [x

[ = 1] _ [x[.
By Corollary 5.2.1, there is a x

such that [x

[ = 1 and x

(x) = [x[. Therefore

[x[ = [x

(x)[ _ sup{[x

(x)[ [ x

. [x

[ = 1].
whence [x[ = sup{[x

(x)[ [ x

. [x

[ = 1].
5.2.2 Theorem
If the dual X

of a normed linear space (X. [ [) is separable, then X is also separable.

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2011 FUNCTIONAL ANALYSIS ALP
Proof. Let S = S(X

) = {x

[ [x

[ = 1]. Since any subset of a separable space is separable, S

is separable. Let {x

n
[ n N] be a countable dense subset of S. Since x

n
S for each n N, we have
that [x

n
[ = 1. Hence, for each n N there is an element x
n
X such that [x
n
[ = 1 and [x

n
(x
n
)[ >
1
2
.
(Otherwise [x

n
(x)[ _
1
2
for all x X and so [x

n
[ _
1
2
, a contradiction.) Let
M = lin({x
n
[ n N]).
Then M is separable since M contains a countable dense subset comprising all linear combinations of the
x
n
s with coefcients whose real and imaginary parts are rational.
Claim: M = X. If M ,= X, then there is an element x
0
X \ M such that d = d(x
0
. M) > 0. By
Theorem 5.2.1, there is an x

such that [x

[ = 1, i.e. x

S, and x

(y) = 0 for all y M. In

particular, x

(x
n
) = 0 for all n N. Now, for each n N,
1
2
< [x

n
(x
n
)[ = [x

n
(x
n
) x

(x
n
)[ = [(x

n
x

)(x
n
)[ _ [x

n
x

[.
But this contradicts the fact that the set {x

n
[ n N] is dense in S. Hence M = X and, consequently, X
is separable.
The converse of Theorem 5.2.2 does not hold. That is, if X separable, it does not follow that its dual
X

is also be separable. Take, for example,

1
. Its dual is (isometrically isomorphic to)
1
. The space
1
is separable whereas
1
is not. This also shows that the dual of
1
is not (isometrically isomorphic to)
1
.
5.2.4 Denition
Let M be a subset of a normed linear space X. The annihilator of M, denoted by M
?
, is the set
M
?
= {x

[ x

(y) = 0 for all y M ].

It is easy to show that M
?
is a closed linear subspace of X

.
5.2.3 Theorem
Let M be a linear subspace of a normed linear space X. Then
X

,M
?
M

.
Proof. Dene : X

,M
?
M

by
(x

M
?
)(m) = x

(m) for all x

and all m M.
We show that is well-dened. Let x

. y

such that x

M
?
= y

M
?
. Then x

M
?
and so x

(m) = y

(m) for all m M. Thus (x

M
?
) = (y

M
?
); i.e., is well-dened.
Clearly, (x

M
?
) is a linear functional on M.
We show that is linear. Let x

. y

and z F. Then, for all m M,

((x

M
?
) z(y

M
?
))(m) = (x

zy

M
?
)(m) = (x

zy

)(m)
= x

(m) zy

(m)
= (x

M
?
)(m) z(y

M
?
)(m)
=
_
(x

M
?
) z(y

M
?
)
_
(m).
Hence, ((x

M
?
) z(y

M
?
)) = (x

M
?
) z(y

M
?
).
We now show that is surjective. Let y

. Then, by Theorem 5.1.4, there is an x

such that
y

(m) = x

(m) for all m M and [y

[ = [x

[. Hence, for all m M,

(x

M
?
)(m) = x

(m) = y

(m).
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2011 FUNCTIONAL ANALYSIS ALP
Thus (x

M
?
) = y

. Furthermore,
[x

M
?
[ _ [x

[ = [y

[ = [(x

M
?
)[.
But for any y

M
?
, x

M
?
= (x

) M
?
. Hence, for all m M,
[(x

M
?
)(m)[ = [(x

)(m)[ _ [x

[[m[.
That is, (x

M
?
) is a bounded linear functional on M and [(x

M
?
)[ _ [x

[ for all
y

M
?
. Thus
[(x

M
?
)[ _ inf
y

2M
?
[x

[ = [x

M
?
[.
It now follows that [(x

M
?
)[ = [x

M
?
[.
5.3 Bidual of a normed linear space and Reexivity
Let (X. [ [) be a normed linear space over F and x X. Dene a functional
x
: X

F by

x
(x

) = x

(x) for all x

.
It is easy to verify that
x
is linear and for each x

,
[
x
(x

)[ = [x

(x)[ _ [x

[[x[.
That is,
x
is bounded and [
x
[ _ [x[. By Corollary 5.2.3,
[x[ = sup{[x

(x)[ [ x

. [x

[ = 1] = sup{[
x
(x)[ [ x

. [x

[ = 1] = [
x
[.
This shows that
x
is a bounded linear functional on X

, i.e.,
x
(X

= X

and [
x
[ = [x[. The
space X

is called the second dual space or bidual space of X. It now follows that we can dene a map
J
X
: X X

by
J
X
x =
x
. for x X. that is, (J
X
x)(x

) = x

(x) for x X and x

.
It is easy to show that J
X
is linear and [x[ = [
x
[ = [J
X
x[. That is, J
X
is a linear isometry of X into
its bidual X

. The map J
X
as dened above is called the canonical or natural embedding of X into its
bidual X

. This shows that we can identify X with the subspace J

X
X = {J
X
x [ x X] of X

.
5.3.1 Denition
Let (X. [ [) be a normed linear space over F. Then X is said to be reexive if the canonical embedding
J
X
: X X

of X into its bidual X

is surjective. In this case X X

.
If X is reexive, we customarily write X = X

. The equality simply means that X is isometrically

isomorphic to X

. Reexivity of X basically means that each bounded linear functional on X

is an
evaluation functional. Since dual spaces are complete, a reexive normed linear space is necessarily a
Banach space. It is therefore appropriate to speak of a reexive Banach space rather than a reexive normed
linear space.
5.3.1 Theorem
(1) Every nite-dimensional normed linear space is reexive.
(2) A closed linear subspace of a reexive space is reexive.
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2011 FUNCTIONAL ANALYSIS ALP
Proof. (1). If dimX < o, then Proposition 4.2.4 implies that dimX = dimX

=dimX

. Since J
X
X is
isometrically isomorphic to X, dim(J
X
X) =dimX =dimX

. Since J
X
X is a subspace of X

, it must
equal to X

.
(2). Let X be reexive and M a closed linear subspace of X. Given y

, it must be shown that

there exists y M such that J
M
y(y

) = y

(y) = y

(y

) for all y

. Dene a functional [ on
X

by
[(x

) = y

(x

[
M
). x

.
Clearly, [ is linear and
[[(x

)[ _ [y

[[x

[
M
[ _ [y

[[x

[
so [ X

. By reexivity of X, there exists y X such that J

X
y = [. That is, [(x

) = x

(y) for each

x

. If y , M, then by Theorem 5.2.1, there exists an x

0
X

such that x

0
(y) = 0 and x

0
(m) = 0
for all m M. Then
0 = x

0
(y) = [(x

0
) = y

(x

0
[
M
) = y

(0) = 0
which is absurd. Thus y M and x

(y) = [(x

) = y

(x

[
M
). x

. By Theorem 5.1.4, every

y

is of the form y

= x

[
M
for some x

. Thus
(J
M
y)(y

) = y

(y) = y

(y

), y

, and the proof is complete.

5.3.2 Theorem
A Banach space X is reexive if and only if its dual X

is reexive.
Proof. Assume that X is reexive. Let J
X
: X X

and J
X

: X

(X

= X

be the
canonical embeddings of X and X

respectively. We must show that J

X
is surjective. To that end, let
x

= (X

and consider the following diagram:

X
J
X
X

F.
Dene a functional x

on X by x

= x

J
X
. It is obvious that x

is linear since both x

and J
X
are
linear. Also, for each x X,
[x

(x)[ = [x

J
X
(x)[ _ [x

[[J
X
x[ = [x

[[x[.
i.e., x

is bounded and [x

[ _ [x

[. Hence x

. We now show that J

X

(x

) = x

. Let
x

be any element. Since J

X
is surjective, there is an x X such that x

= J
X
x. Hence
x

(x

) = x

(J
X
x) = x

(x) = J
X
x(x

) = J
X

(J
X
x) = J
X

(x

).
and therefore J
X

= x

. That is, J
X

is surjective.
Assume that X

is reexive. Then the canonical embedding J

X

: X

is surjective. If
J
X
X = X

, let x

\ J
X
X. Since J
X
X is a closed linear subspace of X

, it follows from
Theorem 5.2.1 that there is a functional X

such that [[ = 1, (x

) = d(x

. J
X
X), and
(J
X
x) = 0 for all x X. Since J
X

is surjective, there is an x

such that J
X

= . Hence, for
each x X,
0 = (J
X
x) = J
X

(J
X
x) = (J
X
x)(x

) = x

(x).
i.e., x

(x) = 0 for all x X. This implies that x

= 0. But then 0 = J
X

= , a contradiction since
= 0. Hence J
X
X = X

; i.e., J
X
is surjective.
5.3.2 Exercise
Show that if X is a non-reexive normed linear space, then the natural inclusions X X

and X

are all strict.

We showed earlier (Theorem 5.2.2) that if the dual space X

of a normed linear space X is separable,

then X is also separable, but not conversely. However, if X is reexive, then the converse holds.
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2011 FUNCTIONAL ANALYSIS ALP
5.3.3 Theorem
If X is a reexive separable Banach space, then its dual X

is also separable.
Proof. Since X is reexive and separable, its bidual X

= J
X
X is also separable. Hence, by Theo-
rem 5.2.2, X

is separable.
5.3.3 Examples
(1) For 1 < p < o, the sequence space
p
is reexive.
(2) The spaces c
0
. c.
1
. and
1
are non-reexive.
(3) Every Hilbert space H is reexive.
5.4 The Adjoint Operator
5.4.1 Denition
Let X and Y be normed linear spaces and T B(X. Y ). The Banach space adjoint (or simply adjoint)
of T, denoted by T

, is the operator T

: Y

dened by
(T

)(x) = y

(T x) for all y

and all x X.
The following diagram helps make sense of the above denition.
X
T
Y
X

.
It is straightforward to show that for any y

, T

is a linear functional on X. Furthermore, for

any y

and x X
[T

(x)[ = [y

(T x)[ _ [y

[[T[[x[.
i.e., T

is a bounded linear functional on X and [T

[ _ [T[[y

[.
5.4.2 Example
Let X =
1
= Y and dene T :
1

1
by
T x = T(x
1
. x
2
. x
3
. . . . ) = (0. x
1
. x
2
. x
3
. . . . ). where x = (x
n
)
1
.
the right-shift operator. Then the adjoint of T is T

:
1

1
is given by
T

y = T

(y
1
. y
2
. y
3
. . . . ) = (y
2
. y
3
. . . . ). where y = (y
n
)
1
.
the left-shift operator.
5.4.1 Theorem
Let X and Y be normed linear spaces over F and let T B(X. Y ).
(a) T

is a bounded linear operator on Y

.
(b) The map : B(X. Y ) B(Y

. X

) dened by T = T

is an isometric isomorphism of
B(X. Y ) into B(Y

. X

).
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2011 FUNCTIONAL ANALYSIS ALP
Proof. (a) Let y

1
. y

2
Y

and F. Then, for all x X

T

(y

1
y

2
)(x) = (y

1
y

2
)(T x) = y

1
(Tx) y

2
(Tx)
= T

1
(x) T

2
(x) = (T

1
T

2
)(x).
Hence, T

(y

1
y

2
) = T

1
T

2
.
Furthermore, as shown above, [T

[ _ [T[[y

[. Hence, T

B(Y

. X

) and [T

[ _ [T[.
(b) We show that [T

[ = [T[, whence [T

[ = [T

[. Indeed,
[T[ = sup
kxkD1
[Tx[ = sup
kxkD1
_
sup
ky

kD1
[y

(Tx)[
_
(by Corollary 5.2.3)
= sup
ky

kD1
_
sup
kxkD1
[y

(T x)[
_
= sup
ky

kD1
[T

[
= [T

[.
5.5 Weak Topologies
We have made the point that a norm on a linear space X induces a metric. A metric, in turn, induces
a topology on X called the metric topology. It now follows that a norm on a linear space X induces a
topology which we shall refer to as the norm topology on X. In this section we dene other topologies on a
linear space X that are weaker than the norm topology. We also investigate some of the properties of these
weak topologies.
5.5.1 Denition
Let (X. [ [) be a normed linear space and F X

. The weak topology on X induced by the family F,

denoted by o(X. F), is the weakest topology on X with respect to which each x

F is continuous.
5.5.2 Remark
The weak topology on X induced by the dual space X

is simply referred to as the weak topology on X

and is denoted by o(X. X

).
What do the basic open sets for the weak topology o(X. X

) look like?
Unless otherwise indicated, we shall denote by .
1
.
2
. . . nite subsets of X

.
Let x
0
X. and c > 0 be given. Consider all sets of the form
V(x
0
: : c) := {x X [ [x

(x) x

(x
0
)[ < c. x

]
=
_
x

2
{x X [ [x

(x) x + (x
0
)[ < c].
5.5.3 Proposition
[1] x
0
V(x
0
: : c).
[2] Given V(x
0
:
1
: c
1
) and V(x
0
:
2
: c
2
), we have
V(x
0
:
1
L
2
: min{c
1
. c
2
]) V(x
0
:
1
: c
1
) V(x
0
:
2
: c
2
).
[3] If x V(x
0
: : c), then there is a > 0 such that V(x: : ) V(x
0
: : c).
Proof. (1) It is obvious that x
0
V(x
0
: : c).
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2011 FUNCTIONAL ANALYSIS ALP
(2) Let x V(x
0
:
1
L
2
: min{c
1
. c
2
]). Then for each x

1
,
[x

(x) x

(x
0
)[ < min{c
1
. c
2
] _ c
1
.
Hence x V(x
0
:
1
: c
1
). Similarly, x V(x
0
:
2
: c
2
). It now follows that
x V(x
0
:
1
: c
1
) V(x
0
:
2
: c
2
) and, consequently
V(x
0
:
1
L
2
: min{c
1
. c
2
]) V(x
0
:
1
: c
1
) V(x
0
:
2
: c
2
).
(3) Let x V(x
0
: : c) and ; = max{[x

(x) x

(x
0
)[ [ x

]. Then 0 _ ; < c. Choose such

that 0 < < c ;. Then, for any y V(x: : ) and any x

, we have
[x

(y) x

(x
0
)[ _ [x

(y) x

(x)[ [x

(x) x

(x
0
)[ < ; < c.
Recall that a collection B of subsets of a set X is a base for a topology on X if and only if
(i) X =
_
{B [ B B]; i.e., each x X belongs to some B B, and
(ii) if x B
1
B
2
for some B
1
and B
2
in B, then there is a B
3
B such that x B
3
B
1
B
2
.
5.5.1 Theorem
Let B = {V(x: : c) [ x X. (nite) X

. c > 0]. Then B is a base for a Hausdorff topology on X.

Proof. (i) It is clear that x V(x: : c) for each x X.
(ii) Let x V(x
1
:
1
: c
1
) V(x
2
:
2
: c
2
). Then x V(x
1
:
1
: c
1
) and x V(x
2
:
2
: c
2
).
By Proposition 5.5.3 (3), there are
1
> 0 and
2
> 0 such that V(x:
1
:
1
) V(x
1
:
1
: c
1
) and
V(x:
2
:
2
) V(x
2
:
2
: c
2
). By Proposition 5.5.3 (2),
V(x:
1
L
2
: min{
1
.
2
]) V(x:
1
:
1
) V(x:
2
:
2
) V(x
1
:
1
: c
1
) V(x
2
:
2
: c
2
).
Hence, B is a base for a topology on X.
Finally, we show that the topology generated by B is Hausdorff. Let x and y be distinct elements of X.
By Corollary 5.2.2, there is an x

such that x

(x) = x

(y). Let 0 < c < [x

(x) x

(y)[. Then
V(x: x

:
e
2
) and V(y: x

:
e
2
) are disjoint neighbourhoods of x and y respectively.
It is easy to see that each x

is continuous with respect to the topology generated by B. Indeed,

let x
0
X, x

and c > 0. Since x

is continuous with respect to the norm topology on X, there is a

norm neighbourhood U of x
0
such [x

(x) x

(x
0
)[ < c for all x U. It now follows V(x
0
: x

: c) is a
neighbourhood of x
0
in the topology generated by B and [x

(x) x

(x
0
)[ < c for all x V(x
0
: x

: c).
One shows quite easily that the topology generated by
B = {V(x: : c) [ x X. (nite) X

. c > 0]
is precisely o(X. X

), the weak topology on X induced by X

. Therefore, a set G is open in the topology

o(X. X

) if and only if for each x G there is a nite set

= {x

. x

1
. x

2
. . . . . x

n
] X

and an c > 0 such that V(x: : c) G.

It now follows that a normed linear space X carries two natural topologies: the norm topology induced
by the norm on X and the weak topology induced by its dual space X

. Topological concepts that are asso-

ciated with the weak topology are usually preceded by the word weak; for example, weak compactness,
weak closure, etc. Those topological concepts pertaining to the topology generated by the norm on X are
usually preceded by the word norm, e.g. norm-closure or by the word strong, e.g. strongly open set.
5.5.4 Lemma
Let {x

. x

1
. x

2
. . . . . x

n
] X

. Then
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2011 FUNCTIONAL ANALYSIS ALP
(1) x

lin{x

1
. x

2
. . . . . x

n
] if and only if
_
n
iD1
ker(x

i
) ker(x

).
(2) If {x

1
. x

2
. . . . . x

n
] is a linearly independent set, then for any set of scalars {c
1
. c
2
. . . . . c
n
],
_
n
iD1
{x X [ x

i
(x) = c
i
] = 0.
Proof. (1) If x

lin{x

1
. x

2
. . . . . x

n
], then x

=
n

iD1

i
x

i
for some scalars
1
.
2
. . . . .
n
. Let
x
n
_
iD1
ker(x

i
). Then x

i
(x) = 0 for each i = 1. 2. . . . . n. Hence,
n

iD1

i
x

i
(x) = 0 and consequently,
x

(x) = 0; i.e., x ker(x

). Therefore
n
_
iD1
ker(x

i
) ker(x

).
Conversely, assume that
n
_
iD1
ker(x

i
) ker(x

). We use induction on n. Let us rst show that if

ker(x

1
) = ker(x

), then x

= x

1
for some nonzero scalar . Let K = ker(x

1
) and z X \ K. Then,
proceeding as in Theorem 5.2.1, each x X is uniquely expressible as x = y zz, where y K and
z F. Hence, since x

(y) = 0 = x

1
(y),
x

(x) = zx

(z) =
zx

(z)
x

1
(z)
x

1
(z) =
_
x

(z)
x

1
(z)
_
zx

1
(z) =
_
x

(z)
x

1
(z)
_
x

1
(x) = x

1
(x).
where =
x

(z)
x

1
(z)
.
Assume that the result is true for n 1. For each i = 1. 2. . . . . n, x

i
is not a linear combination of
the x

j
s for j = 1. 2. . . . . n and i = j. Hence,
_
ji
ker(x

j
) is not contained in ker(x

i
). Therefore there
is an x
i

_
ji
ker(x

j
) such that x

i
(x
i
) = 1. Let
i
= x

(x
i
) for each i = 1. 2. . . . . n. Let x X and
y = x
n

iD1
x

i
(x)x
i
. Then, for each j = 1. 2. . . . . n,
x

j
(y) = x

j
(x)
n

iD1
x

i
(x)x

j
(x
i
) = x

j
(x) x

j
(x) = 0.
Thus, y
n
_
iD1
ker(x

i
). By the assumption, y ker(x

). Therefore
0 = x

(y) = x

(x)
n

iD1
x

i
(x)x

(x
i
) = x

(x)
n

iD1

i
x

i
(x) x

(x) =
n

iD1

i
x

i
(x).
whence x

=
n

iD1

i
x

i
.
(2) Let H
i
= {x X [ x

i
(x) = c
i
] for each i = 1. 2. . . . . n. We want to show that
_
n
iD1
H
i
= 0.
The proof is by induction on n. If n = 1, then, since x

1
= 0, it follows that H
1
= 0. Assume true for n = k
and let H =
_
kC1
iD1
H
i
. By the linear independence of {x

1
. x

2
. . . . . x

kC1
],
_
k
iD1
ker(x

i
) , ker(x

kC1
).
Hence, there is an x
0

_
k
iD1
ker(x

i
) such that x

kC1
(x
0
) ,= 0. Take any x
_
k
iD1
ker(x

i
) and set
y = x x
0
, where = c
kC1

kC1
(x)
x

kC1
(x
0
)
. Then x

i
(y) = x

i
(x) = c
i
for each i = 1. 2. . . . . k and
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2011 FUNCTIONAL ANALYSIS ALP
x

kC1
(y) = c
kC1
. That is, y H.
5.5.2 Theorem
Let t denote the norm topology on X. Then
(a) o(X. X

) t.
(b) o(X. X

) = t if and only if X is nite-dimensional. Thus, if X is innite-dimensional, then the

weak topology o(X. X

) is strictly weaker than the norm topology.

Proof. (a) The topology o(X. X

) is the weakest topology on X making each x

continuous. Hence,
o(X. X

) is weaker than the norm topology t.

(b) Assume that o(X. X

) = t and let x

. Then, since x

is continuous when X is equipped with

the norm topology and, by the hypothesis, it is continuous in the weak topology o(X. X

), it is continous
at 0. Therefore there is a nite set = {x

1
. x

2
. . . . . x

n
] X

and an c > 0 such that [x

(x)[ < 1 for

all x V(0: : c). Let z
n
_
iD1
ker(x

i
). Then x

i
(z) = 0 and so [x

i
(z)[ < c for each i = 1. 2. . . . . n.
That is, z V(0: : c). If x
n
_
iD1
ker(x

i
), then mx
n
_
iD1
ker(x

i
) for each m Z
C
since
n
_
iD1
ker(x

i
) is
a linear subspace of X. It now follows that mx V(0: : c) for each m Z
C
. This, in turn, implies that
1 > [x

(mx)[ = m[x

(x)[ [x

(x)[ <
1
m
.
Since m is arbitrary, x

(x) = 0; i.e., x ker(x

). Hence
n
_
iD1
ker(x

i
) ker(x

). By Lemma 5.5.4,
x

is expressible as x

=
n

iD1

i
x

i
for some scalars
1
.
2
. . . . .
n
. Hence X

is spanned by
the set {x

1
. x

2
. . . . . x

n
]. This shows that X

is nite-dimensional. By Proposition 4.2.4, X is also

nite-dimensional.
Conversely, assume that X is nite-dimensional. Let {x
1
. x
2
. . . . . x
n
] be a basis for X such that
[x
k
[ = 1 for each k = 1. 2. . . . . n. Let U X be open in the norm topology of X. We want to show that
U is open in the weak topology of X. Let x
0
U. Then there is an r > 0 such that B(x
0
. r ) U. For any
x X, x =
n

kD1

k
x
k
. Dene x

i
: X F by x

i
(x) =
i
for each i = 1. 2. . . . . n. Since the
i
s are
uniquely determined, x

i
is well-dened. One shows quite easily that x

i
X

for each i = 1. 2. . . . . n.
Let = {x

1
. x

2
. . . . . x

n
] and c =
r
n
. Then, for any x V(x
0
: : c), we have [x

i
(x) x

i
(x
0
)[ < c
for each i = 1. 2. . . . . n. Hence, if x V(x
0
: : c), then
[x x
0
[ =
_
_
_
_
_
n

kD1
x

k
(x x
0
)x
k
_
_
_
_
_
_
n

kD1
[x

k
(x x
0
)[ < nc = r.
That is, x B(x
0
. r ) U. It now follows that for each x U, there is a V(x: : c) such that
V(x: : c) U. Hence U is open in the weak topology o(X. X

). Thus, o(X. X

) = t.

The following result asserts that the weak topology and the norm topology yield exactly the same
continuous linear functionals. That is, the linear functionals on X that are continuous with respect to
94
2011 FUNCTIONAL ANALYSIS ALP
the topology o(X. X

) are those that are in X

. Therefore weakening the topology does not affect the dual

space of X.
5.5.3 Theorem
Let X be a normed linear space. Then the dual of X under the topologyo(X. X

) is X

; i.e., (X. o(X. X

))

=
X

.
Proof. By denition of the topology o(X. X

), it is clear X

is a subset of the dual of X under the topology

o(X. X

); i.e., X

(X. o(X. X

))

.
Let (X. o(X. X

))

. Then, proceeding as in Theorem 5.5.2, there is a nite set

{x

1
. x

2
. . . . . x

n
] X

and scalars
1
.
2
. . . . .
n
such that =
n

iD1

i
x

i
. Therefore X

.
5.5.4 Theorem
Let K be a convex subset of a normed linear space X. Then the closure of K relative to the weak topology
o(X. X

) is the same as the norm-closure of K, i.e., K

o(X,X

)
= K.
Proof. Since K
o(X,X

)
is closed and K K
o(X,X

)
and K is the smallest closed set containing K, it
follows that K K
o(X,X

)
.
It remains to show that K
o(X,X

)
K. Let x
0
X \ K. Then, by Hahn-Banachs Theorem, there is
an x

and real numbers c

1
and c
2
such that
Re
_
x

(x
0
)
_
_ c
1
< c
2
_ Re
_
x

(x)
_
for all x K.
Consider V = V(x
0
: x

: c
2
c
1
) = {x X [ [x

(x) x

(x
0
)[ < c
2
c
1
]. Then V is a weak neighbour-
hood of x
0
and V K = 0. Thus x
0
, K
o(X,X

)
, and consequently K
o(X,X

)
K.

Since the topology o(X. X

) is weaker than the norm topology, every weakly closed set in X is closed.
However, for convex sets we have the following.
5.5.5 Corollary
A convex subset K of a normed linear space X is closed if and only it is weakly closed.
We now turn our attention to the dual space X

of a normed linear space X. X

carries three natural

topologies: the norm topology, the weak topology o(X

. X

) induced by X

and the weak* topology

o(X

. X) induced by X.
Let J
X
be the canonical embedding of X into its bidual X

. Then X J
X
X X

. A typical basic
open set in the topology o(X

. J
X
X) on X

induced by J
X
X is
V(x

: : c) := {y

[ [(J
X
x)(x

) (J
X
x)(y

)[ < c. c > 0. x (nite) X]

= {y

[ [x

(x) y

(x)[ < c. c > 0. x (nite) X].

It now follows that the weak* topology o(X

. X) on X

is precisely the weak topology on X

induced by
J
X
X. That is, o(X

. X) = o(X

. J
X
X) the weak topology on X

induced by elements of X acting as

continous linear functionals on X

.
Let us observe, in passing, that X

has a weak* topology o(X

. X

) induced by X

. Since X
J
X
X X

, the weak topology o(X. X

) on X turns out to be the relative topology on X induced by

o(X

. X

).
5.5.5 Theorem
Let t

denote the norm topology on X

. Then
95
2011 FUNCTIONAL ANALYSIS ALP
(a) o(X

. X) o(X

. X

) t

.
(b) o(X

. X

) = t

if and only if X is nite-dimensional.

(c) o(X

. X) = o(X

. X

) if and only if X is reexive. Thus, if X is non-reexive, then the weak*

topology o(X

. X) is strictly weaker than the weak topology o(X

. X

).
Proof. (a) Since J
X
X X

, it follows that o(X

. X) = o(X

. J
X
X) o(X

. X

). The containment
o(X

. X

) t

follows from the fact that o(X

. X

) is the weakest topology on X

making each
x

continuous and each x

is continuous with respect to t

.
(b) An argument similar to that used in Theorem 5.5.2(b) shows that o(X

. X

) = t

if and only
if X

is nite-dimensional. But by Proposition 4.2.4, X

is nite-dimensional if and only if X is nite-

dimensional.
(c) X is reexive if and only if J
X
X X

if and only if o(X

. J
X
X) o(X

. X

) if and only if
o(X

. X) o(X

. X

).
5.5.6 Theorem
Let X be a normed linear space. Then the dual of X

under the weak* topology o(X

. X) is X; i.e.,
(X

. o(X

. X))

= X.
Proof. Exercise.
Observe that X

F
X
=

X
F and that the weak* topology o(X

. X) on X

is the relative topology

on X

induced by the product topology on

X
F.
5.5.7 Theorem
(Banach-Alaoglu-Bourbaki Theorem). Let X be a normed linear space over F. Then the closed unit ball
in X

is weak* compact; i.e., the set

B

= B(X

) = {x

[ [x

[ _ 1]
is compact for the topology o(X

. X).
Proof. For each x X, let D
x
= {z F [ [z[ _ [x[]. Then, for each x X, D
x
is a closed interval in R
or a closed disk in C according to whether F = R or F = C. Equipped with the standard topology, D
x
is
compact for each x X. Let D =

{D
x
[ x X]. By Tychonoffs Theorem, D is compact.
The points of D are just functions on X such that (x) D
x
for each x X. If x

B(X

), then
[x

(x)[ _ [x

[[x[ _ [x[ for each x X.

Hence x

(x) D
x
for each x

B(X

) and x X. That is, B(X

{D
x
[ x X]. We observe
that the topology that D induces on B(X

) is precisely the weak* topology on B(X

). It remains to show
that B(X

) is a closed subset of D. To this end, let {x

] be a net in B(X

) and x

D. Then
x

(x) x

(x) for all x X, and for all x. y in X and . in F,

x

(x y) = lim

(x y) = lim

(x) x

(y) | = x

(x) x

(x).
Thus x

is linear. Since
[x

(x)[ = lim

[x

(x)[ _ [x[
for all x X, x

is bounded and [x

[ _ 1. That is, x

B(X

). Therefore B(X

) is closed in D and
hence compact.
96
2011 FUNCTIONAL ANALYSIS ALP
5.5.8 Theorem
(Helly). Let X be a normed linear space over Fand x

. Then, for any nite-dimensional subspace

of X

and any c > 0, there is an x

0
X such that
(i) (J
X
x
0
)(x

) = x

(x

) x

(x
0
) = x

(x

) for each x

, and
(ii) [x
0
[ _ [x

[ c.
Proof. Let {x

1
. x

2
. . . . . x

n
] be a basis for . Then (i) is equivalent to
(i) x

i
(x
0
) = x

(x

i
) for each i = 1. 2. . . . . n.
Let H
i
= {x X [ x

i
(x) = x

(x

i
)] for each i = 1. 2. . . . . n and H =
n
_
iD1
H
i
. Then, by Lemma 5.27,
H = 0. Choose any x
0
H such that [x
0
[ < d(0. H) c. Obviously, x
0
satises (i), hence (i). To
complete the proof, it sufces to show that d(0. H) _ [ x

[. Fix an h H and set h

0
= h x
0
and
H
0
= H x
0
. Then H
0
=
_
n
iD1
ker(x

i
) and d(0. H) = d(x
0
. H
0
) = d(x
0
. H
0
). By the Hahn-Banach
Theorem and Lemma 5.5.4, it follows that
d(0. H) = max{x

(x
0
) [ x

H
?
0
. [x

[ _ 1]
= max{
n

iD1

i
x

i
(x
0
) [
_
_
_
_
_
n

iD1

i
x

i
_
_
_
_
_
_ 1]
= max{
n

iD1

i
x

0
(x

i
) [
_
_
_
_
_
n

iD1

i
x

i
_
_
_
_
_
_ 1]
= max{x

0
_
n

iD1

i
x

i
_
[
_
_
_
_
_
n

iD1

i
x

i
_
_
_
_
_
_ 1]
_ max{x

0
(x

) [ x

. [x

[ _ 1] = [ x

[.
5.5.9 Theorem
(Goldstine). Let X be a normed linear space and J
X
the canonical embedding of X into X

. Let
B = {x X [ [x[ _ 1] and B

= {x

[ [x

[ _ 1]. Then J
X
B is dense in B

relative to the
weak* topology o(X

. X

) on X

. That is,
J
X
B
o(X

,X

)
= B

.
Proof. We must show that for each x

, each nite subset = {x

1
. x

2
. . . . . x

n
] X

and each
c > 0, there is an x B such that J
X
x V(x

: : c); i.e.,
[J
X
x(x

i
) x

(x

i
)[ < c for each i = 1. 2. . . . . n.
Let x

. If [x

[ < 1, then, with c = 1 [x

[, we have, by Theorem 5.5.8, that there is an

x X such that (J
X
x)(x

i
) = x

(x

i
) for each i = 1. 2. . . . . n and [x[ < [x

[ c = 1; i.e., x B.
Hence, x B and 0 = [J
X
x(x

i
) x

(x

i
)[ < c for each i = 1. 2. . . . . n.
If [x

[ = 1, let r = max
1in
[x

i
[ and y

=
_
1
e
2r
_
x

. Then [y

[ < 1, and so by the rst

part, there is an x B such that (J
X
x)(x

i
) = y

(x

i
) for each i = 1. 2. . . . . n. Furthermore, for each
i = 1. 2. . . . . n,
[J
X
x(x

i
) x

(x

i
)[ = [y

(x

i
) x

(x

i
)[ _
c
2r
_
c
2
< c
97
2011 FUNCTIONAL ANALYSIS ALP
5.5.6 Corollary
Let X be a normed linear space over F and let J
X
be the canonical embedding of X into X

. Then J
X
X
is dense in X

relative to the weak* topology o(X

. X

) on X

. That is,
J
X
X
o(X

,X

)
= X

.
Proof. Let x

\ {0]. Then
x

[x

[
B

= J
X
B
o(X

,X

)
J
X
X
o(X

,X

)
.
Since J
X
X
o(X

,X

)
is a linear subspace of X

, it now follows that x

J
X
X
o(X

,X

)
. Hence
X

J
X
X
o(X

,X

)
. Of course, since J
X
X X

, we have that J
X
X
o(X

,X

)
X

, and conse-
quently J
X
X
o(X

,X

)
= X

.
5.5.10 Theorem
Let X be a normed linear space over F and B = {x X [ [x[ _ 1]. Then X is reexive if and only if B
is weakly compact.
Proof. Assume that X is reexive and let J
X
be the canonical embedding of X into X

. Equip B (respec-
tively, B

) with the weak (respectively, weak*) topology and consider the map : B

B dened by
(J
X
x) = x. Now, B

is weak* compact by Banach-Alaoglu-Bourbaki Theorem and (B

) = B. To
prove weak compactness of B, it sufces to show that is continuous. To that end, let (J
X
x

) be a net in
B

that converges to J
X
x in the topology o(X

. X

) on X

. Then, for each x

, we have that
x

_
(J
X
x

)
_
= x

(x

) = (J
X
x

)(x

) J
X
x(x

) = x

(x) = x

( (J
X
x)).
Thus, (J
X
x

) (J
X
x) in the weak topology on B.
Conversely, assume that B is weakly compact. Equip B (respectively, B

) with the weak (respec-

tively, weak*) topology. It follows that J
X
is continuous. Hence, J
X
B is weak* compact in B

. But
J
X
B
o(X

,X

)
= B

. Hence J
X
X = X

and so X is reexive.
98
Chapter 6
Baires Category Theorem and its
Applications
6.1 Introduction
Recall that a subset S of a metric space (X. d) is dense in X if S = X; i.e., for each x X and each
c > 0, there is an element y S such that d(x. y) < c, or equivalently, S B(x. c) = 0.
6.1.1 Theorem
Let (X. d) be a complete metric space. If (G
n
) is a sequence of nonempty, open and dense subsets of X
then G =
_
n2N
G
n
is dense in X.
Proof. Let x X and c > 0. Since G
1
is dense in X, there is a point x
1
in the open set G
1
B(x. c).
Let r
1
be a number such that 0 < r
1
<
c
2
and
B(x
1
. r
1
) G
1
B(x. c).
Since G
2
is dense in X, there is a point x
2
in the open set G
2
B(x
1
. r
1
). Let r
2
be a number such that
0 < r
2
<
c
2
2
and
B(x
2
. r
2
) G
2
B(x
1
. r
1
).
Since G
3
is dense in X, there is a point x
3
in the open set G
3
B(x
2
. r
2
). Let r
3
be a number such that
0 < r
3
<
c
2
3
and
B(x
3
. r
3
) G
3
B(x
2
. r
2
).
Continuing in this fashion, we obtain a sequence (x
n
) in X and a sequence (r
n
) of radii such that for each
n = 1. 2. 3. . . .,
0 < r
n
<
c
2
n
. B(x
nC1
. r
nC1
) G
nC1
B(x
n
. r
n
) and B(x
1
. r
1
) G
1
B(x. c).
It is clear that
B(x
nC1
. r
nC1
) B(x
n
. r
n
) B(x
n1
. r
n1
) B(x
1
. r
1
) B(x. c).
Let N N. If k > N and > N, then both x
k
and x
I
lie in B(x
N
. r
N
). By the triangle inequality
d(x
k
. x
I
) _ d(x
k
. x
N
) d(x
N
. x
I
) < 2r
N
<
2c
2
N
=
c
2
N1
.
99
2011 FUNCTIONAL ANALYSIS ALP
Hence, (x
n
) is a Cauchy sequence in X. Since X is complete, there is a y X such that x
n
y as
n o. Since x
k
lies in the closed set B(x
n
. r
n
) if k > n, it follows that y lies in each B(x
n
. r
n
). Hence
y lies in each G
n
. That is, G =
_
n2N
G
n
,= 0. It is also clear that y B(x. c).
6.1.1 Denition
A subset S of metric space (X. d) is said to be nowhere dense in X if the set X \ S is dense in X; i.e.,
X \ S = X.
6.1.2 Proposition
A subset S of a metric space (X. d) is nowhere dense in X if and only if the closure S of S contains no
interior points.
Proof. Assume that S is nowhere dense in X and that (S)

= 0. Then there is an c > 0 and an x S such

that B(x. c) S. But then X \S X \B(x. c). Since X \B(x. c) is closed, X \B(x. c) = X \ B(x. c).
Therefore
X \ S X \ B(x. c) X.
where the second containment is proper. This is a contradiction. Hence, (S)

= 0.
Conversely, assume that (S)

= 0. Then, for each x S and each c > 0,

B(x. c) X \ S = 0.
This means that each x S is a limit point of the set X \ S. That is, S X \ S. Thus,
X = S L(X \ S) X \ S LX \ S = X \ S X.
Hence X = X \ S and so S is nowhere dense in X.
6.1.3 Example
Each nite subset of R is nowhere dense in R.
6.1.4 Denition
A subset S of a metric space (X. d) is said to be
(a) of rst category or meagre in X if S can be written as a countable union of sets which are nowhere
dense in X. Such sets are also called thin.
(b) of second category or nonmeagre in X if it is not of rst category in X. Such sets are also called
fat or thick.
It is clear that a subset of a set of rst category is itself a set of rst category. Also, a countable union of
sets of rst category is again a set of rst category.
6.1.5 Example
The set Q of rationals is of rst category in R.
6.1.2 Theorem
(Baires Category Theorem). A complete metric space (X. d) is of second category in itself.
Proof. Assume that X is of rst category. Then there is a sequence (G
n
) of sets which are nowhere dense in
X such that X =
_
n
G
n
. Replacing each G
n
by its closure, we get X =
_
n
G
n
. The sets G
n
are closed and
100
2011 FUNCTIONAL ANALYSIS ALP
nowhere dense in X. It follows that the sets U
n
= X \ G
n
are open and dense in X. Since X is complete,
it follows, by Theorem 6.1.1, that U =
_
n
U
n
is dense in X and therefore nonempty since X is nonempty.
However X =
_
n
G
n
implies that
0 =
_
n
U
n
=
_
n
(X \ G
n
) = X \
_
n
G
n
= 0.
which is absurd.
6.2 Uniform Boundedness Principle
We have made the point that if X and Y are normed linear spaces, then B(X. Y ) is a normed linear space.
6.2.1 Denition
A subset F of B(X. Y ) is said to be
(a) norm (or uniformly) bounded if
sup{[T[ [ T F] < o.
(b) pointwise bounded on X if
sup{[Tx[ [ T F] < o
for each x X.
Clearly, a norm bounded set is pointwise bounded on X. Uniform Boundedness Principle (or Banach-
Steinhaus Theorem) says that if X is a Banach space, then the converse also holds.
6.2.1 Theorem
(Uniform Boundedness Principle). Let X be a Banach space, Y a normed linear space and let F be
subset of B(X. Y ) such that sup{[Tx[ [ T F] < ofor each x X. Then sup{[T[ [ T F] < o.
Proof. For each k N, let
A
k
= {x X [ [Tx[ _ k for all T F].
Since T is continuous, A
k
is closed. Indeed, let x A
k
. Then there is a sequence (x
n
) A
k
such that
lim
n!1
x
n
= x. Since x
n
A
k
for each n, [Tx
n
[ _ k for all T F. Hence
[Tx[ _ [Tx T x
n
[ [Tx
n
[ _ [T[[x
n
x[ k k as n o.
That is, [Tx[ _ k and consequently x A
k
.
By the hypothesis, X =
1
_
kD1
A
k
. By Baires Category Theorem, there is an index k
0
such that (A
k
0
)

= 0.
That is, there is an x
0
A
k
0
and an c > 0 such that B(x
0
. c) A
k
0
= A
k
0
.
Let x X \ {0] and set z = x
0
zx, where z =
c
2[x[
. Then [z x
0
[ = z[x[ =
c
2
< c. Hence
z B(x
0
. c) A
k
0
and, consequently, [Tz[ _ k
0
for all T F. It now follows that
[Tx[ =
1
z
[Tz Tx
0
[ _
1
z
([Tz[ [Tx
0
[) _
2k
0
z
=
4k
0
c
[x[.
Hence [T[ _
4k
0
c
for all T F.
101
2011 FUNCTIONAL ANALYSIS ALP
It is essential that X be complete in Theorem 6.2.1. Consider the subset
0

1
of nitely nonzero
sequences in
1
. The set
0
is dense but not closed in
1
. For each n N, let T
n
x = nx
n
, where
x = (x
n
)
0
. For each x
0
, T
n
x = 0 for sufciently large n. Clearly, (T
n
) is pointwise bounded on

0
. On the other hand, for (e
n
)
0
, [e
n
[ = 1 and [T
n
[ _ T
n
e
n
= n for all n N. Thus (T
n
) is not norm
bounded.
6.2.2 Corollary
Let S be a subset of a normed linear space (X. [ [) such that the set {x

(x) [ x S] is bounded for each

x

. Then the set S is bounded.

Proof. Let J
X
be the canonical embedding of X into X

. By the hypothesis, the set {J

X
x(x

) [ x S]
is bounded for each x

. Since X

is a Banach space, it follows from the Uniform Boundedness

Principle that the set {J
X
x [ x S] is bounded. Since [J
X
x[ = [x[, the set S is also bounded.
Let X and Y be normed linear spaces. We remarked earlier that the strong operator limit T of a
sequence (T
n
) B(X. Y ) need not be bounded. However, if X is complete, then T is also bounded. This
is a consequence of the Uniform Boundedness Principle.
6.2.3 Corollary
Let (T
n
) be a sequence of bounded linear operators from a Banach space X into a normed linear space Y .
If T is the strong operator limit of the sequence (T
n
), then T B(X. Y ).
Proof. The proof of linearity of T is straightforward.
We show that T is bounded. Since for each x X, T
n
x Tx as n o, the sequence (T
n
x) is
bounded for each x X. By the Uniform Boundedness Principle, we have that the sequence ([T
n
[) is
bounded. That is, there is a constant M > 0 such that [T
n
[ _ M for all n N. Therefore
[T
n
x[ _ [T
n
[[x[ _ M[x[ for all n N.
By continuity of the norm,
[Tx[ _ [Tx T
n
x[ [T
n
x[ _ [Tx T
n
x[ M[x[ M[x[ as n o.
Hence, [Tx[ _ M[x[ for each x X, i.e., T B(X. Y ).
6.3 The Open Mapping Theorem
6.3.1 Denition
Let X and Y be normed linear spaces over the same eld F and let T : X Y . Then we say that T is an
open mapping if T U is open in Y whenever U is open in X.
6.3.2 Lemma
Let X and Y be Banach spaces over the eld F and let T be a bounded linear operator from X onto Y .
Then there is a constant r > 0 such that
B
Y
(0. 2r ) := {y Y [ [y[ < 2r ] TB
X
(0. 1).
Proof. It is easy to see that X =
1
_
nD1
nB
X
(0. 1). Indeed, if x X, then there is an n N such that
[x[ < n. Hence, x nB
X
(0. 1). Since T is surjective,
Y = TX = T
_
1
_
nD1
nB
X
(0. 1)
_
=
1
_
nD1
nTB
X
(0. 1) =
1
_
nD1
nTB
X
(0. 1).
102
2011 FUNCTIONAL ANALYSIS ALP
By Baires Category Theorem, there is a positive integer n
0
such that (n
0
TB
X
(0. 1))

= 0. This implies
that (TB
X
(0. 1))

= 0. Hence, there is a constant r > 0 and an element y

0
Y such that B
Y
(y
0
. 4r )
TB
X
(0. 1). Since y
0
TB
X
(0. 1), it follows, by symmetry, that y
0
TB
X
(0. 1). Therefore
B
Y
(0. 4r ) = B
Y
(y
0
. 4r ) y
0
TB
X
(0. 1) TB
X
(0. 1).
Since TB
X
(0. 1) is a convex set, TB
X
(0. 1)TB
X
(0. 1) = 2TB
X
(0. 1). Hence, B
Y
(0. 4r ) 2TB
X
(0. 1)
and, consequently, B
Y
(0. 2r ) TB
X
(0. 1).
6.3.3 Lemma
Let X and Y be Banach spaces over the eld F and let T be a bounded linear operator from X onto Y .
Then there is a constant r > 0 such that
B
Y
(0. r ) := {y Y [ [y[ < r ] TB
X
(0. 1).
Proof. By Lemma 6.3.2, there is a constant r > 0 such that B
Y
(0. 2r ) TB
X
(0. 1). Let y B
Y
(0. r ),
i.e., y Y and [y[ < r . Then, with c =
r
2
, there is an element z
1
X such that
[z
1
[ <
1
2
and [y T z
1
[ <
r
2
.
Since y Tz
1
Y and [y Tz
1
[ <
r
2
< r , it follows that y Tz
1
B
Y
(0. r ). Therefore there is an
element z
2
X such that
[z
2
[ <
1
2
2
and [(y Tz
1
) Tz
2
[ <
r
2
2
.
In general, having chosen elements z
k
X, 1 _ k _ n, such that [z
k
[ <
1
2
k
and
[y (T z
1
T z
2
Tz
n
)[ <
r
2
n
.
pick z
nC1
X such that [z
nC1
[ <
1
2
nC1
and
[y T(z
1
z
2
z
n
z
nC1
)[ = [y (T z
1
Tz
2
Tz
n
Tz
nC1
)[ <
r
2
nC1
Claim: The series
1

kD1
z
k
converges to a point x B
X
(0. 1) and Tx = y.
Proof of Claim: Since X is complete, it sufces to show that
1

kD1
[z
k
[ < o. But this is obviously true
since
1

kD1
[z
k
[ <
1

kD1
1
2
k
= 1.
Hence, the series
1

kD1
z
k
converges to some x X with [x[ < 1, i.e., x B
X
(0. 1). Since
lim
n!1
_
_
_
_
_
y T
_
n

kD1
z
k
__
_
_
_
_
= lim
n!1
r
2
n
= 0.
continuity of T implies that
Tx = lim
n!1
T
_
n

kD1
z
k
_
= y.
That is, Tx = y.
103
2011 FUNCTIONAL ANALYSIS ALP
6.3.1 Theorem
(Open Mapping Theorem). Let X and Y be Banach spaces and suppose that T B(X. Y ). If T maps
X onto Y , then T is an open mapping.
Proof. Let U be an open set in X. We need to show that T U is open in Y . Let y T U. Since
T is surjective, there is an x U such that Tx = y. Since U is open, there is an c > 0 such that
B
X
(x. c) = x B
X
(0. c) U. But then y TB
X
(0. c) T U. By Lemma 6.3.3, there is a constant
r > 0 such that B
Y
(0. r ) TB
X
(0. 1). Hence B
Y
(0. r c) TB
X
(0. c). Therefore
B(y. r c) = y B
Y
(0. r c) y TB
X
(0. c) T U.
Hence T U is open in Y .
6.3.4 Corollary
(Banachs Theorem). Let X and Y be Banach spaces and assume T B(X. Y ) is bijective. Then T
1
is
a bounded linear operator from Y onto X, i.e., T
1
B(Y. X).
Proof. We have shown that T
1
is linear. It remains to show that T
1
is bounded. By Theorem 4.1.4, it
sufces to show that T
1
is continuous on Y . To that end, let U be an open set in X. By Theorem 6.3.1,
(T
1
)
1
(U) = T U is open in Y . Hence T
1
is continuous on Y .
6.4 Closed Graph Theorem
6.4.1 Denition
Let X and Y be linear spaces over a eld F and T : X Y . The graph of T, denoted by G(T ), is the
subset of X Y given by
G(T ) = {(x. Tx) [ x X].
Since T is linear, G(T ) is a linear subspace of X Y . Let [ [
X
and [ [
Y
be norms on X and Y
respectively. Then, for x X and y Y , [(x. y)[ := [x[
X
[y[
Y
denes a norm on X Y . If X and
Y are Banach spaces, then so is X Y .
6.4.2 Denition
Let X and Y be normed linear spaces over F. A linear operator T : X Y is closed if its graph G(T ) is
a closed linear subspace of X Y .
6.4.1 Theorem
(Closed Graph Theorem). Let X and Y be Banach spaces and T : X Y a closed linear operator. Then
T is bounded.
Proof. Since X Y , with the norm dened above, is a Banach space, and by the hypothesis G(T) is closed,
it follows that G(T ) is also a Banach space. Consider the map P : G(T ) X given by P(x. Tx) = x.
Then P is linear and bijective. It is also bounded since
[P(x. Tx)[ = [x[ _ [x[ [Tx[ = [(x. Tx)[.
That is, P is bounded and [P[ _ 1. By Banachs Theorem (Corollary 6.3.4), it follows that P
1
: X
G(T ) given by P
1
x = (x. Tx) for x X, is also bounded. Hence [(x. Tx)[ = [P
1
x[ _ [P
1
[[x[.
Therefore
[(x. Tx)[ = [x[ [Tx[ _ [P
1
[[x[ [Tx[ _ [P
1
[[x[.
That is, T is bounded and [T[ _ [P
1
[.
104