Linear Topological Spaces
by
William C. McGu ey
A thesis submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Master of Science
Auburn, Alabama
May 4, 2014
Keywords: topology, linear algebra
Copyright 2014 by William C. McGu ey
Approved by
Michel Smith, Chair, Professor of Mathematics and Statistics
Ulrich Albrecht, Professor of Mathematics and Statistics
Piotr Minc, Professor of Mathematics and Statistics
Abstract
In this thesis several topics from Topology, Linear Algebra, and Real Analysis are com-
bined in the study of linear topological spaces. We begin with a brief look at linear spaces
before moving on to study some basic properties of the structure of linear topological spaces
including the localization of a topological basis. Then we turn our attention to linear spaces
with a metric topology. In particular, we consider problems involving normed linear spaces,
bounded linear transformations, and Hilbert spaces.
ii
Acknowledgments
I would like to thank my advisor, Dr. Michel Smith, for the time and e ort he put into
my development as a mathematician. His style of teaching fostered a constructive learning
environment, and he always encouraged me to try to answer my own questions. I would also
like to thank my other committee members: Dr. Piotr Minc for his help and guidance in
the graduate Topology course sequence, and Dr. Ulrich Albrecht for his encouragement and
willingness to help me in every possible way. Also, I would like to thank Dr. Scott Varagona
who helped me discover my passion for mathematics.
My family deserves many thanks for always being there for me. In particular, thanks to
my parents for all of their love and support during my time at Auburn. Most of all, thanks
to my wife, Danna, for always believing in me and for inspiring me to be the best person I
can be. Before we met there was a hole in my life, but ever since I fell in love with her during
our study group meetings for Dr. Smith?s Topology class, all of the Cauchy sequences in my
life are convergent.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Notation and Terminology . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Preliminary Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2 Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.1 Elementary Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2.2 Base and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.3 Linear Subspaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Linear Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.1 Linear Operations That Are Homeomorphisms . . . . . . . . . . . . . . . . . 19
3.2 Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
3.3 Local Topological Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.4 More Properties of Linear Topological Spaces . . . . . . . . . . . . . . . . . 28
3.5 Elementary Computations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.6 Linear Functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4 Metric Linear Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.1 Totally Bounded Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
4.2 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5.1 Normed Linear Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
5.2 Bounded Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . 47
6 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
iv
6.1 Inner Product Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.2 Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
v
Chapter 1
Introduction
This thesis is a compilation of solutions to problems assigned under the direction of
Professor Michel Smith. The material in this thesis was developed through a one-on-one
Moore method directed study. I was given notes on linear topological spaces which included
de nitions and theorems to be proved. During weekly meetings I presented my solutions
to various problems and we discussed approaches to solving problems. A large portion of
the notes came from Linear Topological Spaces by John L. Kelley and Isaac Namioka, but
other material came about when we were side-tracked during our weekly discussions. All
proofs in this thesis can be assumed to be my own unless otherwise stated. The style of
the thesis re ects the way in which it was developed. It could be used as a framework for
an introductory course on linear topological spaces, and it reads like a set of class notes
with solutions included. A course on linear topological spaces could be useful for upper-
level undergraduate students or new graduate students because it integrates many areas of
mathematics including Topology, Linear Algebra, and Real Analysis.
1.1 Notation and Terminology
Throughout this thesis I try to be as consistent as possible with notation. I use uppercase
English letters such as X, Y, and Z to denote linear spaces, topological spaces, or linear
topological spaces. The elements of these spaces are denoted by lowercase letters such as x,
y, and z. The elements of a linear space are called vectors, but we refer to the elements of
a linear topological space as points. An arbitrary set of vectors or points will be denoted by
uppercase English letters such as A, B, E, F, and M, while the symbols U, V, and W will
usually be reserved for open subsets of a linear topological space.
1
A linear space consists of a set X together with two operations, addition + and scalar
multiplication , and is denoted by (X;+; ). However, the operations are usually omitted,
and the space is simply denoted by X. Moreover, a topological space consists of a set X
together with a collection T of subsets of X and is denoted by (X;T). Again, the topology
is omitted, and we simply denote the space by X.
Associated with a linear space is an underlying eld of (real or complex) numbers,
denoted by F. The elements of the eld are called scalars and are denoted by lowercase
Greek letters such as and . The real part of a complex number will be denoted by 0 such that 2A whenever j j< . Fix 0 2F with 0 n, such that (xi +yi)1i=1
1Y
i=1
Ui U. For
each i2f1;:::;ng, the addition function fi : Xi Xi!Xi, de ned by fi(xi;yi) = xi + yi,
is continuous because Xi is a linear topological space. So, there is an open subset Vi Wi
of Xi Xi containing (xi;yi) such that Vi + Wi Ui. For i>n, let Vi = Xi and Wi = Xi.
30
Then,
V =
1Y
i=1
Vi and W =
1Y
i=1
Wi
are open subsets of X so that V W is an open subset of X X. Moreover, for any
((x0i)1i=1;(y0i)1i=1)2V W, we have
f((x0i)1i=1;(y0i)1i=1) = (x0i +y0i)1i=1 2
1Y
i=1
(Vi +Wi)
1Y
i=1
Ui U
Therefore, coordinate-wise addition on X is a continuous function.
Now, let g denote coordinate-wise scalar multiplication as a function from F X into
X. Fix a scalar 0 2 F and let (xi)1i=1 be a point of X. Given an open subset U of X
containing ( 0xi)1i=1, there is a basis element
1Y
i=1
Ui = U1 U2 Un Xn+1 Xn+2
containing ( 0xi)1i=1 such that
1Y
i=1
Ui U. For each i2f1;:::;ng, the scalar multiplication
gi : F Xi!Xi, de ned by gi( ;xi) = xi, is continuous because Xi is a linear topological
space. For each i2f1;:::;ng, there exists an open subset Vi of F containing 0 and an open
subset Wi of Xi containing (xi)1i=1 such that xi2Ui, for each 2Vi and for each xi2Wi.
Let V =
n\
i=1
Vi. Then, V is an open subset of F containing 0 such that Wi Ui, for all
2V. For i>n, let Wi = Xi so that
1Y
i=1
Vi is an open set in X. As a result, V W is an
open subset of F X such that, for any ( ;(x0i)1i=1)2V W,
g( ;(x0i)1i=1) = ( x0i)1i=1 2
1Y
i=1
Ui U
Therefore, coordinate-wise scalar multiplication on X is continuous. Because both addition
and scalar multiplication are continuous according to their respective product topologies, it
follows that X is a linear topological space.
31
3.5 Elementary Computations
The theorems in this section rely heavily on the fact that the addition operation on a
linear space X is a continuous function from X X into X. In particular, we consider the
relationship between the addition operation and the interiors and closures of sets. We will
also nd conditions to determine if an algebraic sum of sets is closed or compact. Throughout
this section we will assume that X is a linear topological space with subsets A and B. We
will use F to denote the addition operation on X in order to more easily express images and
pre-images of sets under the addition operation.
Theorem 3.14. If each of A and B is a subset of a linear topological space X, then
Int(A+B) = Int(A) + Int(B):
Proof. If U and V are open subsets of X satisfying U A and V B, then U + V is an
open subset of X satisfying U +V A+B. Consequently, we have
Int(A) + Int(B) Int(A+B)
On the other hand, let F : X X !X denote the addition operation on X. Suppose
that x2 Int(A + B) = Int(F(A B)). Then, there is an open subset W of X such that
x2W F(A B). Because F is continuous, F 1(W) is an open neighborhood of F 1(x).
We can write F 1(x) = (a;b), for some a2A and for some b2B. So, there is an open
subset U V of X X such that (a;b)2U V and U V F 1(W). As a result, U is
an open subset of X satisfying a2U and U A, which implies that a2Int(A). Similarly,
b2Int(B). Therefore, F 1(x)2Int(A) Int(B). As a result, we have
x = F(a;b)2F(Int(A) Int(B)) = Int(A) + Int(B)
32
Hence, Int(A+B) Int(A) + Int(B) so that we have both inclusions.
Lemma 3.2. If A and B are subsets of a linear topological space X, then
A+B A+B:
Proof. Let F denote the addition on X as a function from X X into X. Because F is
continuous, the image of the closure of a set is contained in the closure of the image of the
set. Notice that A B = A B in the product space X X. As a result, we have
A+B = F(A B) = F(A B) F(A B) = A+B
Hence, A+B A+B.
Example 3.1. The other inclusion in Lemma 3.2 does not necessarily hold. For instance, let
A =f(x;y)2R2 : x> 0;y 1=xg and B =f(x;y)2R2 : x< 0;y 1=xg:
Both A and B are closed so that A = A and B = B. The algebraic sum is given by
A+B =f(x;y)2R2 : y> 0g, the closure of which is the set A+B =f(x;y)2R2 : y 0g.
Because A+B is not closed, we have A+B6 A+B = A+B.
Theorem 3.15. Let A be a subset of a linear topological space X. For any point x of X,
x+A = x+A:
Proof. The inclusion x+A x+A results from Lemma 3.2. For the other inclusion, suppose
that p2x+A. Write p = x+y so that y = x+p. Let U be an open neighborhood of y.
Then, x + U is an open neighborhood of x + y = p. Since p2x+A, it follows that x + U
intersects x+A. As a result, U must intersect A. Therefore, every open neighborhood of y
intersects A. In other words, y2A so that p = x+y2x+A. Hence, x+A x+A.
33
The previous theorem may be generalized a little bit. Given any subset A of a linear
topological space X and a nite subset F = fx1;:::;xng of X, the set F + A is the nite
union of the translates xi+A, for i2f1;:::;ng. Because the closure of a nite union of sets
is equivalent to the union of the closures of the sets, Theorem 3.15 implies that the closure
of F + A is equal to the union of the translates xi + A. Therefore, F +A = F + A. This
establishes the following corollary.
Corollary 3.6. If A is any subset of a linear topological space X and F is a nite subset of
X, then F +A = F +A.
Theorem 3.16. If each of A and B are compact subsets of X, then A+B is compact.
Proof. Because A and B are compact subsets of X, their product A B is compact. Let
F : X X!X be the addition operation de ned on X. Then,
F(A B) =fa+b : a2A;b2Bg= A+B
So, A+B is the continuous image of the compact set A B. Therefore, A+B is compact.
The proof of the previous theorem uses two basic ideas. One is that the product of
compact sets is compact, and the other is that the image of a compact set under a continuous
map is compact. Given two sets A and B satisfying a topological property that is preserved
under products and continuous maps, the sum A+B will also satisfy that property.
Recall that the sum of two sets is open whenever at least one of the sets is open. The
next example demonstrates that another assumption is necessary to guarantee that the sum
of two sets is closed whenever at least one is closed.
Example 3.2. If A and B are closed subsets of a linear topological space X, then A + B is
not necessarily closed. For instance, let X = R2 and consider the subsets
A =f(x;y) : x> 0;y 1=xg and B =f(x;y) : x<;y 1=xg
34
Each of A and B is closed, but the sum A+B =f(x;y) : y> 0g is not closed.
Although the sum A + B is not necessarily closed whenever both A and B are closed,
it is true that A + B is closed if we include the additional assumption that one of the two
sets is compact. In order to show this, we will rst prove another result that can be useful.
Theorem 3.17. Suppose that A is a closed subset of X and B is a compact subset of X
such that A and B are disjoint. Then there is an open neighborhood V of zero such that A
and B +V are disjoint.
Proof. Let x 2 B. Because A and B are disjoint, it follows that x belongs to the open
set XnA. So, x + XnA is an open neighborhood of zero. By Theorem 3.6, there is a
neighborhood Wx of zero satisfying Wx +Wx x+XnA. For each x2B, x+Wx is an
open neighborhood of x, and therefore the collection fx + Wx : x2Bg is an open covering
of B. Because B is compact, there exist nitely many points x1;:::;xn 2 B such that
B
n[
i=1
(xi + Wxi). Let V =
n\
i=1
Wxi, which is an open set because it is a nite intersection
of open sets. Given x2B, x2xi + Wxi, for some i2f1;:::;ng. In particular, for some
i2f1;:::;ng, we have
x+V x+Wxi (xi +Wxi) +Wxi = xi + (Wxi +Wxi)
By construction, Wxi +Wxi xi +XnA so that xi + (Wxi +Wxi) XnA, and we have
x+V XnA. In other words, A\(x+V) =;, for all x2B. Thus, A\(B +V) =;.
Theorem 3.18. If A is a closed subset of X and B is a compact subset of X, then the
algebraic sum A+B is closed.
Proof. If X = A+B, then the statement holds because X is closed. Suppose that A+B (X
and let x2Xn(A+B). Then, A\(x B) =;because otherwise a point z in this intersection
would satisfy z = a, for some a2A, and it would satisfy z = x b, for some b2B, so
that x could be expressed as x = a+b, contrary to assumption. Notice that B is compact
35
because B is compact and multiplication by a non-zero scalar is a homeomorphism. By
Theorem 3.17, there is an open neighborhood V of 0 such that A\(x+V B) =;, that is,
(A+B)\(x+V) =;. Since x+V is an open neighborhood of x contained in Xn(A+B),
it follows that A+B is closed.
3.6 Linear Functionals
Given a linear space X over a scalar eld F, a linear functional is a linear map from X
into the scalar eld F. Let F(X) denote the collection of linear functionals on X. De ne
addition and scalar multiplication on F(X) as follows: if f;g 2F(X), then f + g is the
linear functional on X such that (f +g)(x) = f(x) +g(x), for each x2X; if f2F(X) and
is any scalar, then f is the element ofF(X) such that ( f)(x) = f(x), for each x2X.
The collection F(X) together with addition and scalar multiplication de ned in this way is
a linear space.
Theorem 3.19. Suppose that X is a linear topological space, and let f2F(X) be a linear
functional which is not identically zero. The following are equivalent:
(1) f is continuous
(2) the null space of f is closed
(3) the null space of f is not dense in X
(4) f is bounded on a neighborhood of zero
Proof.
If f is continuous, then the nullspace of f is closed.
Suppose that f is continuous, and let N denote the nullspace of f. Then, XnN
is non-empty because f is not identically zero. Given a point x0 2XnN, we have
f(x0) 6= 0. Without loss of generality, assume that f(x0) > 0, and let U = (0;1).
36
Since f is continuous, f 1(U) is an open set containing x0. Now, for each x2f 1(U),
we have f(x)2(0;1) so that f(x)6= 0. Therefore, f 1(U) is an open set containing
x0 such that f 1(U) XnN. Hence, XnN is open and the nullspace N is closed.
If the nullspace of f is closed, then the null space of f is not dense in X.
Assume that the nullspace N of f is dense in X, that is, N = X. By assumption, N
is closed so that N = N. However, this implies that N = N = X, which contradicts
the assumption that f is not identically zero.
If the nullspace of f is not dense in X, then f is bounded on a neighborhood of zero.
Suppose that the nullspace N of f is not dense in X. Then, there is a point x and
a neighborhood U of zero such that x + U does not intersect N. Because U is a
neighborhood of zero, it contains a balanced neighborhood V of zero, that is, V U,
for all scalars withj j 1. Assume f is not bounded on V. Then, there exists v2V
such that jf(v)j jf(x)j. Choose 2F with j j 1 such that j j jf(v)j = jf(x)j.
Then,
jf(x)j=j jjf(v)j=j f(v)j=jf( v)j
Because j j 1, we have V U so that there is an element u of U with v = u and
hence jf(x)j=jf(u)j. Without loss of generality, assume that f(u) = f(x) because
we can take as needed. Then,
0 = f(x) +f(u) = f(x+u)
As a result, x + u is an element of the nullspace N of f, contrary to the assumption
that x+U does not intersect N. Thus, f must be bounded on V.
If f is bounded on a neighborhood of zero, then f is continuous.
Suppose that f is bounded on an open set U containing zero. Then, there is a positive
number M such that jf(x)j < M, for all x 2 U. Given > 0, let = M and
37
de ne V = U. For each x2V, there exists u2U such that x = u. As a result,
jf(x)j = jf( u)j = jf(u)j. Because u 2 U, it follows that jf(u)j < M so that
jf(x)j< M = . Therefore, f is continuous at zero. Since f is linear, it follows that
f is continuous at each of its points.
38
Chapter 4
Metric Linear Topological Spaces
In the remaining chapters we will study linear topological spaces for which the topology
is induced by a metric. I will assume some familiarity with metric spaces throughout the
rest of the thesis. Of particular interest will be spaces with a translation-invariant metric d,
meaning that d(x;y) = d(x + z;y + z), for any points x, y, and z in X. Given a point x of
a metric space X, the -ball centered at x will be denoted by B (x).
4.1 Totally Bounded Sets
A subset of a metric space is said to be totally bounded if for each > 0 the set can
be covered by nitely many -balls. The following de nition extends this concept to linear
topological spaces which are not necessarily metric spaces. In this section we show that the
two de nitions are equivalent for metric linear topological spaces.
De nition. A subset B of a linear topological space X is totally bounded if for each neigh-
borhood U of zero there exists a nite set F such that B F +U.
It is clear that the set F in the de nition of a totally bounded set may be assumed to
be a subset of B. It is also clear that a subset of a totally bounded set is totally bounded.
The following theorems demonstrate other properties of totally bounded sets.
Theorem 4.1. The closure of a totally bounded set is totally bounded.
Proof. Suppose that B is a totally bounded subset of a linear topological space X, and let
U be an open neighborhood of zero. Because linear topological spaces are regular, there is a
neighborhood V of zero such that V U. By assumption, there is a nite set F such that
39
B F +V. Then, B F +V. Since F is nite, we have F +V = F +V by Corollary 3.6.
As a result, B F +V F +U. Thus, B is a totally bounded set.
Theorem 4.2. The image of a totally bounded set under a continuous linear map is totally
bounded.
Proof. Suppose that B is a totally bounded subset of X and T : X !Y is a continuous
linear map. Let V be a neighborhood of zero in Y. For any y2T(B), there exists x2B
such that T(x) = y. Since y + V is a neighborhood of y and T is continuous, there is an
open neighborhood U of x in X such that T(U) y + V. Because U is a neighborhood of
x, it follows that x + U is a neighborhood of zero. Since B is totally bounded, there is a
nite set F such that B F + ( x+U). Consequently, we have T(B) T( x+F +U) so
that T(B) T(x) +T(F) +T(U) by linearity of T. Because T(U) y+V and y = T(x),
it follows that T(B) T(F)+V. Moreover, T(F) is nite since F is nite. Therefore, T(B)
is totally bounded.
Theorem 4.3. A subset of a product is totally bounded if and only if each of its projections
is totally bounded.
Proof. Let fX : 2 g be a collection of linear topological spaces. Let denote the
projection map of
Y
2
X onto X . Suppose that B is a totally bounded subset of the
product space
Y
2
X . Fix 2 , and let U be an arbitrary neighborhood of zero in X .
Then,
Y
2
U , where U = X for 6= , is an open neighborhood of zero in
Y
2
X . By
assumption, there is a nite set F such that B F +
Y
2
U . Then, since the projection
map is linear,
(B)
F +
Y
2
U
!
= (F) +U
where (F) is nite. Therefore, (B) is totally bounded.
Conversely, suppose that (B) is totally bounded, for each 2 . Without loss
of generality, we can show that B is contained in a nite translate of a (topological) basis
40
element containing zero. Let
Y
2
U be a neighborhood of zero, where U 6= X for only
nitely many indices, say 1;:::; n, in . For each 2 , U is a neighborhood of zero
in X . By assumption, there exists a nite set F such that (B) F + U . For each
=2f 1;:::; ng, we can take F = f0g since U = X . Let F be the subset of
Y
2
X
de ned by
Y
2
F , that is,
F = (x ) : x j 2F j for j = 1;:::;n;x = 0 otherwise
Then, F is nite since each F j is nite. For each 2 , we have (B) F + U , and
therefore
B
Y
2
(F +U ) =
Y
2
F +
Y
2
U = F +U
Thus, B is totally bounded.
Theorem 4.4. A scalar multiple of a totally bounded set is totally bounded.
Proof. Suppose that B is a totally bounded subset of X. Clearly, 0 B = f0g is totally
bounded. Let be a non-zero scalar. If U is any neighborhood of zero, then 1U is also a
neighborhood of zero. Since B is totally bounded, there is a nite set F =fx1;:::;xngsuch
that B F + 1U. As a result, B (F + 1U) = F +U, where F =f x1;:::; xng
is nite. So, B is totally bounded.
If the metric for a linear topological space is translation-invariant, then the -ball cen-
tered at a point is equvialent to the translate of the -ball centered at zero.
Lemma 4.1. Suppose that X is a metric linear topological space with translation-invariant
metric d. If x2X and > 0, then B (x) = x+B (0).
Proof. Suppose that x 2 x0 + B (0), that is, x = x0 + y, for some y 2 B (0). Because
the metric d is translation-invariant and d(0;y) < , we have d(x0;x0 + y) < . Since
x = x0 +y, it follows that d(x0;x) < so that x2B (x0). On the other hand, suppose that
41
x2B (x0), that is, d(x;x0) < . Because the metric is translation-invariant, d(x x0;0) < .
Consequently, x x0 2B (0) implies that x2x0 +B (0).
Theorem 4.5. Let X be a metric linear topological space with translation-invariant metric
d. A subset B of X is totally bounded if and only if for each > 0 there is a nite covering
of B by open -balls.
Proof. Suppose that B is a totally bounded subset of X. Given > 0, the -ball B (0) is an
open neighborhood of zero. By assumption, there exists a nite set F = fx1;:::;xng such
that B F +B (0). Then, F +B (0) =
n[
i=1
(xi +B (0)). Because the metric is translation-
invariant, we have xi + B (0) = B (xi), for each i2f1;:::;ng, by Lemma 4.1. Therefore,
B F +B (0)
n[
i=1
B (xi).
Now, suppose that for each > 0, there is a nite covering of B by -balls. Let U be
an open neighborhood of zero in X. Then, there exists > 0 such that B (0) U. By
assumption, there exists a nite set F =fx1;:::;xng such that B
n[
i=1
B (xi). Therefore,
B
n[
i=1
B (xi) =
n[
i=1
(xi +B (0))
n[
i=1
(xi +U) =
n[
i=1
xi +U = F +U
Hence, B is totally bounded.
4.2 Completeness
A sequence (xn) in a metric space (X;d) is said to be a Cauchy sequence if for each
> 0 there is an integer N such that d(xm;xn) < , for all m;n N. However, we can
extend this de nition to linear topological spaces which are not necessarily metric spaces.
In this section we show that the two de nitions are equivalent for linear topological spaces
with a translation-invariant metric.
42
De nition. Let X be a linear topological space. A sequence (xn) in X is said to be a
Cauchy sequence if and only if for each neighborhood U of zero, there exists an integer N
such that xm xn2U, for all m;n N.
Theorem 4.6. Suppose that X is a metric linear topological space with translation-invariant
metric d. A sequence (xn) in X is Cauchy if and only if for each > 0 there exists an integer
N such that d(xm;xn) < , for all m;n N.
Proof. Suppose that (xn) is a Cauchy sequence. Given > 0, the -ball B (0) is a neigh-
borhood of zero. By assumption, there exists an integer N such that xm xn 2B (0), for
m;n N. Therefore, xm 2xn + B (0) = B (xn), that is d(xm;xn) < , for all m;n N.
For the converse, let U be any open neighborhood of zero. Then, there is an > 0 such that
B (0) U. By assumption, there exists an N such that d(xm;xn) < , for all m;n N. Be-
cause the metric is translation-invariant, this implies that d(0;xm xn) < , for all m;n N.
Hence, xm xn2B (0) U, for all m;n N.
De nition. A metric space X is complete if and only if every Cauchy sequence of X con-
verges to a point of X.
43
Chapter 5
Normed Linear Spaces
A normed linear space has a natural topology, which is the metric topology induced
by the norm metric. First we will show that the operators on a normed linear space are
continuous with respect to the norm metric, proving that any normed linear space is a linear
topological space. Because we are now working in a metric space, we will be able to utilize
the sequence de nition of continuity and the sequence lemma.
5.1 Normed Linear Spaces
A norm on a linear space can be thought of as a function that gives the length of a
vector in the space. The norm on a linear space leads to a natural metric for which the
distance between two points is the length of the vector joining them.
De nition. A norm on a linear space X is a function k k : X ! [0;1) satisfying, for all
points x;y2X and for all scalars 2F,
(a) (positive de niteness) kxk= 0 if and only if x = 0
(b) (homogeneity) k xk=j jkxk
(c) (triangle inequality) kx+yk kxk+kyk
Theorem 5.1. The function d(x;y) =kx yk is a metric for X.
Proof. Clearly, d(x;y) =kx yk 0 by de nition of the norm. By positive de niteness of
the norm,kx yk= 0 if and only if x y = 0 if and only if x = y. Moreover, by homogeneity
of the norm,
kx yk=k (y x)k=j 1j ky xk=ky xk
44
So, the function d is symmetric. Lastly, the triangle inequality is satis ed since
kx zk=kx y +y zk kx yk+ky zk
Therefore, d is a metric for X.
The norm metric, de ned byd(x;y) =kx yk, on a normed linear spaceX is translation-
invariant because k(x + z) (y + z)k = k(x y) + (z z)k = kx yk, for all x, y, and z
in X. Moreover, the norm metric is homogeneous, that is, d( x; y) = j jd(x;y), for all x
and y in X and for any scalar . This follows directly from homogeneity of the norm since
k x yk=k (x y)k=j jkx yk.
In the following theorem, we need to show that the addition on X is a continuous
function from X X into X and that scalar mutliplication on X is a continuous function
from F X into X. Because the topology of X is induced by the metric d, we can give
both X X and F X a metric topology which is equivalent to the product topology. As
a result, a function on these metric spaces is continuous if and only if the function preserves
sequential limit points.
Theorem 5.2. The normed linear space X together with the topology induced by the norm
metric is a linear topological space.
Proof. Let (xn) and (yn) be convergent sequences in X with xn !x and yn !y. Given
> 0, choose an integer N such thatkxn xk< 2 andkyn yk< 2, for all n N. By the
triangle inequality, for all n N, we have
k(xn +yn) (x+y)k kxn xk+kyn yk< 2 + 2 =
Hence, xn +yn!x+y. Therefore, the addition operation on X is continuous.
Now, let (xn) be a convergent sequence in X as before, and let ( n) be a sequence
of scalars converging to a scalar . Because the sequence ( n) of scalars converges, it is
45
bounded. Let M be a positive number such that j nj M, for all n. Given > 0, choose
an integer N such that kxn xk< 2M and j n j< 2kxk, for all n N. Then, we have
k nxn xk=k nxn nx+ nx xk
k n(xn x)k+k( n )xk
=j nj kxn xk+j n j kxk
0, let x and y be points of X such that kx yk < . Then, we have
jkxk kykj kx yk< by Lemma 5.1. Therefore, the norm is uniformly continuous.
Because the norm k k de ned on a normed linear space is uniformly continuous, the
norm preserves sequential limit points. In other words, if (xn) is a sequence in X such that
xn!x, then kxnk!kxk. The next theorem demonstrates that the operation of addition
on a normed linear space is also uniformly continuous.
46
Theorem 5.3. The addition operation on a normed linear space X is a uniformly continuous
function from X X into X.
Proof. Let denote the metric de ned on X X given by
((x1;y1);(x2;y2)) =kx1 x2k+ky1 y2k
The topology generated by this metric is equvialent to the product topology on X X.
Given > 0 and points (x1;y1) and (x2;y2) in X X satisfying ((x1;y1);(x2;y2)) < , we
have
k(x1 +y1) (x2 +y2)k kx1 x2k+ky1 y2k<
Therefore, the addition on X is a uniformly continuous function from X X into X.
5.2 Bounded Linear Transformations
Linear transformations are essential in the study of linear spaces, and continuous func-
tions are essential in topology. In the study of linear topological spaces, we are concerned
mostly with continuous linear transformations. So, it will be useful to determine conditions
under which linear transformations are continuous. We will see that a linear transformation
is continuous if and only if it is bounded. All linear transformations in a nite-dimensional
space are bounded and therefore continuous.
Let Lin(X) denote the collection of all linear transformations from X into itself. For
any linear transformation T in Lin(X), de ne
jTj= sup
x6=0
kT(x)k
kxk
Notice that jTj can also be expressed as sup
kxk=1
kT(x)k or as sup
kxk 1
kT(x)k.
47
Example 5.1. The set X = C1([0;1]) of smooth (in nitely di erentiable) real-valued func-
tions de ned on the unit interval [0;1] together with the maximum norm
kfk1 = maxff(x) : 0 x 1g
is a normed linear space. Let D denote the derivative operator, that is, D(f) = f0. Consider
the sequence (enx) of functions in X. Then,
kD(enx)k
kenxk =
knenxk
kenxk =
jnjkenxk
kenxk =jnj!1
As a result, jDj is unbounded, that is, jDj=1.
De nition. A linear transformation such that jTj<1 is said to be bounded.
Let LinB(X) denote the set of all bounded linear transformations on the linear topolog-
ical space X. De ne addition and scalar multiplication on LinB(X) as follows: If T1 and T2
are bounded linear transformations on X, then T1 + T2 is the element of LinB(X) so that
(T1 + T2)(x) = T1(x) + T2(x), for all x2X; if T 2 LinB(X) and 2F is any scalar, then
T is the element of LinB(X) so that ( T)(x) = T(x), for all x2X.
Theorem 5.4. The function j j is a norm on LinB(X).
Proof. Given T 2 LinB(X), we have jTj = sup
kxk=1
kT(x)k 0 because the supremum over a
set of non-negative numbers is non-negative. Moreover, jTj = 0 if and only if kT(x)k = 0,
for all x satisfying kxk = 1, if and only if T(x) = 0, for all x2X, i.e., T is the zero map.
Now, for any scalar ,
j Tj= sup
kxk=1
k( T)(x)k= sup
kxk=1
k T(x)k= sup
kxk=1
j jkT(x)k=j j sup
kxk=1
kT(x)k=j j jTj
48
Lastly, for any T1;T2 2LinB(X), we have
jT1 +T2j= sup
kxk=1
k(T1 +T2)(x)k= sup
kxk=1
kT1(x) +T2(x)k
sup
kxk=1
(kT1(x)k+kT2(x)k)
sup
kxk=1
kT1(x)k+ sup
kxk=1
kT2(x)k=jT1j+jT2j
Therefore, (LinB(X);j j) is a normed linear space.
Since (LinB(X);j j) is a normed linear space, it follows that the jT1 T2j de nes the
norm metric on LinB(X). According to Theorem 5.2, the linear space LinB(X) together with
the norm metric forms a linear topological space. This establishes the following corollary.
Corollary 5.1. The set LinB(X) of bounded linear transformations on a normed linear space
is a linear topological space.
Theorem 5.5. A linear transformation T is continuous if and only if T 2LinB(X).
Proof. Suppose that T is a continuous linear map on X. By continuity of T at 0, there exists
> 0 such thatkT(x) T(0)k< 1 wheneverkxk< . Because T is linear, T(0) = 0 so that
kT(x)k< 1 whenever kxk< . Let = 2. For any point x in X satisfying kxk = 1, we
have k xk= kxk= < . By continuity of T at zero, kT( x)k< 1. Because T is linear,
kT( x)k = k T(x)k = kT(x)k. As a result, kT(x)k< 1 so that kT(x)k< 1, for all
x2X satisfyingkxk= 1. Therefore,jTj 1 so that T is bounded, that is, T 2LinB(X).
Conversely suppose that T 2 LinB(X). Because jTj<1, there is a number M > 0
satisfying kT(x)k Mkxk, for each point x of X. Given > 0, let = M. For any
x2X that satis es kxk< , we have Mkxk< . Since T is linear, we have T(0) = 0. By
assumption, kT(x)k Mkxk; therefore, kx 0k< implies that kT(x) T(0)k< . So, T
is continuous at zero and therefore continuous at every point of X.
49
As the next theorem states, all linear transformations on a nite-dimensional space
are bounded. As a result, the previous theorem implies that all linear transformation on a
nite-dimensional space are continuous.
Theorem 5.6. If X is nite-dimensional, then Lin(X) = LinB(X).
50
Chapter 6
Hilbert Spaces
6.1 Inner Product Spaces
Normed linear spaces have nice properties, but they have even better properties when
the norm is induced by an inner product. In spaces with a norm given by an inner product,
we are able to use the Cauchy-Schwarz inequality and the Parallelogram Law. After showing
that an inner product space is a linear topological space, we turn our attention to problems
set in a Hilbert space.
De nition. Let X be a linear space over R. A inner product on X is a function h ; i from
X X into R that satis es, for all x;y;z2X and for all ; 2R,
(a) hx;xi 0, and hx;xi= 0 if and only if x = 0.
(b) hx;yi=hy;xi
(c) h x+ y;zi= hx;zi+ hy;zi
In this case, X is said to be a real inner product space. We can also de ne a complex inner
product space by allowing the scalars and in the de nition to be complex numbers and
by changing part (b) of the de nition to
(d) hx;yi=hy;xi
where z denotes the complex conjugate of the complex number z.
IfX is an inner product space, the inner product norm onX is de ned bykxk= phx;xi.
Before we show that this does in fact de ne a norm on X, we will state a well-known theorem
that holds in inner product spaces. The Cauchy-Schwarz inequality will be used to show
that the inner product norm is actually a norm.
51
Theorem (Cauchy-Schwarz inequality). Let X be an inner product space with inner product
h ; i. For all x;y2X,
jhx;yij2 hx;xi hy;yi
Moreover, equality holds if and only if x and y are linearly dependent.
We will omit the proof of the Cauchy-Scwarz inequality. Notice that the inequality can
be expressed in terms of the inner product norm. By taking the square root of each side we
obtain the inequality jhx;yij kxk kyk. Another useful property of inner product spaces
is the Parallelogram Law. This property will be important in a problem in the next section.
Theorem 6.1 (Parallelogram Law). Let X be an inner product space. For all x;y2X,
kx+yk2 +kx yk2 = 2kxk2 + 2kyk2
Proof. By the properties of the inner product norm, we have
kx+yk2 =hx+y;x+yi=hx;x+yi+hy;x+yi
=hx;xi+hx;yi+hy;xi+hy;yi
=kxk2 +kyk2 + 2hx;yi
Similarly, kx yk2 =kxk2 +kyk2 2hx;yi. Adding together these two equations gives the
Parallelogram Law.
Theorem 6.2. Let X be an inner product space. Then, kxk= phx;xi is a norm on X.
Proof. Clearly, kxk 0 for all x2X. Moreover, kxk = 0 if and only if hx;xi = 0 if and
only if x = 0. Now, for any scalar ,
k xk=
p
h x; ;xi=
p
2hx;xi=j j
p
hx;xi=j j kxk
52
Lastly, for all x and y in X, we have
kx+yk2 =hx+y;x+yi=kxk2 +kyk2 +hx;yi+hy;xi
Now, if X is a real inner product space, then hx;yi = hy;xi. But, if X is a complex inner
product space, then hx;yi = hy;xi. In either case, hx;yi+hy;xi 2jhx;yij. Now, by the
Cauchy-Scwarz inequality, we have jhx;yij kxkkyk. Thus,
kx+yk2 kxk2 +kyk2 + 2kxkkyk= (kxk+kyk)2
Taking the square root of both sides gives the triangle inequality. Thus, kxk = phx;xi
de nes a norm on X, called the inner product norm.
Because kxk= phx;xi de nes a norm on X, the metric induced by the inner product
norm is given bykx yk= phx y;x yiaccording to Theorem 5.1. Moreover, with this
metric the inner product space X is a linear topological space by Theorem 5.2.
Corollary 6.1. An inner product space X together with the topology induced by the inner
product norm metric is a linear topological space.
Theorem 6.3. Let X be an inner product space with inner product h ; i. The inner product
is a continuous function from X X into R.
Proof. Suppose that (xn) and (yn) are sequences in X with xn!x and yn!y. Then,
jhxn;yni hx;yij=jhxn;yni hxn;yi+hxn;yi hx;yij
=jhxn;yn yi+hxn x;yij
jhxn;ynij+jhxn x;yij
kxnk kyn yk+kxn xk kyk
53
Because (xn) converges there is a positive number M such that kxnk M, for all n.
Given > 0, choose N 2N such that kxn xk< 2kyk and kyn yk< 2M, for all n N.
Then, for n N,
jhxn;yni hx;yij kxnk kyn yk+kxn xk kyk<
Hence, hxn;yni!hx;yi. Thus, the inner product is continuous.
6.2 Hilbert Spaces
This nal section includes some problems concerning projections and orthogonality in
Hilbert spaces, including the uniqueness of the projection onto a closed and convex set.
We will also see that a Hilbert space can be written as the direct sum of a closed linear
subspace and its orthogonal complement. Lastly, we consider the relationship between a set,
the closure of its span, the orthogonal complement, and the double orthogonal complement.
De nition. A Hilbert space H is an inner product space that is complete with respect to
the inner product norm.
Theorem 6.4. Suppose that F is a closed and convex subset of a Hilbert spaceH. If x2H,
then there is a unique element P(x) in F satisfying
kx P(x)k= inffkx zk: z2Fg
The mapping P is called the projection of H on F.
Proof. Fix x0 2H, and let = inffkx0 xk : x 2 Fg. If x0 2 F, then kx0 x0k = 0
and therefore = 0. Moreover, x0 is unique because kx0 xk = 0 if and only if x0 = x.
So, we can now assume that x0 =2 F. For each n 2 N, we can choose xn 2 F satisfying
kx0 xnk< + 1n; otherwise, + 1n would be a lower bound of fkx0 xk : x2Fg
that is greater than the greatest lower bound . Without loss of generality, we can assume
54
that x0 = 0 because translation by x0 is a homeomorphism and the metric is translation-
invariant so that our assumptions are not a fected. Then, kxnk< + 1n for each n2N,
which implies thatkxnk! . BecauseHis a Hilbert space, we have the Parallelogram Law,
so that for any m;n2N,
kxm xnk2 = 2kxmk2 + 2kxnk2 kxm +xnk2
or kxm xnk2 = 2kxmk2 + 2kxnk2 4 xm+xn2 2
Now, because F is convex and (xn) is a sequence in F, it follows that xm+xn2 is an element
of F, for every m;n2N. By de nition of , we have
xm +xn
2
which implies
xm +xn
2
2
2:
So, if we replace xm+xn2 2 by 2 in the Parallelogram Law, we will be subtracting a smaller
quantity on the right-hand side so that we get the inequality
kxm xnk2 2kxmk2 + 2kxnk2 4 2
Because kxmk ! and kxnk ! , the right-hand side of this inequality goes to zero.
Therefore, kxm xnk!0, and (xn) is a Cauchy sequence. BecauseH is complete and F is
closed, xn converges to a point x of F satisfying kx k = . For uniqueness, suppose there
is another point y of F such that ky k= . Then, by the Parallelogram Law,
kx y k2 = 2kx k2 + 2ky k2 4
xm +xn
2
2
= 2 2 + 2 2 4
xm +xn
2
2
4 2 4 2 = 0
Thus, x = y and the projection P(x0) = x is unique.
55
De nition. Let A be a subset of a Hilbert space H. A point x in H is orthogonal to A,
written x?A, ifhx;ai= 0 for every a2A. The orthogonal complement of A inH, denoted
by A?, is the set fx : x?Ag.
Theorem 6.5. Suppose that M is a closed linear subspace of a Hilbert space H. Then,
hx P(x);yi= 0, for each x2H and each y2M. Moreover, P(x) is the unique point of
M such that x P(x)2M?.
Proof. Let x be a point of H, and let p denote the projection of x onto the closed, convex
set M, that is, p is the unique point of M that is closer to x than any other point of M.
Given a point y2M and a scalar , we have y2M and p y2M because M is a linear
subspace of H. As a result,
kx (p y)k2 kx pk2 (6.1)
By properties of the inner product norm, we have
kx (p y)k2 =h(x p) + y;(x p) + yi
=kx pk2 +h y;x pi+hx p; yi+k yk2
kx pk2 + 2 0, then we can divide by without a ecting inequality (6.2) to get
2hx p;yi kyk2
As !0+, we have kyk2 !0 so that hx p;yi 0. Now, assuming that < 0, we can
divide both sides of inequality (6.2) by , but the inequality will be reversed:
2hx p;yi kyk2
As !0 , we have kyk2 !0 so that hx p;yi 0. Therefore, hx p;yi= 0. A similar
argument can be used if the scalar eld is the complex numbers. Because y was chosen
arbitrarily, it follows that x p2M?. Now, suppose that q is another point of M such that
x q2M?. For any y2M, we have
hp q;yi=hp x+x q;yi=hx q;yi hx p;yi= 0
In particular, p q2M because M is a linear subspace. So,
kp qk2 =hp q;p qi= 0
Hence, p q = 0 so that p = q.
De nition. A linear space X is a direct sum of linear subspaces A and B if and only if each
element of X can be expressed uniquely as a sum of an element of A and an element of B.
Theorem 6.6. If M is a closed linear subspace of a Hilbert space H, then H= M M?.
Proof. Given x2H, let p denote the projection of x onto the closed, convex set M. Write
x = p+ (x p). By the previous theorem, p is the unique point of M such that x p2M?.
Therefore, p + (x p) is a unique expression for x as a sum of an element of M and an
element of M?.
57
Theorem 6.7. A? is a closed linear subspace of H.
Proof. First, the orthogonal complement A? is non-empty because 0 2A? since h0;ai = 0
for each a2A. Furthermore, for any points x and y in A? and for any scalars and , we
have
h x+ y;ai= hx;ai+ hy;ai= 0 + 0 = 0
Hence, x+ y2A? so that A? is a linear subspace of H.
Now, let (xn) be a sequence in A? which converges to a point x in H. Fix a point a
in A. Given a point a of A, we have hxn;ai!hx;ai by continuity of the inner product.
Because xn 2A?, for each n, we have hxn;ai = 0. Therefore, by continuity of the inner
product,
hx;ai= limn!1hxn;ai= 0:
Hence, the limit x of the sequence (xn) belongs to A? so that A? is closed. Thus the
orthogonal complement of any subset of H is a closed linear subspace of H.
Observation. If A B, then A? B?.
Proof. Suppose that A B, and let x 2 B?. Then, hx;bi = 0, for all b 2 B. Because
A B, it follows that hx;ai= 0, for all a2A. Hence, B? A?.
Theorem 6.8. The orthogonal complement A? of A is identical with the orthogonal com-
plement of the closure of the span of A.
Proof. Let ~A denote the closure of the span of A. Because A span(A) ~A, we have
~A? A? by the previous observation. To get the other inclusion, let x2A?. Given y2 ~A,
there is a sequence (yn) of points in the span of A such that yn !y. For each n, we can
express yn as a nite linear combination
yn =
knX
i=1
(n)i a(n)i where (n)i 2F and a(n)i 2A
58
For each n2N, we have
hx;yni=
*
x;
knX
i=1
(n)i a(n)i
+
=
knX
i=1
hx; (n)i a(n)i i=
knX
i=1
(n)i hx;a(n)i i= 0
where
D
x;a(n)i
E
= 0, for all i2f1;:::;kng, n2N, because x2A?. By continuity of the
inner product, we have
hx;yi= limn!1hx;yni= 0
So, hx;yi= 0, for all y2 ~A. Hence, x2 ~A?. Therefore, A? = ~A?.
Theorem 6.9. A?? is the closure of the span of A.
Proof. Let ~A denote the closure of the span of A. If x2 ~A, then there is a sequence (xn) of
points in the span of A such that xn!x. For each n, xn can be expressed as a nite linear
combination of elements of A:
xn =
knX
i=1
(n)i a(n)i
For any y2A? and for xed n, we have
hxn;yi=
* k
nX
i=1
(n)i a(n)i ;y
+
=
knX
i=1
(n)i
D
a(n)i ;y
E
= 0
By continuity of the inner product, we have hx;yi = limn!1hxn;yi = 0. Since hx;yi = 0 for
any y2A?, it follows that x2A??. Hence, ~A A??. On the other hand, suppose that
x2A??. Because ~A is a closed linear subspace ofH, we haveH= ~A ~A?. So, we can write,
x = p + y, where p is the unique projection of x onto ~A and y2 ~A? = A?. Consequently,
hx;yi= 0 because x2A??. Now, p2 ~A A?? so that hp;yi= 0. Therefore,
kyk2 =hy;yi=hx p;yi=hx;yi hp;yi= 0
Thus, y = 0 from which it follows that x = p2 ~A. Hence ~A = A??.
59
Bibliography
[1] Aliprantis, Charalambos D., and Kim C. Border. \Topological Vector Spaces." In nite
Dimensional Analysis, A Hitchhiker?s Guide. Springer Berlin Heielberg, 1999. 161-170.
Web. 17 Nov. 2013.
[2] Kelley, John L., Isaac Namioka, et al. Linear Topological Spaces. New York, NY: Springer
Berlin Heidelberg, 1963. Print.
[3] Munkres, James R. Topology. 2nd ed. Upper Saddle River, NJ: Pearson Education, Inc.,
2000. Print.
[4] Royden, Halsey L. Real Analysis. 3rd ed. Upper Saddle River, NJ: Prentice Hall, 1988.
217-236. Print.
60