Qamrul Hasan Ansari
Topological Vector Spaces
Topological Vector Spaces
Qamrul Hasan Ansari
Department of Mathematics
Aligarh Muslim University, Aligarh
E-mail: qhansari@gmail.com
Page 1
SYLLABUS
M.A. / M.Sc. III SEMESTER
TOPOLOGICAL VECTOR SPACES
Course Title
Course Number
Credits
Course Category
Prerequisite Courses
Contact Course
Type of Course
Course Assessment
End Semester Examination
Course Objectives
Course Outcomes
Topological Vector Spaces
MMM-3019
4
Compulsory
Functional Analysis, Topology
4 Lecture + 1 Tutorial
Theory
Sessional (1 hour) 30%
(2:30 hrs) 70%
The objective of this course is to teach how one can extend
the results and concepts from normed spaces to
topological vector spaces. This course gives the idea of
topological vector spaces and locally convex topological
vector spaces and their properties.
It also covers several fixed point theorems for set-valued maps
defined on a topological vector space.
After undertaking this course, students can learn the
concepts of Hahn Banach theorem in the setting of vector spaces,
topological vector spaces, locally convex topological vector spaces
and their properties.
Student can also learn several important fixed point results
in the setting of topological vector spaces, namely, KKM theorem,
Browder fixed point theorem, Kakutani fixed point theorem, etc.
2
Qamrul Hasan Ansari
Topological Vector Spaces
Syllabus
UNIT I: Some Concepts from Vector Space
Subspaces, affine sets, convex sets, cones
(pointed cone, convex cones etc), balanced sets, absorbent sets,
hulls (linear hull, affine hull, convex hull, balanced hull)
and their properties and characterizations;
Hahn Banach theorem in vector spaces; Convex functions,
Minkowski function and seminorm with their properties
UNIT II: Topological Vector Spaces
Definition and general properties, product spaces and
quotient spaces, bounded and totally bounded sets;
Topological properties of convex sets, convex cones,
compact sets, convex hull; Hyperplanes, closed half spaces
and separation of convex hulls; Hahn Banach theorem on separation;
Complete topological vector spaces; Metrizable topological vector spaces:
Definition and properties; Normable topological vector spaces
and finite spaces
UNIT III: Locally Convex Spaces
Definition and general properties, subspaces, product spaces
and quotient spaces; Convex and compact sets in
locally convex spaces; Separation theorems in locally convex spaces;
Continuous linear operators: General consideration on continuous
linear operators, open operators and closed operators;
Space of operators: Topologies of uniform convergence,
properties of the space of continuous linear operators
UNIT IV: Dual Vector Spaces and Fixed Point Theory
Definition and properties; Mackey topology; Strong topology:
Definition and properties, semi-reflexive spaces and space
and reflexive spaces, some fixed points theorems in
topological vector spaces; KKM theorem, Browder fixed point theorem,
Kakutani fixed point theorem and related results.
Total
Page 3
No. of Lectures
12
14
14
12
54
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Topological Vector Spaces
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Recommended Books:
1. R. Cristescu, Topological Vector Spaces, Noordhoff International Publishing, Leyden, The
Netherlands, 1977
2. L. Narici and E. Beckenstein, Topological Vector Spaces, Marcel Dekker, Inc., New York
and Basel, 1985
3. Y.-C. Wong, Introductory Theory of Topological Vector Spaces, Marcel Dekker, Inc., New
York Basel and Hong Kong, 1992
4. V.I. Bogachev and O.G. Smolyanov, Topological Vector Spaces and Their Applications,
Springer International Publishing AG, 2017
5. W. Rudin, Functional Analysis, Tata McGraw-Hill Publishing Co. Ltd., New Delhi, 1973
6. S.P. Singh, B. Watson and P. Srivastava, Fixed Point Theory and Best Approximation: The
KKM-map Principle, Kluwer Academic Publishers, Dordrecht, Boston, London, 1997.
7. I.F. Wilde, Basic Analysis - Gently Done: Topological Vector Spaces, Lecture Notes, Department of Mathematics, King’s College, London, 2010.
8. J.R. Giles, Convex Analysis with Applications in the Differentiation of Convex Functions,
Pitman Advanced Publishing Program, Boston, London, Melbourne, 1982.
1
Some Basic Concepts from Vector Spaces
1.1
Subspace and Sublinear Functions
Let X be a vector space, A and B be two subsets of X, x0 ∈ X and λ ∈ R. We use the
following notations:
A ± x0 := {x ± x0 : x ∈ A},
A ± B := {x ± y : x ∈ A, y ∈ B},
λA := {λx : x ∈ A}.
We say that A + x0 is obtained by translating A by x0 .
Warning: 2A 6= A + A.
A set F of functionals defined on a vector space X is called a sufficient set of functionals if
for any x 6= 0 in X, there exists f ∈ F such that f (x) 6= 0.
The set of all linear functionals defined on the vector space X, denoted by X ′ , is a vector
space with respect to the usual operations.
Remark 1.1.1. Let X be a vector space and x ∈ X. The mapping Fx : X ′ → R defined by
for all f ∈ X ′ ,
Fx (f ) = f (x),
is a linear functional on X ′ and the set G = {Fx : x ∈ X} is a vector subspace of (X ′ )′ as
well as a sufficient set of linear functionals on X ′ .
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Topological Vector Spaces
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Definition 1.1.1. A subset E of a vector space X is said to be a subspace if for all x, y ∈ E
and α, β ∈ R, we have αx + βy ∈ E.
Geometrically speaking, a subset E of X is a subspace of X if for all x, y ∈ E, the plane
through the origin, x and y lies in E.
R
E
y
2
in R
e
c
a
bsp
u
S
x
R
Figure 1.1: A subspace in R2 : The line through 0, x and y
Definition 1.1.2. A subspace E of a vector space X is said to be maximal if it is a proper
subspace of X which is not properly contained in any proper subspace of X.
Remark 1.1.2. Planes through the origin in R3 are maximal subspaces.
Definition 1.1.3. Let X be a vector space. A real-valued functional p : X → R is called
(a) subadditive if
p(x1 + x2 ) ≤ p(x1 ) + p(x2 ),
for all x1 , x2 ∈ X;
(b) positively homogeneous if
p(αx) = αp(x),
for all x ∈ X and α ≥ 0;
(c) sublinear if it is subadditive and positively homogeneous;
(d) superlinear if −p is sublinear.
The class of sublinear functionals on X is denoted by X # which is not a vector space
but it is closed under pointwise operations of addition and multiplication by a number
α ≥ 0. We always treat X # as an ordered set, ordered by the pointwise order, that is,
p ≤ q ⇔ p(x) ≤ q(x) for all x ∈ X.
Note 1.1.1. The positive homogeneity of p implies that p(0) = 0.
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Topological Vector Spaces
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Lemma 1.1.1. If p : X → R is a sublinear functional on a vector space X, then the
following assertions are equivalent:
(a) p is linear.
(b) For all x ∈ X, p(x) + p(−x) = 0.
(c) For all x ∈ X, p(x) + p(−x) ≤ 0.
Proof. (a) ⇒ (b) ⇒ (c) are obvious. If (c) holds, then 0 = p(0) = p(x−x) ≤ p(x)+p(−x) ≤ 0,
so p(x) + p(−x) = 0. Hence (b) holds and p(−x) = −p(x). It follows that p(αx) = αp(x) for
all α ∈ R. Then for all x, y ∈ X, p(x) = p(x + y − y) ≤ p(x + y) + p(−y) = p(x + y) − p(y),
that is, p(x) + p(y) ≤ p(x + y). Since p is sublinear, we have p(x) + p(y) ≥ p(x + y) and thus
the additivity of p holds. Consequently, (a) holds.
The above lemma says that a sublinear functional is linear if and only if p(x) + p(−x) ≤ 0
for all x ∈ X.
Definition 1.1.4. Let X be a vector space and f : X → R be a linear functional on X.
The subspace N(f ) := {x ∈ X : f (x) = 0} is called the null space or kernal of f .
Proposition 1.1.1 (pp. 149, Ref. 2). Let X be a vector space and f be a nontrivial
linear functional on X with null space N(f ). Then
(a) f is surjective;
(b) codimension of N(f ) is 1, that is, X/N(f ) is algebraically isomorphic to R;
(c) If λ ∈ R and H = {x ∈ X : f (x) = λ} = f −1 (λ), then for any x ∈ H, N(f ) =
H − x = {y − x : y ∈ H}.
Proposition 1.1.2 (pp. 151, Ref. 2). Let X be a vector space and E be a subspace of
X.
(a) E is maximal if and only if dim X/E = 1.
(b) E is maximal if and only if E = N(f ) = null space of non-trivial linear functional
f on X.
In view of above proposition, the null space N(f ) of a non-trivial linear functional f on a
vector space X is a maximal space of X.
Qamrul Hasan Ansari
1.2
Topological Vector Spaces
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Convex Sets and Cones
Definition 1.2.1. Let X be a vector space, and x and y be distinct elements in X. The set
L := {z : z = λx + (1 − λ)y for all λ ∈ R} is called the line through x and y.
The set [x, y] := {z ∈ X : z = λx + (1 − λ)y for all 0 ≤ λ ≤ 1} is called a line segment with
the endpoints x and y. Similarly, we have the sets
[x, y[ := {z ∈ X : z = λx + (1 − λ)y for all 0 < λ ≤ 1},
]x, y] := {z ∈ X : z = λx + (1 − λ)y for all 0 ≤ λ < 1},
]x, y[ := {z ∈ X : z = λx + (1 − λ)y for all 0 < λ < 1}.
[x
,y
y
]
y
L x
A line through x and y
x
A line segment with endpoints x and y
Figure 1.2: A line and a line segment
Definition 1.2.2. A subset M of a vector space X is said to be an affine set if for all
x, y ∈ M and λ, µ ∈ R such that λ + µ = 1 imply that λx + µy ∈ M, that is, for all x, y ∈ M
and λ ∈ R, we have λx + (1 − λ)y ∈ M.
Geometrically speaking, a subset M of X is an affine set if it contains the whole line through
any two of its points.
Proposition 1.2.1. A subset M of a vector space X is affine if and only if for any
element x0 ∈ M, M − x0 is a subspace of X.
Proof. Let M be an affine set and x0 ∈ M. Then we shall prove that M − x0 is a vector
subspace. For that, let y1 , y2 ∈ M − x0 . Then y1 = x1 − x0 and y2 = x2 − x0 for some
x1 , x2 ∈ M. Let z := 12 x1 + 21 x2 , then z ∈ M. Also, y := 2z + (−1)x0 = 2z − x0 ∈ M.
y1 + y2 = x1 + x2 − 2x0 = [(x1 + x2 ) − x0 ] − x0 = [2z − x0 ] − x0 = y − x0 ,
hence y1 + y2 ∈ M − x0 .
If x ∈ M − x0 , then we can easily prove that αx ∈ M − x0 . Therefore, M − x0 is a subspace.
The converse can be proved easily.
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Topological Vector Spaces
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In view of the above proposition, we have
Corollary 1.2.1. Let M be a set in a vector space X.
(a) M is a subspace if and only if it is affine and contains the origin.
(b) M is affine if and only if either M = ∅ or M is a translate of a subspace, that is,
M can be represented as E + x0 where E is a subspace.
Definition 1.2.3. A subset K of a vector space X is said to be a convex set if for all x, y ∈ K
and λ, µ ≥ 0 such that λ + µ = 1 imply that λx + µy ∈ K, that is, for all x, y ∈ K and
λ ∈ [0, 1], we have λx + (1 − λ)y ∈ K.
y
x
y
x
Figure 1.3: Convex sets
y
x
Figure 1.4: Nonconvex sets
Geometrically speaking, a subset K of a vector space X is convex if it contains the line
segment joining any two of its points.
In R2 , lines, line segments, circular discs, elliptical discs and interior of triangles are all
convex.
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Topological Vector Spaces
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It follows directly from the definition that the image and inverse image of a convex set under
a linear transformation are convex. In particular, if A and B are convex sets and α ∈ R,
then A + B = {x + y : x ∈ A, y ∈ B} and αA = {αx : x ∈ A} are convex.
Note 1.2.1. Evidently, the empty set, each singleton set {x} and the whole space X are
all both affine and convex. A ray, which has the form {x0 + λv : λ ≥ 0}, where v 6= 0, is
convex, but not affine.
Note 1.2.2. If K is a convex set and 0 ∈ K, then λK ⊆ K for all λ ∈ [0, 1]. Hence,
λK ⊆ αK for all 0 ≤ λ ≤ α.
Remark 1.2.1. If K is a convex subset of a vector space X, α > 0 and β > 0, then
αK + βK = (α + β)K. In other words, for any x, y ∈ K, there exists z ∈ K such that
β
α
(α + β)z = αx + βy. In fact this relation can be written as z = α+β
x + α+β
y and we have
β
β
α
α
> 0, α+β > 0 and α+β + α+β = 1 from which the assertion follows.
α+β
Definition 1.2.4. A subset C of a vector space X is said to be a cone if for all x ∈ C and
λ ≥ 0, we have λx ∈ C.
A cone C is said to be pointed if C ∩ (−C) = {0}.
A subset C of X is said to be a convex cone (also called wedge) if it is convex and a cone;
that is, for all x, y ∈ K and λ, µ ≥ 0 imply that λx + µy ∈ C.
A pointed convex cone is said to be proper if it does not contain any line passing through
{0}.
C
−C
Figure 1.5: Pointed Cone
Remark 1.2.2. If C is a cone, then 0 ∈ C. In the literature, it is mostly assumed that the
cone has its apex at the origin. This is the reason why λ ≥ 0 is chosen in the definition of
a cone. However, some references define a set C ⊂ X to be a cone if λx ∈ C for all x ∈ C
and λ > 0. In this case, the apex of the “shifted” cone may not be at the origin, or 0 may
not belong to C.
Topological Vector Spaces
y
Qamrul Hasan Ansari
x+
y
Page 11
λx
x
Figure 1.6: Convex cone
Remark 1.2.3.
cone.
(a) If C is a non-pointed (convex) cone, then C ∪{0} is a pointed (convex)
(b) If C is a pointed cone, then C − {0} is a non-pointed cone.
(c) If C is a pointed convex cone, then C − {0} is not necessarily convex.
Remark 1.2.4.
(a) A cone C may or may not be convex.
(b) If C1 and C2 are convex cones, then C1 ∩ C2 and C1 + C2 are convex cones.
Remark 1.2.5. Let f : X → Y be a linear mapping from a vector space X to another
vector space Y and C be a convex cone in X. Then f (C) is a convex cone in Y .
Note 1.2.3. If K is a convex set in a vector space X and {0} ∈
/ K, then the cone C =
S
λ>0 λK is non-pointed, thus C ∪ {0} is proper.
Remark 1.2.6. It is clear from the above definitions that every subspace is an affine set
as well as a convex cone, and every affine set and every convex cone are convex. But the
converse of these statements may not be true in general.
Type of set
Subspace
Affine
Convex
Cone
Convex cone
Together with x and y
the set also contains:
λx + µy
λx + µy
λx + µy
λx
λx + µy
Whenever λ, µ ∈ R and
No other condition
λ+µ=1
λ + µ = 1, λ, µ ≥ 0
λ≥0
λ, µ ≥ 0
Qamrul Hasan Ansari
Topological Vector Spaces
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R
C
C
R
Figure 1.7: Cone with vertex at origin
Cone but not convex
Definition 1.2.5. Let X be a vector space. Given x1 , x2 , . . . , xm ∈ X, a vector x = λ1 x1 +
λ2 x2 + · · · + λm xm is called
(a) a linear combination of x1 , x2 , . . . , xm if λi ∈ R for all i = 1, 2, . . . , m;
(b) an affine combination of x1 , x2 , . . . , xm if λi ∈ R for all i = 1, 2, . . . , m with
(c) a convex combination of x1 , x2 , . . . , xm if λi ≥ 0 for all i = 1, 2, . . . , m with
(d) a cone combination of x1 , x2 , . . . , xm if λi ≥ 0 for all i = 1, 2, . . . , m.
Pm
i=1
Pm
i=1
λi = 1;
λi = 1;
A set K is a subspace, affine, convex or a cone if it is closed under linear, affine, convex or
cone combination, respectively, of points of K.
Theorem 1.2.1. A subset K of a vector space X is convex (respectively, subspace, affine,
convex cone) if and only if every convex (respectively, linear, affine, cone) combination
of points of K belongs to the set K.
Proof. Since a set that contains all convex combinations of its points is obviously convex, we
only consider K is convex and prove that it contains any convex combination
of its points,
Pm
that is, if K is convex
P and xi ∈ K, λi ≥ 0 for all i = 1, 2, . . . , m with i=1 λi = 1, then we
have to show that m
i=1 λi xi ∈ K. We prove this by induction on the number m of points
of K occurring in a convex combination. If m = 1, the assertion is simply x1 P
∈ K implies
x1 ∈ K, evidently true. If m = 2, then λ1 x1 + λ2 x2 ∈ K for λi ≥ 0, i = 1, 2, 2i=1 λi = 1,
Qamrul Hasan Ansari
Topological Vector Spaces
Page 13
Subspace E
x, y ∈ E ⇒ αx + βy ∈ E
∀λ, µ ∈ R
λ,
1
µ≥
=
µ
0
+
λ
Affine Set M
x, y ∈ M ⇒ λx + µy ∈ M
∀λ, µ ∈ R with λ + µ = 1
λ,
µ≥
Convex Cone C
x, y ∈ C ⇒ λx + µy ∈ C
∀λ, µ ≥ 0
µ=
1
λ+
Convex Set K
x, y ∈ K ⇒ λx + µy ∈ K
∀λ, µ ≥ 0 with λ + µ = 1
0
Figure 1.8: Relationships among Subspaces, Affine Sets, Convex Cones and Convex Sets
holds because K is convex. Now suppose that the result is true for m. Then (for λm+1 6= 1)
m+1
X
λi xi =
i=1
=
m
X
i=1
m
X
λi xi + λm+1 xm+1
(1 − λm+1 )
i=1
= (1 − λm+1 )
m
X
i=1
= (1 − λm+1 )
m
X
λi xi
+ λm+1 xm+1
1 − λm+1
λi
xi + λm+1 xm+1
1 − λm+1
µi xi + λm+1 xm+1 ,
i=1
where µi =
λi
, i = 1, 2, . . . , m. But then µi ≥ 0 for i = 1, 2, . . . , m and
(1 − λm+1 )
m
X
µi =
i=1
so by the result for m, y =
Pm
m+1
X
i=1
Pm
i=1
λi
1 − λm+1
=
1 − λm+1
= 1,
1 − λm+1
µi xi ∈ K. Immediately, by convexity of K, we have
λi xi = (1 − λm+1 )y + λm+1 xm+1 ∈ K.
i=1
The proof for subspace, affine and convex cone cases follows exactly the same pattern.
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Topological Vector Spaces
Page 14
Remark 1.2.7. Arbitrary intersection of convex sets (respectively, subspaces, affine sets,
convex cones) is a convex set (respectively, subspace, affine set, convex cone). However, the
arbitrary union of convex sets (respectively, subspaces, affine sets, convex cones) need not
be convex (respectively, subspaces, affine sets, convex cones).
Definition 1.2.6. Let A be a nonempty subset of a vector space X. The intersection
of all convex sets (respectively, subspaces, affine sets) containing A is called a convex hull
(respectively, linear hull, affine hull) of A, and it is denoted by co(A) (respectively, [A],
aff(A)). Similarly, the intersection of all convex cones containing A is called a conic hull of
A, and it is denoted by cone(A).
A
Convex hull of A
Convex hull of 14 points
Figure 1.9: Convex hull
A
Conic hull of 8 points
O
Conic hull of A
Figure 1.10: Conic hull
By Remark 1.2.7, the convex (respectively, affine, conic) hull is a convex set (respectively,
affine set, convex cone). In fact, co(A) (respectively, aff(A), cone(A)) is the smallest convex
set (respectively, affine set, convex cone) containing A.
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Topological Vector Spaces
Page 15
The cone cone(A) can also be written as
cone(A) = {x ∈ X : x = λy for some λ ≥ 0 and some y ∈ A}.
It is also called a cone generated by A.
Theorem 1.2.2. Let A be a nonempty subset of a vector space X. Then x ∈ co(A) if
and only ifPthere exist xi in A and λiP
≥ 0, for i = 1, 2, . . . , m, for some positive integer
m
m, where i=1 λi = 1 such that x = m
i=1 λi xi .
Proof. Since co(A) is a convex set containing A, therefore, from Theorem 1.2.1, every convex
combination of its points lies in it, that is, x ∈ co(A).
Conversely, let K(A) be the set of all convex combinations of elements of A. We claim that
the set
)
( m
m
X
X
λi = 1, m ≥ 1
λi xi : xi ∈ A, λi ≥ 0, i = 1, 2, . . . , m,
K(A) =
i=1
i=1
Pℓ
Pm
is convex. Indeed, consider y =
j=1 µj zj where yi ∈ A, λi ≥ 0,
i=1 λi yi and z =
Pℓ
Pm
i = 1, 2, . . . , m,
j=1 µj = 1, and let
i=1 λi = 1 and zj ∈ A, µj ≥ 0, j = 1, 2, . . . , ℓ,
0 ≤ λ ≤ 1. Then
m
ℓ
X
X
λy + (1 − λ)z =
λλi yi +
(1 − λ)µj zj ,
i=1
j=1
where λλi ≥ 0, i = 1, 2, . . . , m, (1 − λ)µj ≥ 0, j = 1, 2, . . . , ℓ and
m
X
i=1
λλi +
ℓ
X
j=1
(1 − λ)µj = λ
m
X
λi + (1 − λ)
i=1
ℓ
X
µj = λ + (1 − λ) = 1.
j=1
Also, the set K(A) of convex combinations contains A (each x in A can be written as
x = 1 · x). By the definition of co(A) as the intersection of all convex supersets of A, we
deduce that co(A) is contained in K(A).
Thus the convex hull of A is the set of all (finite) convex combinations from within A.
The above result also holds for an affine set and a convex cone.
Corollary 1.2.2. Let A be a nonempty subset of a vector space X.
(a) The set A is convex if and only if A = co(A).
(b) The set A is affine if and only if A = aff(A).
(c) The set A is a convex cone if and only if A = cone(A).
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Topological Vector Spaces
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(d) The set A is a subspace if and only if A = [A].
Problem 1.2.1. Prove that a set K in a vector space X is convex if and only if for all
λ ∈ [0, 1], λK + (1 − λ)K ⊆ K.
Problem 1.2.2. Prove that the arbitrary intersection of convex sets is convex.
Problem 1.2.3. For Q
each i ∈ I, let Xi be a vector
Q space and Ki be a nonempty subset of
Xi . Prove that K = i∈I Ki is convex in X = i∈I Xi if and only if each Ki is convex in
Xi .
Problem 1.2.4. Let K1 and K2 be two convex subsets of a vector space X and α, β ∈ R.
Prove that the set αA + βB is convex.
Problem 1.2.5. For n ∈ N, let
Kn be a convex subset of a vector space X. If Kn ⊆ Kn+1
S∞
for all n ∈ N, then prove that n=1 Kn is convex.
Problem 1.2.6. If K1 and K2 are convex subsets of a vector space X and α ∈ R, then
prove that K1 + K2 = {x + y : x ∈ K1 , y ∈ K2 } and αK1 = {αx : x ∈ K1 } are convex sets.
Problem 1.2.7. Prove that a subset K of a vector space X is convex if and only if (λ+µ)K =
λK + µK for all λ ≥ 0, µ ≥ 0.
Problem 1.2.8. If K1 and K2 are convex subsets of a vector space X, then prove that
co(K1 ∪ K2 ) = {λK1 + (1 − λ)K2 : 0 ≤ λ ≤ 1}.
Problem 1.2.9. Prove that a pointed convex cone C is proper if and only if the non-pointed
cone C̃, which is the complement of {0} in C, is convex.
Solution. If C contains a line through {0}, then clearly C̃ is not convex. Suppose now that
C is proper and x, y ∈ C̃. The closed segment with end points x, y is contained in C. If
such line segment contains {0}, then λx + (1 − λ)y = {0} for some λ ∈ ]0, 1[. Therefore,
x = µy with µ < 0. Thus C contains the line through {0} and x, contrary to hypothesis.
Problem 1.2.10. Prove that a subset C of a vector space X is a convex cone if and only if
C + C ⊆ C and λC ⊂ C for all λ > 0.
Solution. For the condition λC ⊂ C for all λ > 0 characterizes the cones. If C is convex,
then C + C = 21 C + 21 C = C.
Conversely, if the cone C is such that C +C ⊆ C, then for λ ∈ ]0, 1[, we have λC +(1−λ)C =
C + C ⊆ C which shows that C is convex.
Problem 1.2.11. If C is a convex cone, then prove that the vector space generated by C is
the set C − C (the set of points x − y where x, y vary in C).
Solution. If V = C − C, then V is nonempty, we have λV = V for all λ 6= 0, and V + V =
C + C − (C + C) ⊂ C − C = V , which shows that V is a vector subspace. Finally, every
vector subspace that contains C also contains V .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 17
Problem 1.2.12. If C is a pointed convex cone, then prove that the largest vector subspace
contained in C is the set C ∩ (−C).
Solution. If E = C ∩ (−C), then E is nonempty and λE = E for all λ 6= 0, also
E + E ⊂ (C + C) ∩ (−(C + C)) ⊂ C ∩ (−C) = E,
which shows that E is a vector subspace. Clearly every vector subspace contained in C is
also contained in E.
Problem 1.2.13. If K is a convex setSin a vector space X, then prove that the convex cone
generated by K is identical with C = λ>0 λK.
Solution. The set C is clearly a cone. It is sufficient to show that C is convex. Let αx, βy
be two points of C (α > 0, β > 0, x ∈ K, y ∈ K). Let λ, µ > 0 such that λ + µ = 1. Then
λαx + µβy = (λα + µβ)z with z ∈ K and λα + µβ > 0, hence λαx + µβy ∈ C.
Qamrul Hasan Ansari
1.3
Topological Vector Spaces
Page 18
Hyperplanes
Definition 1.3.1. A proper subset H of a vector space X is said to be a hyperplane if it is
a maximal affine set, that is, H is not a proper subset of any affine set different from the
whole space X.
R
H
ane
l
p
r
e
Hyp
R
Figure 1.11: A hyperplane in R2
Since an affine set is a translation of a subspace and a hyperplane is a translation of a
maximal affine set, a hyperplane is a translation of a maximal subspace. In other words, a
subset H of a vector space X is a hyperplane if and only if H can be represented as E + x0
for some x0 ∈ X and some maximal subspace E of X.
Proposition 1.3.1. If f : X → R is a non-zero linear functional on a vector space X
and λ ∈ R, then
H = {x ∈ X : f (x) = λ}
(1.1)
is a hyperplane.
Conversely, for any hyperplane H, there exist a linear functional f : X → R on X and
a number λ ∈ R such that H is of the form (1.1).
Proof. Let H = {x ∈ X : f (x) = λ} be given for some linear functional f and a number λ.
Then H − x is the null space of f by Proposition 1.1.1 (c). Since null space is a maximal
subspace, H is the translation of a maximal subspace, and hence a hyperplane.
Conversely, if H is a hyperplane, then there exist a maximal vector subspace E and an
element x0 ∈ X such that H = x0 + E. By proposition 1.1.2, there is a linear functional f
Qamrul Hasan Ansari
Topological Vector Spaces
Page 19
on X such that E = N(f ) the null space of f . Let f (x0 ) = λ. Then
H = x0 + E = x0 + N(f ) =
=
=
=
=
x0 + {x ∈ X : f (x) = 0}
{x0 + x ∈ X : f (x) = 0}
{x0 + x ∈ X : f (x0 ) + f (x) = λ}
{x0 + x ∈ X : f (x0 + x) = λ}
{y := x0 + x ∈ X : f (y) = λ}.
Hence H is of the form (1.1).
Definition 1.3.2. Let X be a vector space, λ ∈ R and H = {x ∈ X : f (x) = λ} be a
hyperplane. Consider the following sets
H + := {x ∈ X : f (x) ≥ λ},
H − := {x ∈ X : f (x) ≤ λ},
H ++ := {x ∈ X : f (x) > λ},
H −− := {x ∈ X : f (x) < λ}.
Then H + , H − , H ++ , H −− are convex sets and they are called the half-spaces determined by
H.
In particular,
• H + is called upper closed half-space
• H − is called lower closed half-space
• H ++ is called upper open half-space
• H −− is called lower open half-space
f (x) ≥ λ
f (x) ≤ λ
Figure 1.12: Closed half-spaces
Definition 1.3.3. Let A be a subset of a vector space X. If A ⊂ H + or A ⊂ H − , then we
say that A lies on one side of the hyperplane H.
If A and B are two subsets of a vector space X such that A ⊂ H + and B ⊂ H − , then we
say that H separates A from B.
If A ⊂ H ++ and B ⊂ H −− , then we say that H strictly separates A from B
Qamrul Hasan Ansari
Topological Vector Spaces
Page 20
H
H
B
A
B
A
Hyperplane H strictly separates
A and B
Hyperplane H separates
A and B
Figure 1.13: Separating hyperplanes
Definition 1.3.4. Let A be a subset of a vector space X. A hyperplane H in X is said to
be a supporting hyperplane for A if H contains at least one point of A and all the points of
A lie on the same side of H.
x̄
A
A
A
x̄
b
b
x̄
H
H
H
Figure 1.14: Supporting hyperplanes
b
H
ȳ
Qamrul Hasan Ansari
1.4
Topological Vector Spaces
Page 21
Balanced Sets and Absorbent Sets
Definition 1.4.1. A subset S of a vector space X is said to be symmetric if S = −S.
Example 1.4.1. (a) In R, symmetric sets are intervals of the type (−k, k) with k > 0,
and the sets Z and {−1, 1}.
(b) Any vector subspace in a vector space X is a symmetric set.
(c) For any subset A of a vector space X, S = A ∩ (−A) is a symmetric set.
Definition 1.4.2. A subset B of a vector space X is said to be balanced if |λ| ≤ 1 implies
λB ⊆ B.
Balanced means that the set is symmetric around the origin.
Remark 1.4.1. If B is a balanced set in a vector space X, then
(a) 0 ∈ B,
(b) B is uniform in all directions,
(c) B contains the line segment connecting 0 to another point in B.
Remark 1.4.2. Any balanced set is symmetric. Indeed, let B be a balanced set. Then for
|α| ≤ |β|, we have αB ⊆ βB. So, if |λ| = 1, we have λB = B. Hence B is symmetric
Example 1.4.2. (a) Open and closed spheres with centered at origin in a normed
space are balanced sets.
(b) Any subspace of a vector space is a balanced set.
(c) Any line segment with midpoint at the origin is a balanced set.
(d) B = {(x, y) ∈ R2 : x = 0 or y = 0} is a balanced set in R2 .
Proposition 1.4.1. Let B be the set of balanced subsets of a vector space X such that
for any B ∈ B, there exists B1 ∈ B with 2B1 ⊂ B. Then for any B ∈ B and any α ∈ R,
there exists B0 ∈ B with αB0 ⊂ B.
Proof. Let B ∈ B be arbitrary. Then by hypothesis, there exists B1 ∈ B with 2B1 ⊂ B. Since
B1 ∈ B, again by hypothesis, there exists B2 ∈ B with 2B2 ⊂ B1 , that is, 22 B2 ⊂ 2B1 ⊂ B.
Thus by induction, for any n ∈ N, there exists Bn ∈ B such that 2n Bn ⊂ B. If α ∈ R, then
there will be an n ∈ N with |α| ≤ 2n , and since B is balanced, we have αBn ⊆ 2n Bn . Hence
αBn ⊆ B.
Qamrul Hasan Ansari
Remark 1.4.3.
Topological Vector Spaces
Page 22
(a) Let A be any nonempty subset of a vector space X. Then
B := {λx : |λ| ≤ 1, x ∈ A}
is the smallest balanced set that contains A.
(b) Arbitrary unions and intersections of balanced sets is balanced.
(c) The intersection of all balanced sets containing an arbitrary set A in a vector space is
the smallest balanced set that contains A.
Definition 1.4.3. Let A be a nonempty subset of a vector space X. The intersection of all
balanced sets containing A is called a balanced hull of A and it is denoted by ec(A).
In other words, the set B := {λx : |λ| ≤ 1, x ∈ A} is balanced hull of A as it is the smallest
balanced set that contains A.
Note 1.4.1. Let A be a nonempty subset of a vector space X. The intersection of all convex
balanced sets which contain A is the smallest convex balanced set that contains A. (Prove!)
Definition 1.4.4. Let A be a nonempty subset of a vector space X. The intersection of all
convex balanced sets which contain A is called a convex balanced hull of A and it is denoted
by eco(A).
Definition 1.4.5. Let X be a vector space. Given x1 , x2 , . . . , xm ∈ X, a vector x = λ1 x1 +
λ2 x2 + · · · + λm xm isPcalled a convex balanced combination of x1 , x2 , . . . , xm if λi ∈ R for all
i = 1, 2, . . . , m with m
i=1 |λi | ≤ 1.
Note 1.4.2. Let A be a nonempty subset of a vector space X. The convex balanced hull
eco(A) of A consists all convex balanced combinations of the points of A.
Definition 1.4.6. Let A and B be two nonempty subsets of a vector space X. We say that
B absorbs A (or A is absorbed by B) if there exists ε > 0 such that 0 < |α| ≤ ε implies
αA ⊂ B.
A subset A of X that absorbs any element of the vector space X is called an absorbent set
or absorbing set, that is, for every x ∈ X, there exists α = α(x) > 0 such that x ∈ λA for
all λ ∈ R with |λ| ≥ α, equivalently, λx ∈ A for all |λ| ≤ α.
Remark 1.4.4.
(a) An absorbent set is one such that any vector can be shrunk into it.
(b) A set A is absorbing or absorbent if it absorbs every singleton (and then every finite
set) in A.
(c) Every absorbing set obviously contains 0.
(d) A set which does not absorb itself is R \ {0}.
(e) The arbitrary union of absorbing sets is absorbing.
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Topological Vector Spaces
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(f) Every finite intersection of absorbent sets is an absorbent set. However arbitrary
intersections of absorbent sets may not be an absorbent set. For example, consider a
decreasing sequence of circles about the origin in R2 whose diameters are shrinking to
zero.
(g) Every balanced set absorbs itself (take α = 1).
(h) Let B be a balanced set. If for the set A there is a number ε > 0 such that εA ⊂ B,
then B absorbs A.
(i) If B is balanced and for every x ∈ B, there exists µ ∈ R such that x ∈ µB, then B is
absorbing.
Indeed, if |λ| ≥ |µ|, then |λ−1 µ| ≤ 1 and thus x ∈ µB = λ(λ−1 µ)B ⊂ λB.
Example 1.4.3.
(a) The closed interval [ 12 , 1] is convex but not balanced.
(b) The set E = {(x, y) ∈ R2 : x = 0 or y = 0} is balanced but not convex. There is
no relation between a convex set and a balanced set.
(c) The unit sphere, centered at ( 21 , 0) in R2 , is absorbent but not balanced.
(d) In R3 , closed unit sphere B1 [0] centered at the origin is balanced and absorbent set.
(e) Any subspace of a vector space is balanced. If A is an absorbent subset of a vector
space X, then its linear hull [A] is X. Thus proper subspaces are never absorbent.
Problem 1.4.1. Prove that the convex hull of a balanced set in a vector space is balanced.
Problem 1.4.2. Prove that the finite intersections of absorbent sets are absorbent sets.
Solution. Let A1 , . . . , An be absorbent sets and x be any vector. Then there exist positive
numbers
Tnα1 , . . . , αn such that x ∈ λAi for |λ| ≥ αi , i = 1, 2, . . . , n. For α = mini αi ,
x ∈ α ( i=1 Ai ) for |λ| ≥ α.
Problem 1.4.3. Prove that a subset A of a vector space X is absorbing if and only if
S
n∈N nA = X.
Problem 1.4.4. Let X and Y be vector spaces and f : X → Y be a linear mapping.
(a) If B is a balanced subset of X, then prove that f (B) is balanced in Y .
(b) If A is an absorbent subset of X and f is onto, then prove that f (A) is absorbent in
Y.
Qamrul Hasan Ansari
1.5
Topological Vector Spaces
Page 24
Seminorms, Minkowski Function and Their Properties
Definition 1.5.1. A seminorm on a vector space X is a real-valued function p : X → R
such that
(i) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X,
(ii) p(αx) = |α|p(x) for all x ∈ X and α ∈ R.
Clearly, a seminorm p on X is positive, that is, p(x) ≥ 0 for all x ∈ X; and also p(0) = 0.
A seminorm p is called norm on X if p(x) = 0 implies x = 0.
Definition 1.5.2. A family P of seminorms on a vector space X is said to be separating if
for each 0 6= x ∈ X, there exists at least one p ∈ P such that p(x) 6= 0.
Theorem 1.5.1. Let p be a seminorm on a vector space X. Then
(a) p(0) = 0;
(b) |p(x) − p(y)| ≤ p(x − y) for all x, y ∈ X;
(c) p(x) ≥ 0 for all x ∈ X;
(d) The set E := {x ∈ X : p(x) = 0} is a subspace of X;
(e) The set B := {x ∈ X : p(x) < 1} is convex, balanced and absorbing.
Proof. (a). It follows from p(αx) = |α|p(x) by considering α = 0.
(b). By subadditivity of p, for all x, y ∈ X, we have
p(x) = p(x − y + y) ≤ p(x − y) + p(y)
therefore,
p(x) − p(y) ≤ p(x − y).
By interchanging x and y, we get
p(y) − p(x) ≤ p(y − x),
therefore,
p(x) − p(y) ≥ −p(y − x).
Since p(x − y) = p(y − x), we obtain −p(x − y) ≤ p(x) − p(y) ≤ p(x − y) and thus we get
the conclusion.
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Topological Vector Spaces
Page 25
(c). By taking y = 0 in (b), we obtain the result.
(d). Let x, y ∈ E and α, β be scalars. Then p(x) = p(y) = 0, and by (c), we obtain
0 ≤ p(αx + βy) ≤ |α|p(x) + |β|p(y) = 0,
that is, p(αx + βy) = 0 and hence αx + βy ∈ E. Thus E is a subspace.
(e). Clearly, B is balanced. To prove the convexity of B, let x, y ∈ B and λ ∈ (0, 1). Then
p(λx + (1 − λ)y) ≤ λp(x) + (1 − λ)p(y) < 1,
and hence λx + (1 − λ)y ∈ B. Thus B is convex.
If x ∈ X and p(x) < α, then p(α−1 x) = α−1 p(x) < 1. This shows that B is absorbing
Definition 1.5.3. Let A be an absorbent subset of a vector space X. The real-valued
functional pA : X → R defined on X by
pA (x) := inf{λ > 0 : x ∈ λA}
(1.2)
is called the Minkowski functional associated to A.
Example 1.5.1. Let X be a normed space and B = {x ∈ X : kxk < r} with r > 0.
Then
pB (x) = inf{λ > 0 : x ∈ λB}
= inf{λ > 0 : kx/λk < r}
= inf{λ > 0 : kxk < rλ} = kxkr.
Theorem 1.5.2. Let A be a convex absorbent subset of a vector space X. Then
(a) pA is a sublinear functional;
(b) pA is a seminorm if A is balanced.
Proof. (a). Let x, y ∈ X and ε > 0 be arbitrary. By (1.2) and the property of absorbent of
A, we have
pA (x) + ε ≥ λ for all x ∈ λA and pA (y) + ε ≥ λ for all x ∈ λA,
equivalently,
x ∈ (pA (x) + ε)A and y ∈ (pA (y) + ε)A,
equivalently,
(pA (x) + ε)−1 x ∈ A and (pA (y) + ε)−1 y ∈ A.
Qamrul Hasan Ansari
Topological Vector Spaces
Set
µ :=
Page 26
(pA (x) + ε)
,
(pA (x) + pA (y) + 2ε)
then µ ∈ (0, 1). By convexity of A, we have
µ(pA (x) + ε)−1 x + (1 − µ)(pA (y) + ε)−1 y ∈ A,
or
x + y ∈ (pA (x) + pA (y) + 2ε)A.
By the definition of pA , we have
pA (x + y) ≤ pA (x) + pA (y) + 2ε.
Since ε was arbitrary, we can take ε approaches to 0, then we have
pA (x + y) ≤ pA (x) + pA (y),
that is, pA is subadditive.
Obviously, pA (0) = 0.
Now let x ∈ X and α > 0. Then we have
pA (αx) =
=
=
=
inf{λ > 0 : αx ∈ λA}
inf{λ > 0 : x ∈ α−1 λA}
α inf{α−1 λ > 0 : x ∈ α−1 λA, λ > 0}
αpA (x),
and thus pA is positively homogeneous.
(b). Suppose that A is balanced and 0 6= α ∈ R. Then α = |α|β with |β| = 1, and we have
β −1 A = A. Therefore,
{λ : αx ∈ λA} = {λ > 0 : |α|x ∈ λA}.
Thus p(αx) = |α|p(x) and hence p is a seminorm.
Theorem 1.5.3. Let p be a seminorm on a vector space X and B := {x ∈ X : p(x) < 1}.
Then p = pB .
Proof. If x ∈ X and p(x) < α, then p(α−1 x) = α−1 p(x) < 1. This shows that pB (x) ≤ α.
Hence pB ≤ p.
But if 0 < β ≤ p(x), then p(β −1 x) ≥ 1, and so β −1 x is not in B This implies that p(x) ≤
pB (x) and the proof is complete.
Problem 1.5.1. Let A and B be convex absorbent subset of a vector space X such that
A ⊆ B. Prove that pA (x) ≥ pB (x) for all x ∈ X.
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Topological Vector Spaces
Page 27
Problem 1.5.2. Let A be a convex absorbent subset of a vector space X, B = {x ∈ X :
pA (x) < 1} and C = {x ∈ X : pA (x) ≤ 1}. Then prove that B ⊂ A ⊂ C and pB = pA = pC .
Solution. For each x ∈ X, set
HA (x) = inf{λ > 0 : x ∈ λA}.
Suppose that λ ∈ HA (x) and µ > λ. Since 0 ∈ A and A is convex, we have µ ∈ HA (x).
Hence HA (x) is a half line whose left end point is pA (x).
Let x ∈ B, then pA (x) < 1 and hence 1 ∈ HA (x). This implies that x ∈ A. Similarly, if
x ∈ A, then pA (x) ≤ 1. Thus B ⊂ A ⊂ C. This implies that HB (x) ⊂ HA (x) ⊂ HC (x) for
every x ∈ X, and thus
pC (x) ≤ pA (x) ≤ pB (x),
for all x ∈ X.
To prove the equality holds, suppose that pC (x) < µ < λ. Then µ−1 x ∈ C, and hence
pA (µ−1 x) ≤ 1, so that
µ
pA (λ−1 x) ≤ < 1.
λ
−1
−1
Thus λ x ∈ B, pB (λ x) ≤ 1 and pB (x) ≤ λ.
Problem 1.5.3. Let A be nonempty convex balanced subset of a vector space X. Prove
that the following conditions are equivalent:
(a) The Minkowski functional pA is a norm, that is, pA (x) = 0 ⇒ x = 0.
T
(b) λ>0 λA = {0}.
Problem 1.5.4. Let n ≥ 1 be an integer and consider the sets
A := {(x1 , x2 , . . . xn ) ∈ Rn : x21 + x22 + · · · + x2n < 1}
and
B := {(x1 , x2 , . . . xn ) ∈ Rn : x21 + x22 + · · · + x2n ≤ 1}.
p
For any a = (a1 , a2 , . . . , an ) ∈ Rn , prove that pA (a) = pB (a) = a21 + a22 + · · · + a2n .
Qamrul Hasan Ansari
1.6
Topological Vector Spaces
Page 28
Hahn-Banach Theorem in Vector Spaces
Let us recall that a real-valued functional p : X → R on a vector space X is
(a) subadditive if
p(x1 + x2 ) ≤ p(x1 ) + p(x2 ),
for all x1 , x2 ∈ X;
(b) positively homogeneous if
p(αx) = αp(x),
for all x ∈ X and α ≥ 0;
(c) sublinear if it is subadditive and positively homogeneous;
(d) superlinear if −p is sublinear.
The class of sublinear functionals on X is denoted by X # which is not a vector space but it
is closed under pointwise operations of addition and multiplication by a number α ≥ 0. We
always treat X # as an ordered set, ordered by the pointwise order, that is, p ≤ q ⇔ p(x) ≤
q(x) for all x ∈ X.
Proposition 1.6.1. If p : X → R be a sublinear functional on a vector space X, then
the map q defined by
q(x) = inf{p(x + λw) − λp(w) : λ ≥ 0} for any w ∈ X,
(1.3)
is a sublinear functional on X such that q(x) ≤ p(x) for all x ∈ X.
Proof. We first show that the map q is well-defined. Let p ∈ X # . For all x, w ∈ X and
λ ≥ 0, we have
p(λw) = p(x + λw + (−x)) ≤ p(x + λw) + p(−x).
So −p(−x) ≤ p(x + λw) − p(λw). If we fix w ∈ X and let λ vary, then the numbers
p(x + λw) − p(λw) are bounded below by −p(−x) and so the functional q defined by (1.3)
is a real-valued map.
We now show that q(x) ≤ p(x) for all x ∈ X. By (1.3), q(x) ≤ p(x + λw) − λp(w) for all
λ ≥ 0. Letting λ = 0, we get q(x) ≤ p(x) for all x ∈ X.
We now show that q is sublinear. If α = 0, then q(αx) = inf{p(αx + λw) − λp(w) : λ ≥ 0}.
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Topological Vector Spaces
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Since α = 0, q(αx) = q(0) = 0 = αq(x). For α > 0,
q(αx) = inf{p(αx + λw) − λp(w) : λ ≥ 0}
λ
λ
by homogenuity of p
= inf αp x + w − α p(w) : λ ≥ 0
α
α
= α inf{p(x + µw) − µp(w) : µ ≥ 0}
with µ = αλ
= αq(x).
For subadditivity of q, assume that x, y ∈ X and ε > 0 are given. By the definition of q,
there exist µ, λ ≥ 0 such that
ε
ε
and p(y + λw) − λp(w) ≤ q(y) + .
p(x + µw) − µp(w) ≤ q(x) +
2
2
Hence
q(x) + q(y) ≥ p(x + µw) + p(y + λw) − (µ + λ)p(w) − ε
≥ p(x + y + (µ + λ)w) − (µ + λ)p(w) − ε
≥ q(x + y) − ε.
Since ε was arbitrary, we can take ε tends to 0, so we have q(x) + q(y) ≥ q(x + y).
Lemma 1.6.1 (Zorn’s lemma). Let (A 4) be a pre-ordered (reflexive and transitive) set.
If every nonempty totally ordered subset B of A has an upper bound (lower bound), then
A has at least one maximal element (minimal element).
Definition 1.6.1. An element p ∈ X # is said to be a minimal element of X # if p(x) ≤ q(x)
for all x ∈ X and all q ∈ X # .
Proposition 1.6.2. Let X be a vector space X and p ∈ X # . Then p is a minimal
element of X # if and only if it is linear. In other words, the algebraic dual X ′ of all
linear functionals on X consists precisely of the minimal elements of X # .
Proof. Suppose that f : X → R is a linear functional on X and p ∈ X # such that p(x) ≤ f (x)
for all x ∈ X. Then for all x ∈ X, we have p(x) + p(−x) ≥ p(x + (−x)) = p(0) = 0, and so
p(x) ≤ f (x) = −f (−x) ≤ −p(−x) ≤ p(x)
which implies that f (x) = p(x), that is, f = p.
Conversely, assume that p is a minimal element of X # and let q be a sublinear functional
defined by (1.3) such that q(x) ≤ p(x) for all x ∈ X. By the minimality of p, q(x) = p(x)
for all x ∈ X. If we let λ = 1 and x = −w, then from (1.3), we have
p(−w) = q(−w) ≤ p(−w + w) − p(w) = −p(w).
Hence p(−w) + p(w) ≤ 0. Since w was arbitrary, the linearity of p follows from Lemma
1.1.1.
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Topological Vector Spaces
Page 30
Proposition 1.6.3. For any sublinear functional p : X → R on a vector space X, there
is a linear functional h : X → R on X such that h(x) ≤ p(x) for all x ∈ X.
Proof. Let L := {q ∈ X # : q ≤ p} = {q ∈ X # : q(x) ≤ p(x) for all x ∈ X}. Then clearly
L is nonempty because p ∈ L. We now show that L is a inductively ordered, so that it will
have a minimal element h by Zorn’s lemma; h must then be linear by Proposition 1.6.2.
Let P be a totally ordered subset of L. We claim that for any x ∈ X, the set {q(x) : q ∈ P }
is bounded below. If not, then there exist x ∈ X and pn ∈ P such that pn (x) ≤ −n for all
n ∈ N. Since P is totally ordered, qn = min{p1 , . . . , pn } ∈ P for each n ∈ N. Thus {qn } is
a decreasing sequence of sublinear functionals with the property that qn (x) ≤ −n for all n.
Hence
0 = qn (x + (−x)) ≤ qn (x) + qn (−x) ≤ −n + qn (−x),
so n ≤ qn (−x) ≤ q1 (−x) for all n, which contradicts the fact that q1 is a real-valued. Thus,
p∗ (x) := inf{r(x) : r ∈ P }
is a real number for all x ∈ X. Clearly, p∗ ≤ r ≤ p for all r ∈ P . Hence it only remains to
show that p∗ is sublinear to prove that p∗ is a lower bound for P .
Clearly, p∗ (0) = 0 and p∗ (λx) = λp∗ (x) for all λ ≥ 0 and all x ∈ X. For all x, y ∈ X and
r, q ∈ P ,
p∗ (x + y) ≤ r(x + y) ≤ r(x) + r(y) ≤ r(x) + q(y) if r ≤ q,
whereas
p∗ (x + y) ≤ q(x + y) ≤ r(x) + q(y) if q ≤ r.
Consequently, for all r, q ∈ P , p∗ (x + y) ≤ q(x) + r(y), which implies that p∗ (x + y) ≤
p∗ (x) + p∗ (y).
Theorem 1.6.1 (Hahn-Banach Theorem). Let p : X → R be a sublinear functional
defined on a vector space X and f0 : E → R be a linear functional defined on a subspace
E of X such that
f0 (x) ≤ p(x), for all x ∈ E.
(1.4)
Then f0 can be extended to a linear functional f on the whole space X such that f (x) ≤
p(x) for all x ∈ X.
Proof. For any y ∈ E, f0 (−y) ≤ p(−y). Hence, for every x,
−p(−x) + f0 (−y) ≤ −p(−x) + p(−y),
so
− p(−x) ≤ −p(−x) + p(−y) − f0 (−y).
(1.5)
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Since, by the subadditivity of p, p(−y) − p(−x) ≤ p(x − y), (1.5) yields
−p(−x) ≤ p(x − y) + f0 (y).
The map
q(x) = inf{p(x − y) + f0 (y) : y ∈ E}
is therefore real-valued for all x ∈ X. Moreover, for any x ∈ E, letting y = x, q(x) ≤ f0 (x).
Clearly q ≤ p and we now show that q is sublinear.
Since, for every y ∈ E, f0 (−y) ≤ p(−y), it follows that f0 (y) + p(−y) ≥ 0 for all y ∈ E.
Letting y = 0, it follows that 0 ∈ {f0 + p(−y) : y ∈ E}, so q(0) = 0.
For α > 0,
q(αx) = inf{p(αx − y) + f0 (y) : y ∈ E}
n y
o
y
= inf αp x −
+ αf0
:y∈E
α
α
= αq(x).
Given x, w ∈ X and ε > 0, there exist y, z ∈ E such that
ε
ε
p(x − y) + f0 (y) < q(x) +
and p(w − z) + f0 (z) < q(w) + .
2
2
Therefore,
q(x) + q(w) > p(x − y) + p(w − z) + f0 (y + z) − ε
≥ p(x + w − (y + z)) + f0 (y + z) − ε
≥ q(x + w) − ε,
so the subadditivity of q follows.
By Proposition 1.6.3, there exists a linear functional f on X such that f ≤ q, that is,
f (x) ≤ q(x) for all x ∈ X. Since f ≤ q ≤ f0 on E, it follows that f = f0 on E since f0 must
be a minimal element on E # .
Alternative Proof. Let x0 ∈
/ E and E0 = [E ∪ {x0 }] the subspace spanned by E ∪ {x0 }. By
using (1.4), we have
− p(−x2 − x0 ) − f0 (x2 ) ≤ p(x1 + x0 ) − f0 (x1 ),
for all x1 , x2 ∈ E.
(1.6)
Let α be the supremum of the left hand side of (1.6) and β be the infimum of the right hand
side of (1.6), then α ≤ β. For λ ∈ [α, β], we have
− p(−x − x0 ) − f0 (x) ≤ λ ≤ p(x + x0 ) − f0 (x),
for all x ∈ E.
(1.7)
Let y ∈ E0 . Then y has a unique representation of the form y = X + λx0 with x ∈ E and
λ ∈ R. In this case, we set f1 (y) = f0 (x) + λ λ1 and obatin a linear functional f1 on E0 . One
has
f0 (x) + λ λ1 ≤ p(x + λx1 ).
(1.8)
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Indeed, if λ > 0, then (1.8) follows from the second inequality in (1.7), where λ−1 x is to be
used instead of x.
If λ < 0, then we have to use λ−1 x instead of x in the first inequality (1.7).
If λ = 0, inequality (1.8) is trivial.
Relation (1.8) shows that f0 (y) ≤ p(y) for all y ∈ E0 .
Let F denote the set of all extensions of f of f0 (on the subspaces of X that contain E),
which are linear and such that f (x) ≤ p(x). Now we introduce an order relation on F as
f1 ≤ f2
if and only if f2 is an extension of f1 .
Then it can be easily seen that F satisfies all conditions of Zorn’s lemma (Let (A 4) be
a pre-ordered (reflexive and transitive) set. If every nonempty totally ordered subset B of
A has an upper bound (lower bound), then A has at least one maximal element (minimal
element)) f is a maximal element in F , then f is defined on the whole space. Thus completes
the proof.
Corollary 1.6.1. Let p : X → R be a sublinear functional defined on a vector space X.
Then for any element x0 ∈ X, there exists a linear functional f : X → R on X such
that f (x0 ) = p(x0 ) and f (x) ≤ p(x) for all x ∈ X.
Proof. Let E = [{x0 }] the subspace spanned by {x0 } and f0 : E → R be a functional defined
by
f0 (λx0 ) = λp(x0 ).
Then f0 is a linear functional on E and satisfies condition (1.4). Then the functional f we
are looking for is obtained by extending f0 .
Corollary 1.6.2 (Bonsall’s Theorem). Let X be a vector space, W be a subspace in X,
p : X → R be a sublinear functional on X and q : W → R be a superlinear functional on
W such that q(y) ≤ p(y) for all y ∈ W . Then there exists a linear functional f : X → R
on X such that
f (x) ≤ p(x), for all x ∈ X.
(1.9)
f (y) ≤ p(y),
for all y ∈ W.
(1.10)
Proof. By the hypothesis and subadditivity of p, we have
q(y) ≤ p(y) = p(x + y + (−x)) ≤ p(x + y) + p(−x).
Then −p(−x) ≤ p(x + y) − q(y). Let r(x) := inf{p(x + y) − q(y) : y ∈ W }. This defines a
sublinear functional on X and due to Corollary 1.6.1, there exists a linear functional f on
X such that f (x) ≤ r(x) for all x ∈ X. r(x) ≤ p(x) implies (1.9), and r(−y) ≤ −q(y) for all
y ∈ W implies (1.10).
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Corollary 1.6.3 (Bohnenblust-Sobczyk-Suhomlynow Theorem). Let X be a vector space,
p : X → R be a seminorm on X and f0 : E → R be a linear functional defined on a
subspace E of X such that
|f0 (x)| ≤ p(x),
for all x ∈ E.
(1.11)
Then f0 can be extended on the whole space X to a linear functional f such that |f (x)| ≤
p(x) for all x ∈ X.
Proof. The functional f given by Hahn-Banach Theorem, verifies the following inequalities:
−p(−x) ≤ f (x) ≤ p(x),
for all x ∈ X.
Corollary 1.6.4. If X is a vector space and p : X → R is a seminorm on X, then
for any x0 ∈ X, there exists a linear functional f : X → R defined on X such that
f (x0 ) = p(x0 ) and |f (x)| ≤ p(x) for all x ∈ X.
Proof. Follows immediately from previous corollary.
Definition 1.6.2. Let K be a nonempty convex subset of a vector space X. A point x0 ∈ X
is said to be a boundary point for K if there exist x1 , x2 ∈ X such that ]x1 , x0 [ ⊂ K and
]x0 , x2 [ ⊂ K c (complement of K).
The set of all boundary points of K is called the boundary of K.
Remark 1.6.1. If p : X → R is a sublinear functional on a vector space X and λ > 0, then
H = {x ∈ X : p(x) ≤ λ} is a convex set and its boundary is ∂H = {x ∈ X : p(x) = λ}.
Corollary 1.6.1 to Hahn-Banach theorem asserts that for any x0 belonging to the boundary
of K, one can construct a supporting hyperplane that contains x0 , namely the hyperplane
H = {x ∈ X : p(x) = λ}.
2
Topological Vector Spaces
2.1
Definition and Properties
Definition 2.1.1. A vector space X with a topology T under which the mappings
(x, y) 7→ x + y from X × X → X
and
(α, x) 7→ αx from R × X → X
are continuous, is called a topological vector space. The topology T is called vector topology.
In words, a topological vector space is a vector space which is at the same time a topological
space such that addition and scalar multiplication are continuous.
Note 2.1.1. The conditions
(x, y) 7→ x + y from X × X → X
and
(α, x) 7→ αx from R × X → X
are continuous mean that for every x, y ∈ X
for every neighbourhood Vx+y of x + y, there exist neighbourhoods Vx of x and
Vy of y such that Vx + Vy ⊆ Vx+y ,
and for every x ∈ X and α ∈ R
34
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for every neighbourhood Vαx of αx, there exist neighbourhoods Vx of x and Vα
of α such that Vα · Vx ⊂ Vαx .
Equivalently, for every Vαx of αx, there exist δ > 0 and a neighbourhood Vx of x
such that βVx ⊆ Vαx whenever |β − α| < δ.
Note 2.1.2. The continuity requirements of the mappings
(x, y) 7→ x + y from X × X → X
and
(α, x) 7→ αx from R × X → X,
in terms of net convergence, for a vector topology T on X read as
Whenever {xλ } and {yλ } are nets in X such that xλ → x and yλ → y imply
xλ + yλ → x + y.
Whenever {αλ } and {xλ } are nets in R and X, respectively, such that αλ → α
and xλ → x imply αλ xλ → αx.
Important: For net / sequence convergence, X should be Hausdorff.
Example 2.1.1. Every vector space X is a topological vector space with respect to the
indiscrete (trivial) topology T = {∅, X}. But a vector space X is not a topological vector
space with respect to the discrete topology T consists all subsets of X unless X = {0}.
Indeed assume that X is a topological vector space with respect to the discrete topology
T and 0 6= x ∈ X. The sequence {αn } = { n1 } in R converges to zero in the Euclidean
topology. Therefore, since the scalar multiplication is continuous, αn x → 0 · x = 0, that
is, for any neighbourhood U of 0 in X, there exists N ∈ N such that αn x ∈ U for all
n ≥ N. In particular, we can take U = {0} since it is open in the discrete topology.
Hence, αN x = 0, which implies that x = 0, a contradiction.
Example 2.1.2. Every normed vector space endowed with the topology given by the
metric induced by the norm is a topological vector space.
Note 2.1.3. There are also examples of spaces whose topology cannot be induced by a
norm or a metric but that are topological vector spaces, for example, the space of infinitely
differentiable functions, the space of test functions and the space of distributions.
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In general, a metric vector space may not be a topological vector space. Indeed, there
exist metrics for which both the vector space operations of sum and product by scalars are
discontinuous.
Definition 2.1.2. Let X and Y be two topological vector spaces. A linear mapping f :
X → Y is said to be
(a) homeomorphism from X to Y if f and f −1 are continuous (equivalently, f is continuous
and open);
(b) isomorphism from X to Y if it is a bijective homeomorphism, that is, it is bijective
(one-one and onto) and f is continuous and open.
Proposition 2.1.1. Let X be a Hausdorff topological vector space. For given a ∈ X
and s ∈ R, with s 6= 0, the translation map Ta : X → X defined by
Ta (x) = x + a,
for all x ∈ X,
and the multiplication map (or dilation by s) Ms : X → X defined by
Ms (x) = sx,
for all x ∈ X,
are homeomorphisms of X onto itself.
Proof. Let Φ : X × X → X and Ψ : R × X → X be the continuous mappings defined in
the definition of a topological vector space. Then, by continuity, if {(xα , yα )} is a net which
converges to (x, y) in X × X, we have
xα + yα = Φ((xα , yα)) → Φ((x, y)) = x + y.
Similarly, if (sα , xα ) → (s, x) in R × X, then
sα xα = Ψ((sα , xα )) → Ψ((s, x)) = sx.
Hence, for any net {xα } in X with xα → x, set yα = a for all α and sα = s for all α, where
s 6= 0. Then (xα , yα ) → (x, a) and (sα , xα ) → (s, x) and we conclude that
Ta (xα ) = xα + a → x + a = Ta (x)
and
Ms (x) = sxα → sx = Ms (x).
Thus both Ta and Ms are continuous maps from X into X. Furthermore, Ta has inverse T−a
and Ms has inverse Ms−1 , which are also continuous by the same reasoning. Therefore Ta
and Ms are homeomorphisms of X onto itself.
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Corollary 2.1.1. For any open set G in a Hausdorff topological vector space X, the
sets a + G, A + G and sG are open, for any a ∈ X, A ⊆ X, and s ∈ R with s 6= 0. In
particular, if U is a neighbourhood of 0, then so is tU for any t ∈ R with t 6= 0.
Proof. Suppose that G is open in X. For any a ∈ X and s ∈ R, we have a + G = Ta (G)
and sG = Ms (G). By Proposition 2.1.1, Ta and Ms , for s 6= 0, are homeomorphisms and so
Ta (G) and Ms (G) are open sets.
S
Since A + G = a∈A (a + G) is a union of open sets and so is open.
If U is a neighbourhood of 0, there is an open set G such that 0 ∈ G ⊆ U. Hence 0 ∈ tG ⊆ tU
and tG is open so that tU is a neighbourhood of 0.
Remark 2.1.1. The continuity of the map x 7→ x+a in a topological vector space X implies
the following property.
If V is a neighbourhood of a, then b − a + V is a neighbourhood of b, for any a, b ∈ X.
Proposition 2.1.2. Let X be a topological vector space. For any neighbourhood U of 0
in X, there is a neighbourhood V of 0 such that V + V ⊆ U. Moreover, V can be chosen
to be symmetric, that is, V = −V .
Proof. Let Φ : X × X → X denote the addition mapping defined in the definition of
a topological vector space X. Then Φ is continuous at (0, 0) ∈ X × X. Since U is a
neighbourhood of 0 and Φ is continuous, Φ−1 (U) is a neighbourhood of (0, 0) in the product
topology. In particular, there exist neighbourhoods V1 , V2 of 0 such that V1 × V2 ⊆ Φ−1 (U).
So if we take V = V1 ∩ V2 , then V is still a neighbourhood of 0 satisfying V × V ⊆ Φ−1 (U)
which implies the desired inclusion V + V ⊆ U.
As above, for any neighbourhood U of 0, there are neighbourhoods V1 and V2 of 0 such that
V1 + V2 ⊆ U. The result follows by setting V = V1 ∩ V2 ∩ (−V1 ) ∩ (−V2 ).
Remark 2.1.2. The above proposition can be extended for n number of neighbourhoods.
That is, for any neighbourhood U of 0 and any n ∈ N, there is a (symmetric) neighbourhood
V of 0 such that V
· · + V} ⊆ U.
| + ·{z
n
Corollary 2.1.2. For any topological vector space X and any n ∈ N, the map Φ :
Rn × X n → X defined by
Φ(((t1 , . . . , tn ), (x1 , . . . , xn ))) = t1 x1 + · · · + tn xn ,
for all ((t1 , . . . , tn ), (x1 , . . . , xn )) ∈ Rn × X n , is continuous.
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Proof. Let ((s1 , . . . , sn ), (x1 , . . . , xn )) ∈ Rn × X n and U be any neighbourhood of z = s1 x1 +
· · · + sn xn in X. Then −z + U is a neighbourhood of 0 and so, by Remark 2.1.2, there is a
neighbourhood V of 0 such that
V
· · + V} ⊆ −z + U.
| + ·{z
n−times
Now, for each 1 ≤ i ≤ n, si xi + V is a neighbourhood of si xi and therefore there is δi > 0
and a neighbourhood Ui of xi such that tUi ⊆ si xi + V whenever |t − si | < δi . Set δ =
min{δ1 , . . . , δn }. Then, for any x′i ∈ Ui , 1 ≤ i ≤ n,
t1 x′1 + · · · + tn x′n ∈ s1 x1 + V + · · · + sn xn + V
⊆ s1 x1 + · · · + sn xn + (−z + U)
= U
whenever |ti − si | < δ and the result follows.
Proposition 2.1.3. Let U be a neighbourhood of 0 in a topological vector space X. If
0 < r1 < r2 < · · · and rn → ∞ as n → ∞, then
X=
∞
[
rn U.
n=1
Proof. Fix x ∈ X. Since α 7→ αx is a continuous mapping from R to X, the set of all α with
αx ∈ U is open, and contains 0, hence contains 1/rn for sufficient large n. Thus (1/rn )x ∈ U,
or x ∈ rn U for large n.
Next result shows that the continuity of the linear operations implies that the space is Hausdorff provided every point is closed.
Proposition 2.1.4. Let (X, T ) be a topological vector space.
(a) X is Hausdorff if and only if every one-point set is closed.
(b) X is Hausdorff if and only if for any element 0 6= x ∈ X, there exists a neighbourhood U of 0 that does not contain x.
Proof. (a). If T is a Hausdorff topology, then obviously all one-point sets are closed. For the
converse, let x and y be distinct points in a topological vector space X, and let w = x − y,
so that w 6= 0. Now, if {w} is closed, then X \ {w} is open. Indeed, X \ {w} is an open
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neighbourhood of 0. The continuity of addition (at 0) implies that there are neighbourhoods
U and V of 0 such that
U + V ⊆ X \ {w}.
Since V is a neighbourhood of 0, so is −V (because the scalar multiplication M−1 is a
homeomorphism) and therefore w − V is a neighbourhood of w (because the translation Tw
is a homeomorphism). Any z ∈ w − V has the form z = w − v with v ∈ V , and so w = z + v.
Since w ∈
/ U + V , we must have that z ∈
/ U. Therefore U ∩ (w − V ) = ∅, and so, translating
by y, (y + U) ∩ (y + w − V ) = ∅, that is, (y + U) ∩ (x − V ) = ∅. Hence x − V and y + U are
disjoint neighbourhoods of x and y, respectively, and X is Hausdorff.
(b). If X is a Hausdorff space, then obviously x and 0 would have disjoint neighbourhoods.
To prove the converse, it is sufficient to show that the diagonal of X × X is a closed set.1
Let x 6= 0 and U be a neighbourhood of 0 such that −x ∈
/ U. Then 0 ∈
/ U + x, and as U + x
is a neighbourhood of x, x is not in the closure of the set {0}. Hence {0} is a closed set.
Since the diagonal of X × X is the preimage of {0} by the mapping (x, y) 7→ x − y must be
a closed set.
Proposition 2.1.5. Any neighbourhood of 0 in a topological vector space X is absorbing.
Proof. For any x ∈ X and any neighbourhood U of 0, the continuity of scalar multiplication
(at (0, x)) ensures the existence of α > 0 such that λx ∈ U whenever |λ| < α.
Proposition 2.1.6. Let (X, T ) be a topological vector space. Every neighbourhood of 0
in X contains a balanced neighbourhood of 0, that is, for every neighbourhood U of 0,
there exists a balanced neighbourhood W of 0 such that W ⊆ U.
Proof. Let U be a given neighbourhood of 0 in X. Since scalar multiplication is continuous
(at (0, 0)), there is a δ > 0 and a neighbourhood V of 0 suchSthat for all x ∈ V , βx ∈ U
whenever |β| < δ, that is, βV ⊆ U whenever |β| < δ. Put W = |β|<δ βV . Then 0 ∈ W ⊆ U
and W is a neighbourhood of 0 since δ2 V ⊆ W and 2δ V is a neighbourhood of 0 by Corollary
2.1.1. Let α ∈ R with |α| ≤ 1 and let x ∈ W . Then x ∈ βV for some β ∈ R with |β| < δ.
Hence αx ∈ αβV . But |αβ| ≤ |β| < δ, so that αx ∈ W and therefore W is balanced.
Proposition 2.1.7. The only open linear subspace in a topological vector space X is X
itself, that is, no proper subspace can be open.
Proof. Suppose that E is an open linear subspace in a topological vector space X. Then 0
belongs to E, so there is a neighbourhood U of 0 in X such that U ⊆ E. Let x ∈ X. Since
U is absorbing by Proposition 2.1.5, there is a δ > 0 such that tx ∈ U whenever |t| < δ. In
particular, tx ∈ E for some (suitably small) t with t 6= 0. But E is linear, so it follows that
x = t−1 tx ∈ E, and we have E = X.
1
A topological space X is Hausdorff if and only if its diagonal set D := {(x, x) ∈ X × X : x ∈ X} is
closed.
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Note 2.1.4. In fact, a proper linear subspace E of a topological vector space X can have no
interior points at all. To see this, suppose that x is an interior point of E. Then there is a
neighbourhood V of x with V ⊆ E. But then −x + V belongs to E and is a neighbourhood
of zero. We now argue as before to conclude that E = X.
Proposition 2.1.8. For any nonempty subset A in a topological vector space X and any
x ∈ A, there is a net {aU } in A, indexed by V0 , the set of neighbourhoods of 0 (partially
ordered by reverse inclusion), such that aU → x.
Proof. For any U ∈ V0 , x + U is a neighbourhood of x. Since x ∈ A, (x + U) ∩ A 6= ∅. Let
aU be any element of (x + U) ∩ A. Then {aU } is a net in A such that aU → x. Indeed, for
each U ∈ V0 , aU − x ∈ U and so aU − x → 0 along V0 , that is, aU → x.
Proposition 2.1.9. Let A and B be nonempty subsets of a topological vector space X.
(a) A + B ⊆ A + B;
T
(b) A = U ∈N0 (A + U), where N0 is any neighbourhood basea at 0.
Let (X, T ) be a topological space, x ∈ X and Vx be the collection of all neighbourhoods of x. A
subset Nx of Vx is said to be a (local) neighbourhood base at x (or a fundamental system of neighbourhoods
of x) if any element of Vx contains some elements of Nx
a
Proof. (a). Let x ∈ A and y ∈ B. Then there are nets {aν } and {bν }, indexed by V0 , such
that aν → x and bν → y. By continuity of addition, it follows that aν + bν → x + y. Hence
x + y ∈ A + B.
(b). Assume that x ∈ A and let N0 be any neighbourhood base at 0. For any U ∈ N0 , −U
is a neighbourhood of 0 and so x − U is a neighbourhood
of x. Since x ∈ A, it follows that
T
(x − U) ∩ A 6= ∅, that is, x ∈ A + U, and so A ⊆ U ∈N0 (A + U).
T
On the other hand, suppose that x ∈ U ∈N0 (A + U) and that V is any neighbourhood of
x. Then −x + V is a neighbourhood of 0 and so also is −(−x + V ) = x − V . Therefore
there is some U ∈ N0 such that U ⊆ x − V . Now, x ∈ A + U, by hypothesis, and so
x ∈ A + U ⊆ A + x − V . Hence there is a ∈ A and v ∈ V such that x = a + x − v, that is,
a = v. It follows that every neighbourhood of x contains some element of A and therefore
x ∈ A.
Problem 2.1.1. Let A and B be nonempty subsets of a topological vector space X. Prove
that A + int B ⊆ int (A + B).
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Proposition 2.1.10. If E is a linear subspace of a topological vector space X, then so
is its closure E. In particular, any maximal proper subspace is either dense or closed.
Proof. Let E be a linear subspace of the topological vector space X. We must show that if
x, y ∈ E and t ∈ R, then tx + y ∈ E. There are nets {xν } and {yν } in E, indexed by the
base of all neighbourhoods of 0, such that xν → x and yν → y. It follows that txν → tx and
txν + yν → tx + y and we conclude that tx + y ∈ E, as required.
If E is a maximal proper subspace, the inclusion E ⊆ E implies that either E = E, in which
case E is closed, or E = X, in which case E is dense in X.
Problem 2.1.2. In a topological vector space, prove that the closure of an affine set is affine,
and the closure of a convex balanced set is convex and balanced.
Proposition 2.1.11. Let B be a balanced subset of a topological vector space X. Then
the following assertions hold.
(a) The closure of B (denoted by B) is balanced.
(b) If 0 is an interior point of B, then the interior of B is balanced.
Proof. (a). For any α ∈ R with 0 < |α| ≤ 1, we have
α B = Mα (B) = Mα (B) = α B ⊆ B,
since Mα is a homeomorphism. This clearly also holds for α = 0 and so B is balanced.
(b). Suppose that 0 ∈ int B. Then, as before, for 0 < |α| ≤ 1,
α int B = int(α B) ⊆ α B ⊆ B.
But α int B is open and so α int B ⊆ int B. This is also valid for α = 0 and it follows that
int B is balanced.
Problem 2.1.3. The closed balanced neighbourhoods of 0 form a local neighbourhood base
at 0.
Solution. Let U be any neighbourhood of 0. By Proposition 2.1.2, there is a neighbourhood
V of 0 such that V +V ⊆ U. Also, by Proposition 2.4.3 (a), there is a balanced neighbourhood
of 0, W , say, with W ⊆ V , so that W + W ⊆ U. By Proposition 2.1.9, W ⊆ W + W ⊆ U,
and W is a neighbourhood of 0. We claim that W is balanced. To see this, let x ∈ W and
let t ∈ R with |t| ≤ 1. There is a net {wν } in W such that wν → x. Then twν → tx, since
scalar multiplication is continuous. But twν ∈ W , since W is balanced, and so tx ∈ W . It
follows that W is balanced, as claimed.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 42
Proposition 2.1.12. In any topological vector space X, there exists a basis of neighbourhoods N0 of 0 consisting of closed balanced subsets such that the following conditions
hold.
(i) If U ∈ N0 , then there exists U1 ∈ N0 such that U1 + U1 ⊆ U;
(ii) If U ∈ N0 and 0 6= α ∈ R, then αU ∈ N0 .
Proof. Since the mapping (α, x) 7→ αx is continuous at (0, 0), for any neighbourhood V of
0, there is an ε > 0 and another neighbourhood U of 0 such that αU ⊆ V if |α| ≤ ε. Let
[
U0 :=
αU.
|α|≤ε
Then U0 is a balanced set and also a neighbourhood of 0 as εU ⊆ U0 . Obviously, U0 ⊆ V .
Thus we have shown that there exists a basis of balanced neighbourhood of 0 in X
Any neighbourhood basis N0 of 0 in X has property (i), as (x, y) 7→ (x + y) is continuous
at the point (0, 0) of X × X. (See also Proposition 2.1.2).
Now, let V be a neighbourhood of 0 and U1 be another neighbourhood of 0 such that
U1 + U1 ⊆ V . Let z ∈ U1 . Since the mapping (x, y) 7→ x − y from X × X to X is continuous
at (z, z), there will be a neighbourhood U2 of z such that U2 − U2 ⊆ U1 . As there is a
z0 ∈ U1 ∩ U2 and z − z0 ∈ U2 − U2 , we deduce that z ∈ U1 + U1 and hence z ∈ V . Therefore
U1 ⊆ V .
On the other hand there exists a balanced neighbourhood U3 of 0 such that U3 ⊆ U1 , and
thus U3 ⊆ V . The closure of a balance set is balance too. The set of all closed and balanced
neighbourhoods of 0 will then form a neighbourhood basis of 0.
To complete the proof, we note that if U is a closed balanced neighbourhood of 0, then so
is λU for any λ 6= 0.
Proposition 2.1.13. Let X be a vector space and U be a family of balanced and absorbent
subsets of X satisfying the following properties:
(i) If U1 , U2 ∈ U, then there exists U3 ∈ U such that U3 ⊆ U1 ∩ U2 ;
(ii) If U ∈ U, then there exists U1 ∈ U such that U1 + U1 ⊆ U.
Then there is a vector topology T on X for which U represents a neighbourhood basis of
the origin.
Proof. Let V := {V ⊆ X : there exists U ∈ U such that U ⊆ V }. For an arbitrary element
a ∈ X, we set Va := {V + a : V ∈ V}. Then Va has the following obvious properties:
Qamrul Hasan Ansari
Topological Vector Spaces
Page 43
• If E ∈ Va , then a ∈ E;
• If E1 , E2 ∈ Va , then E1 ∩ E2 ∈ Va ;
• If E1 ∈ Va and E1 ⊆ E2 , then E2 ∈ Va .
Now let E ∈ Va , so that E = V + a with V ∈ V. Consider U ∈ U such that U ⊆ V . Then
by condition (ii), there exists U1 ∈ U with U1 + U1 ⊆ U. If x ∈ U1 + a, then
U1 + x ⊆ U1 + U1 + a ⊆ U + a ⊆ V + a = E.
Hence, for any E ∈ Va , there exists E1 ∈ Va such that E ∈ Vx for all x ∈ E1 .
We conclude that Va is the set of all neighbourhoods of the point a, for a topology T on X.
In particular, U is a neighbourhood basis of the origin.
Condition (ii) ensures that (x, y) 7→ x + y is a continuous mapping.
Finally, we prove that the mapping (α, x) 7→ αx is continuous.
Since every set in U is balanced, the mapping (α, x) 7→ αx must be continuous at (0, 0). On
the other hand, every set in U being absorbent, the mapping α 7→ αx0 is continuous at 0.
It remains to prove that x 7→ α0 x is continuous at the origin. 2 But this immediately follows,
as U ∈ U implies the existence of a U0 ∈ U with the property α0 U0 ⊆ U.
2
Let T be a vector topology on X such that the mapping (x, y) 7→ x + y from X × X to X is continuous.
If the mapping (α, x) 7→ αx from R × X to X is continuous (0, 0) and, for any x0 ∈ X and α0 ∈ R, the
mappings α 7→ αx0 from R → X and x 7→ α0 x from X to X are is continuous at 0 ∈ R and 0 ∈ X,
respectively, then (α, x) 7→ αx is continuous on R × X.
Qamrul Hasan Ansari
2.2
Topological Vector Spaces
Page 44
Product Spaces
Proposition 2.2.1. If {Xi }i∈I is a family of topological vector spaces, then X =
is a topological vector space.
Proof. Set
X=
Y
Xi
and Z =
i∈I
Q
i∈I
Xi
Y
(Xi × Xi ).
i∈I
If endowed with the product topologies, then two spaces X ×X and Z will be homeomorphic
by the mapping
g ({xi }i∈I , {yi }i∈I ) = {(xi , yi )}i∈I .
On the other hand, for any i ∈ I, the function hi : Xi × Xi → Xi defined by
hi (xi , yi ) = xi + yi
is continuous, and therefore, the mapping h : Z → X defined by
h ({(xi , yi )}i∈I ) = {hi (xi , yi )}i∈I = {xi + yi }i∈I
is continuous too.
Since the mapping (x, y) 7→ x + y from X × X to X is the composition of h and g, it is also
continuous.
Similarly, the mapping (α, x) 7→ αx from R × X to X is continuous, due to the continuity of
(α, {xi }i∈I ) 7→ (α, xi ),
(α, xi ) 7→ αxi
for all i ∈ I.
Problem 2.2.1. Let X be a topological vector space and Y be a subspace of X. Then prove
that X/Y is a topological vector space.
Qamrul Hasan Ansari
2.3
Topological Vector Spaces
Page 45
Bounded and Totally Bounded Sets
Definition 2.3.1. A subset B of a topological vector space X is said to be bounded if for
any neighbourhood U of the origin, there exists a (sufficiently large) real number α > 0 such
that B ⊆ αU, that is, there exists α > 0 such that B ⊆ βU for all β > α.
If N0 is a neighbourhood basis of the origin and B is a subset of a topological vector space
X such that for any U ∈ N0 , there is a real number α > 0 with B ⊆ αU (or αB ⊆ U), then
B is bounded.
As there is a basis of balanced neighbourhood of 0 in a topological vector space, we can say
that B is bounded if for any balanced neighbourhood U of 0, there exists α ∈ R such that
B ⊆ αU.
Proposition 2.3.1. Let (X, T ) be a topological vector space. If z ∈ X and A and B are
bounded subsets of X, then the sets {z}, A ∪ B and A + B are also bounded.
Proof. Let U be any neighbourhood of 0. Then by Proposition 2.1.5, U is absorbing and so
z ∈ αU for all sufficiently large α > 0, that is, {z} is bounded.
By Proposition 2.1.2, there is a neighbourhood V of 0 such that V + V ⊆ U. Since A and
B are bounded, there is α1 > 0 such that A ⊆ αV for α > α1 and there is α2 > 0 such that
B ⊆ αV for α > α2 . Therefore,
A ∪ B ⊆ αV,
and
A + B ⊆ αV + αV ⊆ αU,
for all α > max{α1 , α2 }.
Remark 2.3.1. It follows, by induction, that any finite set in a topological vector space is
bounded. Also, taking A = {z}, we see that any translate of a bounded set is bounded.
Proposition 2.3.2. Let (X, T ) be a topological vector space. The following assertions
hold.
(a) If A is a bounded set and B ⊆ A, then B is bounded.
(b) If A and B are bounded sets in X, then A + B is bounded.
(c) If B is a bounded set, then λB is bounded for all λ ∈ R.
(d) The closure B of a bounded set B is bounded.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 46
(e) The bounded hull (intersection of all bounded subsets containing a given set) of a
bounded set is a bounded set.
Proof. Exercise.
Proposition 2.3.3. A subset B of a Hausdorff topological vector space X is bounded if
and only if for any sequence {xn }n∈N of points of B and any sequence {λn }n∈N of positive
real numbers that converges to zero such that limn→∞ λn xn = 0.
Proof. Let B be bounded, xn ∈ B and 0 < λn → 0. Let U be an arbitrary balanced
neighbourhood of 0 and let ε > 0 be such that εB ⊆ U. Let N ∈ N such that λn ≤ ε if
n ≥ N. Since U is balanced, λB ⊆ U for |λ| ≤ ε, hence λn xn ∈ U for n ≥ N. Therefore,
limn→∞ λn xn = 0.
Assume that B is not bounded. Then there exists a balanced neighbourhood U of 0 such
that B 6⊆ nU for all n ∈ N. Hence, for any n ∈ N, there exists xn ∈ B such that xn ∈
/
nU. Thus n−1 xn ∈
/ U for all n ∈ N, and therefore, {n−1 xn }n∈N will not converge to 0, a
contradiction.
Proposition 2.3.4. Any convergent sequence in Hausdorff topological vector space is
bounded.
Proof. Suppose that {xn } is a sequence in a topological vector space (X, T ) such that xn → z.
For each n ∈ N, set yn := xn − z, so that yn → 0. Let U be any neighbourhood of 0. Let V
be any balanced neighbourhood of 0 such that V ⊆ U. Then V ⊆ sV for all s with |s| ≥ 1.
Since yn → 0, there is N ∈ N such that yn ∈ V whenever n > N. Hence yn ∈ V ⊆ tV ⊆ tU
whenever n > N and t ≥ 1. Set A := {y1, . . . , yN } and B := {yn : n > N}. Then A
is a finite set so is bounded and therefore A ⊆ tU for all sufficiently large t. But then it
follows that A ∪ B ⊆ tU for sufficiently large t, that is, {yn : n ∈ N} is bounded and so is
{xn : n ∈ N} = z + (A ∪ B).
Remark 2.3.2. A convergent net in a Hausdorff topological vector space need not be
bounded. For example, let I be R equipped with its usual order and let xα ∈ R be given
by xα = e−α . Then {xα }α∈I is an unbounded but convergent net (with limit 0) in the real
normed space R.
Definition 2.3.2. A subset A of a topological vector space X is called totally bounded
(or precompact) if for any neighbourhood U of the origin,
Sn there exists a finite subset B =
{b1 , . . . , bn } of X such that A ⊆ U + B, that is, A ⊆ i=1 (U + bi ).
Remark 2.3.3. (a) If A totally bounded in a topological vector space X, then for any
neighbourhood U of the origin, S
there exists a finite subset A0 = {a1 , . . . , an } of A such
that A ⊆ U + A0 , that is, A ⊆ ni=1 (U + ai ).
Qamrul Hasan Ansari
Topological Vector Spaces
Page 47
Exercise. Let U0 be a balanced neighbourhood of the origin such that U0 + U0 ⊆ U
and let A0 = {a1 , . . . , an } be a finite subset of X such that A ⊆ U0 + A0 . Then
A ∩ (U0 + ai ) 6= ∅,
for all i = 1, 2, . . . n,
since otherwise we can always diminish the set A0 . If bi ∈ B ∩(U0 +ai ), then bi = xi +ai
with xi ∈ U0 , hence ai ∈ bi + U0 . Setting B0 = {b1 , . . . bn }, we get B ⊆ U + B0 .
(b) Every totally bounded set in a topological vector space is bounded.
Exercise. Let B be a totally bounded set and U be a neighbourhood of the origin.
Choose a balanced neighbourhood V of 0 such that V + V ⊆ U. Since B is totally
bounded, there is a finite subset A of X such that B ⊆ A+V . Since V is absorbent and
balanced, there is an α ≥ 1 such that A ⊆ αV . Then by Theorem “If B is balanced,
then αB = |α|B for any scalar α”, for |β| ≥ α, B ⊆ βV + V ⊆ βV + βV ⊆ βV which
implies that B is bounded.
(c) Any subset of a totally bounded set is totally bounded. The union of totally bounded
sets is totally bounded.
Theorem 2.3.1. If X is a Hausdorff topological vector space such that there exists a
totally bounded neighbourhood of the origin in X, then X is finite dimensional.
Proof. Let U0 be a totally bounded neighbourhood of the origin and A be a finite subset of
X such that
U0 ⊆ 2−1 U0 + A.
Setting E := [A] span of A. Then by induction, we have
U0 ⊆ 2−n U0 + E,
which implies that
U0 ⊆
\
n∈N
−n
for all n ∈ N
2−n U0 + E = E,
since {2 U0 }n∈N forms a neighbourhood basis of the origin in X, by Proposition 2.1.9 (b),
we have that U0 ⊆ E. But E is a closed set (as a finite dimensional vector subspace), hence
U0 ⊆ E. On the other hand
[
λU0 ,
X⊆
λ>0
and therefore, X = E.
Corollary 2.3.1. Every locally compact topological vector space is finite dimensional.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 48
Corollary 2.3.2. If the closed unit ball is compact in a normed vector space X, then X
is finite dimensional.
Qamrul Hasan Ansari
2.4
Topological Vector Spaces
Page 49
Topological Properties of Convex Sets and Compact
Sets
Proposition 2.4.1. Let K be a convex subset of a topological vector space X. Then K
and int K are convex.
Proof. The convexity of K implies that for all λ ∈ [0, 1], we have
λK + (1 − λ)K ⊆ K,
and therefore
λK + (1 − λ)K ⊆ K.
For λ ∈ [0, 1], we also have
λK = λK
and (1 − λ)K = (1 − λ)K
by Proposition 2.1.1. Thus
λK + (1 − λ)K = λK + (1 − λ)K ⊆ λK + (1 − λ)K ⊆ K,
which proves that K is convex.
Since int K ⊆ K and K is convex, we have
λ int K + (1 − λ) int K ⊆ K,
for all λ ∈ (0, 1).
Two sets on the left hand side are open, so is their sum. Since every open subset of K is a
subset of int K, it follows that int K is convex.
Lemma 2.4.1. Let K be a convex subset of a topological vector space X, and let x ∈
int K, y ∈ K and z := λx + (1 − λ)y with λ ∈ (0, 1). Then z ∈ int K
Proof. It is sufficient to assume that z = 0 since otherwise one can translate K. Thus for
−λ
, and hence, αx = y ∈ K for α < 0.
z = 0, we have y = αx for α < 0 with α = 1−λ
Therefore, αx is a limit point of K. Since α(int K) is a neighbourhood of αx and αx is
a limit point of K, we have α(int K) ∩ K 6= ∅. Therefore, there exists a v0 ∈ int K such
that αv0 ∈ K. Let β := α(α − 1)−1 . Then β ∈ (0, 1). Since v0 ∈ int K and αv0 ∈ K, by
hypothesis, βv0 + (1 − β)αv0 = 0. Since v → βv + (1 − β)αv0 is an homeomorphism that
maps v0 → 0, the set
U = {βv + (1 − β)αv0 : v ∈ int K}
is a neighbourhood of the origin. But obviously U ⊆ K, hence 0 ∈ int K.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 50
Proposition 2.4.2. Let K be a convex subset of a topological vector space X. If int K 6=
∅, then K = int K and int K = int K.
Proof. The inclusion int K ⊆ K is obvious. To show the reverse inclusion, let y ∈ K. If
x ∈ int K and λ ∈ (0, 1), then by previous lemma zλ := λx + (1 − λ)y ∈ int K. Taking λ
tends to zero, we have y = limλ→0 zλ and hence y ∈ int K.
Finally, we prove that int K = int K. For this, it is sufficient to show that, if K is convex set
with nonempty interior and 0 ∈ int K, then 0 ∈ int K. Let U be a symmetric neighbourhood
of the origin such that U ⊆ K. The previous result implies that 0 ∈ int K. Thus, there
exists a z ∈ U ∩ int K and in particular we have z ∈ int K, −z ∈ K. By previous lemma
and the fact that 0 = 2−1 z + 2−1 (−z), we conclude that 0 ∈ int K.
Remark 2.4.1. If K is a subset of a topological vector space X, then E = co(K) is the
smallest closed convex set that contains K. E is called the closed convex hull of K and it is
denoted by co(K). Similarly, the closed convex balanced hull of K is the set eco(K), denoted
by eco(K).
Proposition 2.4.3. Let (X, T ) be a topological vector space. Every convex neighbourhood of 0 in X contains a balanced convex neighbourhood of 0, that is, for every convex
neighbourhood U of 0, there exists a balanced convex neighbourhood A of 0 such that
A ⊆ U.
T
Proof. Let U be a convex neighbourhood of 0 in X and set A := |α|=1 αU. Since A is the
intersection of convex sets it is convex. It is balanced because for every |β| ≤ 1,
\
\
βA =
(βα)U =
((|β|)α) U = |β|A,
|α|=1
|α|=1
and by convexity,
|β|A = |β|A + (1 − |β|)0 ⊆ A.
Proposition 2.4.4. Let U be a neighbourhood of the origin in a topological vector space
X. If U is convex and balanced and pU is the Minkowski functional associated to U
(defined as pU (x) = inf{µ > 0 : x ∈ µU}), then
int U = {x ∈ X : pU (x) < 1} and U = {x ∈ X : pU (x) ≤ 1}.
Proof. By Theorem 1.5.2 (b), we know that pU is a seminorm. Set
S0 := {x ∈ X : pU (x) < 1} and S1 := {x ∈ X : pU (x) ≤ 1}.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 51
Then obviously S0 ⊆ U ⊆ S1 . Since U is a neighbourhood of 0, pU is a continuous functional. Therefore, S0 is an open set and S1 is closed one. Let x ∈ S1 , U0 be an arbitrary
neighbourhood of 0 and λ ∈ (0, 1) be such that −λx ∈ U0 . Then clearly (1 − λ)x ∈ S0 and,
as (1 − λ)x ∈ x + U0 , S0 and x + U0 have a common point. Hence x ∈ S0 , that is, S1 ⊆ S0 .
But S0 ⊆ S1 = S1 so S1 = S0 . By Proposition 2.4.2, S0 = int S1 . Since S0 ⊆ U ⊆ S1 , we get
the conclusion.
Corollary 2.4.1. Let U be a neighbourhood of the origin in a topological vector space
X. If U is open, convex and balanced and pU is the Minkowski functional associated to
U (defined as pU (x) = inf{µ > 0 : x ∈ µU}), then U = {x ∈ X : pU (x) < 1}, boundary
of U ∂U = {x ∈ X : pU (x) = 1} and ext U = {x ∈ X : pU (x) > 1}.
Proposition 2.4.5. Every compact subset of a Hausdorff topological vector space X is
totally bounded.
Proof. Let A be a compact subset of X and W be the neighbourhood of the origin in X.
Then the open covering {W + x}x∈A of A will have a finite subcovering.
Problem 2.4.1. Prove that every compact set a Hausdorff topological vector space X is
bounded.
Proposition 2.4.6. Let A and B be two compact subset of a Hausdorff topological vector
space X and E be a closed subset of X. Then the following assertions hold.
(a) αA + βB is a compact set for all α, β ∈ R.
(b) The balanced hull of A, ec(A), is a compact set.
(c) E + A is a closed set.
Proof. (a). Since f (x, y) = αx + βy is a continuous function and A × B is compact in X × X,
f (A × B) = αA + βB
is a compact set.
(b). Let g((α, x)) = αx and S = {α ∈ R : |α| ≤ 1}, then ec(A) = g(S × A) and g being
continuous and S × A being compact in R × X, ec(A) must be compact.
(c). Let x0 ∈
/ E + A. Then we shall prove that x0 is not a limit point of E + A. For any
x ∈ A, E + x is closed so there exists a neighbourhood U(x) of the origin such that
(U(x) + x0 ) ∩ (E + x) = ∅.
(2.1)
Qamrul Hasan Ansari
Topological Vector Spaces
Page 52
We can assume that U(x) is open and symmetric. For any x ∈ A, let U1 (x) be an open and
symmetric neighbourhood of the origin such that
U1 (x) + U1 (x) ⊆ U(x).
Then the system {U1 (x) + x}x∈A form an open covering of A. Therefore there exists a finite
subcover {U1 (xi ) + xi }ni=1 of A. Let
U0 :=
n
\
U1 (xi ).
i=1
Then
U0 + A ⊆
[
(U(x) + x) .
x∈A
By (2.1), x0 ∈
/ U(x) + E + x and therefore x0 ∈
/ U0 + E + A. Hence
(U0 + x0 ) ∩ (E + x) = ∅,
which proves that x0 ∈
/ E + A. Thus E + A is closed.
Problem 2.4.2. Give an example to show that A + B is not closed whenever A and B are
closed.
Proposition 2.4.7. Let K and C be subsets of a topological vector space X such that
K is compact, C is closed and K ∩ C = ∅. Then there exists a neighbourhood U of 0
such that
(K + U) ∩ (C + U) = ∅.
In other words, there exist disjoint open sets that contain K and C.
Proof. If K is empty, then K + U is empty and the conclusion follows. So we assume that
K 6= ∅, and consider a point x ∈ K. Since C is closed and K ∩ C = ∅, then x ∈
/ C
and C c is open. Since the topology on X is invariant under translations, Proposition 2.1.23
implies that 0 has a symmetric neighbourhood Ux such that x + Ux + Ux + Ux ∈ C c , that is,
(x + Ux + Ux + Ux ) ∩ C = ∅. By symmetry of Ux , we have
(x + Ux + Ux ) ∩ (C + Ux ) = ∅.
(2.2)
Every x ∈ K corresponds such a Ux . Since K is compact, there exists a finite collection
(xi , Uxi )ni=1 such that
K ⊆ (x1 + Ux1 ) ∪ · · · ∪ (xn + Uxn ) =
n
[
(xi + Uxi ).
i=1
3
For every neighbourhood U of 0, there exists a symmetric neighbourhood V of 0 such that V +V +V ⊆ U .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 53
Put U := Ux1 ∩ · · · ∩ Ux1 . Then, for every i,
(xi + Uxi + Uxi ) does not intersect (C + Uxi ),
and therefore,
(xi + Uxi + U) does not intersect (C + U).
Taking the union over i:
K +U ⊆
n
[
(xi + Uxi + U) does not inetrsect C + U.
i=1
Note 2.4.1. A topological vector space is regular (a topological space is regular if separates
points from closed sets that do not include that point).
Note 2.4.2. A topological vector space is Hausdorff X if the singleton sets are closed.
Indeed, let x, y ∈ X, x 6= y. If {x} and {y} are closed, then they are compact and by
Proposition 2.4.7, X is Hausdorff.
Proposition 2.4.8. If A1 , . . . , An are convex and compact subsets of a Hausdorff topological vector space X, then
!
n
[
E = co
Ai
i=1
is compact.
P
Proof. Let Λ
the set of all elements (λ1 , λ2 , . . . , λn ) ∈ Rn with λi ≥ 0 and ni=1 λi =
Qdenotes
1. Let A = ni=1 Ai . Consider the subset Λ×A of Rn ×X n and the functional f : Rn ×X n →
X given by
n
X
f (((λ1 , λ2 , . . . , λn ), (x1 , x2 , . . . , xn ))) =
λi xi .
i=1
Since f is continuous, Λ × A is compact (in Rn × X n ) and f (Λ × A) = E, it follows that E
is compact in X.
Corollary 2.4.2. In a Hausdorff topological vector space, the convex hull of a finite set
is compact.
Corollary 2.4.3. In a Hausdorff topological vector space, the convex hull of a finite set
is precompact.
Qamrul Hasan Ansari
2.5
Topological Vector Spaces
Page 54
Continuity and Finite Dimension Property
Proposition 2.5.1. Let X and Y be topological vector spaces and suppose that T : X →
Y is a linear map. Then T is continuous if and only if it is continuous at 0.
Proof. Suppose that T is continuous at 0. Let x ∈ X and suppose that {xν } is a net in X
such that xν → x. Then xν − x → x − x = 0. By hypothesis, T (xν − x) → T (0) = 0. That
is, T (xν ) − T (x) → 0, or T (xν ) − T (x) + T (x) → T (x). Thus T (xν ) → T (x) in Y and hence
T is continuous at x ∈ X. The converse is clear.
Lemma 2.5.1. Let X and Y be topological vector spaces and T : X → Y be a linear
map. Then T (B) is a balanced set in Y if B is a balanced set in X.
Proof. For every |α| ≤ 1, we have
α{T (x) : x ∈ B} =
=
=
⊆
{αT (x) : x ∈ B}
{T (αx) : x ∈ B}
{T (y) : y ∈ αB}
{T (y) : y ∈ B}.
Thus, for all |α| ≤ 0, αT (B) ⊆ T (B) and therefore, T (B) is balanced.
In Banach spaces continuous and bounded linear maps are the same. We need a notion of
bounded map to obtain a generalization for topological vector spaces.
Definition 2.5.1. Let X and Y be topological vector spaces. A linear map T : X → Y is
said to be bounded if it maps bounded sets into bounded sets, that is, for every bounded set
B of X, T (B) is bounded in Y .
The following theorem concerns only scalar-valued linear maps, that is, linear functionals.
Since R is a topological vector space, there is a well-defined notion of continuity for functionals. In particular, we may define:
Definition 2.5.2. Let (X, T ) be a topological vector space. Its dual X ∗ is the space of
continuous linear functionals over X.
As of now, X ∗ is not endowed with any topology, as it was in Banach spaces. We will see
how to topologize X ∗ later when we study weak topologies.
The following theorem establishes the equivalence between continuity and boundedness for
linear functionals.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 55
Theorem 2.5.1. Suppose that f : X → R is a linear functional on a topological vector
space X such that f (x) 6= 0 for some x ∈ X. The following statements are equivalent.
(a) f is continuous.
(b) The null set N(f ) = {x ∈ X : f (x) = 0} of f is closed.
(c) The null set N(f ) of f is not dense in X.
(d) f is bounded on some neighbourhood of 0.
Proof. (a) ⇒ (b): Suppose that f is continuous. Since {0} is closed in R, then N(f ) =
f −1 ({0}) is closed in R. So, clearly (a) implies (b).
(b) ⇒ (c): Suppose that N(f ) is closed. Since we assume that f (x) 6= 0 for some x ∈ X, it
must be that N(f ) 6= X, hence N(f ) cannot be dense in X.
(c) ⇒ (d): Suppose that N(f ) is not dense in X. That is, the complement of N(f ) has
nonempty interior. Hence there is a point x ∈ X and a neighbourhood U of 0 such that
(x + U) ∩ N(f ) = ∅,
(2.3)
that is, f (x + U) 6∋ 0 which means f (y) 6= −f (x) for all y ∈ U. By Proposition 2.4.3 (Every
neighborhood of zero contains a balanced neighborhood of zero) we may assume that U is
balanced. Hence, Lemma 2.5.1, f (U) is balanced.
Balanced set in R are either bounded, in which case we are done, of equal to the whole of
R, which contradicts the requirement that f (y) 6= −f (x) for all y ∈ U.
(d) ⇒ (a): Suppose that f (U) is bounded for some neighbourhood U of 0, that is, there
exists k > 0 such that |f (x)| ≤ k for all x ∈ U. Let ε > 0 be given. Set W := kε U. Then W
is a neighbourhood of 0, and if y ∈ W = kε U we have kε y ∈ U so that |f ( kε y)| < k. Then for
all y ∈ W ,
ε
|f (y)| ≤ sup |f (x)| ≤ ε.
k x∈U
It follows that f is continuous at 0, and hence f is continuous on X, by Proposition 2.5.1.
As for Banach spaces, finite-dimensional topological vector spaces are simpler than infinitedimensional ones.
Lemma 2.5.2. Let (X, T ) be a topological vector space. Then for any x ∈ X, the map
φx : R ∋ α 7→ αx ∈ X
is continuous with respect to T .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 56
Proof. Consider the map ωx : R ∋ α 7→ (α, x) ∈ R × X. Then ωx is continuous when R × X
is equipped with the product topology. Since φx is the composition of continuous maps
multiplication
ω
x
φx : R −→
R × X −−−−−−−→ X,
it is continuous.
Lemma 2.5.3. Let X be a topological vector space. Any linear function T : Rn → X is
continuous.
Proof. Let {e1 , . . . , en } be the standard basis in Rn , so that
(α1 , . . . αn ) = α1 e1 + · · · αn en ,
for all (α1 , . . . αn ) ∈ Rn .
Let
xi = T (ei ),
for all i = 1, 2, . . . , n.
By using the notation of above lemma and linearity of T ,
T ((α1 , . . . αn )) =
n
X
φ x 1 αi ,
for any (α1 , . . . , αn ) ∈ Rn ,
i=1
which means that T can be written as a composition
φx ×···×φxn
addition
· · × X} −−−−→ X.
T :R
· · × R} −−1−−−−−→ |X × ·{z
| × ·{z
n times
n times
Last map is the addition which is continuous and by above lemma first map in this composition is continuous, so T is continuous.
The following proposition shows, in particular, that all finite-dimensional subspaces of a
topological vector space are closed.
Proposition 2.5.2. Let Y be an n-dimensional subspace of a Hausdorff topological vector space X. Then the following assertions hold.
(a) Every isomorphism Rn → Y (equivalence in the category of vector spaces) is a
homeomorphism (equivalence in the category of topological spaces).
(b) Y is closed in X.
Proof. (a) Let T : Rn ⇄ Y be an isomorphism (that is, linear and bijective). We need to
show that T and T −1 are continuous. By Lemma 2.5.3, T is continuous and so only to show
that T −1 .
Qamrul Hasan Ansari
Topological Vector Spaces
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Let K be the surface of the unit ball in Rn , that is,
K = {ξ ∈ Rn : kξk = 1}.
Then K is compact in Rn and so T (K) is compact in X, because T is linear and by Lemma
2.5.3 it is continuous and continuous functions map compact sets into compact sets. Since X
is Hausdorff, T (K) is closed in X. Since 0 ∈
/ K and T is injective, it follows that 0 = T (0)
is not an element of T (K). In other words, 0 is an element of the open set X \ T (K). By
Proposition 2.4.3, there is a balanced neighbourhood W of 0 such that W ⊆ X \ T (K).
In particular, W ∩ T (K) = ∅. Note that W is a neighborhood of 0 in X, but W ∩ Y is
a neighborhood of zero in Y (in the induced topology). Since T −1 is linear, it follows that
E := T −1 (W ∩ Y ) is a balanced set in Rn (as it is linear image of a balanced set and so
balanced) such that T −1 (W ∩ Y ) ∩ K = ∅.
We claim that kξk < 1 for every ξ ∈ T −1 (W ∩ Y ). Indeed, if ξ ∈ T −1 (W ∩ Y ) but kξk ≥ 1.
Then kξk−1ξ ∈ K and also belongs to the balanced set T −1 (W ∩ Y ). This contradicts
T −1 (W ∩ Y ) ∩ K = ∅, and we conclude that every ξ ∈ T −1 (W ∩ Y ) satisfies kξk < 1, as
claimed. It follows that T −1 maps W ∩ Y into S := {ξ ∈ Rn : kξk < 1}.
For each 1 ≤ i ≤ n, let ϕi : Rn → R be the projection map defined as
ϕi ((α1 , . . . , αn )) = αi ,
for all (α1 , . . . , αn ) ∈ Rn .
Then ϕi ◦ T −1 : X → R is a linear functional such that |(ϕi ◦ T −1)(x)| < 1 for all x ∈ W ∩ Y .
By Theorem 2.5.1, ϕi ◦ T −1 is continuous for each 1 ≤ i ≤ n, and therefore,
T −1 : x 7→ (ϕ1 ◦ T −1 )(x), . . . , (ϕn ◦ T −1 )(x)
is continuous.
We conclude that T : Rn → X is a linear homeomorphism.
(b) Let {ui } be a basis for Y , and let T : Rn → Y be a linear map defined by T (ei ) = ui .
It is an isomorphism, hence by the preceding item it is a homeomorphism. Let as above K
and let W be a balanced neighborhood of 0 (in X) that does not intersect T (K).
Let x ∈ Y . By Proposition 2.1.3, there exists a r > 0 for which x ∈ rW , and therefore it is
in the closure of the set4
rW ∩ Y ⊆ rT (S) ⊆ T (rS).
The latter being compact, it is closed. Since x is in the closure of a closed set, x ∈ T (rS) ⊆
Y.
4
It is easy to show that x ∈ W and x ∈ Y implies that x ∈ W ∩ Y .
Qamrul Hasan Ansari
2.6
Topological Vector Spaces
Page 58
Hyperplanes and Hahn-Banach Theorem on Separation
Let us recall the following definitions.
Definition 2.6.1. Two nonempty sets A and B of a topological vector space X are said
to be separated by a closed hyperplane H if A is contained in one of the closed half-spaces
determined by H and B is contained in the other side of closed half-spaces.
Definition 2.6.2. Two nonempty sets A and B of a topological vector space X are said
to be strictly separated by the closed hyperplane H if A is contained in one of the open
half-spaces determined by H and B is contained in the other half-spaces.
Definition 2.6.3. For a subset A of a vector space X, a hyperplane H is called a supporting
hyperplane of A, if H contains at least one point of A and all the points of A lie on the same
side of H.
Lemma 2.6.1. If a Hausdorff topological vector space X is one-dimensional, then it is
isomorphic to R.
Proof. Let 0 6= x ∈ X. Then X = [x] span of {x} as X is one dimensional. Define a mapping
f : R → X by
f (α) = αx.
Then clearly f is linear and one-one and hence onto as dim X = dim R. Also, the mapping
φ : R × [x] → X given by φ(α, x) = αx is continuous by the continuity of the scalar
multiplication function. Also, the mapping ψ : R → R × [x] defined by ψ(α) = (α, x) is
ψ
φ
continuous. Then R −
→ R × [x] −
→ X, and f = φ ◦ ψ is continuous.
Now we show that f −1 is continuous. Since X is Hausdorff and 0 6= x, by Proposition 2.1.4,
there exists a neighbourhood U of origin such that x ∈
/ U. It is enough to show that f −1 is
−1
continuous at origin. As f (0) = 0 because x 6= 0, we need to show that for any ε > 0,
there exists a neighbourhood W of 0 such that f −1 (W ) ⊆ (−ε, ε).
Let ε > 0, then εx 6= 0 and there exists a neighbourhood W of 0 such that εx ∈
/ W . Further
as set of all balanced neighbourhood of 0 form a neighbourhood basis, we can assume W to
be balanced.
Now we claim that f −1 (W ) ⊆ (−ε, ε). Let if possible α ∈ R such that f −1 (αx) = α ∈
/ (−ε, ε),
that is, |α| ≥ ε. Thus as W balanced and αx ∈ W , we have εx = (ε/α)αx ∈ W which
contradicts to εx ∈
/ W . Hence f −1 is continuous.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 59
Proposition 2.6.1. Let X be a topological vector space, f : X → R be a (non-zero)
linear functional on X and λ ∈ R. Then the hyperplane H = {x ∈ X : f (x) = λ} is a
closed set if and only if f is continuous.
Proof. It follows from Theorem 2.5.1.
Lemma 2.6.2. Let X be a topological vector space. A linear functional f : X → R is
continuous if and only if and if only if there exists a nonempty open set A ⊆ X such
that f (A) is bounded.
Proof. Since f is linear, f (0) = 0. If f is continuous, then A = f −1 ((−1, 1)) is required open
set, and also it is nonempty as 0 ∈ (−1, 1).
Conversely, assume that A is an open subset of X such that f (A) is bounded. Then there
exists k > 0 such that |f (x)| ≤ k for all x ∈ A. Let ε > 0 be arbitrary and x0 ∈ A. Set
E := A − x0 . Then E is a neighbourhood of the origin. Since x 7→ µx is a continuous
mapping, for any µ > 0, and 0 7→ µ.0 = 0, there will be a neighbourhood U of 0 such that
µU ⊆ E.5
For µ = 2λ/ε or µε = 2λ, we have |f (x)| ≤ ε for all x ∈ U. Indeed, let x ∈ U be arbitrary.
Then µx ∈ µU ⊆ E = A − x0 . This implies that µx ∈ E = A − x0 and hence µx = a − x0
for some a ∈ A. Therefore, |f (µx)| = |f (a − x0 | ≤ |f (a)| + |f (x0 )| ≤ λ + λ ≤ 2λ, and thus
µ|f (x)| ≤ 2λ = µε. Therefore, |f (x)| ≤ ε for all x ∈ U.
Thus for a given ε > 0, there exists a neighbourhood U of 0 such that |f (x)| ≤ ε for all
x ∈ U, and hence f is continuous at 0. Since f is linear, by Proposition 2.5.1, f is continuous
on X.
Theorem 2.6.1 (Hahn-Banach or S. Mazur). Let K be a nonempty open convex subset
of a topological vector space X and M ⊆ X be an affine set which does not meet K, that
is M ∩ K = ∅. Then there exists a closed hyperplane H which contains M and does not
meet K, that is, H ∩ K = ∅.
Proof. By translation, we reduced the problem to the case when 0 ∈ K. As K is open and
0 ∈ K, we have that K is absorbent by the result “every neighbourhood of 0 is an absorbent
set”.
Let p be the Minkowski functional associated to K. As K is open, then by Corollary 2.4.1,
K = {x ∈ X : pK (x) < 1}. Let V = [M]. Then V is vector subspace of X containing M,
5
For every neighbourhood E of 0 (µ.0 = 0), by continuity, there exists a neighbourhood U of 0 such that
f (U ) ⊆ E (by the continuity of x 7→ µx, we get µU ⊆ E)
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Topological Vector Spaces
Page 60
and M is a maximal subspace in V and hence a hyperplane in V . As 0 ∈ K and M ∩ K = ∅,
then 0 ∈
/ M. Then by Proposition 1.3.1, there exists linear functional f on V such that
M = {y ∈ V : f (y) = 1}.
Since M ∩ K = ∅, we have M ⊆ K c . Then Corollary 2.4.1, we have pK (y) ≥ 1 for all y ∈ K c ,
and so
pK (y) ≥ 1, for all y ∈ M ⊆ K c .
(2.4)
Let y ∈ V be such that f (y) = 1. Then as above y ∈ M. By (2.4), pK (y) ≥ 1.
Now we consider two cases.
y
∈ V as V is a subspace, and so by
Case I. Let y ∈ V such that f (y) > 0. Then f (y)
y
y
y
1
linearity of f f f (y)
= f (y)
≥ 1. By
= 1. Therefore, f (y)
∈ M. Again as above, pK f (y)
sublinearity of pK , we have
1
p (y)
f (y) K
≥ 1, that is, pK (y) ≥ f (y).
Case II. Let y ∈ V such that f (y) ≤ 0. Since pK (y) ≥ 0 for all y (by definition of Minkowski
function), so pK (y) ≥ f (y).
So from Case I and Case II, pK (y) ≥ f (y) for all y ∈ V .
Then by Hahn-Banach Theorem 1.6.1, there exists a linear functional h on X such that
h|V ≡ f,
that is,
h(x) = f (x),
for all x ∈ V,
and
h(x) ≤ pK (x),
for all x ∈ X.
Define H := {x ∈ X : h(x) = 1}, then H is a hyperplane on X such that H ∩ V = M (as
h(x) = f (x) for all x ∈ V and h(x) = f (x) = 1 ⇔ x ∈ M). Since h(x) ≤ pK (x) for all
x ∈ X, and h(x) = 1 for all x ∈ H, we have pK (x) ≥ 1 for all x ∈ H. Then by Corollary
2.4.1, H ⊆ K c , that is, H ∩ K = ∅.
Now, we know that either H is closed or H is dense. If H is dense, then by definition of
denseness, H ∩ U 6= ∅ for all open set U in X. But K is open and H ∩ K = ∅, so H is not
dense. Hence H is closed. Then H is the required closed hyperplane.
Corollary 2.6.1. Let E be a vector subspace of a topological vector space X and K be a
nonempty open convex subset of X such that E ∩ K = ∅. Then there exists a continuous
linear functional f : X → R on X such that f (x) = 0 for all x ∈ E and f (x) > 0 for all
x ∈ K.
Proof. Since E is a vector subspace, we have 0 ∈ E. Then by Hahn-Banach Theorem 2.6.1,
there exists closed hyperplane H such that E ⊆ H and H ∩ K = ∅. But as 0 ∈ E, we have
Qamrul Hasan Ansari
Topological Vector Spaces
Page 61
0 ∈ H and hence H = {x ∈ X : f (x) = 0} for some linear functional f . The closedness of
H implies that f is continuous. Since K is convex and f is continuous, f (K) is an interval
in R and K ∩ H = ∅. Then either K ⊂ {x ∈ X : f (x) > 0}, then f is required function or
K ⊂ {x ∈ X : f (x) < 0}, then we take −f , so (−f )(x) > 0 for all x ∈ K.
Theorem 2.6.2 (E. Eidelheit). If K1 and K2 are two nonempty convex subsets of a
topological vector space X such that int K1 6= ∅ and (int K1 ) ∩ K2 = ∅, then there exists
a closed hyperplane which separates K1 and K2 .
Proof. Let K := (int K1 ) − K2 . Since int K1 and K2 are convex, K is convex. Also, K is
open. Indeed,
[
{int K1 − x}.
K = (int K1 ) − K2 =
x∈K2
Since int K1 is an open set, then by translation, int K1 − x is also open for all x ∈ K2 . Thus
K is the arbitrary union of open sets and hence open.
Also, 0 ∈
/ K. Indeed, if 0 ∈ K = int K1 − K2 , then for some x ∈ int K1 and y ∈ K2 ,
0 = x − y, that is, x = y and thus (int K1 ) ∩ K2 6= ∅, a contradiction to the hypothesis.
By Corollary 2.6.1, there exists a continuous linear functional f : X → R such that f (z) > 0
if z ∈ K, that is, f (z) = f (x − y) > 0 for z = x − y with x ∈ int K1 and y ∈ K2 .
Then by linearity of f , we have f (x) > f (y) for all x ∈ int K1 and y ∈ K2 . That is, f (x)
is bounded below by f (y) for every x ∈ int K1 . Therefore, the infimum of f over int K1
exists. Let λ = inf x∈int K1 f (x). Then f (x) ≥ λ for all x ∈ int K1 and by same argument
as above λ becomes an upper bound of f over K2 . So f (y) ≤ λ for all y ∈ K2 . Then
H = f −1 (λ) = {x ∈ X : f (x) = λ} is the hyperplane separating int K1 and K2 .
Setting H1 := {x ∈ X : f (x) ≥ λ}. Then H1 is closed and int K1 ⊆ H1 , and therefore,
int K1 ⊆ H1 . Since K1 ⊆ K1 and K1 = int K1 , we have K1 = int K1 ⊆ H1 = H1 and so
K1 ⊆ K1 ⊆ H1 , that is, K1 ⊆ H1 . Thus K1 is contained completely in one side of the
hyperplane H, and hence H separates K1 and K2 .
Remark 2.6.1. If K1 and K2 both are open, then H strictly separates them.
Qamrul Hasan Ansari
2.7
Topological Vector Spaces
Page 62
Complete Topological Vector Spaces
Throughout this section, unless otherwise specified, (Γ, ≤) is a directed set and X is assumed
to be a Hausdorff topological vector space.
Definition 2.7.1. A net (also called generalized sequence) {xα }α∈Γ of elements in X is said
to be convergent to a point x ∈ X if for any neighbourhood U of 0, there exists α0 ∈ Γ such
that xα − x ∈ U for any α ≥ α0 .
When {xn }n∈N is an ordinary sequence in X, it is called a convergent sequence if it satisfies
the above condition.
Definition 2.7.2. A net (also called generalized sequence) {xα }α∈Γ of elements in X is
called a Cauchy net (or Cauchy generalized sequence) if for any neighbourhood U of 0, there
exists α0 ∈ Γ such that xα1 − xα2 ∈ U for any α1 , α2 ≥ α0 .
When {xn }n∈N is an ordinary sequence in X, it is called a Cauchy sequence if it satisfies the
above condition.
Remark 2.7.1. It can be easily seen that every Cauchy sequence is bounded, and any
convergence net (or any convergent sequence) is a Cauchy net (or a Cauchy sequence).
Definition 2.7.3. A space X is called complete (or sequentially complete) if any Cauchy
net {xα }α∈Γ (or any Cauchy sequence {xn }n∈N ) of elements in X converges to an element of
X.
A subset A of X is said to be complete (sequentially complete) if any Cauchy net (Cauchy
sequence) of elements in A converges to an element of A.
Remark 2.7.2.
(a) Finite dimensional topological vector spaces are complete.
(b) Any closed subset of a complete topological vector space is complete.
(c) If X is a Hausdorff topological vector space and E is a complete subset of X, then E
is closed.
(d) If {Xi }i∈I is aQ
family of complete topological vector spaces, then the product topological
vector space i∈I Xi is complete.
Theorem 2.7.1. For any Hausdorff topological vector space X, there exists a unique
up to an isomorphism complete Hausdorff topological vector space X̃ such that X is
isomorphic to a topological vector subspace of X̃, dense in X̃.
Proposition 2.7.1. A Hausdorff topological spaces X is compact if and only if every
sequence of elements in X is convergent.
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Topological Vector Spaces
Page 63
Proposition 2.7.2. Let X be a complete Hausdorff topological vector space. Then a
subset A of X is totally bounded if and only if it is relatively compact, that is, A is
compact.
Proof. Assume that A is a totally bounded set. Let {xα }α∈Γ be a universal net (generalized
sequence) in A and U be a neighbourhood of 0. Let U0 be a balanced neighbourhood of 0
such that U0 + U0 ⊆ U. Then by the definition of totally bounded set, there exist ai ∈ X
(i = 1, 2, . . . , n) such that
n
[
A ⊆ (U0 + ai ).
i=1
We write
Eα := {xβ : β ≥ α}.
If Eα * U0 + ai for all i ∈ {1, 2, . . . , n} and all α ∈ Γ, then there exist αi ∈ Γ (i = 1, 2, . . . , n)
such that Eαi ⊆ (U0 + ai )c . If α0 ≥ αi for all i = 1, 2, . . . , n, then Eα0 ⊂ Eαi , hence
!c
n
[
Eαi ⊆
(U0 + ai ) ,
i=1
which is a contradiction. Thus, there will be a k ∈ {1, 2, . . . , n} and a α∗ ∈ Γ with Eα∗ ⊆
U0 + ak , whence xα′ − xα′′ ∈ U0 + U0 ⊆ U for α′ , α′′ ≥ α∗ . Thus {xα }α∈Γ is a Cauchy net.
Since X is complete, there exists x = lim xα and x ∈ A. Hence A is compact.
α∈Γ
Proposition 2.7.3. In a Housdorff topological vector space any compact set is complete.
Definition 2.7.4. A topological vector space is called quasicomplete if any closed bounded
subset is complete.
Qamrul Hasan Ansari
2.8
2.8.1
Topological Vector Spaces
Page 64
Metrizable and Normable Topological Vector Spaces
Metrizable topological vector spaces
Definition 2.8.1. A topological vector space (X, T ) is said to be metrizable if the topology
T on X is defined by means of a metric.
Definition 2.8.2. A metric d in a vector space X is called invariant under translations if
for all x, y, z ∈ X.
d(x + z, y + z) = d(x, y),
Definition 2.8.3. A real-valued function q : X → R on a vector space X is called a
quasinorm (also called pseudonorm) if the following conditions hold: For all x, y ∈ X,
(i) q(x) = 0 if and only if x = 0;
(ii) q(x + y) ≤ q(x) + q(y);
(iii) q(−x) = q(x).
Remark 2.8.1. From (i), (ii) and (iii), it follows that q(x) ≥ 0 for all x ∈ X and also
|q(x) − q(y)| ≤ q(x − y),
for all x, y ∈ X.
Remark 2.8.2. Let d be a metric in a vector space X invariant under translations. Define
q(x) := d(x, 0),
for all x ∈ X.
Then q is a quasinorm on X and this quasinorm is called the quasinorm associated to the
metric d.
Conversely, if q is a quasinorm on a vector space X, then
d(x, y) = q(x − y),
for all x, y ∈ X
defines a metric which is invariant under translations and it is called the metric associated
to the quasinorm. The topology defined by the metric d, associated to the quasinorm q, is
called the topology defined by the quasinorm q.
Definition 2.8.4. A vector space X with a quasinorm q is said to be a quasinormed space
if q satisfies the following conditions:
(i) For all αn ∈ R and x ∈ X, lim αn = 0 implies lim q(αn x) = 0;
n→∞
n→∞
(ii) For all α ∈ R and {xn } ⊆ X, lim q(xn ) = 0 implies lim q(αxn ) = 0.
n→∞
n→∞
If a quasinormed space is a complete metric space (in the topology defined by the quasinorm),
we call it a Fréchet space.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 65
Proposition 2.8.1. Every quasinormed space is a topological vector space with respect
to the topology defined by the quasinorm.
Proof. Let X be a quasinormed space and q be the quasinorm on it. If lim xn = x and
n→∞
lim yn = y in the topology defined by q, then from
n→∞
q((xn + yn ) − (x + y)) ≤ q(xn − x) + q(yn − y),
it follows that x + y = lim (xn + yn ). Therefore vector addition is continuous.6
n→∞
In order to establish that scalar multiplication is also continuous it is sufficient to prove that
for any sequence {xn }n∈N in X and a sequence {αn }n∈N of scalars such that lim q(xn ) = 0
n→∞
and lim αn = 0 imply that lim q(αn xn ) = 0.
n→∞
n→∞
Indeed, this will mean the continuity at (0, 0) of the mapping (α, x) 7→ αx and, since the
mapping α 7→ αx and x 7→ αx are continuous at 0 ∈ R and 0 ∈ X, respectively, the mapping
(α, x) 7→ αx will then be continuous at any point of R × X.
Let lim q(xn ) = 0 and lim αn = 0. Let ε > 0 and for each i ∈ N, let
n→∞
n→∞
Mi := {λ ∈ R : q(λxi ) ≤ ε}.
The set Mi is closed since lim λn = λ implies lim q(λn x) = q(λx) for all x ∈ X. Setting
n→∞
n→∞
Λn :=
∞
\
Mi ,
for all n ∈ N,
i=n
then we obtain
R=
[
Λn .
n∈N
Indeed, if λ ∈ R, then lim q(λxn ) = 0, and hence there exists an n0 ∈ N such that q(λxi ) ≤ ε
n→∞
S
for i ≥ n0 . Thus λ ∈ Mi if i ≥ n0 , that is, λ ∈ Λn0 , and therefore R = n∈N Λn .
Since each Λn is a closed set, there is an n1 ∈ N with int Λn1 6= ∅. Hence there exists λ0 ∈ R
and a number η > 0 such that
q((λ + λ0 )xn ) ≤ ε for |λ| < η and n ≥ n1 .
On the other hand, there exists n2 ∈ N such that |αn | < η for n ≥ n2 . Then
q(αn xn ) ≤ q((αn + λ0 )xn ) + q(λ0 xn ) ≤ 2ε,
if n ≥ max{n1 , n2 }.
We have shown that lim q(αn xn ) = 0, which concludes the proof.
n→∞
6
As X is a metric space, the continuity of vector addition and scalar mutliplication means their sequentially continuity (that is, by usual sequence).
Qamrul Hasan Ansari
Topological Vector Spaces
Page 66
Theorem 2.8.1. A Hausdorff topological vector space X is metrizable if and only if
there exists a countable neighbourhood basis of the origin in X. If X is metrizable, then
its topology is defined by a quasinorm.
Proof. Let X be a Hausdorff topological vector space and there is a countable neighbourhood
basis {Un }n∈N of the origin 0. We can assume that each Un is balanced and
Un+1 + Un+1 ⊆ Un ,
for all n ∈ N.
(2.5)
We denote by F (N) the set of all nonempty finite subsets of N. If J ∈ F (N), that is,
J = {j1 , j2 , . . . , jm }, we set
UJ = Uj1 + Uj2 + · · · + Ujm ,
m
X
1
µJ =
.
ji
2
i=1
We now define a function q : X → R as follows. If there is no J ∈ F (N) such that x ∈ UJ ,
we set q(x) = 1. Otherwise, we write
q(x) = inf{µJ : x ∈ UJ , J ∈ F (N)}.
If for any number ε > 0, we denote
Sε = {x : q(x) ≤ ε},
then obviously,
Un ⊆ S2−n ,
(2.6)
hence S2−n is a neighbourhood of the origin.
Also the relation
S2−n−1 ⊆ Un
(2.7)
holds.
Indeed, if q(x) ≤ 2−n−1 , then there is a J ∈ F (N) with x ∈ UJ and µJ < 2−n , and we get
UJ ⊆ Un by (2.5). Hence (2.7) holds.
Now, (2.6) and (2.7) imply that {S2−n }n∈N is a neighbourhood basis of the origin. Since the
space X is Hausdorff, we have
\
S2−n = {0},
n∈N
and therefore q(x) = 0 if and only if x = 0.
Let us prove that
q(x + y) ≤ q(x) + q(y),
for all x, y ∈ X.
(2.8)
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Topological Vector Spaces
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It is sufficient to consider the case when q(x) + q(y) < 1. In this case, let ε > 0 be such that
q(x) + q(y) + 2ε < 1.
(2.9)
There exist J ′ , J ′′ ∈ F (N) with x ∈ UJ ′ , y ∈ UJ ′′ , µJ ′ < q(x) + ε, µJ ′′ < q(y) + ǫ. There
also exists a J ∈ F (N) such that µJ = µJ ′ + µJ ′′ due to (2.9), µJ ′ + µJ ′′ < 1). Since
UJ ′ + UJ ′′ ⊆ UJ , we have x + y ∈ UJ , and therefore,
q(x + y) ≤ µJ < q(x) + q(y) + 2ε,
for all ε > 0 for which (2.9) holds. Hence, inequality (2.8) is proved.
q(x) = q(−x) holds obviously as any UJ is a balanced set.
We conclude that q is a quasinorm and, since {S2−n }n∈N is a neighbourhood basis of the
origin, the topology of X coincides with the topology defined by the quasinorm q.
Note 2.8.1. The above proposition says that a Hausdorff topological vector space X is
metrizable if and only if the balls with radius 1/n centered at x form a local base at x.
Proposition 2.8.2. For each i = 1, 2, . . . , n, let Xi be a quasinormed
Qn vector space and
qi be the quasinorm of the space Xi . Then the product topology on i=1 Xi is defined by
the quasinorm
n
q((x1 , x2 , . . . , xn )) = max qi (xi ).
i=1
Qn
In particular, if any Xi is a Fréchet space, so is i=1 Xi .
Proposition 2.8.3. Let X be a quasinormed space and p be a seminorm bounded on
any bounded subset of X. Then p is continuous.
Proof. Let q be the quasinorm of X and consider a sequence {xn }n∈N of elements in X
such that lim q(xn ) = 0. Then there exists a sequence {αn }n∈N of natural numbers with
n→∞
lim αn = +∞ and lim αn q(xn ) = 0. Since q(αn xn ) ≤ αn q(xn ), we have lim αn xn = 0 in
n→∞
n→∞
n→∞
the topology of X and {αn xn }n∈N is a bounded sequence. Thus, by assumption, there is a
number µ > 0 such that p(αn xn ) ≤ µ for all n ∈ N. Hence p(xn ) ≤ αn−1 µ for all n ∈ N and
lim p(xn ) = 0.
n→∞
Therefore the seminorm p is continuous at the origin, and thus continuous at any point of
X.
2.8.2
Normable topological vector spaces
Definition 2.8.5. A topological vector space is said to be normable if its topology can be
defined by a norm.
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Topological Vector Spaces
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Theorem 2.8.2 (A. Kolmogorov). A Hausdorff topological vector space X is normable
if and only if there exists a convex bounded neighbourhood of the origin in X.
Proof. If X is a normed space, then the (open or closed) unit sphere (S(0, 1) = {x : kxk < 1}
or S(0, 1) = {x : kxk ≤ 1}) is convex and bounded.
Conversely, let U0 be a convex bounded neighbourhood of the origin in the Hausdorff topological vector space X. We can assume that U0 is also balanced since, otherwise one can
replace U0 by
\
U1 =
λU0 .
|λ|≥1
Indeed, U1 is balanced since for x ∈ U1 and 0 < |α| ≤ 1, we have αx ∈ λU0 for |λ| ≥ 1,
hence αx ∈ U1 .
On the other hand, there exists a balanced neighbourhood U ′ of the origin such that U ′ ⊆ U0
and we obviously have U ′ ⊆ U1 ⊆ U0 . Therefore, U1 is a balanced neighbourhood of the
origin which at the same time is convex (as an intersection of convex sets) and bounded (as
a subset of a bounded set).
Let T be the given topology of the Hausdorff topological vector space X.
Let p0 be the Minkowski functional associated to the balanced, convex and bounded neighbourhood U0 of the origin in X. Then this functional p0 is a seminorm by Theorem 1.5.2. If
p0 (x) = 0, then x ∈ µU0 for any µ > 0. As U0 is bounded, for any neighbourhood U of the
origin, there will be a λ > 0 such that λU0 ⊆ U. Then x ∈ U for any neighbourhood U of 0,
and, since X is Hausdorff, it follows that x = 0. Therefore, p0 is a norm on X.
In the topology defined by the norm p0 , a neighbourhood basis of the origin is given by the
system of all sets λU0 , with λ > 0 since, for the unit balls
S(0, 1) = {x : p0 (x) < 1} and S(0, 1) = {x : p0 (x) ≤ 1},
we have
S(0, 1) ⊆ U0 ⊆ S(0, 1).
Each set λU0 is obviously a neighbourhood of the origin in the topology T of the topological
vector space X. Conversely, if U is a T -neighbourhood of the origin, there exists a λ > 0
such that λU0 ⊆ U, as U0 is T -bounded. Therefore, the topology defined by the norm p0
coincides with the topology T .
3
Locally Convex Spaces
3.1
Basic Definitions and Properties
Definition 3.1.1. A topological vector space (X, T ) is said to be locally bounded if 0 has a
bounded neighbourhood.
Proposition 3.1.1. Let X be a topological vector space and V be a bounded neighbourhood of 0 in X. If α1 > α2 > · · · and αn → 0 as n → ∞, then the collection
{αn V : n = 1, 2, . . .} is a local base for X. In other words, if X is locally bounded, then
X has a countable local base.
Proof. Let U be a balanced neighbourhood of 0 in X. Since V is bounded, there exists s > 0
such that V ⊆ tU for all t > s. Let n be sufficient large such that srn < 1, it follows that
V ⊆ (1/rn )U. Hence U contains all but finitely many of the sets rn V .
Definition 3.1.2. A topological vector space (X, T ) is said to be locally compact if 0 has a
neighbourhood whose closure is compact.
Proposition 3.1.2. Every locally compact topological vector space X has finite dimension.
Proof. Let X be a locally compact topological vector space. Then by definition, 0 has some
neighbourhood U whose closure U is compact. Since every compact set is bounded in a
topological vector space, U is bounded and hence also U. Moreover, B := {(1/2n )U : n ∈ N}
is a local base for X.
Let y ∈ U. By definition y − 12 U intersects U, that is, y ∈ U + 21 U, from which we obtain
[
1
1
x+ U .
U ⊆U+ U =
2
2
x∈U
69
Qamrul Hasan Ansari
Topological Vector Spaces
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Since U is compact, it can be covered by a finite union:
1
1
1
U ⊆ x1 + U ∪ x2 + U ∪ · · · ∪ xm + U .
2
2
2
Let E := Span{x1 , x2 , . . . xm }. Then E is a finite dimensional subspace of X, and hence
closed by Proposition 2.5.2 (b). Since U ⊆ E + 21 U, and since αE = E for all α 6= 0, it
follows that
1
1
1
1
U ⊆ E + U = E + U,
2
2
4
4
and hence,
1
1
1
U ⊆ E + U ⊆ E + E + U = E + U.
2
4
4
Similarly, we may continue in the same way to show that U ⊆ E + 18 U, and so on, namely
∞ \
1
U⊆
E + nU .
2
n=1
Since B is a local base at 0, it follows that U is a subset of every neighbourhood of E, that
is1
U ⊆ E = E.
On the other hand, by Proposition 2.1.3, we have
X=
∞
[
nU ⊆ E,
n=1
which implies that X = E, that is, X is finite dimensional.
Recall the following concepts:
• Let X be a topological space and U be a family of neighbourhood of x ∈ X. A subfamily
B of U is called a neighbourhood base or base of neighbourhoods of x if for every
U ∈ U, there is a B ∈ B such that x ∈ B ⊆ U.
• The vector topology is completely determined by a neighbourhood base of 0, which is
called a local base for the vector topology.
Definition 3.1.3. A topological vector space (X, T ) is said to be locally convex if there
exists a basis of convex neighborhoods (local base)2 of the origin, that is, if there exists a
basis neighbourhood of origin whose members are convex. The topology T is called locally
convex.
Indeed, let x ∈ U , then for every n, x ∈ E + 21n U . Let U be the neighbourhood of the origin. Then
x + U intersects E which proves x ∈ E.
2
A collection B of neighbourhoods of a point x ∈ X is a local base at x if every neighbourhood of x
contains a member of B
1
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Topological Vector Spaces
Page 71
In other words, a topological vector space (X, T ) is locally convex if there exists a local base
of convex neighborhoods.
Proposition 3.1.3. Every locally convex space possesses a neighbourhood basis of the
origin consisting of convex balanced sets.
Proof. Let X be a locally convex space. Then there exists a neighbourhood basis of origin
containing convex sets.
Let U be an arbitrary neighbourhood of 0. Then there exists a convex neighbourhood U0 of
0 such that U0 ⊆ U. We define
\
U1 :=
αU0 .
|α|≤1
Then as U0 is convex, αU0 is convex for all |α| ≤ 1, and hence U1 is convex as it is the
intersection of convex set.
T
Also, U1 is balanced as for all |λ| ≤ 1, λU1 =
λαU0 ⊆ U1 because |αλ| = |α| |λ| ≤ 1.1 =
1.
|α|≤1
Now, since U0 ⊆ U and U1 ⊆ U0 , we have U1 ⊆ U. It shows that there exists a neighbourhood
basis of 0 consisting the convex balanced sets.
Note 3.1.1. The above proposition says that every locally convex space has a convex balanced local base.
Remark 3.1.1. (a) If X is a locally convex space and U is a convex neighbourhood of the
origin, then the set U0 = U ∩ (−U) is a convex balanced neighbourhood of the origin,
contained in U.
(b) In any locally convex space there exists a neighbourhood basis of the origin consisting
of closed convex balanced sets.
Indeed, in any topological vector space there exists a basis of closed neighbourhoods
of the origin and, the closure of a convex balanced set is also convex balanced.
(c) In a locally convex space, the set of all bounded, closed, convex and balanced sets
forms a basis of bounded sets.
Proposition 3.1.4. If ℘ is a set of seminorms on a vector space X, then the system of
all sets
E(p1 , p2 , . . . , pn ; ε) = {x : pi (x) < ε, i = 1, 2, . . . , n}
(3.1)
with pi ∈ ℘ and 0 < ε ∈ R, forms a neighbourhood basis of the origin for a locally convex
topology on X.
In other words, let ℘ = {pi : i ∈ I} be a family of seminorms on a vector space X and
Qamrul Hasan Ansari
Topological Vector Spaces
Page 72
U be the family of all finite intersections of sets of the form {x ∈ X : pi (x) < εi } where
εi > 0 and i ∈ I. Then U forms a neighbourhood basis of the origin (local base of the
origin) for a topology that makes X a locally convex topological vector space.
Proof. We shall check that all the conditions of Proposition 2.1.13 are satisfied.
E(p1 , p2 , . . . , pn ; ε) is convex: Let x, y ∈ E(p1 , p2 , . . . , pn ; ε). Then pi (x) < ε and pi (y) < ε
for all i = 1, 2, . . . , n. Since each pi is seminorm, that is, subadditive and positive homogeneous, we have
pi (λx + (1 − λ)y) ≤ λpi (x) + (1 − λ)pi (y) < λε + (1 − λ)ε = ε.
Hence λx + (1 − λ)y ∈ E(p1 , p2 , . . . , pn ; ε), that is, every set of the form (3.1) is convex.
E(p1 , p2 , . . . , pn ; ε) is balanced: Let α ∈ R such that |α| ≤ 1, and let y ∈ αE(p1 , p2 , . . . , pn ; ε).
Then y = αx with pi (x) < ε. If α = 0, then y = 0·x = 0. Since each pi is a seminorm, pi (0) =
0, and therefore, pi (y) = pi (0) = 0 < ε. Thus 0 · E(p1 , p2 , . . . , pn ; ε) ⊆ E(p1 , p2 , . . . , pn ; ε). If
α 6= 0, then x = α−1 y and so for each i, pi (α−1 y) < ε which further implies that α−1 pi (y) < ε,
that is, pi (y) < αε ≤ ε for all |α| ≤ 1. Then y ∈ E(p1 , p2 , . . . , pn ; ε) and so E(p1 , p2 , . . . , pn ; ε)
is a balanced set.
E(p1 , p2 , . . . , pn ; ε) is absorbent: If x ∈ X but x ∈
/ E(p1 , p2 , . . . , pn ; ε). Setting µ :=
n
max pi (x) and taking α ∈ R such that 0 < |α| < ε. Then µ ≥ ε > 0 and
i=1
pi αµ−1 x = |α|µ−1pi (x)
(because pi (x) is a seminorm)
|α|
|α|pi (x)
=
pi (x) = n
µ
max pi (x)
i=1
≤ |α| < ε
because
pi (x)
n
max p(x)
i=1
!
≤ 1 and |α| < ε .
Therefore, αµ−1x ∈ E(p1 , p2 , . . . , pn ; ε) and hence any set of the form (3.1) is absorbent.
Let
E1 := E(p′1 , p′2 , . . . , p′n ; ε′) and E2 := E(p′′1 , p′′2 , . . . , p′′n ; ε′′ )
(3.2)
and ε = min{ε′ , ε′′}. Then for all x ∈ E1 , we have p′i (x) < ε′ for i = 1, 2, . . . , n and for all
x ∈ E2 , we have p′′i (x) < ε′′ for i = 1, 2, . . . , n.
Let E3 := E(p′1 , p′2 , . . . , p′n , p′′1 , p′′2 , . . . , p′′n ; ε). Then we want to show that E3 ⊆ E1 ∩ E2 . For
that, let z ∈ E3 . Then p′i (z) < ε ≤ ε′ for all i = 1, 2, . . . , n, and p′′i (z) < ε ≤ ε′′ for all
i = 1, 2, . . . , n. Therefore, z ∈ E1 and also z ∈ E2 and so z ∈ E1 ∩ E2 . Thus E3 ⊆ E1 ∩ E2 .
Finally, we prove that there exists E ∗ such that E ∗ + E ∗ ⊆ E. For that, let E ∗ =
E(p1 , p2 , . . . , pn ; 2ε ) and x, y ∈ E ∗ . Then pi (x) < 2ε and pi (y) < 2ε for all x, y ∈ E ∗ ,
Qamrul Hasan Ansari
Topological Vector Spaces
i = 1, 2, . . . , n. Since for each i, pi (x + y) ≤ pi (x) + pi (y) <
x + y ∈ E = E(p1 , p2 , . . . , pn ; ε), that is, E ∗ + E ∗ ⊆ E.
Page 73
ε
2
+
ε
2
= ε, we have
Thus all the conditions of Proposition 2.1.13 are satisfied and hence there is a vector topology T on X for which the system of all sets of the form (3.1) forms a basis of (convex)
neighbourhoods of the origin.
Note 3.1.2. The locally convex topology for which the family of all sets of the form
E(p1 , p2 , . . . , pn ; ε) = {x : pi (x) < ε; i = 1, 2, . . . , n} (with pj ∈ ℘ and 0 < ε ∈ R) is a
neighbourhood basis of the origin is called the topology defined by ℘.
Remark 3.1.2. In any locally convex topological vector space, there is a neighbourhood
base at 0 consisting of convex, balanced (and absorbing) sets.
Proposition 3.1.5. The topology defined by a set ℘ of seminorms is a Hausdorff topology
if and only if ℘ is a sufficient set of seminorms.
Proof. Let X be a Hausdorff space with respect to the topology defined by ℘. Then for any
x 6= 0, there exists a neighbourhood E(p1 , p2 , . . . , pn ; ε) of 0 such that x ∈
/ E(p1 , p2 , . . . , pn ; ε)
(by the property of Hausdorff in topological vector space X). Therefore, pi (x) ≥ ε for some
i, that is, pi (x) 6= 0 for some i. Hence, ℘ is a sufficient set of seminorms.
Conversely, let ℘ be a sufficient set of seminorms. Then there exists p ∈ ℘ such that p(x) 6= 0.
Since E(p; ε) is a neighbourhood of 0, so is p(x)E(p; ε). We claim that x ∈
/ p(x)E(p;
ε).
Suppose contrary that x ∈ p(x)E(p; ε). Then
x
p(x)
∈ E(p; ε). Since p(x) 6= 0, p
x
p(x)
<ε
p(x)
(by the definition of E(p; ε) = {x ∈ X : p(x) < ε}). Therefore p(x)
< ε since p is seminorm,
and thus 1 < ε which is impossible as ε was arbitrary. So x ∈
/ p(x)E(p; ε), that is, p(x)E(p; ε)
is a neighbourhood of 0 such that x ∈
/ p(x)E(p; ε). This implies that X is a Hausdorff
space.
Let p1 and p2 be two seminorms on a vector space X. Then we set
p1 ≤ p2
⇔
p1 (x) ≤ p2 (x) for all x ∈ X.
(3.3)
Let p be a seminorm on a vector space X and α ∈ R. Then we denote by αp the seminorm
given by (αp)(x) = αp(x) for all x ∈ X.
Remark 3.1.3. (a) If ℘ is a directed set of seminorms with respect to order relation (3.3)
on a vector space X, then a neighbourhood basis of the origin in the topology defined
by ℘, is given by the system of the sets E(p; ε), with p ∈ ℘ and ε > 0.
(b) Let ℘ be an arbitrary family of seminorms on a vector space X, and we denote by ℘′
the family of seminorms
n
p′ (x) = max pi (x)
i=1
′
with pi ∈ ℘. Then ℘ is a directed set and it defines the same topology as ℘.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 74
Problem 3.1.1. Let X be a Hausdorff locally convex space with local base B. prove that
the family of seminorms {pU : U ∈ B} is total, that is, pU (x) = 0 for all U ∈ B implies
x = 0.
Solution. If x 6= |bz such that pU (x) = 0 for all U ∈ B, then
x∈
and so,
T
\
{x ∈ X : pU (x) < 1} ⊆
U ∈B
U ∈B
\
U,
U ∈B
U 6= {0}.
Problem 3.1.2. Let ℘ = {pi : i ∈ I} be a family of seminorms which generates a (locally
convex) topology on a locally convex topological space X. Prove that A is a bounded subset
of X if and only if pi (A) is bounded for each i ∈ I.
Solution. If A is bounded, then for each i ∈ I, there exists r ∈ (0, 1) such that λA ⊆ {x ∈
X : pi (x) ≤ 1} for |λ| ≤ r. But then pi (A) ≤ r.
If for each i ∈ I, pi (A) is bounded, then for each i, there exists r > 0 such that pi (A) ≤ r,
that is, pi (x) ≤ r for all x ∈ A which implies that, given 0 < ε < r,
λA ⊆ {x ∈ X : pi (x) ≤ ε},
ε
for all |λ| ≤ .
r
Proposition 3.1.6. Any locally convex topology is defined by a certain directed set of
seminorms.
Proof. Let X be a locally convex space with topology T . Then there exists a basis U of
convex balanced neighbourhoods of the origin.
For any U ∈ U, let pU be the Minkowski functional associated to U. Then pU is a seminorm.
Let T ′ be the topology defined by the system of seminorms {pU }U ∈U . Let U ∈ U, then for
any 0 < ε′ < ε, we have
int U = {x : pU (x) < 1} ⊆ U ⊆ U = {x : pU (x) ≤ 1},
that is,
U ⊆ {x : pU (x) ≤ 1}.
y
∈ U and so pU ( ′ ) ≤ 1 implies
ε
pU (y) ≤ ε′ ). Since 0 < ε′ < ε, we have ε′ U ⊆ {x : pU (x) < ε} and thus ε′ U ⊆ E(pU ; ε), that
Therefore, ε′ U ⊂ {x : pU (x) ≤ ε′ } (because if y ∈ ε′ U, then
y
ε′
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is, for every E(pU ; ε) ∈ {E(pU ; ε)}U ∈U , there exist ε′ U ∈ U such that ε′ U ⊆ E(pU ; ε). Then
by the result3
T′≤T.
(3.4)
Conversely, for any U ∈ U, we know that
int U = {x : pU (x) < 1} ⊆ U
which implies that E(pU ; 1) ⊆ U. That is, for every U ∈ U, there exists E(pU ; 1) ∈
{E(pU ; 1)}U ∈U such that E(pU ; 1) ⊆ U, then by above mention result
T ≤T
′
(3.5)
′
From (3.4) and (3.5), we get T = T .
Definition 3.1.4. Let ℘′ and ℘′′ be two directed sets of seminorms on a vector space X.
We say that ℘′ is weaker than ℘′′ if for any p′ ∈ ℘′ , there exist p′′ ∈ ℘′′ and a number µ > 0
such that p′ ≤ µp′ .
If ℘′ is weaker than ℘′′ and ℘′′ is weaker than ℘′ , we say that ℘′ and ℘′′ are equivalent.
Remark 3.1.4. If p and q are two seminorms on a vector space X and for a certain number
λ > 0, p(x) < λ implies q(x) ≤ 1, then q ≤ λ−1 p.
Indeed, let x ∈ X and ε > 0 be arbitrary. Set y := λ(p(x) + ε)−1 x, then, since p is a
seminorm, we have
p(y) = p(λ(p(x) + ε)−1 x) = λ(p(x) + ε)−1 p(x),
which implies that
p(x)
p(y) = λ
<λ
p(x) + ε
p(x)
∵
<1 .
p(x) + ε
Hence by hypothesis, q(y) ≤ 1. Thus, q(λ(p(x) + ε)−1x) ≤ 1 which implies that λ(p(x) +
ε)−1 q(x) ≤ 1, and therefore
q(x) ≤ λ−1 (p(x) + ε).
Since ε was arbitrary, q(x) ≤ λ−1 p(x), that is, q ≤ λ−1 p.
In particular, if p is a seminorm on a vector space X and we set
S(0, 1) = {x : p0 (x) < 1} and S(0, 1) = {x : p0 (x) ≤ 1},
and E is an absorbent, convex and balanced subset of X such that S(0, 1) ⊆ E ⊆ S(0, 1),
then the Minkowski functional associated to E is identical with p.
3
Let B and B ′ be the bases for the topologies T and T ′ , respectively, on a topological space X. Then the
following are equivalent:
(a) T ′ is finer than T , i.e., T ′ ≥ T .
(b) For each x ∈ X and each basis element B ∈ B containing x, there is a basis element B ′ ∈ B ′ such
that x ∈ B ′ ⊆ B.
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Topological Vector Spaces
Page 76
Proposition 3.1.7. Let ℘′ and ℘′′ be two directed sets of seminorms on a vector space
X. If T ′ and T ′′ are two topologies defined by the ℘′ and ℘′′ , respectively, the condition
T ′ ≤ T ′′ is equivalent to the condition that ℘′ be weaker than ℘′′ .
Proof. Let {E(p′ ; ε)}p′∈℘′ and {E(p′′ ; ε)}p′′∈℘′′ be the neighbourhood bases of the origin for
the directed sets ℘′ and ℘′′ , respectively.
Assume that T ′ ≤ T ′′ . Then for any p′ ∈ ℘′ , there exist p′′ ∈ ℘′′ and a number λ > 0 such
that
E(p′′ ; λ) ⊆ E(p′ ; 1).
Thus p′′ (x) < λ implies p′ (x) < 1, and then by the Remark (3.1.4) we get p′ ≤ λ−1 p′′ . So by
the Definition (3.1.4), ℘′ is weaker than ℘′′ .
Conversely, assume that ℘′ is weaker than ℘′′ . Then by Definition (3.1.4), if p′ ∈ ℘′ and
p′′ ∈ ℘′′ , there exists µ > 0 such that p′ ≤ µp′′ . Then E(p′′ ; µ−1 ε) ⊂ E(p′ ; ε). Indeed,
let x ∈ E(p′′ ; µ−1 ε), then µp′′ (x) < ε and so p′ (x) ≤ µp′′ (x) < ε because p′ ≤ µp′′ . Thus
x ∈ E(p′ ; ε).
Hence for basis element E(p′ ; ε) ∈ {E(p′ ; ε)}p′∈℘′ of topology T ′ , there exists E(p′′ ; µ−1 ε) ∈
{E(p′′ ; ε), ε > 0}p′′ ∈℘′′ such that E(p′′ ; µ−1 ε) ⊆ E(p′ : ε). Then by the result on the footnote
of previous page, T ′ ≤ T ′′ .
Proposition 3.1.8. Let ℘ be a directed set of seminorms on a vector space X. A
seminorm q on X is continuous in the topology T defined by ℘ if and only if there exist
p0 ∈ ℘ and a number µ > 0 such that q ≤ µp0 .
Proof. Let ℘ be a directed set of seminorms. Then the neighbourhood basis of the origin is
given by the sets of form E(p; ε) for all ε > 0 and for all p ∈ ℘.
The continuity of q implies that q is continuous at 0 and thus, for ε = 1, there exists a
neighbourhood U of 0 such that
|q(x) − q(0)| < 1,
for all x ∈ U,
that is,
|q(x)| < 1,
for all x ∈ U.
Since U is a neighbourhood of the origin, there exists E(p0 ; η) ∈ {E(p; ε), ε > 0}p∈℘ such
that E(p0 ; η) ⊂ U (by the property of neighbourhood basis), that is, q(x) < 1 for all
x ∈ E(p0 ; η) ⊆ U. So q(x) < 1 for all p0 < η and hence q ≤ η −1 p0 (by remark (3.1.4)).
Conversely, assume that q ≤ µp0 for some µ > 0 and p0 ∈ ℘. Then for all ε > 0, E(p0 ; µ−1ε)
is such that x ∈ E(p0 ; µ−1ε). Therefore p0 (x) < µ−1 ε, that is, µp0 (x) < µµ−1ε = ε. This
implies that q(x) ≤ µp0 (x) < ε for all x ∈ E(p0 ; µ−1 ε) that is, q(x) is continuous at origin.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 77
Let x0 ∈ X and ε > 0, then for x ∈ x0 + E(p0 ; µ−1 ε) we have (x − x0 ) ∈ E(p0 ; µ−1 ε). This
implies that p0 (x − x0 ) < µ−1 ε and so |q(x) − q(x0 )| ≤ q(x − x0 ) ≤ µp0 (x − x0 ) < µµ−1ε = ε.
Hence q is continuous at x0 .
Corollary 3.1.1. Let ℘ be a set of seminorms on a vector space X and q is a continuous
seminorm in the topology T defined by ℘. Then the topology defined by ℘ ∪ {q} coincide
with T .
Corollary 3.1.2. Let X be a locally convex space. Then its topology is defined by the
set of all continuous seminorms on X.
Corollary 3.1.3. Let X be a locally convex space, ℘ be a set of directed set of seminorms
which defines the topology of X and f : X → R be a linear functional. Then f is
continuous if and only if there exist p0 ∈ ℘ and a number µ > 0 such that |f (x)| ≤ µp0 (x)
for all x ∈ X.
Corollary 3.1.4. Two directed sets of seminorms on the same vector space X define
the same topology if and only if they are equivalent.
Proposition 3.1.9. On any Hausdorff locally convex space X, there exists a sufficient
set of continuous linear functionals.
Proof. By Proposition 3.1.5, for any x0 6= 0, there exists a continuous seminorm p0 such
that p0 (x0 ) 6= 0. By Corollary 1.6.4 of Hahn-Banach Theorem for vector spaces, there exists
a linear functional f : X → R such that with f (x0 ) = p0 (x0 ) and |f (x)| ≤ p0 (x). Since
p0 (x0 ) 6= 0, f (x0 ) 6= 0, that is, sufficient set of continuous linear functional exists.
Proposition 3.1.10. If X is Hausdorff locally convex space, then its completion X̃ is
locally convex too.
Qamrul Hasan Ansari
3.2
Topological Vector Spaces
Page 78
Subspaces and Product Spaces
Any vector subspace G of a locally convex space X is a locally convex space with respect to
induced topology.
If ℘ is a set of seminorms defining the topology on X and ℘0 denotes the restrictions to G
of the seminorms in ℘, then ℘0 defines the induced topology on G.
Proposition 3.2.1. If Y is a topological vector subspace of a locally convex space X,
then for any convex and balanced neighbourhood U0 of the origin in Y , there exists a
convex and balanced neighbourhood U of the origin in X such that U0 = U ∩ Y .
Proof. Since U0 is a neighbourhood of origin in Y , there exists a neighbourhood W0 of origin
in X such that U0 = W0 ∩ Y . Since balanced convex neighbourhood of origin in X forms
a neighbourhood basis, there exists balanced convex neighbourhood U1 of origin such that
U1 ⊆ U0 . This implies that U1 ∩ Y ⊆ U0 .
It can be easily seen that every balanced set in Y is also balanced in X, and balanced union
of two balanced set is balanced. Therefore, U0 is a balanced subset of X, and so U0 ∪ U1 is
a balanced subset of X.4
Setting U := co(U0 ∪ U1 ). Then U is a convex balanced neighbourhood of the origin in X.
Each z ∈ U admits a representation z = λx + (1 − λ)y with x ∈ U0 and y ∈ U1 , 0 ≤ λ ≤ 1.
If z ∈ U ∩ Y and λ = 1, then obviously z = λx ∈ U0 because U0 is balanced. If z ∈ U ∩ Y
but λ 6= 1, then as λx ∈ U0 ⊆ Y and z ∈ Y , we have (1 − λ)y ∈ Y and so y ∈ Y . Therefore,
y ∈ U1 ∩ Y ⊆ U0 . Since U0 is balanced, (1 − λ)y ∈ U0 and thus z = λx + (1 − λ)y ∈ U0
because U0 is convex. Therefore, U ∩ Y ⊆ U0 . Also, since U0 ⊆ U and U0 ⊆ Y , we have
U0 ⊆ U ∩ Y . Hence U0 = U ∩ Y .
Remark 3.2.1. In the above proposition, if we take a closed vector subspace Y and consider
an arbitrary point x0 ∈ X such that x0 ∈
/ Y , then we can find a neighbourhood U of the
origin that, in addition to the above mentioned properties, does not contain x0 .
Lemma 3.2.1. Let X is a vector space, Y be a topological vector space and g : X → Y
be a linear mapping. Then the weakest topology on X for which g is continuous, is the
vector topology.
Proof. Let T be the weakest topology on X for which g is continuous. Let V be a neighbourhood of the origin in X (with respect to T ). Then there exists a neighbourhood U
of the origin in Y such that g −1(U) ⊆ V . But as addition is continuous in Y and U is a
4
x ∈ U0 ∪ U1 ⇒ ∀|α| ≤ 1, αx ∈ U0 and αx ∈ U1 .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 79
neighbourhood of the origin in Y , there exists a neighbourhood U1 of the origin in Y such
that U1 + U2 ⊆ U.
We show that the addition is continuous in X at any point (a′ , a′′ ) ∈ X × X. Since U1 is a
neighbourhood of 0 in Y , g −1(U1 ) is a neighbourhood of 0 in X, and therefore, g −1 (U1 ) +
a′ and g −1 (U1 ) + a′′ are neighbourhoods of a′ and a′′ , respectively, in X. Also for every
neighbourhood V of the origin in X, V + (a′ + a′′ ) is a neighbourhood of a′ + a′′ in X.
We claim (g −1 (U1 ) + a′ ) + (g −1(U2 ) + a′′ ) ⊆ V + (a′ + a′′ ).
Let x′ ∈ g −1(U1 ) + a′ and x′′ ∈ g −1(U2 ) + a′′ . Then x′ − a′ ∈ g −1 (U1 ) and x′′ − a′′ ∈ g −1(U2 ),
and so, g(x′ − a′ ) = g(x′ ) − g(a′ ) ∈ U1 and g(x′′ − a′′ ) = g(x′′ ) − g(a′′ ) ∈ U1 . Therefore,
g(x′ ) − g(a′ ) + g(x′′ ) − g(a′′ ) = g(x′ + x′′ − a′ − a′′ ) ∈ U1 + U2 ⊆ U, which implies that
(x′ + x′′ ) − (a′ + a′′ ) ∈ g −1 (U) ⊆ V , that is, x′ + x′′ ∈ V + (a′ + a′′ ). Hence the mapping
(x′ , x′′ ) 7→ x′ + x′′ is continuous at any (a′ , a′′ ) with respect to T .
Now we show the continuity of scalar multiplication at any point (λ0 , x0 ) ∈ R × X. Let
V be any neighbourhood of 0 in X. Then there exists a neighbourhood U of 0 in Y such
that g −1(U) ⊆ V by continuity of g. Let y0 = g(x0 ) ∈ Y . Since scalar multiplication is
continuous in Y , it is continuous at (λ0 , y0 ). As U + λ0 y0 is a neighbourhood of λ0 y0 in Y ,
there exist ε > 0 and a neighbourhood U0 of 0 in Y such that |λ − λ0 | < ε and y − y0 ∈ U0
imply λy ∈ U + λ0 y0 .
Now g −1 (U0 ) + x0 is a neighbourhood of x0 in X such that for all x ∈ g −1(U0 ) + x0 , we have
x − x0 ∈ g −1 (U0 ) ⇒ g(x) − g(x0 ) ∈ U0
⇒ g(x) ∈ U0 + y0
⇒ λg(x) = g(λx) ∈ U + λ0 y0 for all |λ − λ0 | < ε
⇒ g(λx − λ0 x0 ) ∈ U
⇒ λx − λ0 x0 ∈ g −1 (U) ⊆ V
⇒ λx ∈ V + λ0 x0 for all x ∈ g −1(U0 ) + x0 and all |λ − λ0 | < ε.
Hence, scalar multiplication mapping (λ0 , y0 ) 7→ λ0 y0 is continuous at (λ0 , y0 ) with respect
to T .
Proposition 3.2.2. Let X be a vector space and {Xi }i∈I be a family of locally convex
spaces. For each i ∈ I, let fi : X → Xi be a linear mapping. Then the weakest topology
on X for which all mappings fi are continuous is the locally convex topology.a
a
This topology is called the projective topology on X, relative to the system {(Xi , fi )}i∈I .
Proof. Let T be the weakest topology on X for which all mappings fi are continuous. We
show that T is a locally convex topology, that is, there exists a neighbourhood basis of the
origin whose members are convex.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 80
For each i ∈ I, let Bi be a convex neighbourhood basis of the origin which exists because
each Xi is locally convex. Define
(n−1
)
\
B=
fij (Bij ) : n ∈ N, Bij ∈ Bij .
j=1
Then B forms a neighbourhood basis of the origin for the topology T .
Q
On the other hand, for any x ∈ X, we write g(x) := {fi (x)}i∈I and Y = i∈I Xi . Then g is
a linear mapping from X to Y and T is the weakest topology on X that makes g continuous
(with respect to the product topology on Y ). Then by above lemma, T is a vector topology.
Since each element of B is convex, B is a locally convex topology.
Corollary 3.2.1. IfQ{Xi }i∈I is a family of locally convex spaces, then the product topological vector space i∈I Xi is also locally convex.
Remark 3.2.2. Let {Xi }i∈I be a family of locally convex spaces and,
Qfor each i ∈ I, let
℘i be a family of seminoms that defines the topology of Xi . Let X = i∈I Xi and, for any
pi ∈ ℘i , let us write
p̄i (x) = pi (xi ), for all x = {xi }i∈I ∈ X.
The product topology will then be defined by the system of seminorms {p̄i : pi ∈ ℘i , i ∈ I}.
Problem 3.2.1. If X is a locally convex space and Y is a vector subspace, then prove that
the quotient topological vector space X/Y is also locally convex. If ℘ is a directed set of
seminorms defining the topology of X, then prove that the topology of X/Y will be defined
by the set of all seminorms
p̂(x̂) = inf{p(a) : a ∈ x̂} with p ∈ ℘,
where x̂ denotes the coset of x.
Qamrul Hasan Ansari
3.3
Topological Vector Spaces
Page 81
Separation Theorem in Locally Convex Spaces
Theorem 3.3.1. Let X be a locally convex space, A be a nonempty convex and compact
subset of X and B is a nonempty closed and convex subset of X such that with A∩B = ∅.
Then there exists a closed hyperplane H that separates A from B.
Proof. Let U be the convex neighbourhood basis of the origin in X. Then there exists an
open convex neighbourhood U ∈ U such that U + A and U + B are disjoint sets. Let V be
any neighbourhood of the origin. Since A is compact and B is closed such that A ∩ B = ∅,
we have A ⊆ B c (because B is closed so B c is open). Let x ∈ A, then x ∈ B c (open set).
So for each x ∈ A, there exists a neighbourhood Vx of the origin such that x + Vx ⊆ B c (by
using the definition of openness of B c ). Now by the continuity of addition, there exists a
neighbourhood Ux of 0 such that Ux + Ux ⊆ Vx , that is,
x + Ux + Ux ⊆ x + Vx ⊂ B c
(3.6)
Since {x + Ux }x∈A is an open covering of A and A is compact, there exists a finite subcover
n
T
{xi + Uxi }ni=1 of A. Set W =
Uxi . Now for each x ∈ A, there exists some i such that
x ∈ xi + Uxi and
i=1
x + W ⊆ xi + Uxi + W
⊆ xi + Uxi + Uxi
∵W =
n
\
Ux ⇒ W ⊆ Uxi
i=1
⊆ xi + Vxi
⊆ Bc,
(by (3.6))
!
that is, x + W ⊆ B c for all x ∈ A, and so, A + W ⊆ B c . Hence (A + W ) ∩ B = ∅, that
is, W is a neighbourhood of {0}. So there exists an open neighbourhood U ∈ U such that
U − U ⊆ W . So A + (U − U) ∩ B ⊆ (A + W ) ∩ B = ∅ and thus A + (U − U) ∩ B = ∅.
Therefore, (A + U) ∩ (B + U) = ∅. This is required U. As U is open, A + U and B + U
are open convex disjoint sets. Then by Theorem 2.6.2 and Remark 2.6.1 give a hyperplane
H = {x ∈ x : f (x) = λ, f is linear} which strictly separates U + A and U + B.
Suppose that f (U + A) < f (U + B). Then for all u ∈ U, f (u + A) < f (u + B), that is, U
is a neighbourhood of 0 and so |bz ∈ U. In particular, u = 0 ⇒ f (A) < f (B), that is, H
strictly separates A and B.
Corollary 3.3.1. In any locally convex space, a closed convex set K is the intersection
of all closed half-spaces that contains K.
T
Proof. Let L be the intersection of the closed half spaces that contain K, that is, L = i∈I Hi ,
where each Hi is a closed space such that
T K ⊆ Hi and {Hi }i∈I is family of hyperplanes.
Since K ⊆ Hi for all i ∈ I, we have K ⊆ i∈I Hi = L and therefore, K ⊆ L.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 82
Conversely, let x0 ∈
/ E, then {x0 } is a compact convex set and E is a closed convex set, then
above theorem (3.3.1) ∃ a closed hyperplane Hi , which strictly separates x0 i.e. x0 ∩ Hi = φ,
for some i ∈ I i.e. x0 ∈
/ Hi for some Hi ∈ {Hi }i∈I , x0 ∈
/ ∩i∈I Hi ⇒ x0 ∈
/ L i.e. L ⊆ E
(∵ x ∈
/ E ⇒ X0 ∈
/ L i.e. L ⊆ E) therefore L = E, hence E is the intersection of all closed
hyperplane containing E.
Qamrul Hasan Ansari
3.4
Topological Vector Spaces
Page 83
Convex and Compact Sets in Locally Convex Spaces
Definition 3.4.1 (Extremal point). Let K be a convex subset of a vector space X. A point
x0 ∈ K is called an extremal point of K if x0 ∈ [x1 , x2 ] ⊆ K implies either x0 = x1 or x0 = x2 ,
where [x1 , x2 ] denotes the line segment joining x1 and x2 .
The set of all extremal points of K is denoted by ext K.
Minkowski proved that a compact convex subset of Rn can always be reconstructed as the
convex hull of it’s extreme points, that is if we take any regular polygon, then it always
reconstructed with the help of vertices (that is, extreme points). The following theorem is
the infinite dimensional version of the extreme points that reconstructed the set in infinite
dimensional space.
Theorem 3.4.1 (M.G. Krein - D.P. Milman). Let X be a locally convex Hausdorff
topological vector space and A be a convex compact subset of X. Then A = co(ext A).
Pn
Proof. We set B := co(ext
Pn A). Then B ⊆ A. Indeed, let z ∈ B, then z = i=1 λi xi , where
xi ∈ ext A,Pλi ≥ 0 and i=1 λi = 1. Since each xi ∈ ext A and A is convex and compact, we
have z = ni=1 λi xi ∈ A, that is,
B ⊆ A.
(3.7)
Suppose contrary that B 6= A. Let z ∈ A but z ∈
/ B. Then by Theorem 3.3.1, there exists a
closed hyperplane that strictly separates {z} from B. Let f be continuous linear functional
on X and λ ∈ R be such that f (z) > λ and f (x) < λ for all x ∈ B. Since A is compact and
f is a continuous linear functional, we can let α = sup f (A). Then,
α > λ.
(3.8)
Let H = f −1 (α). Then H is a closed supporting hyperplane for A as
A ⊆ {x : f (x) ≤ α}
(3.9)
and H ∩ A 6= ∅ because A is compact and f is continuous so there exists y ∈ A such that
f (y) = α, i.e y ∈ A ∩ H. On the other hand, H ∩ B = ∅ because f (x) < λ < α for all x ∈ B.
To complete the proof, it will be sufficient to show that
H ∩ (ext A) 6= ∅.
(3.10)
Let G denote the set of all closed affine sets E with the following properties:5
5
A (closed) affine set E with properties (i) and (ii) is called a supporting (closed) affine for the set A. If,
in particular, E is a hyperplane, then A lies on one side of E. Hence E is a supporting hyperlane.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 84
(i) E ∩ A 6= ∅;
(ii) If [x1 , x2 ] ⊆ A and (x1 , x2 ) ∩ E 6= ∅, then [x1 , x2 ] ⊆ E;
(iii) E ⊆ H.
Clearly, the hyperplane H belongs to G .
The set G satisfies all the conditions of Zorn’s lemma with the order relation: E1 4 E2 ⇔
E1 ⊇ E2 .
Indeed, let {Ei }i∈I be a totally
ordered system of elements in G . Then this system is a chain
T
of closed sets and E0 = i∈I Ei is an affine set satisfying (i), (ii) and (iii).
Therefore, there exists a maximal element F in G . The set F contains only one element.
Suppose contrary that a, b ∈ F with a 6= b. Let g be a continuous linear functional on X
such that g(a) 6= g(b). Since F is closed and A is compact, then F ∩ A is compact. As F ∩ A
is compact and g is continuous, we can set
γ := sup g(F ∩ A).
(3.11)
Then γ ∈ g(F ∩ A) and let G := F ∩ g −1 (γ). As g is continuous, then hyperplane g −1(γ)
must be closed and F is closed, so G = F ∩ g −1(γ) is closed affine set (G is affine, as F and
g −1 (γ) are affine set and intersection of affine sets are affine.)
Now we show that G ∈ G , that is, G satisfies (i), (ii) and (iii).
Since F ∩ A is compact and g is continuous, G satisfies (i).
Let [x1 , x2 ] ⊆ A and x′ ∈ (x1 , x2 )∩G. In particular, x′ ∈ (x1 , x2 )∩(F ∩g −1 (γ)) so x′ ∈ F and
hence [x1 , x2 ] ⊆ F . But, at the same time, [x1 , x2 ] ⊆ A, so g(x) ≤ γ for any x ∈ [x1 , x2 ]. Since
g(x′ ) = γ and x′ ∈ (x1 , x2 ), we get g(x) = γ for any x ∈ [x1 x2 ]. Therefore, [x1 , x2 ] ⊆ g −1(γ)
and [x1 , x2 ] ⊆ F which imply that [x1 , x2 ] ⊆ F ∩ g −1 (γ) = G, that is G satisfies condition
(ii).
Moreover G = F ∩ g −1 (γ) and so G ⊆ F (in particular, this means that G satisfies (iii)).
Hence G satisfies all three conditions (i), (ii) and (iii). Therefore, G ∈ G and g(a) 6= g(b),
that is, for G = F ∩ g −1 (γ) both a, b can not belong to G and so G $ F , which is a
contradiction to the maximality of F . Hence F must be a single point c.
This point must be an external.
Indeed, assume that c ∈ [x1 , x2 ] ⊆ A but c 6= x1 , c 6= x2 . As F ∈ G , we have [x1 , x2 ] ⊆ F
which is impossible as F is singleton. Hence c ∈ ext A and c ∈ F ⊆ H. This implies that
c ∈ H ∩ (ext A), that is, H ∩ (ext A) 6= ∅.
Thus H ∩ (ext A) 6= ∅ holds for B 6= A. This contradicts to the fact that H strictly separates
{z} and B = co ext A. Hence B = A.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 85
Problem 3.4.1. Let X be a Hausdorff locally convex space and A be a totally bounded
subset of X. prove that co(A) is totally bounded.
Solution. Without loss of generality, we may work with a convex neighbourhood U of the
origin.
We are given that there exists a finite set C in X such that A ⊆ C + 21 U. Then by Theorem
“In a finite dimensional normed space, the convex full of a compact set is compact”, co(C)
is compact.
P
Pm
If x ∈ co(A), then x = m
i=1 λi xi where {x1 , . . . , xm } ⊆ A and λ1 , . . . , λm ≥ 0 with
i=1 λi =
1.
Now for each i, there exists ci ∈ C such that xi ∈ ci + 12 U. Then
x=
m
X
i=1
λi (xi − ci ) +
m
X
i=1
λi ci , and
m
X
i=1
1
λi (xi − ci ) ∈ U,
2
(because U is convex)
P
1
and m
i=1 λi ci ∈ co(C). So co(A) ⊆ co(C) + 2 U. But co(C) is compact, so there exists a
finite subset D of co(C) such that co(C) ⊆ D + 12 U. Therefore, co(A) ⊆ D + U, that is,
co(A) is totally bounded.
Qamrul Hasan Ansari
3.5
Topological Vector Spaces
Page 86
Bornological spaces
Definition 3.5.1. A subset E of a topological vector space X is called a bornivore set if it
absorbs any bounded subset of X.
Definition 3.5.2. A locally convex space is called a bornological space if any bornivore,
convex and balanced subset is a neighbourhood of the origin.
The topology of a bornological space is called a bornological topology.
Definition 3.5.3. A seminorm p defined on a topological vector space X is called a bounded
seminorm if for any bounded subset A of X, p(A) is a bounded subset too.
Proposition 3.5.1. A locally convex space X is a bornological space if and only if any
bounded seminorm defined on X is continuous.
Proof. Let X be a bornological space and p be a bounded seminorm defined on X. The set
E := {x : p(x) ≤ 1} is bornivore, convex and balanced, hence it is a neighborhood of the
origin. The seminorm p is therefore continuous.
Qamrul Hasan Ansari
3.6
Topological Vector Spaces
Page 87
Continuous Linear, Open and Closed Operators
Note 3.6.1. Let X and Y be topological vector spaces.
• An operator T : X → Y which is additive and continuous (or, sequentially continuous)
at a point x0 , is easily seen to be continuous (or, sequentially continuous) on whole
space.
• An additive and sequentially continuous operator T : X → Y , is homogeneous, that
is, T (αx) = αT (x) for all α ∈ R and all x ∈ X.
• The set of all continuous linear operators from X into Y , is denoted by L (X, Y ). We
also denote X ∗ = L (X, R).
Proposition 3.6.1. Let X and Y be two locally convex spaces and ℘ be a directed set of
seminorms that defines the topology of X. A linear operator T : X → Y is continuous
if and only if for any continuous seminorm q on Y , there exist a p ∈ ℘ and a number
µ > 0 such that
q(T (x)) ≤ µp(x), for all x ∈ X.
(3.12)
Proof. Let T be a continuous linear operator and q be a continuous seminorm on Y . Then
p0 (x) = q(T (x)),
for all x ∈ X
defines a continuous seminorm on X. By Proposition 3.1.8, there is a p ∈ ℘ and a number
µ > 0, for which p0 (x) ≤ µp(x); hence (3.12) follows.
Conversely, if the required condition is satisfied, then T is obviously continuous at the origin
and, being linear, it is continuous in any point.
Remark 3.6.1. (a) If the set ℘ in Proposition 3.6.1 is not directed, then the corresponding
condition is the following:
for any continuous seminorm q on Y , there exist pi ∈ ℘ (i = 1, 2, . . . , n) and
a µ > 0 such that
n
q(T (x)) ≤ µ max pi (x),
i=1
for all x ∈ X.
(b) Proposition 3.6.1 still holds if, instead of arbitrary continuous seminorms on Y , one
considers only seminorm belongings to a directed set Q of seminorms that defines
topology on Y .
Definition 3.6.1. An operator T : X → Y is called bounded if for any bounded set A ⊆ X,
T (A) is also a bounded set.
Any sequentially continuous linear operator T : X → Y is bounded.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 88
Proposition 3.6.2. Let X be a bornological space and Y be a locally convex space. Then
a linear operator T : X → Y is continuous if and only it is bounded.
Proof. We have already addressed that every continuous operator is bounded. So we only
need to prove that every bounded operator is continuous.
Let U be a closed, convex and balanced neighbourhood of the origin in Y and qU be its
associated Minkowski seminorm. Define
p(x) = qU (T (x)),
for all x ∈ X.
Then p is a seminorm on X and one easily verifies that if A is a bounded subset of X, p(A)
is bounded too. Therefore the seminorm p is continuous. Since
T −1 (U) = {x : p(x) ≤ 1},
T is continuous because inverse image of the closed set U is a closed set {x : p(x) ≤ 1}.
Corollary 3.6.1. Let X be a bornological space and Y be a locally convex space. Then
a linear operator T : X → Y is continuous if and only it is sequentially continuous.
3.6.1
Open Operators and Closed Operators
Note 3.6.2. (a) A subset A of a topological space X is said to be nowhere dense in X or
rare if int A = ∅, i.e. A contains no interior point.
(b) A subset A of a topological space X is said to be of first category if it is a countable
union of nowhere dense subsets.
(c) A set A is said to be of second category or Baire second category if it is not of first
category, i.e. if A can not be written as a countable union of nowhere dense sets.
(d) A quasinormed vector space, which is complete (in the topology defined by the quasinorms), is called a Fréchet space.
Problem 3.6.1. Let U0 be a balanced neighbourhood of the origin in a topological vector
space X such that U0 + U0 ⊆ U (a neighbourhood of the origin). Then prove that {nU0 }n∈N
forms a covering of X.
Lemma 3.6.1. Let X and Y be two topological vector spaces and T : X → Y be a linear
operator such that T (X) is a set of second category in Y . Then for any neighborhood U
of the origin in X, the set T (U) is a neighborhood of the origin in space Y .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 89
Proof. Let U be a neighborhood of the origin in X and U0 be a balanced neighborhood of
the origin such that U0 + U0 ⊆ U. Since the system {nU0 }n∈N forms a covering of X, we
have
[
T (X) =
nT (U0 )
n∈N
and, since T (X) is of second category in Y , one deduce that int T (U0 ) 6= ∅. Let x0 ∈ U0 be
such that
y0 = T (x0 ) ∈ int T (U0 ).
The set
U ′ = T (U0 ) − y0 ,
is a neighborhood of the origin in Y . Moreover,
U ′ ⊆ T (U).
(3.13)
Indeed, if x ∈ U0 , then x − x0 ∈ U, hence T (x) − y0 ∈ T (U), that is,
T (U0 ) − y0 ⊆ T (U),
which implies (3.13). The proof is complete.
Definition 3.6.2. Let X and Y be topological spaces. An operator T : X → Y is said to
be open if for any open set A ⊆ X, T (A) is an open set in Y .
Theorem 3.6.1. Let X and Y be Fréchet spaces and T : X → Y be a continuous linear
onto operator. Then T is an open operator.
Proof. Let q, q ′ be the given quasinorms on X and Y , respectively. For any ε > 0, we denote
Sε := {x ∈ X : q(x) ≤ ε} S′ε := {y ∈ Y : q ′ (y) ≤ ε}.
Let εn = 2−n ε for n ∈ N0 . By Lemma 3.6.1, there exists a sequence of numbers {ηn }n∈N
with ηn > 0 and limn→∞ ηn = 0, and such that
S′ηn ⊆ T (Sεn ),
for all n ∈ N0 .
If y ∈ S′η0 , then there is a x0 ∈ Sε0 such that
q ′ (y − T (x0 )) < η1 .
By induction, we find a sequence {xn }n∈N0 , with xn ∈ Sεn , such that
!
n−1
X
T (xi ) < ηn , for all n ∈ N.
q′ y −
By setting x =
P∞
i=0
i=0
xi , we have q(x) ≤ 2ε and y = T (x), hence y ∈ T (S2ε ). Thus
S′η0 ⊆ T (S2ε ).
Consequently, for any neighborhood U of the origin in X, T (U) is a neighborhood of the
origin in Y . This immediately implies that T is an open operator.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 90
Corollary 3.6.2. Let X and Y be Fréchet spaces and T : X → Y be a continuous linear
bijective (one-to-one and onto) operator. Then T −1 is continuous.
Definition 3.6.3. Let X and Y be two topological spaces. An operator T : X → Y is said
to be closed if its graph G(T ) = {(x, y) ∈ X × X : y = T (x)} is a closed subset of X × Y .
Theorem 3.6.2. Let X and Y be Fréchet spaces and T : X → Y be a closed linear
operator. Then T is continuous.
Proof. The graph G(T ) of T is a closed vector subspace of the product space X × Y . If q
and q ′ are the quasinorms given on X and Y , respectively, then X × Y is a Fréchet space
with respect to the quasinorm
q0 ((x, y)) = max (q(x), q ′ (y)) ,
hence G(T ) is a Fréchet space too. Now we define the operators T1 : G(T ) → X and
T2 : G(T ) → Y by
T1 (x, T (x)) = x, T2 (x, T (x)) = T (x).
Then T1 is a continuous linear operator which is a one-one mapping from G(T ) onto X.
Thus T1−1 is continuous, due to the above corollary. If R(T ) denotes the range of T , then T2
maps G(T ) onto R(T ) and is linear and continuous. Since T = T2 T1−1 , it follows that T is
continuous.
Qamrul Hasan Ansari
3.7
Topological Vector Spaces
Page 91
Space of Operators and Topologies of Uniform Convergence
Let X be a nonempty set, Y be a topological vector space and L(X, Y ) be the set of all
mappings from X into Y . Then L(X, Y ) is a vector space With the usual operations6
Let G be a vector subspace of L(X, Y ) and M be a set of subsets of X such that for any
T ∈ G and A ∈ M , T (A) is a bounded set in Y .
Let U be a basis of balanced neighborhood of the origin in Y and, for any A ∈ M and
U ∈ U , let us denote
H(A, U) = {T ∈ G : T (A) ⊆ U}.
(3.14)
The set of all finite intersections of sets of the form (3.14), represents a neighborhood basis
of the origin for a vector topology TM on G by Proposition 2.1.13.
Since U is balanced, H(A, U) is balanced, and moreover H(A, U) is absorbent since T (A) is
a bounded set for any A ∈ M .
The topology TM is called the topology of uniform convergence on M . It does not depend
on the choice of U .
If M is a directed set with respect to inclusion, then the system
H = {H(A, U) : A ∈ M , U ∈ U }
is a neighborhood basis of the origin.
When M represents the set of all finite subsets of X, then TM is called the topology of
simple convergence (or pointwise convergence) on X.
Now, let X and Y be topological vector spaces and L (X, Y ) be the set of all continuous
linear operators from X to Y .
Then L (X, Y ) is a vector space with the usual operations. If M is a set of bounded subsets
of X, one can consider on L (X, Y ) the topology TM and the vector space L (X, Y ) endowed
with TM will be denoted by LM (X, Y ).
Rest of this subsection, we assume that G = L (X, Y ) and M is equal to a set of bounded
subsets of X.
One can easily verify that if M is an arbitrary set of bounded subsets of X and we denote
[
n
M1 = {B : B ⊆ A ∈ M },
M2 =
Ai : Ai ∈ M , n ∈ N ,
i=1
6
If T1 , T2 ∈ L(X, Y ), then (T1 + T2 )(x) = T1 (x) + T2 (x) and (αT )(x) = αT (x) for all x ∈ X and all α ∈ R.
Qamrul Hasan Ansari
Topological Vector Spaces
M3 = {ec(A) : A ∈ M },
Page 92
M4 = {A : A ∈ M },
M5 = {λA : A ∈ M , 0 6= λ ∈ R},
then TMi = TM for all i = 1, 2, 3, 4, 5.
If M is the set of all finite subsets of X, then the topology TM of simple convergence is
denoted by Ts .
The vector space L (X, Y ) endowed with the topology Ts is denoted by Ls (X, Y ).
If M is the set of all bonded subsets of X, then TM is called the topology of bounded
convergence and it is denoted by Tm .
The vector space L (X, Y ) endowed with the topology Tm is denoted by Lm (X, Y ).
If M is an arbitrary set of bounded subsets of X, which covers X, then
Ts ≤ TM ≤ Tm .
Among topologies of uniform convergence, we should also mention: the topology of totally
bounded convergence (that is, the topology TM corresponding to the set M of all totally
bounded subsets of X) and the topology of compact convergence (when M consists of all
compact subsets of X).
If Y is a locally convex space, then so is LM (X, Y ).
If ℘ is a set of seminorms that defines the topology of Y and for any p ∈ ℘ and A ∈ M , we
set
pA (T ) = sup{p(T (x)) : x ∈ A},
then pA is a seminorm and the system
℘M = {pA : p ∈ ℘, A ∈ M }
defines the topology TM .
Indeed, p( A) is obviously a seminorm. If p ∈ ℘, 0 < ε ∈ R and
U = {y ∈ Y : p(y) ≤ ε},
we note that (by taking G = L (X, Y ) in (3.14))
H(A, U) = {T : pA (T ) ≤ ε}.
Let us also remark that, if M ′ is a subset of M with the property that for any A ∈ M , there
exist B ∈ M ′ and λ ∈ R such that A ⊆ λB, then TM is defined by the system of seminorms
℘M ′ .
This is due to the fact that
pλB (T ) = |λ|pB (T ).
Qamrul Hasan Ansari
Topological Vector Spaces
Page 93
Definition 3.7.1. A set M 6= ∅ of bounded subsets of a topological vector space X is called
saturated if it has the following properties:
(i) A ∈ M , 0 6= λ ∈ R ⇒ λA ∈ M ;
(ii) A ∈ M , B ⊆ A ⇒ B ∈ M ;
(iii) Ai ∈ M for i = 1, 2, . . . , n ⇒ eco
n
S
Ai
i=1
∈ M.
f of all saturated sets (of bounded subsets
If M is not saturated, then the intersection M
of X) that contain M , is the smallest saturated set which contains M and it is called the
saturated hull of M .
If X and Y are locally convex spaces and M is a set of bounded subsets of X, then TM = TMf.
Indeed, one may assume that M satisfies the following conditions: if A ∈ M and B ⊆ A,
then B ⊆ M ;
n
S
if Ai ∈ M (i = 1, 2, . . . , n), then Ai ∈ M ;
i=1
if A ∈ M and 0 6= λ ∈ R, then λA ∈ M .
Further, if A ∈ M , then co(A) is bounded as X is locally convex. If U is a convex neighborhood of the origin in Y , then T (A) ⊆ U implies T (co(A)) ⊆ U. Hence, the topology
TM0 associated to the system M0 = {co(A) : A ∈ M }, coincides with TM . According to the
previous remarks (namely, TM3 = TM4 = TM ), we obviously have now TM = TMf.
Remark 3.7.1. If X and Y are normed vector spaces, then the topology of bounded convergence on L (X, Y ) is normable. It is defined by the norm:
kT k = sup{kT (x)k : kxk ≤ 1}.
(3.15)
Whenever no other specification is present, kT k always stands for the norm (3.15), and the
normed vector space L (X, Y ) always stands for the vector space L (X, Y ) endowed with
the norm (3.15).
Qamrul Hasan Ansari
3.8
Topological Vector Spaces
Page 94
Dual Vector Spaces
Definition 3.8.1. Let X and Y be vector spaces and f : X ×Y → R be a bilinear functional
such that
(i) for all x ∈ X, x 6= 0, there exists y ∈ Y such that f (x, y) 6= 0;
(ii) for all y ∈ Y , y 6= 0, there exists x ∈ X such that f (x, y) 6= 0.
Then X and Y are said to be in duality with respect to f and the pair {X, Y } is said to
form a dual system of vector spaces.
Let {X, Y } be a dual system of vectors spaces with respect to a bilinear functional f . We
denote hx, yi = f (x, y).
We also denote by X ′ the set of all linear functionals on a vector space X.
For any element y ∈ Y , the functional ϕy given by
ϕy (x) = hx, yi,
for all x ∈ X,
is linear and the mapping y 7→ ϕy is an isomorphism between the vector space Y and a vector
subspace of X ′ . We identify Y with its image in X ′ , hence we consider Y ⊆ X ′ . Similarly,
we can write X ⊆ Y ′ .
Definition 3.8.2. The weakest topology on X, for which all functionals ϕy are continuous
is called the weak topology on X and it is denoted by σ(X, Y ).
Similarly, one may define the weak topology σ(Y, X) on Y .
σ(X, Y ) is a locally convex topology as it is defined by the system {py }y∈Y of all seminorms
py (x) = |hx, yi|.
A weakly continuous (or σ-continuous) functional on X is, by definition, a continuous functional in the weak topology. We also use the notions of weakly closed (or σ-closed) set, weakly
compact set, etc.
Proposition 3.8.1. Any weakly continuous linear functional f : X → R has a unique
representation of the form
f (x) = hx, yi,
for all x ∈ X.
(3.16)
Qamrul Hasan Ansari
Topological Vector Spaces
Page 95
Proof. For each i = 1, 2, . . . , n, there exists yi ∈ Y such that
n
|f (x)| ≤ max |hx, yi i|.
i=1
We denote by
fi (x) = hx, yi i,
for all i = 1, 2, . . . , n,
then we have f (x) = 0 whenever fi (x) = 0 for all i = 1, 2, . . . , n. Hence f is linear combination of fi (i = 1, 2, . . . , n).7 Now equality (3.16) results immediately. Uniqueness is
obvious.
Definition 3.8.3. The set of all continuous linear functionals on a topological vector space
(X, T ) is called the topological dual (or dual) of the space X and it is denoted by X ∗ or XT∗ .
The space X ∗ is a vector space with respect to the usual operations.
If X is a Hausdorff locally convex space with topology T , then
hx, f i = f (x),
for all x ∈ X and all f ∈ X ∗ ,
(3.17)
defines a bilinear functional on X × X ∗ which turns X and X ∗ into dual spaces.
The topology σ(X, XT∗ ) on X, is also called the weakened topology on X. One also calls T the
initial topology of X. Obviously σ(XT∗ , X) is the simple convergence topology on L (X, R).
7
If x ∈ X ′ and {f1 , . . . , fn } is a linearly independent system in X ′ such that fi (x) = 0 (i = 1, . . . , n)
implies f (x) = 0, then f is a linear combination of fi .
Qamrul Hasan Ansari
3.9
Topological Vector Spaces
Page 96
Strong Topology and Reflexive Spaces
Definition 3.9.1. Let {X, Y } be a dual system of vector spaces. The uniform convergence
topology on the set of all weakly bounded subsets of Y , is called the strong topology on X.
We denote the strong topology on X by B(X, Y ).
Let (X, T ) be a Hausdorff locally convex space.
Let us consider the topology B = B(XT∗ , X) on the dual XT∗ of X. The vector space (XT∗ )∗B
is called the bidual of the space X. For simplicity, the bidual is denoted by X ∗∗ .
If, for any x ∈ X, we set
Fx (f ) = f (x),
for all f ∈ X ∗ ,
then Fx ∈ X ∗∗ . The set Φ(X) = {Fx : x ∈ X} is a vector subspace of X ∗∗ and the mapping
x 7→ Fx is an isomorphism between the vector spaces X and Φ(X).
Definition 3.9.2. If Φ(X) = X, then X is called semireflexive. If X is semireflexive and
T = B(X, XT∗ ), then X is called reflexive.
Proposition 3.9.1. The space X is semireflexive if and only if any weakly bounded
subset of X is relatively weakly compact.
Proposition 3.9.2. With respect to the strong topology, the dual of a reflexive space is
a reflexive space.
Let {X, X ′ } and {Y, Y ′ } be two pairs of dual vector spaces.
Let T : X → Y be a linear operator. Then for y ′ ∈ Y ′ , the relation
fy′ (x) = hT (x), y ′i,
for all x ∈ X,
defines a linear functional on X.
Definition 3.9.3. Let T : X → Y be a linear operator. The operator T ′ : Y ′ → X ′ given
by T ′ (y ′) = fy′ is called the adjoint of the operator T .
Remark 3.9.1. T ′ is a linear operator.
Qamrul Hasan Ansari
3.10
Topological Vector Spaces
Page 97
Polar Sets
Definition 3.10.1. Let {X, Y } be a dual system of vector spaces and A be a subset of X.
The set
A0 = {y ∈ Y : hx, yi ≤ 1 for all x ∈ A}
is called the polar set of A.
The set A00 , the polar set of A0 , is called the bipolar set of A.
Proposition 3.10.1. Polar sets have the following properties:
(a) If B ⊆ A, then A0 ⊆ B 0 ;
(b) If 0 6= λ ∈ R, then (λA)0 = λ−1 A0 ;
(c) If A absorbs a subset B, then B 0 absorbs A0 ;
!0
[
\
(d)
Ai
=
A0i ;
i∈I
i∈I
(e) If A is bounded, then so is A0 and
A0 = {y ∈ Y : |hx, yi| ≤ 1 for all x ∈ A};
(3.18)
(f) If E is a vector subspace of X, then E 0 is a vector subspace of Y and
E 0 = {y ∈ Y : hx, yi = 0 for all x ∈ A}.
Proposition 3.10.2. The polar set of a set A is convex and weakly closed.
Proof. Convexity is immediate.
Since
A0 =
\
{y ∈ Y : hx, yi ≤ 1},
x∈A
the σ-continuity of each functional ψx given by
ψx (y) = hx, yi,
implies that A0 is σ-closed.
for all y ∈ Y,
(3.19)
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Topological Vector Spaces
Page 98
Proposition 3.10.3. For a set A ⊆ X, we have
A00 = co(A ∪ {0}),
where the closure is taken in the weak topology.
Corollary 3.10.1. If E is a vector subspace of X, then E 00 = E.
Corollary 3.10.2. For any A ⊆ X, A000 = A0 .
Problem 3.10.1. Let {X, Y } be a dual system of vector spaces and A be a subset of X.
Prove that
(a) A◦ is a balanced and convex subset of Y ;
(b) A◦◦ contains the convex balanced hull of A;
(c) The set A, the convex hull of A, and the convex balanced hull of A have the same
polar A◦ ;
(d) If A is finite, then A◦ is absorbing.
4
Fixed Point Theory in Topological Vector
Spaces
4.1
Knaster-Kuratowski-Mazurkiewicz Principle and FanKKM Theorem
For each n ∈ N, let N = {0, 1, 2, . . . , n} be the finite set and F (N) be the set of all subsets
of N.
A finite set {v0 , v1 , . . . , vn } in a vector space X is said
Pn if v1 −v0 , v2 −
Pn to be affinely independent
v0 , . . . , vn −v0 are linearly independent, that is, if i=1 λi (vi −v0 ) = 0 and i=1 λi = 0 imply
that λ1 = λ2 = · · · = λn = 0.
In other words, any zero linear combination of these vectors with coefficients that sum to
zero must be trivial.
Definition 4.1.1 (n-Simplex). The convex hull co({v0 , v1 , . . . , vn }) of affinely independent
n + 1 points v0 , v1 , . . . , vn in a vector space X is called an n-simplex . It is denoted by
hv0 v1 . . . vn i. The points v0 , v1 , . . . , vn are called the vertices P
of the simplex. In other words,
n
x belongs to the n-simplex
hv
v
.
.
.
v
i
if
and
only
if
x
=
0 1
n
i=0 λi vi , where λi ≥ 0 for all
Pn
i = 0, 1, . . . , n, and i=0 λi = 1, that is,
)
(
n
n
X
X
hv0 v1 . . . vn i = x =
λi vi : λi ≥ 0 for all i = 0, 1, . . . , n, and
λi = 1 .
i=0
Example 4.1.1.
i=0
• A 0-simplex is a single point.
• A 1-simplex is a line segment.
99
Qamrul Hasan Ansari
Topological Vector Spaces
Page 100
• A 2-simplex is a triangle.
• A 3-simplex is a tetrahedron.
The numbers indicate the dimension of each object. For example, the tetrahedron is a 3dimensional simplex, or 3-simplex. Each N-simplex is generated by taking the convex hull
of the preceding simplex plus one additional point, positioned in N-space so that it is affineindependent with the preceding simplex. The sequence continues into higher dimensions, so
N-simplexes can be defined for arbitrary n ∈ N.
0D
P oint
3D
T etrahedron
1D
Line segment
2D
T riangle
4D
P entachoron
Definition 4.1.2. For 0 ≤ k ≤ n, 0 ≤ i0 < i1 < · · · < ik , the k-simplex hvi0 vi1 . . . vik i is a
subset of the n-simplex hv0 v1 . . . vn i and it called a k-dimensional face (or simply, k-face) of
the simplex hv0 v1 . . . vn i.
Note that an n-simplex is also the union of all faces of an n-simplex with the same vertices.
For the convenience, we define the standard simplex generated by the standard basis of Rn+1 .
Definition 4.1.3 (Standard Simplex). Let {e0 , e1 , . . . , en } be the standard basis for Rn+1 ,
that is, e0 = (1, 0, . . . , 0), e1 = (0, 1, . . . , 0), . . . , en = (0, 0, . . . , 1) in Rn+1 . The convex hull
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Topological Vector Spaces
Page 101
co({e0 , e1 , . . . , en )} of standard basis {e0 , e1 , . . . , en } of Rn+1 is called the standard n-simplex
and it is denoted by ∆N .
The points {e0 , e1 , . . . , en } are called the vertices of the standard n-simplex ∆N , and he0 e1 i,
. . ., he0 en i, . . ., hen en i are called the edges of the standard n-simplex ∆N .
For 0 ≤ k ≤ n, ∆k is called the face in the standard n-simplex ∆N . In particular, for
J ∈ F (N), ∆J = co({ej : j ∈ J}).
Example 4.1.2. The equilateral triangle in R3 with vertices at (1, 0, 0), (0, 1, 0), (0, 0, 1),
is the standard 2-simplex.
z
y
(0, 0, 1)
(0, 1, 0)
(1, 0, 0)
x
The bold edge in the following standard 2-simplex is itself the standard 1-simplex within the
xy-plane.
z
y
y
(0, 0, 1)
(0, 1)
(0, 1, 0)
(1, 0)
(1, 0, 0)
x
x
Definition 4.1.4 (Triangulation). Let T = {τ1 , τ2 , . . . , τm } be finite collection of n-simplexes
in hv0 v1 . . . vn i such that
Qamrul Hasan Ansari
(i)
Sm
i=1 τm
Topological Vector Spaces
Page 102
= hv0 v1 . . . vn i;
(ii) τi ∩ τj is either empty or a common face of any dimension of τi and τj , where i, j ∈
{1, 2, . . . , m}.
Then T is called a triangulation of hv0 v1 , . . . , vn i.
Definition 4.1.5. Let T be a triangulation.
A vertex of a simplex in T is called a vertex of T .
An (n − 1)-face of a simplex from T is called a boundary (n − 1)-simplex of T if it is face of
exactly one n-simplex of T .
v0
v0
v3
v3
v1
v4
v2
v1
v4
v2
Figure 4.1: The 2-simplex to the left is not a triangulation (also known as simplicially
subdivided) because the subsimplexes hv0 v1 v4 i and hv0 v2 v3 i do not have a common face.
The 2-simplex to the right is a triangulation or simplicially subdivided.
Lemma 4.1.1 (Sperner’s Lemma). Let T be a triangulation of an n-simplex ∆ =
hv0 v1 . . . vn i. To each vertex v of T , let ψ(v) ∈ {0, 1, . . . , n} be such that whenever
v ∈ hvi0 vi1 . . . vik i, ψ(v) ∈ {i0 , i1 , . . . , ik } for 0 ≤ i0 < i1 < · · · < ik ≤ n. Then the
number of n-simplexes in T whose vertices receive n + 1 different values is odd.
Theorem 4.1.1 (KKM Principle, 1929). Let C0 , . . . , Cn be closed subsets of the standard
n-dimensional simplex
S ∆N and let
Tn{e0 , . . . , en } be the set of its vertices. If for each
J ∈ F (N), ∆J ⊆ j∈J Cj . Then i=0 Ci 6= ∅.
Definition 4.1.6 (KKM-Map). Let X be a topological vector space. A set-valued map
F : X ⇒ X is said Sto be a KKM-Map if for any finite subset {x0 , x1 , . . . , xn } of X,
co({x0 , x1 , . . . , xn }) ⊆ ni=0 F (xi ).
Qamrul Hasan Ansari
Topological Vector Spaces
Page 103
e2
x0
e1
x1
e3
x2
Figure 4.2: The left diagram shows an equilateral subdivision of ∆2 with m = 4. The right
diagram shows the first barycentric subdivision of the 2-simplex hx0 x1 x2 i.
C0
e0
C1 e1
T2
i=0
Ci 6= ∅
e2
C2
Figure 4.3: The geometric structure of the KKM principle
Theorem 4.1.2 (Fan-KKM Theorem, 1972). Let K be a nonempty set in a Hausdorff
topological vector space X and F : K ⇒ K be a KKM-map such that
(i) for every x ∈ X, F (x) is closed,
(ii) for some x̃, F (x̃) is compact.
Then
T
F (x) 6= ∅.
x∈X
Proof. We first show that
Tn
i=0 F (xi )
6= ∅ for every finite subset {x0 , x1 , . . . , xn } of X.
Let {x0 , x1 , . . . , xn } ⊆ X and {e0 , e1 , . . . , en } be the standard basis of Rn+1 , that is, e0 =
(1, 0, 0, . . . , 0), e1 = (0, 1, 0, . . . , 0), . . ., en = (0, 0, 0, . . . , n). Then the standard n-simplex is
Qamrul Hasan Ansari
Topological Vector Spaces
Page 104
∆N = co({e0 , e1 , . . . , en }). Define a mapping Φ : ∆N → X by
!
n
n
n
X
X
X
Φ
αi ei =
αi xi , for all n and with αi ≥ 0,
αi = 1.
i=0
i=0
i=0
Then clearly Φ is continuous due to the continuity of scalar multiplication and addition in
topological vector space X. For each i = 0, 1, 2, . . . , n, define Ci := Φ−1 F (xi ). Then each
Ci is closed as F (xi ) is closed and Φ is continuous. Since F is a KKM-map,
we have that
S
for each finite subset J = {i0 , i1 , . . . , ik } of I = {0, 1, 2, . . . , n},T∆J ⊆ j∈J Cij . Then the
family {CiT
satisfies the hypothesis of KKM-Principle,
and so, ni=0 Ci 6=
j}
Tn∅. That is, there
T
n
C
),
that
is,
Φ(y)
∈
that
Φ(y)
∈
Φ
(
exists y T
∈ ni=0 Ci which implies
i=0 Φ(Ci ). Hence,
i=0 i
Tn
n
Φ(y) ∈ i=0 F (xi ), and thus i=0 F (xi ) 6= ∅. (That is, the finite intersection of any F (xi )’s
is nonempty).
T
Now we prove
F (x) 6= ∅ by using finite intersection theorem1 .
x∈K
Let x̃ ∈ K be such that F (x̃) is compact. Then for all x ∈ K, F (x̃) ∩ F (x) 6= ∅ (as {x̃, x} is
a finite subset of K) and closed subset of F (x̃). Also, for all x0 , . . . , xn ∈ K,
!
n
n
\
\
(F (x̃) ∩ F (xi )) = F (x̃) ∩
F (xi ) 6= ∅ as {x̃, x0 , . . . , xn } (finite) ⊂ K.
i=0
i=0
So {F (x̃) ∩ F (x)}x∈X is a collection
of closed sets of the compact
set F (x̃) having finite
T
T
intersection property. Hence
(F (x̃) ∩ F (x)) 6= ∅, that is
F (x) =
6 ∅.
x∈X
x∈X
Theorem 4.1.3 (Fan Geometrical Theorem). Let K be a compact convex subset of a
Hausdorff topological vector space X. Let A be a closed subset of K × K such that the
following conditions hold.
(i) (x, x) ∈ A for every x ∈ K;
(ii) For any fixed y ∈ K, the set {x ∈ K : (x, y) ∈
/ A} is convex (or empty).
Then there exists a point y0 ∈ K such that K × {y0} ⊆ A.
Proof. For each x ∈ X, define F (x) = {y ∈ K : (x, y) ∈ A}. Then F (x) is a closed subset of
K. Indeed, let y ∈ F (x). Then there exists yn ∈ F (x) such that yn → y, that is, (x, yn ) ∈ A
such that (x, yn ) → (x, y). As A is closed, we have (x, y) ∈ A, that is, y ∈ F (x). Thus, F (x)
is a closed subset of a compact set K, and hence compact.
1
A topological space X is compact if and only if any collection of its closed sets having finite intersection
property has nonempty intersection.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 105
Pn
Pn
We
claim
that
for
α
x
∈
co({x
,
.
.
.
,
x
})
with
α
≥
0,
i
i
1
n
i
i=1
i=1 αi = 1, we have
Pn
Sn
Pn
α
x
)
∈
A
for
some
1
≤
j
≤
n.
From
(i),
F
(x
)
or
(x
,
α
x
∈
i
j
i=1 i i
i=1
i=1 i i
!
n
n
X
X
αi xi ,
αi xi ∈ A
i=1
i=1
and from (ii),
{x ∈ X :
x,
n
X
i=1
is convex or empty.
αi xi
!
∈
/ A} = U (say)
If U is empty, then wePare done. So assume that U is nonempty,
is convex. Let
P then U P
/ U
αi xi , ni=1 αi xi ) ∈
x1 , . . . , xn ∈ U. Then ni=1 αi xi ∈ U. By the definition of U, ( ni=1P
n
which contradicts
(xj , i=1 αi xi ) ∈ A. This
P to (i). Hence there exists 1 ≤ j ≤ n such that S
implies that ni=1 αi xi ∈ FT(xj ) and therefore co({x1 , . . . , xn }) ∈ ni=1 F (xi ). Then by FanKKM Theorem, we have
F (x) 6= ∅, that is, there exists y0 ∈ F (x) for all x ∈ X. This
x∈X
implies that (x, y0 ) ∈ A for all x ∈ X, and so X × {y0 } ⊆ A.
4.2
4.2.1
Browder’s Fixed Point Theorem
Partition of Unity
Let X be a topological space and φ : X → R be a function. The support of φ is defined by
supportφ = {x ∈ X : φ(x) 6= 0}.
That is, if x lies outside the support of φ, there is some neighborhood of x on which φ
vanishes.
Definition 4.2.1 (Partition of Unity). Let X be a topological space and {Oi }ni=1 be a finite
open covering of X. A collection of continuous functions {βi : X → [0, 1] : i = 1, 2, . . . , n}
is said to be a partition of unity subordinated by {Oi }ni=1 if
conti.
(i) supportβi = {x ∈ X : βi (x) 6= 0} ⊆ Oi for each i = 1, 2, . . . , n;
Pn
(ii)
i=1 βi (x) = 1 for all x ∈ X.
Theorem 4.2.1 (Existence of Finite Partition of Unity). Let {Oi }ni=1 be a finite open
covering of the normal space X. Then there exists a partition of unity subordinated by
the open covering {Oi }ni=1 .
Qamrul Hasan Ansari
Topological Vector Spaces
Page 106
Any compact Hausdorff topological space is paracompact and every paracompact space is
normal. Therefore, every compact Hausdorff topological space is normal. Hence, from above
theorem, we have the following result.
Corollary 4.2.1. Let {Oi }ni=1 be a finite open covering of the compact Hausdorff topological space X. Then there exists a partition of unity subordinated by the open covering
{Oi }ni=1 .
4.2.2
Browder’s Fixed Point Theorem
Definition 4.2.2. Let X and Y be topological vector spaces and F : X ⇒ Y be a set-valued
map. A single-valued map f : X → Y is said to be a selection of F if f (x) ∈ F (x) for all
x ∈ X.
f is called continuous selection of F if f is a selection of F and it is a continuous function
on X.
Example 4.2.1. Let X = Y = R and F : R ⇒ R be define by
(
{x}, if x ≤ 0,
F (x) =
[0, x], if x > 0.
Then F is set-valued map. Define f : R → R by f (x) = x. Then f is continuous and
for every x ∈ R, f (x) ∈ F (x), that is, f is a continuous selection of F .
Example 4.2.2. Let X = R, Y = R2 and F : R ⇒ R2 be define by
F (x) = {(a, b) ∈ R2 : k(a, b)k2 = |x|}, for all x ∈ R.
Define a function f : R → R2 by f (x) = √x2 , √x2 . Then f is continuous and for every
x ∈ R, f (x) ∈ F (x), that is, f is a continuous selection of F .
The following result provides the existence of a continuous selection.
Theorem 4.2.2. Let K be a compact subset of a Hausdorff topological vector space X
and Y be a topological vector space. Let F : K ⇒ Y be a set-valued map such that the
following conditions hold.
(i) For all x ∈ K, F (x) is nonempty and convex.
(ii) For all y ∈ Y , F −1 (y) = {x ∈ K : y ∈ F (x)} is open in K.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 107
Then F has a continuous selection, that is, there exists a continuous function f : K → Y
such that f (x) ∈ F (x) for all x ∈ K.
Proof. We first show that K ⊆
S
F −1 (y). Let x ∈ K, then F (x) ⊆ Y . Since F (x) is
y∈Y
nonempty by assumption
y ∈ Y such that y ∈ F (x) ⇒ x ∈ F −1 (y). This
S −1 (i), so there exists S
implies that x ∈
F (y), that is, K ⊆
F −1 (y). This means that {F −1 (y) : y ∈ Y }
y∈Y
y∈Y
is an open covering of K. Since K is compact, there exists a finite set {y1, y2 , . . . , yn }
n
S
such that K ⊆
F −1 (yi ), that is {F −1 (yi )}ni=1 is a finite open covering of K. We know
i=1
that every compact set gives the guarantee of the partition of unity subordinate to every
finite open covering of K. Let {β1 , . . . , βn } be the partition of unity subordinate to the
finite open covering {F −1 (y1 ), . . . , F −1 (yn )}, that is, βi : K → [0, 1] is continuous such that
n
P
supportβi ⊆ F −1 (yi ) for every 1 ≤ i ≤ n, and βi (x) = 1 for all x ∈ K.
i=1
Define a mapping p : K → Y by p(x) =
n
P
βi (x)yi . Then p is continuous (due to the
i=1
continuity of the scalar multiplication and vector addition in topological vector space). Now
n
P
x ∈ F −1 (yi ) ⇔ yi ∈ F (x) for every 1 ≤ i ≤ n. Since F (x) is convex and βi (x) = 1, p(x)
i=1
is the convex combination of {y1 , . . . , yn }, and therefore,
p(x) =
n
X
βi (x)yi ∈ F (x).
i=1
that is, p is a continuous selection of T .
Theorem 4.2.3 (Brouwer’s Fixed Point Theorem, 1910). Every continuous function
defined on a compact convex subset K of the Euclidean space Rn to itself has a fixed
point.
Remark 4.2.1. A standard n-simplex ∆N in Rn+1 is compact (as a bounded and closed
subset of Rn+1 ) and convex. Then by Brouwer’s fixed point theorem, every continuous
funtion f : ∆N → ∆N has a fixed point. Let D be the finite dimensional simplex spanned
by the finite set {y0 , y1 , . . . , yn } of a topological vector space X. Then there always exists a
homoemorphism between ∆N and D. So, Brouwer’s fixed point theorem is also true for the
simplex D in a topological vector space X.
Theorem 4.2.4 (Browder’s Fixed Point Theorem, 1968). Let K be a nonempty compact
convex subset of a Hausdorff topological vector space X and F : K ⇒ K be a set-valued
map such that the following conditions hold.
(i) For all x ∈ K, F (x) is nonempty and convex.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 108
(ii) For all y ∈ K, F −1 (y) = {x ∈ K : y ∈ F (x)} is nonempty and open in K.
Then there exists a fixed point x̄ ∈ K of F , that is, x̄ ∈ F (x̄).
Proof. Since for every x ∈ S
K, F (x) 6= ∅, therefore there exists y ∈ K such that y ∈ F (x) ⇔
−1
x ∈ F (y), that is, K ⊆
F −1 y. Thus {F −1 (y) : y ∈ K} is an open covering of K. Since
y∈K
K is compact, there exists a finite set {y1, y2 , . . . , yn } in K such that K ⊆
n
S
F −1 (yi ).
i=1
Let {β1 , . . . , βn } be the partition of unity subordinate to the finite open cover {F −1 (yi )}ni=1 .
n
P
Then, supportβi ⊆ F −1 (yi ), that is, βi (x) = 0 for all x ∈
/ F −1 (yi ) for all i; and βi (x) = 1
i=1
for all x ∈ K. Define a mapping p : K → K by
p(x) =
n
X
βi (x)yi ,
for all x ∈ K.
i=1
Then p is continuous (by using the continuity of scalar multiplication and vector addition
n
n
P
P
of topological vector space X). As yi ∈ K and
βi (x) = 1, p(x) = βi (x)yi is a convex
i=1
i=1
combination of yi ’s. Since K is convex, p(x) ∈ K for all x ∈ K. Also, for all i such that,
βi (x) 6= 0, x ∈ F −1 (yi ) ⇔ yi ∈ F (x). Thus p(x) ∈ F (x) for all x ∈ K (as F (x) is convex and
p(x) convex combination of yi ’s which lie in F (x)). Let D be the finite dimensional simplex
spanned by the finite set {y1 , . . . , yn }. Then for all x ∈ D, p(x) ∈ D. Indeed, let x ∈ D,
then x is convex combination of yi ’s. Since K is convex, x ∈ K. Then p(x) is the convex
combination of yi ’s, and hence p(x) ∈ D.
Then p : D → D is a continuous function (being restriction of continuous function p from
set K to D). We know that every finite dimensional simplex is convex and compact. Thus
by Brouwer’s fixed point theorem, there exists a fixed point x̄ ∈ D of p, that is, x̄ = p(x̄).
But as p(x) ∈ F (x) for all x ∈ K and D ⊆ K, we have that x̄ ∈ K such that x̄ ∈ F (x̄).
Remark 4.2.2. Fan-KKM Theorem 4.1.2 implies Browder’s Fixed Point Theorem 4.2.4 in
the following sense that by using Fan-KKM Theorem 4.1.2 we can prove Browder’s Fixed
Point Theorem 4.2.4.
Proof of Browder’s Fixed Point Theorem 4.2.4 by using Fan-KKM Theorem 4.1.2. Define a
set-valued map G : K ⇒ K by G(y) = K \ F −1 (y) for all y ∈ K. Then G(y) is closed in K
since K is closed and F −1 y is open. Also, for all y ∈ K, G(y) 6= ∅, otherwise if G(y) = ∅,
then K = F −1 (y), which implies that y ∈ F (y), and the conclusion holds. Thus for all
y ∈ K, G(y) is a nonempty closed subset of a compact set K, and hence compact.
Given any x ∈ K, choose y ∈ K such that y ∈ F (x). This means that x ∈ F −1 (y). Thus,
Qamrul Hasan Ansari
Topological Vector Spaces
Page 109
by using deMorgan’s law, we have
\
y∈K
G(y) =
\
[
K \ F −1 (y) =
y∈K
F −1 (y)
y∈K
!c
= ∅.
Thus G cannot be a KKM-map (as if G isTa KKM-map, then it satisfies the hypothesis of
Fan-KKM Theorem 4.1.2, which implies
G(y) 6= ∅, a contradiction). Therefore, there
y∈K
exists a finite subset {y1 , y2 , . . . , yn } of K such that
co({y1, y2 , . . . , yn }) *
n
[
G(yi ),
i=1
that is, there exists z ∈ co({y1, y2 , . . . , yn }) such that z =
z∈K\
n
[
G(yi )
i=1
Thus, z ∈ F
λi y i ∈
/
i=1
λi ≥ 0. Hence,
−1
n
P
!
=
n
\
n
S
G(yi ), where
i=1
n
P
λi = 1,
i=1
F −1 (yi ).
i=1
(yi ) for each 1 ≤ i ≤ n. Therefore, yi ∈ F (z) for each 1 ≤ i ≤ n. As F (z) is
n
P
convex, z = λi yi ∈ F (z). Hence z is a fixed point of F .
i=1
Definition 4.2.3. A point x̄ ∈ K is said to be a maximal element of a set-valued map
F : K ⇒ K if F (x̄) = ∅.
Theorem 4.2.5 (Maximal Element Theorem). Let K be a nonempty compact convex
subset of a Hausdorff topological vector space X and F : K ⇒ K be a set-valued map
such that the following conditions hold.
(i) For all x ∈ K, x ∈
/ F (x) and F (x) is convex.
(ii) For all y ∈ K, F −1 (y) = {x ∈ K : y ∈ F (x)} is nonempty and open in K.
Then there exists a maximal element x̄ ∈ K of F , i.e. F (x̄) = ∅.
Proof. Suppose contrary that the conclusion does not hold. Then for any x ∈ K, F (x) 6= ∅.
Then by Theorem 4.2.4, there exists x̄ ∈ K such that x̄ ∈ F (x̄) which contradicts the
assumption that x ∈
/ F (x) for all x ∈ K. Hence we get the required result.
Definition 4.2.4 (Lower Semi-Continuity). Let X be a Hausdorff topological space. A
real-valued function f : X → R is said to be lower semi-continuous at x0 ∈ X if for every
ǫ > 0, there exists a neighborhood U of x0 such that f (x) ≥ f (x0 ) − ǫ for all x in U.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 110
Lemma 4.2.1. Let X be a topological vector space and f : X → R be a lower semicontinuous function. Then, for all α ∈ R
(a) f −1 (α, ∞) is open.
(b) f −1 (−∞, α] is closed.
Proof. Let x0 ∈ f −1 (α, ∞), then f (x0 ) ∈ (α, ∞). Then there exists ǫ > 0 such that
(f (x0 ) − ǫ, f (x0 ) + ǫ) ⊂ (α, ∞). By lower semi-continuity of f at x0 , there exists a neighborhood U of x0 such that f (x) ≥ f (x0 ) − ǫ for all x in U. Then, U ⊆ f −1 (α, ∞), as for
each x ∈ U, we have f (x) ≥ f (x0 ) − ǫ, i.e., f (x) ∈ (f (x0 ) − ǫ, ∞) ⊂ (α, ∞).
Now, as f −1 (−∞, α] = X \ f −1 (α, ∞), we have f −1 (−∞, α] is closed.
Lemma 4.2.2. Let X be a topological vector space, K be a compact convex subset of X
and {x1 , x2 , . . . , xm } be any finite subset of X. Then co (K ∪ {x1 , x2 , . . . , xn }) is compact.
Proof. We prove it by induction on m.
Let x1 ∈ X. Since K is convex,
co (K ∪ {x1 }) = {(1 − λ)k + λx1 : k ∈ K, λ ∈ [0, 1]}.
As K and {x1 } are compact, K1 = K ∪ {x1 } is also compact. Define a mapping
f : K1 × [0, 1] −→ X
((k, x1 ), λ) 7−→ (1 − λ)k + λx1
Then, f is continuous (by the continuity of scalar multiplication and vector addition in X).
As K1 × [0, 1] is compact and f (K1 × [0, 1]) = co(K1 ), hence co(K1 ) = co(K ∪ {x1 }) is
compact. By induction, co(K ∪ {x1 , x2 , . . . , xm }) is compact.
Remark 4.2.3. If X is an infinite dimensional topological vector space and K is any compact
subset of X, then, in general, co(K) need not be compact.
Theorem 4.2.6 (Ky Fan’s Minimax Lemma, 1972). Let K be a nonempty compact
convex subset of a Hausdorff topological vector space X and f : K × K → R be a
function such that the following conditions hold.
(i) For all x ∈ K, f (x, x) ≤ 0.
(ii) For all x ∈ K, {y : f (x, y) > 0} is convex.
(iii) For all y ∈ K, the function f (., y) is lower semicontinuous on K.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 111
Then there exists x̄ ∈ K such that
f (x̄, y) ≤ 0,
for all y ∈ K.
Proof. For each x ∈ K, let F (x) = {y ∈ C : f (x, y) > 0}. Then F (x) is convex for each
x ∈ K by (ii).
For each y ∈ K,
F −1 (y) = {x ∈ K : y ∈ F (x)} = {x ∈ K : f (x, y) > 0} = f (., y)−1(0, ∞)
Therefore, by Lemma 4.2.1, we have F −1 (y) is open in K.
If F (x) 6= ∅ for all x ∈ K, then F satisfies all the conditions of Browder’s Fixed Point
Theorem 4.2.4, hence there exists x0 ∈ K such that F (x0 ) = x0 , that is, f (x0 , x0 ) > 0, a
contradiction to (i).
Thus, F (x̃) = ∅ for some x̃ ∈ K, that is, f (x̃, y) ≤ 0 for all y ∈ K.
4.3
Kakutani Fixed Point Theorem
Let X and Y be topological spaces. Recall that a set-valued map T : X ⇒ Y is upper
semicontinuous if and only if the set {x ∈ X : T (x) ∩ B 6= ∅} is closed for every closed
subset B of Y . Moreover, if Y is a compact Hausdorff space, and if T (x) is closed for
each x ∈ X, then T is upper semicontinuous if and only if T has a closed graph, that is,
GraphF := {(x, y) ∈ X × Y : y ∈ T (x)} is a closed subset of X × Y .
In other words, if Y is a compact Hausdorff space, and if T (x) is closed for each x ∈ X,
then T is upper semicontinuous if and only if for any sequences {xm } in X and {ym } in Y
such that such that xm → x and ym → y with ym ∈ T (xm ) for all m ∈ N imply y ∈ T (x).
Note 4.3.1. As we have seen earlier that a topological vector space X is Hausdorff if and
only if the diagonal set is closed.
Proposition 4.3.1 (Kakutani, 1941). If K is an m-simplex in Rn and T : K ⇒ K is
an upper semicontinuous set-valued mapping with nonempty closed convex values, then
there is a point x0 ∈ K such that x0 ∈ T (x0 ).
Proof. Since K is a m-simplex, that is, K := hv0 . . . vm−1 i with m ≤ n, where vi ’s are the
vertices. If we take yi ∈ T (vi ), then we could set
1
f (x) =
m−1
X
i=0
λi y i ,
(4.1)
Qamrul Hasan Ansari
Topological Vector Spaces
Page 112
Pm−1
where
λ
are
the
barycentric
coordinates
of
x
(that
is,
λ
≥
0,
i
i
i=0 λi = 1 and x =
Pm−1
1
λ
v
).
By
construction,
f
is
a
continuous
function
from
a
compact
set K to itself.
i i
i=0
1
1
Then by Brouwer’s fixed point theorem, there exists a fixed point x of f , that is, x1 =
f 1 (x1 ).
Consider the k-th barycentric subdivision and for each vertex vi in this subdivision pick an
element yi ∈ F (vi ). Define f k (vi ) = yi and extend f k to the interior of each subsimplex as
before. Hence f k is continuous function from K to itself, and therefore, there exists a fixed
point
m−1
m−1
X
X
k
k k
λi vi =
λki yik ,
yik = f k (vik ),
(4.2)
x =
i=0
in the subsimplex
hv0k
k
. . . vm−1
i.
i=0
Since
k
(xk , λk0 , . . . , λkm−1 , y0k , . . . , ym−1
) ∈ K × [0, 1]m−1 × K m−1 ,
0
)
we can assume that this sequence converges to some limit (x0 , λ00 , . . . , λ0m−1 , y00, . . . , ym−1
after passing to a subsequence. Since the simplices shrink to a point, this implies that
vik → x0 and hence yi0 ∈ F (x0 ) since (vik , yik ) ∈ Graph(F ) → (vi0 , yi0) ∈ Graph(F ) by the
closedness of the graph of F . On the other hand, relation (4.3) implies that
0
x =
m−1
X
λ0i yi0 ∈ F (x0 ),
(4.3)
i=0
since F (x0 ) is convex and the claim holds if K is simplex.
To prove the Kakutani fixed point theorem for nonempty compact convex subset of Rn , we
need the following lemma.
Lemma 4.3.1. Let K and D be nonempty compact convex subsets of Rn such that
D ⊆ K. If f : D → K is a continuous function and T : K ⇒ K is a upper semicontinuous set-valued function such that T ◦ f : K → K is upper semi-continuous.
Proof. Consider sequences {xm } and {ym } in K such that xm → x, ym → y with ym ∈
T (f (xm )) for all m ∈ N. We will show that y ∈ T (f (x)). By the continuity of f , we know
that the sequence f (xm ) → f (x) as m → ∞. By the upper semi-continuity of T , we know
then that y ∈ T (f (x)), so T ◦ f is upper semi-continuous, as desired.
A function f : X → Y where Y ⊆ X is retracting if f (y) = y for all y ∈ Y .
Theorem 4.3.1 (Kakutani Fixed Point Theorem, 1941). If K is a nonempty compact
convex subset of Rn and T : K ⇒ K is an upper semicontinuous set-valued mapping
with nonempty closed convex values, then there is a point x0 ∈ K such that x0 ∈ T (x0 ).
Qamrul Hasan Ansari
Topological Vector Spaces
Page 113
e that contains K as a subset (such a simplex
Proof. Consider an arbitrary closed simplex K
must exist because K is compact and convex). We can construct a continuous retracting
e → K (for example, the identity within K, and a function that will fatten
function f : K
e \ K onto the boundary of K). We know that since T ◦ f : K
e ⇒ K and 2K ⊆ 2Ke , we
out K
e to sets in 2Ke . Since f is continuous and T is upper semihave that T ◦ f maps points in K
continuous, by Lemma 4.3.1, we have that T ◦ f is upper semi-continuous. By Proposition
e We know x̄ ∈ T (f (x̄))
4.3.1, it then follows that T ◦ f must have a fixed point, x̄ ∈ K.
by definition of fixed point; however, since T (f (x̄)) ⊂ K, we know that x̄ ∈ K. Since f is
retracting it follows that f (x̄) = x̄), and thus T (f (x̄)) = T (x̄), so x̄ ∈ T (x̄), that is, x̄ is a
fixed point of T .
Remark 4.3.1. If T (x) contains precisely one point for all x, then Kakutani’s fixed point
theorem reduces to the Brouwer’s fixed point theorem.
Kakutani’s fixed point theorem was extended to the locally convex topological vector spaces by
Irving Glicksberg and Ky Fan.
Theorem 4.3.2 (Fan-Glicksberg Fixed Point Theorem, 1952). Let K be a nonempty
convex compact subset of a locally convex topological vector space X and T : K ⇒ K be
an upper semicontinuous set-valued mapping with nonempty closed and convex values.
Then T has a fixed point.
In order to prove this, we need the following lemma.
Lemma 4.3.2. If V is a closed set in a locally convex topological vector space X, then
the sets
Q = {x ∈ K : x ∈ T (x) + V }
and
P = {(x, y) : x ∈ K, y ∈ T (x) + V }
are also closed.
Proof. We show that (K ×X)\P is open. Let (x0 , y0) ∈
/ P , that is, x0 ∈ K and y0 ∈
/ T (x)+V .
Then there exists a neighbourhood U of zero such that
(y0 + U) ∩ (T (x0 ) + V + U) = ∅.
Since T is upper semicontinuous, there exists a neighbourhood V (x0 ) of x0 in K such that
x ∈ V (x0 )
⇒
T (x) ⊆ T (x0 ) + U,
and hence
x ∈ V (x0 )
⇒
(y0 + U) ∩ (T (x) + V ) = ∅.
Therefore a neighbourhood of (x0 , y0 ) in K × X does not belong to P .
Analogous arguments show that Q is also closed.
Qamrul Hasan Ansari
Topological Vector Spaces
Page 114
Proof of Theorem 4.3.2. Let B be a neighbourhood basis of 0 in X which consists closed,
balanced and convex sets. For each V ∈ B, let
SV := {x ∈ K : x ∈ T (x) + V }.
By the above lemma, each SV is closed. Below we will show that SV is nonempty for each
V ∈ B. Then it follows from the choice of B that the intersection of finitely
T many SV is
nonempty. By the finite intersection property, there exists an x with x ∈ V ∈B SV . This
means that x ∈ T (x).
We now have to show that SV =
6 ∅. Since K is compact, we can choose finitely many points
x1 , . . . , xm ∈ K such that (xk + V )m
k=1 form a covering of K. We set M = co{x1 , . . . , xm }
and define
TV (x) := (T (x) + V ) ∩ M.
Because of T (x) ⊆ K and V = −V it follows that TV (x) is nonempty, convex, and closed.
The set M lies in a finite-dimensional subspace of X which can be identified with Rn .
According to above lemma, the map TV : M ⇒ M has a closed graph. The Kakutani Fixed
Point Theorem 4.3.1 implies the existence of a point x with x ∈ TU (x), that is, SV 6= ∅.