Assistant Professor

Department of Mathematics

University of Jammu

Jammu  Jammu and Kashmir, India 

1.1 Introduction

In this chapter, we give a brief review of some basic definitions and important theorems in topological vector spaces and then discuss s-topological vector spaces which are generalized form of topological vector spaces. Many topological vector spaces are spaces of functions, or linear operators acting on topological vector spaces, and the topology is often defined so as to capture a particular notion of convergence of sequences of functions. In this chapter, the scalar field of a topological vector space will be assumed to be either the complex numbers C or the real numbers R, unless clearly stated otherwise.

1.2 Topological Vector Space

Let us begin with the very basic definition of a topological space.

Definition 1.2.1. Let X be a non-empty set. Then a collection τ of subsets that contains X and φ, and is closed under finite intersections and arbitrary union is called a topology on X . This non-empty set X together with topology τ is called a topological space. Members of topology are called open sets. The complement of an open set is a closed set. The intersection of all closed sets that contain A ⊂ X is called closure of A and is denoted by Cl(A) or A. The union of all open sets contained in A ⊂ X is called interior of A and is denoted by Int(A) or A^o.

For a given x ∈ X, we say that N is a neighborhood of x if there exists an open set O containing x such that O ⊆ N. We denote such neighborhoods by Nx.

Definition 1.2.2. Topological Vector Space

 A topological vector space is a vector space X that is endowed with a topology τ such that the vector space operations are continuous. More precisely, this means that the vector addition mapping X × X −→ X and the scalar multiplication mapping

F × X −→ X are continuous.

Equivalently, (X_(F) , τ ) is a topological vector space if

(i) for each x, y ∈ X and for each open neighborhood W of x + y in X, ∃ open neighborhoods U of x and V of y in X such that U + V ⊆ W, and (ii) for each λ ∈ F, x ∈ X and for each open neighborhood W in X containing λ · x, ∃ open neighborhoods U of λ in F and V of x in X such that U · V ⊆ W.

In a geometric sense, topological vector spaces are a special class of path connected spaces. A topological vector space with a non-trivial topology is always Hausdorff. Indeed, topological vector spaces are a category of regular spaces but a topological vector space need not be normal. If X is a Hausdorff topological vector space, then X is an unbounded path connected space. Banach spaces, Hilbert spaces and Sobolev spaces are other well known examples of topological vector spaces.

Let X be a topological vector space. Fix a ∈ X and a non-zero scalar λ. If we define a translation operator

T_a : X → X by

T_a(x) = a + x,

and a multiplication operator,

M_λ : X → X

by M_λ(x) = λx.

Then both T_a and M_λ are homeomorphisms of X onto X. As a consequence of this fact, we have the following results:

1.2. Topological Vector

Space

Theorem 1.2.1. Let X be a topological vector space and A ⊆ X. The following are equivalent:

(1) A is open subset.

(2) x + A is open subset, for each x ∈ X.

(3) λ · A is open subset, for each non-zero scalar λ.

Theorem 1.2.2. If X is a topological vector space and A ⊆ X, then

(1) Cl(x + A) = x + Cl(A) for each x ∈ X,

(2) Int(x + A) = x + Int(A) for each x ∈ X,

(3) Cl(λ · A) = λ · Cl(A) for each non-zero scalar λ, and

(4) Int(λ · A) = λ · Int(A) for each non-zero scalar λ.

Theorem 1.2.3. Let (X, τ ) be a topological vector space and A, B be any two subsets of X. Then A + B is an open set in X when either A or B is an open set. Proof. Let B be an open set and A be any subset of X.

Let a be any element of A, then by Theorem ....., a + B is open in X. Since arbitrary union of open sets is open, so U _a∈A (a + B) is open in X.

Evidently, A + B is open in (X, τ ).

Now we discuss some sets in topological vector spaces.

Definition 1.2.3. Let (X_(F), τ ) be a topological vector space. Then A ⊆ X is said to be:

(1) Bounded if to each neighborhood U of zero, there corresponds a number s > 0 such that A ⊆ tU for each t > s.

(2) Convex if u, v ∈ A then λu + (1 − λ)v ∈ A, ∀ λ ∈ [0, 1].

(3) Balanced if μA ⊂ A, ∀ μ ∈ F such that |μ| ≤ 1.

(4) Compact if each open covering of A has a finite sub-covering.

Some fundamental properties of these sets in topological vector spaces are given below:

Theorem 1.2.4. Let (X, τ ) be a topological vector space and A, B be any two subsets of X. Then

(1) If N ⊆ X is a neighborhood of 0, then the closure A is contained in A + N.

(2) Cl(A) + Cl(B) ⊆ Cl(A + B).

(3) If A, B are bounded, so is A + B.

(4) If A, B are compact, so is A + B.

(5) If A is convex, so are Int(A) and Cl(A).

(6) If A is balanced, so is Cl(A).

(7) If A is balanced and 0 ∈ Int(A), then Int(A) is balanced.

(8) If A is bounded, so is Cl(A).

Theorem 1.2.5. Let (X, τ ) be a topological vector space and A, B be any two subsets of X such that A is compact and B is closed. Then A + B is closed.

Proof. Let x /∈ A + B. Then for each a ∈ A, x /∈ a + B, which is a closed set (since translation of a closed set is a closed set).

Since every topological vector space is regular, there are open sets U_a and V_a such that

x ∈ U_a, a + B ⊆ V_a and U_a ∩ V_a = ∅.

Now V_a − B = U_b∈B (V_a − b) is open and contains a.

Hence A ⊆ U_a∈A(V_a − B).

Since A is compact, there is a finite set {a1, a2, . . . , an} such that A ⊆ Sn i=1 (Vai − B).

Let U =Tn i=1 Ua, then U is a neighborhood of x.

We claim that U ∩ (A + B) = ∅.

If not, then y = a + b ∈ U ∩ (A + B), then y ∈ Vai for some i and y ∈ Uai, which is a contradiction.

Theorem 1.2.6. Let X be a topological vector space. Let K, C ⊂ X satisfy:

K is compact, C is closed and K ∩ C = ∅. Then there exists disjoint open sets in X that contain K and C.

Proof. Let x ∈ K. Since C

c is an open neighborhood of x, then ∃ a Vx ∈ N sym o such that

x + Vx + Vx + Vx ∈ C c, i.e, (x + Vx + Vx + Vx) ∩ C = ∅.

Since Vx is symmetric, we have (x + Vx + Vx) ∩ (C + Vx) = ∅.

For every x ∈ K corresponds such a Vx. Since K is compact, there is a finite collection of (xi, Vxi)

n i=1 such that K ⊆ Sn i=1 (xi + Vxi).

Consider V = Vx1 ∩ Vx2 ∩ · · · ∩ Vxn.

Then, for every i, (x + Vxi + Vxi) does not intersect (C + Vxi).

Then (x + Vxi + V ) does not intersect (C + V ). Taking the union over i: K + V ⊆ Sn i=1

(xi + Vxi + Vxi) does not intersect (C + V ).

The following theorem is an obvious consequence of the translational invari- ance of the topological vector spaces:

Theorem 1.2.7. Let X and Y be topological vector spaces and let T : X → Y be linear. If T is continuous at zero, then it is continuous everywhere. Proof. It is given that a linear map T : X → Y is continuous at 0. Let x ∈ X and V be a neighborhood of T(x) in Y . Then U = V − T(x) be a neighborhood of 0 in Y .

As T is continuous at 0, ∃ a neighborhood W of 0 ∈ X such that T(W) ⊆ U. But then the neighborhood W + x of x satisfies T(W + x) ⊆ U + T(x) = V , and this proves the continuity of T at x.

Theorem 1.2.8. Let X be a topological vector space over the field F and 0 6= f : X → F be a linear functional. If f is continuous, then kerf is closed and not dense in X. Proof. It is given that f : X → F is a continuous linear transformation such that f(x) 6= 0 for some x ∈ X.

Since f is continuous, then K = kerf = f−1({0}) is closed in X as {0} is a closed

set in the scalar field F. By hypothesis, f(x) 6= 0 for some x ∈ X. It follows that kerf is not dense in X.

1.3 s-Topological Vector Spaces

Definition 1.3.1. A subset A of a topological space X is said to be semi-open if A ⊆ Cl(Int(A).

The complement of semi-open set is a semi-closed set and the family of semi-open sets in topological space X is denoted by SO(X). It is clear from the definition that every open set is semi-open, but converse is not true. Example 1.3.1. Let (R, U) be a space of real numbers with the usual topology U.

Then A = [0, 1) is a semi-open set, but it is not open in (R, U).

Definition 1.3.2. A subset N of a topological space X is called semi-open neigh- borhood (or, semi-neighborhood) of a point x ∈ X if there exists a semi-open set U in X such that x ∈ U ⊆ N.

Definition 1.3.3. An s-topological vector space is a vector space X over the field F that is endowed with a topology τ such that (i) for any open neighborhood Wx+y of a vector sum x+y in X, there exist semi-open neighborhoods Ux of x and Vy of y in X such that Ux + Vy ⊆ Wx+y, and (ii) for any open neighborhood Wλx of λ·x in X, there exist semi-open neighborhood Uλ of λ in F and Vx of x in X such that Uλ · Vx ⊆ Wλx.

It follows from the definition that every topological vector space is s-topological vector space. The example below shows that the converse is not true in general.

Example 1.3.2. Let X = R be a vector space over the field F, where F = R with the standard topology and the topology τ on X is generated by the base B = {(a, b), [c, d) : a, b, c and d are real numbers with 0 < c < d}. We show that X(F), τ ) is an s- topological vector space. For this we have to verify the following two conditions:

(i) Let x, y ∈ X. Then, for open neighborhood W = [x + y, x + y + ) (resp. (x + y − , x + y + )) of x+y in X, we can opt for semi-open neighborhoods U = [x, x + δ) (resp. (x − δ, x + δ)) and V = [y, y + δ) (resp. (y − δ, y + δ)) of x and y in X such that U + V ⊆ W for each δ < 2.

(ii) Let x ∈ X and λ ∈ F. We have following cases: Case(1). If λ > 0 and x > 0, then clearly λx > 0. For some  > 0, consider an open neighborhood W = [λx, λx + ) (resp. (λx − , λx + )) of λx in X. We can choose semi-open neighborhoods U = [λ, λ + δ) (resp. (λ − δ, λ + δ)) of λ in F and V = [x, x + δ) (resp. (x − δ, x + δ)) of x in X such that U · V ⊆ W for each δ <  λ+x+1 .

Case(2). If λ < 0 and x < 0, then clearly λx > 0. So, for open neighborhood W = [λx, λx + ) (resp. (λx − , λx + )) of λx in X, we can choose semi-open neighbor-hoods U = (λ − δ, λ] (resp. (λ − δ, λ + δ)) of λ in F and V = (x − δ, x] (resp. (x − δ, x + δ)) of x in X such that U · V ⊆ W for δ ≥−λ+x−1.

Case(3). If λ = 0 and x > 0 (resp. λ > 0 and x = 0), then λx = 0. So, for open neighborhood W = (−, ) of 0 in X, we can opt for semi-open neighborhoods U = (−δ, δ) (resp. (λ − δ, λ + δ)) of λ in F and V = (x − δ, x + δ) (resp. (−δ, δ)) of x in X such that U · V ⊆ W for each δ <  x+1=resp. δ <  λ+1 .

Case(4). If λ = 0 and x < 0 (resp. λ < 0 and x = 0), then λx = 0. So, for open neighborhood W = (−, ) of 0 in X, we can select semi-open neighborhoods U = (−δ, δ) (resp. (λ − δ, λ + δ)) of λ in F and V = (x − δ, x + δ) (resp. (−δ, δ)) of x in X such that U · V ⊆ W for each δ < 1−x resp. δ < 1−λ.

Case(5). If λ = 0 and x = 0. Consider any open neighborhood W = (−, ) of 0 in X, we can find semi-open neighborhoods U = (−δ, δ) of λ in F and V = (−δ, δ) of x in X, we have U · V ⊆ W for each δ < √.

Thus (X(F), τ ) is an s-topological vector space. However(X(F), τ ) is not a topological vector space.

For, choose x = −3 +c, y = 3 and an open neighborhood W = [c, c + ) of x+y = c in τ , we cannot find open neighborhoods U and V containing x and y respectively i τ which satisfy U + V ⊆ W.

Definition 1.3.4. Suppose (X, τ1) and Y, τ2) are topological spaces. A function f : X → X is called semi-continuous if for each open set O in Y , the inverse image f−1 (O) is semi-open in X.

Equivalently, a mapping f : X → Y is semi-continuous if and only if for each x ∈ Xand each open neighborhood V of f(x), there is a semi-open neighborhood U of x with f(U) ⊆ V .

Clearly, continuity implies semi-continuity, but the converse need not be true. Example 1.3.3. Let X = (R, U) and Y = (R, T ), where U is the usual topology on R and T = {∅, {0}, R}. Let f : X → Y be defined as f(x) = (x, if x > 1 0, otherwise Then f is semi-continuous but it is not continuous.

Theorem 1.3.1. Let (X(F) , τ ) be an s-topological vector space. Suppose Tx : X → X is a left translation and Mλ : X → X is multiplication mapping, then Tx and Mλ both are semi-continuous. These spaces are of great importance in functional analysis. Proof. Let y is an arbitrary element in X and W be an open neighborhood of Tx(y) = x + y.

By definition of s-topological vector spaces, there exist semi-open neighborhoods U and V containing x and y respectively such that U + V ⊆ W. In particular, we have x + V ⊆ W which means Tx(V ) ⊆ W.

This shows that Tx is semi-continuous at y. Hence Tx is semi-continuous on X. Next, let λ ∈ F and x ∈ X. Let W be an open neighborhood of Mλ(x) = λx. By definition of s-topological vector spaces, there exist semi-open neighborhoods U of λ in F and V of x in X such that U · V ⊆ W.

In particular, we have λ · V ⊆ W.

⇒ Mλ(V ) ⊆ W.

This shows that Mλ is semi-continuous at x. Hence Mλ is semi-continuous on X.  Theorem 1.3.2. Let (X(F), τ ) be an s-topological vector spaces. If A ∈ τ then

(1) for every x ∈ X, x + A ∈ SO(X),

(2) for every non zero λ ∈ F, λ · A ∈ SO(X).

Proof. (1) Let z ∈ x + A. We have to show that z is a semi-interior point of x + A.

Now z = x + y, where y is some point in A. We know T−x : X → X is semi- continuous at z ∈ X. Thus, for the open set A containing y; y = T(−x)(z), there exists semi-open neighborhood Mz of z such that T−x(Mz) = (−x) + Mz ⊆ A. This implies Mz ⊆ x + A which shows that z is a semi-interior point of x + A. Hence, x + A ∈ SO(X). (2) Let z ∈ λ · A. We have to show that z is a semi-interior point of λ · A. Now z = λ · x, for some x in A. We have multiplication mapping Mλ−1 : X → X is semi-continuous. Thus, for the set A ∈ τ containing Mλ−1 (z) = λ −1 · z = x, there exists semi-open neighborhood Uz of z in X such that Mλ−1 (Uz) = λ−1 · Uz ⊆ A.

This implies Uz ⊆ λ · A. This shows that z is a semi-interior point of λ · A. Hence, λ · A ∈ SO(X). Theorem 1.3.3. Suppose (X(F) , τ ) is an s-topological vector space and S is a sub- space of X. If S contains a non-empty open set, then S is semi-open in (X, τ ). Proof. Suppose U 6= ∅ is an open subset in X such that U ⊆ S. For any y ∈ S, the set Ty(U) = y + U is semi-open in X and is subset of S. Therefore, the subspace S =S y∈S (y + U) is semi-open in X as the union of semi-open sets.

Theorem 1.3.4. Every open subspace S of an s-topological vector space (X(F), τ ) is also an s-topological vector space (called s-topological subspace of X). Proof. Let x, y ∈ S and W be an open neighborhood of x + y in S.

This gives W is an open neighborhood of x + y in X. Hence, there exist semi-open neighborhoods U ⊆ X of x and V ⊆ X of y such that U + V ⊆ W. Now, the sets A = U ∩ S and B = V ∩ S are semi-open neighborhood of x and y respectively in S because S is open in X. Also, A + B ⊆ U + V ⊆ W.

Again, let λ ∈ F and x ∈ S. Let W be an open neighborhood of λ · x in S. Since S is open in X, therefore W is open neighborhood of λ · x in X. Hence, there exist semi-open neighborhood U ⊆ F of λ and V ⊆ X of y such that U · V ⊆ W. Now, the set A = U ∩F is semi-open neighborhood of λ in F and the set B = V ∩S is semi-open neighborhood of y in S. Also, A · B ⊆ U · V ⊆ W, which means that S is an s-topological vector space. 

Definition 1.3.5. A collection {Ui : i ∈ ∆} of semi-open subsets in a topological space X is called semi-open cover of a subset B of X if B ⊆ Si∈Λ Ui .

Definition 1.3.6. A topological space (X, τ ) is said to be semi-compact if every semi-open cover of X has a finite subcover. A subset B of a topological space (X, τ ) is said to be semi-compact relative to X if for every collection {Ui : i ∈ Λ} of semi-open subsets of X such that B ⊆ S i∈Λ Ui , there exists a finite subset Λo of Λ such that B ⊆ S i∈Λo Ui.

Definition 1.3.7. A topological space X is said to be P-regular if for each semi- closed set F and x /∈ F, there exist disjoint open sets U and V in X such that x ∈ U and F ⊆ V .

Theorem 1.3.5. Let X(F, τ ) be a P-regular s-topological vector space. Then the algebraic sum of any semi-compact set and closed set in X is closed. Proof. Let K be a semi-compact subset and C be a closed subset of X. We have to show that K + C is closed set in X. For, let x /∈ K + C. Then for all k ∈ K, x /∈ k + C. Since C is closed in X and the translation of a closed set in s-topological vector space is semi-closed set, we find that k + C is semi-closed in X.

As X is P-regular, there exist open sets Uk and Vk in X such that x ∈ Uk, k+C ⊆ Vk and Uk ∩ Vk = ∅. Therein we find that k ∈ Vk C and hence K ⊆ S k∈K (Vk C). Since any union of semi-open sets is semi-open, Vk C = S c∈C (Vk − c) is semi-open in X.

Consequently, by semi-compactness of K, there exists a finite subset S of K such that K ⊆S s∈S (Vs C). Let U = T s∈S Us, then U is an open neighborhood of x such that U ∩ (K + C) = ∅.

If not, then there is some y ∈ U ∩(K +C) which implies y ∈ Ux∩Vx for some x ∈ S, which is a contradiction to the fact that Uk ∩ Vk = ∅ for each k ∈ K. Thus we find that x /∈ Cl(K + C) and hence K + C is closed in X.  Theorem 1.3.6. Let K be any semi-compact subset of an s-topological vector space X. Then x + K and λ · K are compact sets in X, where x is any element in X and λ be a scalar. Proof. Let {Uα : α ∈ Λ} be an open cover of x + K. Then x + K ⊆ S α∈Λ Uα implies K ⊆ S α∈Λ (−x + Uα). Since Uα is open in X, then −x + Uα is semi-open in X for each α ∈ Λ. By semi-compactness of K, there exists a finite subset Λo of Λ such that K ⊆ S α∈Λo (−x + Uα). Thereby we find that x + K S α∈Λo Uα.

Hence x + K is compact. Similarly, we can prove that λ · K is compact. 

Theorem 1.3.7. Let f : X → Y be a linear map, where X is an s-topological vector space and Y be a topological vector space. If f is continuous at 0, then f is semi-continuous everywhere.

Proof. Suppose that f is continuous at origin. Let 0 6= x ∈ X and W be an open neighborhood of f(x) in Y . Then W − f(x) is an open neighborhood of origin (say 0’) in Y because Y is a topological vector space. By continuity of f at origin, there exists an open neighborhood U of 0 in X such that f(U) ⊆ W − f(x). By linearity of f, f(x + U) ⊆ W.

Since X is an s-topological vector space, x + U is semi-open neighborhood of x in X. Consequently, f is semi-continuous at x ∈ X. Since x is an arbitrary non-zero element of X, it follows that f is semi-continuous at every non-zero element of X. Further, continuity of f at 0 implies that f is semi-continuous at 0.

Hence f is semi-continuous everywhere.  Corollary 1.3.7.1. Let X be an s-topological vector space. Assume 0 6= f : X → F be a linear functional. If f is continuous at origin, then f is semi-continuous.

Theorem 1.3.8. Let f : X → F be a linear functional, where X is an s-topological vector space. If f is continuous at origin, then ker f is semi-closed in X. Proof. Suppose that f is continuous at origin. If f = 0, we are done. Assume f 6= 0. By virtue of above corollary, f is semi-continuous.

Since F (R or C) is endowed with standard topology which is Hausdorff. Therefore, {0} is closed in F. By semi-continuity of f, we have f−1 ({0}) = {x ∈ X : f(x) = 0} = ker f.

Hence proved.