Topological Vector Spaces I: Basic Theory

TVS I c Gabriel Nagy

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

Convention. Throughout this note K will be one of the ﬁelds R or C. All vector spaces mentioned here are over K. Deﬁnitions. Let X be a vector space. A linear topology on X is a topology T such that the maps

X × X 3 (x, y) 7−→ x + y ∈ X , (1) K × X 3 (α, x) 7−→ αx ∈ X , (2) are continuous. (For the map (1) we use the product topology T × T. For the map (2) we

use the product topology TK × T, where TK is the standard topology on K. A topological vector space is a pair (X , T) consisting of a vector space X and a Hausdorﬀ linear topology1 T on X . Example 1. The ﬁeld K, viewed as a vector space over itself, becomes a topological vector space, when equipped with the standard topology TK. Exercise 1. Let X be a vector space. Prove that the trivial topology T = {∅,X} is linear, but the discrete topology T = P(X ) is not. Remark 1. In terms of net convergence, the continuity requirements for a linear topology on X read:

(i) Whenever (xλ) and (yλ) are nets in X , such that xλ → x and yλ → y, it follows that (xλ + yλ) → (x + y).

(ii) Whenever (αλ) and (xλ) are nets in K and X , respectively, such that αλ → α (in K)) and xλ → x (in X )), it follows that (αλxλ) → (αx). Exercise 2. Show that a linear topology on X is Hausdorff, if and only if the singleton set {0} is closed. Example 2. Let I be an arbitrary non-empty set. The product space KI (defined as the space of all functions I → K) is obviously a vector space (with pointwise addition and scalar multiplication). The product topology turns KI into a topological vector space. This can be easily verified using Remark 1. Remark 2. If X is a vector space, then the following maps are continuous with respect to any linear topology on X :

(i) The translations Ty : X → X , y ∈ Y, deﬁned by Tyx = x + y.

(ii) The dilations Dα : X → X , α ∈ K, deﬁned by Dαx = αx. 1 Many textbooks assume linear topologies are already Hausdorﬀ.

1 The translations are in fact homeomorphisms. So are the non-zero dilations Dα, α 6= 0. Notations. Given a vector space X , a subset A ⊂ X , and a vector x ∈ X , we denote the translation Tx(A) simply by A + x (or x + A), that is,

A + x = x + A = {a + x : a ∈ A}.

Likewise, for an α ∈ K we denote the dilation Dα(A) simply by αA, that is, αA = {αa : a ∈ A}.

Given another subset B ⊂ X , we deﬁne [ [ A + B = {a + b : a ∈ A, b ∈ B} = (a + B) = (A + b). a∈A b∈B Warning! In general we only have the inclusion 2A ⊂ A + A. Proposition 1. Let T be a linear topology on the vector space X .

(i) For every T-neighborhood V of 0, there exists a T-neighborhood W of 0, such that W + W ⊂ V.

(ii) For every T-neighborhood V of 0, and any compact set C ⊂ K, there exists a T- neighborhood W of 0, such that γW ⊂ V, ∀ γ ∈ C.

Proof. (i) Let A : X × X → X denote the addition map (1). Since A is continuous at (0, 0) ∈ X × X , the pre-image A−1(V) is a neighborhood of (0, 0) in the product topology. −1 In particular, there exists T-neighborhoods W1, W2 of 0, such that W1 × W2 ⊂ A (V), so if −1 we take W = W1 ∩ W2, then W is still a T-neighborhood of 0, satisfying W × W ⊂ A (V), which is precisely the desired inclusion W + W ⊂ V. (ii) Let M : K × X → X denote the multiplication map (2). Since M is continuous at (0, 0) ∈ K × X , the pre-image M −1(V) is a neighborhood of (0, 0) in the product topology. In particular, there exists a neighborhood M of 0 in K and a T-neighborhood W0 of 0 in X , −1 such that M × ×W0 ⊂ M (V). Let then ρ > 0 be such that M contains the closed “disk” −1 Bρ(0) = {α ∈ K : |α ≤ ρ}, so that we still have the inclusion Bρ(0) × W0 ⊂ A (V) i.e.

α ∈ K & |α| ≤ ρ ⇒ αW0 ⊂ V. (3)

Since C ⊂ K is compact, there is some R > 0, such that |γ| ≤ R, ∀ γ ∈ C. (4)

Let us then deﬁne W = (ρ/R)W0. First of all, since W is a non-zero dilation of W0, it is a T-neighborhood of 0. Secondly, if we start with some γ ∈ C and some w ∈ W, written as w = (ρ/R)w0 with w0 ∈ W0, then

γw = (ρα/R)w0.

By (4) we know that |ρα/R| ≤ ρ, so by (3) we get γw ∈ V.

2 Deﬁnitions. Let X be a vector space. (i) A subset A ⊂ X is said to be absorbing, if for every x ∈ X , there exists some scalar ρ > 0, such that ρx ∈ A.

(i) A subset A ⊂ X is said to be balanced, if for every α ∈ K with |α| ≤ 1, one has the inclusion αA ⊂ A. Remark 3. Given T a linear topology of a vector space X , all T-neighborhoods of 0 are 1 absorbing. Indeed, if we start with some x ∈ X , the sequence xn = n x clearly converges to 0, so every T-neighborhood of 0 will contain (many) terms xn. Notation. Given a vector space X , and some non-empty subset A ⊂ X , the set [ bal A = αA α∈K |α|≤1 is clearly balanced. Furthermore, if B ⊂ X is a balanced subset that contains A, then B ⊃ bal A. In other words, bal A is the smallest balanced subset in X , that contains A, so we will call it the balanced hull of A. Remark 4. If X is equipped with a linear topology, and if A is an open set that contains 0, then bal A is also open. Indeed, since q ∈ A, we also have the equality [ bal A = αA, α∈K 0<|α|≤1 which is now a union of (non-zero dilations of) open sets. Proposition 2. Let X be a vector space and let T be a linear topology on X . A. If V is a basic system of T-neighborhoods of 0, then: (i) For every V ∈ V, there exists W ∈ V, such that W + W ⊂ V.

(ii) For every V ∈ V and every compact set C ⊂ K, there exists W ∈ V, such that γW ⊂ V, ∀ γ ∈ C.

(iii) For every x ∈ X , the collection Vx = {V + x : V ∈ V} is a basic system of T- neighborhoods for x. T (iv) The topology T is Hausdorff, if and only if V∈V V = {0}. B. There exists a basic system of T-neighborhoods of 0, consisting of T-open balanced sets. Proof. A. Statements (i) and (ii) follow immediately from Proposition 1. Statement (iii) is clear, since translations are homeomorphisms. T (iv). Denote for simplicity the intersection V∈V V by J , so clearly 0 ∈ J . Assume first T is Hausdorff. In particular, for each x ∈ X r{0}, the set X r{x} is an open neighborhood of 0, so there exists some Vx ∈ V with Vx ⊂ X r {x}. We then clearly have the inclusion

\ x \ J ⊂ V ⊂ (X r {x}) = {0}, x6=0 x6=0

3 so J = {0}. Conversely, assume J = {0}, and let us show that T is Hausdorﬀ. Start with two points x, y ∈ X with x 6= y, so that x − y 6= 0, and let us indicate how to construct two disjoint neighborhoods, one for x and one for y. Using translations, we can assume y = 0. T Since 0 6= x 6∈ V∈V V, there exists some V ∈ V, such that x 6∈ V. Using (i), there is some W ∈ V, such that W + W ⊂ V, so we still have x 6∈ W + W. This clearly forces2

x + ((−1)V) ∩ V = ∅. (5) Since V is a neighborhood of 0, so is (−1)V (non-zero dilation), thus by (iii) the left-hand side of (5) is indeed a neighborhood of x. B. Let us take the V to be the collection of all open balanced sets that contain 0. All we have to prove is the following statement: for every neightborhood V of 0, there exists some B ∈ V, such that B ⊂ V. Using (ii) there exists some open set W 3 0, such that

γW ⊂ V, ∀ γ ∈ K s.t. |γ| ≤ 1. (6) In particular, the balanced hull B = bal W is contained in V. By Remark 4, B is also open, so B ∈ V. The following definition is a generalization of a well known notion in Calculus. Definition. Assume T is a linear topology on a vector space X . A subset B ⊂ X is said to be T-bounded, if it satisfies the following condition:

(b) for every T-neighborhood V of 0, there exists ρ > 0, such that B ⊂ ρV. Proposition 3 (“Zero · Bounded” Rule). Suppose T is a linear topology on a vector space X . If the net (αλ)λ∈Λ ⊂ K converges to 0, and the net (xλ)λ∈Λ ⊂ X is T-bounded, then (αλxλ)λ∈Λ is T-convergent to 0.

Proof. Start with some T-neighborhood V of 0. We wish to construct an index λV ∈ Λ, such that αλxλ ∈ V, ∀ λ λV . (7) Using Proposition 2.B, we can assume that V is balanced (otherwise we replace it with a balanced open set V0 ⊂ V). Using the boundedness condition (b) we ﬁnd ρ > 0, such that

xλ ∈ ρV, ∀ λ ∈ Λ. (8)

Using the condition αλ → 0, we then choose λV ∈ Λ, so that 1 |α | ≤ , ∀ λ λ . (9) λ ρ V

To check (7), start with some λ λV and apply (8) to write xλ = ρv, for some v ∈ V. Now we have αλxλ = (αλρ)v ∈ (αλρ)V, with |αλρ| ≤ 1, so using the fact that V is balanced, it follows that αλxλ ∈ V.

2 We deliberately avoid the shotrcut notation −V in place of (−1)V.

4 Before we continue, the reader is urged to solve the following: Exercises 3-22. Assume (X , T) is a topological vector space. (In some of the exercises below, T need not be Hausdorﬀ. Identify them!)

3. Prove that if A ⊂ X is open, then A + S is open, for any subset S ⊂ X .

4. Prove that if V ⊂ X is a neighborhood of 0, then for any subset S ⊂ X , the closure S is contained in S + V.

5. Prove that the collection of all balanced closed neighborhoods of 0 constitutes a basic neighborhood system for 0. In other words, show that for any neighborhood V of 0, there exists a balanced closed neighborhood of W of 0, such that W ⊂ V.

6. Consider the set C = {0} – the closure of the singleton {0}. Prove that

(i) C equals the intersection of all neighborhoods of 0; (ii) C is a closed linear subspace; (iii) C is compact.

7. Prove that if A ⊂ X is closed and C ⊂ X is compact, then A + C is closed. S 8. Prove that if A ⊂ X and C ⊂ K are compact, then so is γ∈C γA. In particular, the balanced hull bal A is compact. S 9. Prove that, A ⊂ X is closed and C ⊂ K r {0} is compact, then γ∈C γA is closed. Give a counterexample that shows that the condition C 63 0 is essential

10. Prove that if A, B ⊂ X are compact, then so is A + B.

11. Prove that if A ⊂ X is closed, C ⊂ X is compact, and A ∩ C = ∅, then there exists a neighborhood V of 0, such that: (A + V) ∩ (C + V) = ∅.

12. Prove that, if in Proposition 3, the net (xλ) is only assumed to be eventually bounded

(i.e. there exists λ0, such that (xλ)λ λ0 is bounded), then again (αλxλ) converges to 0.

13. Prove that if a net (αλ)λ ⊂ K is eventually bounded (i.e. there exists λ0, such that

supλ λ0 |αλ| < ∞) and the net (xλ)λ ⊂ X converges to 0, then (αλxλ)λ converges to 0.

14. Prove that, if B ⊂ X is not bounded, there exists a net (xλ) in B, and a net (αλ) ∈ K, such that αλ → 0, but (αλxλ) does not converge to 0. 15. Prove that if B ⊂ X is bounded, then it satisﬁes the following stronger condition:

(b’) for every T-neighborhood V of 0, there exists ρ > 0, such that B ⊂ rV, ∀ r ≥ ρ. 16. Prove that all compact subsets in X are bounded.

17. Prove that every convergent sequence in X is bounded.

5 18. Give an example of a topological vector space in which there exists nets convergent to 0, that are not eventually bounded, as defined in Exercise 3. (Hint: Try KI , with I i I i infinite. For every F ∈ Pfin(I) let xF = (xF )i∈I ∈ K be defined by xF = 0, if i ∈ I, i and xF = card F , otherwise. Analyze the net (xF )F ∈Pfin(I).) 19*. Prove that, if B ⊂ X is bounded, then its closure B is also bounded. (Hint: Use Exercise 4.)

20. Prove that if A, B ⊂ X are bounded, then so is A + B. S 21. Prove that if A ⊂ X and C ⊂ K are bounded, then so is γ∈C γA. In particular, the balanced hull bal A is bounded. 22. Assume Y is a topological vector space and T : X → Y is linear. Prove that the following are equivalent:

(i) T is continuous; (ii) T is continuous at 0.

23. Suppose Y is a topological vector space and T : X → Y is linear and continuous. Show that, if A ⊂ X is bounded, then T (A) is bounded in Y.

The next two results (Theorem 1 and its Corollary) provide the main method for con- structing linear topologies. Theorem 1. Suppose X is a K-vector space and V is a ﬁlter3 in X , with the following properties:

(i) V 3 0, ∀ V ∈ V;

(ii) every V ∈ V is absorbing;

(iii) for every V ∈ V, there exists W ∈ V, such that W + W ⊂ V;

(iv) for every V ∈ V, there exist W ∈ V and r > 0, such that ρW ⊂ V, for all ρ ∈ K with |ρ| ≤ r.

Then there exists a unique linear topology T on X , so that V is a basic system of T- T neighborhoods of 0. Moreover, T is Hausdorﬀ, if and only if V∈V V = {0}. Proof. Declare a set A ⊂ X open, if it has the following property:

(∗) For every a ∈ A, there exists V ∈ V, such that V + a ⊂ A,

and let T be the collection of all open sets. In order to show that T is a topology, we need to check the following axioms:

(t0) T contains the empty set ∅ and the total set X ;

3 This means that all sets in V are non-empty, and for any V1, V2 ∈ V, there exists V ∈ V, such that V ⊂ V1 ∩ V2.

6 S (t1) for any collection (Ai)i∈I ⊂ T, the union i∈I Ai also belongs to T;

(t2) if A1 and A2 belong to T, then the intersection A1 ∩ A2 also belongs to T.

The axioms (t0) and (t1) are trivially verified. To check (t2) one uses the filter property. One useful feature of the topology T is translation invariance, i.e. (**) for any given x ∈ X , a subset A ⊂ X is open, if and only if A + x is open. (This is pretty obvious from the definition of T.) The next important step in the proof is contained in the following statement: Claim 1. For every M ∈ V, there exists an open set A, such that M ⊃ A 3 0. To prove Claim 1, we start off by defining A = x ∈ X : there exists V ∈ V, such that V + x ⊂ M . Since by condition (i) we know that V + x 3 x, ∀ x ∈ X , V ∈ V, it is clear that A ⊂ M. Since M + 0 = M ∈ V, it follows that A contains 0. To prove that A is open, start with some point a ∈ A and check condition (∗). On the one hand, by construction, there exists V0 ∈ V, such that V0 + a ⊂ M. (10)

On the other hand, by (iii) there exists V ∈ V such that V + V ⊂ V0, so by (10) we also get V + V + a ⊂ M. (11) In particular, for every x ∈ V + a we have the inclusion V + x ⊂ M, which means that x ∈ A. Of course, this means that V + a ⊂ A, and we are done.

Claim 2. For every x ∈ X , the collection V(x) = (V + x)V∈V constitutes a basic system of T-neighborhoods of x. Indeed, by translation invariance (∗∗), it suﬃces to consider only the case x = 0, when V(0) = V. On the one hand, by Claim 1, it follows that every V ∈ V is indeed a T- neighborhood of 0. On the other hand, by the deﬁnition (∗), for every open set A ⊂ X the condition A 3 0, implies the existence of some V ∈ V with 0 3 V ⊂ A, so V is indeed a basic system of T-neighborhoods of 0. We now proceed with the proof of the continuity of the addition map (1) and the multiplication map (2). To prove the continuity of addition, we start with two nets (xλ)λ∈Λ and (yλ)λ∈Λ in X , xλ → x, and yλ → y, and we prove that (xλ + yλ) → (x + y). By Claim 2, all we need to prove is that:

• for every V ∈ V, there exists λV ∈ Λ, such that:

xλ + yλ ∈ V + (x + y), ∀ λ λV . (12)

To get this we ﬁrst choose, using (iii), some W ∈ V, such that W + W ⊂ V, and then using the hypotheses xλ → x and yλ → y, we choose µ, ν ∈ Λ, such that

xλ ∈ W + x, ∀ λ µ,

yλ ∈ W + y, ∀ λ ν.

7 Now we are done by choosing λV ∈ Λ to be any element such that λV µ, ν, because the condition λ λV will force

xλ + yλ ∈ (W + x) + (W + y) = (W + W) + (x + y) ⊂ V + (x + y).

To prove the continuity of multiplication, we start with two nets (xλ)λ∈Λ ⊂ X and (αλ)λ∈Λ ⊂ K, xλ → x, and αλ → α, and we prove that (αλxλ) → (αx). Using the continuity of addition, it suﬃces to prove that

(a) αλ(xλ − x) → 0 and

(b) (αλ − α)x → 0.

Start with some V ∈ V, and let us indicate how to produce indices λV , νV ∈ Λ, such that

αλ(xλ − x) ∈ V, ∀ λ λV , (13)

(αλ − α)x ∈ V ∀ λ νV . (14)

For both conditions we start by ﬁxing r > 0 and W ∈ V as in condition (iv). For (a) we notice that, since αλ → α, there is some index λ0, such that the quantity

M = supλ λ0 |αλ| is ﬁnite. Let n ∈ N be such that nr ≥ M, and let Z ∈ V be such that nZ ⊂ Z + Z + ··· + Z ⊂ W. (15) | {z } n times

(The ﬁrst inclusion is always true, for any subset Z.) By the choice of λ0 (which implies |αλ/n| ≤ r, ∀ λ λ0), by the construction of r and n, and by (15), we now have

αλZ = (αλ/n)(nZ) ⊂ (αλ/n)W ⊂ V, ∀ λ λ0. (16)

Since xλ → x, there exists λ1 ∈ Λ, such that

xλ − x ∈ Z, ∀ λ λ1, so using (16) we see that any λV λ1, λ2 will satisfy (a) For (b) we use condition (ii) (for the ﬁrst time!) to choose some t > 0, such that tx ∈ W and we choose νV such that |αλ − α| ≤ rt, ∀ λ νV . (17) −1 To check (b) we simply notice that, if λ νV , then |t (αλ − α)| ≤ r, so by condition (iv) −1 we have t (αλ − α)W ⊂ V, ∀ λ νV . Of course, since tx ∈ W, this will imply

−1 −1 (αλ − α)x = t (αλ − α)(tx) ∈ t (αλ − α)W ⊂ V, ∀ λ νV .

The uniqueness part, as well as the characterization of the Hausdorﬀ property, are quite obvious.

Corollary. Suppose X is a K-vector space and V is a ﬁlter in X , with the following properties:

8 (i) V 3 0, ∀ V ∈ V;

(ii) every V ∈ V is absorbing;

(iii) for every V ∈ V, there exists W ∈ V, such that W + W ⊂ V;

(iv) every V ∈ V is balanced. Then there exists a unique linear topology T on X , so that V is a basic system of T- T neighborhoods of 0. Moreover, T is Hausdorﬀ, if and only if V∈V V = {0}. Proof. Simply notice that condition (iv) in the statement implies condition (iv) from Theo- rem 1, by taking W = V and r = 1. Example 3. Fix some positive real number p, and an inﬁnite set S. Consider the space X `p (S) = x = (x ) : |x |p < ∞ . K s s∈S s s∈S

(When K = C, the subscript is omitted from the notation. When S = N the set is also omitted from the notation.) It is not too hard to show, using for instance the inequalities

(a + b)p ≤ 2p(ap + bp), ∀ a, b ≥ 0, (18)

that `p (S) is a -vector space. Deﬁne now, for every r > 0, the set K K X V = x = (x ) ∈ `p (S): |x |p < r . r s s∈S K s s∈S

It is not hard to check that the collection V = {Vr}r>0 satisﬁes conditions (i)-(iv) from the above Corollary. The only non-trivial condition is (iii). Using the inequality (18), however, we immediately get the inclusions

Vr/2p+1 + Vr/2p+1 ⊂ Vr,

and (iii) is now clear. The unique Hausdorﬀ linear topology T on `p (S), given by the K Corollary, is referred to as the standard metric topology. The reason for this terminology is the fact that this topology can be in fact deﬁned by a metric Dp, given by

 P |x − y |p , if 0 < p < 1  s∈S s s Dp(x, y) =  P p1/p s∈S |xs − ys| , if 1 ≤ p < ∞ (The case 0 < p < 1 relies on the inequality: (a + b)p ≤ ap + bp, ∀ a, b ≥ 0. The case p ≥ 1 will be discussed in anothe section. It relies on the Minkowski Inequality.) Example 4. Fix some positive real number p, and a measure space (X, A, µ). (This means that A is a σ-algebra on X and µ is a measure on A.) Consider the space Z Lp (X, A, µ) = f : X → : f A-measurable, with |f|pdµ < ∞ . K K X

9 (When K = C, the subscript is omitted from the notation.) Exactly as above, one can show that that Lp (X, A, µ) is a -vector space. Deﬁne now, for every r > 0, the set K K Z V = f ∈ Lp (X, A, µ): |f|pdµ < r . r K X

Exactly as above, the collection V = {Vr}r>0 satisﬁes conditions (i)-(iv) from the above Corollary. The unique (non-Hausdorﬀ!) linear topology T on Lp (X, A, µ), given by the K Corollary, is referred to as the standard semi-metric topology. As before, the Dp, given by

 R |f − g|pdµ , if 0 < p < 1  X Dp(f, g) =  R p 1/p X |f − g| dµ , if 1 ≤ p < ∞ is a semi-metric, in the sense that

• Dp(f, g) = Dp(g, f) ≥ 0,

• Dp(f, h) ≤ Dp(f, g) + Dp(g, h), for all f, g, h ∈ Lp (X, A, µ). (The implication D (f, g) = 0 ⇒ f = g fails! The correct K p implication is Dp(f, g) = 0 ⇒ f = g, µ-a.e.) Comment. The `p- and Lp-spaces for 0 < p < 1 are quite pathological, as will be shown in another section. The case p ≥ 1 is much nicer, as will be explained later.