TVS I c Gabriel Nagy Topological Vector Spaces I: Basic Theory 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 .