Topological Vector Spaces III: Finite Dimensional Spaces

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1 TVS III c Gabriel Nagy Topological Vector Spaces III: Finite Dimensional Spaces Notes from the Functional Analysis Course (Fall 07 - Spring 08) Convention. Throughout this note K will be one of the fields R or C, equipped with the standard topology. All vector spaces mentioned here are over K. In this section we take a closer look at finite dimensional topological vector spaces, and we will learn that they are uninteresting from the topological point of view. The main tool in our investigation will be the study of linear continuous maps, either defined on, or taking values in finite dimensional spaces. Lemma 1. If T is a linear topology on some vector space X, then for any vector x X, the map φ x : K α αx X is continuous with respect to T. Proof. Consider the map ω x : K α (α, x) K X. When we equip K X with the product topology, ω x is clearly continuous. The continuity of φ x is then clear, since we can write it as a composition of continuous maps φ x : K ωx K X multiplication X. Proposition 1. Let T be a linear topology on some vector space X and let n be a positive integer. If we equip the product space K n with the product topology, then all linear maps T : K n X are continuous with respect to T. Proof. Let {e 1,..., e n } be the standard basis in K n, so that (α 1,..., α n ) = α 1 e α n e n, (α 1,..., α n ) K n. Fix some linear map T : K n X, and let x k = T e k, k = 1,..., n. Using the notations from Lemma 1, we clearly have T (α 1,..., α n ) = φ x1 (α 1 ) + + φ xn (α n ), (α 1,..., α n ) K n, which means that T can be written as a composition T : K K }{{} n factors φ x1 φ xn X X }{{} n factors addition X. The addition map is continuous. By Lemma 1, the map φ x1 φ xn is continuous, so T is also continuous. 1

2 Lemma 2. Suppose X is a vector space, equipped with a linear topology T. For a linear map φ : X K, the following are equivalent: (i) φ is continuous; (i ) φ is continuous at 0; (ii) the linear subspace Ker φ = {x X : φ(x) = 0} is closed in X ; Proof. The implications (i ) (i) (ii) are pretty clear 1. (ii) (i ). Without any loss of generality, we can assume that φ is not identically 0, so there exists some e X with φ(e) = 1. Denote for simplicity Ker φ by Y. Since φ ( x φ(x)e ) = φ(x) φ(x)φ(e) = 0, x X, we see that X = Y +Ke (direct sum). Suppose now we have a net (x λ ) λ Λ in X, with x λ 0, and let us prove that φ(x λ ) 0. If we write each x λ as x λ = y λ + α λ e, with y λ Y and α λ K, then all we have to show is that α λ 0. Assume the contrary, which means that there exists some ε > 0, such that the set is cofinal in Λ, i.e. Σ = {σ Λ : α λ ε} ( ) for every λ Λ, there exists σ Σ with σ λ. In particular, restricting to Σ we get two subnets (y σ ) σ Σ in Y and (α σ ) σ Σ in K, such that (a) (y σ + α σ e) 0 (in X ); (b) α σ ε, σ Σ. Consider then the net γ σ = ε/α σ. By (b) we have γ σ 1, σ Σ, so using Exercise?? from TVS I we still have γ σ (y σ + α σ e) 0. This simply reads: (γ σ y σ + εe) 0, or equivalently (γ σ y σ ) e. Since Y is closed, this would force e to belong to Y, which is impossible, since φ( e) = 1. Proposition 2. The only Hausdorff linear topology on K is the natural topology T K. Proof. Let T be an arbitrary Hausdorff linear topology on K, so that we need to prove that both identity maps Id : (K, T K ) (K, T), (1) Id : (K, T) (K, T K ) (2) are continuous. The continuity of the map (1) follows immediately from Lemma 1. (In the notation used there, Id = φ 1.) Since Id is linear and Ker Id = {0}, which is closed in (K, T), by Lemma 2 it follows that the map (2) is continuous. 1 See Exercise 10 from TVS I. 2

3 Exercise 1. Show that the only other linear (non-hausdorff) topology on K is the trivial topology T = {, K}. Theorem 1. A. On every finite dimensional vector space X there is a unique topological vector space structure. In other words, any two Hausdorff linear topologies on X coincide. B. If Y is a topological vector space, then any finite dimensional linear subspace X Y is closed. Proof. To prove statement A it suffices to prove a special case: X = K n. (Indeed, if n = dim X, and we fix some linear isomorphism φ : X K n, then any Hausdorff linear topology S on X is of the form φ T, where T is a Hausdorff linear topology on K n, namely T = {φ(s) : S S}.) Both statements A and B will be proved simultaneously, by showing that for any positive integer n, the following statements hold: (a n ) The product topology T prod is the only Hausdorff linear topology on K n. (b n ) Given a topological vector space X, all n-dimensional linear subspaces in X are closed. We prove (a n ) and (b n ) by simultaneous induction on n. To begin, first notice that statement (a 1 ) is already contained in Proposition 2. To prove (b 1 ), fix some topological vector space (X, T) and a non-zero vector x X, and let us prove that Kx is closed in X. Suppose we have a net (α λ ) λ Λ in K, and some vector y X, such that α λ x y. Consider the double net (α λ α µ ) (λ,µ) Λ Λ, which has the property that (α λ α µ )x 0, relative to T. (3) If we consider the map φ x : K α αx X, and the resulting pull-back topology S = φ xt, then (3) simply states that (α λ α µ ) 0, in (K, S). (4) But by (a 1 ) the topology S (which is Hausdorff) coincides with the natural topology T K on K, and then condition (4) simply states that the net (α λ ) λ Λ is Cauchy in K. Therefore α λ α, for some α K, and then by Lemma 1 it follows that α λ x αx. Since T is Hausdorff, this forces y = αx Kx, and we are done. We now proceed with the inductive step. Assume statements (a n ) and (b n ) are true, and let us prove statements (a n+1 ) and (b n+1 ). To prove statement (a n+1 ), let T be some Hausdorff linear topology on K n+1, and let us show the continuity of the identity maps Id : (K n+1, T prod ) (K n+1, T), (5) Id : (K n+1, T) (K n+1, T prod ). (6) 3

4 The continuity of (5) is immediate from Proposition 1. To prove the continuity of (6) we notice that, by the definition of the product topology (see TVS II), all we need to do is prove the continuity of the coordinate maps π i : (K n+1, T) (α 1,..., α n+1 ) α i K, i = 1, 2,..., n + 1. But this follows immediately from Lemma 2 and the inductive hypothesis (b n ), since Ker π i are all n-dimensional, hence closed in (K n+1, T). To prove property (b n+1 ) start with some topological vector space X, an (n + 1)- dimensional linear subspace Y X, and let us show that Y is closed. Fix some non-zero vector y Y and consider the quotient space X /(Ky), together with the quotient map π : X X /(Ky). We know (see TVS II; by (b 1 ) the space Ky is closed in X ) that X /(Ky) comes naturally equipped with a topological vector space structure, which makes π continuous. Since 0 y Y, it follows that π(y) is an n-dimensional linear subspace in X /(Ky), so by the inductive hypothesis (b n ) it follows that π(y) is closed. But now by the continuity of π, the pre-image π 1( π(y) ) = Y is closed in X. Exercises Suppose X is a topological vector space, and let Y, Z X be two linear subspaces. Show that, if Y is closed and Z is finite dimensional, then Y + Z is closed. 3. Suppose X and Y are topological vector spaces and T : X Y is a linear map with finite dimensional range. Prove that the following are equivalent: (i) T is continuous; (ii) Ker T is closed. 4*. Suppose X is equipped with a linear topology, n is a positive integer, and φ : X K n is a surjective linear continuous map. Show that φ is open, i.e. A open in X φ(a) open in K n. (Hint 2 Endow K n with the quotient topology.) Comment. In connection with Exercise 2, it is possible to construct (infinite dimensional) topological vector spaces X, together with (infinite dimensional) closed subspaces Y, Z X, so that the subspace Y + Z is not closed. Such constructions will be clarified when we will study Banach spaces. We conclude this section with an interesting characterization of finite dimensionality. Theorem 2. for a topological vector space K, the following are equivalent: (i) X is finite dimensional; (ii) X is locally compact. 2 Exercises marked with an asterisk are non-trivial, but important. 4

5 Proof. By Theorem 1, the implication (i) (ii) is trivial, since the product topology on K n is locally compact. (ii) (i). Fix V a compact neighborhood of 0 in X. If we consider the open set A = 1Int(V) which contains 0 then by compactness, there exists x 2 1,..., x k V, such that V n i=1 (A + x i). In particular, we also have the inclusion V n i=1 ( 1V + x 2 i), so if we consider the space Y = Span{x 1,..., x k }, we also have the inclusion V 1 V + Y. (7) 2 Claim 1: V 1 V + Y, n N. 2 n This can be shown by induction on n. The case n = 1 is just (7). Assuming that the inclusion is valid for n, we get 1 ( 1 1 V + Y ) n 1 V + Y, 2 n+1 and using (7) we also get V ( 1 V + Y ) + Y 1 V + Y. 2 n+1 2 n+1 Claim 2: ( 1 n=1 V + Y ) Y. 2 n Indeed, if we take some x in the intersection, then for every n N, there are y n Y and v n V, such that x = y n + 1 v 2 n n. (8) Since all v n s belong to a compact set V, by Exercise?? from TVS I, it follows that 1 2 n v n 0, and then (8) forces y n x. Since Y is finite dimensional thus closed by Theorem 1 this forces x Y. Having proven Claim 2, we now notice that by Claim 1 we now have the inclusion V Y. Since V is absorbing, this inclusion forces in fact the equality Y = X, and we are done. 5

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