Topological Vector Spaces III: Finite Dimensional Spaces

TVS III 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, 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 φx : K 3 α 7−→ αx ∈ X is continuous with respect to T.

Proof. Consider the map ωx : K 3 α 7−→ (α, x) ∈ K × X . When we equip K × X with the product topology, ωx is clearly continuous. The continuity of φx is then clear, since we can write it as a composition of continuous maps

ωx multiplication φx : K −→ K × X −−−−−−−→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 Kn with the product topology, then all linear maps T : Kn → X are continuous with respect to T. n Proof. Let {e1, . . . , en} be the standard basis in K , so that

n (α1, . . . , αn) = α1e1 + ··· + αnen, ∀ (α1, . . . , αn) ∈ K .

n Fix some linear map T : K → X , and let xk = T ek, k = 1, . . . , n. Using the notations from Lemma 1, we clearly have

n T (α1, . . . , αn) = φx1 (α1) + ··· + φxn (αn), ∀ (α1, . . . , αn) ∈ K , which means that T can be written as a composition

φx1 ×···×φxn addition T : K × · · · × K −−−−−−−→X × · · · × X −−−−→X . | {z } | {z } n factors n factors

The addition map is continuous. By Lemma 1, the map φx1 × · · · × φxn is continuous, so T is also continuous.

1 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 (i0) ⇒ (i) ⇒ (ii) are pretty clear1. (ii) ⇒ (i0). 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

φx − 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 (xλ)λ∈Λin X , with xλ → 0, and let us prove that φ(xλ) → 0. If we write each xλ as xλ = yλ + αλe, with yλ ∈ Y and αλ ∈ K, then all we have to show is that αλ → 0. Assume the contrary, which means that there exists some ε > 0, such that the set

Σ = {σ ∈ Λ: |αλ| ≥ ε}

is coﬁnal in Λ, i.e. (∗) for every λ ∈ Λ, there exists σ ∈ Σ with σ λ.

In particular, restricting to Σ we get two subnets (yσ)σ∈Σ in Y and (ασ)σ∈Σ in K, such that

(a) (yσ + ασe) → 0 (in X );

(b) |ασ| ≥ ε, ∀ σ ∈ Σ.

Consider then the net γσ = ε/ασ. By (b) we have |γσ| ≤ 1, ∀ σ ∈ Σ, so using Exercise ?? from TVS I we still have γσ(yσ + ασe) → 0. This simply reads: (γσyσ + εe) → 0, or equivalently (γσyσ) → −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 TK. Proof. Let T be an arbitrary Hausdorﬀ linear topology on K, so that we need to prove that both identity maps

Id : (K, TK) → (K, T), (1)

Id : (K, T) → (K, TK) (2) are continuous. The continuity of the map (1) follows immediately from Lemma 1. (In the notation used there, Id = φ1.) 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 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 = Kn. (Indeed, if n = dim X , and we fix some linear isomorphism φ : X → Kn, then any Hausdorff linear topology S on X is of the form φ∗T, where T is a Hausdorff linear topology on Kn, 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:

n (an) The product topology Tprod is the only Hausdorﬀ linear topology on K .

(bn) Given a topological vector space X , all n-dimensional linear subspaces in X are closed.

We prove (an) and (bn) by simultaneous induction on n. To begin, ﬁrst notice that statement (a1) is already contained in Proposition 2. To prove (b1), 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 (αλ)λ∈Λ in K, and some vector y ∈ X , such that αλx → y. Consider the double net (αλ − αµ)(λ,µ)∈Λ×Λ, which has the property that

(αλ − αµ)x → 0, relative to T. (3)

If we consider the map φx : K 3 α 7−→ αx ∈ X , ∗ and the resulting pull-back topology S = φxT, then (3) simply states that

(αλ − αµ) → 0, in (K, S). (4)

But by (a1) the topology S (which is Hausdorff) coincides with the natural topology TK on K, and then condition (4) simply states that the net (αλ)λ∈Λ is Cauchy in K. Therefore αλ → α, for some α ∈ K, and then by Lemma 1 it follows that αλx → αx. Since T is Hausdorff, this forces y = αx ∈ Kx, and we are done. We now proceed with the inductive step. Assume statements (an) and (bn) are true, and let us prove statements (an+1) and (bn+1). n+1 To prove statement (an+1), let T be some Hausdorff linear topology on K , and let us show the continuity of the identity maps

n+1 n+1 Id : (K , Tprod) → (K , T), (5) n+1 n+1 Id : (K , T) → (K , Tprod). (6)

3 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

n+1 πi :(K , T) 3 (α1, . . . , αn+1) 7−→ αi ∈ K, i = 1, 2, . . . , n + 1.

But this follows immediately from Lemma 2 and the inductive hypothesis (bn), since Ker πi are all n-dimensional, hence closed in (Kn+1, T). To prove property (bn+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 (b1) 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 6= y ∈ Y, it follows that π(Y) is an n-dimensional linear subspace in X /(Ky), so by the inductive hypothesis (bn) it follows that π(Y) is closed. But now by the continuity of π, the pre-image π−1π(Y) = Y is closed in X . Exercises 2-4.

2. 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 → Kn is a surjective linear continuous map. Show that φ is open, i.e.

n A open in X ⇒ φ(A) open in K .

(Hint2 Endow Kn 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 Proof. By Theorem 1, the implication (i) ⇒ (ii) is trivial, since the product topology on Kn is locally compact. (ii) ⇒ (i). Fix V a compact neighborhood of 0 in X . If we consider the open set 1 A = 2 Int(V) – which contains 0 – then by compactness, there exists x1, . . . , xk ∈ V, such Sn Sn 1 that V ⊂ i=1(A + xi). In particular, we also have the inclusion V ⊂ i=1( 2 V + xi), so if we consider the space Y = Span{x1, . . . , xk}, we also have the inclusion

1 V ⊂ 2 V + Y. (7)

1 Claim 1: V ⊂ 2n V + Y, ∀ n ∈ 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 1 2 V ⊂ 2 2n V + Y ⊂ 2n+1 V + Y, and using (7) we also get

1 1 V ⊂ 2n+1 V + Y + Y ⊂ 2n+1 V + Y.

T∞ 1 Claim 2: n=1 2n V + Y ⊂ Y. Indeed, if we take some x in the intersection, then for every n ∈ N, there are yn ∈ Y and vn ∈ V, such that 1 x = yn + 2n vn. (8) 1 Since all vn’s belong to a compact set V, by Exercise ?? from TVS I, it follows that 2n vn → 0, and then (8) forces yn → 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.