Topological Vector Spaces IV: Completeness and Metrizability

Home , Embedding, Homeomorphism, Metric (mathematics), Net (mathematics), Sequentially complete, Space (mathematics)

TVS IV c Gabriel Nagy

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

In this section we isolate two important features of topological vector spaces, which, when present, are very useful. With few exceptions, the material from sub-section A is is optional.

Convention. Throughout this note K will be one of the ﬁelds R or C, and all vector spaces are over K. A. Completeness for Topological Vector Spaces

This sub-section is aimed at treating completeness for most general types of topological vector spaces, beyond the traditional metric framework. Our approach here is essentially and adaptation of the the theory of uniform spaces1. As it turns out, a topological vector space is always a uniform space. Deﬁnitions. Suppose (X , T) is a topological vector space.

A. A net (xλ)λ∈Λ ⊂ X is said to be T-Cauchy, if:

This condition is equivalent to the fact that the double net (xλ −xµ)(λ,µ)∈Λ×Λ converges to 0.

B. The linear topology T is said to be complete, if every T-Cauchy net in X is convergent.

Example 1. Let I be an arbitrary non-empty set. If (Xi)i∈I is a collection of topological Q vector spaces, then the product space X = i∈I Xi is complete, when equipped with the product topology, if and only if all the Xi, i ∈ I, are complete. This follows from the fact i that a net (xλ)λ∈Λ in X , written in coordinates as xλ = (xλ)i∈I , ∀ λ, is Cauchy in X , if and i only if, for every i ∈ I, the net (xλ)λ∈Λ is Cauchy in Xi. In particular, all ﬁnite dimensional topological vector spaces are complete.

1 A uniform space is a pair (X, U) consisting of a non-empty set X and a ﬁlter U of subsets U ⊂ X × X, T subject to the following conditions: (i) U∈U U = {(x, x): x ∈ X} – the diagonal of X; (ii) if U belongs to U, then its inverse (as a relation) U −1 = {(y, x):(x, y) ∈ U} also belongs to U; (iii) for every U ∈ U there exists V ∈ U, such that the composition (as relations) V ◦ V is contained in U. The ﬁlter U is then called a uniformity on X.

1 Exercise 1. Prove that a topological vector space X is complete, if and only if every Cauchy net has a convergent sub-net. Comment. If one attempts to construct the completion of a topological vector space in a manner similar to the one used in the metric framework, then Definition B above is a bit useless, since there is no such thing as “the set of all nets.” Therefore, one would like to replace (for now) nets with filters2, thus the following definition is natural. Definition. Suppose (X , T) is a topological vector space. A filter F in X is called Cauchy, if for every T- neighborhood V of 0, there exists F ∈ F, such that: x − y ∈ V, ∀ x, y ∈ F. With all the preparations in place, the exercises below will guide the reader towards build- ing the completion of a topological vector space (X , T). Except for Notations, Definitions, and Lemma 1, the remainder of this sub-section is optional. Exercises 2-8. These three exercises clarify the connections between Cauchy filters and Cauchy nets. Fix a topological vector space X .

2. Prove that, if a ﬁlter F is convergent, that is, there exists x ∈ X , such that, for every neighborhood V of x, there exists F ∈ F with F ⊂ V, then F is Cauchy.

3. Prove that, for a net (xλ)λ∈Λ in X , the following are equivalent:

(i) (xλ)λ∈Λ is Cauchy;

(ii) the tail filter F = {Fλ}λ∈Λ, defined by Fλ = {xµ : µ λ}, is Cauchy. 4*. Prove that, for a filter F on X , the following are equivalent:

(i) F is Cauchy; (ii) all F-section nets3 are Cauchy.

(Hint: For (ii) ⇒ (i), equip each F ∈ F with a well ordering F , and consider the set Λ = {(F, z): F ∈ F, z ∈ F}, equipped with the following order: (F1, z1) (F2, z2),

if either F1 ( F2, or F1 = F2 and z1 F1 z2. Show that Λ is directed. Consider the obvious map Λ 3 (F, z) 7−→ z ∈ X as a net (xλ)λ∈Λ. Show that this net is an F-section, and prove that by condition (ii) the ﬁlter F is Cauchy.)

5. Prove that if F is a Cauchy filter, then so is any filter F0 ⊃ F. In particular, all ultrafilters that contain F are Cauchy.

6. Prove that the following conditions are equivalent:

(i) X is complete; (ii) every Cauchy ﬁlter in X is convergent;

2 Recall that, given some non-empty set X, a ﬁlter in X is a collection F of non-empty subsets, which is directed with respect to the reverse inclusion, that is, for every F1,F2 ∈ F, there exists F ∈ F, with F ⊂ F1 ∩ F2. 3 Recall that a net (xλ)λ∈Λ is called an F-section, if for every F ∈ F, there exists λF ∈ Λ, such that xλ ∈ F, ∀ λ λF .

2 (iii) every Cauchy ultraﬁlter in X is convergent.

(Hints: Prove the implications (i) ⇒ (ii) ⇒ (iii) ⇒ (i). For (i) ⇒ (ii) use Exercise 4. For (iii) ⇒ (i), start with a Cauchy net (xλ) and choose an ultrafilter U that contains the tail filter. By Exercises 3 and 5, U is Cauchy, so assuming (iii), the ultrafilter U converges to some x. Prove that xλ → x.)

7. Prove that if F1 and F2 are Cauchy filters, then {F1 + F2 : F1 ∈ F1, F2 ∈ F2} is a Cauchy filter. By an abuse of notation, this filter will be denoted by F1 + F2.

8. Prove that if F is a Cauchy filter, then for every α ∈ K, {αF : F ∈ F} is a Cauchy filter. By an abuse of notation, this filter will be denoted by αF. Notation. We denote by cf(X ) the set of all Cauchy filters in (X , T). Equip cf(X ) with the following relation (use the notations from Exercises 5 and 6):

F ≈ G ⇔ F + (−1)G converges to 0.

Exercise 9. Prove the following (i) ≈ is an equivalence relation on cf(X ). (ii) If F is a Cauchy ﬁlter, and F0 is a ﬁlter that contains F (which is automatically Cauchy by Exercise 5), then F ≈ F0.

(iii) If F1 ≈ G1 and F2 ≈ G2, then F1 + F2 ≈ G1 + G2.

(iv) If F ≈ G, then αF ≈ αG, ∀ α ∈ K. Notations. For each F ∈ cf(X ), let [F] denote its equivalence class. Given a non-empty set S ⊂ X , deﬁne

S˜ = {[F]: F ∈ cf(X ), F 3 S}.

Using the preceding Exercise, we see that one has two operations

X˜ × X˜ −−−−→addition X˜, multiplication K × X˜ −−−−−−−→ X˜, correctly defined on equivalence classes by [F1] + [F2] = [F1 + F2] and α[F] = [αF]. Exercise 10. Prove that the above two operations define a vector space structure on the quotient space X˜. Exercise 11. Define, for every x ∈ X , the trivial filter Fx = {x} . Prove that:

(i) For every x ∈ X , the ﬁlter Fx is convergent, thus Cauchy. ˜ (ii) The map J : X 3 x 7−→ [Fx] ∈ X is an injective linear map. Exercise 12. Consider the collection B = {N˜ : N ⊂ X balanced T-neighborhood of 0}. Prove that:

3 (i) B is a ﬁlter;

(ii) all sets in B are balanced and absorbing;

(iii) for any V ∈ B, there exists W ∈ B, such that W + W ⊂ V; T (iv) V∈B V = {0}. Conclude, using Corollary 1 from TVS I, that there exists a unique Hausdorﬀ linear topology T˜ on X˜, so that B is a basic system of T˜-neighborhoods of 0. Exercise 13. Consider the injective linear map J : X → X˜ deﬁned in Exercise 10.

(i) Prove that the linear subspace R = Range J ⊂ X˜ is dense in X˜.

(ii) Prove that, if (xλ)λ∈Λ is a Cauchy net in X , and we consider its tail ﬁlter F deﬁned in ˜ ˜ Exercise 1, then J(xλ) → [F] in (X , T). (iii) Prove that any point in X˜ can be realized using the construction from (ii), namely, for ˜ every v ∈ X , there exists a Cauchy net (xλ)λ∈Λ in X , such that J(xλ) → v. ˜ ∼ (iv) Prove that, when we equip R with the induced topology T R, the map J : X −→R becomes a linear homeomorphism.

(v) Prove that, if (X , T) is already complete, then J :(X , T) → (X˜, T˜) is a linear homeomorphism.

(Hint: For (iv) show that if V is a neighborhood of 0 in X , then V ⊂ J −1(V˜) ⊂ V.) Exercise 14*. Prove that (X˜, T˜) is a complete topological vector space. (Hint: Con- sider the filter V = {V˜ : V neighborhood of 0 in X}. Show that if G is a Cauchy filter in X˜, then G + V is again a Cauchy filter in X˜, equivalent to G. Using the preceding Exercise, it follows that the collection F = {J −1(A): A ∈ G0} is a filter in X . Show that F is Cauchy, and then prove that G0 converges to v = [F] in X˜. Conclude that, since G is equivalent to G0, it also converges to v.) Definition. The topological vector space (X˜, T˜) is called the completion of (X , T). Comment. The construction of the completion is borrowed (word for word) from the theory of uniform spaces in Topology. Very few of the properties of topological vector spaces are employed here. In particular, the same construction works for topological groups. (A topological group is a pair (G, T) consisting of a group G and a topology T on G, such that the maps G × G 3 (g, h) 7−→ gh ∈ G and G 3 g 7−→ g−1 ∈ G are continuous.) Exercise 15-16. Suppose (X1, T1) and (X2, T2) are topological vector spaces. Declare a (not necessarily linear) map Φ : X1 → X2 uniformly continuous, if it satisfies the following condition.

(uc) for every T2-neighborhood V of 0 in X2, there exists a T1-neighborhood W of 0 in X1, so that: whenever x, y ∈ X1 are such that x − y ∈ W, it follows that Φ(x) − Φ(y) ∈ V. It is not hard to show that such a Φ is indeed continuous.

4 ˜ 15. Assume that Φ is uniformly continuous. Let Jk : Xk → Xk, k = 1, 2 denote the corresponding injections. ˜ ˜ ˜ ˜ (i) Prove that there exists a unique continuous map Φ: X1 → X2, such that Φ ◦ J1 = J2 ◦ Φ. (ii) Prove that Φ˜ is in fact uniformly continuous.

(Hint: Either use Exercise 13, by showing that, if (xλ)λ∈Λ is Cauchy in X1, then Φ(xλ) λ∈Λ is Cauchy in X2, or, for an alternative approach, see the Comment below.)

16. Prove that, for any linear continuous map T :(X1, T1) → (X2, T2), is uniformly continuous, so the above Exercise can be used to obtain a uniformly continuous map ˜ ˜ ˜ ˜ ˜ ˜ T :(X1, T1) → (X2, T2). Prove that T is linear. Exercise 17. Suppose X is a topological vector space, and S ⊂ X is non-empty. Let J : X → X˜ denote the map described in Exercise 11. Show that S˜ = J(S) (the closure of J(S) in X˜). Exercise 18. Suppose X is a topological vector space, and Y ⊂ X is a linear subspace, equipped with the induced topology, and let X˜ and Y˜ be the completions of X and Y. Let T : Y ,→ X denote the inclusion map. Prove the following:

(i) The continuous extension T˜ : Y˜ → X˜ is injective.

(ii) If we equip the linear subspace Z = Range T˜ with the induced topology (from X˜), then T˜ : Y˜ → Z is a linear homeomorphism.

(iii) If J : X → X˜ denotes the map described in Exercise 11, then Z = J(Y) (the closure of J(Y) in X˜).

Comments. A. In connection with Exercise 15, perhaps the most optimal way to construct Φ˜ is as follows. Start with some v ∈ X˜, represented as v = [F], by a Cauchy filter F in X1, and simply notice that the collection Φ(F) = {Φ(F): F ∈ F} is a Cauchy filter in ˜ X2, so one can define Φ(v) = [Φ(F)]. Of course, one has to check that this construction is independent on the choice of F, but this is quite obvious. B. The map Φ˜ can also be constructed in an analogous way to the traditional metric ˜ version, as follows. Start with some v ∈ X1, so by Exercise 13 there exists a Cauchy net ˜ T1 (xλ)λ∈Λ ⊂ X1, such that J1(xλ) −→ v. By uniform continuity the net Φ(xλ) λ∈Λ is Cauchy ˜ ˜ in (X2, T2), hence the net J Φ(xλ) λ∈Λ is convergent in X2, T2) to some unique vector, which we declare to be Φ(˜ v). Of course, this definition has a slight ambiguity built in (it appears to depend on the choice of the net), but a posteriori this is easily removed. C. Using Exercises 14-15, the completion of a topological vector space is unique up to a linear homeomorphism. The result below provides a direct way of recognizing the completion.

Theorem 1. Suppose (Y, S) is a complete topological vector space, and X is a linear subspace. Let T denote the induced topology S X , and let T denote the induced topology S X on the closure of X . If we consider the completion (X˜, T˜) and the standard inclusion map

5 J : X ,→ X˜, there exists a unique linear homeomorphism T :(X˜, T˜) −→∼ (X , T), such that T (J(x)) = x, ∀ x ∈ X . Proof. Let I :(X , T) ,→ (Y, S) be the inclusion map, which is obviously continuous, and let T = I˜ :(X˜, T˜) → (Y, S) be the extension to the completion (as indicated in Exercises 15-16). By construction, T is linear, continuous, and satisfies T ◦ J = I, Strictly speaking, T = I˜ takes values in the completion of Y, which is in turn identified with Y, as indicated in Exercise 13. Using Comment B above, T can be implicitly defined by the implication

T˜ ˜ S J(xλ) −→ v (in X ) =⇒ xλ −→ T (v) (in Y), (1) ˜ where (xλ)λ∈Λ is a net in X , and v is a vector in X . It is clear that Range T ⊂ X , so we can think from now on T as a linear continuous map T :(X˜, T˜) → (X , T). Claim 1: T is injective. Start with some v ∈ X˜, with T (v) = 0, and let us show that v = 0. Choose a net T˜ ˜ S (xλ)λ∈Λ ⊂ X , with J(xλ) −→ v (in X ), so that by (1) it follows that xλ −→ 0, in Y. By the T deﬁnition of the induced topology this forces xλ −→ 0, in X , so by the continuity of J we also T˜ ˜ ˜ get J(xλ) −→ 0, in X , which yields (X is, of course, Hausdorﬀ) v = 0. ˜ Claim 2: For a vector v ∈ X and a net (xλ)λ∈Λ ⊂ X , the following are equivalent:

T˜ ˜ (i) J(xλ) −→ v, in X ;

S (ii) xλ −→ T (v), in Y;

T (ii’) xλ −→ T (v), in X . The equivalence (ii) ⇔ (ii0) is clear, since T is the induced topology. The implication (i) ⇒ (ii) – which is the same as (1) – was already discussed. To prove the implication (ii) ⇒ (i), we simply observe that condition (ii), combined with the fact that T is the induced topology, implies the fact that (xλ)λ∈Λ is Cauchy in (X , T), so the net J(xλ) λ∈Λ ˜ ˜ ˜ S will converge in (X, T) to some vector w ∈ X . Of course, by (1) this forces xλ −→ T (w), in Y. By the uniqueness of limits (Y is, of course, Hausdorﬀ), this yields T (v) = T (w), and then by Claim 1 we get w = v, thus proving (i). Based on Claims 1 and 2 it follows that T is a homeomorphism, when regarded as a map T : X˜ → Range T , where the target space is equipped with the induced topology (by T), so the proof will be ﬁnished, once we prove that Range T = X . Start with some y ∈ X , and let us indicate how a vector v ∈ X˜ can be found, such that T (v) = y. We know that there exists S a net (xλ)λ∈Λ ⊂ X , such that xλ −→ y. As argued before, it follows that (xλ)λ∈Λ is Cauchy ˜ ˜ ˜ T˜ in (X , T), and J(xλ) λ∈Λ is Cauchy in (X, T), so there exists v ∈ X , such that J(xλ) −→ v, ˜ S in T. By (1) it follows that xλ −→ T (v) in Y, so by the uniqueness of limits, it follows that T (v) = y.

6 We conclude this sub-section with an important technical result, which will be used in DT II.

Lemma 1 (Bourbaki’s Embedding Lemma). Suppose T1 ⊃ T2 are two Hausdorff linear topologies on a vector space X . Let I :(X , T1) → (X , T2) denote the identity map, and let ˜ ˜ (Xk, Tk) denote the completion of (X , Tk), k = 1, 2. If there exists a fundamental system V of T1-neighborhoods of 0, such that all sets V ∈ V ˜ ˜ ˜ ˜ ˜ are T2-closed, then the map I :(X1, T1) → (X2, T2) is injective. Proof. We know already that I˜ is linear (and continuous), so all we need to do is to prove ˜ ˜ ˜ that Ker I = {0}. Fix some vector v ∈ X1, such that I(v) = 0, and let us prove that v = 0. ˜ Since the topology T1 is Hausdorff, it suffices to prove the following ˜ ˜ Claim: For every T1-neighborhood U of 0 (in X1), the set v + U contains 0. For the proof of the above statement, we start off by fixing some T1-Cauchy net (xλ)λ∈Λ ˜ ˜ ˜ ˜ ˜ in X that represents v, i.e. J1(xλ) → v in (X1, T1). Since I(v) = 0 in (X2, T2), it follows T2 that xλ −→ 0. ˜ ˜ First of all, by the construction of the completion (X1, T1) we can assume that U = ˜ ˜ W(⊂ X1), for some T1-neighborhood W of 0 in (X , T1). Secondly, by the hypothesis, upon replacing W with a smaller T1-neighborhood, we can also assume that W is also T2-closed. Thirdly, since the net (xλ) is T1-Cauchy, there exists λW , such that xλ−xµ ∈ W, ∀ λ, µ λW . If we fix for the moment µ λW , this means that for every λ λW , there exists zλ ∈ W, T2 T2 such that xλ − xµ = zλ. Of course, since xλ −→ 0, this means that zλ −→ (−xµ). But the fact that W is T2-closed will then force

−xµ ∈ W, ∀ µ λW . (2) ˜ ˜ ˜ Since the net (−xµ)µ λW represents −v in (X1, T1), condition (2) forces −v ∈ W = U, and the Claim is proven.

B. Metrizability

As it turns out, for a large class of topological vector spaces it is possible to carry on the completion operation by traditional (i.e. metric) means. In preparation for that approach, we discuss an important topological property: metrizability. As a matter of terminology, we will say that a topological (vector) space (X , T) is metrizable, if there exists a metric d on X , so T that T agrees with the metric topology deﬁned by d. In other words, xλ −→ x ⇔ d(xλ, x) → 0. With this terminology, one has the following fundamental result. Theorem 2 (Metrizability Theorem). For a topological vector space (X , T), the following conditions are equivalent: (ii) The topology T is metrizable, i.e. there exists a metric d on X , so that T coincides with the metric topology deﬁned by d.

(ii) The topology T is ﬁrst countable4.

4 This means that every point x ∈ X has a countable fundamental (or basic) T-neighborhood system. Because of translation invariance, it suﬃces to check this condition only at x = 0.

7 Moreover, in the case when one of the above conditions is satisﬁed, the metric d from (i) can be chosen to be translation invariant, i.e. d(x + z, y + z) = d(x, y), ∀ x, y, z ∈ X . Proof. The implication (i) ⇒ (ii) is trivial. To prove the reverse implication, we start oﬀ by ∞ choosing basic neighborhood system (Vn)n=0 of 0, with V0 = X , such that:

(a) (−1)Vn = Vn,

(b) Vn−1 ⊃ Vn + Vn + Vn, for all n ∈ N. (This can be easily accomplished using Proposition 2 from TVS I. Condition (a) is obviously satisfied, if for instance all the Vn’s are balanced.) Since T is Hausdorff, we also have: ∞ \ Vn = {0}. (3) n=1 Define now ρ : X → [0, 1] by

−k ρ(x) = inf{2 : x ∈ Vk}.

−n ∞ Obviously, ρ takes values in {0} ∪ {2 }n=0, and using (3) we see that ρ(x) = 0 ⇔ x = 0. (4)

Moreover, by construction, for a net (xλ)λ in X , one has the equivalence:

ρ(xλ) → 0 ⇔ xλ → 0. (5) Using condition (a) it also follows that: ρ(−x) = ρ(x), ∀ x ∈ X . (6)

Claim 1: ρ(x + y + z) ≤ 2 · max{ρ(x), ρ(y), ρ(z)}, ∀ x, y, z ∈ X . To prove this we can assume (by symmetry) that the max{ρ(x), ρ(y), ρ(z)} = ρ(z). The cases when ρ(z) = 0 or ρ(z) = 1, are trivial, so the remaining one is when ρ(z) = 2−k, for −k some integer k ≥ 1. But then the inequalities ρ(x), ρ(y) ≤ ρ(z) = 2 will force x, y, z ∈ Vk, −(k−1) so by (b) we get x + y + z ∈ Vk−1, so ρ(x + y + z) ≤ 2 = 2ρ(z). Claim 2: ρ(x1 + ··· + xn) ≤ 2 ρ(x1) + ··· + ρ(xn) , ∀ x1, . . . , xn ∈ X . We prove the inequality by induction on n. The cases n = 1, 2, 3 are trivial by Claim 1. Assume the inequality is true for all n with 1 ≤ n ≤ N, for some N ≥ 3, and let us prove it for n = N + 1. Fix x1, . . . , xN+1, and let us prove that ρ(x1 + ··· + xN+1) ≤ 2 ρ(x1) + ··· + ρ(xN+1) . (7)

If ρ(xj) = 0, for some j, then everything is clear, by the inductive hypothesis. Likewise, if ρ(xj) = 1, for some j, then the inequality is also trivial, since ρ takes values in [0, 1]. Assume now 0 < ρ(xj) < 1, for all j, and also (re-index if necessary), that: ρ(x1) ≤ · · · ≤ ρ(xN+1). −kj Equivalently, one has integers k1 ≥ k2 ≥ · · · ≥ kN+1 such that ρ(xj) = 2 , for all j. Let us consider now two possibilities:

8 (i) There exists j ∈ {2,...,N}, such that ρ(xj) = ρ(xj+1). In this case by Claim 1 (note that ρ(xj+1) = max{ρ(xj−1), ρ(xj), ρ(xj+1)}) we have:

ρ(xj−1) + ρ(xj) + ρ(xj+1) ≥ 2ρ(xj+1) ≥ ρ(xj−1 + xj + xj+1),

so we get

ρ(x1)+···+ρ(xN+1) ≥ ρ(x1)+···+ρ(xj−2)+ρ(xj−1+xj +xj+1)+ρ(xj+2)+···+ρ(xN+1),

so we can use the inductive hypothesis (for n = N − 1) to get (7)

(ii) ρ(x1) ≤ ρ(x2) < ρ(x3) < ··· < ρ(xN+1). In this case we know that xj ∈ Vkj , with k1 ≥ k2 > k3 > ··· > kN+1, and then an easy recursive argument shows that

x1 + ··· + xm ∈ Vkm−1 ⊂ Vkm+1 , ∀ m = 1,...,N.

In particular x1 + ··· + xN+1 ∈ VkN+1 + VkN+1 ⊂ VkN+1−1, which forces

−(kN+1−1) ρ(x1 + ··· + xN+1) ≤ 2 = 2ρ(xN+1),

and (7) follows immediately.

Deﬁne now the map δ : X → [0, 1] by:

n n X n n X δ(x) = inf ρ(xj): n ∈ N, (xj)j=1 ∈ X , xj = x}. j=1 j=1

By construction and by Claim 2, we clearly have the inequalities:

1 2 ρ(x) ≤ δ(x) ≤ ρ(x), ∀ x ∈ X , (8) thus δ, up to this point, has the following features:

δ(x) = 0 ⇔ x = 0; (9)

δ(xλ) → 0 ⇔ xλ → 0; (10) δ(−x) = δ(x), ∀ x ∈ X . (11)

One additional feature, pretty obvious from the deﬁnition of δ, is:

δ(x + y) ≤ δ(x) + δ(y), ∀ x, y ∈ X . (12)

The proof is now finished, by defining d(x, y) = δ(x − y). Comment. The proof of Theorem 2 is again borrowed (word for word) from the theory of uniform spaces in Topology, and works equally well for topological groups. (See the Comment from sub-section A). Definition. Given a metrizable topological vector space (X , T) a translation invariant metric d on X , whose metric topology coincides with T, will be declared uniformly compatible

9 with T. In this case, when convenient, we deﬁne the d-gauge | · |d : X → [0, ∞) by |x|d = d(x, 0), so that by translation invariance we have

d(x, y) = |x − y|d, ∀ x, y ∈ X . Comment. There are three “generic” methods of producing metrizable topological vector spaces, which are pretty easy to justify (see the three Remarks below). Remark 1. If X is a metrizable topological vector space, then any linear subspace Y ⊂ X , equipped with the induced topology is metrizable.

Remark 2. If (Xj)j∈J is a family of metrizable topological vector spaces, then the Q product space j∈J Xj, equipped with the product topology, is metrizable, if and only if the index set J is countable. If we identify, for instance, J as a subset of N, and we pick, for each j ∈ J, a metric dj on Xj, whose metric topology agrees with the given topology on Xj, then the metric

X 1 dj(xj, yj) Y d(x, y) = j · , ∀ x = (xj)j∈J , y = (yj)j∈J ∈ Xj 2 1 + dj(xj, yj) j∈J j∈J deﬁnes the product topology. (See P II.) In particular, all ﬁnite dimensional topological vector spaces are metrizable. Remark 3. If X is a metrizable topological vector space, and Y ⊂ X is a closed linear subspace, then the quotient space X /Y, equipped with the quotient topology, is metrizable. Indeed, if we consider the quotient map π : X → X /Y, then we know (see TVS II) that

π is continuous and open. In particular, if we choose a basic system {Vn}n∈N of open neighborhoods of 0 in X , then {π(Vn)}n∈N constitutes a basic system of (open) neighborhoods of 0 in X /Y, thus the quotient topology (which is Hausdorff anyway, since Y is closed) is first countable. In connection with the issue of metrizability for quotient spaces, we prove a technical result below, which will be useful later. Lemma 2. Let X be a metrizable topological vector space and let d be a uniformly compatible metric on X Let Y ⊂ X be a closed linear subspace, and let π : X → X /Y denote the quotient map. The map dˆ:(X /Y) × (X /Y) → [0, ∞), defined by

dˆ(v, w) = inf{d(x, y): π(x) = v, π(y) = w}, v, w ∈ X /Y, is a metric on X /Y, that is uniformly compatible with the quotient topology. Proof. Let us show first that dˆ is indeed a metric. The symmetry property dˆ(v, w) = dˆ(w, v) is trivial, and so is the triangle inequality dˆ(u, w) ≤ dˆ(u, v) + dˆ(v, w). Suppose v, w ∈ X /Y are such that dˆ(v, w) = 0, and let us show that v = w (in X /Y). By construction, there exists ∞ ∞ two sequences (xn)n=1, (yn)n=1, such that π(xn) = v, π(yn) = w, ∀ n, and d(xn, yn) → 0. By translation invariance, the sequence zn = xn − yn satisfies d(zn, 0) = d(xn, yn), so zn → 0. In particular, since π is linear and continuous, and we clearly have v−w = π(xn)−π(yn) = π(zn), ∀ n, it follows that v − w = 0, i.e. v = w. To prove the translation invariance of dˆ, it suffices to show that

dˆ(v, w) ≥ dˆ(u + v, u + w), ∀ u, v, w ∈ X /Y. (13)

10 But this is quite trivial, because if we ﬁx z ∈ X such that π(z) = u, then for any x, y ∈ X that satisfy π(x) = v and π(y) = w, it follows that π(z + x) = u + v and π(z + y) = w, so by the translation invariance of d we have

d(x, y) = d(z + x, z + y) ≥ dˆ(u + v, u + w), so if we take the inﬁmum (over all x and y, satisfying π(x) = v and π(y) = w), we get (13) ˆ Finally, in order to prove that the metric topology Tdˆ on X /Y, deﬁned by d, coincides with the quotient topology, all we must show (see TVS II) is that the quotient map π : X →

(X /Y, Tdˆ) is both continuous and open. The continuity is trivial, since by deﬁnition we have the inequality dˆπ(x), π(y) ≤ d(x, y), ∀ x, y ∈ X (so not only π is continuous, but in fact Lipschitz.) To prove that π is open, start with some open set A, and show that π(A) is open in the metric topology Tdˆ. Fix v ∈ π(A), and let us indicate how to construct r > 0, so that

dˆ(v, w) < r ⇒ w ∈ π(A). (14)

First, choose a ∈ A with π(a) = v, and choose r > 0, such that

d(a, y) < r ⇒ y ∈ A. (15)

To prove (14) start with some w ∈ X /Y with dˆ(v, w) < r, so there exist x, y ∈ X with π(x) = v, π(y) = w, and d(x, y) < r. Since π(x) = π(a) = v, it follows that a − x ∈ Y, so the vector y0 = y + a − x will also satisfy π(y0) = w. But now by translation invariance we have d(a, y0) = d(a, y + a − x) = d(0, y − x) = d(y, x) < r, so using (15) it follows that y0 ∈ A, thus the vector w = π(y0) indeed belongs to π(A). In the remainder of this sub-section we revisit the results from sub-section A, under the metrizability assumption.

Proposition 1. Suppose X1 and X2 are metrizable topological vector spaces. For a (not necessarily linear) map Φ: X → Y, the following conditions are equivalent.

(i) Φ is uniformly continuous, in the sense that it has property (u) introduced in the previous sub-section.

(ii) there exist uniformly compatible metrics d1 on X1 and d2 on X2 with the property

(muc) For every ε > 0, there exists δ > 0, so that, whenever x, y ∈ X1 are such that

|x − y|d1 < δ, it follows that |Φ(x) − Φ(y)|d2 < ε.

(ii’) Any uniformly compatible metrics d1 (on X1) and d2 (on X2) satisfy condition (muc).

Condition (muc) is referred to as the metric uniform continuity condition.

11 0 Proof. (i) ⇒ (ii ). Let d1 and d2 be two uniformly compatible metrics on X1 and X2, respectively. Fix ε > 0, and let us indicate how δ > 0 can be constructed, satisfying the

requirement in (muc). First of all, the set V = {v ∈ X2, : |v|d2 < ε} is a neighborhood of 0 in X2, so by (i) there exists a neighborhood W of 0 in X1, such that: whenever x, y ∈ X1

are such that x − y ∈ W, it follows that Φ(x) − Φ(y) ∈ V, i.e. |Φ(x) − Φ(y)|d2 < ε. Now we see that δ > 0 can be chosen (use again translation invariance) simply by the condition

W ⊃ {w : |w|d1 < δ}, which is always possible, since the metric topology deﬁned by d1 coincides with T1. The implication (ii0) ⇒ (ii) is trivial. To prove the implication (ii) ⇒ (i) we ﬁx d1 and d2, and argue pretty much as above.

Start with V a neighborhood of 0 in X2, then there exists ε > 0, such that V ⊃ {v : |v|d2 < ε},

so if we choose δ > 0 as in (muc), the set W = {w : |w|d1 < δ} will do the job.

Corollary 1. If X1 and X2 are as above, and Φ: X1 → X2 is linear, then the conditions (i), (ii), (ii0), from Proposition 1, are also equivalent to the condition that Φ is continuous. Proof. Immediate from the observation that Φ(x)−Φ(y) ∈ V is equivalent to Φ(x−y) ∈ V, so condition (i) is obviously equivalent to the fact that Φ is continuous at 0, and this condition in turn is equivalent to the (global) continuity of Φ. Proposition 2. Suppose (X , T) is a metrizable topological vector space. For a net (xλ)λ∈Λ in X , the following are equivalent.

(i) (xλ)λ∈Λ is Cauchy, in the sense introduced in sub-section A. (ii) There exists a uniformly compatible metric d, such that:

(mc) for every ε > 0, there exists λε ∈ Λ, such that |xλ − xµ|d < ε, ∀ λ, µ λε.

(ii’) Condition (mc) is satisﬁed. for every uniformly compatible metric d.

Condition (mc) is referred to as the metric Cauchy condition.

0 Proof. (i) ⇒ (ii ). Fix a uniformly compatible metric d, and assume the net (xλ)λ∈Λ is Cauchy, that is, i.e. it satisfies the condition (c) from sub-section A. The fact that (xλ)λ∈Λ satisfies condition (mc) is trivial, if we use condition (c) with V = {x : |x|d < ε}, which is clearly a neighborhood of 0 (for every ε > 0). The implication (ii0) ⇒ (ii) is trivial. (ii) ⇒ (i). Suppose (xλ)λ∈Λ satisfies condition (mc) for a uniformly compatible metric d. condition (c) is pretty obvious, since for every neighborhood V of 0, there exists ε > 0, such that V ⊃ {v : |vd < ε}, so if we choose λε as in (c), then for λ, µ λε, we have |xλ − xµ|d < ε, which forces xλ − xµ ∈ V. Proposition 3. For a metrizable topological vector space (X , T), the following are equivalent.

(i) (X , T) is complete.

(ii) Every Cauchy sequence in X is convergent.

12 (iii) Every Cauchy sequence in X has a convergent subsequence.

(iv) There exists a uniformly compatible metric d, such that:

∞ P∞ (st) every sequence (xn)n=1 ⊂ X , with the property n=1 |xn+1 − xn|d < ∞, is convergent.

(iv’) Every uniformly compatible metric has property (st).

Condition (st) is referred to as the Summability Test.

|xm − xn|d ≤ tk, ∀ m, n ≥ k,

∞ and then the condition tk → 0, combined with Proposition 2, clearly shows that (xn)n=1 is Cauchy (as a net), thus convergent. The implication (iv0) ⇒ (iv) is trivial. (iv) ⇒ (iii). Fix a uniformly compatible metric d with property (st), and a Cauchy ∞ ∞ sequence (xn)n=1 ⊂ X . By Proposition 2, we know that (xn)n=1 satisfies the metric Cauchy ∞ condition (mc). In particular, there exists a sequence of (jk)k=1 of positive integers, such −k ∞ that |xm − xn|d ≤ 2 , m, n ≥ jk. If we define the sequence (kn)n=1 recursively by k1 = j1, ∞ and kn+1 = max{k1, . . . , kn, jn} + 1, then (kn)n=1 is strictly increasing, and the sub-sequence ∞ −n P∞ (xkn )n=1 satisfies the inequalities |xkn+1 −xkn |d ≤ 2 , ∀ n, so clearly n=1 |xkn+1 −xkn |d < ∞. ∞ Now we are done, because (st) forces (xkn )n=1 to be convergent. ∞ ∞ (iii) ⇒ (ii). Assume condition (iii) and let (xn)n=1 be a Cauchy sequence. Let (xkn )n=1 be a subsequence which converges to some point x. By the cauchy condition (xn − xkn ) → 0, so xn → x. (ii) ⇒ (i). Assume condition (ii) holds, and let us show that every Cauchy net in X is convergent. Again, as above, this condition is equivalent to the fact that every Cauchy net in X has a convergent sub-net. Fix then a Cauchy net (xλ)λ∈Λ, and choose a uniformly compatible metric d, so that, by Proposition 2, we know that the metric Cauchy condition ∞ (mc) is satisfied. In particular, one has a sequence (σn)n=1 ⊂ Λ, such that

−n |xλ − xµ|d ≤ 2 , ∀ λ, µ σn. (16)

∞ Using directedness one can assume that (σn)n=1 is increasing, i.e. σn+1 σn, ∀ n. It is −k ∞ obvious then that |xσm − xσn |d ≤ 2 , ∀ m, n ≥ k, so the sequence (xσn )n=1 is Cauchy, thus convergent to some x ∈ X . Taking limit in the preceding inequalities also yields:

−n |xσn − x|d ≤ 2 , ∀ n ∈ N. (17)

Consider now the directed set Λ×N, and the subset Γ = {(µ, k) ∈ Λ×N : λ σk} ⊂ Λ×N. Remark that Γ is coﬁnal in Λ × N, which means that, for any (λ, n) ∈ Λ × N, there exists (µ, k) ∈ Γ, such that µ λ and k ≥ n. (Simply take k = n, and µ ∈ Λ such that

13 µ λ, σk.) If we now consider the map φ :Γ 3 (µ, k) 7−→ µ ∈ Λ, then φ is directed map, so the net (yγ)γ∈Γ given by yγ = xφ(γ), is a sub-net of (xλ)λ∈Λ. Notice now that, if we set γn = (σn, n) ∈ Γ, then for every γ = (µ, k) ∈ Γ with γ γn, using (16) and (17), we have −n |xµ − x|d ≤ |xµ − xσn |d + |xσn − x|d ≤ 2 · 2 . In other words, we have

−(n−1) |yγ − x|d ≤ 2 , ∀ γ γn,

which clearly forces yγ → x. Comment. Condition (ii) from Proposition 3 is referred to as sequential completeness. There are many examples of topological vector spaces which are sequentially complete, but not complete. A general construction of such spaces will be outlined in section B ??. Remark 4. The completion of a metrizable topological vector space X is constructed in the “traditional” way (see P II). One fixes a uniformly compatible metric d and constructs the metric completion (X 0, d0), where X 0 consists of equivalence classes of Cauchy sequences ∞ ∞ in X . Here, two Cauchy sequences x = (xn)n=1 and y = (yn)n=1 are declared equivalent, if ∞ limn→∞ d(xn, yn) = 0. If we denote, for each Cauchy sequence x = (xn)n=1, its equivalence 0 0 0 0 class (in X ) by [x], the metric d on X is defined by d ([x], [y]) = limn→∞ d(xn, yn). One can show with little difficulty that X 0 is a vector space5, and the addition and scalar multiplication are continuous.

C. Property (F)

As a result of the discussion from the previous sub-section, an important class of topological vector spaces is identified (by the Definition below). As sub-section D will show, these spaces enjoy some remarkable properties. Definitions. A topological space X is said to be an (F)-space, it it is complete and metrizable. (The letter “F” was suggested by Stefan Hanach, in recognition of Maurice Frechet. As it turns out, a special class of (F)-spaces consists of the so-called Frechet spaces, which will be introduced in LCVS III. Banach also introduced “property (B)” named after himself!) Remarks 1’-3’. In connection with the three “generic” methods discussed in the previous sub-section (Remarks 1-3), we have the following:

1’. If X is an (F)-space, then: a linear subspace Y ⊂ X , equipped with the induced topology is and (F)-space, if and only if Y is closed in X . This follows from the trivial observation that any Cauchy sequence in Y is convergent in X . Q 2’. If (Xj)j∈J is a family of metrizable spaces, then: the product space j∈J Xj, equipped with the product topology, is an (F)-space, if and only if all Xj’s are (F)-spaces, and the index set J is countable. (See P II.) In particular, all ﬁnite dimensional topological vector spaces are (F)-spaces.

5 Simply note that the space cs(X ) of all Cauchy sequences is a linear subspace in Q X , and furthermore, N X 0 = cs(X )/N , where N is the linear subspace of sequences convergent to 0.

14 3’. If X is an (F)-space, and Y ⊂ X is a closed linear subspace, then the quotient space X /Y, equipped with the quotient topology, is an (F)-space. To prove this, we use Lemma 1 (and the notations used there) and the Summability ∞ Test. Fix d a uniformly compatible metric on X , let (vn)n=1 be a sequence in X /Y, P∞ such that n=1 |vn+1 −vn|dˆ < ∞, and let us show that (vn) is convergent in X /Y. Start with an arbitrary x1 ∈ X with π(x1) = v1, and construct recursively a whole sequence ∞ (xn)n=1 in X as follows: assuming xn is chosen (such that π(xn) = vn), we choose −n yn, zn ∈ X such that π(yn) = vn, π(zn) = vn+1, and |zn − yn|d ≤ 2 + |vn+1 − vn|dˆ, and we set xn+1 = xn + zn − yn, so that we have:

−n |xn+1 − xn|d = |zn − yn|d < 2 + |vn+1 − vn|dˆ, as well as (by linearity of π):

π(xn+1) = π(xn) + π(zn) − π(yn) = vn + vn+1 − vn = vn+1.

∞ −n So now our sequence (xn)n=1 satisﬁes: (a) π(xn) = vn, and (b)] |xn+1 − xn|d ≤ 2 + P∞ |vn+1 − vn|dˆ, for all n. On the one hand by (b) it follows that n=1 |xn+1 − xn|d < ∞, so (xn) is convergent. On the other hand by (a), combined with the continuity of π, it follows that (vn) is convergent too. Exercise 19*. Suppose X is a metrizable topological vector space, and Y is a linear subspace. Assume Y is an (F)-space, when equipped with the induced topology. (i) Prove that Y is closed in X , so in particular the quotient space X /Y is metrizable as well, when equipped with the quotient topology.

(ii) Prove that, if X /Y is an (F)-space, when equipped with the quotient topology, then X is also an (F)-space.

(Hint for (ii): Fix a uniformly compatible metric d on X , and let dˆbe the associated metric on X /Y defined in Lemma 1. As seen in the proof of Proposition 3, it suffices to show that ∞ −n if a sequence (xn)n=1 ⊂ X satisfies |xn+1 − xn|d ≤ 2 , ∀ n ∈ N, the (xn) is convergent. Let −n vn = π(xn) ∈ X /Y, and note that |vn+1 − vn|dˆ ≤ 2 , ∀ n, so in particular (vn) converges in −n X /Y to some v, and moreover we have |vn − v|dˆ ≤ 2 , ∀ n. Fix w ∈ X , such that π(w) = v, −n and choose, for every n, some yn ∈ Y, such that |xn − w + yn|d ≤ |vn − v|dˆ + 2 , so that we −n −n have in fact the inequalities |xn − w + yn| ≤ 2 · 2 , ∀ n. Prove that |yn+1 − yn|d ≤ 5 · 2 , ∀ n, so (yn) is convergent to some y ∈ X . Conclude that xn → y + w.) D. Four Principles of Functional Analysis

The four results discussed in this sub-section sit at the foundation of Functional Analysis. All of them are applications of Baire’s Theorem, concerning complete metric spaces, so these results hold only for (F)-spaces. Theorem 3 (Equi-continuity Principle). Suppose X is an (F)-space, Y is an arbitrary topological vector space, and Θ is a collection of linear continuous maps from X into Y, which is pointwise bounded, in the sense that:

15 (pb) for every x ∈ X , the set Θx = {T (x): T ∈ Θ} is bounded (in Y). Then the collection Θ is equi-continuous6, in the sense that:

(ec) for every neighborhood W of 0 in Y, there exists a neighborhood V of 0 in X , such that T (V) ⊂ W, ∀ T ∈ Θ.

Proof. Fix W a neighborhood of 0 in Y. Using Proposition 2 and Exercise 5 from TVS I, there exists a closed balanced neighborhood U of 0 in Y, such that U + U ⊂ W. Using property (pb), for every x ∈ X , there exists ρx > 0, such that ρxT (x) ∈ U, ∀ T ∈ Θ. Since U is balanced, we also have rT (x) ∈ U, ∀ T ∈ Θ, r ∈ [0, ρx]. Therefore, we have the following property:

(∗) For every x ∈ X , there exists a positive integer n , such that 1 T (x) ∈ U, ∀ T ∈ Θ. x nx Deﬁne now, for every n ∈ N, the set

1 An = {x ∈ X : n T (x) ∈ U, ∀ T ∈ Θ}.

Since U is closed, and all T ∈ Θ are continuous, it follows that An is closed. Furthermore, S by (∗) it follows that n∈ An = X . Using Baire’s Theorem (here it is essential that X is N 1 an (F)-space), there exists n ∈ N, such that Int(An) 6= ∅. Let then V0 = n Int(An), so that V0 is open, and T (V0) ⊂ U. If we deﬁne then the set V = V0 + (−1)V0, then V is an open neighborhood of 0, and satisﬁes:

T (V) ⊂ U + (−1)U ⊂ W.

(The last inclusion follows from the fact that U + U ⊂ W, combined with the fact that U is balanced, which implies (−1)U = U.) Comment. A particular case of Theorem 3, known as the Uniform Boundedness Prin- ciple, will be treated in B II. Theorem 4 (Open Mapping Principle). Suppose X and Y are (F)-spaces. Then any surjective linear continuous map T : X → Y is open, that is:

• whenever A is open in X , it follows that T (A) is open in Y.

Proof. It is pretty clear that the desired conclusion is equivalent to the fact that, for every x ∈ X , the following condition is satisﬁed: whenever V is a neighborhood of x in X , it follows that T (V) is a neighborhood of T (x) in Y. By linearity we only need to check this for x = 0, that is:

(∗) If V is a neighborhood of 0 in X , then T (V) is a neighborhood of 0 in Y.

6 Condition (ec) implies the fact that: whenever xλ → x (in X ), it follows that T (xλ) → T (x), uniformly for all T ∈ Θ.

16 The proof of (∗) will be done in several steps. Claim 1: If V is a neighborhood of 0 in X , then the closure T (V) has non-empty interior. Indeed, since for every n ∈ N the dilation y 7−→ ny is a homeomorphism, it follows that nB = nB, ∀ B ⊂ Y, so by the linearity of T we have:

∞ ∞ ∞ ∞ [ [ [ [ nT (V) = nT (V) ⊃ T (nV) = T nV = T (X ). n=1 n=1 n=1 n=1 S∞ (The last equality follows from the equality n=1 nV = X , which is a consequence of the fact 1 that limn→∞( n x) = 0, ∀ x ∈ X .) Since T is surjective (i.e. T (X ) = Y), the above inclusion S∞ yields the equality n=1 nT (V) = Y. Using Baire’s Theorem (for which the hypothesis that Y is an (F)-space is essential), there exists n such that nT (V) has non-empty interior, so T (V) itself has non-empty interior. Claim 2: If V is a neighborhood of 0 in X , then T (V) is a neighborhood of 0 in Y. To prove Claim 2, we choose U a neighborhood of 0 in X , such that U + (−1)U ⊂ V, and we use Claim 1, which tells us that T (U) has non-empty interior, which we denote by A. Of course, A ∩ T (U) 6= ∅, so there exists x ∈ U, such that T x ∈ A. In particular (since A is open), there exists a neighborhood W of 0 in Y, such that (T x) + W ⊂ A. Observe now that, since V ⊃ (−1)U + U ⊃ −x + U, and translations are homeomorphisms, we obtain:

T (V) ⊃ T (−x + U) = −T x + T (U) = −T x + T (U) ⊃ −T x + (T x + W) = W, so T (V) is indeed a neighborhood of 0 in Y. Having proven Claim 2, we now proceed with the proof of (∗). Fix uniformly compatible metric d on X . Since we only need to prove (∗) for V in a basic neighborhood system, we can −n assume that V = {x ∈ X : |x|d ≤ r}, for some r > 0. Deﬁne Vn = {x ∈ X : |x|d < 2 r}, for all n ∈ N, so that all Vn’s are again neighborhoods of 0. By Claim 2, we see that, in order to prove that T (V) is a neighborhood of 0 in Y, it suﬃces to prove the inclusion

T (V) ⊃ T (V1). (18)

0 Fix now a uniformly compatible metric d on Y, as well as some point w ∈ T (V1). We ∞ ∞ construct recursively two sequences: (vn)n=1 ⊂ X ,(yn)n=1 ⊂ Y, with y1 = w, such that:

(i) yn ∈ T (Vn),

(ii) vn ∈ Vn,

(iii) yn+1 = yn − T vn,

−(n−1) (iv) |yn|d0 ≤ 2 |w|d0 , for all n ∈ N. This construction is carried on as follows. Assume yn ∈ T (Vn) and vn−1 ∈ Vn−1 7 are given , and let us indicate how vn and yn+1 are constructed. Since yn ∈ T (Vn), every neighborhood of yn intersects T (Vn). In particular, using Claim 2, the set

−n B = T (Vn+1) ∩ {y ∈ Y : |y|d0 , 0 ≤ 2 |w|d0 }

7 When n = 1 we do not assume anything: we are simply given y1 = w.

17 is a neighborhood of 0, so yn + (−1)B is a neighborhood of yn, therefore the intersection (yn + (−1)B) ∩ T (Vn) is non-empty, hence it contains some point z. On the one hand, since z ∈ T (Vn), it follows that z = T vn, for some vn ∈ Vn. On the other hand, since z ∈ yn + (−1)B, it follows that z = yn − yn+1, for some yn+1 ∈ B, which means that −n yn+1 ∈ T (Vn+1), so we have (iii), and also |yn+1|d0 ≤ 2 |w|d0 , so we also have (iv). Having constructed our two sequences, let us remark ﬁrst that, by (iii), we have

w − yn+1 = T v1 + ··· + T vn = T xn, (19)

where xn = v1 + ··· + vn. Secondly, since xn+1 − xn = vn+1, by (ii) we know that −(n+1) |xn+1 − xn|d ≤ 2 r, ∀ n ∈ N. (20) P∞ On the one hand, since clearly this yields n=1 |xn+1 − xn|d < ∞, by the Summability Test it follows that (xn) converges to some point x ∈ X . (The condition that X is an (F)-space is essential here.) On the other hand, by (ii), the inequalities (20) also give r r r |x | ≤ |x | + |x − x | + ··· + |x − x | ≤ + + ··· + ≤ r, ∀ n ∈ , n d 1 d 2 1 d n n−1 d 2 22 2n N

so taking the limit (we know that xn → x) we obtain |x|d ≤ r, which means that x belongs to V. Going back to (19) it follows that yn = w − T xn−1 is convergent to w − T x. But by condition (iv) we also know that yn → 0, therefore we must have the equality w = T x, which proves that w indeed belongs to T (V). Corollary 1 (Inverse Mapping Principle). If X and Y are (F)-spaces, and T : X → Y is a continuous linear isomorphism, then its inverse T −1 : Y → X is also continuous. Proof. The continuity of T −1 is equivalent to the fact that T is open, so everything follows from Theorem 4. Corollary 2 (Closed Graph Principle). Suppose X and Y are (F)-spaces. For a linear map T : X → Y, the following are equivalent: (i) T is continuous;

(ii) The graph of T , i.e. the set GT = {(x, y) ∈ X × Y : T x = y}, is closed in X × Y, in the product topology. Proof. The implication (i) ⇒ (ii) is trivial. (It holds for arbitrary topological Hausdorﬀ spaces X and Y and arbitrary continuous maps T : X → Y.) (ii) ⇒ (i). Assume GT is closed in X ×Y. First of all, by linearity, GT is a linear subspace. Secondly, since X ×Y is an (F)-space (by Remark 2’), it follows that GT is also an (F)-space, when equipped with the induced topology. Thirdly, since both the coordinate maps

PX : X × Y 3 (x, y) 7−→ x ∈ X ,

PY : X × Y 3 (x, y) 7−→ y ∈ Y

are continuous, so will be their restrictions L = PX : GT → X and M = PY : GT → Y. GT GT The point is that L is a linear isomorphism, so by the Inverse Mapping Principle, the inverse −1 L : X → GT is also continuous. Now we are done, since T can be written as a composition T = M ◦ L−1 of continuous maps.