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LCVS III c Gabriel Nagy

Locally Convex Vector Spaces III: The Point of View

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

Warning! Some proofs are based on Exercises from previous lectures. The reader is urged to solve all Exercises from CW I-III and LCVS I-II. (See Remarks 1-6 below.) Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K.

A. Locally convex defined by

In this sub-section we outline a method of constructing locally convex topologies using seminorms. Notation. Given a vector X and a p on X , we define the two unit p-balls as the sets:

BX (p) = {x ∈ X : p(x) < 1},

BX (p) = {x ∈ X : p(x) ≤ 1}.

(When there is no danger of confusion, the subscript X will be omitted from the notation.) Remarks 1-7. Most of the properties listed below are essentially contained in CW II (see the brief references provided)

1. The sets B(p) and B(p). are convex and balanced. The B(p) is openly absorbing, and its is: qB(p) = p. (Exercise 8 from CW II.) 2. If A ⊂ X is convex and openly absorbing, then A can be recovered from its Minkowski functional as: A = {x ∈ X : qA(x) < 1}.

In particular, if A is also balanced, then qA is a seminorm, and one has the A = B(qA). (Proposition 2 from CW II). 3. A sufficient condition for a 0 ∈ A ⊂ X to be openly absorbing is that A is open with respect to some linear on X . (Exercise 6 from CW II.)

4. For two seminorms p and q, the following are equivalent:

(i) p(x) ≤ q(x), ∀ x ∈ X ; (ii) B(p) ⊃ B(q).

1 The implication (i) ⇒ (ii) is trivial. For the implication (ii) ⇒ (i), observe that for 1  q(x) every x ∈ X and every ε > 0, one has q q(x)+ε x = q(x)+ε < 1, so if (ii) holds then this 1  implies also p q(x)+ε x ≤ 1, i.e. p(x) ≤ q(x) + ε, ∀ ε > 0, hence (i) follows. 5. If p is a seminorm on X then

εB(p) = {x ∈ X : p(x) < ε}, εB(p) = {x ∈ X : p(x) ≤ ε}.

Clearly one has x ∈ εB(p) ⇔ ε−1x ∈ B(p) ⇔ p(ε−1x) < 1 ⇔ p(x) < ε, and similar equivalences with ≤ in place of <.

6. If p is a seminorm on X then

|p(x) − p(y)| ≤ p(x − y), ∀ x, y ∈ X .

Use p(x − y) = p(y − x), and the triangle inequalities p(y) + p(x − y) ≥ p(x) and p(x) + p(y − x) ≥ p(y).

7. If X is equipped with a linear topology T, and p is a seminorm on X , then the following are equivalent:

(i) p is continuous; (ii) p is continuous at 0; (iii) B(p) is a neighborhood of 0; (iii’) B(p) is a neighborhood of 0.

First of all, the implications (i) ⇒ (iii) ⇒ (iii0) are trivial. For the implication (iii0) ⇒ (ii), we must show that, assuming (iii’), for every ε > 0, the set V = {x ∈ ε X : p(x) < ε} is a neighborhood of 0. But this is trivial, since V ⊃ 2 B(p). Finally, for the implication (ii) ⇒ (i), we notice that if p is continuous at 0, and xλ → x, then (xλ − x) → 0, and then the inequalities 0 ≤ |p(xλ) − p(x)| ≤ p(xλ − x) will clearly force p(xλ) → p(x). The starting observation in this section is contained in the following result. Proposition 1. Let X be a locally convex .

(i) For every balanced convex neighborhood A of 0, the Minkowski functional qA is a con- tinuous seminorm on X .

(ii) The correspondence     A open, convex, q continuous A ⊂ X 3 A 7−→ qA ∈ q : X → [0, ∞) (1) balanced, A 3 0 seminorm on X is bijective. Its inverse is the map:

2     q continuous A open, convex, q : X → [0, ∞) 3 p 7−→ B(p) ∈ A ⊂ X . (2) seminorm on X balanced, A 3 0 Proof. (i). Fix some balanced convex neighborhood A of 0. We already know (CW II) that qA is a seminorm on A, and A = B(qA). By Remark 7, qA is continuous. (ii). For simplicity, let B denote the source set of (1), and let P denote the collection of all continuous seminorms. First of all, for every p ∈ P, the unit B(p) is convex and openly absorbing (by Remark 1). Furthermore, B(p) is also open, by continuity, so the correspondence (2) indeed takes values in B. Secondly, by Remarks 1 and 2, we also know that

qB(p) = p, ∀ p ∈ P;

B(qA) = A, ∀ A ∈ C,

which clearly show that (1) and (2) are inverses of each other. Theorem-Definition 1. Let X be a vector space, and let P be a non-empty collection of seminorms on X . If we define, for every ε > 0 and p ∈ P the set

Up.ε = {x ∈ X : p(x) < ε},

there exists a unique locally convex topology T on X , such that the collection

U = {Up.ε : p ∈ P, ε > 0}

constitutes a fundamental system of T-neighborhoods of 0. Moreover: (i) all p ∈ P are T-continuous;

(ii) T is the weakest among all locally convex topologies on X which make all p ∈ P are continuous. The topology T is referred to as the locally convex topology defined by P, as is denoted by T(P). If P is a singleton {p}, we denote this topology simply by T(p). Proof. Consider the collection C = {B(p): p ∈ P}. Since all sets in C are convex, balanced, and (openly) absorbing, by Theorem 1 from LCVS I, there exists a unique locally convex topology T, for which the collection1

U = {εB(p): ε > 0, p ∈ P} (3)

is a fundamental system of T-neighborhoods of 0. To prove (i) we simply notice that, since every p ∈ P can be written (by Remark 1) as the Minkowski functional of its unit ball p = qB(p), and B(p) is by construction a convex balanced T-neighborhood of 0, the contituity of p follows from Proposition 1(i). To prove (ii), fix another locally convex topology T0 such that all p ∈ P are T0-continuous, and let us show that T0 ⊃ T. Since both T and T0 are linear, all we need to show is that:

1 By Remark 5, εB(p) = Up,ε.

3 (∗) every T-neighborhood of 0 is also a T0-neighborhood of 0.

Since T has U as a fundamental system of T-neighborhoods for 0, the above condition is equivalent to the condition that every set in U is a T0-neighborhood of 0. Using (3) and the fact that dilations are in linear topologies we now see that (∗) is in fact equivalent to the following.

(∗∗) For every p ∈ P, the unit ball B(p) is a T0-neighborhood of 0.

But (∗∗) is clearly true, since all p ∈ P are T0-continuous (so the B(p)’s are in fact open neighborhoods of 0.) Remark 8. With note notations as above, if we consider, for every p ∈ P, ε > 0, the set

Wpε = {x ∈ X : p(x) ≤ ε} = εB(p), then the collection W = {Wpε : p ∈ P, ε > 0} also constitutes a fundamental system of T-neighborhoods of 0. This is pretty obvious from the inclusions Upε ⊂ Wpε ⊂ Up,2ε. Remark 9. Every locally convex topology on a vector space X can be constructed as T(P), for a suitably chosen collection P of seminorms on X . More precisely, if S is a locally convex topology, we can simply take P to be the collection of all S-continuous seminorms. On the one hand, by Theorem-Definition 1, since all p ∈ P are S-continuous, we automatically have the inclusion S ⊃ T(P). On the other hand,

(∗) every S-neighborhood of 0 is also a T(P)-neighborhood of 0, so we also have the inclusion S ⊂ T(P). Property (∗) can be proven as follows. Start with some S-neighborhood V of 0, and choose a balanced open A ⊂ V. By Proposition 1(i), the Minkowski functional qA is S-continuous, hence qA belongs to P. By Remark 2, A = B(qA), so by the definition of T(P), it follows that A is a T(P)-neighborhood of 0, and so will be V ⊃ A. Remark 10. If a locally convex space X has its topology defined by a family P of semi- norms, the for a (xλ)λ∈Λ in X and some x ∈ X , the following conditions are equivalent:

(i) xλ → x;

(ii) p(xλ − x) → 0, ∀ p ∈ P. Of course, the implication (i) ⇒ (ii) is trivial. Conversely, if condition (ii) holds, then for every p ∈ P and every ε > 0, there exists λp,ε ∈ Λ, such that p(xλ − x) < ε, ∀ λ λp,ε. Equivalently, using the notations from Theorem-Definition 1, it follows that for every U ∈ U, there exists λU such that xλ ∈ U + x, ∀, λ λU , and since by construction the collection {U + x : U ∈ U} is a fundamental system of neighborhoods for x, it follows that (xλ) indeed converges to x.

4 Remark 11. Given a family P of seminorms on X , the locally convex topology T(P) is Hausdorff, if and only if the “null set” N (P) = {x ∈ X : p(x) = 0, ∀ p ∈ P} is equal to the singleton set {0}. This follows from the obvious equality T εB(p) = {x ∈ X : p(x) = 0}, T ε>0 which implies: N (P) = p∈P εB(p), and then the desired equivalence is a consequence of ε>0 Remark 1 from LCVS I. It should be noted that, without any additional assumptions on P the null set N (P) is a T(P)-closed , namely the {0} of 0. Comment. The whole point of Theorem-Definition 1 (and of Theorem 1 from LCVS I) is that, in many instances, the collection P (or C from the Theorem) can be chosen to be (very) small. Therefore the following question is important: Given two collections P and Q of seminorms on X , when does one have the inclusion T(Q) ⊂ T(P)? Of course, by Theorem-Definition 1, the following conditions are equivalent:

• T(Q) ⊂ T(P);

• every q ∈ Q is T(P)-continuous.

What we want, however, is to give a characterization which is intrinsic to P and Q. One such formulation is contained in the following result. Proposition 2. Let X be a vector space and let T = T(P) be the locally convex topology defined by a family P of seminorms on X . For a seminorm q on X , the following are equivalent:

(i) q is T-continuous;

(ii) there exists p1, . . . , pn ∈ P and t1, . . . , tn ≥ 0, such that:

q(x) ≤ t1p1(x) + ··· + tnpn(x), ∀ x ∈ X . (4)

Proof. (i) ⇒ (ii). Assume p is continuous, so that the unit ball B(p) is an open convex balanced T- neighborhood of 0. Using Theorem-Definition 1 (and Remark 2), there exists p1, . . . , pn ∈ P, and ε1, . . . , εn > 0, such that

ε1B(p1) ∩ · · · ∩ εnB(pn) ⊂ B(q).

−1 −1 Consider the seminorm p = ε1 p1 + ··· + εn pn. The proof of (ii) will be finished once we show that q(x) ≤ p(x), ∀ x ∈ X . By Remark 4 this inequality is equivalent to the inclusion B(p) ⊂ B(q). Start then with some x ∈ B(p), i.e.

−1 −1 ε1 p1(x) + ··· + εn pn(x) < 1.

−1 In particular, for every k = 1, . . . , n we have the inequality εk pk(x) < 1, which is equivalent to pk(x) < εk, i.e. x ∈ εkB(pk). In other words, it follows that x ∈ ε1B(p1) ∩ · · · ∩ εnB(pn) ⊂ B(q), and we are done. (ii) ⇒ (i). Assume q satisfies (4), and let us show that q is T-continuous. We use Remark 7, so we start with a net (xλ)λ∈Λ in X which converges to 0, and let us prove that

q(xλ) → 0. (5)

5 With p1, . . . , pn as in (4), we know (by continuity) that pk(xλ) → 0, ∀ k = 1, . . . , n. The desired conclusion (5) then follows immediately from the inequalities

0 ≤ q(xλ) ≤ t1p1(xλ) + ··· + tnpn(xλ). Comments. A. Using an intrinsic terminology, one could say that a seminorm q is dominated by the seminorm family P, if it satisfies condition (ii) from Proposition 2. With this language, the above result simply says that q is T(P)-continuous, if and only if q is dominated by P. B. Suppose one starts with a collection P of seminorms on X . It is obvious that, for any sub-collection P0 ⊂ P, one has the inclusion T(P0) ⊂ T(P). If one wants then to replace P with P0, without changing the locally convex topologies they define, a necessary and sufficient condition is: every p ∈ P is dominated by P0. This observation is our main tool in producing small families of seminorms that define a given locally convex topology.

Example 1. If P = {p1, . . . , pn} is a finite collection of seminorms on X , then there exists one seminorm q, so that T(P) = T(q), namely q = p1 + ··· + pn. Obviously q is T(P)-continuous. Conversely, since pk ≤ q, it follows that all pk, k = 1, . . . , n are T(q) continuous. Comment. Going back to Theorem-Definition 1, it is useful to ask when the collection U introduced there is in fact a basic T(P)-neighborhood system for 0. As before, we want an intrinsic characterization. For this purpose, we introduce the following pre-order2 relation relation on the set of all seminorms: p q ⇔ ∃t > 0, such that tp ≥ q. Using this pre-order relation one can also introduce an defined by: p ∼ q ⇔ p q and q p. Using this terminology, one has the following result. Proposition 3. Using the notations from Theorem-Definition 1, the collection U is a basic system of T(P)-neighborhoods of 0, if and only if P is directed, in the sense that:

(d) for any p1, p2 ∈ P, there exists p ∈ P with p p1 and p p2. Proof. We already know from Theorem-Definition 1 that U = {εB(p): p ∈ P, ε > 0} is a fundamental system of neighborhood of 0. This means that the condition that U is basic is equivalent to the condition that U is a filter (i.e. for any U1, U2 ∈ U, there exists U ∈ U with U ⊂ U1 ∩ U2). Suppose U is a filter and let us show that P is directed. Start with two seminorms p1.p2 ∈ P. In particular the unit balls B(p1) and B(p2) both belong to U, so there exist p ∈ P and ε > 0, such that εB(p) ⊂ B(p1) ∩ B(p2) (6) Since, for any x ∈ X one has the equivalences x ∈ εB(p) ⇔ p(x) < ε ⇔ ε−1p(x) ⇔ x ∈ B(ε−1p),

2 What is missing from the definition of an honest order relation is: (p q and q p) ⇒ p = q.

6 −1 −1 we see that (6) forces the inclusions B(ε p) ⊂ B(p1) and B(ε p) ⊂ B(p2), so using Remark −1 4, we get p1, p2 ≤ ε p, thus p p1, p2. Conversely, assume P is directed, and let us prove that U is a filter. Start with two sets U1, U2 ∈ U, represented as U1 = ε1B(p1) and U2 = ε2B(p2), with p1, p2 ∈ P and ε1, ε2 > 0. Use the directedness of P to produce some p ∈ P and some t > 0, such that p1, p2 ≤ tp, and −1 −1 define ε = min{t ε1, t ε3}. To finish the proof, we will show that

εB(p) ⊂ ε1B(p1) ∩ ε2B(p2).

−1 Start now with some x ∈ εB(p). We have, of course the inequality p(x) < ε ≤ t εk, which implies tp(x) < εk, k = 1, 2. By the choice of p this yields pk(x) < εk, k = 1, 2, so x indeed belongs to the intersection ε1B(p1) ∩ ε2B(p2). Example 2. Given an arbitrary collection P of seminorms, we can always find a directed  P one, which defines the same topology. Define Q = p∈F p : F ⊂ P finite . Clearly Q is directed, it contains P, hence T(P) ⊂ T(Q), and furthermore, every q ∈ Q is dominated by P, so we also have the other inclusion T(Q) ⊂ T(P). Let us also remark that, if P is infinite, then card Q = card P. Of course, when P is finite, Q can be replaced by a singleton, as discussed in Example 1. Exercises 1-3 In the following three problems (X , T) is a locally convex space.

1♥. Assume T is defined by a collection P of seminorms.

(i) Prove that a set S ⊂ X is bounded, if and only if

sup p(x) < ∞, ∀ p ∈ P. x∈S

T (ii) Prove that, if S ⊂ X is bounded, then the closure conv(bal S) is also bounded. 2. Let Y ⊂ X be a linear subspace and let π : X → X /Y denote the quotient map.

(i) Show that, if p is a seminorm on X , the functionalp ˆ : X /Y → [0, ∞), defined by

pˆ(v) = inf{p(x): x ∈ X , π(x) = v}, v ∈ X /Y

is a seminorm on the quotient space X /Y. (ii) Show that, if p is continuous, thenp ˆ is continuous with respect to the quotient topology. (iii) Show that, if the topology T (on X ) is defined by a directed family P of seminorms (on X ), then the quotient topology (on X /Y) is defined by the family Pˆ = {pˆ : p ∈ P}.

Exercises 3-5 In the following three problems we start with a family (Xi)i∈I of locally Q convex vector spaces. We denote the product space Xi by Y and we equip it with the L i∈I . We denote the direct sum i∈I Xi by X and we equip it with the locally convex sum topology.

7 3. Suppose for each i ∈ I the topology on Xi is defined by a family Qi of seminorms (on Xi). Prove that all the maps in the set [ Q = {q ◦ πi : q ∈ Qi} i∈I are continuous seminorms on the product space Y, and furthermore, the product topol- ogy is T(Q).

4. Denote, for each i ∈ I, by the collection of all continuous seminorms on Xi, and let Q P = i∈I Pi. For each p = (pi)i∈I ∈ P, definep ˜ : X → [0, ∞) by X p˜(x) = pi(xi), ∀ x = (xi)i∈I ∈ X . i∈I

(By the definition of the direct sum, xi 6= 0 only for finitely many i’s, so the above sum has in fact only finitely many non-zero terms.) Show that

(i) for every p ∈ P the mapp ˜ is a semniorm on X ; (ii) the collection P˜ = {p˜ : p ∈ P} defines the locally convex sum topology; (iii) P is directed.

5*. Assume all Xi, i ∈ I, are Hausdorff. Use the notations from Exercise ?? from LCVS II. Prove that a subset S ⊂ X is bounded, if and only if there exists a finite set F ⊂ I, such that S ⊂ X (F ), and S is bounded in X (F ). (Hint: Argue by contradiction, assuming that, for every finite set F ⊂ I, there is some x = (xi)i∈I ∈ S and some j ∈ I r F , such that xj 6= 0. Reduce the statement to the case when I = N, as follows. Using the above assumption, there exists a countable infinite subset J ⊂ I, such that, for every j ∈ J, there exists x = (xi)i∈I ∈ S with that xj 6= 0. Use L L the restriction map R : i∈I Xi → j∈J Xj and replace S with R(S). Assuming now I = N, construct, for each k ∈ N, a continuous seminorm pk on Xk, such that sup{pk(xk): x = (xn)n∈N ∈ S} ≥ k. If one considers the system p = (pn)n∈N ∈ P, then sup{p˜(x): x ∈ S} = ∞, thus a contradiction is reached.) L Exercises 6-9. Let I be some non-. Consider the spaces X = i∈I K and Q I Y = i∈I K(= K ).

6. For every i ∈ I, let pi : Y → [0, ∞) be the map

qi(x) = |xi|, ∀ x = (xi)i∈I ∈ Y.

Show that qi are seminorms on Y.

7. Show that the collection Q = {qi}i∈I defines the product topology Tprod on Y. 8. Show that X is dense in Y in the product topology.

8 I 9. Define, for every I-tuple t = (ti)i∈I ∈ [0, ∞) , the map pt : X → [0, ∞) by X pt(x) = ti|xi|, ∀ x = (xi)i∈I ∈ X . i∈I

(As in Exercise 5, only finitely many terms in the above sum are non-zero.) Show that:

I (i) pt is a seminorm on X , for every t ∈ [0, ∞) ; I (ii) the collection P = {pt : t ∈ [0, ∞) } defines the locally convex sum topology Tsum on X ; (iii) P is directed.

We conclude this sub-section with a discussion of completeness for locally convex case. These exercise rely on the material covered in TVS IV. Exercises 10-12. Assume (X , T) is a locally convex .

10. Suppose p is a continuous seminorm on X .

(i) Prove that p is uniformly continuous (see TVS for the definition), when regarded as a map p : X → R. (ii) Consider the mapp ˜ : X˜ → R constructed in Exercise 14 from TVS IV (upon identifying the completion R˜ with R). Show thatp ˜ is a continuous seminorm on X˜.

Assume the topology on X is defined by a family P of seminorms.

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

(i) (xλ)λ∈Λ is Cauchy;

(ii) for every p ∈ P, and every ε > 0, there exists λp,ε ∈ Λ, such that p(xλ − xµ), ∀ λ, µ λp,ε. 12. With the notations as in Exercise 10, show that, the locally convex topology on X˜, ˜ ˜ ˜ defined by the collection P = {p˜}p∈P, coincides with T. In particular T is a locally convex topology.

B. Metrizability The special case, when a Hausdorff locally convex topology is defined by a countable collection of seminorms, is particularly interesting in the light of the following result. Theorem 2 (Metrizability Theorem). For a locally convex 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.

9 (ii) The topology T is first countable3.

(iii) There exists a countable collection P of seminorms which defines T.

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. We already know from TVS IV B that (i) ⇔ (ii), so we only need to prove the equivalence (ii) ⇔ (iii). ∞ (ii) ⇒ (iii). Assume 0 has a countable basic T-neighborhood system A = {An}n=1. 4 Without any loss of generality we can assume that all An’s are open, convex, and balanced. ∞ Consider then the Minkowski functionals P = {qAn }n=1. By Proposition 1, every seminorm in P is T-continuous, so by Theorem-Definition 1 we have the inclusion T ⊃ T(P). To prove the other inclusion, we must show that (see Remark 9 and the Comment preceding Propo- sition 2): all T-continuous seminorms are dominated by P. Start with some T-continuous seminorm p. The unit ball B(p) = {x ∈ X : p(x) < 1} is then an open convex balanced neighborhood of 0, so by construction there is some n ∈ N such that An ⊂ B(p). Of course

(by Proposition 1), An coincides with the unit ball B(qAn ), and then (by Remark 4) the inclusion B(qAn ) ⊂ B(p) yields the inequality p ≤ qAn . (iii) ⇒ (ii). This is quite trivial, since given a family P of seminorms that define the topology, the collection  1 V = n B(p): p ∈ P, n ∈ N clearly constitutes a fundamental system of neighborhoods of 0. Of course, if P is countable, then so is V. The two exercises below provide an explicit construction of a metric d with the properties discussed in the Metrizability Theorem. Exercises 13-14. Suppose (X , T) is metrizable locally convex , whose ∞ topology is defined by the family of seminorms P = {pn}n=1.

13. Prove that, for every n, the map dn : X × X → [0, 1) defined by:

pn(x − y) dn(x, y) = , ∀ x, y ∈ X , 1 + pn(x − y)

is a transaltion invariant contractive semi-metric5, in the sense that: Note that:

(i) dn(x, y) = dn(y, x) ≥ 0, ∀ x, y ∈ X ;

(ii) dn(x, y) ≤ dn(x, z) + dn(z, y), ∀ x, y, z ∈ X ;

3 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. 4 0 0 0 0 ∞ If each An is replaced by another neighborhood An of 0, with An ⊂ An, then A = {An}n=1 is again a basic system of neighborhoods. 5 The only condition that is missing from the definition of a metric is the equivalence dn(x, y) = 0 ⇔ x = y.

10 (iii) dn(x, y) = dn(x + z, y + z), ∀ x, y, z ∈ X ;

(iv) if |α| ≤ 1, then dn(αx, αy) ≤ dn(x, y), ∀ x, y ∈ X .

t (Hint: Consider the f(t) = 1+t , t ∈ [0, ∞] – with the convention f(∞) = 1, and show that f(s + t) ≤ f(s) + f(t), ∀ s, t ∈ [0, ∞]. Note also that f : [0, ∞] → [0, 1] is a strictly increasing .) 14. Use the notations as above, and define the the map d : X × X → [0, 1), by

∞ X dn(x, y) d(x, y) = , x, y ∈ X . 2n n=1 (i) Prove that d satisfies properies (i)-(iv) from the preceding exercise. (ii) Prove that d is a translation invariant metric on X , whose metric topology coin- cides with T. The next two exercises are somehow related to the previous two. As we shall see, besides the metric constructed in Exercise 13-14, there is another natural “candidate,” which fails, however, due to some very subtle obstructions. Exercises 15-16. Suppose (X , T) is a metrizable locally convex topological space. ∞ Therefore its topology can be defined by a countable family P = {pn}n=1 of seminorms. De- ∞ fine, for every n ∈ N the seminorm qn = p1 +···+pn. (Obviously the collection Q = {qn}n=1 defines the same locally convex topology.) Define the map δ : X → X → [0, 1] by q (x − y) δ(x, y) = lim n . n→∞ 1 + qn(x − y) 15. Prove that: (i) δ is a translation invariant metric on X ;

(ii) the metric topology Tδ is stronger than T;

(iii) the addition X × X 3 (x, y) 7−→ x + y ∈ X is Tδ-continuous.

16. Give an example in which the multiplication K × X 3 (α, y) 7−→ αx ∈ X is not Tδ-continuous. Conclude that, in general, (X , Tδ) may fail to be a topological vector space. Metrizability appears naturally in the presence of bounded sets, as indicated in the fol- lowing two exercises. Exercises 17-18 Assume (X , T) is a locally convex topological vector space (in particular T is Hausdorff). 17♥. Prove that the following are equivalent: (i) There exists a bounded neighborhood of 0;

(ii) There exists some neighborhood V of 0, such that {εV}ε>0 is a basic neighborhood system for 0.

11 (iii) There exists one seminorm p, so that T = T(p). Moreover, any seminorm with this property is a , i.e. one has the implication: p(x) = 0 ⇒ x = 0.

In particular, (X , T) is metrizable. Prove that d(x, y) = p(x − y) defines a translation invariant metric, whose metric topology coincides with T.

18♥. Suppose B ⊂ X is non-empty, convex and balanced. S (i) Prove that Z = span B can also be represented as: Z = t∈[0,∞) tB = S∞ n=1 nB. In particular, B is absorbing in Z, so we can consider its associated Minkowski functional qB on Z, defined by

qB(z) = inf{t > 0 : z ∈ tB}.

(ii) Prove that, if B is bounded, then qB is a norm on Z, which from now on we will denote by k . kB. norm (iii) Prove that the locally convex topology T on Z, defined by k . kB is stronger B than the T Z . In other words, for a net (zλ)λ∈Λ ⊂ Z, one has T the implication kzλkB → 0 ⇒ zλ −→ 0.

C. Frechet spaces

As we have already seen in TVS IV, the metrizability problem is often associated with the completeness problem. Definition. A locally convex topological space X is called a Frechet space, if it is an (F)-space, that is, X is metrizable and complete. Remarks 12-14. Based on the results from TVS IV and LCVS II, the following state- ments yield three general methods of constructing Frechet spaces:

12. If X is a Frechet space, then: a linear subspace Y ⊂ X , equipped with the induced topology is a Frechet space, if and only if Y is closed in X . (See Remark 1’ from TVS IV and Example 1 from LCVS II.)

13. If (Xj)j∈J is a family of metrizable locally convex topological vector spaces, then: the Q product space j∈J Xj, equipped with the product topology, is an Frechet space, if and only if all Xj’s are Frechet spaces, and the index set J is countable. (Remark 2’ from TVS IV and Example 2 from LCVS II.) In particular, all finite dimensional topological vector spaces are Frechet spaces.

14. If X is a Frechet space, and Y ⊂ X is a closed linear subspace, then the quotient space X /Y, equipped with the quotient topology, is a Frechet space. (See Remark 3’ from TVS IV and Proposition 1 from LCVS II.)

12 One other method of constructing normed Frechet spaces6 is based on Exercise 18, in connection with which we introduce the following terminology. Definition. Suppose X is a vector space, and B is a non-empty, convex, balanced subset, so that by Exercise 18 (i), B is absorbing in Z = span B. In particular the Minkowski 0 0 functional qB is defined on Z. Define dB : Z × Z → [0, ∞), by dB(z, z ) = qB(z − z ). We say that B is self-complete, if it satisfies the following two conditions.

(i) qB is a norm on Z, or equivalently, dB is a metric on Z. In this case, qB will be denoted by k . kB, and will be referred to as the B-norm.

(ii) (B, dB) is a complete .

norm In this case, the locally convex topology on Z, defined by it, will be denoted by TB , and will be referred to as the B-norm topology. Remarks 15-16. Use the notations as above, and assume that B ⊂ X is non-empty, convex and balanced. T 15. Condition (i) above is equivalent to: t>0 tB = {0} (see CW II). By Exercise 18 (ii), this can be achieved, for instance, if B is bounded relative to some Hausdorff linear topology on X .

16. Assuming now condition (i), we see that condition (ii) is equivalent to either one of the following conditions:

(iii) (Z, dB) is a , and B is closed in Z, relative to the metric topology;

(iii’) (Z, dB) is a complete metric space, and

B = {z ∈ Z : kzkB ≤ 1}. (7)

norm In particular, (Z, TB ) is a Frechet space. It is obvious that (iii0) ⇒ (iii) ⇒ (ii), so we only need to justify the implication (ii) ⇒ (iii0). This follows from the well known features of the Minkowski functional (see CW II), which using the above notations yield the inclusions:

{z ∈ Z : kzkB < 1} ⊂ B ⊂ {z ∈ Z : kzkB ≤ 1}.

If we start with some element z ∈ Z, such that kzkB, then clearly the (zn) 1 given by zn = (1 − n )z belongs to B and is convergent in norm to z, so by the completeness of B, it follows that z belongs to B, thus proving (7). To prove that ∞ (Z, dB) is complete, start with some sequence (zn)n=1 ⊂ Z is Cauchy, relative to dB, 7 and notice that there exists some t > 0, such that kznkB ≤ t, ∀ n. Obviously the 1 ∞ sequence ( t zn)n=1 is Cauchy in (B, dB), thus convergent to some b ∈ B, and then (zn) will clearly converge to tb in (Z, dB).

6 Later on, this terminology will be replaced: normed Frechet spaces will be called Banach spaces 7 ∞ Since kzmkB − kznkB ≤ kzm − znkB = dB(zm, zn), we see that the sequence (kznkB)n=1 ⊂ [0, ∞) is Cauchy in R, thus bounded.

13 Exercise 19♥. Let (X , T) be a locally convex topological vector space. Prove that, if B is a non-empty, convex, balanced, and T-compact, then B is self-complete. Comment. In connection with the preceding exercise, the reader is warned that on the space Z = span B we now have two topologies: (i) the B-norm topology Tnorm, and (ii) the B induced topology T , which, in general, is strictly weaker than the norm topology. One Z way to see this is by observing that B is compact in (Z, T Z ), but cannot be compact in norm (Z, TB ), unless Z is finite dimensional. (See TVS III.) L Exercise 20*. Consider the space X = i∈I K, equipped with the locally convex sum topology Tsum. (See Exercise 6-9.) Prove that, if I is infinite, then (X , Tsum) is not metrizable. (Hint: Without any loss of generality, one can assume that I = N. This follows from LCVS II. Show that every in (X , Tsum) is bounded, thus by Exercise 5* convergent. Now if (X , Tsum) were metrizable, it would be a Frechet space, and at the same time a countable of finite dimensional subspaces. Use Baire’s Theorem to reach a contradiction.)

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