Locally Convex Vector Spaces II: Natural Constructions in the Locally Convex Category

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1 LCVS II c Gabriel Nagy Locally Convex Vector Spaces II: Natural Constructions in the Locally Convex Category Notes from the Functional Analysis Course (Fall 07 - Spring 08) Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. This section is divided into two parts. In the first part we review those general constructions from TVS II and we see that those fit nicely within the framework of locally convex spaces. In the second part we introduce some new constructions, that are analogous to some standard ones from Topology (push-forwards and direct sums), but were not available in the general framework of linear topologies (see the discussion at the end of TVS II). As it turns out, Theorem 1 from LCVS I will be an essential ingredient in these constructions. The first construction in TVS II was the quotient topology, for which we now have the following result (see Proposition-Definition 1 from TVS II for notations): Proposition 1. If is a locally convex vector space and Y is a linear subspace, then, when equipped with the quotient topology, the quotient space /Y is also locally convex. Proof. Let π : /Y denote the quotient map. Start with some neighborhood V of 0 in /Y, and let us indicate how to construct an open convex set B /Y, with 0 A V. First of all, since V is a neighborhood of 0 and π is continuous, the pre-image π 1 (V) will be a neighborhood of 0 in. Secondly, since is locally convex, there exists an open convex set A, with 0 A π 1 (V). Now we are done, by taking B = π(a), which is convex and open (since π is open). The next construction from TVS II was the joint pull-back (see Proposition-Definition 3 from TVS II for notation and terminology). Proposition 2. Suppose (Y i ) i I is a family of locally convex vector spaces, and a vector space is given, together with a system Ψ = ( Y i ) i I of linear maps. Let T be the topology obtained as the joint Ψ-pull-back, that is, T is the weakest among all topologies on, that make all maps ψ i : Y i continuous. Then T is locally convex. Proof. We know from the construction joint pull-back that S = i I ψ i { ψ 1 (A) : A open in Y i } constitutes a sub-base for T. As in the previous proof, we start with some T-neighborhood V of 0 in, and we wish to construct some convex T-open subset B in, with 0 B V. 1

2 Since S defined above is a sub-base, there exists a set S, which is a finite intersection of sets in S, such that 0 S V. In other words, there exist indices i 1,..., i n I, and open sets A k ik, k = 1,..., n, such that 0 ψ 1 (A k ) V. Obviously, since the ψ ik s are linear, each A k is an open neighborhood of 0 in ik, so by local convexity, for each k = 1,..., n, there exists an open convex set C k in ik, with 0 C k A k. By linearity, the sets ψ 1 (C k ) ψ 1 (A k ) are then convex and T-open, and so will be their intersection B = ψ 1 (C k ) ψ 1 (A k ) V, which is also a neighborhood of 0. Example 1. Suppose (Y, T) is locally convex, and Y is a linear subspace. If we apply the (joint) pull-back construction to the collection consisting of a single map the inclusion ι : Y we obtain the induced topology T = {A : A T-open }, which is again locally convex. We also know that if T is Hausdorff, then so is T. Example 2. If we start with a collection ( i ) i I of locally convex spaces, the product topology on = i I i is constructed as the joint Π-pull-back, where Π = ( π i i ) i I are the coordinate maps. By Proposition 2, the product topology is locally convex. Moreover, is Hausdorff, when equipped with this topology, if and only if all the i are (see TVS II). As noticed in TVS II (the end comments) one important construction from Topology does not carry on directly to the case of linear topologies. As we shall shortly see, it is possible to correct this deficiency if we work with locally convex topologies. Theorem-Definition 1. Suppose ( i ) i I is a family of locally convex vector spaces, and a vector space Y is given, together with a system Φ = ( i Y)i I of linear maps. Let { C = C Y C balanced, absorbing, convex (C) neighborhood of 0 in i, i I φ 1 i φ i }. (1) (i) There exists a unique locally convex topology T on, which has C as a fundamental system of neighborhoods for 0. (ii) T (defined above) is the strongest among all locally convex topologies on, that make all maps φ i : i Y continuous. (iii) Given a locally convex space space Z, a linear map θ : Y Z is continuous relative to T (defined above), if and only if all the compositions θ φ i : i Y are continuous. The topology T is referred to as the joint locally convex Φ-push-forward topology. 2

3 Proof. (i). By linearity of the φ i s and the definition of C, it is quite obvious that C C, then εc C, ε > 0, so the existence and uniqueness of T follows immediately from Theorem 1 from LCVS I. (ii). Let us show first that all φ i : i Y, i I are T-continuous. By linearity we only need to show that φ i is continuous at 0. Since C is a fundamental system of T-neighborphoods of 0, all we need is the fact that, for each C C, the pre-image φ 1 i (C) is a neighborhood of 0, for each i I. But this is trivial by construction. Let S be another locally convex topology on Y, which makes all the φ i s continuous, and let us prove the inclusion S T. Start with some S-open set A, and let us show that A is also T-open. To get this, we are going to prove that ( ) A is a T-neighborhood of every one of its points. Fix some a A. Since A is S-open, there exists some convex, balanced, S-open set C, such that C + a A. (2) Since all φ i : i Y are continuous with respect to S, it follows that φ 1 i (C) is open in i, ini. Therefore, by (1), C belongs to C. In particular C is a T-neighborhood of 0, so by (2), A is indeed a T-neighborhood of a. (iii). Let Z be a locally convex space, and let θ : Y toz be linear. Assume first θ is T-continuous. Then by (ii) the compositions θ φ i, i I, are continuous. Conversely, assume all these compositions are continuous, and let us prove that θ is T-continuous. Let S be the θ-pull-back of the topology from Z, that is S = {θ 1 (A) : A open in Z}. (3) By Proposition 2, S is a locally convex topology on Y. Notice that, by construction, if we start with some B S, written as B = θ 1 (A), for some open set A Z, we have φ 1 i (B) = (θ φ i ) 1 (A), so by our assumption, it follows that φ 1 i (B) is open in i, for every i I. In other words, S is a locally convex topology on Y, which makes all the φ i s continuous, so by (ii) we must have the inclusion S T. But now (3) yields which means that θ is indeed continuous. θ 1 (A) T, for every open set A Z, Example 3. The above construction is interesting even in the case when the system Φ consists of a singleton, that is, we start with one linear map φ : Y and with a locally convex topology S on. If we apply the locally convex push-forward, we get a locally convex topology T, which is the strongest among all locally convex topologies on Y which makes φ continuous. The reader is warned that there exists another topology (not necessarily locally convex!) which is the strongest topology that makes φ continuous, namely the topological push-forward φ S = {A Y : φ 1 (A) S}. 3

4 In general one only has the inclusion T φ S. (Any set A disjoint from φ( ) is in φ S, but not all such A s are in T. For instance A can be a singleton!) Exercises With the notations as in the Example above, prove that φ S = T, if and only if φ is surjective. (In this case T is just the quotient topology.) 2. Suppose (, S) is locally convex, and is contained as a linear subspace in a (bigger) vector space Y. Let T be the locally convex topology on obtained by the locally convex (single) push-forward described in Example 3, using the inclusion map φ : Y, so that by construction T is the strongest locally convex topology on Y such that T S. (i) Show there exists a locally convex topology T 0 on Y, such that T 0 = S. Conclude that T = S. (ii) Show that, if S is Hausdorff, then so is T. Example 4. Let ( i ) i I be a collection of locally convex spaces and consider their direct sum 1 = i I i, described as the vector space of all I-tuples x = (x i ) i I i I i, for which the support set supp x = {i I : x i 0} is finite. For every i, j I, let φ ij : i j be either the identity (if i = j), or the zero (if i j) map. Using these maps we then define the standard inclusions φ i : i by φ i (x) = ( φ ij (x) ) j I, x i. The locally convex joint push-forward topology on, associated with the system Φ = (φ i ) i I is called the locally convex sum topology. Exercises 3-7. Let ( i ) i I,, and φ i : i be as above, and let T sum denote the locally convex sum topology. Denote the product space i I i by Y, and let T prod denote the product topology. 3. Prove that T prod T sum. Conclude that, if all the i are Hausdorff, then so is. 4. Assume, for each i I, an open set A i i is given, and let A = i I A i Y. (i) Show that A T sum. (i) Show that, if A i i, for infinitely many i s, then A T prod. Therefore, it this case, one has a strict inclusion T prod T sum 5. Suppose (E i ) i I is another collection of locally convex spaces. Let E = i I E i be equipped with the locally convex sum topology T sum, and let the product space F = i I E i be equipped with the product topology T prod. Show that, if for every i I a linear continuous map φ i : E i i is given, then (i) the product map φ = i I φ i : (F, T prod ) (Y, T prod ) is continuous; 1 See TVS II. 4

5 (ii) the restriction φ E : (E, T sum ) (, T sum ) is also continuous. 6. Define, for every subset J I, the set (J) = {(x i ) i I i I i : x i = 0, i J}. (i) Show that, when we equip (J) with the induced topology T sum (J) and the sum j J j with the locally convex sum topology, the obvious linear isomorphism (J) j J j is a homeomorphism. (ii) Show that, if all i are Hausdorff, then (J) is a closed linear subspace both in (, T sum ) and in (Y, T prod ). (iii) Show that the restriction map R J : i I i (x i ) i I (x j ) j J j J j is continuous. 7. Show that, when I is finite, in which case = Y, the two topologies T sum and T prof coincide. 5