Noncomplete Mackey Topologies on Banach Spaces
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Bull. Aust. Math. Soc. 81 (2010), 409–413 doi:10.1017/S0004972709001154 NONCOMPLETE MACKEY TOPOLOGIES ON BANACH SPACES JOSÉ BONET ˛ and BERNARDO CASCALES (Received 28 March 2009) Abstract Answering a question of W. Arendt and M. Kunze in the negative, we construct a Banach space X and a norm closed weak* dense subspace Y of the dual X 0 of X such that X, endowed with the Mackey topology µ.X; Y / of the dual pair hX; Y i, is not complete. 2000 Mathematics subject classification: primary 46B10; secondary 46B50, 46A03. Keywords and phrases: Banach spaces, Mackey topologies, norming subspaces, Krein–Smulyan theorem. The following problem appeared in a natural way in connection with the study of Pettis integrability with respect to norming subspaces developed by Kunze in his PhD thesis T5U. This question was put to the authors by Kunze himself and his thesis advisor W. Arendt. PROBLEM. Suppose that .X; k·k/ is a Banach space and Y is a subspace of its topological dual X 0 which is norm closed and weak* dense. Is there a complete topology of the dual pair hX; Y i in X? We use the locally convex space (lcs) notation as in T4, 6, 7U. In particular, σ .X; Y / and µ.X; Y / denote the weak topology and the Mackey topology in X associated with the dual pair hX; Y i. For a Banach space X with topological dual X 0, the weak* topology is σ .X 0; X/. By the Bourbaki–Robertson lemma T4, Section 18.4.4U there is a complete topology in X of the dual pair hX; Y i if and only if the space .X; µ.X; Y // is complete. Therefore, the original question and the following are equivalent. PROBLEM A. Let .X; k·k/ be a Banach space. Is .X; µ.X; Y // complete for every norm closed weak* dense subspace Y of the dual space X 0? Let .X; k·k/ be a normed space. A subspace Y of X 0 is said to be norming if 0 0 the function p of X given by p.x/ D supfjx .x/j V x 2 Y \ BX0 g is a norm equivalent The research of Bonet was partially supported by FEDER and MEC Project MTM2007-62643 and by GV Prometeo/2008/101. The research of Cascales was supported by FEDER and MEC Project MTM2008- 05396 and by Fundacion´ Seneca´ de la CARM, project 08848/PI/08. c 2010 Australian Mathematical Publishing Association Inc. 0004-9727/2010 $16.00 409 Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.76, on 02 Oct 2021 at 11:54:59, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972709001154 410 J. Bonet and B. Cascales [2] to k·k. We notice that Problem A is not affected by changing the given norm of X to any equivalent one. Thus, to study Problem A for some norming subspace Y ⊂ X 0 we can and will assume that Y is indeed 1-norming, that is, kxk D supfjx0.x/j V x0 2 Y \ BX0 g. Let us observe that under the conditions of Problem A, if .X; µ.X; Y // is quasi-complete (in particular, complete), then the Krein–Smulyan theorem (see T4, Section 24.5(4)U) implies that for every σ .X; Y /-compact subset H of X, the σ .X; Y /- σ .X;Y / closed absolutely convex hull M VD aco H of H is also σ .X; Y /-compact. There are several papers dealing with the validity of the Krein–Smulyan theorem for topologies weaker than the weak topology; see, for instance, T1, 2U where it is proved that for every Banach space X not containing `1.T0; 1U/ and for every 1-norming subspace Y ⊂ X 0, if H is a norm bounded σ .X; Y /-compact subset of σ .X;Y / X then aco H is σ .X; Y /-compact. It was proved in T3U that the hypothesis `1.T0; 1U/ 6⊂ X is also necessary for the latter. The following useful observation will be used several times later. PROPOSITION 1. Let .X; k·k/ be a Banach space and let Y be a 1-norming subspace of X 0. If .X; µ.X; Y // is quasi-complete, then every σ .X; Y /-compact subset of X is norm bounded. PROOF. Let H ⊂ X be σ .X; Y /-compact. As already noted, the Krein–Smulyan theorem T4, Section 24.5(4)U implies that the σ .X; Y /-closed absolutely convex hull σ .X;Y / M VD aco H is σ .X; Y /-compact. Therefore, M is an absolutely convex, bounded and complete subset of the lcs .X; σ .X; Y //. Now we can apply T4, Section 20.11(2)U to obtain that M is a Banach disc, that is, X VD S nM is a M n2N Banach space with the norm kxkM VD inffλ ≥ 0 V x 2 λMg; x 2 X M : Since M is bounded in .X; σ .X; Y //, the inclusion J V X M ! .X; σ .X; Y // is continuous, therefore J V X M ! .X; k·k/ has a closed graph, hence it is continuous by the closed graph theorem. In particular, the image of the closed unit ball M of X M is bounded in .X; k·k/, and the proof is complete. 2 The following example is an immediate consequence of the foregoing. EXAMPLE 2. Let X D C.T0; 1U/ be endowed with its supremum norm and take 0 Y VD spanfδx V x 2 T0; 1Ug ⊂ X : Then .X; µ.X; Y // is not quasi-complete. PROOF. Notice that σ .X; Y / coincides with the topology τp of pointwise convergence on C.T0; 1U/. Since there are sequences τp-convergent to zero which are not norm bounded, .X; µ.X; Y // cannot be quasi-complete by Proposition 1. 2 Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.76, on 02 Oct 2021 at 11:54:59, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972709001154 [3] Noncomplete Mackey topologies on Banach spaces 411 The subspace Y of X 0 in Example 2 is weak* dense in X 0 but not norm closed. Another example of the same nature is the following: take X D c0, Y D ', the space of 0 sequences with finitely many nonzero coordinates, which is norm dense in X D `1. In this case µ.X; Y / D σ .X; Y /, since every absolutely convex σ .Y; X/-compact subset of Y is finite-dimensional by the Baire category theorem. In this case .X; σ .X; Y // is not even sequentially complete. The following example, taken from T3, Lemma 11U, provides the negative solution to Problem A. 1 EXAMPLE 3. Take X D .` .T0; 1U/; k·k1/ and consider the space Y D C.T0; 1U/ of continuous functions on T0; 1U as a norming subspace of the dual X 0 D `1.T0; 1U/. Then .X; µ.X; Y // is not quasi-complete. 1 PROOF. Let H VD fex V x 2 T0; 1Ug be the canonical basis of ` .T0; 1U/. The set H σ .X;Y / is clearly σ .X; Y /-compact but we will prove that aco H is not σ .X; Y /- compact, and therefore .X; µ.X; Y // cannot be quasi-complete. Indeed, proceeding σ .X;Y / by contradiction, let us assume that W VD aco H is σ .X; Y /-compact. We 0 write M.T0; 1U/ D .C.T0; 1U/; k·k1/ to denote the space of Radon measures in T0; 1U endowed with its variation norm. The map φ V X ! M.T0; 1U/ P ∗ given by φ((ξx /x2T0;1U/ D x2T0;1U ξx δx is σ .X; Y /-w -continuous. We notice that: (1) φ.W/ ⊂ φ.`1.T0; 1U//; (2) φ.W/ is an absolutely convex w∗-compact subset of M.T0; 1U/; (3) fδx V x 2 T0; 1Ug ⊂ φ.W/. From the above we obtain that ∗ w 1 BM.T0;1U/ D acofδx V x 2 T0; 1Ug ⊂ φ.W/ ⊂ φ.` .T0; 1U//; which is a contradiction because there are Radon measures on T0; 1U which are not of P the form x2T0;1U ξx δx . The proof is complete. 2 PROPOSITION 4. If X is a Banach space containing an isomorphic copy of `1.T0; 1U/, then there is a subspace Y ⊂ X 0 norm closed and norming such that .X; µ.X; Y // is not quasi-complete. PROOF. In the proof of T3, Proposition 3U the authors construct a norming subspace σ .X;E/ E ⊂ X 0 and H ⊂ X norm bounded σ .X; E/-compact such that aco H is not σ .X; E/-compact. If we take Y D E ⊂ X 0, norm closure, then norm bounded σ .X; E/-convergent nets in X are σ .X; Y /-convergent; hence we obtain that: (i) H ⊂ X is σ .X; Y /-compact; and σ .X;E/ σ .X;Y / (ii) aco H D aco H . Downloaded from https://www.cambridge.org/core. IP address: 170.106.35.76, on 02 Oct 2021 at 11:54:59, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0004972709001154 412 J. Bonet and B. Cascales [4] σ .X;Y / Consequently H is σ .X; Y /-compact and aco H is not.