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DUAL LOCAL COMPLETENESS 1. Introduction a Sequence Σ PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 125, Number 4, April 1997, Pages 1063{1070 S 0002-9939(97)03864-1 DUAL LOCAL COMPLETENESS STEPHEN A. SAXON AND L. M. SANCHEZ´ RUIZ (Communicated by Dale Alspach) Abstract. The 1971 articles in which Saxon-Levin and Valdivia indepen- dently proved their Theorem feature two conditions equivalent to dual local completeness. One became Ruess’ property (LC). The other is among new characterizations previously known only as necessary conditions. 1. Introduction A sequence σ = An : n N of absolutely convex subsets of a locally convex space E ( )isabsorbing{ in∈E if} it is increasing and each x E is absorbed by T ∈ some An;thenwedenoteby σ the finest locally convex topology on E that induces the same topology as on eachT A . (Cf. [1], [8].) The topology is defined by T n Tσ the family of those seminorms whose restrictions to the sets An are continuous for the topology induced on An by . In Valdivia’s development of the Saxon-Levin- Valdivia Theorem, he showed thatT = always holds when E ( ) is barrelled T Tσ T ([16], Theorem 5). Ruess [9] defined a space E ( )tohaveproperty (L) if = σ holds for each absorbing sequence σ, and to haveT the weaker property (LC)Tif eachT is compatible with the dual pair (E;E0). Thus Tσ barrelled property (L) property (LC). ⇒ ⇒ Furthermore, every barrelled space has its Mackey topology, and for any Mackey space E, it is clearly true that E has property (L) if and only if E has property (LC). Valdivia explicitly pointed out that [ barrelled property (LC)], and the latter property, according to Ruess (see below), is equivalent⇒ to dual local completeness. The crux of the Saxon-Levin proof was to show that every barrelled space E has the property that any countable-codimensional subspace spanned by a closed absolutely convex subset of E is, itself, closed ([15], 3, Lemma). We will show this property, also, is equivalent to dual local completeness,§ which thus provides an unexpected common footing for both papers. This and other new equivalences unify vital ideas already in book form (e.g., [5], [6], [7], [19], [22]). Every space will be assumed a Hausdorff locally convex space over the scalar field of real or complex numbers; “countable” means “finite or denumerable”. Received by the editors September 15, 1995. 1991 Mathematics Subject Classification. Primary 46A08. Key words and phrases. Dual locally complete, barrelled, properties (L) and (LC). This paper was started while the second author stayed at the University of Florida supported by DGICYT PR95-182, later by PR94-204 and IVEI 003/033. c 1997 American Mathematical Society 1063 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 1064 STEPHEN A. SAXON AND L. M. SANCHEZ´ RUIZ 2. Dual local completeness When σ = A , we will denote σ as A. Ruess ([11], Propositions 2.2, 2.3) asserts without{ proof} that a space ET ( ) hasT property (L) (respectively, (LC)) if T and only if, given any absorbing absolutely convex set A (respectively, any f E∗ ∈ such that f A is continuous), then = A (respectively, f E0). We prove Ruess’ characterizations| with new ones involvingT T `1. ∈ Recall that a space E is locally complete if and only if every bounded closed absolutely convex subset of E is a Banach disk [7]. E is `1-complete if, for each λ `1 and bounded sequence x E,theseries λ x converges in E. { n}n ∈ { n}n ⊂ n n n Theorem 2.1. AspaceEis locally complete if and onlyP if it is `1-complete. Proof. Suppose E is locally complete and we are given λ `1 and a bounded { n}n ∈ sequence xn n in E. The closed absolutely convex hull A of xn n is a Banach disk. Now{q>p} implies { } λ x λ x ( λ + :::+ λ )A: n n − n n ∈ | p| | q| n q n<p X≤ X Thus n λnxn converges in the Banach space EA, and so must also converge in the relatively coarser topology of E. Conversely,P suppose E is `1-complete and A is a closed disk in E, with y a { n}n Cauchy sequence in the normed space EA. There is an increasing sequence nk k k { k} ⊂ N such that yq yp 2− whenever p; q; k N with p; q nk.Ifλk=2− and k k − k≤ ∈ ≥ xk =2 ynk+1 ynk ,then xk k Ais bounded and n λnxn converges to some x in E.Givenj>k− ,wehave{ } ⊂ P k j k+1 λ x λ x 2− + :::+2− A 2− A: n n− n n ∈ ⊂ n j n<k X≤ X k+1 For k fixed, 2− A is closed in E,and k+1 x λnxn = lim λnxn λnxn 2− A: − j − ∈ n<k n j n<k X X≤ X Therefore as k increases, ynk = yn1 + n<k λnxn tends to yn1 + x in EA,asdoes, then, yn . We conclude that A is a Banach disk and E is locally complete. { }n P Obviously, every sequentially complete space is locally complete (cf. [7]). Theorem 2.2. Given a space E ( ), the following assertions are equivalent: T (1) (Ruess) E has property (L). (2) (Ruess) = A for any absorbing absolutely convex subset A of E. T 1T (3) If λn ` and pn : n N is a pointwise bounded sequence of continu- { }n ∈ { ∈ } ous seminorms on E,then n λn pn, pointwisely defined, is a continuous seminorm on E. | | P Proof. [(1) (2)] is clear. [(2) (3)].⇒ Let A = x E : p (x) 1 .Toseethatp= λp ⇒ n { ∈ n ≤ } n|n|n is continuous, we need only see that p A is continuous because of (2), and for T | P this we just check that p A is continuous at the origin (Garling; cf. [9], Lemma 3.3(1)). Given >0, choose| k such that λ <=2. Continuity supplies a n>k | n| 0-neighborhood W in E with n k λn pn (x) <=2forx W. Hence for each ≤ | | P ∈ P License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use DUAL LOCAL COMPLETENESS 1065 x W A we have p (x) n k λn pn (x)+ n>k λn <=2+/2=; i.e., p A is∈ continuous∩ at the origin.≤ ≤ | | | | | P P [(3) (1)]. Let σ = An : n N be an absorbing sequence of E ( ); then ⇒n { ∈ } T σ∗ = 2 An : n N is an absorbing sequence of E, also, and σ = σ∗ .Abaseof { ∈ } nT T 0-neighborhoods for σ is given by all the sets W = U0 2 An + Un : n N , T ∗ ∩ ∈ where Un : n =0;1;2;::: runs through all the sequences of absolutely convex neighborhoods{ of the origin} in E ( ). (See Lemma 1 of [8],T or 8.1.12 of [7].) We need T n show that any such W is a -neighborhood of the origin. If pn is the gauge of 2 An+ n T Un,then 2 pn:n N is a family of continuous seminorms defined on E,andis { ∈ } 1 n ∞ pointwise bounded since σ is absorbing. By (3), we have p = n=1 2n 2 pn is a 1 continuous seminorm on E. Therefore p− ([0; 1]) is a -neighborhood contained in 1 n T P pn− ([0; 1]) : n N 2 An + Un : n N ,sothatW is a -neighborhood of the origin. ∈ ⊂ ∈ T T T Recall that a space E is dual locally complete [17] if E0 (σ (E0;E)) is locally 1 1 complete. E is dual ` -complete if E0 (σ (E0;E)) is ` -complete. Ruess’ Final Remarks [10] imply that ([7], 8.1.29(i)) dual local completeness is equivalent to property (LC). And [E is barrelled] [E has property (S); i.e., E0 is 1 ⇒ σ (E0;E)-sequentially complete] E is dual ` -complete .DeWildeprovedthat every metrizable space with property⇒ (S) is barrelled. Saxon’s proof ([13], Theorem 2.7) actually used dual `1-completeness instead of property (S). Theorem 2.3. For any space E ( ) the following assertions are equivalent: T (1) (Ruess) E is dual locally complete. (2) E is dual `1-complete. (3) (Ruess) E has property (LC). (4) (Ruess) If A is an absorbing absolutely convex set and f E∗ with f ∈ |A continuous, then f E0. ∈ Proof. [(1) (2)]. By Theorem 2.1. ⇔ [(2) (3)]. Let Bn : n N be an increasing sequence of closed absolutely ⇒ { ∈ } convex sets covering E (cf. 8.1.17(i) of [7]) and suppose that f E∗ has a con- ∈ tinuous restriction on each B . We must show that f E0.Iff=0,thisis n ∈ trivial; otherwise, choose x E such that f (x) = 1. Then, for each n N,we 1 ∈ ∈n have Cn = Bn f − ( 0 ) is closed and absolutely convex and misses 2− x;the ∩ { } n bipolar theorem yields fn E0 with fn (x)=1and fn(y) <2− for every y Cn. ∈ | | n ∈ Since Bn : n N increases and covers E, the sequence 2 (fn+1 fn):n N { ∈ } { n n − ∈ } is σ (E0;E)-bounded, so that by (2) we have f = f + 2− [2 (f f )] is in 1 n n+1 − n E0. P [(3) (4)]. The sequence An = A n is absorbing since A is. ⇒ { 1 } [(4) (2)]. Given λ ` and a σ (E0;E)-bounded sequence f ,let ⇒ { n}n ∈ { n}n f E∗be the pointwise limit of λ f . To show that f E0, it suffices to show ∈ n n n ∈ that the restriction of f to A = fn : n N ◦ is continuous at the origin, using (4).
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