On the barrelledness of spaces of bounded vector functions

Arch. Math., Vol. 63, 449-458 (1994) 0003-889X/94/6305-0449 $ 3.50/0 9 1994 Birkhfiuser Verlag, Basel On the barreUedness of spaces of bounded vector functions By LECH DREWNOWSKI, MIGUEL FLORENCIO and PEDRO s PA(JL 1. Introduction. If S is a set and E a locally convex space, we denote by l~ (S, E) the locally convex space of all bounded functions from S to E equipped with the topology of uniform convergence on S. Of course, if E is a normed or metrizable space, so is lo~(S, E). In general, if E is barrelled, then lo~(S, E) need not even be quasi-barrelled; see Section 3 for more details. The origin of the results presented here was the following question: If E is a barrelled normed space, is also t~(S, E) barrelled? From the work of Lurje [16] it was known that the answer is 'yes' if S = N. Our initial idea of coping with this question for arbitrary sets S was to develop or modify a divide and conquer technique similar to those used in [8, 3.3 and 4.1], [5], and [6] to solve analogous problems for various spaces of measurable vector functions, usually defined over nonatomic measure spaces. As it turned out, however, this idea works only partially, namely when the cardinals of S or E are not too big. More explicitly, we obtained the following Theorem 1. Let S be a set and E a barrelled normed space. If i S l or IE t is nonmeasurable, then the space l~(S, E) is barrelled. Let us recall at this point that a cardinal number m is said to be (Ulam) measurable if, given a set Twith m = IT ], there exists a countably additive measure/~ : ~ ( T ) ~ {0, 1} such that #(T) = t and #({t}) = 0 for all t ~ T. This is equivalent to the existence of a free (or non-principal, Le. with c~q/= ~ ) ultrafilter q / o n Twhich is closed under the formation of countable intersections. (The map # ~ {A :/z(A) = 1} is a bijection between the set of #'s and the set of q/'s.) It is not known whether there exists a set with measurable cardinal; if such a set exists, then its cardinal must be strongly inaccessible. For more information on measurable cardinals, see [4] and [15]. At it will become clear later on, in Lemma I and the proof of Theorem 3, these nonmeasurability assumptions in Theorem 1, and in other results as well, are by no means incidental. On the contrary, they pop up quite naturally once we assume that 1~ (S, E) is not barrelled and look for a contradiction: We show that then S must contain a subset So with ]Sol measurable (hence IS I itself must be measurable) and, in addition, we produce quite a peculiar pointwise bounded set of functionals on Io~(So, E) which is not normArchivder Mathematik 63 29 450 L. DREWNOWSKt,M. FLORENCIOand P. L PA~rL ARf2H.NATtt. bounded (see the remark foilowing the proof of Theorem 3). The only way ou~ o~~this pit (to get a contradiction) we could find was to assume that [SI or [El is non-measurable or, when dealing with subspaces X c l| (S, E) rather than all of l~ (S, E), to assume that the functions in X have ranges of non-measurable cardinalities. This is not the first time that non-measurable cardinals appear in the context of permanence properties of locally convex spaces; the Mackey-Ulam theorem (see [14, w28.8] or [17, 6;2.23]) says that N_s is bornological if and only if ISt is non-measurable. Now, a normed space is barrelled if and only if the weak*-bounded sets in its dual are norm-bounded. For general locally convex spaces E this corresponds to what is sometimes called the Banach-Mackey property of E, and which requires every weak* (i.e., a(E', E)-) bounded subset of E ' to be strongly (i.e., fl(E', E)-~ bounded. Spaces E with this property are also called Banach-Mackey spaces [18, p. 158], Thus a locally convex space is barrelled if and only if it is Banach-Mackey and quasi-barrelled. Recall in this context that metrizable spaces are always quasi-barrelled. With this in mind, and also because the original proof of Theorem ~ did not require much changes, we have chosen to state our main results, in Section 2, as results on the inheritance of the Banach-Mackey property from E to l~ (S, E) (or some subspaces of I~o(S, E)). In Section 3 we return to the problem of barrelledness and give some positive answers as welt as examples and remarks that, hopefully, might be of some use in attempts to solve the general question as stated above. We refer the reader to [14], [17] and [18] for the concepts and notation of the theory of locally convex spaces used in this paper. 2. Banaeh-Mackeysubspaeesof 1~ (S, E). Theorem 1 above is a direct consequence of the following Theorem 2. Let S be a set and E a Banach-Mackey space. I f IS[ or l El is non-measurable, then the space lo~(S, E) is Banach-Mackey. Actually, our main result, of which the above is a particular case, is the following Theorem 3. Let S be a set and E a locally convex space. I f E is Banach-Mackey, so is every subspace X of I~ (S, E) satisfying the following conditions: (1) whenever f e X and A c S, then f zA E X; (2) whenever (A,) is a decreasing sequence of subsets of S with an empty intersection and (f,) is a bounded sequence in X such that supp f, ~ A n for every n ~ N , then the pointwise sum ~ ctnfn is in X Jot every (c~,)~ I1; n=l (3) X contains all the constant functions x Z~(x ~ E); (4) lf(S)l is non-measurable for every f e X. Most of the work needed to prove Theorem 3 is done in L e m m a 1 below. The idea of part (a) of this lemma, and of its proof as well, comes from [2], [5] and [6]. In particular, concerning condition (2) which plays a crucial role in the proof, cf. condition (*)in T h . 1 of [6], Also note that part (a) is valid without the assumption that H is countable. Vol. 63, 1994 Barrelled spaces of bounded functions 451 Lemma 1. Let S be a set, E a locally convex space, and assume that a subspace X of loo(S, E) satisfies conditions (1) and (2) above. Let B be an absolutely convex bounded subset of E, and H a weak*-bounded countable subset of X', and define a set function ~l = qB, n : ~ ( S ) --+~R+ by ~/(A) = sup { l ( u , f ) ] : u ~ H , f e bx(A, B)}, where b x(A, B) : = { f E X : f (S) ~ B and supp f ~ A} for all A c S . Then rl is a submeasure on ~ ( S ) ; that is, it is subadditive, nondecreasing, and ~l(~) = O. Moreover: co (a) q( {'~ A,) = oo for every decreasing sequence (A,) of subsets of S such that ~l(A,) = oo tl=l for every n. (b) Every family d consisting of pairwise disjoint sets A c S with tl(A ) > 0 is (at most) countable. (c) I f q(S) = o% then there exists a subset S O of S such that tl(So) = oo and Ygo : = {A: A c S o and tl(A ) = oo} is an ultrafilter on S O which is closed under countable intersections. P r o o f. It is easy to verify that t/is indeed a submeasure. (a): Suppose that there is a decreasing sequence (A,) such that t/(A,) = oo for every n and q(A~o) < 0 % where A~ = (~ A,. Then the sets A', = A , \ A ~ form a decreasing n=l sequence with an empty intersection and t/(A'.) = oe for all n. F o r every n e N , as t/(A',)= 0% we can find f , ~ bx(A',, B) and u, e H such that I ( u , , f , ) [ > n. In view of (2), we may define a linear operator V : l l - ~ X by setting V(e,) = ~ e . f . . Since V m a p s the unit ball of 11 into bx(S, B), a bounded subset of X, it is continuous. Then the functionats u o V (u e H) are pointwise bounded on ll, hence norm-bounded by the classic Banach-Steinhaus theorem. But I(u,o V,e,)l = I(u,,f.)[ > n for every n; a contradiction. (b): Suppose, to the contrary, that d is uncountable. For each A e d choose UA e H andfA e bx(A, B) so that (uA,fA) > 0. Since d is uncountable, there exists ~ > 0 and an uncountable subfamily ~r c d such that (uA,fA) > e for every A e d ' . As H is countable, there must be a u e H such that u = u.4 for every A from an infinite (even uncountable) subfamily ~ " c ~ ' . Now, for every n pick any n distinct sets A t ..... A, in d " and observe that fa, e bx(S , B) i=l and (u, fa,} > ne , i=l which obviously contradicts the continuity (or even boundedness) of u. 29* L. DREWNOWSK1,M. FLORENClOand P. J. PAOL 452 ARCH. MATH. (c): We start by proving that there exists a subset S' of S such that ~(S') < ~ , and for every A ~ S" : = S \ S' one has t~(A)= 0 or oo. Indeed, consider a maximal family consisting of pairwise disjoint sets D c S with 0 < q(D) < oo. By (b), @ is (at mostj countable. Let S' : = w N. If t/(S') = o o , then @ is infinite, say @ = {D1, D2 . . . . }, and setting A, = ~ D~ we arrive at a contradiction with (a). Hence, by the maximality of @, i=n S' and its complement S" are as required. Now, since ~/(S) = co, we also have q(S") = oo. Observe that from (a) it follows that there is no infinite disjoint sequence (A,) in N (S") with all t/(A,) = oo. N o w apply the following dyadic splitting procedure: Start with S" and then, at each step, split, if that's still possible, one of the two sets obtained in the preceding step into two sets on which ~ is infinite~ In view of the observation that we have just made, this procedure must terminate after a finite number of steps so that we will eventually arrive at a desired set So ~ S" such that for every A ~ So one of the values q(A) and ~/(So\ A} is 0 and the other is oo. This tells us that ~//o is an ultrafilter which, by (a), is closed under countable intersections. [] L e m m a 2. Let S and T be two sets, and let ~ be a free uItrafilter on S closed under countable intersections. Then every map f : S--+ T with [f (S)[ non-measurable is eventually constant relative to ~[. That is, there exists A ~ og such that f [ A is constam. P r o o f. Denote R = f (S) and let ~ : = { C ~ R: f - 1 (C) e q/}. Then ~ is an ultrafilter on R closed under countable intersections. Since IR[ is non-measurable, there must exist r e R with {r} e-r and f is constant on f - l({r}) ~ ~//. [] P r o o f o f T h e o r e m 3. Suppose X is not a Banach-Mackey space. Thus there is a weak*-bounded countable subset H of X', and an absolutely convex b o u n d e d subset B of E such that H is not (uniformly) bounded on bx(S, B). In other words, defining = t/B,n as in L e m m a 1, we have ~/(S) = ~ . Let So c S and the ultrafilter ~ o on So be as given by L e m m a t (e). Since E is Banach-Mackey, ~/({s}) < ~ for every s e S. Hence {s} ~q/o for every s ~ S o a n d so ~//o is a free ultrafilter on S O which is closed under countable intersections. It follows that [So] is measurable. By L e m m a 2 and condition (4), i f f e X t h e n f [ S0 is eventually constant relative to ~/oo In particular, the limit limbo f (s) exists for every f e X. Consequently, we can define a continuous linear o p e r a t o r P from X to E by setting P ( f ) - lim~of (s). In view o f ( l ) a n d (3), P is onto. In addition, it is d e a r that P maps the set bx(S o, B) onto B. Now, note that x ~ XXso is an isomorphic embedding of E into X. Hence we can assign to every u ~ H a functional ~7e E ' by setting (~, x ) = (u, XZso). Remembering that q ( S o \ A) = 0 for every A e ~ o , it should be clear that i f f e bx(S o, B) and u e H, then the value ( u , f z a ) is the same for all A ~ ~r In fact, (u,f)~A) = (u, P(f))~so) = (u, P ( f ) ) . It follows that sup {[(~, x)]: u ~ H, x ~ B} = sup { l ( u , f ) J : u e H, f e b x(So, B)} = n (So) = oo. Thus the s e t / t : = {fi: u ~ H} ~ E', though obviously pointwise b o u n d e d on E, is not uniformly b o u n d e d on B. A contradiction because E is a B a n a c h - M a c k e y space. [] Vol. 63, 1994 Barrelled spaces of bounded functions 453 R e m a r k . With the same notation as in the proof above, note that if E is BanachMackey but X is not, then the set Ho := {UZso : u ~ H} c X ' is weak*-bounded but not strongly bounded. Besides, there is a free ultrafilter ~ o on So, closed under countable intersections, such that if f e X is supported on the complement of a set in q/o, then (v,f> = 0 for all v e H o. The corollary below gives a sample of subspaces X ~ l~ (S, E) satisfying the hypotheses of Theorem 3. Corollary 1. Let S be a set and m an infinite non-measurable cardinal number, l f E is a Banach-Mackey space, so are the subspaces ofl~ (S, E) consisting of all those f which satisfy one (or more) of the following conditions: I) [ supp f [ < m; 2) I f (S)[ < m; 3) f (S) is separable; 4) I f (S)] is non-measurable; 5) f (S) is precompact and metrizable. P r o o f. Let X be the subspace defined by condition 1). To show that X is BanachMackey, fix a weak*-bounded set H in X', and a bounded sequence (f,) in X. Let T = supp f,. Then [ T[ < m and (f,) c l~ (T, E) c l~ (S, E). Since l~ (T, E) is BanachMackey by Theorem 2, H is uniformly bounded on (fn). The assertion for the subspaces defined by one of the conditions 2)-4) is an easy consequence of Theorem 3. Concerning condition 3) let us only explain that if f (S) is separable, then If(S)[ < 2 c, where c = 2 ~~ (see [10, 1.5.3]), and 2 c is non-measurable (see [4, Theorem on p. 193] or [15, Ch. IX.3, Th. 5]). An alternative argument for the subspace X defined by condition 4) runs as follows: Fix a weak*-bounded set H in X ' and a bounded sequence (f,) in X. Define a mapping ~p: S -+ [I f,(S) by r put T = q~(S). Then the composition operator ,= 1 = (f,(s)), and C~: l~(T, E) ~ l~(S, E); g ~ g o r is an isomorphic embedding, its range is contained in X, and there is a sequence (g,) in l~ (T, E) such that C~(g,) = f , for every n. Now, clearly, l T[ is non-measurable so that by Theorem 2 the space lo~(T, E) is Banach-Mackey. Hence, since G : = {u o C~, : u E H} is obviously a weak*-bounded subset of (l~ (T, E))', we have sup {l<u,f~>l : u ~ H, n ~ N} = sup {[<v, g,)]" v ~ G, n ~ N} < ~ . It follows that X is Banach-Mackey. Finally, consider the set of functions isolated by condition 5). It is indeed a vector subspace of l~ (S, E) because the sum of two precompact and metrizable sets in E is again precompact and metrizable. (Use the fact that continuous images of metrizable compact spaces are of the same type; cf. [10, 4.4.15]). Also note that every precompact and metrizable set is of (non-measurable) cardinality < 2 ~~ To verify that condition (2) of Theorem 3 is satisfied in the present case, let (f~) be a bounded sequence of functions as required by condition (2), each with a precompact and metrizable range. Fix (a,)~ 11 and let f = 5Z a~ f~ be the corresponding pointwise sum. We have to check that f (S) is precompact and metrizable. For each n the set K , :=f~ (S), where the closure is taken in the oo completion/~ of E , is compact and metrizable. Hence also K := IF[ K~, in its product n=l 454 L. D~CeWNOWSKI,M. FLORENCIOand P. J. PAI)L ARCH.MATH. topology, is compact and metrizable. Moreover, since the functions j~ are uniformly bounded, the union of the K~'s is a bounded subset of/~. Therefore, the series Z ~. x, n=l converges in/~ for every (x~) ~ K, and the map (x,) --. ~ e~ x, from K to E thus obtained is easily seen to be continuous. Hence the image of K under this map is compact and metrizable and, obviously, contains f ( S ) . The subspaces defined by combinations of conditions 1)-5) are treated analogously. [] 3. Barrelled subspaces of Iv (& E). As we pointed out in the introduction, a locally convex space E is barrelled if and only if it is quasi-barrelled and Banach-Mackey. So, if we want to use the results of the previous section to deduce barrelledness conclusions for the space l~ (S, E), we must know in advance that l~ (S. E) is quasi-barrelled No extra work is needed for that if E is metrizable: it is automatically quasi-barrelled, and so are l~ (S, E) and all of its subspaces. Consequently, if the space E in the results of Section 2 is assumed barrelled and metrizable, then in the conclusions one may replace 'BanachMackey' with 'barrelled'. For instance, Theorem 2 and Corollary 1 take the following form. Theorem 2'. Let S be a set and E a metrizable barrelled space. I f tSt or lEt ~s non-measurable, chen the space l~ (S, E) is barrelled. Corollary 1". Let S be a set and m an infinite non-measurable cardinal number. I r E is a metrizable barrelled space, so are the subspaces of l~ (S, E) considered in Corollary I. In particular, this corollary includes the case of the subspace of loo(S, E) consisting of functions with countable support which, independently of the present work, was shown to be barrelled (when E is a barrelled normed space) by K~kol and Roelcke [13I. The same remark applies to the subspace of those functions having countable range studied by Ferrando [11]. Now, let Es (s e S) be a family of normed spaces. Fix any i __<p < cc and define Fp := lp(E~, s ~ S) to be the normed space consisting of all those elements x = (x~) in [-I E~ for which IIx lip := ( 5Z [Ix, [f)x/p < oo, with the norm II lip given by this equality, sr sES Define l~ (E~, s ~ S) analogously. Then (*) l~ (Es, s ~ S) is isomorphic to a complemented subspace of l~ (S, Fp). To see this let P~be the natural norm one projections from Fp onto E~ (s ~ S). Then the operator P" loo(S, Fp) --* l~o(E~, s e S) defined by the equality P 9 = (P~g(s))~s is a desired (norm one) projection. Thus, as an immediate consequence of Theorem 1 (or 2'), observation (*), and the known fact that the space Fv is barrelled whenever all the spaces E~ are barrelled (see [9, paragraph following 5.6], [13] or [16]), we have the following Corollary 2. Let S be a set with t S[ non-measurable, and let Es(s ~ S) be a family of barrelled normed spaces. Then also the space l~ (Es, s ~ S) is barrelled. Vol. 63, 1994 Barrelled spaces of bounded functions 455 If we deal with general locally convex spaces, the question of when i| (S, E) is quasibarrelled is still open, even when S = N. As we mentioned in the introduction, there are even barrelled spaces E such that loo(S, E) is not quasi-barrelled. The first example of such a space is even the dual of a separable reflexive Fr6chet space, and was given by S. Dierolf [7, w5]. For the important class of (D F)-spaces, to which S. Dierolf's example belongs, the problem has been solved by Bierstedt and Bonet [2, 1.8] for S = N, and by Bierstedt, Bonet and Schmets [3] for arbitrary S. In this latter paper, they study the space CB(S, E) of all continuous bounded functions from a completely regular Hausdorff topological space S to a (D F)-space E. Now, if we endow a set S with the discrete topology, then CB(S, E) = t~ (S, E), so their results [3, Thins. 5 and 6] can be stated in our particular case as parts (a)-also given originally by S. Dierolf [7, 5.13] - and (b) of the following theorem, whereas its part (c) follows from Theorem 2 and part (b). Theorem 4. Let S be an infinite set and E a (DF)-space. Then: (a) l~(S, E) is a (DF)-space. (b) l~ (S, E) is quasi-barrelled if and only if the bounded subsets of E are metrizable. (c) I f l S[ or lE[ is non-measurable, then the space l~(S, E) is barrelled if and only if E is barrelled and its bounded subsets are metrizable. R e m a r k. Let us go back to our original question if l~ (S, E) is barrelled when E is a barrelled normed space. We have shown in Theorem 1 (and Theorem 2') that the answer is affirmative provided that iS] or [Ef is non-measurable, but we have been unable to solve the general case. Conclusion (c) of Lemma 1 tells us that any technique used to solve the general case must bear some relationship with the measurability of the cardinals of the sets or spaces involved. However, the following observations tell us that, even for normed spaces E, Theorem 1 is still not the end of the story of the barreltedness problem for the spaces Iv (S, E). That is, lo~(S, E) can be barrelled even if both ISI and ]EI are measurable. 1) First of all, and trivially, for every set S and every Banach space E, the space l~o(S, E) is Banach, hence barrelled. 2) Let S and T be two sets with measurable cardinals, and let E be a barrelled normed space with l E t non-measurable. Then, by Theorem 1, the space F : = t~ (T, E) is barrelled and, clearly, IF t is measurable. Now, I~ (S, F) ~ l~ (S x T, E) so that, again by Theorem 1, l~ (S, F) is barrelled although both ISI and IFf are measurable. 3) Let S be a set with IS I measurable, and T be any set. It is well-known (see e.g. [8] and the references therein) that the subspace E : = mo (T, E), consisting of all simple (i.e., finitely valued) functions from T to the space of scalars E , is barrelled. Now, it is easy to see that mo(Sx T,K) = I~(S,E) c I~(Sx T , K ) , where mo(S x T, E ) is barrelled and dense in I~ (S x T, E). It follows that I~ (S, E) is barrelled. However, 1SI is measurable and, if we choose T with ] T I measurable, also IE I will be measurable. We now give a positive answer (Theorem 5) for the case when E has property (K) and finish with a reformulation of the problem in terms of ultrapowers of E (Theorem 6). L. DREWNOWSKI,M. FLORENCIOand P. L PAI~IL 456 ARCH. MATH. A locally convex space E is said to have property ( K ) i f every null sequence (x,) in E contains a subsequence (x,~) for which the series x ~ converges in E. Property (K) has k=l been used, mainly by Antosik and Swartz, to give non-categorical proofs Of several classical theorems, as well as some new results, about open mappings, closed graphs a n d barrelledness (see [1] or [17, 1,2.15 and 1.4] and references therein), tn particular, e v e r y metrizable locally convex space with property (K)is barrelled; moreover, b y a result of Burzyk, Kli~ and Lipecki [17, 12.171, it is even a Baire space. Theorem 5. Let S be any set and E a normed space with property(K), Then I| E) is barrelled. P r o o f. According to condition (c) of L e m m a 1 and the remark following the proof of Theorem 3, if X = l| (S, E) is not barrelled, then there is a set S o c S and a countable subset H ~ X ' such that Ho : = {u ZSo : u ~ H} is weak*-bounded but not norm-bounded. Moreover, there is a free ultrafilter ~ o on S 0, d o s e d under countable intersections; such that i f f e X is supported on the complement of a set in ~ o, then (v, f ) = 0 for all v ~ Ho. Take a null sequence (f,) in X and a pointwise bounded sequence (v,) c Ho such that (v,, f , ) > n and consider the infinite matrix ((v,, f,,)). The rows of this matrix converge to zero, its columns are bounded, and its diagonal is not bounded. Hence, by [12, Thin. I ] , there is an infinite set I ~ N such that ~ I(V,, f,~)t < co for each n e I; and for every infinite set J ~ ! ,~x <v,, fro> >=~ [<v,, f , ) f > ~, (**) for all n e J . Clearly, we may assume that I = N. Now, for each s e S apply p r o p e r t y (K) to the sequence (f,(s)) c E to obtain an infinite set M(s) c N such that the series ~. f~(s) mEM(s) converges. Since the mapping s ~ M ( s ) e N ( N ) has a non-measurable range, it follows from L e m m a 2 that it is eventually constant relative to q/o, i.e. there is a set Ao ~ ~a'o and an infinite set M c. N such that Z f,,(s) converges for all s e A o. Define a function m~M f : S ~ E by setting f ( s ) : = ~ f~(s) if s s A o, and f ( s ) : = 0 otherwise. ( f need not be meM bounded.) Write M = {m~, rn 2.... }, where m~ < m 2 < .... By L e m m a 2 again, the mapping s~ f(s)-- f,.,( s ) f= j=l from S to ~ must be eventually constant relative to ~o- Hence there is A e Ogo with A c Ao such that f (s)= ~ f,,(s) uniformly for s ~ A. In other w o r d s , the series m~M J~,)~a converges i n l~(S, E) to g : = f x a . Since (vn, fro> = (v., f,,XA>, by the property m~M stated at the end of the first paragraph of the proof, we can apply inequality (**) with J = M to deduce that l<v,, g>l > n/2 for all n ~ M. Thus the sequence (v,) is not b o u n d e d at the point g of l~o(S, E); a contradiction. D Vol. 63, 1994 Barrelled spaces of bounded functions 457 Theorem 6. Let S be a set and E a barrelled normed space. Then the following are equivalent. (a) loo(S, E) is barrelled. (b) For every free ultrafilter og on S the ultrapower (E)~ is barrelled. P r o o f. That ( a ) ~ (b) is obvious because (E)~ is, by definition, a quotient space of l~ (S, e). (b) =~ (a): Suppose the space X = l~o(S, E) is not barrelled. Then there exists a countable weak*-bounded set H in X ' which is not norm bounded, i.e., not uniformly bounded on bx(S, B), where B is the closed unit ball of E. Hence, with t / = ~/B,n defned as in Lemma 1, we have tl(S ) = or. Let S o c S and q/o the ultrafilter on So be as in Lemma 1 (c). Since E is barrelled, ~ contains no finite sets; thus ~ o is a free ultrafilter on S Owhich is closed under countable intersections. Then, clearly, also the ultrafilter q / : = {A : A c S and A o c A for some A o ~ q/0} on S has the same properties. F r o m the definition of q and the properties of So it is obvious that every u ~ H vanishes on all f ~ l o o ( S , E ) with supp f ~ S o \ A o for some A o ~ ~ . In consequence, every v ~ K : = {u ZSo : u ~ H } vanishes on all f ~ lo~(S, E) with S \ supp f E q/. Since q/is closed under countable intersections, this is, by Lemma 2, the same as saying that every v ~ K vanishes on the subspace Lo : = { f ~ lo~(S, E): lim i[f (s)[l = 0} of lo~(S, E). Hence these functionals v induce corresponding funcfionals ~ on the ultrapower (E)~ = lo~(S, E)/Lo of E. In this way we get a pointwise bounded family /~ = {~ : v ~ K} in the dual of (E)ou that is not norm bounded. Therefore, the ultrapower (E)~ is not barrelled. [] Acknowledgements. We want to thank La Consejeria de Educaci6n y Ciencia de la Junta de Andalucia for partially supporting the visits of L. Drewnowski to Seville and P. J. Pafil to Poznafl. The research of the first named author was also partially supported by Komitet Badafl Naukowych (State Committee for Scientific Research), Poland, grant no. 2 1083 91 01. References [1] P. ANTOSIKand C. SWARTZ,Matrix Methods in Analysis. LNM 1113, Berlin-Heidelberg-New York 1985. [2] K. D. BIERSTEDTand J. BOYET,Dual density conditions in (DF)-spaces, I. 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J~ PAOL, Uniform boundedness of operators and barrelledness in spaces with Boolean algebras of projections. Atti Sem. Mat. Fis Univ Modena 41, 317-329 (1993). [10] R. ENGELK~NG,General Topology. Warszawa, 1977. [I 1] J. C. FERRANDO,On the barretledness of the vector-valued bounded function space. To appear. [12] M. FLO~NClO, P. J. PA6L and J. M. Vmu~s, On the Mikusifisky-Antosik diagonal theorem. Bull Polish Acad. Sci. Math. 40, 189-195 (1992). [13] J. K#KOL and W. ROELCV~, On the barrelledness of/e-direct sums of seminormed spaces for 1 =<p =<oo. Arch. Math 62, 331-334 (1994). [14] G. K6vrm, Topological Vector Spaces, I and II. Berlin-Heidelberg-New York 1969 and 1979. [I 5] K. KURATOWSK!and A. MOSTOWSKI,Set Theory. Amsterdam-Warszawa t967. [16] P. LtrgJE, Tonnelierheit in lokalkonvexen Vektorgruppen. Manuscripta Math. 14, 107-t21 (1974). [17] P. P~REZ CARRERASand J. BOnnET, Barrelled Locally Convex Spaces. Notas Mat. 131, North. Holland, Amsterdam-New York-O• 1987. [18] A. WILANSKY,Modern Methods in Topological Vector Spaces. New York 1978. Eingegangen am 28.12o 1993 Anschriften d e r Autoren: L. Drewnowski Faculty of Mathematics and Informafics, Adam Miekiewicz University, UI. Matejki 48/49, 60-769 Poznafi Poland M. F10rencio ?. J. Pafil E. S. Ingenieros tndustriales, Avda. Reina Mercedes s/n, 4t012 Sevilla Spa~n