ON X-KOTHE¨ ECHELON SPACES AND APPLICATIONS
By Fernando Blasco U. D. Matematicas,´ ETSI Montes, Universidad Politecnica´ de Madrid
(Communicated by S. Dineen, m.r.i.a.)
[Received 18 June 1997. Read 29 January 1998. Published 30 December 1998.]
Abstract Properties such as Heinrich’s density condition or being Montel for X-Kothe¨ echelon spaces are considered and applications to the study of reflexivity of spaces of n-homogeneous continuous polynomials on Frechet´ spaces are given.
Introduction In this article we study a class of Frechet´ spaces, the X-Kothe¨ echelon spaces, that generalises the class of Kothe¨ echelon spaces λp(A). The spaces λp(A), 1 ≤ p ≤∞, and many of their properties have been widely studied (see, for instance, [3], [4], [22], [30]) because they are a good source of examples and counterexamples in functional analysis. The spaces studied here are more general than Kothe¨ echelon spaces and are also a good source of getting examples, as we shall see in § 4. Bellenot defines X-Kothe¨ echelon spaces and obtains some of their properties in [2]. This ‘new’ class of echelon spaces has already been used to give examples and counterexamples [27]. We describe properties of X-Kothe¨ echelon spaces, point out similarities with ‘classical’ Kothe¨ echelon spaces and give applications. The paper consists of four sections: in § 1 we study basic properties of λX (A) and describe the dual space. In § 2 we characterise X-Kothe¨ echelon spaces that have Heinrich’s density condition. Section 3 describes λX (A) spaces that are Montel in terms of their Kothe¨ matrix. Finally, in § 4, examples and applications are given, which show the importance of these spaces in constructing new examples in functional analysis.
1. Basic properties
∞ Let X be a Banach space with a basis {en}n=1 and let a =(an)n∈N be a sequence of positive numbers. Suppose that k·kdenotes the norm in X. We define ( ) X∞ N Xa = (xn)n∈N ∈ C : anxnen ∈ X n=1
AMS Subject Class. (1991): 46G20,46A03
Mathematical Proceedings of the Royal Irish Academy, 98A (2), 191–208 (1998) c Royal Irish Academy 192 Mathematical Proceedings of the Royal Irish Academy
P∞ and, for (xn)n∈N ∈ Xa, we define k(xn)n∈Nka = k n=1 anxnenk. It is easy to check that (Xa, k·ka) is a Banach space. Moreover, the mapping X∞ ja:(xn)n∈N 3 Xa 7→ anxnen ∈ X n=1 is an isometry between Xa and X.
∞ Definition 1.1 [2]. Let X be a Banach space, {en}n=1 a basis for X and A aKothe¨ matrix defined over N (i.e. A =(am)m∈N, where am is a sequence am =(am,k)k∈N, satisfying that am+1,k ≥ am,k > 0 for each pair (m, k) ∈ N × N). Then ( ) X∞ N λX (A)= x =(xn)n∈N ∈ C : am,nxnen ∈ X for all m ∈ N n=1 is called an X-Kothe¨ echelon space (with X as the associated Banach space). We give a Frechet´ space structure to λX (A) when we endowP it with the topology defined by the seminorms {k·k } , where k(x ) k = k ∞ a x e k . am m∈N n n∈N am n=1 m,n n n Classical Kothe¨ echelon spaces are a particular case of X-Kothe¨ echelon spaces.
In the following, X will always be a Banach space with 1-unconditional Schauder basis, which sometimes will be shrinking and/or boundedly complete. We recall that an unconditionalP basis {en}n∈N for a Banach space X over C is called 1-unconditional ∞ ∞ if, for every x = n=1 xnen ∈ X and every complex sequence (θn)n=1 with |θn|≤1 P∞ P∞ for each n ∈ N, k n=1 θnxnenk ≤ k n=1 xnenk . For the definitions of shrinking and boundedly complete basis see [11].
Proposition 1.2. Let A =(am)m∈N beaKothe¨ matrix over N and let X be a Ba- nach space with a 1-unconditional basis {e }∞ . Then λ (A)=limX is a reduced n n=1 X ←− am projective limit. m
Proof. By definition 1.1 and the definition of projective limit λ (A)=limX . To X am ←−m prove that it is reduced, fix m ∈ N and consider, for each n ∈ N, the sequence u =(δ ) ∈ λ (A). Since (u ) is total in X ,λ (A) is dense in X . Thus, the n n,k k∈N X n n∈N am X am projective limit is reduced. (The particular case where X is reflexive can be seen in [21, lemma 1 and theorem 1].)
To obtain more information about X-Kothe¨ echelon spaces we need to define co-echelon spaces. For this purpose we adopt the following notations.
Definition 1.3 [4, definitions 1.2 and 1.4]. For a Kothe¨ matrix A =(am)m∈N we de- 1 ¯ note by V the matrix (vm)m∈N, where vm =(vm,n)n∈N, with vm,n = , and by V the am,n set ¯ V = {¯v =(¯vn)n∈N: ¯vn ≥ 0 for each n ∈ N and sup am,n¯vn < ∞ for each m ∈ N}. n Blasco—On X-Kothe¨ echelon spaces 193
Remark 1.4. Every v ∈ V is a sequence of positive numbers, but this is not true for ¯ ¯ ¯v ∈ V. In fact, every sequence µ =(µn)n∈N, where µn =0forn large belongs to V. It is also important to note that, in the case we are considering, it is always possible to find positive elements in V.¯ We construct some of those elements by fixing a sequence (Rk)k∈N of positive numbers and defining
¯vn = min{Rkvk,n:1≤ k ≤ n}.
By the unconditionality of the basis, it is easy to verify that ¯vn > 0 for all n and that ¯ ¯v =(¯vn)n∈N ∈ V.
Definition 1.5. Let λX (A)beanX-Kothe¨ echelon space. The co-echelon spaces are defined as k (V )=∪ X and K (V¯ )=∩ X with the natural topologies X m∈N vm X ¯v∈V¯ ¯v
(i) k (V )=limX , X vm −→m ¯ (ii) KX (V )=limX¯v. −→¯v It is well known that the inductive limit used in the definition of classical co- echelon spaces is a regular inductive limit [4, proof of theorem 2.7]. Results due to Komatsu ([21]) show that this is also true in our case when X is a reflexive Banach space:
Proposition 1.6 [21, lemma 1 and theorem 6]. Let (Xj )j∈N be a sequence of reflexive Banach spaces. Then limXj is a complete reflexive space. Moreover, for every bounded −→j subset B in limXj there exists an index k such that B is bounded as a subset of Xk, −→j that is the inductive limit is regular.
¯ Both kX (V ) and KX (V ), which often coincide, can be considered as ‘dual spaces’ of λX (A). The following proposition extends the result due to Bierstedt, Meise and Summers for classical Kothe¨ echelon spaces [4, theorem 2.3].
Proposition 1.7. Let X be a Banach space with 1-unconditional boundedly ∞ ¯ complete basis {en}n=1. Then kX (V ) and KX (V ) are algebraically and topologically isomorphic.
¯ ¯ Proof. First we prove KX (V ) ⊂ kX (V ). If not, then there is x ∈KX (V ) such that x 6∈ X for every m ∈ N. Since the basis {e }∞ is boundedly complete, this is vm n n=1 n o PN equivalent to saying that, for each m ∈ N, the set n=1 vm,nxnen : N ∈ N is not bounded. Since {e }∞ is 1-unconditional basis there exists a strictly increasing n n=1 P Nm sequence (Nk)k∈N of natural numbers such that n=1 vm,nxnen >mfor every m ∈ N. 194 Mathematical Proceedings of the Royal Irish Academy
Next we define the sequence ¯v =(¯vn)n∈N by ¯vn = vk,n, for Nk−1 am,n¯vn ≤ max {1, max{am,nvk,n: k X∞ X∞ X∞ ¯vn xn¯vnen = xn vm,nen ≤ sup am,n¯vn xnvm,nen < ∞, vm,n n=1 n=1 n∈N n=1 ¯ which implies that x ∈ X¯v and then kX (V ) ⊂ X¯v for every ¯v ∈ V. Hence we have the ¯ inclusion kX (V ) ⊂KX (V ). ¯ Finally we prove that kX (V ) and KX (V ) are topologically isomorphic. The ¯ above inequality gives the continuity of the inclusion kX (V ) ⊂KX (V ). To see the continuity of the opposite inclusionS let us take a neighbourhood U of0inkX (V ). We can assume that U = Γ¯ ( ρ B ), where (ρ ) is a sequence of positive m Xvm m m∈N m∈N numbers. For each m ∈ N, let 2m αm = max max{am,n:1≤ n ≤ m}, . ρm For each n ∈ N, choose j ∈ N, 1 ≤ j ≤ n, such that α v = min{α v :1≤ n n jn jn,n m m,n m ≤ n}. The sequence w¯ =(w¯ ) , where w¯ = α v is also in V¯ and w¯ > 0for n n∈N n jn jn,n n all n ∈ N (see Remark 1.4). Fix x =(xn)n∈N such that there exists N ∈ N with xn =0forn>Nand P P kxk ≤ 1, then we have N α v x e = w¯ x e = kxk . w¯ n=1 jn jn,n n n n∈N n n n w¯ Define from x a sequence (g ) ⊂ X , where g =(g ) , with g = m m∈N w¯ m m,n n∈N jn,n α ρ x and g =0form 6= j . jn jn n m,n n Then ∞ X X kgmk = vm,ngm,nen = vj ,nαmρmxnen vm n n=1 jn=m n≤N N X ≤ ρ v α x e = ρ kxk ≤ ρ . m m,n m n n m w¯ m n=1 Blasco—On X-Kothe¨ echelon spaces 195 XN 1 For n>Nwe have gm,n = 0 while, for n ≤ N, we have gm,n =0ifm 6= jn. αmρm m=1 XN 1 1 XN 1 X∞ 1 Thus g = g = x and then g = x. Since ≤ m,n jn,n n m αmρm αj ρj αmρm αmρm m=1 n n m=1 ! m=1 X∞ 1 XN 1 [ =1, we obtain x = g ∈ Γ¯ ρ B = U. m m m Xvm 2 αmρm m=1 m=1 m∈N If y =(y ) ∈K (V¯ ) ∩ B then y ∈ k (V ), and there is m ∈ N such that n n∈N X Xw¯ X y ∈ X . If yN =(yN ) , where yN = y if n ≤ N and yN =0ifn>N,then vm n n∈N n n n X∞ N N ky − y kv = ynvm,nen . Hence lim y = y in Xv and in kX (V ). m N→∞ m n=N+1 N Since U is closed in kX (V ) and y ∈ U for each N ∈ N, we have that y belongs ¯ to U and so {y ∈KX (V ): kykw¯ ≤ 1}⊂U. This proves that the canonical topology ¯ on KX (V ) (the projective topology) is coarser than the canonical topology on kX (V ) (the inductive topology). ∞ We recall that if {en}n=1 is a 1-unconditional shrinking basis for the Banach 0 ∞ 0 0 ∞ space X and {en}n=1 is the associated basis in the strong dual X , then {en}n=1 is an unconditional basis for X0 [20, theorem 3]. To prove that the unconditionality constant is 1, take complex sequences {yn}n∈N and {θn}n∈N such that |θn|≤1for each n ∈ N and fix N ∈ N, then ! ! ! N N ∞ N X X X X y θ e0 (x) = y θ e0 x e = y θ x n n n n n n m m n n n n=1 n=1 m =1 ! n=1 ! N N ∞ X X X = y (θ x ) = y e0 θ x e n n n n n m m m n=1 n=1 m=1 N ∞ N X X X ≤ y e0 θ x e ≤ y e0 kxk n n m m m n n n=1 m=1 n=1 X∞ P P N 0 N 0 for each x = xnen ∈ X, so n=1 ynθnen ≤ n=1 ynen . n=1 Proposition 1.8. Let X be a Banach space, {en}n∈N a 1-unconditional shrinking basis 0 0 for X and a =(an)n∈N a sequence of positive numbers. Then (Xa) =(X ) 1 , where a 1 =(1 )∞ . a an n=1 Proof. We have already noted at the beginning that the Banach spaces X and Xa are isomorphic by means of a mapping ja: Xa → X. Then, the transpose mapping X∞ 0 0 0 0 0 0 0 0 ja: X → (Xa) is an isomorphism. Given y ∈ (Xa) choose y = ynen ∈ X such n=1 196 Mathematical Proceedings of the Royal Irish Academy X∞ 0 0 0 0 0 that y = ja(y ). Then if x =(xn)n∈N ∈ Xa,y(x)=y (ja(x)) = ynxnan. Hence (Xa) n=1 X∞ 0 0 0 0 can be identified with the space of sequences (ynan)n∈N satisfying ynen ∈ X , that n=1 0 is with (X ) 1 . a ∞ Corollary 1.9. Let X be a Banach space and {en}n=1 a 1-unconditional shrinking basis 0 for X. Then (λX (A)) = kX0 (V ) (as vector spaces) for every Kothe¨ matrix A (and its associated set V ). Proof. By [25, IV.4.4, and proposition 1.8] we have 0 0 0 0 (λ (A)) =(limX ) = lim(X ) = lim(X ) = k 0 (V ). X am am vm X ←−m −→m −→m The duality between these spaces is the following: if x =(xn)n∈N ∈ λX (A) and P∞ y =(yn)n∈N ∈ kX0 (V ), then hx, yi = n=1 xnyn. The identity given in Corollary 1.9 is merely algebraic, but the topological identity remains true for spaces X with a boundedly complete basis, when we consider the 0 0 strong topologies on (λX (A)) and X . We first need to characterise the bounded sets in λX (A). Proposition 1.10. Let X be a Banach space and let {en}n∈N be a 1-unconditional boundedly complete basis for X. Then B ⊂ λX (A) is bounded if and only if there is a strictly positive ¯v ∈ V¯ such that B ⊂ B . X1/¯v Proof. We begin by proving the necessity. If B is bounded in λX (A), we may assume there exists a sequence (Mm)m∈N of positive numbers such that B = {y ∈ λ (A): kyk ≤ M for each m ∈ N}. X am m m+1 We define the sequence ¯v =(¯vn)n∈N as ¯vn = min{2 Mmvm,n:1≤ m ≤ n}. By Remark 1.4, we have ¯v ∈ V.¯ Next for each n ∈ N define j ∈ N, such that 1 ≤ j ≤ n and 2jn+1M v ≤ n n jn jn,n m+1 2 Mmvm,n for each m ∈ N with 1 ≤ m ≤ n. a We get a ¯v = a 2jn+1M v =2jn+1M and then jn,n = 1 . Fix y ∈ B, jn,n n jn,n jn jn,n jn jn+1 2 Mjn ¯vn for each N ∈ N we have N N X 1 X a y e = y jn,n e . n n n j +1 n ¯vn 2 n Mj n=1 n=1 n Blasco—On X-Kothe¨ echelon spaces 197 Since for each n ≤ N there is m ≤ N with jn = m, we obtain N N X a X X a y jn,n e = y jn,n e n j +1 n n j +1 n 2 n Mj 2 n Mj n=1 n m=1 jn=m n n≤N N N ∞ X 1 X X 1 X = y a e ≤ y a e m+1 n m,n n m+1 n m,n n 2 Mm 2 Mm m=1 jn=m m=1 n=1 n≤N XN 1 XN 1 1 ≤ M ≤ < < 1. m+1 m m+1 2 Mm 2 2 m=1 m=1 Since {en}n∈N is boundedly complete, we have that y ∈ X 1 and kyk1/¯v ≤ 1. Hence B ⊂ B . ¯v X1/¯v ¯ The sufficiency of the condition is easier. Fix ¯v ∈ V. Given m ∈ N, let Mm = sup ¯v a . If y ∈ B then n m,n X1/¯v n∈N ∞ ∞ ∞ X X 1 X 1 am,nynen = am,n¯vnyn en ≤ Mmyn en ≤ Mm. ¯vn ¯vn n=1 n=1 n=1 Hence B is a bounded subset of λ (A). X1/¯v X The above proposition allows us to establish the topological identity between 0 ¯ (λX (A)) and KX0 (V ). Corollary 1.11. Let X be a reflexive Banach space with 1-unconditional basis. Then 0 ¯ (i) (λX (A))β = KX0 (V ), ¯ 0 (ii) (KX (V ))β = λX0 (A), with the usual topologies for echelon and co-echelon spaces. Proof. Since X is reflexive with 1-unconditional basis, the basis is shrinking and boundedly complete [11, chapter V], so we can apply the above results. (i) By proposition 1.10, {B : ¯v ∈ V,¯ ¯v is strictly positive} is a fundamental X1/¯v system of bounded sets for λ (A). So the polars of these sets, B ; ¯v ∈ V,¯ ¯v strictly X X1/¯v 0 positive, form a basis of neighbourhoods of 0 in (λX (A))β. Since ( ) o X∞ _ B = y ∈K 0 (V¯ ): y x ≤ 1 for every x ∈ B X1/¯v X n n X1/¯v ( n=1 ) X∞ 0 = y ∈K 0 (V¯ ): ¯v y e ≤ 1 X n n n n=1 we obtain the result. 198 Mathematical Proceedings of the Royal Irish Academy (ii) Recall that K (V¯ )=k (V )=limX and this inductive limit is regular. X X vm −→m Hence {B } is a fundamental system of bounded sets for this space. Since Xvm m∈N 0 (kX (V )) is isomorphic to λX0 (A) as linear spaces [25] and ( ) o X∞ _ 0 BX = y =(yn)n∈N ∈ λX0 (A): am,nyne ≤ 1, , vm n n=1 we see that the usual topology on λX0 (A) and the strong topology β(λX0 (A),kX (V )) coincide. Remark 1.12. (a) Corollary 1.11 can also be obtained from theorems 11 and 12 in [21], but we prefer to give a proof using arguments similar to those used by Bierstedt et al. in [4]. (b) From Corollary 1.11 we obtain that λX (A) is reflexive for any election of the Kothe¨ matrix A, when X is a reflexive Banach space with a 1-unconditional basis, since 0 0 0 (λ (A)) =(K 0 (V¯ )) = λ 00 (A)=λ (A). X β β X β X X Note that λX (A) may be reflexive without X being reflexive. See, for instance, [5] ¯ and [16]. Under these hypotheses we can also prove that KX (V ) is reflexive. 2. X-Kothe¨ spaces satisfying the density condition Bierstedt and Bonet [3] obtained a characterisation of the Kothe¨ echelon spaces that satisfy Heinrich’s density condition (DC) [19] in terms of their Kothe¨ matrices. This condition plays an important role when we consider spaces of polynomials defined on Frechet´ spaces and, in particular, on Kothe¨ echelon spaces ([28], [14], [6], [5]). General properties of X-Kothe¨ echelon spaces are described in this section and the next section as they will be used in the last section to give new examples and counterexamples to some questions in functional analysis. With this in mind, it is reasonable to wonder about the relation between the λX (A) spaces that satisfy the density condition and their Kothe¨ matrices A. We recall that a λp(A) space has the density condition if and only if its matrix A satisfies the so-called condition (D) [3, theorem 2.6]. This condition (D) is independent of p. We shall prove here that λX (A) satisfies Heinrich’s density condition if and only if its Kothe¨ matrix A satisfies condition D. Definition 2.1 [3]. A Kothe¨ matrix A satisfies condition (D) if and only if there is an increasing sequence J =(Ik)k∈N of subsets Ik of N such that a mk ,n (N,J): for each k ∈ N, there exists mk ∈ N with inf > 0, for m>mk, am,n n∈Ik and (M,J): for each m ∈ N and I0 ⊂ N with I0 ∩ (I \ Ik) 6= ∅ for every k ∈ N, there exists m0 = m0(m, I ) >mwith inf am,i =0. 0 a 0 i∈I0 m ,i Blasco—On X-Kothe¨ echelon spaces 199 To characterise the λX (A) spaces with the density condition we first prove the following. Proposition 2.2. Let X be a Banach space with a boundedly complete 1-unconditional ∞ basis and let J =(Ik)k=1 be an increasing sequence of index sets Ik ⊂ N. The following are equivalent. (i) For each m ∈ N and I0 ⊂ N such that I0 ∩ (N \ Ik) 6= ∅ for every k ∈ N there 0 0 am,n exists m = m (m, I0) with inf =0. n∈I0 am0,n ¯ (ii) On every bounded subset of KX (V ) its topology is defined by the system of X ¯ seminorms {k·k¯v,k} ¯v∈V¯ , where kxk¯v,k = ¯vnxnen for every x ∈KX (V ). k∈N n∈Ik Proof. For m ∈ N let ( ) ∞ X B = (x ) ∈ X : x v e ≤ 1 . m n n∈N vm n m,n n n=1 ¯ By Proposition 1.10, (Bm)m∈N is a fundamental system of bounded sets in KX (V ). (1) ⇒ (2). Fix m ∈ N and ¯v ∈ V.¯ As pointed out in [3, proposition 2.9], there is k ∈ N such that ¯v a ≤ 1 when n 6∈ I . n m,n 2 k ( ) ∞ X Take the 0-neighbourhood in K (V¯ ),U= x ∈K (V¯ ): kxk = x ¯v e ≤ 1 X X ¯v n n n n=1 ˜ ¯ 1 ˜ and define U = x ∈KX (V ): kxk¯v,k ≤ 2 . It is clear that U ⊂ 2U. ˜ If x ∈ U ∩ Bm, we have ∞ X X X ¯v x e ≤ ¯v x e + ¯v x e n n n n n n n n n n=1 n∈I n∈Ic k k 1 X 1 1 X ≤ + ¯v a v x e ≤ + v x e ≤ 1. 2 n m,n m,n n n 2 2 m,n n n n∈Ic n∈Ic k k ˜ So U ∩ Bm ⊂ U. By [18, lemma 3.5], we see that the topology induced on Bm by ¯ the seminorms {k·k¯v,k} ¯v∈V¯ coincides with the topology induced by KX (V )onthe same set. k∈N c (2) ⇒ (1). If (1) is not satisfied, there are m0 ∈ N and I0 ⊂ N such that I0 ∩Ij 6= ∅ a m0,n for each j ∈ N, and for every m, inf > 0. Once we fix a sequence (ij )j∈N with n∈I0 am,n i ∈ I ∩ Ic we define for each j ∈ N a sequence xj =(xj ) , where xj = a and j 0 j n n∈N ij m0,ij 200 Mathematical Proceedings of the Royal Irish Academy j xn = 0 when n 6= ij . We have ∞ X sup ¯v xj e = sup ¯v a ≤ sup ¯v a < ∞, n n n ij m0,ij n m0,n j∈N n=1 j∈N n∈N ¯ j ¯ for every ¯v ∈ V, so (x )j∈N is a bounded sequence in KX (V ). Moreover, for every X ¯v ∈ V¯ and each k ∈ N, kxj k = ¯v xj e = 0 for each j ∈ N. This is equivalent ¯v,k n n n n∈Ik j to saying that (x )j∈N is a null sequence for the topology generated by {k·k¯v,k} ¯v∈V¯ . k∈N If (2) holds then, since it is a bounded sequence, it should be a null sequence for the ¯ N canonical topology in KX (V ), but this does not happen: define w¯ =(w¯n)n∈N ∈ R by w¯ = v if n ∈ I and by w¯ =0ifn 6∈ I . Since for every m ∈ N n m0,n 0 n 0 a −1 m0,n am,nw¯n ≤ sup vm ,nam,n = inf < ∞, 0 n∈I n∈I0 0 am,n we have w¯ ∈ V¯ and X kxj k = w¯ xj e = v a =1, w¯ n n n m0,ij m0,ij n∈N which contradicts (2). ∞ Theorem 2.3. Let X be a reflexive Banach space and let {en}n=1 be a boundedly com- plete 1-unconditional basis. The following are equivalent. (i) A satisfies condition (D). 0 ¯ (ii) Every bounded subset of (λX (A))β 'KX0 (V ) is metrisable, when it is endowed ¯ with the topology induced by KX0 (V ). (iii) λX (A) satisfies the density condition. Proof. Since λX (A) is metrisable we see that (2) is equivalent to (3) [3, theorem 1.4]. Suppose A satisfies condition (D). By condition (M,J ) and Proposition 2.2, the ¯ topology induced by KX0 (V ) on a bounded subset coincides with the topology ¯ induced by the seminorms k·k¯v,k. For each k ∈ N we define ¯vk ∈ V as follows: a mk ,n ¯vk,n = vm ,n if n ∈ Ik and ¯vk,n =0ifn 6∈ Ik, where we choose mk such that inf > 0 k am,n n∈Ik ¯ for m>mk. We also note that using condition (N,J) we obtain (¯vk)k∈N ⊂ V . We have X X X kxk = ¯v x e0 = ¯v a v x e0 ≤ ρ v x e0 = ρ kxk , ¯v,k n n n n mk ,n mk ,n n n k mk ,n n n k ¯vk ,k n∈Ik n∈Ik n∈Ik where ρ = sup ¯v a < ∞. So the topology on every bounded subset of K 0 (V¯ )is k n mk ,n X n∈N Blasco—On X-Kothe¨ echelon spaces 201 ¯ defined by a countable family of seminorms. Thus every bounded subset of KX0 (V ) is metrisable and (1) ⇒ (2). ¯ To prove (2) ⇒ (1) we define, for each m ∈ N,Bm = {x ∈KX0 (V ): kxkv ≤ 1}. ¯ m Fix m ∈ N. Since Bm is metrisable for the topology induced by KX0 (V ), there is ¯ a countable family of seminorms defining the topology of KX0 (V ), so there is a m ¯ ¯ sequence (w¯k )k∈N ⊂ V such that for every ¯v ∈ V there exists k ∈ N with ¯ ¯ {x ∈KX0 (V ): kxkw¯ m ≤ 1}∩Bm ⊂{x ∈KX0 (V ): kxk¯v ≤ 1}∩Bm. k We can define j ¯vk = max{(v1,1, 0, 0,...), max{w¯i : i, j ∈ N,i,j≤ k}}. We have that (¯v ) is an increasing sequence in V.¯ Moreover, {x ∈ B : kxk ≤ k k∈N m ¯vmax{k,m} 1}⊂{x ∈ Bm: kxk¯v ≤ 1}. Hence, the seminorms {k·k¯v }k∈N define on Bm the same ¯ k topology as KX0 (V ), for each m ∈ N. n o ¯vk,n ∞ Define, for each k ∈ N, the sets Ik = n ∈ N: ≥ 1 . The sequence (Ik) is vk,n k=1 increasing and satisfies condition (N,J) [3, theorem 2.6]. ¯vk,n Let m ∈ N,x∈ Bm and k ∈ N,k≥ m, such that kxk¯v ,k ≤ 1. Since < 1for k vk,n each n 6∈ Ik, we obtain ∞ X X X ¯v x e0 ≤ ¯v x e0 + ¯v x e0 k,n n n k,n n n k,n n n n=1 n∈I n6∈I k k X ¯ X X∞ vk,n 0 0 0 ≤ 1+ vk,nxnen ≤ 1+ vk,nxnen ≤ 1+ vm,nxnen ≤ 2. vk,n n6∈Ik n6∈Ik n=1 Hence {x ∈K 0 (V¯ ): kxk ≤ 1}∩B ⊂{x ∈K 0 (V¯ ): kxk ≤ 2}∩B . If we X ¯vk ,k m X ¯vk m denote by TS the topology induced on Bm by a system of seminorms S, we have T(k·k ) ≤T(k·k ) ≤T(k·k ) ≤T(k·k ) . On the bounded sets, T(k·k ) ¯vk k∈N ¯vk ,k k∈N ¯v,k ¯v∈V¯ ¯v ¯v∈V¯ ¯v ¯v∈V¯ k∈N coincides with T . Hence the topology induced on every bounded subset of (k·k¯v )k∈N ¯ k ¯ KX0 (V ) is defined by the seminorms {k·k¯v,k: ¯v ∈ V, k ∈ N} and, by Proposition 2.2, we obtain that condition (M,J ) is satisfied. 3. X-Kothe¨ spaces which are Montel spaces We study properties on the Kothe¨ matrices A that make λX (A) a Montel space. This extends work of Bierstedt et al. in [4]. We prove that λX (A) is a Montel space if and only if its Kothe¨ matrix A satisfies condition (M). Definition 3.1 [4, definition 4.1]. A Kothe¨ matrix A satisfies condition (M) if and only if there is no infinite subset I0 ⊂ N for which there exists m0 ∈ N such that v a m,n m0,n inf = inf > 0, for m>m0. n∈I v n∈I a 0 m0,n 0 m,n 202 Mathematical Proceedings of the Royal Irish Academy J If J ⊂ N, the spaces λX (A)={(xn)n∈N ∈ λX (A): xn =0ifn 6∈ J} are called sectional subspaces. These subspaces are complemented in λX (A). Lemma 3.2. Let X be a Banach space and let {en}n∈N be a boundedly complete 1- unconditional basis. If A isaKothe¨ matrix that does not satisfy (M), then there exists an infinite dimensional Banach space Y complemented in X isomorphic to a sectional subspace of λX (A). Proof. Suppose that A does not satisfy condition (M). Then there exist an infinite index set I0 ⊂ N, an integer m0 ∈ N and a sequence (εm)m∈N of positive numbers a m0,n such that infn∈I = εm > 0 for every m ∈ N. Define Y = [en: n ∈ I0]. Since 0 am,n {en}n∈N is unconditional, Y is complemented in X. Consider the linear mapping X ϕ:(x ) 3 λI0 (A) 7→ a x e ∈ Y. n n∈N X m0,n n n n∈I0 P The equality kϕ(x)kY = k am ,nxnenk = kxkX implies that ϕ is continuous. n∈I0 0 am0 P Since the basis is boundedly complete, every y ∈ I can be written as a sum x e . We define a mapping n∈I0 n n X I0 ψ: xnen 3 Y 7→ (yn)n∈N ∈ λX (A), n∈I0 xn where yn = a if n ∈ I0 and yn =0ifn 6∈ I0. Obviously, ψ is a linear mapping and m0,n it is easy to prove that it is well-defined. Moreover, if m ∈ N, we have ! X X a X 1 1 ψ x e = m,n x e ≤ x e = kxk , n n a n n ε n n ε Y n∈I n∈I m0,n n∈I m m 0 am 0 0 I0 so ψ is continuous. Since ψ ◦ ϕ is the identity mapping on λX (A) and ϕ ◦ ψ is the identity mapping on Y,these spaces are isomorphic, which completes the proof. Theorem 3.3. Let X be a Banach space with a boundedly complete 1-unconditional basis. The following are equivalent (i) λX (A) is Montel. (ii) A satisfies the property (M). Proof. Suppose λX (A) is Montel. If (2) is not satisfied, λX (A) has a sectional subspace isomorphic to a infinite dimensional Banach space (Lemma 3.2). So λX (A)isnot Montel, which contradicts our hypothesis. By Proposition 1.10, {B : ¯v ∈ V¯ } is a fundamental system of bounded subsets X1/¯v ¯ in λX (A). Thus, in order to prove the converse, fix ¯v ∈ V, strictly positive. We must prove that B is a relatively compact subset in λ (A). We have that for each X1/¯v X Blasco—On X-Kothe¨ echelon spaces 203 m ∈ N, lim am,n¯vn = 0 [4, proposition 4.3]. Hence there exists a finite subset I1 of n→∞ ¯ ε N (note that I1 depends on m) such that am,nvn ≤ 2 if n 6∈ I1. Define −1 X ε Um = y =(yn)n∈N ∈ λX (A): |yn|≤ ej for each n ∈ I1 . 2am,n j∈I1 Then Um is a 0-neighbourhood for the topology on λX (A) induced by the pointwise convergence topology on CN. If y ∈ B ∩ U , then X1/¯v m ∞ X X X y kyk = a y e ≤ a y e + a ¯v n e am m,n n n m,n n n m,n n n ¯vn n=1 n∈I n6∈I 1 1 X ε X y X 1 ε n ≤ am,nynen + en ≤ am,n P en 2 ¯vn am,n 2 e n∈I1 n6∈I1 n∈I1 j∈I1 j ε ε ε + = + = ε 2 2 2 So U ∩ B ⊂{y ∈ B : kyk ≤ ε}, which proves that X induces on B m X1/¯v X1/¯v am am X1/¯v a topology coarser than the pointwise topology. Hence the topology of pointwise convergence on B is coarser than the induced topology from λ (A). Since B is X1/¯v X X1/¯v bounded for the topology of the pointwise convergence on the Montel space CN it is compact. So it is compact for the other topology (which is coarser). Hence λX (A) is a Montel space. Remark 3.4. (1) It is well known that non-Montel Kothe¨ echelon spaces λp(A) contain complemented copies of `p. Bellenot proves that, for non-Montel λX (A) spaces, there is a subsequence of the canonical basis of λX (A) that spans an infinite dimensional Banach space [2, proposition 3.7]). The technique he used is different from the one we used in Lemma 3.2. We obtained the existence of an infinite dimensional Banach space Y that is complemented both in X and in λX (A). Since every complemented subspace of `p is isomorphic to `p [26, lemma 18.6]), Lemma 3.2 generalises the result for λp(A). (2) The technique used in the proof of Theorem 3.3 is similar to that used by Bierstedt et al. [4] when they give an analogous result for λp(A) spaces. Charac- terisation of Montel X-Kothe¨ spaces in terms of their Kothe¨ matrices can also be obtained from results of Dıaz´ and Minarro˜ [10]. 4. Polynomials on X-Kothe¨ spaces: examples Reflexivity and barrelledness of spaces of homogeneous polynomials defined on Kothe¨ echelon spaces have recently been studied by several authors ([10], [16], [5]). In this paper we study reflexivity and barrelledness of spaces of polynomials on 204 Mathematical Proceedings of the Royal Irish Academy X-Kothe¨ echelon spaces and we point out similarities and differences of spaces of polynomials defined on the classical Kothe¨ echelon spaces and on X-Kothe¨ echelon spaces. We develop a method that allows us to extend Banach space results to X-Kothe¨ echelon spaces (which are Frechet´ spaces). In particular, if we take the original Tsirelson’s space as X, it is a reflexive space with a 1-unconditional basis, so we can consider the space λX (A), for a suitable Kothe¨ matrix A. For instance, we study the n reflexivity of P( λX (A)), pointing out a difference with a known result on reflexivity of spaces of homogeneous polynomials on classical Kothe¨ echelon spaces ([16], [9], [5]). For a locally convex space E, the space of n-homogeneous continuous poly- nomials on E is isomorphic to the topological dual of the completion of the n-symmetric tensor product of E (endowed with the projective topology), i.e. P(nE) ' ( ⊗ˆ E)0 [24]. We denote by β the topology, recently introduced in [15], n,s,π induced on P(nE) ' ( ⊗ˆ E)0 by the topology β on ( ⊗ˆ E)0 of uniform convergence n,s,π n,s,π on the bounded subsets of ⊗ˆ E. On P(nE) we consider, apart from β, the following n,s,π three natural topologies: the compact-open topology τ0, the topology of uniform convergence on bounded subsets τb and the Nachbin ported topology τω [13]. When E has the (BB)n,s property, i.e. when every bounded subset in ⊗ˆ E is n,s,π contained in the closed convex hull of the n-symmetric tensor product of a bounded n ˆ 0 n subset of E, (P( E),τb) is isomorphic to the strong dual ( ⊗ E)β (= (P( E),β)) [15]. n,s,π The spaces we are considering, X-Kothe¨ echelon spaces, have the (BB)n,s property for each n ∈ N [14]. We need a preliminary lemma Lemma 4.1. Let X be a Banach space with a 1-unconditional basis and let A be a Kothe¨ matrix on N. Then ⊗ˆ λ (A)=lim ⊗ˆ X X am n,s,π ←−m n,s,π and this is a reduced projective limit. Proof. The projective topology π on ⊗ λX (A) is generated by the seminorms n,s P P N n N {⊗ α: α ∈ sc(λX (A))} [24], where ⊗ α(θ) = inf i=1[α(xi)] : θ = i=1 ⊗ xi ,or, n,s n,s n equivalently, by the seminorms {πm}m∈N, where ( ) N ∞ n N X X X πm(θ) = inf am,rxi,rer : θ = ⊗ xi n ( i=1 r=1 ) i=1 XN XN n = inf kxika : θ = ⊗ xi = kθk ⊗ X , m n am i=1 i=1 n,s,π so ⊗ˆ λ (A)=lim ⊗ˆ X . X am n,s,π ←−m n,s,π Blasco—On X-Kothe¨ echelon spaces 205 We now prove that it is a reduced limit, i.e. we prove that (⊗ ia )( ⊗ˆ λX (A)) is n m n,s,π dense in ⊗ˆ Xa , where ia is the canonical inclusion of λX (A) into Xa . n,s,π m m m n ^ Fix (r1,...,rn) ∈ D (D is the set N × ···×N endowed with the ‘square order’ i) [17], [24], [12]). We have that ur ⊗ ···⊗ur ∈⊗λX (A) (where ui =(0,...,1, 0,...)). 1 s s n n,s Thus (⊗ ia )( ⊗ˆ λX (A)) ⊃ [{ur ⊗ ···⊗ur :(r1,...,rn) ∈ D}] and so (⊗ ia ) n m n,s,π 1 s s n n m ( ⊗ˆ λX (A)) is dense in ⊗ˆ Xa for each m ∈ N, since it is a linear subspace, n,s,π n,s,π m which contains a Schauder basis of ⊗ˆ Xa . n,s,π m Theorem 4.2. Let X be a Banach space with a 1-unconditional basis and A beaKothe¨ matrix on N. If ⊗ˆ X is reflexive, then ⊗ˆ λX (A) is reflexive for every Kothe¨ matrix n,s,π n,s,π n n A. Equivalently, if (P( X),τb) is reflexive, then (P( λX (A)),τb) is reflexive, for every Kothe¨ matrix A. Proof. (a) Since ⊗ˆ X is reflexive and for every mXa is isomorphic to X isomorphic n,s,π m to ⊗ˆ Xa , we have that ⊗ˆ Xa is reflexive for each m ∈ N. Thus ⊗ˆ λX (A)isa n,s,π m n,s,π m n,s,π projective limit of reflexive spaces and hence is reflexive [21, lemma 1 and theorem 1]). (b) As we have noted, the space λX (A) satisfies (BB)n,s for each n ∈ N and n every Kothe¨ matrix A. So the topologies τb and β coincide on P( λX (A)). On n n the other hand, recall that (P( X),τb)=(P( X),β) is reflexive if and only if ⊗ˆ X is reflexive [22, §23.5 (7)]. By (a) we have that ⊗ˆ λX (A) is reflexive, so n,s,π n,s,π n n (P( λX (A)),β)=(P( λX (A)),τb) is reflexive. Next consider as X the original Tsirelson space T 0. The space T 0 was introduced by Tsirelson [29] as a reflexive sequence space without any copies of c0 and `p. This space has been used to answer different questions in Banach spaces theory (see [8]). n 0 Alencar et al. [1] prove that (P( T ),τb) is reflexive for each n ∈ N. In fact, this is n the first example of an infinite dimensional Banach space E for which (P( E),τb) is reflexive for each n ∈ N. Although there are similarities between classical Kothe¨ echelon spaces and X-Kothe¨ echelon spaces, the next proposition shows that, for suitable choices of X, the structure of these spaces is different. It is known that n (P( λp(A)),τb) is a reflexive space for each n ∈ N if and only if A satisfies condition n (M) ([16], [5]), while (P( λT 0 (A)),τb) is reflexive, independent of A. From these results and Theorem 4.2 we get the following. Proposition 4.3. If T 0 is the original Tsirelson space and A is an arbitrary Kothe¨ matrix then n (P( λT 0 (A)),τb) is reflexive for each n ∈ N. 206 Mathematical Proceedings of the Royal Irish Academy The problem of the coincidence of the topologies of uniform convergence on bounded sets (τb) and Nachbin’s ported topology (τω) on spaces of n-homogeneous polynomials is an old question in holomorphy ([13], [23]). The results on reflexivity that we have obtained here give a new example of a space E for which the topologies n τb and τω coincide on P( E). We state it in the following. Corollary 4.4. For each n ∈ N and every Kothe¨ matrix A we have that n n (P( λT 0 (A)),τb)=(P( λT 0 (A)),τω). n Proof. Since (P( λT 0 (A)),τb) is reflexive, it is barrelled. Moreover, since λT 0 (A)is metrisable, τω is the barrelled topology associated with τ0 [13, example 1.37]), so τb = τω. A further application of these spaces is given next. If E isaFrechet´ Montel space n that satisfies (BB)n,s then (P( E),β) is reflexive [15]. Until now, the known examples of Frechet´ spaces E, neither Banach nor Montel, for which (P(nE),β) is reflexive for each n ∈ N can be obtained from Proposition 4.5 and all satisfy Heinrich’s density condition. Proposition 4.5 [7, proposition 5]. Let G1 beaFr´echet space of one of the following types: (a) a (FG) decomposable space that is Montel, (b) aFr´echet Schwartz space with the bounded approximation property, (c) a nuclear Fr´echet space and let G2 beaFr´echet space such that ⊗ˆ G2 is Montel (resp. reflexive, has the density n,π condition). Then ⊗ˆ (G1 ⊗ˆ G2) and ⊗ˆ (G1 × G2) are Montel (resp. reflexive, have the n,π π n,π density condition). The results we have obtained allow us to go a step further in the theory and give an example of a Frechet´ space E without the density condition such that (P(nE),β) is reflexive for each n ∈ N. 0 Proposition 4.6. Let T be the original Tsirelson space and let A0 beaKothe¨ matrix n that does not satisfy condition (D). Then (P( λT 0 (A0)),τb) is reflexive for each n ∈ N and λT 0 (A0) does not satisfy the density condition. Proof. Our choice of A0 implies that λT 0 (A0) does not satisfy Heinrich’s density condition. Corollary 4.4 lets us complete the proof. 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