Some New Results on Frechet Spaces with Nuclear Kothe Quotients

INTRODUCTION

In 1936, M. Eidelheit showed that any non-Banach Frechet space has a quotient Isomorphic to . In 1957, C. Bessaga and A.Pelczynski showed that a Frechet space fails to admit a continuous norm iff it has a subspace isomorphic to. Again in 1959, Bessaga, Pelczynki and S.Rolewicz showed that a Frechet space which admits continuous norm has a nuclear kothe subspace iff it is not Banach. A Frechet space which admits continuous norm can have a quotient. If we consider only nuclear Frechet spaces, then we are looking for kothe quotients and the problem has a positive solution: every nuclear Frechet space not isomorphic to have a kothe quotients.

Our characterization of Frechet spaces with nuclear kothe quotients will be in terms o the following condition which we lable(*) .If E is a Frechet space and (|| .||k ) a fundamental sequence of seminorms we denote by the banach space determined by the unit ball of the dual norm .Moreover ,in view of the open mappings theorem ,the condition has the following equivalent formulation: In this form our condition is very close to being a dual to he following condition used by Bessaga, Pelczynski and Rolewicz in their determination of those Frechet spaces which have nuclear kothe subspaces. every subspaces Y of E with Finite co dimension. The role of condition (*) in our characterization is contained in the following result. : Frechet space E satisfies condition (*) iff it has a quotient which admits continuous norm and satisfies condition (*). Suppose That E satisfies condition (*).We may assume for this condition that and , Let which by the bipolar theorem ,is the closure of in the weak topology from E .we will show that is the desired quotient. First we check that admits continuous norm. Let and suppose that the seminorm in induced by 1 annihilates .This means that ⊂.

Hence we have Since it follows that lim so . This shows that ,so the seminorm induced by 1 is a norm. Now we verify condition (*) for . Now we fix k and let Vk be the unit ball of k in E . Snce E satisfies condition (*) we have a sequence with and .This implies that so a fortiori 0 . Moreover , if , then since Hence 0 . Thus we have shown that 0 0 . But (= M and the unit ball of the dual norm of the norm in induced by k is . This shows that satisfies condition (*). let be a qutient of E. If ( is a fundamental sequence of nhds.of O for E then by general duality, can be represented as a vector subspace of and a fundamental sequence of

equicontinuous sets for( )ř

Is given by k . Hence if satisfies condition (*) (say with and then and a sequence of constants with ). then so k ř. Hence E satisfies condition (*). Theorem (1) A seperable Frechet space E has a nuclear Kothe quotient iff it satisfies condition (*). Proof Suppose that E satisfies condition (*) .By proposition (1) we may suppose that E admits continuous norm. Let ( be a dense sequence in E and the vector subspace it generates .we will establish the existence of a nuclear kothe quotient by constructing a biorthogonal sequence (. Let be a fundamental sequence of norms for E .The space defined in the beginning of this section is the dual of the normed space (k) We may assume that condition (*) is satisfied with and .Then we have for each kN. řk/||g||řk+1: g Our construction is by induction. We can choose with . Since is dense in E . Before passing to the induction .We must prove that the above statement of condition (*) remains true if the requirement is replaced by the weaker requirement that where for some finite . We may assume that L is a linearly independent set say .Since is a norm , is dense in with respect to the weak topology from E so we can find is biorthogonal. Then the map defined by is a projection whose kernel is G and P is continuous for every Hence if we have with And , since H is finite dimensional, řk řk+1 (h Therfore we have,for each , we may conclude that Turning to our induction step we assume that (i=1,.........,) have been selected. We apply the above condition with Then we select inductively so as to annihilate , , and satisfy, for , we set annihilates , as desired. and nduction is completed. A Frechet space E is a quojection iff every quotient of E which admits continuous norm is a Banach space. Proof Let E be a quojection and f be a quotient of E. If E is the projective limit of the surjections then it is easy to see from the definition of projective limit that the canonical projections (given by are also surjections .Let be the quotient map,|.| a continuous norm on F and the norm on . By the ontinuity of and k ,x E.Hence from an algebraic point of view is a unique map Since T is surjection S is also .Moreover ,it follows from the losed graph theorem and the fact that |.| is a norm that S is continuous.Hence ,F is a quotient of a Banach space so it is a Banach space. Coversely let be aa fundamental sequence of seminorms for E ,the kernel of and the quotient Fechet space. Since is increasing we have the canonical maps which are surjections and it is easy to see that E is isomorphic to the projective limit of this sequence of maps.On the other hand ,the seminorm induced on by is clearly a norm so by assumption , is aBanach space.Thus ,E is aquojection. Corollary(1) If E is a quojection then E does not have a nuclear Kothe quotient.

space then E does not have a nuclear kothe quotient. It would be nice to know that the quojections are precisely those Frechet spaces (amongst separable spaces) which fail to have nuclear kothe quotients . It would be nice to know that the quojections are precisely those Frechet spaces(amongst separable spaces) which fail to have nuclear kothe quotients .Unfortunately ,we are unable to prove the converse of corollary 1 so this remains open. We can obtain this result ,however ,if we restrict our considerations to reflexive spaces. In order to investigate this situation we consider , for an arbitrary Frechet spaces E, the vector space of all linear functional on which are bounded on bounded sets .Obviously , is a dual system and we indicate thepolar of a set A by . We call consider the topology on of uniform convergence on bounded sets,that is,a fundamental system of neighbourhood of o is given by the sets B is a bounded subset of . Proposition (3) If E is a separable Frechet space which does not have a nuclear Kothe quotient then is quojetion.

We can apply theorem(2) to conclude that E does not satisfy condition(*) and so we can find a fundamental sequence of seminorms for for every k, the closure of is closed in each Banach space ,.This implies that if is the closure of in and each is equipped with the norm then is closed subspace of . But the unit balls of the form a fundamental sequence of bounded sets for and it is easy to check that is isomorphic to the projective limit of the sequence of maps , adjoint to the inclusions of it follows that the maps are surjections so is quojetion. If E is a separable reflexive Frechet space,then E has a nuclear Kothe quotient iff E is not a quojection. Corollary(4) Every Frechet Montel space not isomorphic to has a nuclear kothe quotient.

If E is a Frechet Montel space then, as is well known ,E is separable and reflexive. We will show that E is not a quojection. Suppose that E is the projective limit of the surjections We may assume that maps the unit ball of onto the unit ball of .Also ,since E is not isomorphic to we may assume that is infinite dimensional. Then, viewing E as the projective limit, it is easy to see that is a closed,boundedsubset of E.But the projection of this set in is the unit ball so it is not compact.Hence ,the set is not compact in E which is a contradiction. Frechet spaces which admits continuous norm.Again the situation with quotients seems more difficult than in the case of subspaces.We do not know whether a separable non-banach Frechet spac which admits a continuous norm (or is even countable normed) necessarily has nuclear kothe quotient. Actually, this is the same as the above question regarding the converse of corollary (1). In fact, if E is not a quojection, then by proposition (2), E has a non-Banach quotient which admits a continuous norm, so if this question had a positive answer, E would have a nuclear kothe quotient and thus the converse of corollary (1) would hold. Conversely, if E is a non-Banach Frechet space which admits continuous norm then clearly E is not a quojection so if the converse of corollary (1) held ,E would have nuclear kothe quotient.

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Department of Mathematics, B.N. College, Patna-800005, Bihar, India abhik51@gmail.com