Some remarks on operator ranges

Abstract

Let X be a Banach space and let X* be its conjugate space. A system (x. e, ~), 7 E F, xy ~ X, f7 ~ X*, F being some set, is said to be biorthogonal if fr(xfl) = 0 for ~ ~ fl and = 1 for 7 = fl, and fundamental if the closed linear hull [xr: 7 E F] = X. A fundamental biorthogonal system in which for each x E X, x ~ 0, there exists an index 7 E F such that f~,(x) ~ 0 is called a Markushevich basis (abbreviated: an M-basis). We say that a linear subspace Y CX is an operator range if there exist a Banach space Z and a bounded linear operator T: Z -, X such that TZ = Y. By et(F ) we shall denote the space of absolutely summable sequences x(7), 7 E F. The space el(FO), F 0 C F, will be identified with a subspace of el(F ). All the spaces are assumed to be real. THEOREM 1. Assume that there exists a continuous linear injective operator T: et(F ) --X with a dense range. Then X has two dense operator ranges, intersecting only at zero. Proof. Let Fi: i E J, be a partition of F into countable subsets: F = Ui~JFi. In each space el(Fi) there exist dense operator ranges Yi, Zi, Yi N Z i = 0 [1]. We denote by Y the subspace consisting of the absolutely convergent series ~iEJYi' Yi E Yi, and by Z the subspace consisting of the absolutely convergent series ZiEJZi, z i ~ Z i. Then TY and TZ will be the required operator ranges. COROLLARY 1. A Banach space X with a fundamental biorthogonal system (xe, fy, 7 E F) has two dense operator ranges, intersecting only at zero. Proof. Obviously, we can assume that Ix.rH = 1. Let e r be the unit vectors of the space el(F ). We set Te e = ~. The mapping T extends to a bounded linear operator from el(F ) into X. It is easy to see that it is injective. COROLLARY 2. In the space e| there exist two dense operator ranges, intersecting only at zero. Indeed, e~(F) has a fundamental biorthogonal system [2]. Theorem 1 and the corollaries are connected with the question from [3] regarding the existence in an arbitrary Banach space, in particular in e,o, of two operator ranges, intersecting only at zero. This question is related to the investigation of supportless convex sets. We recall the definitions. A subset A of a normed space is said to be supportless if no nonzero continuous linear functional attains its supremum and infimum on A. A normed space is said to be supportless if it contains a closed, bounded, convex, supportless set. We say that a Banach space is weakly compactly generated if it is the closed linear hull of a weakly compact subset. By the symbol c0(F ) we shall denote the space of those sequences x(7 ) such that for any e > 0 there exist only a finite number of coordinates exceeding e in absolute value, with the supremum norm. LEMMA 1. In the space et(F ) there exists a dense operator range which intersects the linear hull lin(e~,: 7 E F) of the unit vectors of el(F ) at zero. Proof. Let Fi: i E J, be a partition of F into countable subsets. In each el(Fi) there exists a dense operator range ~5 i, U i n fin(%: ), E Fi) = 0; this follows from the results of [1]. Then the subspace U, consisting of the absolutely convergent series ~iEjui, u i E Ui, is the required one. THEOREM 2. In each weakly compactly generated space X there exists a dense, supportless operator range. Proof. Let Y be a reflexive space and let R be a dense, injective operator from Y into X [4]. Let (Yr' ~)~,~r be an Mbasis in Y, lY~,I = 1 [5, p. 693] and let % be the standard unit vectors of the space el(F ). We set S(eT) = Yr The mapping S can be extended to a continuous linear injective operator from el(F ) into Y. According to Lemma 1, we select a dense operator range U C el(F ), U n l ine,re r = 0. Then the operator T = RS is weakly compact, iniective and, since S is a conjugate operator, A = T(Bu) (B U is the el-ball of the subspace U) is a closed subset of the normed space Z = (TU, I IX). We mention that the operator S* maps Y* into the space c0(F ) C eo.(F) = el(F)*. Therefore, if a is a support point of the set A with support