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Geometric Properties on Non-Complete Spaces 11 GEOMETRIC PROPERTIES ON NON-COMPLETE SPACES FRANCISCO J GARC´IA-PACHECO AND BENTUO ZHENG This paper is dedicated to the beloved memory of Prof. Antonio Aizpuru Abstract. The purpose of this paper is to study certain geometrical proper- ties for non-complete normed spaces. We show the existence of a non-rotund Banach space with a rotund dense maximal subspace. As a consequence, we prove that every separable Banach space can be renormed to be non-rotund and to contain a dense maximal rotund subspace. We then construct a non- smooth Banach space with a dense maximal smooth subspace. We also study the Krein-Milman Property on non-complete normed spaces and provide a suf- ficient condition for an infinite dimensional Banach space to have an infinite dimensional, separable quotient. 1. Introduction Various geometric properties of normed linear spaces have been studied for many years. An important question is to find out which geometric properties of non- complete normed spaces extend to their completion. It is well known [7] that a Banach space is reflexive if and only if every bounded linear functional on X is norm attaining. In the 1960's, Frank Deutsch asked whether every normed linear space is complete if every bounded linear functional on it is norm attaining. If this was true, then the space is automatically reflexive. However James [7] gave a negative answer by constructing the following example. This is probably the first result on the weakness of geometrical properties under the lack of completeness. Theorem 1.1 (James, 1971). There exists a non-complete normed space on which every functional is norm-attaining. In 1976 Blatter toughened the condition that \all functionals are norm-attaining" accomplishing something positive (see [1]). Theorem 1.2 (Blatter, 1976). Let X be a normed space. If every closed, convex subset of X has a minimum-norm element, then X is complete (and thus reflexive). Some of the geometric properties pass to dense subspaces. Two of the most important such properties are (uniform) rotundity and (uniform) smoothness. The modulus of rotundity of a normed linear space X is the function : [0; 2] ! [0; 1] kx+yk defined by () = inff1 − 2 : x; y 2 BX ; kx − yk ≥ g, where BX is the closed unit ball of X. If () > 0 for all 0 < ≤ 2, then X is said to be uni- formly rotund. By approximation, it is easy to see from the definition that if a 2000 Mathematics Subject Classification. Primary 46B20, 46B15, 46A08. Key words and phrases. rotund, smooth, Markushevich basis, dense maximal subspace, cs- closed. Bentuo Zheng is supported in part by NSF.. 1 2 FRANCISCO J GARC´IA-PACHECO AND BENTUO ZHENG dense subspace of a Banach space is uniformly rotund (uniformly smooth), then the Banach space itself is also uniformly rotund (uniformly smooth). So uniform convexity (uniform smoothness) extends from non-complete normed spaces to their completion. Considering the above discussion, a natural question we have here is that whether rotundity and smoothness extend from non-complete normed spaces to their completion. It is quite surprising that the answer turns out to be negative. In Section 2, we construct a non-complete normed space X so that X is rotund but the completion X¯ of X is non-rotund. Moreover, X as a dense subspace of X¯ is also maximal. As a consequence, we have the following theorem. Theorem A. Let X be an infinite dimensional Banach space that admits a rotund equivalent norm. Then, X admits an equivalent renorming such that X is non- rotund but has a rotund, dense, and maximal subspace. A subspace of a Banach space is said to be maximal if it is of codimension one. Since every separable Banach space can be renormed to be rotund by Clarkson [3], we get the following corollary immediately. Corollary A. Every infinite dimensional separable Banach space can be renormed to be non-rotund and to contain a rotund, dense, and maximal subspace. Locally uniformly rotund spaces can be viewed as objects better than rotund spaces but worse than uniformly rotund spaces (the definition is contained in Sec- tion 2). Given Theorem A and the discussion above, a natural question is whether there exists a non-rotund Banach space with a locally uniformly rotund dense sub- space. Section 3 is devoted to the study of this question. Although we believe that the answer is positive, we only get the following partial result. Theorem B. If there exists a three dimensional normed space so that the unit sphere is symmetric with respect to the z-coordinate and contains exactly two sym- metric maximal segments, then every infinite dimensional Banach space that admits a locally uniformly rotund norm can be renormed to be non-rotund and to contain a locally uniformly rotund dense maximal subspace. We also give a closed surface in R3 which is a possible candidate to satisfy the hypothesis of Theorem B. In Section 4, we construct a non-complete normed space Y so that Y is Fr´echet- smooth but the completion is not Gateaux-smooth. To be more precise, we have Theorem C. Let Y be an infinite dimensional Banach space that admits a Fr´echet- smooth equivalent norm. Then Y admits an equivalent renorming such that Y is non-Gateaux smooth but has a Fr´echet-smooth, dense, and maximal subspace. The study of the conditions for the equivalence of RNP and Krein-Milman prop- erty (KMP) is an important and interesting subject. There are many positive results for certain classes of Banach spaces. However, it is shown in [10] that RNP and KMP are incomparable properties in non-complete spaces. So an open problem is whether the KMP for a normed space implies the completeness of the normed GEOMETRIC PROPERTIES ON NON-COMPLETE SPACES 3 space. In Section 5, we show that the KMP can substitute completeness in cer- tain settings. Using Baire Category theorem, it is easy to see that every infinite dimensional Banach space does not admit a countable Hamel basis. By a simple construction, we prove Theorem D. Every infinite dimensional normed space with the KMP does not have a countable Hamel basis. The failing of KMP in non-complete spaces exists almost everywhere in virtue of the following theorem. Theorem E. Every infinite dimensional separable Banach space has a dense sub- space lacking the KMP. In Section 6, we revisit the characterization of the separable quotient problem by Rosenthal [13, 14]. By introducing the so called cs-complemented subspaces of a Banach space, we release a little bit of the conditions given by Rosenthal. 2. A non-rotund Banach space with a dense and maximal rotund subspace Definition 2.1. Let X be a normed space. Let x 2 SX . Then: (1) The point x is said to be a rotund point of BX if x is not contained in any x+y non-trivial segment of SX , that is, if y 2 SX and 2 = 1, then x = y. The set of rotund points of BX is usually denoted as rot (BX ). (2) The point x is said to be a locally uniformly rotund point of BX if every x+yn sequence (yn) ⊂ BX such that converges to 1 is convergent n2N 2 n2N to x. The set of locally uniformly rotund points of BX is usually denoted as rotu (BX ). As the reader may consult in the previous three references, a normed space is rotund or locally uniformly rotund if every element in its unit sphere is a rotund point or a locally uniformly rotund point. Let X be a normed space. We use SX to denote its unit sphere and BX to denote its unit ball. Lemma 2.2. Let X be an infinite dimensional Banach space enjoying the following property: there exist x 6= y 2 SX such that [x; y] and [−x; −y] are the only maximal segments in SX . Then, X is non-rotund but has a rotund dense and maximal subspace. Proof. Let f : X ! R be a non-continuous linear functional such that f (x) = 0 and f (y) = 1. We know that ker (f) is a dense and maximal subspace of X and ((x; y] [ (−x; −y])\ker (f) = ?. By hypothesis, rot (BX ) = SX n([x; y] [ [−x; −y]). We will show that ker (f) is rotund. Observe that Sker(f) n fx; −xg ⊆ SX n ([x; y] [ [−x; −y]) \ ker (f) = rot (BX ) \ ker (f) ⊆ rot Bker(f) : It remains to show that x; −x 2 rot Bker(f) . Assume not. Then, there exists z 2 Sker(f) n fx; −xg such that [z; x] ⊂ Sker(f) ⊂ SX and [−z; −x] ⊂ Sker(f) ⊂ SX . By hypothesis, z 2 [y; x) and −z 2 [−y; −x), which is a contradiction. □ 4 FRANCISCO J GARC´IA-PACHECO AND BENTUO ZHENG Remark 2.3. In the proof of the previous lemma it would have been much easier to take f : X ! R a non-continuous linear functional such that f (x) = f (y) = 1. However, we wanted to show that the dense maximal rotund subspace can be chosen to intersect a segment of the superspace. We will show in the next section that this situation is not possible with dense locally uniformly rotund subspaces. In the next section we will discuss about the existence of infinite dimensional Banach spaces exists enjoying the hypothesis of Lemma 2.2. Lemma 2.4. There exists a 2-dimensional Banach space Y with the following property: there exist x 6= y 2 SY such that [x; y] and [−x; −y] are the only maximal segments in SX .
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