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Characterizations of the Reflexive Spaces in the Spirit of James' Theorem

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Characterizations of the Reflexive Spaces in the Spirit of James' Theorem Characterizations of the reflexive spaces in the spirit of James' Theorem Mar¶³aD. Acosta, Julio Becerra Guerrero, and Manuel Ruiz Gal¶an Since the famous result by James appeared, several authors have given char- acterizations of reflexivity in terms of the set of norm-attaining functionals. Here, we o®er a survey of results along these lines. Section 1 contains a brief history of the classical results. In Section 2 we assume some topological properties on the size of the set of norm-attaining functionals in order to obtain reflexivity of the space or some weaker conditions. Finally, in Section 3, we consider other functions, such as polynomials or multilinear mappings instead of functionals, and state analogous versions of James' Theorem. Hereinafter, we will denote by BX and SX the closed unit ball and the unit sphere, respectively, of a Banach space X. The stated results are valid in the complex case. However, for the sake of simplicity, we will only consider real normed spaces. 1. James' Theorem In 1950 Klee proved that a Banach space X is reflexive provided that for every space isomorphic to X, each functional attains its norm [Kl]. James showed in 1957 that a separable Banach space allowing every functional to attain its norm has to be reflexive [Ja1]. This result was generalized to the non-separable case in [Ja2, Theorem 5]. After that, a general characterization of the bounded, closed and convex subsets of a Banach space that are weakly compact was obtained: Theorem 1.1. ([Ja3, Theorem 4]) In order that a bounded, closed and convex subset K of a Banach space be weakly compact, it su±ces that every functional attain its supremum on K: See also [Ja3, Theorem 6] for a version in locally convex spaces. James gave a wider list of characterizations of reflexivity in [Ja4, Theorems 1, 2 and 3]. A simpler proof of this result for the unit ball appeared in [Ja5, Theorem 2]. Since then, some authors have tried to obtain easier proofs for this result. For instance, Pryce [Pr] simpli¯ed in some aspects the proof of James' Theorem. In the 1991 Mathematics Subject Classi¯cation. Primary 46A25, 47A07 ; Secondary 47H60, 47A12. Key words and phrases. Reflexivity, James' Theorem, norm attaining functional, numerical radius attaining operator, norm attaining polynomial. The ¯rst and third author were supported in part by D.G.E.S., project no. BFM 2000-1467. The second author was partially supported by Junta de Andaluc¶³aGrant FQM0199. 1 2 MAR¶IA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALAN¶ text by Holmes [Ho, Theorem 19.A], one can ¯nd Pryce's proof in the case that the dual unit ball is weak-¤ sequentially compact, which enables us to eliminate some steps. Another proof, owing to Simons [Si1], makes use of a minimax inequality. We will show a proof of Theorem 1.1 for those spaces that do not contain an isomorphic copy of `1 or are separable. The key result in this proof is another result by Simons, the so-called \Simons' inequality" [Si2]. We also refer to J. Diestel [Di] and K. Floret [Fl]. No proof of James' Theorem, in its general version, can be considered elemen- tary. However, in the text by Deville, Godefroy and Zizler [DGZ] an easy proof is obtained from Simons' inequality in the separable case. Proposition 1.2. ([Si2]) Suppose that X is a Banach space, B ½ A are ¤ bounded subsets of X , fxng is a bounded sequence in X, C is the closed convex hull of fxn : n 2 Ng and it is satis¯ed that 8x 2 C 9 b¤ 2 B : b¤(x) = sup a¤(x): a¤2A Then ¤ ¤ sup lim sup b (xn) = sup lim sup a (xn): b¤2B n a¤2A n Let us recall that a Banach space has theGrothendieck property provided that each weak-¤ null sequence in X¤ is actually weak null. It is very easy to check that a separable Banach space satisfying the Grothendieck property is reflexive. Proposition 1.3. Let X be a Banach space such that every functional attains its norm. Then X has the Grothendieck property. Proof. Let us assume that X does not satisfy the Grothendieck property. ¤ Then there exist a weak-¤ null sequence fxng in SX¤ and a functional ' 2 BX¤¤¤ nf0g ¤ ¤ ¤¤¤ such that ' is a cluster point of fxng in the w -topology of X and 8x 2 X; '(x) = 0: (1) ¤¤ ¤¤ Let us ¯x x0 2 BX¤¤ with '(x0 ) > 0: If we take as B := BX and A := BX¤¤ in Simons' inequality, we obtain ¤ ¤¤ ¤ sup lim sup xn(x) = sup lim sup x (xn); ¤¤ x2BX n x 2BX¤¤ n which is impossible, since ¤ sup lim sup xn(x) = 0 x2BX n and ¤¤ ¤ ¤¤ sup lim sup x (xn) ¸ '(x0 ) > 0: ¤¤ x 2BX¤¤ n It is also known that a Banach space is reflexive if it does not contain `1 and has the Grothendieck property [Va]. A di®erent proof of James' Theorem under some restrictions can be found in [FLP, Theorem 5.9]. Here, the authors use the following key result: CHARACTERIZATIONS OF THE REFLEXIVE SPACES 3 If a convex and w¤-compact subset K ½ X¤ has a boundary B which is norm sep- arable, then K is the closure (in the norm topology) of the convex hull of B [FLP, Theorem 5.7]. The above result can be easily deduced from Simons' inequality (see also [Go, Theorem I.2]). Azagra and Deville proved that in any in¯nite-dimensional Banach space X, there is a bounded and starlike body A ½ X (subset containing a ball centered at zero such that every ray from zero meets the boundary of A once at the most) such that every functional attains in¯nitely many local maxima on A [AD]. 2. Other results relating the size of the set of norm attaining functionals with properties of the Banach space In the following, we will write A(X) for the set of norm attaining functionals on X. Let us recall that this set is always dense in the dual space (Bishop-Phelps Theorem [BP1]). Apart from James' Theorem, some other results imply isomorphic properties of X by assuming a weaker condition on the size of A(X). In order to state the ¯rst of these results, let us note that for a dual space X, say Y ¤, then Y ½ Y ¤¤ = X¤ is a closed subspace of X¤, which is w¤-dense (Goldstine's Theorem) and such that Y ½ A(Y ¤) = A(X). Petunin and Plichko got the converse result under additional conditions. Theorem 2.1. ([PPl]) Assume that X is separable and there is a Banach space Y ½ X¤ such that Y ½ A(X) and Y is w¤-dense in X¤. Then X is isometric to a dual space. The previous result was also shown to hold in the case that X is weakly com- pactly generated [Ef]. In order to motivate the next result, let us begin with a simple example. For the space c0, the set A(c0) is the subset of ¯nite supported sequences, so A(c0) ½ `1 is of the ¯rst Baire category. Of course, the unit ball of c0 has no slices of small diameter, since any slice of it has diameter equal to 2. That is, the unit ball of c0 is not dentable. For separable spaces sharing with c0 this isometric property of the unit ball, as a consequence of the techniques developed by Bourgain and Stegall, we obtain a similar result. Before stating it, let us introduce the following technical result: Lemma 2.2. ([Bou, Lemma 3.3.3]) Let X be a Banach space, x0; y 2 X; ¤ ¤ ¤ ¤ 2 x ; y 2 S ¤ and t > 0: Suppose that x (y) < x (x ); kx ¡ yk · 1 and X ¡ ¢ 0 t 0 ¤ ¤ ¤ y (x0) > supfy (x): x 2 y + ker x \ tBX g: Then 2 kx¤ ¡ y¤k · kx ¡ yk: t 0 Theorem 2.3. ([Bou, Theorem 3.5.5]) Let X be a Banach space and C ½ X be a closed, bounded and convex subset which is separable and non-dentable. Then A(C) := fx¤ 2 X¤ : x¤ attains its maximum on Cg is of the ¯rst Baire category in X¤. 4 MAR¶IA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALAN¶ Proof. Since C is not dentable, there is ± > 0 such that any slice of C has diameter at least 3±. Since C is separable, there is a dense sequence fxng in C. We write, for all n ¸ 1; Cn = C \ (xn + ±BX ) and © ¤ ¤ ¤ ª On = x 2 X : S(C; x ; ´) \ Cn = ;; for some ´ ; ¤ ¤ ¤ where S(C; x ; ´) = fx 2 C : x (x) > sup x (C)¡´g. Since fxn : n 2 Ng is dense in ¤ C, it holds that C = [nCn and so it is immediate to check that A(C) ½ X n(\nOn). It su±ces to prove that On is an open and dense set, for every n. Assume ¤ ¤ that x 2 On, therefore S(C; x ; ´) \ Cn = ; for some ´ > 0. For " > 0 such that ´ " supc2C kck < , it is satis¯ed that 4 ³ ´ ¤ ¤ ¤ ´ ¤ y 2 x + "B ¤ ) S C; y ; ½ S(C; x ; ´) (1) X 2 since jx¤(x) ¡ y¤(x)j · " sup kck; 8x 2 C: c2C Therefore, ³ ´ ´ x 2 S C; y¤; ) x¤(x) ¸ y¤(x) ¡ " sup kck > 2 c2C ´ > sup y¤(C) ¡ ¡ " sup kck ¸ 2 c2C ´ ¸ sup x¤(C) ¡ ¡ 2" sup kck > sup x¤(C) ¡ ´: 2 c2C ³ ´ ¤ ¤ ´ Since S(C; x ; ´) \ Cn = ;, then by (1), we know that S C; y ; 2 \ Cn = ; and ¤ y 2 On, so On is open.
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