Characterizations of the reflexive spaces in the spirit of James Theorem

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1 Characterizations of the reflexive spaces in the spirit of James Theorem María D. Acosta, Julio Becerra Guerrero, and Manuel Ruiz Galán Since the famous result by James appeared, several authors have given characterizations of reflexivity in terms of the set of norm-attaining functionals. Here, we offer 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 B X and S X 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 suffices 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] simplified in some aspects the proof of James Theorem. In the 1991 Mathematics Subject Classification. Primary 46A25, 47A07 ; Secondary 47H60, 47A12. Key words and phrases. Reflexivity, James Theorem, norm attaining functional, numerical radius attaining operator, norm attaining polynomial. The first and third author were supported in part by D.G.E.S., project no. BFM The second author was partially supported by Junta de Andalucía Grant FQM

2 2 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN text by Holmes [Ho, Theorem 19.A], one can find 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 l 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 elementary. 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 A are bounded subsets of X, {x n } is a bounded sequence in X, C is the closed convex hull of {x n : n N} and it is satisfied that Then x C b B : b (x) = sup a (x). a A sup b B lim sup b (x n ) = sup n a A lim sup a (x n ). 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 {x n} in S X and a functional ϕ B X \{0} such that ϕ is a cluster point of {x n} in the w -topology of X and x X, ϕ(x) = 0. (1) Let us fix x 0 B X with ϕ(x 0 ) > 0. If we take as B := B X and A := B X in Simons inequality, we obtain which is impossible, since and sup lim sup x n(x) = x B X n sup x B X sup x B X sup lim sup x n(x) = 0 x B X n lim sup x (x n), n lim sup x (x n) ϕ(x 0 ) > 0. n It is also known that a Banach space is reflexive if it does not contain l 1 and has the Grothendieck property [Va]. A different proof of James Theorem under some restrictions can be found in [FLP, Theorem 5.9]. Here, the authors use the following key result:

3 CHARACTERIZATIONS OF THE REFLEXIVE SPACES 3 If a convex and w -compact subset K X has a boundary B which is norm separable, 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 infinite-dimensional Banach space X, there is a bounded and starlike body A 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 infinitely 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 first of these results, let us note that for a dual space X, say Y, then Y Y = X is a closed subspace of X, which is w -dense (Goldstine s Theorem) and such that Y 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 X such that Y 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 compactly generated [Ef]. In order to motivate the next result, let us begin with a simple example. For the space c 0, the set A(c 0 ) is the subset of finite supported sequences, so A(c 0 ) l 1 is of the first Baire category. Of course, the unit ball of c 0 has no slices of small diameter, since any slice of it has diameter equal to 2. That is, the unit ball of c 0 is not dentable. For separable spaces sharing with c 0 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, x 0, y X, x, y S X and t > 0. Suppose that x (y) < x 2 (x 0 ), t x 0 y 1 and y (x 0 ) > sup{y (x) : x y + ( ) ker x tb X }. Then x y 2 t x 0 y. Theorem 2.3. ([Bou, Theorem 3.5.5]) Let X be a Banach space and C X be a closed, bounded and convex subset which is separable and non-dentable. Then A(C) := {x X : x attains its maximum on C} is of the first Baire category in X.

4 4 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN Proof. Since C is not dentable, there is δ > 0 such that any slice of C has diameter at least 3δ. Since C is separable, there is a dense sequence {x n } in C. We write, for all n 1, C n = C (x n + δb X ) and O n = { x X : S(C, x, η) C n =, for some η }, where S(C, x, η) = {x C : x (x) > sup x (C) η}. Since {x n : n N} is dense in C, it holds that C = n C n and so it is immediate to check that A(C) X \( n O n ). It suffices to prove that O n is an open and dense set, for every n. Assume that x O n, therefore S(C, x, η) C n = for some η > 0. For ε > 0 such that ε sup c C c < η 4, it is satisfied that since Therefore, y x + εb X S ( C, y, η ) S(C, x, η) (1) 2 x (x) y (x) ε sup c, x C. c C ( x S C, y, η ) 2 x (x) y (x) ε sup c > c C > sup y (C) η 2 ε sup c c C sup x (C) η 2 2ε sup c C c > sup x (C) η. ( C, y, η 2 Since S(C, x, η) C n =, then by (1), we know that S ) C n = and y O n, so O n is open. Now we will check that O n is dense in X. Let us fix x X and 0 < ε < 1. Since S(C, x, η) = S(C, tx, tη) for any η, t > 0, then R + O n O n and we can assume that x = 1. C is bounded and x 0; hence we can choose y 0 X such that x (x) > x (y 0 ), x C. Let M = sup{ x y 0 : x C}, fix t > 2M ε, and write V = ker x tb X, A = V +y 0. By the choice of y 0, it holds that C n \A, since x (a) = x (y 0 ) for any a A and C n C, and so x (x) > x (y 0 ) = x (a) for any x C. Since diam C n 2δ < 3δ and C n \A, by the Hahn-Banach separation Theorem, there is a functional whose supremum on C n is greater than the supremum on A. An appropriate slice of co(c n A) determined by this functional is essentially contained in C n (see [Bou, Theorem 3.4.1]). Hence, there is a slice S of the set co(c n A) with diameter at most 3δ and such that S C n. Then in the case that C co(c n A) since C n C, S C would be a slice of C with diameter less than 3δ, which is not possible because of the choice of δ. Thus, there is an element c 0 C\co(C n A). Hence, there is a functional y S X such that y (c 0 ) > sup y (C n A). Finally, in view of 2 t c 0 y 0 2 t M ε < 1, from Lemma 2.2, it follows that x y 2 t c 0 y 2 t M < ε. Since sup y (C) y (c 0 ) > sup y (C n ), then S(C, y, η) C n = for η < sup y (C) sup y (C n ), that is, y O n and O n is dense.

5 CHARACTERIZATIONS OF THE REFLEXIVE SPACES 5 As far as the authors know, it remains open whether the previous result also holds by dropping the separability of the space. However, Kenderov, Moors and Sciffer proved the following result: Theorem 2.4. ([KMS]) If K is any infinite compact and Hausdorff topological space, then A(C(K)) is of the first Baire category. By considering the weak- topology in X, instead of the norm topology, Debs, Godefroy and Saint Raymond proved that a separable and non reflexive Banach space X satisfies that A(X) does not contain weak- open sets [DGS]. The analogous result for any Banach space was proven to hold by Jiménez- Sevilla and Moreno [JiM]. Lemma 2.5. ([JiM, Lemma 3.1]) If the dual unit ball of a Banach space X contains a slice of norm attaining functionals, then X is reflexive. Proof. Assume that x 0 S X determines a slice such that S S(B X, x 0, η) A(X). Let us fix x 0 B X with x 0 < 1 satisfying that x 0 (x 0) > 1 η 2. We define B := (S x 0) ( S + x 0) (B X x 0) (B X + x 0). By the choice of x 0, we know that an element in (B X x 0) (B X + x 0) is in fact in B, since x x 0 = y + x 0 ( x, y 1) implies that x y 2 = x 0 and we know that 1 η 2 < x 0 (x 0) = x 0 (x ) x 0 (y ) 2 1 x 0 (y ), 2 that is, x 0 ( y ) > 1 η y S, and by a similar argument x S also, so x x 0 = y + x 0 (S x 0) ( S + x 0), as we wanted to check. Hence the balanced and convex set B = (B X x 0) (B X + x 0) is a w -closed set of X and it is immediate that rb X B for 0 < r < 1 x 0. That is, B is the dual unit ball of an equivalent norm on X. By assumption, for any x X with x = 1, then x + x 0 S and x + x 0 S. Also, x + x 0 = 1 or x x 0 = 1. In the first case, there is x 0 X\{0} with (x + x 0)(x 0 ) y (x 0 ), y X, y 1 x (x 0 ) z (x 0 ), z B X x 0 x (x 0 ) z (x 0 ), z B, that is, x x attains its (new) norm at the element 0 x 0. In the second case, we argue in a similar way to arrive at the same conclusion. We checked that for the norm any functional in the unit sphere attains the norm, and so, by James Theorem, the space is reflexive. Theorem 2.6. ([JiM, Proposition 3.2]) Let X be a Banach space; then either X is reflexive or S X A(X) contains no (non empty) open set in (S X, w ). Also in the second case, the convex hull of S X \A(X) is (norm) dense in the dual unit ball.

6 6 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN Proof. If S X A(X) contains a (non-empty) w -open set in (S X, w ), we can assume without losing generality that there is a subset W X that can be expressed in the form W = {x X : x (x i ) δ i, i = 1,, n}, where x i S X, δ i > 0 (1 i n), such that W S X has a non-empty interior in the relative w -topology and such that W S X A(X). An element x in the border (norm topology) of V := W B X has to satisfy that x = 1 or x (x i ) = δ i for some 1 i n. In the first case, x S X W, so x attains its norm. As a consequence, there is an element x S X such that x (x) y (x), y V. (1) In the second case, the above inequality holds for x = x i. If y0 satisfies that y0 < 1 and y0(x i ) > δ i for every 1 i n, then y0 is interior (norm topology) to V and so the subset B := (V y 0) (y 0 V ) contains a ball centered at zero. It is clear that B is a bounded, convex, balanced and w -closed subset of X. Hence B is the dual unit ball for an equivalent norm on X. By using that every element x in the border of V satisfies (1), it is easy to check that any element in the border of B that can be written as w y 0 = y 0 z, where w or z belongs to the border of V, attains its (new) norm. By James Theorem, X is reflexive. Finally, if the convex hull of S X \A(X) were not dense in B X, then by the Hahn-Banach separation Theorem, there is a slice S of B X such that S S X A(X). It follows that S A(X), and by Lemma 2.5, X is reflexive. By using Lemma 2.5 again then, for spaces with special geometric properties of smoothness, it can be stated that the space is reflexive or A(X) contains no open subset for the norm topology. Let us recall that a Banach space X has the Mazur intersection property if, and only if, every bounded closed convex subset of X is an intersection of closed balls [Ma]. For instance, spaces with a Fréchet differentiable norm satisfy the Mazur intersection property. Theorem 2.7. ([JiM, Proposition 3.3]) A Banach space X satisfying the Mazur intersection property is either reflexive or A(X) contains no (non-empty) open set. Proof. If A(X) contains an open ball, since A(X) is a cone, then A(X) contains an open ball centered at an element x 0 in S X of radius r > 0. Since X has the Mazur intersection property, then the subset of w -denting points of the unit sphere is dense in S X [GGS1, Theorem 2.1]. Therefore, there is an element y0 S X with y0 x 0 < r 2 such that y 0 has w -slices of small diameter. Then, for every 0 < ε < r 2, there is an element y 0 X determining a slice on B X with diameter less than ε such that the slice contains y0. Hence for some η > 0 we have S(B X, y 0, η) y 0 + εb X.

7 CHARACTERIZATIONS OF THE REFLEXIVE SPACES 7 Therefore, the above slice is contained in By Lemma 2.5, X is reflexive. y 0 + r 2 B X x 0 + rb X A(X). By using a different idea of proof, we will show a new result along the same lines. Before that, let us note that every Banach space can be renormed such that the set of norm attaining functionals contains an open ball [AR1, Corollary 2]. Therefore, no trivial isomorphic assumption implies reflexivity in the case that the set of norm attaining functionals contains a ball. Let us recall that a space X is weak Hahn-Banach smooth if every x A(X) has a unique Hahn-Banach extension to X. Theorem 2.8. ([AR2, Theorem 1]) A weak Hahn-Banach space X, such that A(X) has a non-empty interior, is reflexive. Proof. Every weak Hahn-Banach space is Asplund [GGS2, Theorem 3.3], so X does not contain l 1. We will assume that X is not reflexive, therefore by [Va, Theorem 5], X does not have the Grothendieck property. That is, there is a sequence {x n} in S X which is w -convergent to zero, but not weakly convergent. Therefore, there is a non zero w -cluster point of {x n} in X, x 0 B X, also satisfying x 0 X 0. If we fix x 0 S X and ε > 0, by passing to a subsequence we will assume that x 0 (x n) > x 0 (x 0 ) ε, n N. By assumption, there is x 0 S X, r > 0 with x 0 +rb X A(X). By using Simons inequality (Proposition 1.2) for the sequence x 0 + rx n and the subsets B = B X and A = B X, we will get that x 0 x 0(x 0 ) + r ( x 0 (x 0 ) ε The previous inequality holds for any positive number ε, so Since x 0 is any element in S X, then x 0 (x 0 + rx 0 )(x 0 ). x 0 x 0 + rx 0, where x 0 X \{0}. Since x 0 (x) = 0, for every x X, the previous inequality says that the functional x 0 + rx 0 is a Hahn-Banach extension of x 0 A(X) to X, which is impossible since X is weak Hahn-Banach smooth. Therefore, X is reflexive. It is known that the previous result is not true for smooth spaces (spaces such that every element at S X has a unique normalized support functional). In fact, every separable Banach space can be renormed in order to be smooth while preserving the subset of norm attaining functionals [DGS]. Therefore, by using that every Banach space is isomorphic to a new one satisfying that the set of norm attaining functionals has a non-empty interior, in the separable case we obtain a new norm which is smooth and satisfies that the set of norm attaining functionals contains a ball. ).

8 8 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN Let us observe that Theorem 2.8 was obtained by Jiménez Sevilla and Moreno for the weak topology [JiM, Corollary 3.7]. In relation to the weak topology we can state the following result: Theorem 2.9. Let X be a separable Banach space such that A(X) S X contains a (non-empty) open set in the topological space (S X, w). Then X is quasireflexive. Proof. Suppose that there is a (relative) weak open and convex subset U B X such that its closure W satisfies W S X A(X). We may assume that 0 U. If not, take x U, x < 1, and consider the new equivalent dual norm. whose unit ball is B = ( x +B X ) (x +B X ). Observe that W 1 = ( x +W ) (x W ) satisfies W 1 S (X, ) A(X, ). We will write for the norm instead of and from now on we assume that U is of the form U = {x B X : x i (x ) < δ, i = 1,..., n} where {x i } n i=1 S X and δ is a positive number. Suppose that the linear space generated by X {x i } n i=1 is a proper subspace of X. Then there is a functional φ S X such that φ(z) = 0 for all z X {x i } n i=1. Given ε > 0, there exists y S X, such that φ(y ) > 1 ε. We are assuming that X is separable, so the restriction of the σ(x, X {x i } n i=1 {y })-topology to bounded sets is metrizable. Hence, in view of the w -denseness of B X in B X, we can find a sequence {x n} in S X that converges to φ in the σ(x, X {x i } n i=1 {y })- topology while satisfying and x n U, n 1 x n(y ) > 1 ε, n N. Now fix n 0 N. Since U is open, U S X A(X) and A(X) is a cone, we can consider r > 0, such that the ball of radius r centered at x n 0 is contained in A(X). Clearly we are in conditions to apply Proposition 1.2, taking B = B X, A = B X and {x n 0 + rx n}, so we get But, and sup lim sup(x n 0 + rx n)(x) = sup x B X n sup x B X x B X lim sup(x n 0 + rx n)(x ) (1) n sup lim sup(x n 0 + rx n)(x) = x n 0 = 1 x B X n lim sup(x n 0 + rx n)(x ) (x n 0 + rx n)(y ) (1 ε)(1 + r). n Since ε > 0 is arbitrary, the above equality (1) implies r, a contradiction. By using essentially the same idea of proof as in Theorem 2.8, but combining it with other results it was shown:

9 CHARACTERIZATIONS OF THE REFLEXIVE SPACES 9 Theorem ([ABR2, Theorem 1]) A Banach space X is reflexive if it does not contain l 1 and for some r > 0, where we denote by co w B X = co w {x S X : x + rb X A(X)}, the w -closure of the convex hull. For the space l 1 the dual unit ball satisfies the previous assumption for r = 1 2. In fact, every Banach space can be renormed such that the convex hull of the interior of the norm attaining functionals in the unit sphere is (norm) dense in the dual unit ball [ABR2, Proposition 3]. Finally, let us observe that there is also a version of Theorem 2.10 for operators instead of functionals (see [ABR1]) and a result along the same lines for the space of operators (or compact operators) between two Banach spaces. 3. James-type results for polynomials, multilinear forms and numerical radius This section is devoted to the study of some versions of James Theorem in which polynomials and multilinear forms are considered instead of linear functionals. Moreover, we give a James-type result in terms of numerical radius attaining operators. In spite of the variety of problems, it is possible to offer a unifying treatment of them, thanks to a reformulation of James Theorem stated below. Let us recall that, given a Banach space X and a bounded subset C of X, a subset B of C is said to be a boundary for C if x X b B : b (x) = sup c (x). c C Bauer s optimization Principle provides some examples: the set of extreme points of the dual unit ball of a Banach space is a boundary for that ball. The reformulation above mentioned is established in these terms: Theorem 3.1. ([Ru]) Let X be a Banach space, let C X and let B be a boundary for C. If B X, then C X. Observe that James Theorem follows easily from this result. Let us recall that a continuous homogeneous polynomial P on a Banach space X attains the norm when x 0 B E : P (x 0 ) = P := sup x B X P (x). In order to pose the right version of James Theorem for polynomials, let us note that even for reflexive spaces there are polynomials not attaining the norm. For instance, consider the continuous 2-homogeneous polynomial on l 2 given by P (x) := n 1 ( 1 1 n ) x(n) 2, (x l 2 ) which does not attain the norm, since P = 1 and if x 0 then P (x) < x 2. On the other hand, notice that, given n functionals x i X (1 i n), the polynomial P x 1 x defined by n P x 1 x n (x) := x 1(x) x n(x), (x X)

10 10 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN attains the norm provided that X is reflexive. These polynomials are precisely those that allow us to establish a polynomial James Theorem: Theorem 3.2. ([Ru]) A Banach space X is reflexive if (and only if) there exist n 1 and x 1,..., x n X \{0} such that, for all x X, the polynomial attains the norm. P x 1 x n x Let us emphasize that Theorem 3.2 follows from Theorem 3.1 for suitable subsets B of X and C of X. A different question from the one considered here, the reflexivity of the space of all continuous n-homogeneous polynomials on a Banach space, has been studied by several authors (see [JM] and the references therein). Next we exhibit a James Theorem by means of multilinear forms. A continuous n-linear form ϕ on a Banach space X attains its norm when there exist x 1,..., x n B X such that ϕ(x 1,..., x n ) = ϕ := sup{ ϕ(y 1,..., y n ) : y 1,..., y n B X }). The multilinear version of Theorem 3.2 can be easily derived form James Theorem, but this does not happen in the symmetric case. If x 1,..., x n X, we will write for the symmetrization of the multilinear form S x 1 x n i.e., (x 1,..., x n ) x 1(x 1 ) x n(x n ), (x 1,..., x n X), S x 1 x (x n 1,..., x n ) = 1 x σ(1) n! (x 1) x σ(n) (x n) σ n ( n is the set of all permutations of n elements). For this kind of multilinear forms we have, making use of Theorem 3.1, Theorem 3.3. ([Ru]) A Banach space X is reflexive if (and only if) there are n 1 and x 1,..., x n X \{0} such that for all x X the symmetric (n + 1)-linear form S x 1 x n x attains the norm. Let us observe that it is not possible to derive Theorem 3.2 or 3.3 directly from James Theorem: we will give two simple examples of a polynomial and a symmetric bilinear form of the finite type attaining the norm and such that this does not happen for all the functionals involved. Example. Let us fix an element x B l1 with x (e 1 ) = 0 and consider the polynomial P on c 0 given by x (e 1 x )(x)(e 1 + x )(x). If x B c0 is a fixed element and we write α := x (x), then P (x) = (x(1) α)(x(1) + α) = x(1) 2 α 2 max{1, α 2 } = 1, and P (e 1 ) = 1, so P attains its norm and the functionals e 1 + x, e 1 + x do not attain the norm if the support of x is not finite.

11 CHARACTERIZATIONS OF THE REFLEXIVE SPACES 11 For the symmetric bilinear form, we will take X = l 1 and a functional x = i=1 α ne n l where α 1 = 1 2, 0 < α n < 1 and lim {α n } = 1. The symmetric bilinear form S on l 1 given by ( ) (x, y) α n x(n) y(1) + n=1 ( n=1 ) α n y(n) x(1) attains the norm at (e 1, e 1 ) since for x, y B l1, if we write a = x(1), b = y(1) and we assume that x(n), y(n) 0, we get ( 1 ) ( 1 ) S(x, y) 2 a + (1 a) b + 2 b + (1 b) a = = a + b ab = a(1 b) + b 1. Since x = 1 and {α n } satisfies 0 < α n < 1 for all n, x does not attain the norm. We now consider a version of James Theorem for numerical radius. The numerical range of an operator T L(X) (Banach algebra of all bounded and linear operators on X) is given by where and its numerical radius by V (T ) := {x (T x) : (x, x ) Π(X)}, Π(X) := {(x, x ) X X : x = x = x (x) = 1} v(t ) := sup{ µ : µ V (T )}. An operator T L(X) is said to attain its numerical radius if (x 0, x 0) Π(X) : x 0(T x 0 ) = v(t ). Theorem 3.1 makes it possible to obtain a James-type result in terms of numerical radius: Theorem 3.4. ([AR3]) A Banach space is reflexive provided that each rankone operator attains its numerical radius. The converse is false; in fact, in [AR3] it was proven that a separable Banach space is finite-dimensional if, and only if, for any equivalent norm, every rank-one operator attains the numerical radius. It is not known if every reflexive Banach space has an equivalent norm for which every rank-one operator attains its numerical radius. A partial answer was given in [ABR3]. Theorem 3.5. A Banach space X is reflexive if, and only if, there exist x 0 S X and an equivalent norm for which each rank-one operator of the form x x 0 (x X ) attains its numerical radius. By using the refinement of James Theorem proven by Jiménez Sevilla and Moreno (Theorem 2.6), some slight improvements to the results we mentioned can be given: Theorem 3.6. ([ABR3]) Let X be a Banach space. Then the following conditions are equivalent: i) X is reflexive.

12 12 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN ii) There is n 1 and x 1,..., x n X \{0} such that the set {x X : P x 1 x nx attains its norm} has a non-empty weak- interior. iii) For some n 1 and x 1,..., x n X \{0} the set {x X : S x 1 x nx attains its norm} has a non-empty weak- interior. If there is an element x 0 S X satisfying that the set {x X : x x 0 attains its numerical radius} has a non-empty weak- interior, the space X is reflexive. 4. Open questions 1. If X is a Banach space such that X contains a w -dense linear subspace which is norm closed and contained in A(X), is X isometric to a dual space? 2. If B X is non-dentable, is A(X) of the first Baire category (norm topology)? 3. If for any equivalent norm on X, the subset A(X) has a non-empty interior, is X reflexive? 4. If A(X) S X W S X for some w-open set W, is X reflexive? 5. Given a Banach space X and ε > 0, is there a Banach space Y with d(y, X) < 1 + ε and such that B Y = co w {x S X : x in the interior of A(Y )}? 6. Suppose that X has the Mazur intersection property and NA(X, Y ) has a non-empty interior (Y {0}), is X reflexive? 7. If X is separable and the unit ball is non-dentable, is NA(X, Y ) of the first Baire category, for any Banach space Y? 8. Can we assure that any reflexive Banach space has an equivalent norm for which every rank-one operator attains its numerical radius?

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14 14 MARÍA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALÁN [Ma] S. Mazur, Über schwache Konvergenz in den Raumen lp. Studia Math. 4 (1933), [PPl] J.I. Petunin and A.N. Plichko, Some properties of the set of functionals that attain a supremum on the unit sphere. Ukrain. Math. Zh. 26 (1974), , MR 49:1075. [Pr] J.D. Pryce, Weak compactness in locally convex spaces. Proc. Amer. Math. Soc. 17 (1966), [Ru] M. Ruiz Galán, Boundaries and weak compactness, preprint. [Si1] S. Simons, Maximinimax, minimax, and antiminimax theorems and a result of R.C. James. Pacific J. Math. 40 (1972), [Si2] S. Simons, An eigenvector proof of Fatou s lemma for continuous functions. Math. Intelligencer, 17 (1995), [Va] M. Valdivia, Fréchet spaces with no subspaces isomorphic to l 1. Math. Japon. 38 (1993), Departamento de Análisis Matemático, Universidad de Granada, Granada, Spain address: dacosta@ugr.es Departamento de Matemática Aplicada, Universidad de Granada, Granada, Spain address: juliobg@ugr.es Departamento de Matemática Aplicada, Universidad de Granada, Granada, Spain address: mruizg@ugr.es