Characterizations of the reflexive spaces in the spirit of James’ Theorem

Characterizations of the reflexive spaces in the spirit ofJames’ Theorem

Marıa D. Acosta, Julio Becerra Guerrero, and Manuel Ruiz Galan

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 offer a survey of results along these lines. Section 1 contains a brief history ofthe classical results. In Section 2 we assume some topological properties on the sizeof the set of norm-attaining functionals in order to obtain reflexivity of the spaceor some weaker conditions. Finally, in Section 3, we consider other functions, suchas polynomials or multilinear mappings instead of functionals, and state analogousversions of James’ Theorem.

Hereinafter, we will denote by BX and SX the closed unit ball and the unitsphere, respectively, of a Banach space X. The stated results are valid in thecomplex case. However, for the sake of simplicity, we will only consider real normedspaces.

1. James’ Theorem

In 1950 Klee proved that a Banach space X is reflexive provided that for everyspace isomorphic to X, each functional attains its norm [Kl]. James showed in1957 that a separable Banach space allowing every functional to attain its normhas 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, closedand 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 convexsubset K of a Banach space be weakly compact, it suffices that every functionalattain its supremum on K.

See also [Ja3, Theorem 6] for a version in locally convex spaces. James gavea wider list of characterizations of reflexivity in [Ja4, Theorems 1, 2 and 3]. Asimpler 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. Forinstance, 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 2000-1467.The second author was partially supported by Junta de Andalucıa Grant FQM0199.

1

2 MARIA D. ACOSTA, JULIO BECERRA GUERRERO, AND MANUEL RUIZ GALAN

text by Holmes [Ho, Theorem 19.A], one can find Pryce’s proof in the case that thedual unit ball is weak-∗ sequentially compact, which enables us to eliminate somesteps. Another proof, owing to Simons [Si1], makes use of a minimax inequality. Wewill show a proof of Theorem 1.1 for those spaces that do not contain an isomorphiccopy 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 isobtained from Simons’ inequality in the separable case.

Proposition 1.2. ([Si2]) Suppose that X is a Banach space, B ⊂ A arebounded subsets of X∗, {xn} is a bounded sequence in X, C is the closed convexhull of {xn : n ∈ N} and it is satisfied that

∀x ∈ C ∃ b∗ ∈ B : b∗(x) = supa∗∈A

a∗(x).

Thensupb∗∈B

lim supn

b∗(xn) = supa∗∈A

lim supn

a∗(xn).

Let us recall that a Banach space has theGrothendieck property provided thateach weak-∗ null sequence in X∗ is actually weak null. It is very easy to check thata separable Banach space satisfying the Grothendieck property is reflexive.

Proposition 1.3. Let X be a Banach space such that every functional attainsits 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 SX∗ and a functional ϕ ∈ BX∗∗∗\{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 ∈ BX∗∗ with ϕ(x∗∗0 ) > 0. If we take as B := BX and A := BX∗∗

in Simons’ inequality, we obtain

supx∈BX

lim supn

x∗n(x) = supx∗∗∈BX∗∗

lim supn

x∗∗(x∗n),

which is impossible, since

supx∈BX

lim supn

x∗n(x) = 0

andsup

x∗∗∈BX∗∗lim sup

nx∗∗(x∗n) ≥ ϕ(x∗∗0 ) > 0.

It is also known that a Banach space is reflexive if it does not contain `1 andhas 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:

CHARACTERIZATIONS OF THE REFLEXIVE SPACES 3

If a convex and w∗-compact subset K ⊂ 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 (seealso [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 atzero such that every ray from zero meets the boundary of A once at the most) suchthat every functional attains infinitely many local maxima on A [AD].

2. Other results relating the size of the set of norm attainingfunctionals with properties of the Banach space

In the following, we will write A(X) for the set of norm attaining functionalson X. Let us recall that this set is always dense in the dual space (Bishop-PhelpsTheorem [BP1]).

Apart from James’ Theorem, some other results imply isomorphic properties ofX by assuming a weaker condition on the size of A(X). In order to state the firstof 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 thatY ⊂ A(Y ∗) = A(X). Petunin and Plichko got the converse result under additionalconditions.

Theorem 2.1. ([PPl]) Assume that X is separable and there is a Banach spaceY ⊂ X∗ such that Y ⊂ A(X) and Y is w∗-dense in X∗. Then X is isometric to adual 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. Forthe space c0, the set A(c0) is the subset of finite supported sequences, so A(c0) ⊂ `1is of the first Baire category. Of course, the unit ball of c0 has no slices of smalldiameter, 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 theunit 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 technicalresult:

Lemma 2.2. ([Bou, Lemma 3.3.3]) Let X be a Banach space, x0, y ∈ X,x∗, y∗ ∈ SX∗ and t > 0. Suppose that x∗(y) < x∗(x0), 2

t ‖x0 − y‖ ≤ 1 andy∗(x0) > sup{y∗(x) : x ∈ y +

(kerx∗ ∩ tBX

)}. Then

‖x∗ − y∗‖ ≤ 2t‖x0 − y‖.

Theorem 2.3. ([Bou, Theorem 3.5.5]) Let X be a Banach space and C ⊂ Xbe 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∗.


Proof. Since C is not dentable, there is δ > 0 such that any slice of C has diameterat least 3δ. Since C is separable, there is a dense sequence {xn} in C. We write,for all n ≥ 1,

Cn = C ∩ (xn + δBX)and

On ={x∗ ∈ X∗ : S(C, x∗, η) ∩ Cn = ∅, for some η

},

where S(C, x∗, η) = {x ∈ C : x∗(x) > sup x∗(C)−η}. Since {xn : n ∈ N} is dense inC, it holds that C = ∪nCn and so it is immediate to check that A(C) ⊂ X∗\(∩nOn).

It suffices to prove that On is an open and dense set, for every n. Assumethat x∗ ∈ On, therefore S(C, x∗, η) ∩ Cn = ∅ for some η > 0. For ε > 0 such thatε supc∈C ‖c‖ < η

4 , it is satisfied that

y∗ ∈ x∗ + εBX∗ ⇒ S(C, y∗,

η

2

)⊂ S(C, x∗, η) (1)

since|x∗(x)− y∗(x)| ≤ ε sup

c∈C‖c‖, ∀x ∈ C.

Therefore,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‖ ≥

≥ supx∗(C)− η

2− 2ε sup

c∈C‖c‖ > sup x∗(C)− η.

Since S(C, x∗, η) ∩ Cn = ∅, then by (1), we know that S(C, y∗, η

2

)∩ Cn = ∅ and

y∗ ∈ On, so On is open.Now we will check that On 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+On ⊂ On and we canassume that ‖x∗‖ = 1.

C is bounded and x∗ 6= 0; hence we can choose y0 ∈ X such that

x∗(x) > x∗(y0), ∀x ∈ C.

Let M = sup{‖x−y0‖ : x ∈ C}, fix t > 2Mε , and write V = kerx∗∩tBX , A = V +y0.

By the choice of y0, it holds that Cn\A 6= ∅, since x∗(a) = x∗(y0) for any a ∈ Aand Cn ⊂ C, and so x∗(x) > x∗(y0) = x∗(a) for any x ∈ C.

Since diam Cn ≤ 2δ < 3δ and Cn\A 6= ∅, by the Hahn-Banach separationTheorem, there is a functional whose supremum on Cn is greater than the supremumon A. An appropriate slice of co(Cn∪A) determined by this functional is essentiallycontained in Cn (see [Bou, Theorem 3.4.1]). Hence, there is a slice S of the setco(Cn ∪ A) with diameter at most 3δ and such that S ∩ Cn 6= ∅. Then in the casethat C ⊂ co(Cn ∪A) since Cn ⊂ C, S ∩C would be a slice of C with diameter lessthan 3δ, which is not possible because of the choice of δ. Thus, there is an elementc0 ∈ C\co(Cn ∪A). Hence, there is a functional y∗ ∈ SX∗ such that

y∗(c0) > sup y∗(Cn ∪A).

Finally, in view of 2t ‖c0 − y0‖ ≤ 2

t M ≤ ε < 1, from Lemma 2.2, it follows that‖x∗ − y∗‖ ≤ 2

t ‖c0 − y‖ ≤ 2t M < ε.

Since sup y∗(C) ≥ y∗(c0) > sup y∗(Cn), then S(C, y∗, η) ∩ Cn = ∅ for η <sup y∗(C)− sup y∗(Cn), that is, y∗ ∈ On and On is dense.


As far as the authors know, it remains open whether the previous result alsoholds by dropping the separability of the space. However, Kenderov, Moors andSciffer proved the following result:

Theorem 2.4. ([KMS]) If K is any infinite compact and Hausdorff topologicalspace, 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 Banachspace X satisfies that A(X) does not contain weak-∗ open sets [DGS].

The analogous result for any Banach space was proven to hold by Jimenez-Sevilla and Moreno [JiM].

Lemma 2.5. ([JiM, Lemma 3.1]) If the dual unit ball of a Banach space Xcontains a slice of norm attaining functionals, then X is reflexive.

Proof. Assume that x∗∗0 ∈ SX∗∗ determines a slice such that

S ≡ S(BX∗ , x∗∗0 , η) ⊂ A(X).

Let us fix x∗0 ∈ BX∗ with ‖x∗0‖ < 1 satisfying that x∗∗0 (x∗0) > 1− η2 . We define

B := (S − x∗0) ∩ (−S + x∗0) ⊂ (BX∗ − x∗0) ∩ (BX∗ + x∗0).

By the choice of x∗0, we know that an element in (BX∗ −x∗0)∩ (BX∗ +x∗0) is in factin B, since x∗ − x∗0 = y∗ + x∗0 (‖x∗‖, ‖y∗‖ ≤ 1) implies that x∗−y∗

2 = x∗0 and weknow 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 = (BX∗−x∗0)∩ (BX∗ +x∗0) is a w∗-closed

set of X∗ and it is immediate that rBX∗ ⊂ B for 0 < r < 1− ‖x∗0‖. That is, B isthe 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 isx0 ∈ X\{0} with

(x∗ + x∗0)(x0) ≥ y∗(x0), ∀y∗ ∈ X∗, ‖y∗‖ ≤ 1

⇒ x∗(x0) ≥ z∗(x0), ∀z∗ ∈ BX∗ − x∗0⇒ x∗(x0) ≥ z∗(x0), ∀z∗ ∈ B,

that is, x∗ attains its (new) norm at the element x0|||x0 ||| . In the second case, we

argue in a similar way to arrive at the same conclusion. We checked that for thenorm ||| ||| 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 eitherX is reflexive or SX∗ ∩A(X) contains no (non empty) open set in (SX∗ , w∗). Alsoin the second case, the convex hull of SX∗\A(X) is (norm) dense in the dual unitball.


Proof. If SX∗ ∩ A(X) contains a (non-empty) w∗-open set in (SX∗ , w∗), wecan assume without losing generality that there is a subset W ⊂ X that can beexpressed in the form

W = {x∗ ∈ X∗ : x∗(xi) ≥ δi, i = 1, · · · , n},where xi ∈ SX , δi > 0 (1 ≤ i ≤ n), such that W ∩SX∗ has a non-empty interiorin the relative w∗-topology and such that W ∩ SX∗ ⊂ A(X).

An element x∗ in the border (norm topology) of V := W ∩BX∗ has to satisfythat ‖x∗‖ = 1 or x∗(xi) = δi for some 1 ≤ i ≤ n. In the first case, x∗ ∈ SX∗ ∩W,so x∗ attains its norm. As a consequence, there is an element x ∈ SX such that

x∗(x) ≥ y∗(x), ∀y∗ ∈ V. (1)

In the second case, the above inequality holds for x = −xi. If y∗0 satisfies that‖y∗0‖ < 1 and y∗0(xi) > δi for every 1 ≤ i ≤ n, then y∗0 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, balancedand w∗-closed subset of X∗. Hence B is the dual unit ball for an equivalent normon X∗.

By using that every element x∗ in the border of V satisfies (1), it is easy tocheck 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 SX∗\A(X) were not dense in BX∗ , then by theHahn-Banach separation Theorem, there is a slice S of BX∗ such that S ∩ SX∗ ⊂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 ofsmoothness, it can be stated that the space is reflexive or A(X) contains no opensubset for the norm topology. Let us recall that a Banach space X has the Mazurintersection property if, and only if, every bounded closed convex subset of X is anintersection of closed balls [Ma]. For instance, spaces with a Frechet differentiablenorm satisfy the Mazur intersection property.

Theorem 2.7. ([JiM, Proposition 3.3]) A Banach space X satisfying theMazur 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) con-tains an open ball centered at an element x∗0 in SX∗ of radius r > 0. Since Xhas the Mazur intersection property, then the subset of w∗-denting points of theunit sphere is dense in SX∗ [GGS1, Theorem 2.1]. Therefore, there is an elementy∗0 ∈ SX∗ with ‖y∗0 − 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 y0 ∈ X determining a sliceon BX∗ with diameter less than ε such that the slice contains y∗0 . Hence for someη > 0 we have

S(BX∗ , y0, η) ⊂ y∗0 + εBX∗ .


Therefore, the above slice is contained in

y∗0 +r

2BX∗ ⊂ x∗0 + rBX∗ ⊂ A(X).

By Lemma 2.5, X is reflexive.

By using a different idea of proof, we will show a new result along the samelines. Before that, let us note that every Banach space can be renormed such thatthe set of norm attaining functionals contains an open ball [AR1, Corollary 2].Therefore, no trivial isomorphic assumption implies reflexivity in the case that theset of norm attaining functionals contains a ball. Let us recall that a space X isweak Hahn-Banach smooth if every x∗ ∈ A(X) has a unique Hahn-Banach extensionto X∗∗.

Theorem 2.8. ([AR2, Theorem 1]) A weak Hahn-Banach space X, such thatA(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 `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 asequence {x∗n} in SX∗ 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 ∈ BX∗∗∗ , alsosatisfying x∗∗∗0 |X ≡ 0.

If we fix x∗∗0 ∈ SX∗∗ and ε > 0, by passing to a subsequence we will assumethat

x∗∗0 (x∗n) > x∗∗∗0 (x∗∗0 )− ε, ∀n ∈ N.

By assumption, there is x∗0 ∈ SX∗ , r > 0 with x∗0+rBX∗ ⊂ A(X). By using Simons’inequality (Proposition 1.2) for the sequence x∗0 +rx∗n and the subsets B = BX andA = BX∗∗ , 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

‖x∗0‖ ≥ (x∗0 + rx∗∗∗0 )(x∗∗0 ).

Since x∗∗0 is any element in SX∗∗ , then

‖x∗0‖ ≥ ‖x∗0 + rx∗∗∗0 ‖,where x∗∗∗0 ∈ X∗∗∗\{0}. Since x∗∗∗0 (x) = 0 , for every x ∈ X, the previous inequalitysays that the functional x∗0 + rx∗∗∗0 is a Hahn-Banach extension of x∗0 ∈ A(X) toX∗∗, which is impossible since X is weak Hahn-Banach smooth. Therefore, X isreflexive.

It is known that the previous result is not true for smooth spaces (spacessuch that every element at SX∗ has a unique normalized support functional). Infact, every separable Banach space can be renormed in order to be smooth whilepreserving the subset of norm attaining functionals [DGS]. Therefore, by usingthat every Banach space is isomorphic to a new one satisfying that the set of normattaining functionals has a non-empty interior, in the separable case we obtain anew norm which is smooth and satisfies that the set of norm attaining functionalscontains a ball.


Let us observe that Theorem 2.8 was obtained by Jimenez Sevilla and Morenofor the weak topology [JiM, Corollary 3.7]. In relation to the weak topology wecan state the following result:

Theorem 2.9. Let X be a separable Banach space such that A(X)∩ SX∗ con-tains a (non-empty) open set in the topological space (SX∗ , w). Then X is quasi-reflexive.

Proof. Suppose that there is a (relative) weak open and convex subset U ⊆BX∗ such that its closure W satisfies W∩SX∗ ⊆ A(X). We may assume that 0 ∈ U .If not, take x∗ ∈ U , ‖x∗‖ < 1, and consider the new equivalent dual norm ||| . ||| whoseunit ball is B = (−x∗+BX∗)∩(x∗+BX∗). Observe that W1 = (−x∗+W )∩(x∗−W )satisfies W1 ∩ S(X∗,||| |||) ⊆ A(X, ||| |||).

We will write ‖ ‖ for the norm instead of ||| ||| and from now on we assume thatU is of the form

U = {x∗ ∈ BX∗ : |x∗∗i (x∗)| < δ , i = 1, ..., n}where {x∗∗i }n

i=1 ⊆ SX∗∗ and δ is a positive number. Suppose that the linear spacegenerated by X ∪ {x∗∗i }n

i=1 is a proper subspace of X∗∗. Then there is a functionalφ ∈ SX∗∗∗ such that φ(z) = 0 for all z ∈ X ∪ {x∗∗i }n

i=1. Given ε > 0, there existsy∗∗ ∈ SX∗∗ , such that φ(y∗∗) > 1 − ε. We are assuming that X is separable, sothe restriction of the σ(X∗∗∗, X ∪ {x∗∗i }n

i=1 ∪ {y∗∗})-topology to bounded sets ismetrizable. Hence, in view of the w∗-denseness of BX∗ in BX∗∗∗ , we can find asequence {x∗n} in SX∗ that converges to φ in the σ(X∗∗∗, X ∪ {x∗∗i }n

i=1 ∪ {y∗∗})-topology while satisfying

x∗n ∈ U, ∀n ≥ 1

andx∗n(y∗∗) > 1− ε , ∀n ∈ N .

Now fix n0 ∈ N. Since U is open, U ∩ SX∗ ⊂ A(X) and A(X) is a cone, we canconsider r > 0, such that the ball of radius r centered at x∗n0

is contained in A(X).Clearly we are in conditions to apply Proposition 1.2, taking B = BX , A = BX∗∗

and {x∗n0+ rx∗n}, so we get

supx∈BX

lim supn

(x∗n0+ rx∗n)(x) = sup

x∗∗∈BX∗∗lim sup

n(x∗n0

+ rx∗n)(x∗∗) (1)

But,sup

x∈BX

lim supn

(x∗n0+ rx∗n)(x) = ‖x∗n0

‖ = 1

and

supx∗∗∈BX∗∗

lim supn

(x∗n0+ rx∗n)(x∗∗) ≥ (x∗n0

+ rx∗n)(y∗∗) ≥ (1− ε)(1 + r).

Since ε > 0 is arbitrary, the above equality (1) implies 1 ≥ 1 + r, a contradiction.

By using essentially the same idea of proof as in Theorem 2.8, but combiningit with other results it was shown:


Theorem 2.10. ([ABR2, Theorem 1]) A Banach space X is reflexive if it doesnot contain `1 and for some r > 0,

BX∗ = cow∗{x∗ ∈ SX∗ : x∗ + rBX∗ ⊂ A(X)},where we denote by cow∗ the w∗-closure of the convex hull.

For the space `1 the dual unit ball satisfies the previous assumption for r = 12 .

In fact, every Banach space can be renormed such that the convex hull of theinterior of the norm attaining functionals in the unit sphere is (norm) dense in thedual unit ball [ABR2, Proposition 3].

Finally, let us observe that there is also a version of Theorem 2.10 for operatorsinstead of functionals (see [ABR1]) and a result along the same lines for the spaceof operators (or compact operators) between two Banach spaces.

3. James-type results for polynomials, multilinear forms and numericalradius

This section is devoted to the study of some versions of James’ Theorem inwhich polynomials and multilinear forms are considered instead of linear function-als. Moreover, we give a James-type result in terms of numerical radius attainingoperators. In spite of the variety of problems, it is possible to offer a unifyingtreatment 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∗, asubset B of C is said to be a boundary for C if

∀x ∈ X ∃ b∗ ∈ B : b∗(x) = supc∗∈C

c∗(x).

Bauer’s optimization Principle provides some examples: the set of extreme pointsof 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 aboundary 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 spaceX attains the norm when

∃ x0 ∈ BE : |P (x0)| = ‖P‖ := supx∈BX

|P (x)|.

In order to pose the right version of James’ Theorem for polynomials, let us notethat even for reflexive spaces there are polynomials not attaining the norm. Forinstance, consider the continuous 2-homogeneous polynomial on `2 given by

P (x) :=∑

n≥1

(1− 1

n

)x(n)2, (x ∈ `2)

which does not attain the norm, since ‖P‖ = 1 and if x 6= 0 then |P (x)| < ‖x‖2.On the other hand, notice that, given n functionals x∗i ∈ X∗ (1 ≤ i ≤ n), the

polynomial Px∗1 ···x∗n defined by

Px∗1 ···x∗n(x) := x∗1(x) · · ·x∗n(x), (x ∈ X)


attains the norm provided that X is reflexive. These polynomials are preciselythose that allow us to establish a polynomial James’ Theorem:

Theorem 3.2. ([Ru]) A Banach space X is reflexive if (and only if) thereexist n ≥ 1 and x∗1, . . . , x

∗n ∈ X∗\{0} such that, for all x∗ ∈ X∗, the polynomial

Px∗1 ···x∗nx∗ attains the norm.

Let us emphasize that Theorem 3.2 follows from Theorem 3.1 for suitable sub-sets B of X and C of X∗∗.

A different question from the one considered here, the reflexivity of the spaceof all continuous n-homogeneous polynomials on a Banach space, has been studiedby several authors (see [JM] and the references therein).

Next we exhibit a James’ Theorem by means of multilinear forms. A con-tinuous n-linear form ϕ on a Banach space X attains its norm when there existx1, . . . , xn ∈ BX such that

|ϕ(x1, . . . , xn)| = ‖ϕ‖ := sup{|ϕ(y1, . . . , yn)| : y1, . . . , yn ∈ BX}).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

Sx∗1 ···x∗n for the symmetrization of the multilinear form

(x1, . . . , xn) 7→ x∗1(x1) · · ·x∗n(xn), (x1, . . . , xn ∈ X),

i.e.,

Sx∗1 ···x∗n(x1, . . . , xn) =1n!

∑

σ∈∆n

x∗σ(1)(x1) · · ·x∗σ(n)(xn)

(∆n is the set of all permutations of n elements). For this kind of multilinear formswe have, making use of Theorem 3.1,

Theorem 3.3. ([Ru]) A Banach space X is reflexive if (and only if) thereare n ≥ 1 and x∗1, . . . , x

∗n ∈ X∗\{0} such that for all x∗ ∈ X∗ the symmetric

(n + 1)-linear formSx∗1 ···x∗nx∗

attains the norm.

Let us observe that it is not possible to derive Theorem 3.2 or 3.3 directlyfrom James’ Theorem: we will give two simple examples of a polynomial and asymmetric bilinear form of the finite type attaining the norm and such that thisdoes not happen for all the functionals involved.

Example. Let us fix an element x∗ ∈ B`1 with x∗(e1) = 0 and consider thepolynomial P on c0 given by

x 7→ (e∗1 − x∗)(x)(e∗1 + x∗)(x).

If x ∈ Bc0 is a fixed element and we write α := x∗(x), then

|P (x)| = |(x(1)− α)(x(1) + α)| = |x(1)2 − α2| ≤ max{1, α2} = 1,

and P (e1) = 1, so P attains its norm and the functionals e∗1 + x∗, e∗1 + x∗ do notattain the norm if the support of x∗ is not finite.


For the symmetric bilinear form, we will take X = `1 and a functional x∗ =∑∞i=1 αne∗n ∈ `∞ where α1 = 1

2 , 0 < αn < 1 and lim {αn} = 1. The symmetricbilinear form S on `1 given by

(x, y) 7→( ∞∑

n=1

αnx(n))y(1) +

( ∞∑n=1

αny(n))x(1)

attains the norm at (e1, e1) since for x, y ∈ B`1 , if we write a = x(1), b = y(1) andwe assume that x(n), y(n) ≥ 0, we get

|S(x, y)| ≤(1

2a + (1− a)

)b +

(12b + (1− b)

)a =

= a + b− ab = a(1− b) + b ≤ 1.

Since ‖x∗‖ = 1 and {αn} satisfies 0 < αn < 1 for all n, x∗ does not attain thenorm.

We now consider a version of James’ Theorem for numerical radius. The nu-merical range of an operator T ∈ L(X) (Banach algebra of all bounded and linearoperators on X) is given by

V (T ) := {x∗(Tx) : (x, x∗) ∈ Π(X)},where

Π(X) := {(x, x∗) ∈ X ×X∗ : ‖x‖ = ‖x∗‖ = x∗(x) = 1}and its numerical radius by

v(T ) := sup{|µ| : µ ∈ V (T )}.An operator T ∈ L(X) is said to attain its numerical radius if

∃ (x0, x∗0) ∈ Π(X) : |x∗0(Tx0)| = v(T ).

Theorem 3.1 makes it possible to obtain a James-type result in terms of numericalradius:

Theorem 3.4. ([AR3]) A Banach space is reflexive provided that each rank-one operator attains its numerical radius.

The converse is false; in fact, in [AR3] it was proven that a separable Banachspace is finite-dimensional if, and only if, for any equivalent norm, every rank-oneoperator attains the numerical radius. It is not known if every reflexive Banachspace has an equivalent norm for which every rank-one operator attains its numer-ical radius. A partial answer was given in [ABR3].

Theorem 3.5. A Banach space X is reflexive if, and only if, there existx0 ∈ SX and an equivalent norm for which each rank-one operator of the formx∗ ⊗ x0 (x∗ ∈ X∗) attains its numerical radius.

By using the refinement of James’ Theorem proven by Jimenez Sevilla andMoreno (Theorem 2.6), some slight improvements to the results we mentioned canbe given:

Theorem 3.6. ([ABR3]) Let X be a Banach space. Then the following con-ditions are equivalent:

i) X is reflexive.


ii) There is n ≥ 1 and x∗1, . . . , x∗n ∈ X∗\{0} such that the set

{x∗ ∈ X∗ : Px∗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∗ : Sx∗1 ···x∗nx∗ attains its norm}has a non-empty weak-∗ interior.

If there is an element x0 ∈ SX satisfying that the set

{x∗ ∈ X∗ : x∗ ⊗ x0 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 subspacewhich is norm closed and contained in A(X), is X isometric to a dualspace?

2. If BX 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) ∩ SX∗ ⊃ W ∩ SX∗ 6= ∅ 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

BY ∗ = cow∗{x∗ ∈ SX∗ : 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 6= {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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Departamento de Analisis Matematico, Universidad de Granada, 18071 Granada,Spain

E-mail address: [email protected]

Departamento de Matematica Aplicada, Universidad de Granada, 18071 Granada,Spain


Departamento de Matematica Aplicada, Universidad de Granada, 18071 Granada,Spain


Characterizations of the reflexive spaces in the spirit of James’ Theorem

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