DOI: 10.1007/s00209-004-0748-7

Math. Z. 250, 267–277 (2005) Mathematische Zeitschrift

On interpolation of Asplund operators

Fernando Cobos1, Luz M. Fernandez-Cabrera2, Antonio Manzano3,Anton Martınez4

1 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad Complu-tense de Madrid, 28040 Madrid, Spain (e-mail: cobos@mat.ucm.es)

2 Seccion Departamental de Matematica Aplicada, Escuela de Estadıstica, UniversidadComplutense de Madrid, 28040 Madrid, Spain(e-mail: luz fernandez-c@mat.ucm.es)

3 Departamento de Matematicas y Computacion, Escuela Politecnica Superior, Univers-idad de Burgos, 09001 Burgos, Spain (e-mail: amanzano@ubu.es)

4 Departamento de Matematica Aplicada, E.T.S. Ingenieros Industriales, Universidad deVigo, 36200 Vigo, Spain (e-mail: antonmar@uvigo.es)

Received: 20 February 2003 /Published online: 7 January 2005 – © Springer-Verlag 2005

Abstract. We study the interpolation properties of Asplund operators by the com-plex method, as well as by general J - and K-methods.

Mathematics Subject Classification (2000): 46B70, 47B10

1. Introduction

Let A, B be Banach spaces. A bounded linear operator T ∈ L(A,B) is said to be aRadon-Nikodym operator if for any probability measure µ, T maps each µ-contin-uousA-valued measure of finite variation into aµ-differentiableB-valued measure(see [22] and [11]). An operator T ∈ L(A,B) is called an Asplund operator if T ∗ isa Radon-Nikodym operator. Asplund operators have been studied widely. See, forexample, the papers by Edgar [12], Stegall [24] and the book by Diestel, Jarchowand Tonge [10]; see also the book by Pietsch [22] and the paper by Heinrich [15]where they are referred to as decomposing operators. Asplund operators A form aclosed injective and surjective operator ideal in the sense of Pietsch [22].

A Banach space A is said to be an Asplund space if its identity operator IAbelongs to A. This class of Banach spaces has attracted considerable attention inrecent years. It originated in the work of Asplund [1] where these spaces are calledstrong differentiability space. In fact, a Banach space A is Asplund if every con-tinuous convex function on A is Frechet differentiable at each point of a dense Gδsubset ofA.Any reflexive space isAsplund. Further, c0 and some other non-reflexivespaces are Asplund, but �1 and �∞ are not Asplund. We refer to the books by Giles

268 F. Cobos et al.

[14], Bourgin [3] and Fabian [13] for further details and relevant references. Letus only point out here that every closed subspace of an Asplund space is also Aspl-und, and that every Asplund operator factors through an Asplund space. This lastresult is due to Heinrich [15], Cor. 2.4, and independently to Reınov [23] and Ste-gall [24], Thm. 1.14. The proofs of Heinrich and Stegall use certain interpolationtechniques based on ideas developed by Davis, Figiel, Johnson and Pelczynski intheir famous paper on the factorization property of weakly compact operators [9].More precisely, focus on Heinrich’s paper, first he showed that if the embeddingi : A0 ∩A1 −→ A0 +A1 is an Asplund operator, then the real interpolation space(A0, A1)θ,q is Asplund for 0 < θ < 1, 1 < q < ∞, and from this result he derivedeasily the factorization property of Asplund operators.

In this paper we continue the research on interpolation properties of Asplundspaces and Asplund operators. In Section 2 we prove that the complex interpola-tion space (A0, A1)[θ ] is Asplund when any of A0 or A1 has this property. How-ever, similarly to the case of weak compactness, it is not true in general that ifi : A0 ∩ A1 −→ A0 + A1 is Asplund then (A0, A1)[θ ] is Asplund. Afterwards, inSection 3, we investigate the behaviour of Asplund operators under general J - andK-methods. Many authors have worked on these interpolation methods, which areextensions of the real method (see, for example, Peetre [20], Brudnyı and Krugljak[4], Cwikel and Peetre [8] and Nilsson [19]). These methods arise when one wantsto describe all interpolation spaces with respect to the couple (L1, L∞). The usualweighted Lq norm of the real method is not enough and one should work withgeneral lattice norms. Here we prove that if the lattice � that defines the methods isan Asplund space and T : A0 ∩ A1 −→ B0 + B1 is an Asplund operator, then theinterpolated operators by the K- and the J -method are Asplund operators as well.We end the paper by returning to the complex method to show that if T : A0 −→ B0is Asplund and T : A1 −→ B1 is bounded, then T : (A0, A1)[θ ] −→ (B0, B1)[θ ]is Asplund.

2. Complex interpolation

Let A = (A0, A1) be a Banach couple and let F(A) be the space of all functions ffrom the closed strip S = {z ∈ C : 0 ≤ Re z ≤ 1} into A0 + A1 such that

(a) f is bounded and continuous on S and analytic on the interior of S, and(b) the functions t −→ f (j + it) (j = 0, 1) are continuous from R into Aj and

tend to zero as |t | → ∞.

We put ‖f ‖F = maxj=0,1

{supt∈R ‖f (j + it)‖Aj

}. The complex interpolation

space (A0, A1)[θ ] (where 0 < θ < 1) consists of all a ∈ A0 + A1 such that

a = f (θ) for some f ∈ F(A). The norm of (A0, A1)[θ ] is ‖a‖[θ ] = inf{‖f ‖F :

f (θ) = a , f ∈ F(A0, A1)}.

Full details on complex interpolation can be found in [2] and [25]. We onlyrecall that A0 ∩ A1 is dense in (A0, A1)[θ ], and that, if A0 ∩ A1 is dense in A0and in A1, then the dual space (A0, A1)

∗[θ ] of (A0, A1)[θ ] coincides with the upper

complex space (A∗0, A

∗1)

[θ ].

On interpolation of Asplund operators 269

In order to determine whether (A0, A1)[θ ] is Asplund, we shall need the follow-ing characterization of Asplund operators. Given any Banach space A, we denoteby UA the closed unit ball of A. If D ⊆ A is a bounded set, let µ

Ddenote the

seminorm on A∗ given by

µD(f ) = sup {|f (x)| : x ∈ D} , f ∈ A∗.

Theorem 2.1 (see [3], Thm. 5.2.11). Let T ∈ L(A,B) be a bounded linear opera-tor from the Banach spaceA into the Banach spaceB. Then T is Asplund if and onlyif the seminormed space (B∗, µ

T (D)) is separable whenever D ⊆ UA is countable.

Now we are ready to describe the behaviour of Asplund spaces under complexinterpolation.

Theorem 2.2. Let A = (A0, A1) be a Banach couple and let 0 < θ < 1. Then thecomplex interpolation space (A0, A1)[θ ] is Asplund wheneverA0 orA1 is Asplund.

Proof. Let A◦j be the closure of A0 ∩ A1 in Aj . Since (A0, A1)[θ ] = (A◦

0, A◦1)[θ ]

and any closed subspace of an Asplund space is Asplund, without loss of generalitywe may assume that A0 ∩ A1 is dense in A0 and in A1. Hence

(A0, A1)∗[θ ] = (A∗

0, A∗1)

[θ ] = (A∗0, A

∗1)[θ ].

The last equality holds because A∗j has the Radon-Nikodym property for j = 0 or

j = 1 (see [21]). In particular, we get that

(A0 + A1)∗ = A∗

0 ∩ A∗1 is dense in (A0, A1)

∗[θ ]. (2.1)

On the other hand, in the following diagram

(A0, A1)[θ ]

A0

A0 + A1

A1

� �

������

� �������

any of these two embeddings is Asplund. Whence, by [15], Prop. 1.7, we obtainthat the embedding

i : (A0, A1)[θ ] −→ A0 + A1 is Asplund. (2.2)

Consequently, given any D ⊆ U(A0,A1)[θ ] with D countable, it follows from (2.2)that

((A0 + A1)

∗, µD

)is separable. Combining this fact with (2.1), we conclude

that((A0, A1)

∗[θ ], µD

)is separable. Using Theorem 2.1, this implies that (A0, A1)[θ ]

is Asplund. �Remark 2.1. If we only assume that the embedding i : A0 ∩ A1 −→ A0 + A1 isAsplund, then it is not true in general that (A0, A1)[θ ] is Asplund. Indeed, Mastylohas constructed in [18], page 161, a couple of Lorentz spaces (�ϕ,�ψ) such thatthe embedding i : �ϕ ∩�ψ −→ �ϕ+�ψ is weakly compact and so it is Asplund,but (�ϕ,�ψ)[θ ] contains a subspace isomorphic to �1 and therefore (�ϕ,�ψ)[θ ]is not an Asplund space.

We will come back to Theorem 2.2 at the end of the next section and we willshow that it can be extended to general couples A, B and any operator T withT : A0 −→ B0 Asplund and T : A1 −→ B1 bounded.

270 F. Cobos et al.

3. Real interpolation

In this section we will show that the behaviour under interpolation of Asplundoperators improves when one works with the extensions of the real method.

By a Z-lattice � we mean a Banach space of real valued sequences with Z

as index set, such that it contains all sequences with only finitely many non-zerocoordinates, and moreover � satisfies that whenever |ξm| ≤ |µm| for each m ∈ Z

and {µm} ∈ �, then {ξm} ∈ � and ‖{ξm}‖� ≤ ‖{µm}‖� .A Z-lattice � is said to be regular if for any {τn}n∈N ⊆ � with τn ↓ 0 it follows

that ‖τn‖� −→ 0 . The associated space �′ of � is formed by all sequences {ηm}for which

‖{ηm}‖�′ = sup

{ ∞∑m=−∞

|ηmξm| : ‖{ξm}‖� ≤ 1

}< ∞.

The space �′ is also a Z-lattice.The Z-lattice � is said to be K-non-trivial if {min(1, 2m)} ∈ �. It is called

J -non-trivial if

sup

{ ∞∑m=−∞

min(1, 2−m)|ξm| : ‖ξ‖� ≤ 1

}< ∞.

Let A = (A0, A1) be a Banach couple and let� be aK-non-trivial Z-lattice. TheK-space A�;K = (A0, A1)�;K consists of all a ∈ A0 +A1 such that {K(2m, a)} ∈�. Here

K(2m, a) = K(2m, a;A0, A1)

= inf{‖a0‖A0 + 2m‖a1‖A1 : a = a0 + a1, aj ∈ Aj }.The norm in (A0, A1)�;K is given by ‖a‖A�;K = ‖{K(2m, a)}‖� .

If � is J -non-trivial, the J -space A�;J = (A0, A1)�;J consists of all sums a =∞∑

m=−∞um (convergence inA0 +A1) where {um} ⊆ A0 ∩A1 and {J (2m, um)} ∈ �.

Here

J (2m, um) = J (2m, um;A0, A1) = max{‖um‖A0 , 2m‖um‖A1}.We put

‖a‖A�;J = inf

{‖{J (2m, um)}‖� : a =

∞∑m=−∞

um

}.

Spaces A�;K and A�;J are Banach spaces.Let B = (B0, B1) be another Banach couple. We write T ∈ L(A, B) to mean

that T is a linear operator from A0 + A1 into B0 + B1 whose restriction to eachAj defines a bounded operator from Aj into Bj (j = 0, 1). We put ‖T ‖A,B =max

{‖T ‖A0,B0 , ‖T ‖A1,B1

}.

On interpolation of Asplund operators 271

If T ∈ L(A, B), then it is clear that the restriction of T to A�;K defines abounded operator T : A�;K −→ B�;K with ‖T ‖A�;K,B�;K ≤ ‖T ‖A,B . A simi-lar estimate holds for the J -method. In other words, K- and J -method are exactinterpolation methods.

If � is K- and J -non-trivial Z-lattice, then A�;K ↪→ A�;J . But it is not truein general that A�;K coincides with A�;J . We refer to [19] and [4] for full detailson these interpolation methods. For � = �q(2−θm), the space �q with the weight{2−θm}, K- and J -methods coincide with the real method

(A0, A1)θ,q = (A0, A1)�q(2−θm);K = (A0, A1)�q(2−θm);J (see [2] and [25]).

Here 0 < θ < 1, 1 ≤ q ≤ ∞. In a more general way, if f is a function parameterand � = �q(1/f (2m)) then we recover the real method with a function parameter

(A0, A1)f,q=(A0, A1)�q(1/f (2m));K =(A0, A1)�q(1/f (2m));J (see [20] and [16]).

For t > 0, let tR be R with the norm ‖λ‖tR = t |λ|. The characteristic functionϕK

of the K-method is defined by

(R, (1/t)R)�;K = (1/ϕK(t))R (see [16]).

The characteristic function ϕJ

of the J -method is defined analogously. It turns outthat ϕ

K(t) = ‖{min(1, 2m/t)}‖−1

� and

ϕJ(t) = sup

{ ∞∑m=−∞

min(1, t/2m) |ξm| : ‖{ξm}‖� ≤ 1

}, t > 0,

(see [6], Lemma 2.4). Characteristic functions are quasiconcave. Later on we willneed to know the behaviour of these functions at 0 and ∞. We write ϕ ∈ P0 if

min{1, 1/t}ϕ(t) −→ 0 as t → 0 or as t → ∞.

We put ϕ∗(t) = 1/ϕ(1/t) . We also recall that the functions (see [5])

ψA�;K

(t) = sup{K(t, a) : ‖a‖A�;K = 1},ρA�;J

(t) = inf{J (t, a) : a ∈ A0 ∩ A1, ‖a‖A�;J = 1},are related with the characteristic functions by the inequalities

ψA�;K

(t) ≤ ϕK(t) for all t > 0, (3.1)

ρ∗A�;J

(t) ≤ ϕ∗J(t) for all t > 0, (3.2)

(see [6], Lemma 2.1).Given any sequence of Banach spaces {Em}, we put

�(Em) ={{xm} : xm ∈ Em and ‖{xm}‖�(Em) = ‖ {‖xm‖Em

} ‖� < ∞}.

We denote by Qk : �(Em) −→ Ek the projection Qk{xm} = xk , and by Pr :Er −→ �(Em) the embedding Prx = {δrmx} where δrm is the Kronecker delta.

In the following results we use again the characterization of Asplund operatorsgiven in Theorem 2.1.

272 F. Cobos et al.

Theorem 3.1. Let � be a K-non-trivial Asplund Z-lattice with ϕK

∈ P0, let A =(A0, A1), B = (B0, B1) be Banach couples and let T ∈ L(A, B). Then T :A�;K −→ B�;K is Asplund if and only if T : A0 ∩ A1 −→ B0 + B1 is Asplund.

Proof. Factorization

A0 ∩ A1 A�;K B�;K B0 + B1� � �� ��T � � ��

yields that T : A0 ∩A1 −→ B0 +B1 is Asplund if T : A�;K −→ B�;K is Asplund.Conversely, assume that T : A0 ∩A1 −→ B0 +B1 is Asplund. Since ϕ

K∈ P0

it follows from (3.1) that

limt→0

ψA�;K

(t) = 0 = limt→∞(1/t)ψA�;K

(t).

Then, by [7], Thm. 3.3, we get that T : A�;K −→ B0 + B1 is Asplund. Let Fm bethe space B0 + B1 normed by K(2m, ·), m ∈ Z, and let T : A�;K −→ �(Fm) bethe operator defined by T a = {· · · , T a, T a, T a, · · · }. Since any two norms of thefamily {K(2m, ·)} are equivalent on B0 + B1, and since QmT = T , we have thatQmT : A�;K −→ Fm is Asplund for each m ∈ Z. We claim that

T : A�;K −→ �(Fm) is Asplund. (3.3)

Indeed, letD ⊆ UA�;K be a countable set. For eachm ∈ Z, writeWm = QmT (D).We know that (F ∗

m,µWm ) is separable. Let �m be a countable set that is dense in(F ∗m,µWm ). To establish (3.3) it suffices to show that the countable set

� ={

N∑k=−N

Pkgk : gk ∈ �k , N ∈ N

}

is dense in(�(Fm)

∗, µT (D)

). Since � is Asplund, � does not contain a copy either

of �∞ or �1. Hence, by [26], Thms. 117.3 and 117.2, the spaces � and �′ are reg-ular. By regularity of � and [18], Prop. 3.1, we get that the dual space �(Fm)∗ of�(Fm) is isometrically isomorphic to �′(F ∗

m). Take any h ∈ �(Fm)∗ = �′(F ∗

m)

and put hm = Qmh ∈ F ∗m. Using that �′ is regular, given any ε > 0, we can find

N ∈ N such that∥∥∥h−

∑|k|≤N

Pkhk

∥∥∥�′(F ∗

m)≤ ε/2‖T ‖A,B . Choose gk ∈ �k such that

µWk(hk − gk) ≤ ε/(4N + 2). Then

µT (D)

(h−

∑|k|≤N

Pkgk

)≤ µ

T (D)

(h−

∑|k|≤N

Pkhk

)+

∑|k|≤N

µT (D)

(Pk(hk − gk))

≤ ‖T ‖A,B∥∥∥h−

∑|k|≤N

Pkhk

∥∥∥�′(F ∗

m)+

∑|k|≤N

µWk(hk − gk) ≤ ε.

This establishes (3.3).The operator T can be factorized as T = jT , where j : B�;K −→ �(Fm) is

the isometric embedding defined by jb = {· · · , b, b, b, · · · }. Using that Asplundoperators forms an injective operator ideal, it follows from (3.3) that T : A�;K −→B�;K is Asplund and completes the proof. �

On interpolation of Asplund operators 273

The corresponding result for J -spaces reads:

Theorem 3.2. Let � be a J -non-trivial Asplund Z-lattice with ϕ∗J

∈ P0, let A =(A0, A1), B = (B0, B1) be Banach couples and let T ∈ L(A, B). Then T :A�;J −→ B�;J is Asplund if and only if T : A0 ∩ A1 −→ B0 + B1 is Asplund.

Proof. Clearly, if T : A�;J −→ B�;J is Asplund then T : A0 ∩ A1 −→ B0 + B1is Asplund because A0 ∩ A1 ↪→ A�;J and B�;J ↪→ B0 + B1.

Conversely, assume that T : A0 ∩A1 −→ B0 +B1 is Asplund. Since ϕ∗J

∈ P0,(3.2) implies that ρ∗

A�;J∈ P0. Using [7], Thm. 3.1, we obtain that T : A0 ∩A1 −→

B�;J is Asplund. Moreover, since � is an Asplund space, we know by [26], Thms.117.3 and 117.2, that � and �′ are regular. LetGm be the space A0 ∩A1 endowedwith the norm J (2m, ·),m ∈ Z, and let T : �(Gm) −→ B�;J be the operator given

by T {um} = T( ∞∑m=−∞

um). For each m ∈ Z, the operator T Pm = T : Gm −→

B�;J is Asplund because any two norms of the family {J (2m, ·)} are equivalent onA0 ∩ A1. We claim that

T : �(Gm) −→ B�;J is Asplund. (3.4)

Indeed, take any D ⊆ U�(Gm) countable. For each m ∈ Z, put Dm = Qm(D) ⊆UGm . We have that

(T ∗(B∗

�;J ), µPm(Dm))

is separable, so there is a countable set�m that is dense. Put

� =

∑|k|≤N

PkQkgk : gk ∈ �k , N ∈ N

.

We are going to show that the countable set � ⊆ �′(G∗m) = �(Gm)

∗ is dense in(T ∗(B∗

�;J ), µD). Take any f ∈ B∗

�;J and any ε > 0. Using the regularity of �′ wecan find N ∈ N such that

∥∥∥f T −∑

|k|≤NPkQkf T

∥∥∥�′(G∗

m)≤ ε/2.

Now, for each |k| ≤ N , choose gk ∈ �k such thatµPkQk(D)

(f T −gk) ≤ ε/(4N+2).Then we obtain

µD

(f T −

∑|k|≤N

PkQkgk

)

≤∥∥∥f T −

∑|k|≤N

PkQkf T

∥∥∥�′(G∗

m)+

∑|k|≤N

µD(PkQk(f T − gk))

≤ ε/2 +∑

|k|≤NµPkQk(D)

(f T − gk) ≤ ε.

This proves (3.4).

274 F. Cobos et al.

The operator T is the composition of T with the metric surjectionπ :�(Gm) −→A�;J defined by π{um} =

∞∑m=−∞

um. Consequently, using surjectivity of Asplund

operators, we derive that T : A�;J −→ B�;J is Asplund. �

Corollary 3.1. Let � be an Asplund Z-lattice and let A = (A0, A1) be a Banachcouple with the embedding i : A0 ∩ A1 −→ A0 + A1 being an Asplund operator.

(i) If � is K-non-trivial with ϕK

∈ P0, then the space A�;K is Asplund.(ii) If � is J -non-trivial with ϕ∗

J∈ P0, then the space A�;J is Asplund.

If one of the spaces in the couple is Asplund, then we can weaken the assump-tions on ϕ

Kand ϕ

J. Indeed, repeating the proofs of Theorems 3.1 and 3.2 but using

now [5], Thms. 3.1 and 3.2 instead of [7], Thms. 3.3 and 3.1, we obtain:

Corollary 3.2. Let � be an Asplund Z-lattice and let A = (A0, A1) be a Banachcouple. Assume that A0 is Asplund.

(i) If � isK-non-trivial and limt→∞ϕK (t)/t = 0, then the space A�;K is Asplund.

(ii) If � is J -non-trivial and limt→0

t/ϕJ(t) = 0, then the space A�;J is Asplund.

The behaviour at 0 and ∞ of the functions ϕK

, ϕJ

can be controlled by thenorms of shift operators on �. For k ∈ Z, the shift operator τk is defined byτk{ξm}m∈Z = {ξm+k}m∈Z. It turns out (see [6], Lemma 2.5) that if

2−n‖τn‖�,� −→ 0 and ‖τ−n‖�,� −→ 0 as n → ∞,

then ϕK

∈ P0 and ϕ∗J

∈ P0. In particular, these conditions are satisfied for � =�q(2−θm) or � = �q(1/f (2m)), f being a function parameter.

Writing down Corollary 3.1 for� = �q(2−θm)with 0 < θ < 1 and 1 < q < ∞,we recover a result of Heinrich [15], Cor. 2.5/(ii):

Corollary 3.3. Let 0 < θ < 1, 1 < q < ∞ and let A = (A0, A1) be a Banachcouple. Then the following are equivalent:

(a) (A0, A1)θ,q is Asplund.(b) The embedding i : A0 ∩ A1 −→ A0 + A1 is an Asplund operator.

As we have seen in Remark 2.1, a similar result to Corollary 3.3 does not holdfor the complex method.

The next result refers to the limit cases q = 1 and q = ∞.

Proposition 3.1. Let 0 < θ < 1 and q = 1 or q = ∞. Let A = (A0, A1) be aBanach couple. Then the following are equivalent:

(a) (A0, A1)θ,q is Asplund.(b) The embedding i : A0 ∩ A1 −→ A0 + A1 is an Asplund operator and its

range is closed.

On interpolation of Asplund operators 275

Proof. If (A0, A1)θ,q is Asplund then it does not contain a subspace isomorphic to�q (because q = 1 or q = ∞) and so, using [17], Thm. 1, we have that A0 ∩A1 isclosed inA0+A1. In other words, i : A0∩A1 −→ A0+A1 has closed range. More-over, i factorizes through the identity of (A0, A1)θ,q , therefore i : A0 ∩ A1 −→A0 + A1 is Asplund. This shows that (a) implies (b). The converse implicationfollows from [6], Prop. 4.7. �

We end the paper by returning to the complex method to establish Theorem 2.2in its general form.

Corollary 3.4. Let 0 < θ < 1, let A = (A0, A1), B = (B0, B1) be Banach couplesand let T ∈ L(A, B). If T : A0 −→ B0 is Asplund, then T : (A0, A1)[θ ] −→(B0, B1)[θ ] is also Asplund.

Proof. We can factorize T : A0 −→ B0 as

A0/Ker(T )

A0

B0

B0�

��

�IB0

T

�

j0

where�(x) = [x] is the quotient mapping and j0[x] = T x. PutE = A0/Ker(T ).The map j0 : E −→ B0 is a continuous embedding, and so (E,B0) is a Banachcouple. Since the ideal of Asplund operators is surjective, it follows from the factthat T : A0 −→ B0 is Asplund that j0 : E −→ B0 is Asplund. Hence, by Corollary3.3, we get thatW = (E,B0)1/2,2 is anAsplund space. The operatorT : A0 −→ B0admits the factorization

A0 B0 .

W�

��

���

�

��

��

��T

T

I

Consequently, the interpolated operator by the complex method can be factorizedas

276 F. Cobos et al.

(A0, A1)[θ ] (B0, B1)[θ ]

(W,B1)[θ ]

��

��

��

�

��

��

��T

T

I

and Theorem 2.2 implies that T : (A0, A1)[θ ] −→ (B0, B1)[θ ] is Asplund. �Combining Corollary 3.4 with the reiteration theorem for the complex method,

we obtain:

Corollary 3.5. Let A = (A0, A1), B = (B0, B1) be Banach couples and let T ∈L(A, B). If there exits 0 < θ0 < 1 such that T : (A0, A1)[θ0] −→ (B0, B1)[θ0] isAsplund, then T : (A0, A1)[θ ] −→ (B0, B1)[θ ] is Asplund for all 0 < θ < 1.

Interpolation properties of other operator ideals have been investigated by theauthors in [6]. There, we have studied the case of weakly compact operators,Rosenthal operators and Banach-Saks operators.

Acknowledgements. Authors have been supported in part by Ministerio de Ciencia yTecnologıa (BFM2001-1424).

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