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    J. Math. Anal. Appl. 338 (2008) 407415www.elsevier.com/locate/jmaa

    The strong approximation property

    Eve Oja

    Faculty of Mathematics and Computer Science, Tartu University, J. Liivi 2, EE-50409 Tartu, Estonia

    Received 7 November 2006

    Available online 24 May 2007

    Submitted by Richard M. Aron

    Abstract

    We introduce and investigate the strong approximation property of Banach spaces which is strictly stronger than the approxima-tion property and at least formally weaker than the weak bounded approximation property. Among others, we show that the weakbounded approximation property is equivalent to a quantitative strengthening of the strong approximation property. Some recentresults on the approximation property of Banach spaces and their dual spaces are improved. 2007 Elsevier Inc. All rights reserved.

    Keywords: Banach spaces; (Bounded) approximation properties; (Weakly) compact operators

    1. Introduction

    Let X and Y be Banach spaces. We denote by L(X,Y) the Banach space of all bounded linear operators from Xto Y, and by F(X,Y), K(X,Y), and W(X,Y) its subspaces of finite-rank, compact, and weakly compact operators.

    Let IX denote the identity operator on X. If there exists a net (S ) F(X,X) such that S IX uniformly oncompact subsets ofX, then X is said to have the approximation property (AP). The AP was thoroughly studied byGrothendieck [6]. Among others, he proved that a Banach space X has the AP if and only ifK(Y,X) =F(Y,X) forevery Banach space Y (see, e.g., [12, p. 32]). Grothendieck [6] also discovered that if one reverses the roles ofX and Y,then one obtains a similar criterion of the AP for the dual space: X has the AP if and only ifK(X,Y)=F(X,Y) forevery Banach space Y (see, e.g., [12, p. 33]).

    In [9,10,15], the AP ofX

    andX

    was characterized in terms of the approximability of operators in the strongoperator topology. Here the same phenomenon occurs: the AP of X characterizes through the approximability ofoperators with values in X, and the AP of X characterizes through the approximability of operators from X. Forinstance, the following result was proven.

    Theorem 1.1. (See [10, Theorem 3.1], [15, proof of Theorem 2 and Theorem 5].) Let X be a Banach space. The

    following three assertions are equivalent.

    This research was partially supported by Estonian Science Foundation Grant 5704.E-mail address: [email protected].

    0022-247X/$ see front matter

    2007 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2007.05.038

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    408 E. Oja / J. Math. Anal. Appl. 338 (2008) 407415

    (a) X has the AP.(b) For every Banach space Y and for every operator T W(Y,X), there exists a net (T ) F(Y,X) such that

    supT T andT

    T in the strong operator topology.

    (c) For every separable reflexive Banach space Z and for every operator T K(Z,X), there exists a bounded net(T )F(Z,X) such thatT T in the strong operator topology.

    The following two assertions are equivalent.

    (a) X has the AP.(b) For every Banach space Y and for every operator T W(X,Y), there exists a net (T ) F(X,Y) such that

    supT T andT

    T in the strong operator topology.

    Here (b) has been obtained by reversing the roles ofX and Y in condition (b). Reversing the roles of X and Yin (c) yields the following condition.

    (c) For every separable reflexive Banach space Z and for every operator T K(X,Z), there exists a bounded net

    (T )F(X,Z) such thatTx T x for all x X.

    We shall say that a Banach space X has the strong approximation property (strong AP) provided condition (c)holds.

    In Section 2, we prove that the strong AP of X lies strictly between the approximation properties of X and X.Relying on a fundamental result of Grothendieck and Reinov, we show among others that the strong AP and the APfor X are equivalent whenever X is complemented in its bidual X.

    On the other hand, being strictly stronger than the AP, the strong AP is (at least formally) weaker than the boundedapproximation property (BAP) (the definition is recalled in Section 3). We conjecture that it is strictly weaker. If thestrong AP and the BAP were equivalent, then the AP of the dual space X of a Banach space X would imply the BAPofX. Recall that this is a long-standing famous open problem.

    Problem 1.2. (See, e.g., [1, p. 289].) Does the AP of the dual space X of a Banach space X imply the BAP or even

    the metric AP ofX?

    The strong AP is even (at least formally) weaker than the weak BAP, a recent concept introduced and studiedin [11]. We show (see Theorem 3.6 in Section 3) that the weak BAP is actually equivalent to a natural quantita-tive strengthening of the strong AP. This result is proven in Section 4 relying on a lemma on finite-rank operatorsof special form (Lemma 4.2). This lemma complements the famous DavisFigielJohnsonPeczynski factorizationlemma [2]. As application, among others, some results on the AP of Banach spaces and their dual spaces from [10,15](in particular, Theorem 1.1 above) are improved.

    Our notation is standard. We consider normed linear spaces (Banach spaces) over the same field of real or complexnumbers. In a linear normed space X, we denote BX(r) = {x X: x r} and BX = BX(1). The closure of a setAX is denoted by A. A Banach space X will always be regarded as a subspace of its bidual X under the canonicalembedding.

    2. The strong approximation property and the approximation property

    It is well known that the AP of a Banach space X follows from the AP of its dual space X. Our first result assertsthat the strong AP ofX lies strictly between the AP ofX and the AP ofX. We retain the notation of Section 1.

    Theorem 2.1. LetX be a Banach space. Then

    (a) (c) (a).

    Neither of the two implications is reversible.

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    E. Oja / J. Math. Anal. Appl. 338 (2008) 407415 409

    Proof. (a) (c). By Grothendiecks classics, K(X,Z) =F(X,Z). Hence, for T K(X,Z), we have a sequence(Tn)F(X,Z) such that Tn T 0. In particular, (Tn) is bounded and Tnx T x for all x X.

    (c) (a). We shall use that (a) is equivalent to the following well-known condition due to Grothendieck [6] (see,e.g., [12, p. 32]): for all sequences (xn)X and (xn )X

    such that

    n=1 xnx

    nN

    xn 0. Then T =nk=1

    k y

    kwith

    k V

    Kand y

    k Y. We may

    assume thatn

    k=1 yk = 1. Using (4.3), choose xk X such that k J

    K jxk< . Put S=

    nk=1 xk yk . Then

    SJK =n

    k=1 JK j

    xk yk and

    T SJK =

    n

    k=1

    k J

    K j

    xk yk

    n

    k=1

    k JK jxkyk< . 2Considering X (X), we immediately get the following result which is to compare with Lemma 4.1(i).

    Corollary 4.3. LetX and Y be Banach spaces. IfK is a weakly compact subset ofX, then

    BF(XK ,Y ) B{SJK : SF(X,Y)}.

    The next result should be compared with Lemma 4.1(ii).

    Corollary 4.4. LetX and Y be Banach spaces. IfK is a weakly compact subset ofX, then

    BF(Y,Z) B{J S: SF(Y,X)}

    where Z = (XK ) andJ= JK |X .

    Proof. If T F(Y,Z), then T F(XK , Y) and, by Lemma 4.2, T can be approximated in norm with opera-

    tors SJK with S X Y. Consequently, T = T|Y can be approximated with operators J S with S Y X =F(Y,X). 2

    Our first application of Lemma 4.2 concerns conditions (b) and (b) (see Theorem 1.1).

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    414 E. Oja / J. Math. Anal. Appl. 338 (2008) 407415

    Proposition 4.5. Let X be a Banach space. Then condition (b) (respectively, (b)) is equivalent to its strengtheningwhere the net(T ) is replaced by a net(S T ) (respectively, (T S)) with (S)F(X,X).

    Proof. Assume that condition (b) holds. Let Y be a Banach space and let T W(Y,X). We may clearly assume thatT = 1. Put K = T (BY). Then K is a weakly compact subset ofBX and we can apply the isometric version of the

    DavisFigielJohnsonPeczynski construction from [9]. We get that T factorizes through XK as T = JK t, wheret L(Y,XK ) and t = JK = 1. Since JK W(XK , X), by condition (b), there is a net (T) F(XK , X) suchthat supT JK = 1 and T

    x

    JK x for all x X. By Corollary 4.3, we may assume (by passing to the set

    of indices (,), where > 0, ordered in the natural way and choosing S(,) F(X,X) such that S(,)JK 1 andT S(,)JK < ) that T = SJK with some S F(X,X). Now we have ST = SJK t= T t and tT x

    tJK x = Tx for all x X. Hence supST supTt 1 and (ST )

    x Tx for all x X, asdesired.

    Assuming that condition (b) holds, we proceed similarly to the above. Let Y be a Banach space and let T W(X,Y) with T = 1. Denoting K = T(BY) and factorizing T through XK as T

    = JK t with t = JK = 1,we find, for J := JK |X W(X,Z), where Z = (X

    K )

    , a net (T ) F(X,Z) as in (b). Corollary 4.4 allows us toassume without loss of generality that T = J S for some S F(X,X). But now (T S ) = SJK t= T

    t, implying

    that supT S 1 and (T S)

    y

    = T

    ty

    J

    ty

    = T

    y

    for all y

    Y

    , as desired. 2

    Remark 4.1. Proposition 4.5 improves [10, Theorem 3.1] and [15, Theorem 5] strengthening respectively their con-ditions (c) and (b1).

    Similarly to Proposition 4.5, the strong AP can be reformulated in a seemingly stronger way.

    Proposition 4.6. A Banach space X has the strong AP if and only if for every Banach space Y and for every operator

    T K(X,Y), there exists a net(S)F(X,X) such that the net(T S) is bounded andT Sx T x for all x X.

    Proof. Assume that X has the strong AP. Let Y be a Banach space and let T K(X,Y) with T = 1. Denoting

    K = T

    (BY), we proceed as in the second part of the proof of Proposition 4.5. The only difference is that nowK is a compact subset of BX , and therefore (see [9]) the reflexive space Z is separable and J K(X,Z) (sinceJK is a compact operator). For J, choose a net (T ) as in (c), the definition of the strong AP. As in the proof ofProposition 4.5, we may assume without loss of generality that T = J S for some S F(X,X) and therefore(T S )

    = T t. This yields that supT S < and T Sx = (T S )x = tTx t

    J x = Tx = T x for allx X. 2

    By the above, we have essentially got a proof of Theorem 3.6.

    Proof of Theorem 3.6. Let 1 < and let X be a Banach space satisfying condition (c). By the proof ofProposition 4.6, condition (c) implies that for every Banach space Y and for every operator T K(X,Y), there

    exists a net (S) F

    (X,X) such that supT S T and T S x T x for all x X. But this condition isequivalent to the weak -BAP ofX (see [11, Theorem 2.4]).Conversely, condition (c) immediately follows from the definition of the weak -BAP. 2

    Let us conclude with one more application of Lemma 4.2. Our final result complements and improves [10, Theo-rem 3.2] giving the seemingly strongest known condition which is still equivalent to the AP ofX.

    IfX is a closed subspace of a Banach space Y, then an operator L(X, Y) is called a HahnBanach extensionoperator if(x)(x) = x(x) and x = x for all x X and x X. The existence of means, according tothe terminology of [5], that X is an ideal in Y. For instance, every Banach space X is an ideal in X (consider thecanonical embedding from X to X).

    Theorem 4.7. Let X be a Banach space. Then X has the AP if and only if, for every Banach space Y containing

    X as an ideal, for every HahnBanach extension operator L(X, Y), for every Banach space Z, and for every

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    operator T W(Z,Y), there exists a net (S ) F(Y,X) such that supST T and TS x

    Tx for

    all x X.

    Proof. Assume that X has the AP. We start as in the proof of Proposition 4.5. Let T W(Z,Y) with T = 1.Denoting K = T (BZ ), we factorize T through YK as T = JK t with t = JK = 1. By the proof of [10, Theo-

    rem 3.2, (a) (b)], there exists a net (J)F(YK , X) such that supJ 1 and J x JK x for all x X.Corollary 4.3 allows us to assume without loss of generality that J = SJK for some S F(Y,X). Now we haveS T= S JK t= Jt. Hence, supS T 1 and T

    S x = tJ x

    tJK x = Tx for all x X.

    The if part is obvious from [10, Theorem 3.2]. 2

    Acknowledgment

    The author is grateful to the referee for useful comments and suggestions.

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