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PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 134, Number 3, Pages 689–695 S 0002-9939(05)08338-3 Article electronically published on October 17, 2005 EQUICOMPACT SETS OF OPERATORS DEFINED ON BANACH SPACES E. SERRANO, C. PI ˜ NEIRO, AND J. M. DELGADO (Communicated by Jonathan M. Borwein) Abstract. Let Xand Ybe Banach spaces. We say that a set M⊂K(X, Y ) (K(X, Y ) denotes the space of all compact operators from Xinto Y)isequicompact if there exists a null sequence (x∗ n)nin X∗such that Tx≤supn|x∗ n(x)| for all x∈Xand all T∈M. It is easy to show that collectively compactness and equicompactness are dual concepts in the following sense: Mis equicompact iff M∗={T∗:T∈M}is collectively compact. We study some properties of equicompact sets and, among other results, we prove: 1) a set M⊂K(X, Y ) is equicompact iff each bounded sequence (xn)nin Xhas a subsequence (xk(n))nsuch that (Tx k(n))nis a converging sequence uniformly for T∈M;2)ifYdoes not have finite cotype and M⊂K(X, Y ) is a maximal equicompact set, then, given ε>0 and a finite set {x1,...,x n}in X,thereis an operator S∈Msuch that Tx i≤(1 + ε)Sxifor i=1,...,n and all T∈M. 1. Introduction Let us consider (real or complex) Banach spaces Xand Y.AsusualK(X,Y ) will denote the vector space of all compact linear maps endowed with the operator norm. We say that a set M⊂K(X,Y )isequicompact if there exists a null sequence (x∗ n)nin X∗so that Tx≤sup n|x∗ n(x)| for all x∈Xand all T∈M. We recall that a set M⊂K(X,Y ) is called collectively compact iff T∈MT(BX) has compact closure. A standard proof using separations theorems and the well-known fact that a compact set in a Banach space is contained in the closure of the convex hull of a null sequence of the space allows us to state that Mis equicompact iff M∗={T∗∈K(Y∗,X∗): T∈M}is collectively compact. We study some properties of equicompact sets. Among other results we have proved the following: (1) A set M⊂K(X,Y ) is equicompact iff Msatisfies the next property (invoked as property (P)): (P) “For every bounded sequence (xn)nin Xthere exists a subsequence (xk(n))nso that (Txk(n))nis a converging sequence uniformly for T∈M.” Received by the editors April 20, 2004. 2000 Mathematics Subject Classification. Primary 47B07. Key words and phrases. Compact operators, equicompact set, collectively compact set. 689
690 E. SERRANO, C. PI ˜ NEIRO, AND J. M. DELGADO (2) Given a null sequence (x∗ n)nin X∗, we will denote by M((x∗ n)n) the equicompact set of all compact operators T∈K(X,Y ) satisfying Tx≤ supn|x∗ n(x)|for all x∈X. These sets are absolutely convex and closed in the strong operator topology. If Ydoes not have finite cotype we prove that, given ε>0 and a finite set {x1,...,x n}in X, there exists an operator Q∈M((x∗ n)n)sothatTxi≤(1 + ε)Qxifor i=1,...,n and all T∈M((x∗ n)n). We also have obtained a partial converse of (2): if M⊂K(X,Y ) is countably compact for the strong operator topology (for short, SOT), then Mis equicompact when it verifies the following property (invoked as property (F)): (F) There exists a positive constant Csuch that, for every finite set {x1,...,x n}⊂Xthere is an operator Qin the closed absolutely convex hull of Msatisfying Txi≤ CQxifor i=1,...,n and all T∈M. Our notation is standard. If Xis a Banach space, BXwill denote its closed unit ball. If Iis an arbitrary index set, we will write 1 a(I,X) (respectively, ∞(I,X)) for the Banach space of all absolutely summable (respectively, bounded) X-valued functions defined on I, endowed with the norm ξ=i∈Iξ(i)(respectively, ξ=sup{ξ(i):i∈I})foreachξ∈1 a(I,X) (respectively, ξ∈∞(I,X)). 2. Equicompact sets If M⊂K(X,Y ) is bounded, we can consider the continuous linear map U:1 a(M,X)−→ Ydefined by U(ξ)=T∈MT(ξ(T)) for all ξ∈1 a(M,X). Proposition 2.1. The following statements are equivalent for a set M⊂K(X, Y ): a) Mis collectively compact. b) The operator Uis compact. Proof. Given S∈Mand x∈BX,denotebyξS,x the element of 1 a(M,X) defined by ξS,x(T)=0ifT=S, xif T=S. It is clear that U(ξS,x)=Sx;thenH=U{ξS,x :S∈M,x∈BX}=T∈MT(BX). On the other hand, for every ξ∈B1 a(M,X)we have: U(ξ)= T∈M T(ξ(T)) = T∈M ξ(T)=0 ξ(T)Tξ(T) ξ(T). Therefore, U(B1 a(M,X))⊂co (H). So we have obtained H⊂U(B1 a(M,X))⊂co (H) and this concludes the proof. Now we consider the operator V:X−→ ∞(M,Y) defined by (Vx)(T)=Tx for all T∈Mand x∈X. For the proof of the following proposition, notice that the equivalence a)⇔b)isobviousandb)⇔c) is straightforward. Proposition 2.2. The following statements are equivalent for a set M⊂K(X, Y ): a) Mis equicompact. b) The operator Vis compact. c) Mhas property (P).
EQUICOMPACT SETS OF OPERATORS DEFINED ON BANACH SPACES 691 Now we are ready to face our main result: Theorem 2.3. If M⊂K(X,Y )is bounded, the following statements are equivalent: a) Mis collectively compact. b) M∗has property (P). Proof. The adjoint map of U:1 a(M,X)−→ Yis the operator U∗:Y∗−→ ∞(M,X∗) defined by (U∗y∗)(T)=T∗y∗, for all T∈Mand y∗∈Y∗.Infact,wehave: ξ,U∗y∗=Uξ,y∗= T∈MT(ξ(T)),y∗= T∈Mξ(T),T∗y∗ for all ξ∈1 a(M,X)andy∗∈Y∗. A call to Propositions 2.1 and 2.2 concludes the proof. The next lemma easily yields the dual equivalence Mhas property (P)⇐⇒ M∗is collectively compact. Lemma 2.4. If (x∗ n)nis a null sequence in X∗and Mis a subset of K(X, Y )such that Tx≤sup n|x∗ n(x)| for all x∈Xand all T∈M,then T∗∗x∗∗≤sup n|x∗∗(x∗ n)| for all x∗∗ ∈X∗∗ and T∈M. Proof. Given T∈M,ε>0andx∗∗ ∈BX∗∗ ,choosey∗∈BY∗so that T∗∗x∗∗= |y∗(T∗∗x∗∗)|. By hypothesis, there exists n0∈Nsuch that x∗ n<ε/4 for all n≥n0. Now we consider the weak∗neighborhood W=W(x∗ 1,...,x ∗ n0,T∗y∗;ε/2) of 0; there exists x∈BXsatisfying x∈x∗∗ +W.Thenwehave: T∗∗x∗∗=|T∗y∗,x ∗∗| ≤|T∗y∗,x ∗∗−T∗y∗,x|+|T∗y∗,x| <ε 2+sup n|x∗ n,x| ≤ε 2+sup n|x∗ n,x−x∗ n,x ∗∗|+sup n|x∗ n,x ∗∗| <ε+sup n|x∗ n,x ∗∗| for all ε>0. Letting ε→0, we obtain T∗∗x∗∗≤supn|x∗∗(x∗ n)|for all x∗∗ ∈X∗∗ and all T∈M. Remark 2.5.AsetM⊂L(X, Y )issequentially weak–norm equicontinuous (or uniformly completely continuous) if, for every weakly null sequence (xn)nin X, limnTxn= 0 uniformly for T∈M. It is obvious that every equicompact set is uniformly completely continuous (for short, u.c.c.). If Xdoes not contain a copy of 1, then Rosenthal’s theorem about 1tells us that each bounded sequence in X has a weakly Cauchy subsequence. So, in this case, every u.c.c. set has property (P) and, therefore, it is equicompact. That is to say, if X1, then the following
692 E. SERRANO, C. PI ˜ NEIRO, AND J. M. DELGADO statements are equivalent for a bounded set M⊂K(X,Y ): a) Mis equicompact. b) Mis u.c.c. c) M∗is collectively compact. The equivalence stated in Remark 2.5 and [4, theorem 2.2] yields directly the recent Mayoral’s theorem [3]: Theorem (F. Mayoral, 2001).If Xdoes not contain a copy of 1,asetM⊂ K(X,Y )is relatively compact iff Mis u.c.c. and, for every x∈X,thesetM(x)= {Tx:T∈M}is relatively compact in Y. Nevertheless, for an arbitrary Banach space X, a u.c.c. set M⊂K(X,Y ) is equicompact if, in addition, every 1-sequence (xn)nin Xhas a subsequence (xk(n))nsuch that (Txk(n))nis uniformly convergent for T∈M. 3. Dominated sets of operators The simplest examples of equicompact sets are the sets dominated by a compact operator, that is to say, the sets for which there exists an operator S∈K(X,Y ) such that Tx≤Sxfor all x∈Xand all T∈M. We are going to prove that a maximal equicompact set Mis dominated by an operator QH∈Mon every finite set H={x1,...,x n}⊂X. But a more general class of sets M⊂L(X,Y )enjoys this property. That is why we now consider the class of (Z, S)-dominated sets. Given a Banach space Zand a linear map S:X−→ Z,wesaythataset M⊂L(X,Y )is(Z, S)-dominated if Tx≤Sxfor all x∈Xand all T∈M. Examples. (1) If M⊂K(X,Y ) is an equicompact set satisfying Tx≤supn|x∗ n(x)|for all x∈Xand all T∈M,with(x∗ n)na null sequence in X∗, consider the map S:X−→ c0defined by Sx =(x∗ n(x))n. Then, Mis (c0,S)-dominated. (2) Let Πp(X,Y )bethespaceofp-summing operators from Xinto Yendowed with the norm πp(T)=sup{(nTxnp)1/p :(xn)n∈Bp w(X)},where p w(X) is the Banach space of the weakly p-summable sequences in X.A set M⊂Πp(X,Y ) is said to be uniformly p-dominated if there exists a positive Radon measure µon BX∗such that Txp≤BX∗|x∗(x)|pdµ(x∗) for all x∈Xand all T∈M.PutZ=Lp(µ, BX∗) and define S:X−→ Z by (Sx)(x∗)=x∗(x) for all x∗∈BX∗and all x∈X. Then every uniformly p-dominated set is (Z, S)-dominated. If Xand Yare Banach spaces, we will denote by M(Z, S) the set of all operators T∈L(X,Y ) satisfying Tx≤Sxfor all x∈X.NotethatM(Z, S) is absolutely convex and closed in K(X,Y ) for the SOT topology. The next theorem shows that M=M(Z, S) is dominated by an operator Q∈Mon every finite set {x1,...,x n}⊂ Xwhen Ydoes not have finite cotype.
EQUICOMPACT SETS OF OPERATORS DEFINED ON BANACH SPACES 693 Theorem 3.1. Let Ybe a Banach space that does not have finite cotype. Given ε>0, for every finite set {x1,...,x n}⊂Xthere exists Q∈M(Z, S)such that Txi≤(1 + ε)Qxi for i=1,...,n and all T∈M(Z, S). Proof. Since Ydoes not have finite cotype, Ycontains ∞ nuniformly (∞ n= (Rn,.∞)). By [2, theorem 14.1], for every ε>0andn∈N, there is an isomorphism Jnfrom ∞ nonto a subspace of Ysatisfying J−1 n=1andJn≤1+ε for all n∈N. Given {x1,...,x n}⊂X,choosez∗ i∈BZ∗so that |z∗ i(Sxi)|=Sxifor i= 1,...,n.Putyi=Jnei,(ei)n i=1 being the unit basis of ∞ n. We define an operator Q:X−→ Yby Qx =1 1+εJn((Sx,z∗ i)n 1). Then we have: Qx≤(1 + ε)−1Jn(Sx,z∗ i)n 1∞≤Sxsup iz∗ i≤Sx andthisprovesthatQ∈M(Z, S). Finally, we need to prove that Txi≤(1 + ε)Qxifor i=1,...,n and all T∈M(Z, S). Put y∗ i=e∗ i◦J−1 n,(e∗ i)n i=1 being the unit basis of (∞ n)∗≃1 n.Note that y∗ i≤1fori=1,...,n. Wealsodenotebyy∗ ia Hahn-Banach extension of e∗ i◦J−1 nto Y. It is easy to show that yi,y∗ j=δij.Wehave: Qxj≥|Qxj,y∗ j| =(1+ε)−1 n i=1Sxj,z∗ iyi,y∗ j =(1+ε)−1Sxj,z∗ j =(1+ε)−1Sxj ≥(1 + ε)−1Txj for all T∈M(Z, S)andj=1,...,n. Corollary 3.2. If the Banach space Ydoes not have finite cotype, every equicompact subset of K(X,Y )may be uniformly dominated, on each finite subset of X,by a compact operator. Now we give a partial converse of the last theorem in case M⊂K(X,Y )isaSOTcountably compact set. Note that the absolutely convex hull of an equicompact set is equicompact, too. So, without loss of generality, we can suppose from now on that Mis absolutely convex. Theorem 3.3. Let Mbe an absolutely convex subset of K(X, Y )enjoying property (F).IfMis countably compact for the strong operator topology, then Mis equicompact. Proof. (a) First, we prove the theorem in case Xis a separable Banach space. Let (xn)nbe a dense sequence in X. By hypothesis, for every n∈N,thereexists Qn∈Mso that Txi≤CQnxi
694 E. SERRANO, C. PI ˜ NEIRO, AND J. M. DELGADO for i=1,...,n and all T∈M. The sequence (Qn)nhas a cluster point Q∈Mfor the SOT. Proceeding by contradiction, it is easy to show that (1) Txi≤CQxi for i=1,...,n and all T∈M.Now,givenx∈Xand ε>0, choose xisuch that x−xi<ε/K,whereK=sup T∈MT. Using (1), we have Tx≤Tx−Txi+Txi <ε+CQxi ≤ε+C(Qxi−Qx+Qx) ≤ε+Cε+CQx for all T∈M. Letting ε→0 we conclude Tx≤CQxfor all T∈M. (b) Consider now an arbitrary Banach space X. In order to show that Mis equicompact, we will prove that Mhas property (P). Let (xn)nbe a bounded sequence in Xand put H=span {xn:n∈N}.IfiHdenotes the inclusion map from Hinto X, then part (a) applied to the set N={T◦iH:T∈M}⊂K(H, Y ) yields the equicompactness of Nand, therefore, also of M. Examples. 1. A maximal equicompact set of operators valued in 2that fails property (F).LetMbe the set of all operators Tfrom c0into 2satisfying Tα2≤sup n 1 √ne∗ n(α) for all α∈c0((e∗ n)ndenotes the unit basis of 1). By contradiction, suppose there exists a positive constant Csuch that, for every finite set {α1,...,α n}⊂c0,thereis an operator Q∈Msuch that Tαk2≤CQαk2for k=1,...,n and all T∈M. In particular, for every n∈N,thereexistsQn∈Mso that Tek2≤CQnek2 for k=1,...,n and all T∈M(here (en)ndenotes the unit basis of c0). Then we have (2) n k=1 Tkek2 2≤C2 n k=1 Qnek2 2 for all n∈Nand all {T1,...,T n}⊂M. Now we consider the operators Tk∈M defined by Tkα=1 √ke∗ k(α)ukfor all α∈c0,where(un)nis the unit basis of 2. Notice that Tkek2=1 √k; then (2) yields C−2 n k=1 1 k≤ n k=1 Qnek2 2 for all n∈N. Finally, recall that every bounded operator Tfrom c0into 2is 2-summing and π2(T)≤λT,forsomeconstantλ>0 (see [2, theorem 3.5]). This fact allows us to obtain π2(Qn)2≥ n k=1 Qnek2 2≥C−2 n k=1 1 k for all n∈N. This is a contradiction because Mis bounded for the π2-norm. 2. It is interesting to show that the countable compactness of Mcannot be omitted in Theorem 3.3. For example, consider the closed unit ball of L(c0,c 0).
EQUICOMPACT SETS OF OPERATORS DEFINED ON BANACH SPACES 695 This ball is the same that M(c0,I), Ibeing the identity map on c0. The proof of Theorem 3.1 shows that M(c0,I) has the following property: “There exists a positive constant Csuch that, for every finite set {α1,...,α n}⊂c0, there is an operator of finite rank Q∈M(c0,I) satisfying Tαk≤CQαkfor k=1,...,nand all T∈M(c0,I).” This implies that the closed unit ball of K(c0,c 0), M, has property (F). Nevertheless, we are going to show that Mis not equicompact. For each δ=(δn)n belonging to the unit ball of c0,wedenotebyTδthe operator defined by Tδ(αn)n= (αn·δn)nfor all (αn)n∈c0.ItisobviousthatTδ∈Mfor all δ∈Bc0. For every n∈N,wehaveTen=en=1. Since(en)nis weakly null in c0, this shows that Mis not u.c.c. Acknowledgement The authors thank the referee for his interesting suggestions. References [1] N. Dunford, J. T. Schwartz, Linear operators. Part I: General theory, Wiley Interscience, New York and London, 1958. MR0117523 (22:8302) [2] J. Diestel, H. Jarchow, A. Tonge, Absolutely summing operators, Cambridge studies in advanced Mathematics 43, Cambridge University Press, Cambridge, 1995. MR1342297 (96i:46001) [3] F. Mayoral, Compact sets of compact operators in absence of 1, Proc Amer. Math. Soc. 129 (2001), no. 1, 79–82. MR1784015 (2001e:46026) [4]T.W.Palmer,Totally bounded sets of precompact operators, Proc. Amer. Math. Soc. 20 (1969), 101–106. MR0235425 (38:3734) Departamento de Matem´ aticas, Facultad de Ciencias Experimentales, Campus Universitario del Carmen, Avda. de las Fuerzas Armadas s/n, 21071 Huelva, Spain E-mail address:[email protected] Departamento de Matem´ aticas, Facultad de Ciencias Experimentales, Campus Universitario del Carmen, Avda. de las Fuerzas Armadas s/n, 21071 Huelva, Spain E-mail address:[email protected] Departamento de Matem´ aticas, Facultad de Ciencias Experimentales, Campus Universitario del Carmen, Avda. de las Fuerzas Armadas s/n, 21071 Huelva, Spain E-mail address:[email protected]