Some properties and applications of equicompact sets of operators

STUDIA MATHEMATICA 181 (2) (2007) Some properties and applications of equicompact sets of operators by E. Serrano,C. Pi˜ neiro and J. M. Delgado (Huelva) Abstract. Let Xand Ybe Banach spaces. A subset M of K(X, Y ) (the vector space of all compact operators from Xinto Yendowed with the operator norm) is said to be equicompact if every bounded sequence (xn) in Xhas a subsequence (xk(n))nsuch that (T xk(n))nis uniformly convergent for T∈M. We study the relationship between this concept and the notion of uniformly completely continuous set and give some applications. Among other results, we obtain a generalization of the classical Ascoli theorem and a compactness criterion in Mc(F, X), the Banach space of all (finitely additive) vector measures (with compact range) from a field Fof sets into Xendowed with the semivariation norm. 1. Introduction. Throughout this paper Xand Ywill be Banach spaces. As usual, we will denote by K(X, Y ) the Banach space of all compact operators from Xinto Yendowed with the operator norm. In [9] the authors introduced the notion of an equicompact set of operators. A set M⊂K(X, Y ) is said to be equicompact if every bounded sequence (xn) in Xhas a subsequence (xk(n))nsuch that (Txk(n))nis uniformly convergent for T∈M. They proved that the notions of equicompact set and collectively compact set are dual in the following sense: M ⊂K(X, Y ) is equicompact (respectively, collectively compact) iff M∗={T∗:T∈M}is collectively compact (respectively, equicompact). We recall that M is called collectively compact if the set ST∈MT(BX) is relatively compact. Thus, the well known Palmer theorem [7] takes the following new form: Theorem A. If Mis a subset of K(X, Y ), then the following statements are equivalent: (i) M is relatively compact. (ii) M is equicompact and Mx={Tx :T∈M}is relatively compact for every x∈X. 2000 Mathematics Subject Classification: 47B07, 46G10. Key words and phrases: compact operators, equicompact sets of operators, collectively compact set, vector measures, Ascoli’s theorem. [171] c Instytut Matematyczny PAN, 2007 172 E. Serrano et al. (iii) M is collectively compact and M∗y∗={T∗y∗:T∈M}is relatively compact for every y∗∈Y∗. In particular, the authors of [9] have obtained the following characterization of compactness in a dual Banach space that we will use throughout this paper. Corollary B. Let Xbe a Banach space and A⊂X∗a bounded set. Then Ais relatively compact iff every bounded sequence (xn)in Xhas a subsequence (xk(n))nso that (hxk(n), ai)nis uniformly convergent for a∈A. In [9], the authors proved that a set M ⊂K(X, Y ) is equicompact iff there exists a null sequence (x∗ n) in X∗such that kTxk ≤ supn|hx, x∗ ni| for all x∈Xand T∈M. They also proved that equicompact sets are uniformly completely continuous, that is, kTxnk → 0 uniformly for T∈M whenever (xn) is a weakly null sequence in X. Actually, if the Banach space Xdoes not contain a copy of ℓ1, equicompact sets and uniformly completely continuous sets are the same (see Proposition 2.2 below). In this paper we deepen the study of the relationships between equicompact and uniformly completely continuous sets. Moreover, we obtain a generalization of the classical Ascoli theorem and a characterization of compactness in Mc(F, X), the Banach space of all (finitely additive) vector measures from Finto Xwith compact range, Fbeing a field of subsets of a set Ω. We use the classical notation in Banach space theory. If Xis a Banach space, X∗denotes its dual space, BXits closed unit ball and SXits unit sphere. For a subset Aof X,co(A) is the closed convex hull of A. As usual, ℓ1(I, X) (respectively ℓ∞(I, X)) stands for the Banach space of all functions bx:I→Xsatisfying Pi∈Ikbx(i)k<∞(respectively sup{kbx(i)k:i∈I}<∞) endowed with its natural norm. We will use the following version of the well known Vala compactness criterion in ℓc ∞(I, X) (the Banach space of all functions bx:I→Xwith relatively compact range endowed with the supremum norm). Theorem 1.1 (K. Vala [10, Theorem 1]).Let M⊂ℓc ∞(I, X)be bounded. The following statements are equivalent: (i) M is relatively compact. (ii) M has the following properties: (a) For every ε > 0there exists a finite partition {D1,...,Dp}of I such that 1≤k≤p, i, j ∈Dk⇒ kbx(i)−bx(j)k< ε for all bx∈M. (b) M(i) = {bx(i) : bx∈M}is relatively compact for all i∈I. Our notation from vector measure theory follows [3]. We only consider vector measures defined on fields of sets. If Fis a field of subsets of a set Ω, Equicompact sets of operators 173 Xis a Banach space and m:F→Xis such a measure, we denote by kmk(A) the semivariation of A∈F: kmk(A) = sup{|x∗◦m|(A) : x∗∈BX∗}. The range of mis denoted by rg(m), that is, rg(m) = {m(A) : A∈F}. Finally, we denote by B(F) the Banach space of all scalar-valued functions on Ωthat are uniform limits of simple functions modeled on F. 2. Relationships between equicompact and uniformly completely continuous sets Theorem 2.1.Let Mbe a bounded subset of K(X, Y ). The following statements are equivalent: (i) M is equicompact. (ii) M has the following properties: (a) M is uniformly completely continuous. (b) For every seminormalized sequence (xn)in Xequivalent to the ℓ1unit vector basis and every ε > 0, there exists a finite partition {D1,...,Dp}of Nsuch that kTxn−Txmk< ε for all T∈M whenever m, n ∈Diand i= 1,...,p. Proof. (i)⇒(ii). We only have to prove (b). Let φ:ℓ1→span{xn:n∈N} be an isomorphism with φ(en) = xnfor all n∈N. Obviously, the set M◦φ= {T◦φ:T∈M}is equicompact and, therefore, φ∗◦M∗is collectively compact. That is, the set [ T∈M φ∗(T∗(BY∗) = {(hxn, T∗y∗i) : y∗∈BY∗, T ∈M} is relatively compact in ℓ∞. According to Theorem 1.1, given ε > 0, there exists a finite partition {D1,...,Dp}of Nso that n, m ∈Di⇒ |hxn−xm, T∗y∗i| < ε for all y∗∈BY∗and T∈M, for i= 1,...,p. This yields kTxn−T xmk< ε for all T∈M and i= 1,...,p. (ii)⇒(i). Let (xn) be a bounded sequence in X. By Rosenthal’s ℓ1- theorem, (xn) has a subsequence which is either weakly Cauchy or equivalent to the unit basis of ℓ1(for simplicity, we will go on denoting it by (xn)). In the first case, (Txn) is uniformly convergent for T∈M because M is uniformly completely continuous. In the second case, by hypothesis, there exists a partition {D1,...,Dp}of Nso that n, m ∈Di⇒ kTxn−Txmk<1 for all T∈M and i= 1,...,p. Some of the Di’s must be infinite, so we can choose i≤p such that Diis infinite. If k1:N→Diis an increasing bijection, then (xk1(n)) is a sequence equivalent to the unit basis of ℓ1; so repeating this process 174 E. Serrano et al. inductively, we can determine a sequence of subsequences (kp(n))nsuch that (kp+1(n))nis a subsequence of (kp(n))nand kTxkp(n)−Txkp(m)k<1/p for all n, m ∈Nand T∈M for all p∈N. Now it is easy to deduce that (Txkp(p))pis uniformly convergent for T∈M. The next proposition proves that all uniformly completely continuous sets are equicompact iff Xdoes not contain a copy of ℓ1. We denote by V(X, Y ) the vector space of all completely continuous operators from X into Yendowed with the operator norm. Proposition 2.2.Let Xbe a Banach space. The following statements are equivalent: (i) For every Banach space Yand every M⊂V(X, Y ), M is equicompact whenever Mis uniformly completely continuous. (ii) There exists a Banach space Ysuch that every uniformly completely continuous set M⊂K(X, Y )is equicompact. (iii) Xdoes not contain copy of ℓ1. Proof. If X6←֓ ℓ1, then V(X, Y ) = K(X, Y ) for all Banach spaces Y and (iii)⇒(i) can be deduced using Theorem 2.1; so we only have to prove (ii)⇒(iii). Assuming (ii), to prove that Xdoes not contain a copy of ℓ1, we show that every uniformly completely continuous subset Aof X∗is relatively compact [5, Th. 2]. Take y0∈SYand put M = A⊗y0. It is obvious that M is a uniformly completely continuous subset of K(X, Y ). So, by hypothesis, M is equicompact, which yields the equicompactness of Aas a subset of K(X, R). Finally, a call to Corollary B tells us that Ais relatively compact. Recall that an operator T:X→Yis said to be conditionally weakly compact if every bounded sequence (xn) in Xadmits a subsequence (xk(n))n so that (Txk(n))nis weakly Cauchy. We denote by CW(X, Y ) the vector space of all conditionally weakly compact operators from Xinto Y. Proposition 2.3.For an operator Q∈L(X, Z), the following statements are equivalent: (i) Q∈CW(X, Z). (ii) If A⊂Z∗is uniformly completely continuous,then Q∗(A)is relatively compact. (iii) For every Banach space Yand every N⊂V(Z, Y ), N◦Qis equicompact whenever Nis uniformly completely continuous. Proof. (i)⇒(ii). Let A⊂Z∗be uniformly completely continuous and Q∈CW(X, Z). Given a bounded sequence (xn) in X, there exists a subsequence (xk(n))nsuch that (Qxk(n))nis weakly Cauchy. Then (hQxk(n), ai)n= Equicompact sets of operators 175 (hxk(n), Q∗ai)nis uniformly convergent for a∈A. Now, Corollary B concludes the proof. (ii)⇒(iii). Let Ybe a Banach space and N ⊂V(Z, Y ) uniformly completely continuous. We prove that Q∗◦N∗is collectively compact. For this, take a sequence ((Q∗◦S∗ n)y∗ n)nin SS∈NQ∗◦S∗(BY∗) and put A={S∗ ny∗ n: n∈N}. The set Ais uniformly completely continuous. In fact, if (zn) is a weakly null sequence in Z, we have |hzn, S∗ my∗ mi| =|hSmzn, y∗ mi| ≤ kSmznk. Then, by hypothesis, the set Q∗(A) is relatively compact and, therefore, ((Q∗◦S∗ n)y∗ n)nhas a convergent subsequence. (iii)⇒(i). By hypothesis, S(Q(BX)) is relatively compact for all Banach space Yand all S∈V(Z, Y ). According to [8, p. 377], the set Q(BX) is conditionally weakly compact. The next theorem shows that every equicompact set M admits a representation of the form M = N◦Q, where N is uniformly completely continuous and Qis conditionally weakly compact. Theorem 2.4.Let Mbe a subset of L(X, Y ). The following statements are equivalent: (i) M is equicompact. (ii) There exist a closed subspace Zof c0,Q∈K(X, Z)and N⊂ K(Z, Y )such that Nis equicompact and M = N ◦Q. (iii) There exist a Banach space Z,Q∈CW(X, Z)and N⊂V(Z, Y ) such that Nis uniformly completely continuous and M = N ◦Q. Proof. Only (i)⇒(ii) needs to be proved. According to [9, Prop. 2.2], the equicompactness of M implies that there exists a null sequence (x∗ n) in X∗ so that kTxk ≤ sup n |hx, x∗ ni| for all x∈Xand T∈M. For each n∈N, we define λn=pkx∗ nkand b∗ n=λ−1 nx∗ n(we can assume that λn6= 0 for all n∈N). Obviously, λn→0 and kb∗ nk → 0. Now, in a similar way to the proof of [6, Th. 17.1.4], we find a closed subspace Z of c0and operators Q∈K(X, Z) and ST∈K(Z, Y ) satisfying T=ST◦Q, for all T∈M (Q:x∈X7→ (hx, b∗ ni)∈c0,Z={Qx :x∈X}and ST(hx, b∗ ni) = Tx). Put N = {ST:T∈M}. Since Z ֒→c0, we have Z∗≈ℓ1/Z⊥. If (en) denotes the unit vector basis of ℓ1, it is clear that h(hx, b∗ ni), λm[em]i=hx, λmb∗ mi=hx, x∗ miand kλn[en]k → 0. 176 E. Serrano et al. Then, for all T∈M, we have kST(hx, b∗ ni)k=kTxk ≤ sup m |hx, x∗ mi| = sup m |h(hx, b∗ ni), λm[em]i|, that is, N is equicompact. Finally, notice that M = N ◦Q. 3. A generalization of the classical Ascoli theorem. In this section we generalize the notion of equicompact set to a wider class of functions. Let Jbe an arbitrary set and Za complete metric space. If M is a set of functions from Jinto Zwith relatively compact range, we say that M is equicompact if every sequence (jn) in Jhas a subsequence (jk(n))nsuch that (f(jk(n)))nis uniformly convergent for f∈M. If M⊂K(X, Y ), where Xand Yare Banach spaces, then Mis equicompact (in the original sense) iff M = {T|BX:T∈M}is equicompact. Throughout this section Xwill be a Banach space and Ian infinite index set. The mapping ψ:bx∈ℓc ∞(I, X)7→ ψ(bx) = Tbx∈K(ℓ1(I), X) defined by Tbx(ξi)i∈I=Pi∈Iξibx(i) is an isometric isomorphism. Using a similar argument to the proof of Theorem 2.1, it is easy to prove the next lemma: Lemma 3.1.Let Mbe a bounded subset of K(ℓ1(I), X). Then Mis equicompact iff for every ε > 0there exists a finite partition {D1,...,Dp} of Isuch that 1≤k≤p, i, j ∈Dk⇒ kTei−Tejk< ε for all T∈M. Remark 3.2.If ψ(M) ⊂K(ℓ1(I), X) is equicompact, then M is equicompact and bounded, but, in general, an equicompact set M in ℓc ∞(I, X) is not necessarily bounded. To see this, take an equicompact and bounded sequence (bxk) in ℓc ∞(I, X) and choose x0∈SX. Now, for each k∈N, denote by bzk∈ℓc ∞(I, X) the function defined by bzk(i) = bxk(i) + kx0for all i∈I. It is easy to prove that (bzk) is an equicompact sequence but, nevertheless, it is not bounded. Proposition 3.3.Let Mbe a bounded subset of ℓc ∞(I, X). The following statements are equivalent: (i) M is equicompact. (ii) ψ(M) is equicompact. (iii) There exists a null sequence (b βn)in ℓ∞(I)so that kbx(i)−bx(j)k ≤ sup n |b βn(i)−b βn(j)|for all i, j ∈Iand bx∈M. Proof. (i)⇒(ii). Consider the operator U:ℓ1(I)→ℓ∞(M, X) defined by U(ei) = (bx(i))bx∈Mfor all i∈I((ei)i∈Iis the canonical basis of ℓ1(I)). By (i), Uis compact and, therefore, the set ψ(M) is equicompact [9, Prop. 2.2]. Equicompact sets of operators 177 (ii)⇒(iii). According to [9, Prop. 2.2], there exists a null sequence (b βn) in ℓ∞(I) satisfying kTbx(ξ)k ≤ supn|hξ, b βni| for all ξ∈ℓ1(I) and bx∈M. In particular, for i, j ∈Iwe have kbx(i)−bx(j)k ≤ supn|b βn(i)−b βn(j)|for all bx∈M. (iii)⇒(i). Given a sequence (in) in I, there exists a subsequence (ik(n))n such that (heik(n),b βmi)nis uniformly convergent for m∈Nbecause of the compactness of {b βm:m∈N}(Corollary B). Now, from the uniform convergence of (b βm(ik(n)))nfor m∈Nand (iii), it is easy to obtain (i). Remark 3.4.According to Lemma 3.1 and Proposition 3.3, a bounded subset M of ℓc ∞(I, X) is equicompact iff it satisfies condition (ii)(a) in Vala’s theorem (Th. 1.1). As usual, if Ωis a compact topological space, C(Ω, X) is the Banach space of all continuous functions φ:Ω→Xendowed with the supremum norm. Obviously, C(Ω, X) is a subspace of ℓc ∞(Ω, X). Now we are ready to show the main result of this section: a generalization of the classical Ascoli theorem [4, Th. 7.5.7]. Theorem 3.5.Let Mbe a subset of C(Ω, X), Ωbeing an arbitrary compact topological space. The following statements are equivalent: (i) M is relatively compact. (ii) M has the following properties: (a) M is equicompact. (b) M(ω) = {φ(ω) : φ∈M}is relatively compact for all ω∈Ω. Proof. (i)⇒(ii). follows directly from Theorem 1.1, Proposition 3.3 and Lemma 3.1. According to Remark 3.4, to prove (ii)⇒(i) we only need to show that M is bounded. To see this, consider the function F:ω∈Ω7→ F(ω) = (φ(ω))φ∈M∈ℓ∞(M, X). Then Fis well defined and has compact range since M is equicompact. The next proposition lists some elementary properties of equicompact sets of functions that allow us to consider the above theorem as a generalization of the classical Ascoli–Arzel`a theorem. Proposition 3.6.Let Ωbe an arbitrary compact topological space and Ma bounded subset of C(Ω, X). (1) If Mis equicompact,then it is sequentially equicontinuous. (2) If ,in addition,Ωis metrizable,then Mis equicompact iff it is equicontinuous. Proof. (1) Suppose (ωn) is a sequence in Ωwith limit ω0∈Ω. By continuity, for each φ∈M, we have φ(ωn)→φ(ω0). As M is equicompact, by contradiction, it is easy to prove that φ(ωn)→φ(ω0) uniformly in φ∈M. 178 E. Serrano et al. (2) In case Ωis metrizable, equicontinuity and sequential equicontinuity are the same. So, we only have to prove the sufficiency. Assume M is equicontinuous. Given a sequence (ωn) in Ω, as Ωis metrizable and compact, there is a convergent subsequence (ωk(n))n. Suppose that ωk(n)→ω0∈Ω. Since M is equicontinuous it follows that φ(ωk(n))→φ(ω0) as n→ ∞ uniformly in φ∈M. 4. Compactness in Mc(F, X).As in Section 3, we say that a set M ⊂ Mc(F, X) is equicompact if every sequence (An) in Fhas a subsequence (Ak(n))nsuch that (m(Ak(n)))nis uniformly convergent for m∈M. By [3, I.5.3], all vector measures in Mc(F, X) are strongly additive. To start, we prove that equicompact sets of vector measures are uniformly strongly additive. We recall that a set M of strongly additive vector measures is called uniformly strongly additive if, for every sequence (An) of pairwise disjoint members of F, limn→∞ kP∞ k=nm(Ak)k= 0 uniformly in m∈M. Proposition 4.1.If Mis an equicompact set of vector measures,then it is uniformly strongly additive. Proof. According to [3, Proposition I.1.17], we have to prove that limn→∞ km(An)k= 0 uniformly in m∈M whenever (An) is a sequence of pairwise disjoint members of F. Arguing by contradiction, suppose there exist ε > 0, a sequence (mn) in M and a subsequence (Ak(n))nso that (1) kmn(Ak(n))k> ε for all n∈N. By hypothesis, (Ak(n))nhas a subsequence (Ah(n))nsuch that limnm(Ah(n)) = 0 uniformly in m∈M, which contradicts (1). If m:F→Xis a finite additive measure with compact range, then the integration map Im:f∈B(F)7→ T Ωf dm ∈Xis compact. In fact, in [3, p. 263] it is proved that the sums of the form Pn i=1 αim(Ai), 0 ≤α1≤ · · · ≤ αn≤1, Ai∩Aj=∅for i6=j, belong to co(rg(m)). This yields the inclusion n \ Ω f dm :f∈BB(F)o⊂co(rg(m)) −co(rg(m)). Then the operator Imis compact. Now we are ready to state our main result. Theorem 4.2.Let Mbe a subset of Mc(F, X). The following statements are equivalent: (i) M is relatively compact. (ii) M is equicompact and M(A)is relatively compact for all A∈F. Proof. (i)⇒(ii). Put b M = {Im:m∈M}. As there is an isometry between Mc(F, X) and K(B(F), X) defined by m↔Im,b M is a relatively Equicompact sets of operators 179 compact subset of K(B(F), X). By Theorem A, b M is equicompact and b M(f) = { T Ωf dm :m∈M}is relatively compact, for all f∈B(F). So, in particular, (ii) holds. (ii)⇒(i). We consider the vector measure G:A∈F7→ (m(A))m∈M∈ℓ∞(M, X), which has compact range since M is equicompact. Thus the integration map IG:B(F)→ℓ∞(M, X) is compact and defined by IG(f) = (Im(f))m∈Mfor f∈B(F). From the compactness of IGit follows that b M = {Im:m∈M} is equicompact. To prove that b M is relatively compact in K(B(F), X), we only have to show that b M(f) = { T Ωf dm :m∈M}is relatively compact for all f∈B(F). Given f∈B(F), choose a sequence (φn)nof simple functions so that f= limn→∞ φnin B(F). Fix ε > 0, and take n∈Nsuch that kf−φnk< ε/s, where s= sup{kmk(Ω) : m∈M}. For all m∈M, we have \ Ω f dm = \ Ω (f−φn)dm + \ Ω φndm ∈M(φn) + εBX, since k T Ω(f−φn)dmk ≤ εfor all m∈M. It is obvious that M(φn) is relatively compact, so we have proved that b M(f) is relatively compact for all f∈B(F). Corollary 4.3.Let (mn)be an equicompact sequence in Mc(F, X). If limn→∞ mn(A)exists for all A∈F,then (mn)is convergent. Remark 4.4.A uniformly strongly additive set is not necessarily equicompact. For an example, take a noncompact and weakly compact subset Wof L1(µ), µbeing Lebesgue measure on [0,1]. Denote by M(W) the set of indefinite integrals λf= T (·)fdµ with frunning over W. By [1, Th. VII.13], M(W) is uniformly countably additive, nevertheless, it is not equicompact in view of Theorem 4.2. Examples of equicompact sets can be obtained in the following way: take a uniformly completely continuous set N ⊂V(Z, X), Zbeing an arbitrary Banach space, and a vector measure m∈Mc(F, X). It is easy to prove either directly or using Proposition 2.4 that the set M = N ◦mis equicompact. Theorem 4.5.Let Mbe a bounded subset of Mc(F, X). Then Mis equicompact iff there exist a Banach space Z,a vector measure m∈Mc(F, X) and a uniformly completely continuous set N⊂V(Z, X)so that M = N◦m. Proof. We only have to prove the necessity. So, let M ⊂Mc(F, X) be bounded and equicompact. As in the proof of the above theorem, we can consider the vector measure G:A∈F7→ (m(A))m∈M∈ℓ∞(M, X).