Uniformly Summing Sets of Operators on Spaces of Continuous Functions
Abstract
Let X and Y be Banach spaces. A set ℳ of 1-summing operators from X into Y is said to be uniformly summing if the following holds: given a weakly 1-summing sequence (xn) in X, the series ∑n‖Txn‖ is uniformly convergent in T∈ℳ. We study some general properties and obtain a characterization of these sets when ℳ is a set of operators defined on spaces of continuous functions.
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IJMMS 2004:63, 3397–3407 PII. S0161171204403585 http://ijmms.hindawi.com © Hindawi Publishing Corp. UNIFORMLY SUMMING SETS OF OPERATORS ON SPACES OF CONTINUOUS FUNCTIONS J. M. DELGADO and CÁNDIDO PIÑEIRO Received 30 March 2004 Let Xand Ybe Banach spaces. A set ᏹof 1-summing operators from Xinto Yis said to be uniformly summing if the following holds: given a weakly 1-summing sequence (xn)in X, the series nTxnis uniformly convergent in T∈ᏹ.Westudysomegeneralproperties and obtain a characterization of these sets when ᏹis a set of operators defined on spaces of continuous functions. 2000 Mathematics Subject Classification: 47B38, 47B10. 1. Introduction. Throughout this paper, Xand Ywill be Banach spaces. If Xis a Banach space, BX={x∈X:x≤1}will denote its closed unit ball and X∗will be the topological dual of X. Given a real number p∈[1,∞), a (linear) operator T:X→Yis said to be p-summing if there exists a constant C>0 such that n i=1 Txi p1/p ≤C·sup n i=1 x∗,xi p1/p :x∗∈BX∗ (1.1) for every finite set {x1,...,xn}⊂X. The least Cfor which the above inequality always holds is denoted by πp(T) (the p-summing norm of T). The linear space of all p-summing operators from Xinto Yis denoted by Πp(X,Y) which is a Banach space endowed with the p-summing norm. As usual, p w(X) will be the Banach space of weakly p-summable sequences in X, that is, the sequences (xn)⊂Xsatisfying n|x∗,xn|p<∞for all x∗∈X∗;the norm in p w(X) is p(xn)=sup{(n|x∗,xn|p)1/p :x∗∈BX∗}. The set of all strongly p-summable sequences in Xis denoted by p a(X); the norm in this space is πp(xn)= (nxnp)1/p.IfT∈Πp(X,Y), the correspondence T:(xn)(Txn)always induces a bounded operator from p w(X) into p a(Y) with T=πp(T) [5, Proposition 2.1]. Families of operators arise in different applications: equations containing a parameter, homotopies of operators, and so forth. In these applications, it may be very interesting to know that, given a set ᏹ⊂Πp(X,Y) and (xn)∈p w(X),theseriesnTxnp is uniformly convergent in T∈ᏹ. The main purpose of this paper is to study uniformly p-summing sets, that is, those sets ᏹ⊂Πp(X,Y) for which, given (xn)∈p w(X),the series nTxnpis uniformly convergent in T∈ᏹ. These sets also enjoy some properties that justify their study; the next proposition lists some of them.
3398 J. M. DELGADO AND C. PIÑEIRO Proposition 1.1.(a) Let (Tk)be a sequence in Πp(X,Y). Then, Tk k →0pointwise if and only if Tk k →0pointwise and (Tk)is uniformly p-summing. (b) Let ᏹ⊂Πp(X,Y) be a uniformly p-summing set. If ᏹis endowed with the strong operator topology, then the map T∈ᏹnTxnp∈Ris continuous for every (xn)∈ p w(X). A basic argument shows that uniformly p-summing sets are bounded for the psumming norm. In fact, if Xdoes not contain any copy of c0, bounded sets and uniformly 1-summing sets are the same. That is the reason for which we only consider operators defined on a Ꮿ(Ω)-space, Ωbeing a compact Hausdorff space. We recall that every weakly compact operator T:Ꮿ(Ω)→Yhas a representing measure mT:Σ→ Ydefined by mT(B) =T∗∗(χB)for all B∈Σ, where Σdenotes the Borel σ-field of subsets of Ωand χBdenotes the characteristic function of B. The vector measure mTis regular and countably additive [6, Theorem VI.2.5 and Corollary VI.2.14]. If we denote by Tthe operator T∗∗ restricted to B(Σ)(the space of all bounded Borel-measurable scalar-valued functions defined on Ω), then Tϕ=Ω ϕdmT,(1.2) for all ϕ∈B(Σ)(the integral is the elementary Bartle integral [6, Definition I.1.12]). It is well known that every p-summing operator defined on a Banach space Xis weakly compact. In Section 2, we consider 1-summing operators Tdefined on Ꮿ(Ω); these operators are characterized as those with representing measure mThaving finite variation and π1(T) =|mT|(Ω)[6, Theorem VI.3.3]. We show that a set ᏹ⊂Π1(Ꮿ(Ω),Y) is uniformly 1-summing if and only if the family of all variation measures {|mT|:T∈ᏹ}is uniformly bounded and there is a countably additive measure µ:Σ→[0,∞)such that {|mT|:T∈ᏹ}is uniformly µ-continuous. In Section 3, we mention a special class of uniformly p-summing operators: uniformly dominated sets. The relationship between uniformly summing sets and relatively weak compactness is also studied. Finally, we give some examples and open problems. 2. Uniformly 1-summing sets in Π1(Ꮿ(Ω),Y).Before facing our main theorem, we include three results which correspond to the vector measure theory. These results will be usually invoked along the following lines. Proposition 2.1 [6, Proposition I.1.17].The following statements about a collection {mi:i∈I}of Y-valued measures defined on a σ-field Σare equivalent: (a) the set {mi:i∈I}is uniformly countably additive, that is, if (En)is a sequence of pairwise disjoint members of Σ, then limnk≥nmi(Ek)=0uniformly in i∈I, (b) the set {y∗◦mi:i∈I, y∗∈BY∗}is uniformly countably additive, (c) if (En)is a sequence of pairwise disjoint members of Σ, then limnmi(En)=0 uniformly in i∈I, (d) if (En)is a sequence of pairwise disjoint members of Σ, then limnmi(En)=0 uniformly in i∈I, where midenotes the semivariation of mi, (e) the set {|y∗◦mi|:i∈I, y∗∈BY∗}is uniformly countably additive.
UNIFORMLY SUMMING SETS OF OPERATORS … 3399 Theorem 2.2 [6, Theorem I.2.4].Let {mi:Σ→Y:i∈I}be a uniformly bounded (with respect to the semivariation) family of countably additive vector measures on a σ-field Σ. The family {mi:i∈I}is uniformly countably additive if and only if there exists a positive real-valued countably additive measure µon Σsuch that {mi:i∈I}is uniformly µ-continuous, that is, lim µ(E)→0 mi(E) =0 (2.1) uniformly in i∈I. If Ωis a compact Hausdorff space and Σdenotes the σ-field of the Borel subsets of Ω, a vector measure mon Σis regular if for each Borel set Eand ε>0 there exists a compact set Kand an open set Osuch that K⊂E⊂Oand m(O\K) < ε. Proposition 2.3 [6, Lemma VI.2.13].Let be a family of regular (countably additive) scalar measures defined on Σ. Each of the following statements implies all the others: (a) for each pairwise disjoint sequence (On)of open subsets of Ω,limnµ(On)=0 uniformly in µ∈, (b) for each pairwise disjoint sequence (On)of open subsets of Ω,limn|µ|(On)=0 uniformly in µ∈, (c) is uniformly countably additive, (d) is uniformly regular, that is, if E∈Σand ε>0, then there exists a compact set Kand an open set Osuch that K⊂E⊂Oand supµ∈|µ|(O\K) < ε. Now, we are able to show our main result. In the proof, we will use the fact that |mT| is regular when T:Ꮿ(Ω)→Yis 1-summing [7, Proposition 15.21]. Theorem 2.4.Let ᏹ⊂Π1(Ꮿ(Ω),Y) be a bounded set. The following statements are equivalent: (a) ᏹis uniformly 1-summing, (b) the family of nonnegative measures {|mT|:T∈ᏹ}is uniformly countably additive, (c) given ε>0and a disjoint sequence (En)of Borel subsets of Ω, there exists n0∈N such that n≥n0 mTEn <ε, (2.2) for all T∈ᏹ. Proof. (a)⇒(b). According to [6, Lemma VI.2.13], it suffices to show that limn→∞ |mT|(On)=0 uniformly in T∈ᏹ, for all disjoint sequences (On)of open subsets of Ω. By contradiction, suppose that there exists ε>0, a sequence (Tn)in ᏹ, and a strictly increasing sequence (kn)of natural numbers such that mTn Okn>2ε, ∀n∈N.(2.3)
3400 J. M. DELGADO AND C. PIÑEIRO Now we consider the operators Sn:Ꮿ(Ω,Okn)→Ydefined by Snϕ=Okn ϕdmTn,(2.4) for all ϕ∈Ꮿ(Ω,Okn), where Ꮿ(Ω,Okn)is the closed subspace of Ꮿ(Ω)formed by all continuous functions ϕon Ωsuch that ϕvanishes in Ω\Okn. It is known that π1(Sn)= |mTn|(Okn), for all n∈N[7, Theorem 19.3]. For each n∈N, we can choose a finite set {ϕn 1,...,ϕn pn}⊂Ꮿ(Ω,Okn)satisfying 1(ϕn i)pn i=1≤1and pn i=1 Snϕn i >π 1Sn−ε. (2.5) Since the open sets Oknare disjoint, it follows that the sequence (ϕ1 1,...,ϕ1 p1,ϕ2 1,..., ϕ2 p2,...) belongs to 1 w(Ꮿ(Ω)). Nevertheless, for all n∈N, we have m≥n pm i=1 Tnϕm i ≥ pn i=1 Tnϕn i = pn i=1 Snϕn i >π 1Sn−ε= mTn Okn−ε>ε. (2.6) This denies (a) and proves that (a) implies (b). (b)⇒(c). Again we proceed by contradiction. Suppose (En)is a disjoint sequence of Borel subsets of Ωfor which there exists ε>0, a sequence (Tn)in ᏹ, and a strictly increasing sequence (kn)of natural numbers so that kn+1 i=kn+1 mTnEi >ε, ∀n∈N.(2.7) If we put Bn=kn+1 i=kn+1Ei, the above inequality yields |mTn|(Bn)>ε. So, in view of [6, Proposition I.1.17], the family {|mT|:T∈ᏹ}is not uniformly countably additive. (c)⇒(b). We need to prove lim n→∞ mT En=0 uniformly in T∈ᏹ,(2.8) for all disjoint sequences (En)of Borel subsets of Ω. Suppose (b) fails. Then, there exists ε>0, a sequence (Tn)in ᏹ, and a strictly increasing sequence (kn)of natural numbers satisfying mTn Ekn>ε, ∀n∈N.(2.9) For each n∈N, we choose a finite partition {En 1,...,En pn}of Eknfor which pn i=1 mTnEn i >ε. (2.10) Then, the disjoint sequence (E1 1,...,E1 p1,E2 1,...,E2 p2,...) does not satisfy (c).
UNIFORMLY SUMMING SETS OF OPERATORS … 3401 (b)⇒(a). According to [6, Theorem I.2.4] there exists a countably additive measure µ:Σ→[0,∞)so that lim µ(E)→0 mT (E) =0 uniformly in T∈ᏹ.(2.11) Hence, given ε>0, there exists δ>0 such that, if E∈Σverifies µ(E) < δ,then |mT|(E) < ε/2, for all T∈ᏹ. Next, given (ϕn)∈1 w(Ꮿ(Ω)) with 1(ϕn)≤1, notice that the series ∞ n=1|ϕn(t)|is convergent for all t∈Ω.Putfn(t) =n k=1|ϕk(t)|and f(t)=limn→∞ fn(t). By Egorov’s theorem, the sequence (fn)is quasi-uniformly convergent to f. Then, there exists E∈Σ such that µ(E) < δ and fn|Ω\E→f|Ω\E(2.12) uniformly. If C=sup{|mT|(Ω):T∈ᏹ}, there exists n0∈Nso that n≥n0 ϕn(t) <ε 2C,∀t∈Ω\E. (2.13) Now, n≥n0 Tϕn = n≥n0 Ω ϕn(t)dmT ≤ n≥n0 E ϕn(t)dmT + n≥n0 Ω\E ϕn(t)dmT ≤ n≥n0E ϕn(t) d mT + n≥n0Ω\E ϕn(t) d mT =E n≥n0 ϕn(t) d mT +Ω\E n≥n0 ϕn(t) d mT ≤ mT (E)+ε 2C mT (Ω\E) <ε. (2.14) We denote by ᐂ(X,Y) the class of completely continuous operators from Xinto Y, that is, the class of operators which map weakly convergent sequences in Xinto norm-convergent sequences in Y. A set ᏹ⊂ᐂ(X,Y) is said to be uniformly completely continuous if, given a weakly convergent sequence (xn)in X,(Txn)is norm convergent uniformly in T∈ᏹ. The following result gives some characterizations of uniformly completely continuous sets in ᐂ(Ꮿ(Ω),Y). Recall that an operator Tdefined on Ꮿ(Ω) is completely continuous if and only if Tis weakly compact [6, Corollary VI.2.17], so mTis countably additive and regular, too. Theorem 2.5.Let ᏹ⊂ᐂ(Ꮿ(Ω),Y) be a bounded set for the operator norm. The following statements are equivalent: (a) ᏹis uniformly completely continuous, (b) the family {mT:T∈ᏹ}is uniformly countably additive,
3402 J. M. DELGADO AND C. PIÑEIRO (c) ᏹ∗={T∗:T∈ᏹ}is collectively weakly compact, that is, the set T∈ᏹT∗(BY∗) is relatively weakly compact in Ꮿ(Ω)∗. Proof. (a)⇒(b). By [6, Proposition I.1.17], the family {mT:T∈ᏹ}is uniformly countably additive if and only if ᏺ={y∗◦mT:T∈ᏹ,y ∗∈BY∗}is. According to [6, Lemma VI.1.13], we have to prove that lim n→∞ y∗◦mTOn=0 uniformly in ᏺ,(2.15) for all disjoint sequences (On)of open subsets of Ω. By contradiction, suppose there exists such a sequence (On)for which limn→∞ y∗◦mT(On)=0 but not uniformly in ᏺ. Then, there exists ε>0 and sequences (y∗ n)⊂BY∗,(Tn)∈ᏹ, and (Okn)⊂(On)such that y∗ n◦mTnOkn >ε, ∀n∈N.(2.16) Now, using the regularity of each mTn, we can find a sequence of compact sets (Hn) with Hn⊂Oknand mTn Okn\Hn<ε 2,∀n∈N,(2.17) (mdenotes the semivariation of m, that is, m(E) =sup{|y∗◦m|(E) :y∗∈BY∗}). By Urysohn’s lemma, for every n∈Nthere exists a continuous function ϕn:Ω→ [0,1]such that ϕn(Hn)=1 and ϕn(Ω\Okn)=0. Obviously, the series ∞ n=1ϕnis unconditionally convergent in Ꮿ(Ω).Sinceᏹis uniformly completely continuous, there exists n0∈Nsuch that Tϕn <ε 2,∀n≥n0,∀T∈ᏹ.(2.18) Then, we have mTnOkn ≤ mTnOkn−Tnϕn + Tnϕn = Ω χOkndmTn−Ω ϕndmTn + Tnϕn = Okn1−ϕndmTn + Tnϕn = Okn\Hn1−ϕndmTn + Tnϕn ≤ mTn Okn\Hn+ Tnϕn <ε, (2.19) for all n≥n0. This is in contradiction with (2.16). (b)⇒(a). By [6, Theorem I.2.4], there exists a scalar countably additive measure µ:Σ→ [0,∞)such that {mT:T∈ᏹ}is uniformly µ-continuous. Then, if (ϕn)is a sequence
UNIFORMLY SUMMING SETS OF OPERATORS … 3403 that tends to zero weakly in Ꮿ(Ω), it is obvious that zero is the pointwise limit of the sequence (ϕn(t)). Now, using Egorov’s theorem and proceeding along similar lines as the proof of (b)⇒(a) in Theorem 2.4, the proof concludes. (b)(c). The set T∈ᏹT∗(BY∗)={y∗◦mT:T∈ᏹ,y ∗∈BY∗}⊂Ꮿ(Ω)∗is relatively weakly compact if and only if it is bounded and uniformly countably additive [4,Theorem VII.13]. A call to [6, Proposition I.1.17] makes clear that T∈ᏹT∗(BY∗)is uniformly countably additive if and only if condition (b) is satisfied. Corollary 2.6.If ᏹ⊂Π1Ꮿ(Ω),Yis uniformly 1-summing, then ᏹis uniformly completely continuous. The converse of the last result is not true in general. Proposition 2.7.Suppose that the cardinal of Ωis infinite. The following statements are equivalent: (a) each subset of Π1(Ꮿ(Ω),Y) uniformly completely continuous is uniformly 1-summing, (b) Yis finite-dimensional. Proof. (a)⇒(b). By contradiction, suppose there is an unconditionally summable serie kykin Ysuch that kyk=∞.Let(ωk)be a sequence in Ωwith ωk≠ωl when k≠l. For each m∈Nconsider the operator Tm:Ꮿ(Ω)→Ydefined by Tmϕ= m k=1 ϕωkyk.(2.20) It is not difficult to show that ᏹ=(Tm)is uniformly completely continuous. Nevertheless, π1Tm= m k=1 yk m →∞,(2.21) so ᏹcannot be uniformly 1-summing because it is not π1-bounded. (b)⇒(a). This follows easily in view of conditions (b) in Theorems 2.4 and 2.5. We have showed that the converse of Corollary 2.6 is not true in general. However, a direct argument using Theorems 2.4 and 2.5 leads up to conclude that every uniformly completely continuous set ᏹ⊂Π1(Ꮿ(Ω),Y) verifying the following condition is uniformly 1-summing: (i) given T∈ᏹand a finite subset {(ϕ1,y∗ 1),...,(ϕm,y∗ m)}of Ꮿ(Ω)×BY∗,there exist S∈ᏹand z∗∈BY∗such that |y∗ n,Tϕn| ≤ |z∗,Sϕn|,n=1,...,m. 3. Final notes and examples. The Grothendieck-Pietsch domination theorem states that an operator T:X→Yis p-summing if and only if there exists a positive Radon measure µdefined on the (weak∗) compact space BX∗such that Tx p≤BX∗ x∗,x|pdµx∗,(3.1)
3404 J. M. DELGADO AND C. PIÑEIRO for all x∈X[5, Theorem 2.12]. Since the appearance of this theorem, there is a great interest in finding out the structure of uniformly p-dominated sets. A subset ᏹof Πp(X,Y) is uniformly p-dominated if there exists a positive Radon measure µsuch that the inequality (3.1) holds for all x∈Xand all T∈ᏹ.In[3,8,9], the reader can find some of the most recent steps given on this subject. Now we are going to show that these sets are uniformly p-summing. Proposition 3.1.If ᏹ⊂Πp(X,Y) is a uniformly p-dominated set, then ᏹ∗∗ = {T∗∗ :T∈ᏹ}is uniformly p-summing. Proof. Let µbe a measure for which ᏹis uniformly p-dominated. In a similar way as in the Pietsch factorization theorem [5, Theorem 2.13], we can obtain, for all T∈ᏹ, operators UT:Lp(µ) →∞(BY∗),UT≤µ(BX∗)1/p, and an operator V:X→L∞(µ) such that the following diagram is commutative: XT V y iY ∞BY∗ L∞(µ) ipLp(µ) UT (3.2) Here, ipis the canonical injection from L∞(µ) into Lp(µ) and iYis the isometry from Yinto ∞(BY∗)defined by iY(y) =(y∗,y)y∗∈BY∗. Notice that i∗∗ pcan be viewed as ipcomposed with the canonical projection P:L∞(µ)∗∗ →L∞(µ) which is simply the adjoint of the usual embedding L1(µ) →L1(µ)∗∗. By weak compactness, we may and do consider T∗∗ as a map from X∗∗ into Yfor which iY◦T∗∗ =UT◦ip◦P◦V∗∗.(3.3) Given ε>0 and (x∗∗ n)∈p w(X∗∗), we can choose n0∈Nso that n≥n0 ip◦P◦V∗∗x∗∗ n p<ε µBX∗,(3.4) because ip◦P◦V∗∗ is p-summing. Then, we have n≥n0 T∗∗x∗∗ n p= n≥n0 iY◦T∗∗x∗∗ n p= n≥n0 UT◦ip◦P◦V∗∗x∗∗ n p ≤µBX∗ n≥n0 ip◦P◦V∗∗x∗∗ n p<ε, (3.5) for all T∈ᏹ.So,ᏹ∗∗ is uniformly p-summing.
UNIFORMLY SUMMING SETS OF OPERATORS … 3405 It is easy to show that the study of uniformly p-summing sets can be reduced to the behavior of its sequences. Indeed, a bounded set ᏹin Πp(X,Y) is uniformly psumming if and only if every sequence (Tn)in ᏹadmits a uniformly p-summing subsequence. Thus, it seems to be interesting to make clear the relationship between uniformly p-summing sets and relatively weakly compact sets. For p=1, we have the following result. Proposition 3.2.Every relatively weakly compact set in Π1(X,Y) is uniformly 1summing. Proof. Let ᏹbe a relatively weakly compact set in Π1(X,Y).Givenˆ x=(xn)∈ 1 w(X), consider the (weak-weak) continuous operator Uˆ x:Π1(X,Y) →1 a(Y) defined by Uˆ x(T) =(Txn). Then, Uˆ x(ᏹ)is relatively weakly compact in 1 a(Y); according to [2, Theorem 2], we can conclude that ᏹis uniformly 1-summing. Proposition 3.2 does not remain true if p=2. For example, for each β=(βn)∈2 consider the operator Tβ:c0→2defined by T(αn)=(αn·βn)and put ᏹ={Tβ:β∈ B2}⊂Π2(c0,2)[5, Theorem 3.5]. If we consider 2as a subspace of Π2(c0,2), the set ᏹ=B2is relatively weakly compact. Nevertheless, no matter how we choose k∈N, n≥k Teken 2=1,(3.6) so ᏹcannot be uniformly 2-summing. Now we show that there are uniformly p-summing sets failing to be relatively weakly compact. Proposition 3.3.If every uniformly p-summing set is relatively weakly compact in Πp(X,Y), then Yis reflexive. Proof. Fixing x∗ 0∈X∗with x∗ 0=1, the isometry y∈Yx∗ 0⊗y∈x∗ 0⊗Yallows us to see Yas a subspace of Πp(X,Y).Ifε>0 and (xn)∈p w(X), there exists n0∈N so that n≥n0 x∗ 0,xn p<ε; (3.7) hence, for every y∈BY, n≥n0 x∗ 0⊗yxn p= n≥n0 x∗ 0,xn pyp<ε. (3.8) This yields that BYis uniformly p-summing and, by hypothesis, weakly compact. The converse of Proposition 3.3 is not always true. By contradiction, suppose every uniformly 1-summing set in Π1(1,2)is relatively weakly compact. Because 1does not contain any copy of c0, every bounded set in Π1(1,2)is relatively weakly compact. Then, we conclude that Π1(1,2)is reflexive, which is not possible since ∗ 1,viewedas a subspace of Π1(1,2),isnot. However, if p=1 and X=Ꮿ(Ω), the reflexivity of Yis a sufficient condition for a uniformly 1-summing set to be relatively weakly compact. Indeed, if rbvca(Σ,Y) denotes