Uniformly Summing Sets of Operators on Spaces of Continuous Functions

Let X and Y be Banach spaces. A set ᏹ of 1-summing operators from X into Y is said to be uniformly summing if the following holds: given a weakly 1-summing sequence (x n) in X, the series n T x n is uniformly convergent in T ∈ ᏹ. We study some general properties and obtain a characterization of these sets when ᏹ is a set of operators defined on spaces of continuous functions.

1. Introduction.Throughout this paper, X and Y will be Banach spaces.If X is a Banach space, B X = {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 → Y is said to be p-summing if there exists a constant C > 0 such that for every finite set {x 1 ,...,x n } ⊂ X.The least C for which the above inequality always holds is denoted by π p (T ) (the p-summing norm of T ).The linear space of all p-summing operators from X into Y is denoted by Π p (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 ( The set of all strongly p-summable sequences in X is denoted by p a (X); the norm in this space is π p (x n ) = ( n x n p ) 1/p .If T ∈ Π p (X, Y ), the correspondence T : (x n ) (T x n ) always induces a bounded operator from p w (X) into p a (Y ) with T = π p (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 ⊂ Π p (X, Y ) and (x n ) ∈ p w (X), the series n T x n p is uniformly convergent in T ∈ ᏹ.The main purpose of this paper is to study uniformly p-summing sets, that is, those sets ᏹ ⊂ Π p (X, Y ) for which, given (x n ) ∈ p w (X), the series n T x n p is uniformly convergent in T ∈ ᏹ.These sets also enjoy some properties that justify their study; the next proposition lists some of them.A basic argument shows that uniformly p-summing sets are bounded for the psumming norm.In fact, if X does not contain any copy of c 0 , 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, Ω being a compact Hausdorff space.We recall that every weakly compact operator T : Ꮿ(Ω) → Y has a representing measure m T : Σ → Y defined by m T (B) = T * * (χ B ) for all B ∈ Σ, where Σ denotes the Borel σ -field of subsets of Ω and χ B denotes the characteristic function of B. The vector measure m T is regular and countably additive [6, Theorem VI.2.5 and Corollary VI.2.14].If we denote by T the operator T * * restricted to B(Σ) (the space of all bounded Borel-measurable scalar-valued functions defined on Ω), then 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 X is weakly compact.In Section 2, we consider 1-summing operators T defined on Ꮿ(Ω); these operators are characterized as those with representing measure m T having finite variation and π 1 (T ) = |m T |(Ω) [6, Theorem VI.3.3].We show that a set ᏹ ⊂ Π 1 (Ꮿ(Ω), Y ) is uniformly 1-summing if and only if the family of all variation measures {|m T | : T ∈ ᏹ} is uniformly bounded and there is a countably additive measure µ : Σ → [0, ∞) such that {|m T | : 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.

Uniformly 1-summing sets in Π 1 (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.
If Ω is a compact Hausdorff space and Σ denotes the σ -field of the Borel subsets of Ω, a vector measure m on Σ is regular if for each Borel set E and ε > 0 there exists a compact set K and an open set O such that K ⊂ E ⊂ O and 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 (a) ᏹ is uniformly 1-summing, (b) the family of nonnegative measures {|m T | : T ∈ ᏹ} is uniformly countably additive, (c) given ε > 0 and a disjoint sequence (E n ) of Borel subsets of Ω, there exists n 0 ∈ N such that for all T ∈ ᏹ.

Proof. (a)⇒(b).
According to [6, Lemma VI.2.13], it suffices to show that lim n→∞ |m T |(O n ) = 0 uniformly in T ∈ ᏹ, for all disjoint sequences (O n ) of open subsets of Ω.By contradiction, suppose that there exists ε > 0, a sequence (T n ) in ᏹ, and a strictly increasing sequence (k n ) of natural numbers such that m Tn O kn > 2ε, ∀n ∈ N. ( Since the open sets O kn are disjoint, it follows that the sequence Nevertheless, for all n ∈ N, we have This denies (a) and proves that (a) implies (b).
(b)⇒(c).Again we proceed by contradiction.Suppose (E n ) is a disjoint sequence of Borel subsets of Ω for which there exists ε > 0, a sequence (T n ) in ᏹ, and a strictly increasing sequence (k n ) of natural numbers so that (2.7) So, in view of [6, Proposition I.1.17],the family {|m T | : T ∈ ᏹ} is not uniformly countably additive.
(c)⇒(b).We need to prove for all disjoint sequences (E n ) of Borel subsets of Ω. Suppose (b) fails.Then, there exists ε > 0, a sequence (T n ) in ᏹ, and a strictly increasing sequence (k n ) of natural numbers satisfying (2.11) Hence, given ε > 0, there exists δ > 0 such that, if (t).By Egorov's theorem, the sequence (f n ) is quasi-uniformly convergent to f .Then, there exists E ∈ Σ such that µ(E) < δ and (2.14) We denote by ᐂ(X, Y ) the class of completely continuous operators from X into Y , that is, the class of operators which map weakly convergent sequences in X into norm-convergent sequences in Y .A set ᏹ ⊂ ᐂ(X, Y ) is said to be uniformly completely continuous if, given a weakly convergent sequence (x n ) in X, (T x n ) 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 T defined on Ꮿ(Ω) is completely continuous if and only if T is weakly compact [6, Corollary VI.2.17], so m T is countably additive and regular, too.
that tends to zero weakly in Ꮿ(Ω), it is obvious that zero is the pointwise limit of the sequence (ϕ n (t)).Now, using Egorov's theorem and proceeding along similar lines as the proof of (b)⇒(a) in Theorem 2.4, the proof concludes.

Proof. (a)⇒(b). By contradiction, suppose there is an unconditionally summable serie
(2.20) It is not difficult to show that ᏹ = (T m ) is uniformly completely continuous.Nevertheless, .21) so ᏹ cannot be uniformly 1-summing because it is not π 1 -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 ⊂ Π 1 (Ꮿ(Ω), Y ) verifying the following condition is uniformly 1-summing: (i) given T ∈ ᏹ and a finite subset

Final notes and examples.
The Grothendieck-Pietsch domination theorem states that an operator T : X → Y is p-summing if and only if there exists a positive Radon measure µ defined on the (weak * ) compact space B X * such that 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 ᏹ of Π p (X, Y ) is uniformly p-dominated if there exists a positive Radon measure µ such that the inequality (3.1) holds for all x ∈ X and 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.Proof.Let µ be a measure for which ᏹ is 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 and an operator V : X → L ∞ (µ) such that the following diagram is commutative: Here, i p is the canonical injection from L ∞ (µ) into L p (µ) and i Y is the isometry from Y into ∞ (B Y * ) defined by i Y (y) = ( y * ,y ) y * ∈B Y * .Notice that i * * p can be viewed as i p composed with the canonical projection P : L ∞ (µ) * * → L ∞ (µ) which is simply the adjoint of the usual embedding L 1 (µ) → L 1 (µ) * * .By weak compactness, we may and do consider T * * as a map from X * * into Y for which because i p • P • V * * is p-summing.Then, we have for all T ∈ ᏹ.So, ᏹ * * is uniformly p-summing.
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 ᏹ in Π p (X, Y ) is uniformly psumming if and only if every sequence (T n ) in ᏹ admits 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 Π 1 (X, Y ) is uniformly 1summing.
Proof.Let ᏹ be a relatively weakly compact set in ], we can conclude that ᏹ is uniformly 1-summing.Proposition 3.2 does not remain true if p = 2.For example, for each Theorem 3.5].If we consider 2 as a subspace of Π 2 (c 0 , 2 ), the set ᏹ = B 2 is relatively weakly compact.Nevertheless, no matter how we choose k ∈ N, so ᏹ cannot be uniformly 2-summing.Now we show that there are uniformly p-summing sets failing to be relatively weakly compact.x * 0 ,x n p y p < ε. (3.8)This yields that B Y is 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 Π 1 ( 1 , 2 ) is relatively weakly compact.Because 1 does not contain any copy of c 0 , every bounded set in Π 1 ( 1 , 2 ) is relatively weakly compact.Then, we conclude that Π 1 ( 1 , 2 ) is reflexive, which is not possible since * 1 , viewed as a subspace of Π 1 ( 1 , 2 ), is not.
However, if p = 1 and X = Ꮿ(Ω), the reflexivity of Y is a sufficient condition for a uniformly 1-summing set to be relatively weakly compact.Indeed, if r bvca(Σ,Y ) denotes the set of all regular, countably additive, Y -valued measures m on Σ with bounded variation, recall that relatively weakly compact sets ᏹ in r bvca(Σ,Y ) are those verifying the following conditions: (i) ᏹ is bounded; (ii) the family of nonnegative measures {|m| : m ∈ ᏹ} is uniformly countably additive; and (iii) for each E ∈ Σ, the set {m(E) : m ∈ ᏹ} is relatively weakly compact in Y [6, Theorem IV.It is well known that a linear operator T is 1-summing if and only if T * * is.So, it is natural to ask if a set ᏹ is uniformly 1-summing whenever ᏹ * * = {T * * : T ∈ ᏹ} is.Unfortunately, we are going to show that this is not true in general.It suffices to take X as the separable ᏸ ∞ -space of Bourgain and Delbaen [1].This space has the Radon-Nikodym property, so it does not contain any copy of c 0 .Nevertheless, X * is isomorphic to 1 and, therefore, X * * contains a copy of c 0 .Let (e n ) be the canonical basis of 1 and J : 1 → X * an isomorphism.Put T n = Je n ∈ Π 1 (X, R); the set ᏹ = {T n : n ∈ N} is uniformly 1-summing since it is bounded and X does not contain any copy of c 0 .Notice that the elements of ᏹ * * are the linear forms x * * ∈ X * * x * * ,Je n ∈ R, for all n ∈ N. If (e * n ) is the canonical basis of c 0 , then ((J * ) −1 (e * n )) ∈ 1 w (X * * ); hence, no matter how we choose k ∈ N, it turns out that and ᏹ * * cannot be uniformly 1-summing.Nevertheless, if ᏹ is a set of operators defined on c 0 , then it is true that ᏹ is uniformly 1-summing if and only if ᏹ * * is too.To see this, notice that for a 1-summing operator T : When ᏹ is a set of operators defined on a Ꮿ(Ω)-space, we do not know whether ᏹ * * inherits the property or not.Anyway, we are going to prove the following weaker result.We inject isometrically B(Σ) into Ꮿ(Ω) * * in the natural way.
Proof.The argument is similar to the one used in the proof of (b)⇒(a) in Theorem 2.4.
Finally, we give an example to show that Corollary 2.6 is not true if Ꮿ(Ω) is replaced by a general Banach space X.It suffices to take X = 2 and ᏹ = {e * n : n ∈ N}, where (e * n ) is the unit basis of *

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Proposition 1 . 1 .
(a) Let (T k ) be a sequence in Π p (X, Y ).Then, T k k → 0 pointwise if and only if T k k → 0 pointwise and (T k ) is uniformly p-summing.(b) Let ᏹ ⊂ Π p (X, Y ) be a uniformly p-summing set.If ᏹ is endowed with the strong operator topology, then the map T ∈ ᏹ n T x n p ∈ R is continuous for every (x n ) ∈ p w (X).

Theorem 2 . 4 .
then there exists a compact set K and an open set O such that K ⊂ E ⊂ O and sup µ∈ |µ|(O\K) < ε.Now, we are able to show our main result.In the proof, we will use the fact that |m T | is regular when T : Ꮿ(Ω) → Y is 1-summing [7, Proposition 15.21].Let ᏹ ⊂ Π 1 (Ꮿ(Ω), Y ) be a bounded set.The following statements are equivalent:

Corollary 3 . 4 .
2.5].Having in mind the identification between Π 1 (Ꮿ(Ω), Y ) and r bvca(Σ,Y ), and making use of the characterization of uniformly 1-summing sets obtained in Theorem 2.4, we conclude the next characterization.The following statements are equivalent: (a) Y is reflexive, (b) every set ᏹ in Π 1 (Ꮿ(Ω), Y ) is uniformly 1-summing if and only if ᏹ is relatively weakly compact.

Theorem 2.2 [
where m i denotes the semivariation of m i , (e) the set {|y * • m i | : i ∈ I, y * ∈ B Y * } is uniformly countably additive.6, Theorem I.2.4].Let {m i : 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 {m i : i ∈ I} is uniformly countably additive if and only if there exists a positive real-valued countably additive measure µ on Σ such that {m 1.17], the family {m T : T ∈ ᏹ} is uniformly countably additive if and only if ᏺ = {y * • m T : T ∈ ᏹ, y * ∈ B Y * } is.According to [6, Lemma VI.1.13],we have to prove that ) of open subsets of Ω.By contradiction, suppose there exists such a sequence (O n ) for which lim n→∞ y * •m T (O n ) = 0 but not uniformly in ᏺ.Then, there exists ε > 0 and sequences (y * n ) ⊂ B Y * , (T n ) ∈ ᏹ, and (O kn ) ⊂ (O n ) such that y n * n • m Tn O kn > ε, ∀n ∈ N. (2.16) Now, using the regularity of each m Tn , we can find a sequence of compact sets (H n ) with H n ⊂ O kn and m Tn O kn \H n < ε 2 , ∀n ∈ N, (2.17) ( m denotes the semivariation of m, that is, m (E) = sup{|y * •m|(E) : y * ∈ B Y * }).By Urysohn's lemma, for every n ∈ N there exists a continuous function ϕ n : Ω → [0, 1] such that ϕ n (H n ) = 1 and ϕ n (Ω\O kn ) = 0. Obviously, the series for all n ≥ n 0 .This is in contradiction with (2.16).(b)⇒(a).By [6, Theorem I.2.4], there exists a scalar countably additive measure µ 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]makesclear that T ∈ᏹ T * (B Y * ) is uniformly countably additive if and only if condition (b) is satisfied.The converse of the last result is not true in general.