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
LetXandYbe Banach spaces. A setᏹof 1-summing operators fromXintoYis said to be uniformly summingif the following holds: given a weakly 1-summing sequence(xn)inX, the seriesnT xnis uniformly convergent inT∈ᏹ. We study some general properties and obtain a characterization of these sets whenᏹis a set of operators defined on spaces of continuous functions.
2000 Mathematics Subject Classification: 47B38, 47B10.
1. Introduction. Throughout this paper,X and Y will be Banach spaces. IfX is a Banach space,BX= {x∈X:x ≤1}will denote its closed unit ball andX∗will be the
topological dual ofX. Given a real numberp∈[1,∞), a (linear) operatorT:X→Y is said to bep-summingif there exists a constantC >0 such that
n
i=1
T xip
1/p
≤C·sup
n
i=1
x∗,x
i p
1/p
:x∗∈BX∗
(1.1)
for every finite set {x1,...,xn} ⊂X. The least C for which the above inequality
al-ways holds is denoted by πp(T )(thep-summing normofT). The linear space of all
p-summing operators fromX intoY is denoted byΠp(X,Y )which is a Banach space
endowed with thep-summing norm.
As usual,pw(X)will be the Banach space of weakly p-summable sequences in X,
that is, the sequences (xn)⊂X satisfying n|x∗,xn |p<∞ for all x∗ ∈X∗; the
norm inwp(X)isp(xn)=sup{(n|x∗,xn |p)1/p:x∗∈BX∗}. The set of all strongly
p-summable sequences inXis denoted bypa(X); the norm in this space isπp(xn)=
(nxnp)1/p. IfT∈Πp(X,Y ), the correspondenceT:(xn)(T xn)always induces a
bounded operator frompw(X)intopa(Y )withT =πp(T )[5, Proposition 2.1].
Families of operators arise in different applications: equations containing a parame-ter, homotopies of operators, and so forth. In these applications, it may be very inter-esting to know that, given a setᏹ⊂Πp(X,Y )and(xn)∈pw(X), the seriesnT xnp
is uniformly convergent inT∈ᏹ. The main purpose of this paper is to studyuniformly
p-summingsets, that is, those setsᏹ⊂Πp(X,Y )for which, given(xn)∈wp(X), the
seriesnT xnpis uniformly convergent inT∈ᏹ. These sets also enjoy some
Proposition1.1. (a)Let(Tk)be a sequence inΠp(X,Y ). Then,Tk→k 0pointwise if and only ifTk→k 0pointwise and(Tk)is uniformlyp-summing.
(b)Letᏹ⊂Πp(X,Y )be a uniformlyp-summing set. Ifᏹis endowed with the strong operator topology, then the mapT∈ᏹnT xnp∈Ris continuous for every(xn)∈
pw(X).
A basic argument shows that uniformlyp-summing sets are bounded for the p -summing norm. In fact, ifXdoes not contain any copy of c0, bounded sets and
uni-formly 1-summing sets are the same. That is the reason for which we only consider operators defined on aᏯ(Ω)-space,Ωbeing a compact Hausdorff space. We recall that every weakly compact operator T :Ꮿ(Ω)→Y has a representing measure mT :Σ→
Y defined bymT(B)=T∗∗(χB) for all B ∈Σ, where Σdenotes the Borel σ-field of
subsets ofΩandχBdenotes the characteristic function ofB. The vector measuremTis
regular and countably additive [6, Theorem VI.2.5 and Corollary VI.2.14]. If we denote byTthe operatorT∗∗restricted toB(Σ)(the space of all bounded Borel-measurable scalar-valued functions defined onΩ), then
T ϕ=
Ωϕ dmT, (1.2)
for allϕ∈B(Σ)(the integral is the elementary Bartle integral [6, Definition I.1.12]). It is well known that everyp-summing operator defined on a Banach spaceXis weakly compact. InSection 2, we consider 1-summing operatorsTdefined onᏯ(Ω); these op-erators are characterized as those with representing measuremThaving finite variation
andπ1(T )= |mT|(Ω)[6, Theorem VI.3.3]. We show that a setᏹ⊂Π1(Ꮿ(Ω),Y )is
uni-formly 1-summing if and only if the family of all variation measures{|mT|:T∈ᏹ}is
uniformly bounded and there is a countably additive measureµ:Σ→[0,∞)such that {|mT|:T∈ᏹ}is uniformlyµ-continuous.
InSection 3, we mention a special class of uniformlyp-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. Uniformly1-summing sets inΠ1(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.
Proposition2.1[6, Proposition I.1.17]. The following statements about a collection
{mi:i∈I}of Y-valued measures defined on aσ-fieldΣare equivalent:
(a) the set{mi:i∈I}is uniformly countably additive, that is, if(En)is a sequence of pairwise disjoint members of Σ, thenlimnk≥nmi(Ek) =0uniformly ini∈I,
(b) the set{y∗◦m
i:i∈I, y∗∈BY∗}is uniformly countably additive,
(c) if(En)is a sequence of pairwise disjoint members ofΣ, thenlimnmi(En) =0 uniformly ini∈I,
(d) if(En)is a sequence of pairwise disjoint members ofΣ, thenlimnmi(En)=0 uniformly ini∈I, wheremidenotes the semivariation ofmi,
(e) the set{|y∗◦m
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 measureµonΣsuch that{mi:i∈I}is uniformlyµ-continuous, that is,
lim
µ(E)→0
mi(E)=0 (2.1)
uniformly ini∈I.
IfΩis a compact Hausdorff space andΣdenotes theσ-field of the Borel subsets of Ω, a vector measuremonΣis regular if for each Borel setEandε >0 there exists a compact setKand an open setOsuch thatK⊂E⊂Oandm(O\K) < ε.
Proposition2.3[6, Lemma VI.2.13]. Letbe a family of regular (countably additive) scalar measures defined onΣ. Each of the following statements implies all the others:
(a) for each pairwise disjoint sequence(On)of open subsets ofΩ, limnµ(On)=0 uniformly inµ∈,
(b) for each pairwise disjoint sequence(On)of open subsets ofΩ,limn|µ|(On)=0 uniformly inµ∈,
(c) is uniformly countably additive,
(d) is uniformly regular, that is, ifE∈Σandε >0, then there exists a compact set
Kand an open setOsuch thatK⊂E⊂Oandsupµ∈|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 whenT:Ꮿ(Ω)→Y is 1-summing [7, Proposition 15.21].
Theorem2.4. Letᏹ⊂Π1(Ꮿ(Ω),Y )be a bounded set. The following statements are equivalent:
(a) ᏹis uniformly1-summing,
(b) the family of nonnegative measures{|mT|:T∈ᏹ}is uniformly countably addi-tive,
(c) givenε >0and a disjoint sequence(En)of Borel subsets ofΩ, there existsn0∈N such that
n≥n0
mT
En< ε, (2.2)
for allT∈ᏹ.
Proof. (a)⇒(b). According to [6, Lemma VI.2.13], it suffices to show that limn→∞|mT|(On)=0 uniformly inT∈ᏹ, for all disjoint sequences(On)of open
sub-sets ofΩ. By contradiction, suppose that there existsε >0, a sequence(Tn)inᏹ, and
a strictly increasing sequence(kn)of natural numbers such that
mT
n Okn
Now we consider the operatorsSn:Ꮿ(Ω,Okn)→Y defined by
Snϕ=
Oknϕ dmTn, (2.4)
for allϕ∈Ꮿ(Ω,Okn), whereᏯ(Ω,Okn)is the closed subspace ofᏯ(Ω)formed by all
continuous functionsϕonΩsuch thatϕvanishes inΩ\Okn. It is known thatπ1(Sn)=
|mTn|(Okn), for alln∈N[7, Theorem 19.3]. For eachn∈N, we can choose a finite set
{ϕn
1,...,ϕnpn} ⊂Ꮿ(Ω,Okn)satisfying1(ϕni)pi=n1≤1 and
pn
i=1
Snϕn
i> π1Sn−ε. (2.5)
Since the open setsOkn are disjoint, it follows that the sequence(ϕ 1
1,...,ϕ1p1,ϕ
2 1,...,
ϕ2
p2,...)belongs to
1
w(Ꮿ(Ω)). Nevertheless, for alln∈N, we have
m≥n pm
i=1
Tnϕm i ≥
pn
i=1
Tnϕn i=
pn
i=1
Snϕn i> π1
Sn
−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 ofΩfor which there existsε >0, a sequence(Tn)inᏹ, and a strictly
increasing sequence(kn)of natural numbers so that
kn+1
i=kn+1
mT
n
Ei> ε, ∀n∈N. (2.7)
If we putBn=ki=n+kn1+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 inT∈ᏹ, (2.8)
for all disjoint sequences(En)of Borel subsets ofΩ. Suppose (b) fails. Then, there exists
ε >0, a sequence(Tn)inᏹ, and a strictly increasing sequence(kn)of natural numbers
satisfying
mTn Ekn
> ε, ∀n∈N. (2.9)
For eachn∈N, we choose a finite partition{En
1,...,Epnn}ofEknfor which
pn
i=1
mT
n
En
i> ε. (2.10)
Then, the disjoint sequence(E1
1,...,Ep11,E
2
(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 inT∈ᏹ. (2.11)
Hence, given ε >0, there exists δ > 0 such that, if E ∈ Σ verifies µ(E) < δ, then |mT|(E) < ε/2, for allT∈ᏹ.
Next, given(ϕn)∈w1(Ꮿ(Ω))with1(ϕn)≤1, notice that the series
∞
n=1|ϕn(t)|is
convergent for allt∈Ω. Putfn(t)=nk=1|ϕk(t)|andf (t)=limn→∞fn(t). By Egorov’s
theorem, the sequence(fn)is quasi-uniformly convergent tof. Then, there existsE∈Σ
such thatµ(E) < δand
fn|Ω\E →f|Ω\E (2.12)
uniformly. IfC=sup{|mT|(Ω):T∈ᏹ}, there existsn0∈Nso that
n≥n0
ϕn(t) < ε
2C, ∀t∈Ω\E. (2.13)
Now,
n≥n0
T ϕn=
n≥n0
Ωϕn(t)dmT
≤
n≥n0
Eϕn(t)dmT
+
n≥n0
Ω\
Eϕn(t)dmT
≤
n≥n0
E
ϕn(t) d mT +
n≥n0
Ω\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 X into Y, that is, the class of operators which map weakly convergent sequences inX into norm-convergent sequences inY. A setᏹ⊂ᐂ(X,Y )is said to beuniformly completely continuousif, given a weakly convergent sequence(xn)inX,(T xn)is norm convergent
uniformly inT ∈ᏹ. The following result gives some characterizations of uniformly completely continuous sets inᐂ(Ꮿ(Ω),Y ). Recall that an operatorT defined onᏯ(Ω) is completely continuous if and only ifT is weakly compact [6, Corollary VI.2.17], so mT is countably additive and regular, too.
Theorem 2.5. Letᏹ⊂ ᐂ(Ꮿ(Ω),Y ) be a bounded set for the operator norm. The following statements are equivalent:
(a) ᏹis uniformly completely continuous,
(c) ᏹ∗= {T∗:T∈ᏹ}is collectively weakly compact, that is, the set
T∈ᏹT∗(BY∗) is relatively weakly compact inᏯ(Ω)∗.
Proof. (a)⇒(b). By [6, Proposition I.1.17], the family {mT :T ∈ᏹ} is uniformly
countably additive if and only ifᏺ= {y∗◦m
T :T ∈ᏹ, y∗∈BY∗}is. According to
[6, Lemma VI.1.13], we have to prove that
lim
n→∞y ∗◦m
T
On
=0 uniformly inᏺ, (2.15)
for all disjoint sequences(On)of open subsets ofΩ. By contradiction, suppose there
exists such a sequence(On)for which limn→∞y∗◦mT(On)=0 but not uniformly inᏺ.
Then, there existsε >0 and sequences(yn∗)⊂BY∗,(Tn)∈ᏹ, and(Okn)⊂(On)such
that
y∗
n◦mTn
Okn > ε, ∀n∈N. (2.16)
Now, using the regularity of eachmTn, we can find a sequence of compact sets(Hn)
withHn⊂Oknand
mT
nOkn\Hn
<2ε, ∀n∈N, (2.17)
(mdenotes the semivariation ofm, that is,m(E)=sup{|y∗◦m|(E): y∗∈B
Y∗}).
By Urysohn’s lemma, for every n∈Nthere exists a continuous function ϕn :Ω→
[0,1] such thatϕn(Hn)= 1 and ϕn(Ω\Okn)=0. Obviously, the series
∞
n=1ϕn is
unconditionally convergent inᏯ(Ω). Sinceᏹis uniformly completely continuous, there existsn0∈Nsuch that
T ϕn< ε
2, ∀n≥n0,∀T∈ᏹ. (2.18)
Then, we have mT
n
Okn≤mTn
Okn
−Tnϕn+Tnϕn
=
ΩχOkndmTn−
ΩϕndmTn
+Tnϕn
=
Okn
1−ϕn
dmTn
+Tnϕn
=
Okn\Hn
1−ϕndmTn
+Tnϕn
≤mTnOkn\Hn
+Tnϕn
< ε,
(2.19)
for alln≥n0. This is in contradiction with (2.16).
that tends to zero weakly inᏯ(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) inTheorem 2.4, the proof concludes.
(b)(c). The setT∈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, Theo-rem VII.13]. A call to [6, Proposition I.1.17] makes clear thatT∈ᏹT∗(BY∗)is uniformly
countably additive if and only if condition (b) is satisfied.
Corollary2.6. Ifᏹ⊂Π1Ꮿ(Ω),Yis uniformly1-summing, thenᏹis uniformly completely continuous.
The converse of the last result is not true in general.
Proposition2.7. Suppose that the cardinal ofΩis infinite. The following statements are equivalent:
(a) each subset ofΠ1(Ꮿ(Ω),Y )uniformly completely continuous is uniformly1 -sum-ming,
(b) Y is finite-dimensional.
Proof. (a)⇒(b). By contradiction, suppose there is an unconditionally summable seriekykin Y such that
kyk = ∞. Let(ωk)be a sequence inΩwithωk≠ωl
whenk≠l. For eachm∈Nconsider the operatorTm:Ꮿ(Ω)→Y defined by
Tmϕ= m
k=1
ϕωk
yk. (2.20)
It is not difficult to show thatᏹ=(Tm)is uniformly completely continuous.
Neverthe-less,
π1
Tm
=
m
k=1
yk m
→ ∞, (2.21)
soᏹcannot be uniformly 1-summing because it is notπ1-bounded.
(b)⇒(a). This follows easily in view of conditions (b) inTheorems 2.4and2.5.
We have showed that the converse ofCorollary 2.6is not true in general. However, a direct argument usingTheorems 2.4and 2.5 leads up to conclude that every uni-formly completely continuous setᏹ⊂Π1(Ꮿ(Ω),Y )verifying the following condition is
uniformly 1-summing:
(i) givenT ∈ᏹand a finite subset{(ϕ1,y1∗),...,(ϕm,ym∗)}ofᏯ(Ω)×BY∗, there
existS∈ᏹandz∗∈BY∗such that|yn∗,T ϕn | ≤ |z∗,Sϕn |, n=1,...,m.
3. Final notes and examples. The Grothendieck-Pietsch domination theorem states that an operatorT:X→Y is p-summing if and only if there exists a positive Radon measureµdefined on the (weak∗) compact spaceBX∗ such that
T xp
≤
BX∗
for allx∈X[5, Theorem 2.12]. Since the appearance of this theorem, there is a great interest in finding out the structure of uniformlyp-dominated sets. A subset ᏹ of Πp(X,Y )is uniformly p-dominated if there exists a positive Radon measure µ such
that the inequality (3.1) holds for allx∈Xand allT ∈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 uniformlyp-summing.
Proposition 3.1. If ᏹ ⊂ Πp(X,Y ) is a uniformly p-dominated set, then ᏹ∗∗ =
{T∗∗: T∈ᏹ}is uniformlyp-summing.
Proof. Letµbe a measure for whichᏹis uniformlyp-dominated. In a similar way as in the Pietsch factorization theorem [5, Theorem 2.13], we can obtain, for allT∈ᏹ, operatorsUT:Lp(µ)→∞(BY∗),UT ≤µ(BX∗)1/p, and an operatorV:X→L∞(µ)such
that the following diagram is commutative:
X T
V
y
iY
∞BY∗
L∞(µ) ip Lp(µ) UT
(3.2)
Here,ipis the canonical injection fromL∞(µ)intoLp(µ)andiY is the isometry from
Y into∞(BY∗)defined byiY(y)=(y∗,y )y∗∈BY∗. Notice thati∗∗p can be viewed as
ipcomposed with the canonical projectionP:L∞(µ)∗∗→L∞(µ)which is simply the
adjoint of the usual embeddingL1(µ)→L1(µ)∗∗. By weak compactness, we may and
do considerT∗∗as a map fromX∗∗intoY for which
iY◦T∗∗=UT◦ip◦P◦V∗∗. (3.3)
Givenε >0 and(x∗∗n )∈ p
w(X∗∗), we can choosen0∈Nso that
n≥n0
ip◦P◦V∗∗x∗∗
n
p
< ε µBX∗
, (3.4)
becauseip◦P◦V∗∗isp-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)
It is easy to show that the study of uniformlyp-summing sets can be reduced to the behavior of its sequences. Indeed, a bounded setᏹ inΠp(X,Y ) is uniformlyp
-summing if and only if every sequence(Tn)inᏹadmits a uniformlyp-summing
sub-sequence. Thus, it seems to be interesting to make clear the relationship between uni-formly p-summing sets and relatively weakly compact sets. Forp =1, we have the following result.
Proposition 3.2. Every relatively weakly compact set in Π1(X,Y ) is uniformly1 -summing.
Proof. Let ᏹ be a relatively weakly compact set inΠ1(X,Y ). Given ˆx =(xn)∈
1
w(X), consider the (weak-weak) continuous operatorUxˆ:Π1(X,Y )→1a(Y )defined
byUxˆ(T )=(T xn). Then,Uxˆ(ᏹ)is relatively weakly compact in1a(Y ); according to [2,
Theorem 2], we can conclude thatᏹis uniformly 1-summing.
Proposition 3.2does not remain true ifp=2. For example, for eachβ=(βn)∈2
consider the operatorTβ:c0→2defined byT (αn)=(αn·βn)and putᏹ= {Tβ:β∈
B2} ⊂Π2(c0,2)[5, Theorem 3.5]. If we consider2as a subspace ofΠ2(c0,2), the set
ᏹ=B2is relatively weakly compact. Nevertheless, no matter how we choosek∈N,
n≥k
Te
ken 2
=1, (3.6)
soᏹcannot be uniformly 2-summing.
Now we show that there are uniformlyp-summing sets failing to be relatively weakly compact.
Proposition3.3. If every uniformlyp-summing set is relatively weakly compact in Πp(X,Y ), thenY is reflexive.
Proof. Fixingx0∗∈X∗withx0∗ =1, the isometryy∈Yx0∗⊗y∈x∗0⊗Yallows
us to seeY as a subspace ofΠp(X,Y ). Ifε >0 and(xn)∈pw(X), there existsn0∈N
so that
n≥n0
x∗
0,xn p< ε; (3.7)
hence, for everyy∈BY,
n≥n0
x∗
0⊗y
xnp=
n≥n0
x∗
0,xn pyp< ε. (3.8)
This yields thatBY is uniformlyp-summing and, by hypothesis, weakly compact.
The converse ofProposition 3.3is not always true. By contradiction, suppose every uniformly 1-summing set inΠ1(1,2)is relatively weakly compact. Because 1 does
not contain any copy ofc0, every bounded set inΠ1(1,2)is relatively weakly compact.
Then, we conclude thatΠ1(1,2)is reflexive, which is not possible since∗1, viewed as
a subspace ofΠ1(1,2), is not.
the set of all regular, countably additive,Y-valued measuresm onΣ with bounded variation, recall that relatively weakly compact setsᏹinr bvca(Σ,Y )are those ver-ifying the following conditions: (i)ᏹis bounded; (ii) the family of nonnegative mea-sures{|m|:m∈ᏹ}is uniformly countably additive; and (iii) for eachE∈Σ, the set {m(E):m∈ᏹ}is relatively weakly compact inY [6, Theorem IV.2.5]. Having in mind the identification betweenΠ1(Ꮿ(Ω),Y )andr bvca(Σ,Y ), and making use of the
char-acterization of uniformly 1-summing sets obtained in Theorem 2.4, we conclude the next characterization.
Corollary3.4. The following statements are equivalent:
(a) Y is reflexive,
(b) every setᏹinΠ1(Ꮿ(Ω),Y )is uniformly1-summing if and only ifᏹis relatively weakly compact.
It is well known that a linear operatorT is 1-summing if and only ifT∗∗is. So, it is natural to ask if a setᏹis uniformly 1-summing wheneverᏹ∗∗= {T∗∗:T ∈ᏹ}is. Unfortunately, we are going to show that this is not true in general. It suffices to take Xas the separableᏸ∞-space of Bourgain and Delbaen [1]. This space has the Radon– Nikodym property, so it does not contain any copy ofc0. Nevertheless,X∗is isomorphic
to1and, therefore,X∗∗contains a copy ofc0. Let(en)be the canonical basis of1
and J:1→X∗ an isomorphism. PutTn=Jen∈Π1(X,R); the setᏹ= {Tn:n∈N}
is uniformly 1-summing since it is bounded andX does not contain any copy ofc0.
Notice that the elements ofᏹ∗∗are the linear formsx∗∗∈X∗∗x∗∗,Je
n ∈R, for
alln∈N. If(e∗n)is the canonical basis ofc0, then((J∗)−1(e∗n))∈1w(X∗∗); hence, no
matter how we choosek∈N, it turns out that
n≥k
T∗∗
k
J∗−1e∗n =
n≥k
J∗−1en∗
,Jek =
n≥k
e∗
n,ek =1, (3.9)
andᏹ∗∗cannot be uniformly 1-summing.
Nevertheless, ifᏹis a set of operators defined onc0, then it is true thatᏹis uniformly
1-summing if and only ifᏹ∗∗is too. To see this, notice that for a 1-summing operatorT: (αn)∈c0∞n=1αnxn∈X, the second adjointT∗∗:∞→Xis defined byT∗∗(βn)=
∞
n=1βnxn, for all(βn)∈∞.
Whenᏹis a set of operators defined on aᏯ(Ω)-space, we do not know whetherᏹ∗∗ inherits the property or not. Anyway, we are going to prove the following weaker result. We inject isometricallyB(Σ)intoᏯ(Ω)∗∗in the natural way.
Proposition3.5. Ifᏹ⊂Π1(Ꮿ(Ω),X)is uniformly1-summing, thenᏹ = T:B(Σ)→
X:T∈ᏹ}is uniformly1-summing too.
Proof. The argument is similar to the one used in the proof of (b)⇒(a) in Theo-rem 2.4.
Finally, we give an example to show thatCorollary 2.6is not true ifᏯ(Ω)is replaced by a general Banach spaceX. It suffices to takeX=2andᏹ= {en∗:n∈N}, where(e∗n)
is the unit basis of∗22. The setᏹis bounded inΠ1(2,R)and, therefore, uniformly
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J. M. Delgado: Departamento de Matemáticas, Facultad de Ciencias Experimentales, Campus Universitario del Carmen, Avda. de las Fuerzas Armadas, 21071 Huelva, Spain
E-mail address:[email protected]
Cándido Piñeiro: Departamento de Matemáticas, Facultad de Ciencias Experimentales, Campus Universitario del Carmen, Avda. de las Fuerzas Armadas, 21071 Huelva, Spain
