L-Dunford-Pettis property in Banach spaces

0.12 MB·

by  A. Retbi

MethodsofFunctional AnalysisandTopology
Vol. 22(2016), no. 4,pp. 387–392
L-DUNFORD-PETTIS PROPERTY IN BANACH SPACES
A.RETBIANDB.ELWAHBI
7
1 Abstract. In this paper, we introduce and study the concept of L-Dunford-Pettis
0 setsandL-Dunford-PettispropertyinBanachspaces. Next,wegiveacharacteriza-
2 tion of the L-Dunford-Pettis property with respect to some well-known geometric
propertiesofBanachspaces. Finally,somecomplementabilityofoperatorsonBanach
n
spaceswiththeL-Dunford-Pettispropertyarealsoinvestigated.
a
J

2

] 1. Introduction and notation
A
F A norm bounded subset A of a Banach space X is called Dunford-Pettis (DP for
h. short) if every weakly null sequence (fn) in X′ convergeuniformly to zero on A, that is,
t limn→∞supx∈A|fn(x)|=0.
a An operator T between two Banach spaces X and Y is completely continuous if T
m
maps weakly null sequences into norm null ones.
[ Recallfrom[11],thatanoperatorT :X →Y betweentwoBanachspacesisDunford-

1 Pettiscompletely continuous(abb. DPcc)if itcarriesaweaklynullsequence,whichis a
v DPsetinX tonormnullonesinY. Itisclearthateverycompletelycontinuousoperator
2 is DPcc. Also every weakly compact operator is DPcc (see Corollary 1.1 of [11]).
5 A Banach space X has:
5
0 – arelativelycompactDunford-Pettisproperty(DPrcPforshort)ifeveryDunford-
0 Pettis set in X is relatively compact [5]. For example, every Schur spaces have
. the DPrcP.
1
0 – aGrothendieckproperty(oraBanachspaceX isaGrothendieckspace)ifweak⋆
7 and weak convergence of sequences in X′ coincide. For example, each reflexive
1 space is a Grothendieck space.
:
v – aDunford-Pettisproperty(DPpropertyforshort)ifeveryweaklycompactopera-
i torT fromX intoanotherBanachspaceY iscompletelycontinuous,equivalently,
X
if every relatively weakly compact subset of X is DP.
r
a – a reciprocal Dunford-Pettis property (RDP property for short) if every com-
pletely continuous operator on X is weakly compact.
A subspace X of a Banach space X is complemented if there exists a projection P
1
from X to X (see page 9 of [2]).
1
Recall from [1], that a Banach lattice is a Banach space (E,k·k) such that E is a
vector lattice and its norm satisfies the following property: for each x,y ∈ E such that
|x|≤|y|, we have kxk≤kyk.
We denote by c , ℓ1, and ℓ∞ the Banach spaces of all sequences converging to zero,
0
all absolutely summable sequences, and all bounded sequences, respectively.
Let us recall that a norm bounded subset A of a Banach space X′ is called L-set
if every weakly null sequence (x ) in X converge uniformly to zero on A, that is,
n
2010 Mathematics Subject Classification. 46A40,46B40,46H42.
Key words and phrases. Dunford-Pettis set, Dunford-Pettis relatively compact property, Dunford-
Pettiscompletelycontinuous operator.
387
388 A.RETBIANDB.ELWAHBI
limn→∞supf∈A|f(xn)| =0. Note also that a Banach space X has the RDP property if
and only if every L-set in X′ is relatively weakly compact.
In his paper, G. Emmanuelle in [4] used the concept of L-set to characterize Banach
spaces not containing ℓ1, and gave several consequences concerning Dunford-Pettis sets.
Later, the idea of L-set is also used to establish a dual characterizationof the Dunford-
Pettis property [6].
The aim of this paper is to introduce and study the notion of L-Dunford-Pettis set
in a Banach space, which is related to the Dunford-Pettis set (Definition 2.1), and note
thateveryL-setinatopologicaldualofaBanachspaceisL-Dunford-Pettisset(Proposi-
tion2.3). Afterthat,weintroducetheL-Dunford-PettispropertyinBanachspacewhich
is sharedby those Banachspaces whose L-Dunford-Pettissubsets of his topologicaldual
are relatively weakly compact (Definition 2.6). Next, we obtain some important con-
sequences. More precisely, a characterizations of L-Dunford-Pettis property in Banach
spaces in terms of DPcc and weakly compact operators (Theorem 2.7), the relation be-
tweenL-Dunford-PettispropertywithDPandGrothendieckproperties(Theorem2.8),a
new characterizationsof Banachspace with DPrcP (resp, reflexive Banach space) (The-
orem 2.5) (resp, Corollary 2.10). Finally, we investigate the complementability of the
class of weakly compact operators from X into ℓ∞ in the class of DPcc from X into ℓ∞
(Theorem 2.13 and Corollary 2.14).
Thenotationsandterminologiesarestandard. WeusethesymbolsX,Y forarbitrary
Banach spaces. We denoted the closed unit ball of X by B , the topological dual of X
X
by X′ and T′ : Y′ → X′ refers to the adjoint of a bounded linear operator T : X → Y.
We refer the reader for undefined terminologies to the references [1, 8, 9].
2. Main results
Definition 2.1. LetX be aBanachspace. Anormbounded subsetAofthe dualspace
X′ is called an L-Dunford-Pettis set, if every weakly null sequence (x ), which is a DP
n
set in X converges uniformly to zero on A, that is, limn→∞supf∈A|f(xn)|=0.
For a proof of the next Proposition, we need the following Lemma which is just
Lemma 1.3 of [11].
Lemma 2.2. A sequence (x ) in X is DP if and only if f (x )→0 as n→∞for every
n n n
weakly null sequence (f ) in X′.
n
The following Proposition gives some additional properties of L-Dunford-Pettis sets
in a topological dual Banach space.
Proposition 2.3. Let X be a Banach space. Then
(1) every subset of an L-Dunford-Pettis set in X′ is L-Dunford-Pettis,
(2) every L-set in X′ is L-Dunford-Pettis,
(3) relatively weakly compact subset of X′ is L-Dunford-Pettis,
(4) absolutely closed convex hull of an L-Dunford-Pettis set in X′ is L-Dunford-
Pettis.
Proof. (1) and (2) are obvious.
(3) Suppose A ⊂ X′ is relatively weakly compact but it is not an L-Dunford-Pettis
set. Then, there exists a weakly null sequence (x ), which is a DP set in X, a sequence
n
(f ) in A andan ǫ>0 such that |f (x )|>ǫ for all integer n. As A is relatively weakly
n n n
compact, there exists a subsequence (g ) of (f ) that converges weakly to an element g
n n
in X′. But from
|g (x )|≤|(g −g)(x )|+|g(x )|
n n n n n
L-DUNFORD-PETTIS PROPERTY IN BANACH SPACES 389
and Lemma 2.2, we obtain that |g (x )|→0 as n→∞. This is a contradiction.
n n
(4)Let A be a L-Dunford-Pettisset inX′, and(x ) be a weaklynull sequence,which
n
is a DP set in X. Since
sup |f(x )|=sup |f(x )|
f∈aco(A) n f∈A n
n n
for each n, where aco(A) = {P λ x :x ∈A,∀i,P |λ |≤1} is the absolutely
i=1 i i i i=1 i
closed convex hull of A (see [1, pp. 148, 151]), then it is clear that aco(A) is L-Dunford-
Pettis set in X′. (cid:3)
We need the following Lemma which is just Lemma 1.2 of [11].
Lemma 2.4. A Banach space X has the DPrcP if and only if any weakly null sequence,
which is a DP set in X is norm null.
FromLemma2.4,weobtainthefollowingcharacterizationofDPrcPinaBanachspace
in terms of an L-Dunford-Pettis set of his topological dual.
Theorem 2.5. A Banach space X has the DPrcP if and only if every bounded subset of
X′ is an L-Dunford-Pettis set.
Proof. (⇐) Let (x ) be a weakly null sequence, which is a DP set in X. As
n
kx k=sup |f(x )|
n f∈BX′ n
for each n, and by our hypothesis, we see that kx k →0 as n →∞. By Lemma 2.4 we
n
deduce that X has the DPrcP.
(⇒) Assume by way of contradiction that there exist a bounded subset A, which is
notanL-Dunford-PettissetofX′. Then,thereexistsaweaklynullsequence(x ),which
n
is a Dunford-Pettis set of X such that sup |f(x )| > ǫ > 0 for some ǫ > 0 and each
f∈A n
n. Hence, for every n there exists some f in A such that |f (x )|>ǫ.
n n n
On the other hand, since (f ) ⊂ A, there exist some M > 0 such that kf k ≤ M
n n X′
for all n. Thus,
|f (x )|≤Mkx k
n n n
for each n, then by our hypothesis and Lemma 2.4, we have |f (x )| → 0 as n → ∞,
n n
which is impossible. This completes the proof. (cid:3)
Remark 1. Note by Proposition2.3 assertion(3) that every relatively weakly compact
subsetofatopologicaldualBanachspaceisL-Dunford-Pettis. Theconverseisnottruein
general. Infact,theclosedunitballBℓ∞ ofℓ∞isL-Dunford-Pettisset(seeTheorem2.5),
but it is not relatively weakly compact.
We make the following definition.
Definition 2.6. A Banach space X has the L-Dunford-Pettis property, if every L-
Dunford-Pettis set in X′ is relatively weakly compact.
As is known a DPcc operator is not weakly compact in general. For example, the
identity operator Idℓ1 :ℓ1 →ℓ1 is DPcc, but it is not weakly compact.
Inthe followingTheorem,wegiveacharacterizationsofL-Dunford-Pettispropertyof
Banach space in terms of DPcc and weakly compact operators.
Theorem 2.7. Let X be a Banach space, then the following assertions are equivalent:
(1) X has the L-Dunford-Pettis property,
(2) for each Banach space Y, every DPcc operator from X into Y is weakly compact,
(3) every DPcc operator from X into ℓ∞ is weakly compact.
390 A.RETBIANDB.ELWAHBI
Proof. (1) ⇒(2) Suppose that X has the L-Dunford-Pettis property and T : X → Y is
DPcc operator. Thus T′(BY′) is an L-Dunford-Pettis set in X′. So by hypothesis, it is
relatively weakly compact and T is a weakly compact operator.
(2)⇒(3) Obvious.
(3)⇒(1)IfX doesnothavetheL-Dunford-Pettisproperty,thereexistsanL-Dunford-
Pettis subset A of X′ that is not relatively weakly compact. So there is a sequence
(f ) ⊆ A with no weakly convergent subsequence. Now, we show that the operator
n
T : X → ℓ∞ defined by T(x) = (f (x)) for all x ∈ X is DPcc but it is not weakly
n
compact. As (f ) ⊆ A is L-Dunford-Pettis set, for every weakly null sequence (x ),
n m
which is a DP set in X we have
kT(x )k=sup |f (x )|→0, as m→∞,
m n n m
so T is a Dunford-Pettis completely continuous operator. We have T′((λ )∞ ) =
n n=1
P∞ λ f for every (λ )∞ ∈ ℓ1 ⊂ (ℓ∞)′. If e′ is the usual basis element in ℓ1
n=1 n n n n=1 n
then T′(e′ )=f , for all n∈N. Thus, T′ is not a weakly compact operator and neither
n n
is T. This finishes the proof. (cid:3)
Theorem 2.8. Let E be a Banach lattice.
If E has both properties of DP and Grothendieck, then it has the L-Dunford-Pettis pro-
perty.
Proof. Suppose that T :E →Y is DPcc operator. As E has the DP property, it follows
from Theorem 1.5 [11] that T is completely continuous.
On the other hand, ℓ1 is not a Grothendieck space and Grothendieck property is
carried by complemented subspaces. Hence the Grothendieck space E does not have
any complemented copy of ℓ1. By [10], E has the RDP property and so the completely
continuous operator T is weakly compact. From Theorem 2.7 we deduce that E has the
L-Dunford-Pettis property. (cid:3)
Remark 2. Since ℓ∞ has the Grothendieck and DP properties, it has the L-Dunford-
Pettis property.
LetusrecallthatK isaninfinitecompactHausdorffspaceifitisacompactHausdorff
space, which contains infinitely many points.
ForaninfinitecompactHausdorffspaceK,wehavethefollowingresultfortheBanach
space C(K) of all continuous functions on K with supremum norm.
Corollary 2.9. If C(K) contains no complemented copy of c , then it has L-Dunford-
0
Pettis property.
Proof. Since C(K) contains no complemented copy of c , it is a Grothendieck space [3].
0
On the other hand, C(K) be a Banach lattice with the DP property, and by Theorem
2.8 we deduce that C(K) has L-Dunford-Pettis property. (cid:3)
Corollary 2.10. A DPrc space has the L-Dunford-Pettis property if and only if it is
reflexive.
Proof. (⇒)IfaBanachspaceX hastheDPrcP,thenbyTheorem1.3of[11],theidentity
operator Id on X is DPcc. As X has the L-Dunford-Pettis property, it follows from
X
Theorem 2.7 that Id is weakly compact, and hence X is reflexive.
X
(⇐) Obvious. (cid:3)
Remark 3. Note that the Banach space ℓ1 is not reflexive and has the DPrcP, then
from Corollary 2.10, we conclude that ℓ1 does not have the L-Dunford-Pettis property.
Theorem 2.11. If a Banach space X has the L-Dunford-Pettis property, then every
complemented subspace of X has the L-Dunford-Pettis property.
L-DUNFORD-PETTIS PROPERTY IN BANACH SPACES 391
Proof. Consider a complemented subspace X of X and a projection map P : X →X .
1 1
Suppose T : X → ℓ∞ is DPcc operator, then TP : X →ℓ∞ is also DPcc. Since X has
1
L-Dunford-Pettis, by Theorem 2.7, TP is weakly compact. Hence T is weakly compact,
also from Theorem 2.7 we conclude that X has L-Dunford-Pettis, and this completes
1
the proof. (cid:3)
Let X be a Banach space. We denote by L(X,ℓ∞) the class of all bounded linear
operators from X into ℓ∞, by W(X,ℓ∞) the class of all weakly compact operators from
X into ℓ∞, and by DPcc(X,ℓ∞) the class of all Dunford-Pettis completely continuous
operators from X into ℓ∞.
RecallthatBahreiniin[2]investigatedthecomplementabilityofW(X,ℓ∞)inL(X,ℓ∞),
and she proved that if X is not a reflexive Banach space, then W(X,ℓ∞) is not com-
plemented in L(X,ℓ∞). In the next Theorem, we establish the complementability of
W(X,ℓ∞) in DPcc(X,ℓ∞).
We need the following lemma of [7] .
Lemma 2.12. Let X be a separable Banach space, and φ:ℓ∞ →L(X,ℓ∞) is a bounded
linear operator with φ(e )=0 for all n, where e is the usual basis element in c . Then
n n 0
there is an infinite subset M of N such that for each α ∈ ℓ∞(M), φ(α) = 0, where
ℓ∞(M) is the set of all α=(α )∈ℓ∞ with α =0 for each n∈/ M.
n n
Theorem 2.13. If X does not have the L-Dunford-Pettis property, then W(X,ℓ∞) is
not complemented in DPcc(X,ℓ∞).
Proof. ConsiderasubsetAofX′ thatis L-Dunford-Pettisbutitis notrelativelyweakly
compact. So there is a sequence (f ) in A such that has no weakly convergent subse-
n
quence. Hence S : X → ℓ∞ defined by S(x) = (f (x)) is an DPcc operator but it is
n
not weakly compact. Choose a bounded sequence (x ) in B such that (S(x )) has no
n X n
weakly convergent subsequence. Let X = hx i, the closed linear span of the sequence
1 n
(x ) in X. It follows that X is a separable subspace of X such that S/X is not a
n 1 1
weakly compact operator. If g = f /X , we have (g ) ⊆ X′ is bounded and has no
n n 1 n 1
weakly convergent subsequence.
Now define the operator T : ℓ∞ → DPcc(X,ℓ∞) by T(α)(x) = (α f (x)), where
n n
x∈X and α=(α )∈ℓ∞. Then
n
kT(α)(x)k=sup |α f (x)|≤kαk.kf k.kxk<∞.
n n n n
We claim that T(α)∈DPcc(X,ℓ∞) for each α=(α )∈ℓ∞.
n
Let α = (α ) ∈ ℓ∞ and let (x ) be a weakly null sequence, which is a DP set in X.
n m
As (f ) is L-Dunford-Pettis set sup |f (x )|→0 as m→∞. So we have
n n n m
kT(α)(x )k=sup |α f (x )|≤kαk.sup |f (x )|→0,
m n n n m n n m
as m → ∞. Then this finishes the proof that T is a well-defined operator from ℓ∞ into
DPcc(X,ℓ∞).
Let R:DPcc(X,ℓ∞)→DPcc(X ,ℓ∞) be the restriction map and define
1
φ:ℓ∞ →DPcc(X ,ℓ∞) by φ=RT.
1
Now suppose that W(X,ℓ∞) is complemented in DPcc(X,ℓ∞) and
P :DPcc(X,ℓ∞)→W(X,ℓ∞)
is a projection. Define ψ : ℓ∞ →W(X ,ℓ∞) by ψ = RPT. Note that as T(e ) is a one
1 n
rank operator, we have T(e )∈W(X,ℓ∞). Hence
n
ψ(e )=RPT(e )=RT(e )=φ(e )
n n n n
392 A.RETBIANDB.ELWAHBI
for all n∈N. From Lemma 2.12, there is an infinite set M ⊆N such that ψ(α)=φ(α)
for all α∈ℓ∞(M). Thus φ(χ ) is a weakly compactoperator. Onthe other hand, if e′
M n
is the usual basis element of ℓ1, for each x∈X and each n∈M, we have
1
(φ(χ ))′(e′ )(x)=f (x).
M n n
Therefore (φ(χ ))′(e′ ) = f /X = g for all n ∈ M. Thus (φ(χ ))′ is not a weakly
M n n 1 n M
compact operator and neither is φ(χ ). This contradiction ends the proof. (cid:3)
M
As a consequence of Theorem 2.7 and Theorem 2.13, we obtain the following result.
Corollary 2.14. LetX beaBanach space. Then thefollowing assertions areequivalent:
(1) X has the L-Dunford-Pettis property,
(2) W(X,ℓ∞)=DPcc(X,ℓ∞),
(3) W(X,ℓ∞) is complemented in DPcc(X,ℓ∞).
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Universit´e Ibn Tofail, Facult´e des Sciences, D´epartement de Math´ematiques, B.P. 133,
K´enitra,Morocco
E-mail address: [email protected]
Received 01/03/2016; Revised 15/04/2016