Note that a semi-compact operator is not necessary weakly compact

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OF SEMI-COMPACT OPERATORS

BELMESNAOUI AQZZOUZ and AZIZ ELBOUR Communicated by the former editorial board

We characterize Banach lattices on which each positive semi-compact operator becomes weakly compact.

AMS 2010 Subject Classification: Primary 46B42; Secondary 46B40, 46B42.

Key words: semi-compact operator, weakly compact operator, order continuous norm, KB-space, reflexive Banach lattice.

1. INTRODUCTION AND NOTATION

One of the famous result of Dodds and Fremlin said that if E and F are two Banach lattices such that the norms of the topological dualE⁰and ofF are order continuous, then each semi-compact and AM-compact operator T from E intoF is compact ([6], Theorem 125.5). Starting from this result of Dodds–

Fremlin, Banach lattices on which each positive semi-compact operator (not necessary AM-compact) is compact were investigated in [2]. In what follows, we generalize these results by studying the weak compactness of semi-compact operators.

Note that a semi-compact operator is not necessary weakly compact. In fact, the identity operator Id_l^∞ : l^∞ −→ l^∞ is semi-compact, but it is not weakly compact. And conversely, a weakly compact operator is not necessary semi-compact. In fact, the identity operatorId_l² :l² −→l² is weakly compact, but it is not semi-compact.

Let us recall, from [1], that an operator T from a Banach space E into a Banach lattice F is said to be semi-compact if for each ε > 0, there exists some u ∈ F⁺ such that T(BE) ⊂ [−u, u] +εBF where BH is the closed unit ball of H = E or F and F⁺ = {y ∈ F : 0 ≤ y}. Note that the class of semi-compact operators has two properties that are opposed to that of weakly compact operators. On the one hand, it satisfies the domination property, and on the other hand, it does not satisfy the duality property.

In this paper, we will prove that if E and F are two Banach lattices and X is a Banach space and if the norm ofF is order continuous, then each

MATH. REPORTS16(66),1(2014), 1–6

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semi-compact operator fromX intoF is L-weakly compact (in particular, it is weakly compact). And conversely, if each positive semi-compact operator from E into F is weakly compact, then the norm of E⁰ or F is order continuous.

Next, if we assume that F is Dedekind σ-complete (resp. the norm of E is order continuous), we shall show that each positive semi-compact operator T :E −→ F is weakly compact if and only if E is reflexive or the norm of F is order continuous.

To state our results, we will use the term operator T :E −→F between two Banach lattices to mean a bounded linear mapping. It is positive ifT(x)≥ 0 inF wheneverx≥0 inE. The operatorT is said to be regular ifT =T1−T₂ where T₁ and T₂ are positive operators from E into F. It is well known that each positive linear map on a Banach lattice is continuous.

For a terminology concerning Banach lattice theory and positive operators, we refer the reader to the excellent book of Aliprantis–Burkinshaw [1].

2. MAIN RESULTS

Recall from [1] that an operator T from a Banach spaceE into a Banach lattice F is said to be L-weakly compact if for each disjoint sequence (y_n), in the solid hull of T(B_E), we have lim_nky_nk = 0. Note that by Theorem 5.71 of [1] each L-weakly compact operator is semi-compact. But a semi-compact operator is not necessary L-weakly compact. In fact, the identity operator Id_l^∞ :l^∞−→l^∞ is semi-compact, but it is not L-weakly compact.

In ([2], Theorem 2.2), we established that if E and F are two Banach lattices such that the norm of F is order continuous, then each positive semi- compact operator fromE intoF is weakly compact.

The following result generalizes Theorem 2.2 of [2] and gives also the converse, in addition its demonstration is short.

Theorem2.1. LetEandF be two Banach lattices and letXbe a Banach space.

(1) If the norm of F is order continuous, then each semi-compact operator from X into F is L-weakly compact (and hence, is weakly compact).

(2) Conversely, if each positive semi-compact operator from E into F is weakly compact, then one of the following conditions holds:

i) The norm ofE⁰ is order continuous.

ii) The norm ofF is order continuous.

Proof. (1) Let T be a semi-compact operator from X into F. Since the norm ofF is order continuous, it follows from Proposition 3.6.2 of [4] thatT is L-weakly compact. Hence, by Proposition 3.6.12 of [4], T is weakly compact.

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(2) Assume by way of contradiction that neither E⁰ nor F has an order continuous norm. By Proposition 3.5.12 of [4] there exist a positive rank 1 operator T :E −→ F and a non-weakly compact operator S :E −→ F such that 0 ≤ S ≤ T. Clearly T is semi-compact. So, by Theorem 5.72 of [1], S is semi-compact but it is not weakly compact. This is a contradiction and completes the proof.

Recall from [1] that a Banach lattice E is said to be lattice embeddable into another Banach lattice F whenever there exists a lattice homomorphism T :E→F and there exist two positive constants K and M satisfying

Kkxk ≤ kT(x)k ≤Mkxk for allx∈E.

In this case T is called a lattice embedding fromE intoF, andT(E) is a closed sublattice ofF which can be identified withE.

A Banach lattice E is said to be a KB-space whenever every increasing norm bounded sequence of E⁺ is norm convergent. Note that each KB-space has an order continuous norm, but a Banach lattice with an order continuous norm is not necessary a KB-space. In fact, the Banach latticec₀ has an order continuous norm but it is not a KB-space. However, if E is a Banach lattice, the topological dualE⁰is a KB-space if and only if its norm is order continuous.

On the other hand, recall that Dodds [3] introduced a class of operators that he called order weakly compact operators. An operatorT from a Banach latticeE into a Banach space F is said to beorder weakly compact if for each x∈E⁺, the subsetT[0, x] is a relatively weakly compact subset ofF.It is clear that each weakly compact operator from E into F is order weakly compact, but the converse is false in general. In fact, the identity operatorId_c₀ :c₀ →c₀ is order weakly compact but it is not weakly compact.

Theorem2.2. LetEandF be two Banach lattices such thatF is Dedekind σ-complete. Then the following assertions are equivalent:

(1) Each semi-compact operator T :E−→F is weakly compact.

(2) Each positive semi-compact operator T :E−→F is weakly compact.

(3) One of the following conditions holds:

i) E is reflexive.

ii) the norm of F is order continuous.

Proof. (1) =⇒(2) Obvious.

(2) =⇒ 3) It suffices to establish that if the norm of F is not order continuous, then E is reflexive. Indeed, suppose that the norm of F is not order continuous. By Theorem 2.4.15 of [4] we need to show that E and E⁰ are KB-spaces.

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Step 1: E⁰ is a KB-space. In fact, since the norm of F is not order continuous then Theorem 2.1 (2) implies that the norm ofE⁰ is order continuous, i.e. E⁰ is a KB-space.

Step 2: E is a KB-space. Otherwise, it follows from Theorem 2.4.12 of [4] that E contains a closed sublattice order isomorphic to c₀. Let (x_n)⊂E⁺ correspond to the natural basis (e_n) of c₀. We may assume that kx_nk= 1 for all n. Since (en) is disjoint and order bounded in (c0)⁰⁰ =l^∞ then (xn) is also disjoint and order bounded in E⁰⁰. It follows from Theorem 116.3 (iii) of [6]

that there exists a disjoint sequence (φ_n) of positive elements in the unit ball of E⁰ such that

(∗) 1

2 ≤φn(xn)≤1 for allnand φn(xm) = 0 f orm6=n.

Now, define an operator T₁ :E −→ l^∞ by T₁(x) = (φ_n(x))^∞_n=1. Clearly T1 is well defined and is positive.

On the other hand, since the norm of F is not order continuous then by Theorem 4.51 of [1], the Banach lattice l^∞ is lattice embeddable in F. Let T₂:l^∞−→F be a lattice embedding. Then there exist two positive constants K andM satisfying

Kkxk_∞≤ kT₂(x)k ≤Mkxk_∞ for allx∈l^∞.

Since the closed unit ball ofl^∞is an order interval, the positive operator T₂ is semi-compact (compare Exercise 15 of Section 5.3 of [1]). Hence, the composed operator T =T2◦T1 is semi-compact. To finish the proof we have just to show that T is not weakly compact. Assume by way of contradiction that T is weakly compact. Then the second adjoint T⁰⁰ : E⁰⁰ −→ F⁰⁰ is also weakly compact and hence, T⁰⁰ is order weakly compact. As (xn) is an order bounded disjoint sequence inE⁰⁰, it follows from Theorem 5.57 (2) of [1] that

limn kT(xn)k= lim

n

T⁰⁰(xn)

= 0.

But (∗) implies T1(xn) =φn(xn)en for each n, and hence, kT(xn)k=φn(xn)kT₂(en)k ≥ K

2 ke_nk_∞= K 2 >0

for eachn, which contradicts with limnkT(xn)k= 0. So,E is a KB-space and we are done.

3,i) =⇒1) In this case, each operator T :E −→F is weakly compact.

3,ii) =⇒1) Follows from Theorem 2.1 (1).

In particular, whenever E =F in Theorem 2.2, we obtain the following characterization of the order continuity of the norm of a Banach lattice:

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Corollary 2.3. Let E be a Dedekind σ-complete Banach lattice. Then the following assertions are equivalent:

(1) Each semi-compact operator T :E−→E is weakly compact.

(2) Each positive semi-compact operator T :E−→E is weakly compact.

(3) The norm ofE is order continuous.

As a consequence of Theorem 2.2 we obtain the following result:

Corollary 2.4. Let E be a Banach lattice. Then the following state- ments are equivalent:

(1) Each operator T :E −→l^∞ is weakly compact.

(2) Each positive operator T :E −→l^∞ is weakly compact.

(3) E is reflexive.

Now, if we replace in Theorem 2.2 the hypothesis “F is Dedekind σ- complete” by “the order continuity of the norm in E”, we obtain the same necessary and sufficient conditions:

Theorem 2.5. Let E and F be two Banach lattices such that the norm of E is order continuous. Then the following assertions are equivalent:

(1) Each semi-compact operator T :E−→F is weakly compact.

(2) Each positive semi-compact operator T :E−→F is weakly compact.

(3) One of the following conditions holds:

i) E is reflexive.

ii) the norm of F is order continuous.

Proof. (1) =⇒(2) Obvious.

(2) =⇒ (3) It suffices to establish that if the norm of F is not order continuous, then E is reflexive. Indeed, suppose that the norm of F is not order continuous. It suffices to show that E and E⁰ are KB-spaces ([4], The- orem 2.4.15). Clearly E⁰ is a KB-space (Theorem 2.1 (2)). Now assume by way of contradiction thatE is not a KB-space. By Theorem 2.4.12 of [4], the Banach lattice E contains a closed sublattice U order isomorphic to c0 and we denote this isomorphism by i : U −→ c0. Since the norm of E is order continuous then Corollary 2.4.3 xi) of [4] implies that there exists a positive projection P :E −→U.

On the other hand, since the norm of F is not order continuous then by Theorem 117.1 of [6] we may assume that c₀ is a closed sublattice of F such that the closed unit ball of c₀ is order bounded in F. Then the composed operator T = i◦P : E −→ c0 ⊂ F is a positive semi-compact but it is not weakly compact (the composed operator restricted to U is an isomorphism).

This is a contradiction with (2). Therefore, E is a KB-space and we are done.

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3,i) =⇒1) In this case, each operator T :E −→F is weakly compact.

3,ii) =⇒1) Follows from Theorem 2.1 (1).

Remark. The assumption “F is Dedekind σ-complete” (resp. “the norm of E is order continuous”) is essential for Theorem 2.2 (resp. Theorem 2.5).

For instance, if we take E = l^∞ and F = c. It is clear that each operator T :l^∞ −→ c is semi-compact (because c is an AM-space with unit). On the other hand, each operator T : l^∞ −→ c is weakly compact (see the proof of Proposition 1 of Wnuk [5]). Then the class of semi-compact operators and the class of weakly compact operators from l^∞ intoc coincide. But the condition (3) of Theorem 2.2 (resp. Theorem 2.5) is not satisfied.

REFERENCES

[1] C.D. Aliprantis and O. Burkinshaw, Positive operators. Reprint of the 1985 original.

Springer, Dordrecht, 2006.

[2] B. Aqzzouz, R. Nouira and L. Zraoula,Compactness of positive semi-compact operators on Banach lattices. Rend. Circ. Mat. Palermo55(2006),3, 305–313.

[3] P.G. Dodds, o-weakly compact mappings of vector lattices. Trans. Amer. Math. Soc.

214(1975), 389–402.

[4] P. Meyer-Nieberg,Banach lattices. Universitext. Springer-Verlag, Berlin, 1991.

[5] W. Wnuk, Remarks on J.R. Holub’s paper concerning Dunford-Pettis operators. Math.

Japon.38(1993),6,1077–1080.

[6] A.C. Zaanen,Riesz spaces II, North Holland Publishing Company 1983.

Received 27 June 2011 Universit´e Mohammed V-Souissi, Facult´e des Sciences Economiques,

Juridiques et Sociales, D´epartement d’Economie, B.P. 5295, SalaEljadida, Morocco

baqzzouz@hotmail.com Universit´e Moulay Isma¨ıl,

Facult´e des Sciences et Techniques,

D´epartement de Math´ematiques, B.P. 509, Errachidia, Morocco

azizelbour@hotmail.com