1. IntroductionIn 1992, Saab and Smith [1] studied Banach spaces with the property that every un- conditionally converging operator from such a space to an arbitrary Banach space is weakly completely continuous. It is well known that every completely continuous operator is an unconditionally converging operator. In this paper, the converse of this fact by using some localized properties, e.g., -sets, -sets will be studied.

First, the property for Banach spaces is presented, then spaces which have property are considered. Next, it is demonstrated that has property if and only if every unconditionally converging operator is completely continuous, for every Banach space.

In [2] , it has shown that a Banach space has the Reciprocal Dunford-Pettis property whenever it has property but the converse is false. We investigate under what conditions this implication could be reversed.

Also, it will verified that property implies the Dunford-Pettis property. In addition, we show that a Banach space with property and the Dunford-Pettis property, as well has the property.

The notions of a relatively compact sets, -sets, -sets have a significant role in this study.

Definitions and NotationIn this article, real Banach spaces will be symbolized by and, the unit ball of will be denoted by, and the continuous linear dual of will be denoted by. A continuous and linear map from to is called an operator, and the adjoint of is denoted by. Also will stand for the set of all operators .

If is an operator from to and is an unconditionally con- vergent series in whenever the series in is weakly unconditionally converging, then is called an unconditionally converging operator. A completely continuous operator is an operator which takes weakly null sequences in into norm convergent sequences in.

Recall that a series is weakly unconditionally converging if and only if for each and the series is unconditionally converging if and only if every rearrangment converges in the norm topology of.

Suppose is a bounded set in, then it is stated that is weakly precompact if every single sequence from includes a weakly Cauchy subsequence. Rosentlal’s Theorem states that “A Banach space does not contain copies of if and only if is a weakly precompact set in.”

The reader is referred to Diestel [3] or Dunford-Schwartz [4] for undefined notation and terminology.

2. Main ResultsLet be a bounded set in, then is said to be an -set, if for all weakly null sequence in we have the following ,

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e.g., see [5] [6] [7] . Moreover, any -subset of is relatively compact if and only if does not embed in [5] [6] .

Closely related to the notion of -set is the idea of a -set. Let be a bounded set in, then is called a -set, if for all weakly unconditionally converging series in, we have the following

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It can easily be derived directly from the definitions that every -subset of is a -subset of. Next, equivalent characterizations of Banach spaces such that the converse statement holds are defined.

Initially, new property for Banach spaces will be introduced.

Definition 2.1. Let be a Banach space. Then we say that has property if each -set in is also an -set in.

Theorem 3.1 (ii) in [2] , plays a consistent and vital position in this research. It states that; A necessary and sufficient condition for an operator to be completely continuous is that is an -subset of.

In the following, necessery and sufficient condition are given that every -subset of is an -subset of.

Now, the main theorem, which establishes Banach spaces with property will be determined as the resulting.

Theorem 2.2. The following statements are equivalent about a Banach space.

(i) has property.

(ii) Let be any Banach space, then an operator from to is completely continuous whenever is unconditionally converging.

(iii) If an operator from to is unconditionally converging, then it is also completely continuous.

Proof. Assume that is a Banach space and operator is unconditionally converging. Now let the series in be weakly unconditionally converging. Thus is an unconditionally converging in and. Hence for each weakly unconditionally converging series in X, , and thus we have the following,

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This follows that is a -set in, and as has property, then it is also an -set in. Therefore the operator is completely continuous [2] .

It is obvious.

Assume that is a -set in and let be some sequence in A. Let us define by and suppose that the series is weakly unconditionally converging in X. See that, and. As is a -set in, we have the following,

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So, which implies that the operator is unconditionally converging. Finally, by assumption we conclude that is a completely continuous operator. Now assume is a weakly null sequence in and. Then. Hence is an -set in which follows that is also an -set in. Therefore has property and the proof is complete.

Remembrance that if for every Banach space, each operator is completely continuous whenever is weakly compact operator, then is said to have the Dunford-Pettis property. Equivalently, has Dunford-Pettis property if and only if whenever is weakly null in X* and is weakly Cauchy in [8] . A Banach space has property if each operator is weakly compact for every Banach space whenever is unconditionally con- verging [9] .

Examples of Banach spaces which satisfy both property and the Dunford-Pettis property are and.

Theorem 2.2 has some corollaries which they are proved at this time. The first part is connected to Dunford-Pettis property.

Corollary 2.3. Let be a Banach space. Then the following is given:

(i) has the Dunford-Pettis property provided that has property.

(ii) has the Dunford-Pettis property if has property.

(iii) has property whenever has property and the Dunford- Pettis property.

Proof. (i) First we assume that the operator is weakly compact. Then it is easily seen that is an unconditionally converging operator. Now Theorem 2.2 implies that the operator is completely continuous, as has property. Thus the proof is complete and has the Dunford-Pettis property.

(ii) Note that (i) concludes that has the Dunford-Pettis property, as it has property. Hence has the Dunford-Pettis property [8] and we get the result.

(iii) Let the operator be unconditionally converging. Now as has property and the Dunford-Pettis property, then is weakly compact which implies that it is also completely continuous. Finally Theorem 2.2 gives the result, that is, has property.

Note that has both the Dunford-Pettis property and property which implies that it also has property. The converse of Corollary 2.3 (iii) does not hold. In general, it is not true that if has property, then it also has property. For example, does not have property although it has property (otherwise its dual would be weakly sequentially complete [9] which contradicts that contains a copy of).

A Banach space is said to have the Reciprocal Dunford-Pettis property if for every Banach space, every completely continuous operator is weakly compact. In [2] , Bator, Lewis, and Ochoa, showed that property implies the Reciprocal Dunford-Pettis property, but generally, the converse of this fact is not true. To see, this inference could be reversed, we begin with a space containing property.

Corollary 2.4. If has property and the Reciprocal Dunford-Pettis property, then has property.

Proof. Suppose is an unconditionally converging operator. Since has property, then by Theorem 2.2, is completely continuous. As has the Reciprocal Dunford-Pettis property, then is weakly compact. Hence has property.

Soyabs, in [10] , introduced a property called the property. A Banach space has the property if every linear operator is weakly compact for every Banach space whose dual does not contain an isomorphic copy of. If Banach space has property, then it has the property. Also, E. Saab and P. Saab [11] have introduced the property. A Banach space has the property if every operator is weakly compact. If Banach space has property, then it has the property [10] , therefore, we immediatly have the following result.

Corollary 2.5. Let be a Banach space with the Reciprocal Dunford-Pettis property and the property (MB). Then we have the following:

(i) Every operator is weakly compact for every Banach space whose dual does not contain an isomorphic copy of.

(ii) Every operator is weakly compact.

Remebering that, Banach spaces which does not contain a copy of, have the Reciprocal Dunford-Pettis property. Certainly, Odell’s result ( [12] , p. 377) implies that, for any Banach space, each operator from to is compact provided it is completely continuous if and only if does not embed in. Also, Rosenthal and Dor's Theorem [3] , if is a Banach space and is a sequence in such that fails to have a weak Cauchy subsequence, then has a subsequence which is equivalent to the unit vector of, is required for the following consequence. Note that Banach space has the property but does not contain, see [13] (Corollary 2.3 (i)).

Theorem 2.6. Suppose that has property. Then we have the following:

(i) If does not contain, then every unconditionally converging operator is compact for every Banach space.

(ii) If does not contain, then every operator is completely continuous.

Proof. (i) Suppose is an unconditionally converging operator. By Theo- rem 2.2, is completely continuous, since has property. As does not contain, is a compact operator ( [12] , p. 377).

(ii) Suppose by contradiction that is not completely continuous operator. Let be a weakly null sequence in and for some. Choose so that. Since, by Rosenthal-Dor Theo- rem, we may assume that is weakly Cauchy. By corollary 2.3, has the Dunford-Pettis property, hence, that is a contradiction.

Theorem 2.6 has an immediate consequence; thus the proof is omitted.

Corollary 2.7. Suppose that has property, then every operator is completely continuous for every separable dual space.