-Smooth Points in Some Banach Spaces

Abstract

We characterize the -smooth points in some Banach spaces. We will deal with injective tensor product, the Bochner space of (equivalence classes of) -essentially bounded measurable -valued functions, and direct sums of Banach spaces.

1. Introduction

For a unit vector in a Banach space , consider the state space . The point is a smooth point if consists exactly of one point. The set of all smooth points is denoted by . Smooth points are important tools in the study of the geometry of Banach spaces. For two Banach spaces Heinrich, [1], gave a description of smooth points of the unit ball in the space of compact operators from into . The research then turned to the space of bounded operators. Kittaneh and Younis, [2], were the first to deal with this problem. They characterized smooth points in . Their result was then generalized in [3] to the space ; . For smooth points in see [4, 5]. In [6] Werner gave a description of smooth points in under some conditions on and . Smooth points in certain vector valued function spaces were given in [7].

In [8] the authors generalize the notion of smoothness by calling a unit vector in a Banach space a , or a multismooth point of order if has exactly linearly independent vectors, equivalently, if . For a natural number , the set of -smooth points in is denoted by . Note that is a -compact convex set and hence it is easy to see that if and only if . Multismoothness in Banach spaces was extensively studied by Lin and Rao in [9]. In paricular, they showed that, in a Banach space of finite dimension , any -smooth point is unitary and hence a strongly extreme point. The aim of this paper is to characterize multismoothness in some Banach spaces. Indeed, we will deal with injective tensor product, the Bochner space of (equivalence classes of) essentially bounded measurable valued functions, and direct sums of Banach spaces.

The set of all extreme points of the unit ball of a Banach space is denoted by .

2. Multismoothness in Injective Tensor Products

In [9] the authors characterized multismoothness in the completed injective tensor product when is an predual space and is a smooth Banach space. We generalize their result to any Banach space . Note that whenever either of and has the approximation property. Here is the space of all compact and to weakly continuous operators from to , endowed with usual operator norm.

Recall that if and then as follows:

Theorem 1. Let be an predual space and any Banach space. Let with . Then is a multismooth point of finite order in if and only if attains its norm at exactly finitely many independent vectors, say at such that is a multismooth point of finite order in ; .
In this case the order of smoothness of is , where is the order of smoothness of , .

Proof. One can easily prove that if attains its norm at infinitely many independent vectors in then is not a multismooth point of any finite order. The same conclusion will be obtained if attains its norm at some and is not a multismooth point of finite order in . So, suppose attains its norm at exactly finitely many independent vectors, say at such that is a multismooth point of finite order in ; and let . We will prove that is a multismooth point of order in .
For each there are exactly linearly independent functionals in attaining their norm at , say . Since is an predual space, then there are distinct atoms with . Set , where and . These are extreme functionals in attaining their norms at : We claim that the ’s are linearly independent. Indeed, if for some scalars then . But since the correspond to distinct atoms then for all . Since are linearly independent, then , .
Finally, Let with . We will show that . We can suppose that (see Section 1). Now, by a result of Ruess and Stegall [10], , where and . Then So, and hence . But since the correspond to distinct atoms then for some and consequently . Therefore . Hence . This proves that the ’s form a maximal linearly independent set in .
Therefore .

As a corollary, we get the following.

Corollary 2 (see [9]). Let be an predual space and a smooth Banach space. Let with . Then is a multismooth point of finite order in if and only if   attains its norm at exactly independent vectors in .

Open Problem. For Banach spaces and let with . Is it true that is a multismooth point of finite order in if and only if attains its norm at only finitely many independent vectors, say at such that each is a multismooth point of finite order, say , in , where

Theorem 1 above tells us that the answer is yes when is an predual space, since in this case .

3. Multismoothness in Bochner Spaces

Let be a Banach space. In this section we discuss multismoothness in the Bochner space of (equivalence classes of) -essentially bounded measurable -valued functions. Recall that measurable functions are constants on the atoms.

Lemma 3. Let and suppose that there is no atom such that . Then is not a multismooth point of any finite order.

Proof. Fix . We will prove that is not a multismooth point of order . Write , where ’s are disjoint measurable sets of positive measure, with . For , define Then and . Moreover, for all . This shows that is not a smooth point of any order ; see [8]. In particular, is not a multismooth point of order .

Lemma 4. Let and suppose that there are exactly atoms such that . If , then is not a multismooth point of any finite order.

Proof. Fix and write as a disjoint union of measurable sets of positive measure and proceed as in the above proof.

Theorem 5. Let with . Then is a multismooth point of finite order if and only if there are exactly finitely many distinct atoms such that and and each is a multismooth point of finite order, say , in . In this case the order of smoothness of is .

Proof. The above two lemmas prove the “only if” part. For the converse, we choose, for any , linearly independent set . So, . For and define by . These are linear functionals attaining their norm at . Suppose that for some scalars . Then , . Choosing ; if and ; if , where , we get for all and . Consequently, for all . Since is linearly independent, then for all and . Therefore, the ’s are linearly independent.
For an atom and let . We will prove that if with then for all . Without loss of generality, say . We claim that there is such that . Indeed, if such does not exist, we would have a sequence outside such that . But then Hence, for all , a contradiction to our assumption. Thus, and therefore for all such that , . This proves that for all . Let sum) and let . Then by Krein-Millman Theorem we have with . So, any has the form for some . Note that if, for example, has the form , then . Since , then any ’s of the above form must be linearly dependent. Consequently, any elements must be linearly dependent.
Now, let such that . We will prove that the ’s are linearly dependent. By the argument above we see that for all . For define by , where . Then with so that for all and therefore is linearly dependent. Hence, there are scalars , not all zeros, with . The proof will be complete if we show that . Indeed, if then . This shows that and completes the proof of the “if” part.

4. Multismoothness in Direct Sums of Banach Spaces

Lin and Rao characterized in [9] multisoothness in -direct sums and proved the following theorem.

Theorem 6 (see [9]). Let be an infinite family of nonzero Banach spaces. Let and let be a unit vector in . Let , let , and let . Then, is a multismooth point of finite order if and only if is finite, and . In this case the order of smoothness of is , where is the order of smoothness of in .

In this section we deal with -direct sums. Indeed we prove the following result.

Theorem 7. Let be any family of nonzero Banach spaces. Let and let be a unit vector in . Let , , , , and . Then is a multismooth point of finite order if and only if and is finite. In this case the order of smoothness of is , where is the order of smoothness of in and is the number of elements in .

For the sake of completeness, let us first state and prove the characterization of smoothness.

Theorem 8. Let be any family of nonzero Banach spaces. Let and let be a unit vector in . Then is a smooth point if and only if for any ,  .

Proof. Suppose and , or but is not a smooth point in . Then there are distinct unit functionals with . For , choose any unit functional with . Let and where and . Clearly, and are distinct elements in . So is not a smooth point.
Conversely, suppose is the unique element in . Let . Then with and . Now if then . Since then for all and hence .

Lemma 9. Let and be nonzero Banach spaces and . Let where . If then .

Proof. Let be linearly independent and choose linearly independent unit functionals . Let Clearly ; . They are linearly independent. Indeed if then and . Since is lineraly independent then ; and thus . But is also lineraly independent. So ; . Hence is lineraly independent.
Now suppose ; , where and with . Thus . Since then we see that must be linearly dependent. Therfore .

Lemma 10. Let and be nonzero Banach spaces and . Let where and . Then .

Proof. Say . Let and let be linearly independent sets. Let Clearly ; . They are linearly independent. Indeed if then and . Since and are linearly independent sets then ; and . This makes for all . Hence is linearly independent.
Now to prove that it is clearly enough to show that any must be in . So suppose , where and . Then So and and hence there are scalars ; and ; such that and . Thus . Similarly .
let with for all . Then ; and ; . Thus = since . This proves that . Therefore .

We now can easily prove Theorem 7.

Proof of Theorem 7. First note that if , , is infinite, or is infinite, then one can easily construct an infinite linearly independent subset of . On the other hand, if and is finite, then writing as , where , , and applying Theorem 8 and Lemmas 9 and 10 we get the result.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.