- About this Journal ·
- Abstracting and Indexing ·
- Aims and Scope ·
- Annual Issues ·
- Article Processing Charges ·
- Articles in Press ·
- Author Guidelines ·
- Bibliographic Information ·
- Citations to this Journal ·
- Contact Information ·
- Editorial Board ·
- Editorial Workflow ·
- Free eTOC Alerts ·
- Publication Ethics ·
- Reviewers Acknowledgment ·
- Submit a Manuscript ·
- Subscription Information ·
- Table of Contents

Journal of Function Spaces and Applications

Volume 2013 (2013), Article ID 187536, 19 pages

http://dx.doi.org/10.1155/2013/187536

## Köthe-Bochner Spaces and Some Geometric Properties Related to Rotundity and Smoothness

Department of Mathematics, Freie Universität Berlin, Arnimallee 6, 14195 Berlin, Germany

Received 24 May 2013; Accepted 27 June 2013

Academic Editor: T. S. S. R. K. Rao

Copyright © 2013 Jan-David Hardtke. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

In 2000, Kadets et al. introduced the notions of acs, luacs, and uacs spaces, which form common generalisations of well-known rotundity and smoothness properties of Banach spaces. In a recent paper, the author introduced some further related notions and investigated the behaviour of these geometric properties under the formation of absolute sums. The present paper is in a sense a continuation of this work. Here we will study the behaviour of the said properties under the formation of Köthe-Bochner spaces, thereby generalising some results of G. Sirotkin on the acs, luacs, and uacs properties of -Bochner spaces.

#### 1. Introduction

We begin with some notation and definitions. Throughout this paper, denotes a real Banach space, its dual, its unit ball and its unit sphere.

In the next definition, we summarise the most important rotundity properties.

*Definition 1. *A Banach space is called(i)*rotund* () if for any two elements the equality implies , (ii)*locally uniformly rotund* (LUR) if for every , the implication
holds for every sequence in , (iii)*weakly locally uniformly rotund* (WLUR) if for every and every sequence in one has
(iv)*uniformly rotund* (UR) if for any two sequences and in the implication
holds, (v)*weakly uniformly rotund* (WUR) if for any two sequences and the following implication holds:

Figure 1 shows the obvious implications between these notions. No other implications are valid in general (see the examples in [1]). Note, however, that all these notions coincide in finite-dimensional spaces, by the compactness of .

The modulus of convexity of the space is defined by for every in the interval . Then is UR if and only if for all .

For the local version one defines for every and each . Then is LUR if and only if for all and all .

Let us also recall some notions of smoothness. The space is called *smooth* () if its norm is Gâteaux-differentiable at every nonzero point (equivalently at every point of ), which is the case if and only if for every there is a unique functional with (cf. [2, Lemma 8.4 (ii)]). is called *Fréchet-smooth* (FS) if the norm is Fréchet-differentiable at every nonzero point. The norm of the space is said to be *uniformly Gâteaux-differentiable* (UG) if for each the limit exists uniformly in . Finally, is called *uniformly smooth* (US) if , where denotes the modulus of smoothness of defined by for every .

In [3] the following notions were introduced (in connection with the so called Anti-Daugavet property).

*Definition 2. *A Banach space is called (i)*alternatively convex or smooth* (acs) if for every with and every with one has as well, (ii)*locally uniformly alternatively convex or smooth* (luacs) if for every , every sequence in and every functional one has
(iii)*uniformly alternatively convex or smooth* (uacs) if for all sequences , in and in , one has
The author introduced the following related notions in [4].

*Definition 3. *A Banach space is called(i)*strongly locally uniformly alternatively convex or smooth* (sluacs) if for every and all sequences in and in , one has
(ii)*weakly uniformly alternatively convex or smooth* (wuacs) if for any two sequences , in and every functional , one has

The obvious implications between the acs properties and the rotundity properties are indicated in Figure 2. No other implications are generally valid (see the examples in [4]), but note again that the properties acs, luacs, sluacs, wuacs and uacs coincide in finite-dimensional spaces, by compactness.

The connection between some of the acs properties to smoothness properties is illustrated in Figure 3.

Let us mention that if we replace the condition by for every in the definitions of the properties uacs and sluacs, respectively we still obtain the same classes of spaces. For uacs spaces this was first proved by Sirotkin in [5] using the fact that uacs spaces are reflexive (see below). For sluacs spaces this characterisation can be proved by means of the Bishop-Phelps-Bollobás theorem (see [4, Proposition 2.1]).

This characterisation enables us to define the following “uacs-modulus” of a given Banach space (cf. [4, Definition 1.4]).

*Definition 4. *For a Banach space one defines
Then is uacs if and only if for every and one clearly has for each .

The above characterisation shows that the class of uacs spaces coincides with the class of -spaces introduced by Lau in [6] and our modulus is the same as the modulus of -convexity from [7]. Also, the notion of -spaces which was introduced in [8] coincides with the notion of acs spaces.

Recall that a Banach space is said to be uniformly nonsquare if there is some such that for all we have or . It is easily seen that uacs spaces are uniformly nonsquare and hence by a well-known theorem of James (cf. [9, page 261]) they are superreflexive, as was observed in [3, Lemma 4.4]. For a proof of the superreflexivity of uacs spaces that does not rely on James’ result on uniformly nonsquare spaces, see [4, Proposition 2.8].

Let us also restate here the following auxiliary result [4, Lemma 2.30] (it is the generalisation of [10, Lemma 2.1] to sequences, with a completely analogous proof).

Lemma 5. *Let and be sequences in the (real or complex) normed space such that .**Then for any two bounded sequences , of nonnegative real numbers one also has . *

Finally, we will need two more definitions from [4].

*Definition 6. *A Banach space is called (i)a space if for every , every sequence in with , and all one has
(ii)a space if for every , every sequence in with , and all sequences in , one has
Obviously, every WLUR space is , and every LUR space is .

In the next section we will recall some facts on Köthe-Bochner spaces.

#### 2. Preliminaries on Köthe-Bochner Spaces

If not otherwise stated, will denote a complete, -finite measure space. For we denote by the characteristic function of .

A Köthe function space over is a Banach space of real-valued measurable (i.e., -Borel-measurable) functions on modulo equality -almost everywhere (we will henceforth abbreviate this by -a.e. or simply a.e. if is tacitly understood) such that (i) for every with , (ii)for every and every set with is -integrable over , (iii)if is measurable and such that -a.e., then and .

The standard examples are of course the spaces for .

Every Köthe function space is a Banach lattice when endowed with the natural order if and only if -a.e.

Recall that a Banach lattice is said to be order complete (-order complete) if for every net (sequence) in which is order bounded, the supremum of said net (sequence) in exists. A Banach lattice is called order continuous (-order continuous) provided that every decreasing net (sequence) in whose infimum is zero is norm-convergent to zero.

It is easy to see that a Köthe function space is always -order complete, and thus by [11, Proposition ] is order continuous if and only if is -order continuous if and only if is order complete and order continuous. Also, reflexivity of implies order continuity, for any -order complete Banach lattice which is not -order continuous contains an isomorphic copy of (cf. [11, Proposition 3.1.4]).

Let us also mention the following well-known fact that will be needed later.

Lemma 7. *If is a Köthe function space, a sequence in , and such that , then there is a subsequence of which converges pointwise almost everywhere to . *

For a Köthe function space we denote by the space of all measurable functions (modulo equality -a.e.) such that Then is again a Köthe function space, the so called Köthe dual of . The operator defined by is well-defined, linear and isometric. Moreover, is surjective if and only if is order continuous (cf. [11, page 149]), thus for order continuous we have .

We refer the reader to [12] or [11] for more information on Banach lattices in general and Köthe function spaces in particular.

Now recall that if is a Banach space, a function is called simple if there are finitely many disjoint measurable sets such that for all , is constant on each and for every . The function is said to be Bochner-measurable if there exists a sequence of simple functions such that -a.e. and weakly measurable if is measurable for every functional . According to Pettis’ measurability theorem (cf. [11, Theorem 3.2.2]) is Bochner-measurable if and only if is weakly measurable and almost everywhere separably valued (i.e., there is a separable subspace such that -a.e.).

For a Köthe function space and a Banach space we denote by the space of all Bochner-measurable functions (modulo equality a.e.) such that . Endowed with the norm becomes a Banach space, the so called Köthe Bochner space induced by and . The most prominent examples are again the Lebesgue-Bochner spaces for .

Next we recall how the dual of can be described provided that is order continuous. A function is called weak*-measurable if is measurable for every . We define an equivalence relation on the set of all weak*-measurable functions by setting if and only if for every a.e., and we write for the space of all (equivalence classes of) weak*-measurable functions such that there is some with a.e.

A norm on can be defined by Then the following deep theorem holds.

Theorem 8 (cf. [13]). *Let be an order-continuous Köthe function space over the complete, -finite measure space , and let be a Banach space. Then the map defined by
**
is an isometric isomorphism, and, moreover, every equivalence class in has a representative such that and . *

There are a number of papers on various rotundity and smoothness properties in Köthe-Bochner spaces in general and Lebesgues-Bochner spaces in particular, see for example [14–17] and references therein.

Sirotkin proved in [5] that for the Lebesgue-Bochner space is acs, respectively luacs, and uacs, whenever has the respective property. In the next section we will study the more general case of Köthe-Bochner spaces.

#### 3. Results and Proofs

We begin with the acs spaces, for which we have the following result.

Proposition 9. *If is an order-continuous acs Köthe function space and is an acs Banach space, then is acs as well. *

*Proof. *The proof is similar to that of [4, Proposition 3.3]. First we fix two elements such that and a functional with .

Since is order continuous, by Theorem 8, can be represented via an element such that and . It follows that
and hence
We also have
and thus
Since is acs it follows from (19) and (22) that
In a similar way as we have obtained (22) we can also show
Because is acs this together with (19), (22) and (23) implies
From (19), (23) and (25) we get
Now we will show that
To this end, let us denote by and the null sets on which the equality from (20) and (26), respectively does not hold. Let .

Put and . We claim that is a null set.

To see this, define by for and for . Then is measurable and since for all we have with . We also have for every and hence by (19)
which also implies . Together with (22) we now get
since is acs. Taking into account (23) we arrive at
Hence a.e. and thus must be a null set.

Now if then , and and as well as
By [10, Lemma 2.1] this implies
Since is acs it follows that .

So is a null set with for every and (27) is proved.

Now combining (23) and (27) we obtain
which finishes the proof.

Before we turn to the case of luacs spaces, let us recall Egorov’s theorem (cf. [18, Theorem A, page 88]), which states that for any finite measure space and every sequence of measurable functions on which converges to zero pointwise -a.e. and each there is a set with such that is uniformly convergent to zero on .

Now we are ready to prove the following theorem.

Theorem 10. *Let be an order-continuous Köthe function space over the complete -finite measure space and an luacs Banach space. If *(a)* is WLUR or *(b)* is and is also order continuous,**then is also luacs. *

*Proof. *Suppose that we are given a sequence in and an element such that as well as a functional such that . As before, we can represent by an element . We then have
and hence
By passing to a subsequence we may also assume that
We further have
and thus
An analogous argument also shows
Moreover, the inequality
holds for every . It follows that
Analogously one can see that
Finally, we have
Consequently
Since is in particular luacs we get from (35) and (38) that
Because is in any case , it follows from (41), (44), and (45) that
and thus
So by passing to a further subsequence we may assume
Next we will show that
Since is -finite there is an increasing sequence in such that for every and .

Denote by and the null sets on which the convergence statement from (36) and (48), respectively does not hold, and let . Put and . We will see that is a null set.

First we define for every a function by setting for and for . Note that each is measurable and since for every we have .

We have for every and every , so by Egorov’s theorem we can find for every an increasing sequence in with such that converges uniformly to zero on each .

It follows that is a null set for every .

Let us now first suppose that (b) holds, so is order continuous. We have
and moreover this sequence is decreasing, so the order continuity of implies
So if and are given, we can find an index such that , and then, by uniform convergence, an index such that for every and every .

Then we have
for each .

In conclusion we have
Now if (a) holds, that is, if is WLUR, then by (38) the sequence must be weakly convergent to in , and hence
for all . Since decreases to zero a.e., the order continuity of gives us for every .

A similiar argument as before now easily yields that also holds in case (a). But is nothing else than
Combining this with (35) leaves us with
Since is luacs and because of (38), it follows that
Taking into account (45) we get
and hence for every we have a.e. Consequently, is a null set for every , and thus is also a null set.

Now suppose that . Then we have , and , as well as and
By passing to a subsequence we may assume that is bounded away from zero. Then it follows from Lemma 5 that
Also, we have
Since is luacs we can conclude that .

So is a null set with for every and (49) is proved.

From (45) and (49) it follows that
and we are done.

Recall that a subset is said to be equi-integrable if for every there is some such that It is well known that for a finite measure a bounded subset is relatively weakly compact in if and only if is equi-integrable (see, e.g., [19, Theorem 13.6]). One ingredient for the usual proof of this fact is the following lemma (see [19, Proposition 13.4]), which we will also need in the sequel.

Lemma 11. *For a finite measure space , a sequence in is equi-integrable whenever the sequence is convergent for each . *

We will also need Vitali’s lemma, which reads as follows (see, e.g., [11, Lemma 3.1.13] for an even more general version).

Lemma 12. *Let be a finite measure space, and let be a sequence in such that is equi-integrable. Let be a measurable function on such that -a.e. Then and . *

Finally, let us recall that a Banach space is said to have the Kadets-Klee property (also known as property ) if for every sequence in and each the implication holds. For example, every LUR space and every dual of a reflexive, FS space has the Kadets-Klee property.

It is known that every Banach lattice with the Kadets-Klee property is order continuous, (cf. [12, page 28]). With this in mind we can prove the following result concerning spaces.

Theorem 13. *If the measure is finite and is LUR, then is a space whenever is . If in addition is order continuous then the assertion also holds if is merely -finite. *

*Proof. *By the previous theorem, is luacs, so we only have to show the implication “” in Definition 6 (i). To this end, let be a sequence in and such that , and let such that . It will be enough to show that a subsequence of converges to one.

Since is order continuous, we can as before represent by some and conclude
Also, just as we have done in the previous proof, we find that
Since is LUR, it follows that
Hence, by passing to a subsequence, we may assume that (cf. Lemma 7)
By (67) and (64) we also have
Since is , it follows from (65), (69) that
From (67) we also get
Thus by Lemma 11 the sequence and hence also the sequence are equi-integrable with respect to for every with . This combined with Vitali’s lemma and (71) implies
So if , we immediately get
because of (70).

If is merely -finite but is order continuous, we can fix an increasing sequence in such that and for every . Then the sequence decreases pointwise to zero, and, by the order continuity of , we can conclude that .

Thus given any , we find an such that . Since , there exists such that
It follows that for every
So we have
and because of (70) it follows as before that
finishing the proof.

Now we turn to the sluacs spaces. An easy normalisation argument shows that a Banach space is sluacs if and only if for every , every sequence in , and all sequences in with , and , we have . In view of this characterisation, is sluacs if and only if for every and every the number is strictly positive, where Next we will prove an easy lemma on the continuity of .

Lemma 14. *For all , and all , one has
**
that is is -Lipschitz continuous with respect to the norm of . *

*Proof. *First we fix and . Put and take , such that . It follows that .

Now let be arbitrary. We can find with . Define . Then
and hence
But we have and as well as
Thus we get
and since was arbitrary, it follows that
Again, since was arbitrary we can conclude that
and by symmetry it follows that
Analogously one can prove that
for all and all , . An application of the triangle inequality then yields the result.

In the paper [16] Kamińska and Turett proved various theorems concerning different rotundity properties of Köthe-Bochner spaces. For example, by [16, Theorem 5] if has the so-called Fatou property and is LUR, then is LUR whenever is LUR. We will adopt the technique of proof from [16, Theorem 5] to show the following result.

Theorem 15. *If is LUR and is sluacs smooth, then is also sluacs.*

*Proof. *Since is LUR it is order continuous.

Let and be arbitrary and let
for every . Since by the previous lemma is continuous, it follows that the sets are measurable. Also, the sequence is increasing and because is sluacs, we have ; hence decreases pointwise to zero. The order continuity of implies , and thus we can find with
Now let us take and with and . Let be represented by . As in the proof of Proposition 9 we can conclude that
Next we define
Then is measurable, and
Since , it follows that
Let us fix such that
Now consider the following sets:
Then are measurable, pairwise disjoint, and . Thus by (96) there exists some such that
If , then, since for , it follows that
and again by definition of we obtain
and hence
In the case of one can obtain the same statement by an analogous argument. To treat the remaining cases we need some preliminary considerations.

Let us denote by the null set on which the equality from (93) does not hold and suppose that . Then in particular and , and hence
Moreover, by the definitions of and and the choice of we have
Since , we also have
So by definition of we must have
Once more by the definition of this implies
where .

Now suppose that . Then
Consequently
Since , the definition of implies that
where the latter inequality holds because of . It follows that
where which by (97) is greater than zero. Because of , it follows that
So if we put and , then
Now we will show that if and , respectively then
Let us first assume that , that is,
Since for it follows that
Because is a null set, we have
where the second last inequality holds because of (91).

Now assume that ; that is,
It follows that
and hence as before we get