Quantitative weakly compact sets and Banach-Saks sets in $\ell _1$

Abstract

In this paper, we show a quantitative version of the theorem stating that relatively weakly compact sets in $\ell _1$ coincide with those having the Banach-Saks property. Namely, we prove that the measure of the weak noncompactness based on the Eberlein double limit criterion is equal to the measure of the non-Banach-Saks property defined by the arithmetic separation of sequences.

1 Introduction

In this article, we aim to quantify the relationship of compact sets and the Banach-Saks sets in the Banach space $\ell _1$ by using measures of weak noncompactness and the Banach-Saks property. Measures of noncompactness and weak noncompactness have been widely applied in functional analysis, both in applications and Banach space theory. In the area of differential and integral equations, they become indispensable to characterize compact sets and weakly compact sets, and then to get fixed points and further solutions to equations, see [6, 12, 13, 25] for example. On the other hand, they are widely used in Banach space theory to get deeper understanding of the implications through quantitative means. The quantitative methods provide different angles to view the theoretical results. There is a new trend to investigate the quantified properties of Banach spaces, see, e.g., [9, 10, 14, 17] and their references. From the applications in both equations and theories, we may infer that representations of the measures are always crucial. Every representation has its own advantages. To satisfy different goals, there have appeared many different measures of noncompactness and weak noncompactness, and the relationship between these measures is of interest, see [1, 7, 22] for example. In view of [3, 15, 17], it is specially interesting to study on which spaces the measures are equal. In the sequel, we work on De Blasi’s [11] measure of weak noncompactness $\omega$ and the measure $\gamma$ based on the Eberlein double limit criterion for weakly compact sets. Results in [3, 4] showed that the two measures are not equivalent in general. We will prove that $\gamma$ is exactly 2$\omega$ in $\ell _1$.

A Banach space X is said to have the Banach-Saks property if every bounded sequence $(x_n)$ in X has a subsequence $(x'_n)$ such that the Cesàro means $(x'_1 + \dots + x'_n)/n$ converge. As a weaker form, the weak Banach-Saks property of Banach spaces has been introduced. It means that every weakly convergent sequence has a subsequence whose Cesàro means converge in norm. For example, the spaces $c_0$, $\ell _1$, and $L_1[0,1]$ have the weak Banach-Saks property. As for localization, a bounded subset A of a Banach space is said to be a Banach-Saks set if every sequence in A has a subsequence whose Cesàro sum converges. An analogue of the weak Banach-Saks set could be defined. The Banach-Saks property connects closely to reflexivity and weak compactness. Any Banach space with the Banach-Saks property was shown to be reflexive by a so-called summability method [23]. Meanwhile there are reflexive spaces without the Banach-Saks property [5]. Via the Rosenthal $\ell _1$ theorem, every Banach-Saks set has been proved to be relatively weakly compact [21]. But in general, the reverse is not true because of the counterexample by Schreier [24] (see also, Baernstein [5]). Many mathematicians keep trying to quantify the involving Banach-Saks properties, see, e.g., [9, 18, 19].

Suppose that a sequence $(x_n)$ is contained in a relatively weakly compact set of a Banach space X, then there is a subsequence $(x_{n_k})$ weakly converging to some point. If additionally the space X has the weak Banach-Saks property, then we could get a subsequence of $(x_{n_k})$, whose Cesàro sum converges in norm. Consequently, we may see that Banach-Saks sets coincide with relatively weakly compact sets in Banach spaces having the weak Banach-Saks property. Recently Kryczka [19] proved a quantitative Szlenk theorem which states the equivalence of the relatively weak compactness and the Banach-Saks property in $L_1[0,1]$. Inspired by their works, in Section  2, we prove a similar quantitative equivalence result in $\ell _1$ by proving an equality in terms of measures of the Banach-Saks property and weak noncompactness.

2 Quantitative Banach-Saks property

Let X be a real infinite dimensional Banach space normed with $\Vert \cdot \Vert$, and let $X^*$ be its dual. $B_X$ is the unit ball of X. For any subset $E\subset X$, $\mathrm {co}(E)$ is the convex hull of E. Denote by ${\mathcal {B}}(X)$ the collection of all nonempty bounded subsets of X. By ${\mathbf {N}}$ we understand the set of positive integers, and |A| is the cardinality of a subset $A\subset {\mathbf {N}}$. Recall the Hausdorff measure of noncompactness $\chi : {\mathcal {B}}(X) \rightarrow {\mathbf {R}}$ on X.

$$\begin{aligned} \chi (A) {:}{=} \inf \{t>0\mid A \subset K + tB_X, K\subset X \text { is compact}\}, \end{aligned}$$

(1)

where $A\in {\mathcal {B}}(X)$. The Hausdorff measure is widely applied and the representation of the measure is of interest. In the Banach space $\ell _1$, $\chi$ may be expressed by the following formula (see [2, p. 5]).

$$\begin{aligned} \chi (A) = \lim _{n\rightarrow \infty } \sup _{x\in A} \sum _{k=n}^\infty |x(k)| \end{aligned}$$

(2)

for any $A\in {\mathcal {B}}(\ell _1)$.

$\chi$ characterizes compact sets in X as $\chi (A)=0$ if and only if A is relatively compact. Replacing the compact set K in formula (1) by a weakly compact one, it is then the De Blasi measure of weak noncompactness $\omega$. By the well-known Schur theorem, we observe easily that $\omega (A) = \chi (A)$. Recall that the measures $\omega$ and $\chi$ satisfy all the axiomatic principles for regular measures, see [6].

Another measure of weak noncompactness we are interested in is the measure $\gamma$ based on the classical double-limit criterion of Eberlein. For any $A\in {\mathcal {B}}(X)$,

$$\begin{aligned} \gamma (A) {:}{=} \sup | \lim _n\lim _m \langle f_n, x_m \rangle - \lim _m\lim _n \langle f_n, x_m \rangle |, \end{aligned}$$

where the supremum is taken over all sequences $f_n \in B_{X^*}$ and $x_m \in A$ such that the double limits exist. It has been proved in [4] that the measures $\omega$ and $\gamma$ are not equivalent, and they are shown in [3] to have the relationship $\gamma (A)\le 2\omega (A)$ for any $A\in {\mathcal {B}}(X)$.

Kryczke et al. [20] found that the measure $\gamma$ can be expressed exactly in terms of the James convex separation criterion of weak compactness (see [16]). In detail, they proved

$$\begin{aligned} \gamma (A) = \sup \{\mathrm {csep}(x_n)\mid (x_n) \subset \mathrm {co}(A)\},\ \end{aligned}$$

where

$$\begin{aligned} \mathrm {csep}(x_n) \, {:}{=} \, \inf _m d\{\mathrm {co}\{x_n\}_{n=1}^m, \mathrm {co}\{x_n\}_{n=m+1}^\infty \}. \end{aligned}$$

(3)

Applying this result, we will see in next theorem that the reverse relationship of $\omega$ and $\gamma$ could also be verified particularly in $\ell _1$.

Theorem 1

For any nonempty bounded subset A of $\ell _1$, $\gamma (A) = 2 \omega (A)$.

Proof

With the comments above in mind, we only need to prove the inequality $\gamma (A)\ge 2\omega (A)$. Without loss of generality, we may suppose $\omega (A) = \chi (A) =\theta > 0$ since the case is trivial when $\theta = 0$. By fomula (2), for any $\varepsilon >0$, there exists an integer $K_0 \in {\mathbf {N}}$ such that for any $x\in A$,

It is easy to see that $\sup _{x\in A} \sum _{k=n}^\infty |x(k)|$ is decreasing in n. Then for any $K\in {\mathbf {N}}$, there is $x\in A$ with $\theta -\varepsilon < \Vert x\Vert _K$. Thus there exist $x_1\in A$ and $K_1>K_0$ such that

Proceeding this process, we inductively produce a sequence $(x_n)\subset A$ and an increasing sequence $(K_n)\subset {\mathbf {N}}$ such that for any $m\in {\mathbf {N}}$,

We observe that the sequence $(x_n)$ satisfies a nice property. To specify that, let us take just two elements $x_1$ and $x_2$ as an example. Clearly,

To explain the last inequality, we may tell that $\Vert x_2\Vert _{K_1}^{K_2} - \Vert x_2\Vert _{K_0}^{K_1} - \Vert x_2\Vert _{K_2} \ge \theta - 3\varepsilon$ since

$$\begin{aligned} \Vert x_2\Vert _{K_1}^{K_2} > \theta - \varepsilon \text {\ and\ } \Vert x_2\Vert _{K_0}^{K_1} + \Vert x_2\Vert _{K_2} < 2\varepsilon . \end{aligned}$$

The latter is true because additionally

$$\begin{aligned} \Vert x_2\Vert _{K_0} = \Vert x_2\Vert _{K_0}^{K_1} + \Vert x_2\Vert _{K_1}^{K_2} + \Vert x_2\Vert _{K_2} < \theta + \varepsilon . \end{aligned}$$

By the same process, we have

$$\begin{aligned} \Vert x_1\Vert _{K_0}^{K_1} - \Vert x_1\Vert _{K_1} \ge \theta - 3\varepsilon . \end{aligned}$$

In fact, this property could be valid not only for $x_1$ and $x_2$, but also for any element in the set $\{x_n\}$ and its multiplication with a real number. Namely, for any $x_i \in \{x_n\}$ and $\lambda \in {\mathbf {R}}$,

$$\begin{aligned} \begin{aligned} \Vert \lambda x_i\Vert _{K_0}&\ge |\lambda |\Vert x_i\Vert _{K_{i-1}}^{K_i} - \sum _{j\ne i} |\lambda |\Vert x_i\Vert _{K_{j-1}}^{K_j}\\&\ge |\lambda |(\theta - \varepsilon - 2\varepsilon )\\&= |\lambda |(\theta - 3\varepsilon ). \end{aligned} \end{aligned}$$

Applying the similar calculation for $x_1$ and $x_2$ to arbitrary finite elements, we have that for any fixed $n\in \text {N}$ and any $\lambda _i \in \mathbf{R}$ with $i=1,\dots ,n$,

(4)

For any $y\in \mathrm {co}\{x_i\}_{i=1}^m$ and $z \in \mathrm {co}\{x_i\}_{i=m+1}^\infty$ with $m\in {\mathbf {N}}$, there exist $n>m$, $a_i\ge 0$ with $i=1,\ldots ,m$ and $b_j\ge 0$ with $j = m+1, \ldots ,n$, such that

$$\begin{aligned} \begin{aligned} \sum _{i=1}^m a_i =1, \quad \sum _{j=m+1}^n b_j =1,\\ y = \sum _{i=1}^{m} a_i x_i, \quad z= \sum _{j=m+1}^n b_j x_j. \end{aligned} \end{aligned}$$

By formula (4), it is easy to see that

$$\begin{aligned} \Vert y-z\Vert \ge (\theta - 3\varepsilon )\left( \sum _{i=1}^m|a_i| + \sum _{j=m+1}^n |-b_j|\right) = 2(\theta - 3\varepsilon ). \end{aligned}$$

Thus by (3), we have $\mathrm {csep}(x_n) \ge 2(\theta - 3\varepsilon )$.

Now we have proved that for any $\varepsilon >0$, there is a sequence $(x_n)$ in A such that $\mathrm {csep}(x_n) \ge 2(\theta - 3\varepsilon )$. It further means that $\gamma (A)\ge 2 \omega (A)$, and the proof is completed. $\square$

Beauzamy in [8] characterized spaces having the Banach-Saks property by spreading models, i.e., a Banach space X does not have the Banach-Saks property if and only if there exist $\theta >0$ and a bounded sequence $(x_n)\subset X$ such that for any subsequence $(x'_n)$,

$$\begin{aligned} \left\| \frac{1}{m} \left( \sum _{n=1}^k x'_n - \sum _{n=k+1}^m x'_n \right) \right\| \ge \theta . \end{aligned}$$

for any positive integers $k\le m$. Kryczka [18] modified Beauzamy’s condition and introduced a deviation $\varphi$ as the following to denote whether a set of a Banach space is a Banach-Saks set. We may call it the deviation measure of non-Banach-Saksness. For any $A\in {\mathcal {B}}(X)$,

$$\begin{aligned} \varphi (A) = \sup \{ \mathrm {asep}(x_n)\mid (x_n)\in A\}, \end{aligned}$$

where

$$\begin{aligned} \mathrm {asep}(x_n) {:}{=} \inf \Big \Vert \frac{1}{m}\left( \sum _{n\in C}x_n - \sum _{n\in D}x_n \right) \Big \Vert \end{aligned}$$

with the infimum taken over all $m\in {\mathbf {N}}$ and finite $C, D\subset {\mathbf {N}}$ having $|C| = |D|=m$ and $\max C < \inf D$. The measure $\varphi$ satisfies (see [18]) that for any $A,B\in {\mathcal {B}}(X)$,

• (i) $\varphi (A)=0$ if and only if A is a Banach-Saks set;

• (ii) $\varphi (A)\le \varphi (B)$ whenever $A\subset B$;

• (iii) $\varphi (tA) = |t|\varphi (A)$ for $t\in {\mathbf {R}}$;

• (iv) $\varphi (A+B) \le \varphi (A) + \varphi (B)$ if A and B are convex.

From the definitions (3) and (), we may get $\gamma (A)\le \varphi (A)$, which quantifies the result that every Banach-Saks set is weakly compact. A glimpse on the unit bases $(e_n)$ of $\ell _1$ gives $\varphi (B_{\ell _1}) = 2$. We will use a quantitative method to state that the compact sets, weakly compact sets, and Banach-Saks sets coincide with each other in $\ell _1$, and moreover the measures of these properties are equal.

Theorem 2

For any nonempty bounded subset A of $\ell _1$, $\varphi (A) = \gamma (A) = 2\omega (A) = 2\chi (A)$.

Proof

It is sufficient to prove $\gamma (A) \ge \varphi (A)$. Suppose that $t>\omega (A)$, then there exists a weakly compact set $K\subset \ell _1$ such that $A\subset K + tB_{\ell _1}$. By Krein’s theorem, it is reasonable to assume that K is convex. Noticing the properties of $\varphi$ and $\varphi (K)=0$ since $\ell _1$ has the weak Banach-Saks property, we get

$$\begin{aligned} \varphi (A) \le \varphi (K) + tB_{\ell _1} = 2t. \end{aligned}$$

It means $\gamma (A) = 2\omega (A) \ge \varphi (A)$, and the proof is completed. $\square$