Ces\`{a}ro convergent sequences in the Mackey topology

TL;DR

This paper investigates Cesàro convergence in the Mackey topology within Banach spaces, establishing new properties and stability results, especially for subspaces generated by operators with Banach-Saks adjoints and applications to Banach lattices and Lebesgue-Bochner spaces.

Contribution

It introduces property (μ^s) for Banach spaces, proves its validity for certain subspaces, and explores its stability and applications in Banach lattices and Lebesgue-Bochner spaces.

Findings

Property (μ^s) holds for subspaces strongly generated by operators with Banach-Saks adjoint.

Property (μ^s) is stable under ℓ^p-sums.

Banach lattices with order continuous norm have property (μ_𝒜^s) for all L-weakly compact sets.

Abstract

A Banach space $X$ is said to have property ($\mu^s$) if every weak$^*$-null sequence in $X^*$ admits a subsequence such that all of its subsequences are Ces\`{a}ro convergent to $0$ with respect to the Mackey topology. This is stronger than the so-called property (K) of Kwapie\'{n}. We prove that property $(\mu^s)$ holds for every subspace of a Banach space which is strongly generated by an operator with Banach-Saks adjoint (e.g. a strongly super weakly compactly generated space). The stability of property $(\mu^s)$ under $\ell^p$-sums is discussed. For a family $\mathcal{A}$ of relatively weakly compact subsets of $X$, we consider the weaker property $(\mu_\mathcal{A}^s)$ which only requires uniform convergence on the elements of $\mathcal{A}$, and we give some applications to Banach lattices and Lebesgue-Bochner spaces. We show that every Banach lattice with order continuous norm and weak unit has property $(\mu_\mathcal{A}^s)$ for the family of all $L$-weakly compact sets. This sharpens a result of de Pagter, Dodds and Sukochev. On the other hand, we prove that $L^1(\nu,X)$ (for a finite measure $\nu$) has property $(\mu_\mathcal{A}^s)$ for the family of all $\delta\mathcal{S}$-sets whenever $X$ is a subspace of a strongly super weakly compactly generated space.

Equations

$$\big{\{}T(C):\,T:L^{1}[0,1]\to X\mbox{ operator, }C\in wk(L^{1}[0,1])\big{\}},$$

$$g_k=\sum _{n\in I_k}{a}_n{f}_n$$

$$v_k:=\frac{1}{2^k}\sum _{n=1}^{2^k}{u}_n\text{mbox}and{w}_k:=\frac{1}{2^{k-1}}\sum _{n=2^{k-1}+1}^{2^k}{u}_n$$

$$\lim_{N\to\infty}\sup_{y\in B_{Y}}\Big{|}\Big{\langle}\frac{1}{N}\sum_{k=1}^{N}x_{n_{k}}^{*},T(y)\Big{\rangle}\Big{|}=\lim_{N\to\infty}\Big{\|}\frac{1}{N}\sum_{k=1}^{N}T^{*}(x_{n_{k}}^{*})\Big{\|}_{Y^{*}}=0.$$

$$\Big{(}\bigoplus_{i\in I}X_{i}\Big{)}_{\ell^{p}}.$$

$$\pi_{j}:\Big{(}\bigoplus_{i\in I}X_{i}\Big{)}_{\ell^{p}}\to X_{j}\quad\mbox{and}\quad\rho_{j}:\Big{(}\bigoplus_{i\in I}X^{*}_{i}\Big{)}_{\ell^{q}}\to X^{*}_{j}$$

$$\bigcap _{i\in J}{\pi }_i^{-1}(W_i)\cap \bigcap _{i\in I\setminus J}{\pi }_i^{-1}(\{0\}),$$

$${\tilde{\xi }}_j\ge {\xi }_j:=(\Vert {\rho }_i(y_j^{\ast }){\Vert }_{X_i^{\ast }}{)}_{i\in I}\text{mbox}pointwisein{\ell }^q(I)$$

$$\sup_{j>j_{0}}\Big{(}\sum_{i\in I\setminus I_{0}}\psi_{i}(\tilde{\xi}_{j})^{q}\Big{)}^{\frac{1}{q}}\leq\varepsilon,$$

$$\Vert {\rho }_i(y_j^{\ast }){\Vert }_{X_i^{\ast }}={\psi }_i({\xi }_j)\le {\psi }_i({\tilde{\xi }}_j)\text{mbox}foreveryi\in I\text{mbox}andj\in \mathbb{N},$$

$$\sup_{j>j_{0}}\Big{(}\sum_{i\in I\setminus I_{0}}\|\rho_{i}(y_{j}^{*})\|_{X_{i}^{*}}^{q}\Big{)}^{\frac{1}{q}}\leq\varepsilon.$$

$$\underset{j>j_0}{\sup }\sum _{i\in I\setminus I_0}|b_i|\cdot \Vert {\rho }_i(y_j^{\ast }){\Vert }_{X_i^{\ast }}\le \epsilon \text{mbox}forevery(b_i{)}_{i\in I}\in B_{{\ell }^p(I)}.$$

$$\big{|}\langle\rho_{i}(y_{j}^{*}),\pi_{i}(x)\rangle\big{|}\leq\frac{\varepsilon}{|I_{0}|}\quad\mbox{for every }j>j_{1},\,i\in I_{0}\mbox{ and }x\in L.$$

$$\big{|}\langle y_{j}^{*},x\rangle\big{|}\leq\sum_{i\in I}\big{|}\langle\rho_{i}(y_{j}^{*}),\pi_{i}(x)\rangle\big{|}\stackrel{{\scriptstyle\eqref{eqn:yk-coordinates}}}{{\leq}}\varepsilon+\sum_{i\in I\setminus I_{0}}\big{|}\langle\rho_{i}(y_{j}^{*}),\pi_{i}(x)\rangle\big{|}\\ \leq\varepsilon+\sum_{i\in I\setminus I_{0}}\|\pi_{i}(x)\|_{X_{i}}\cdot\|\rho_{i}(y_{j}^{*})\|_{X_{i}^{*}}\stackrel{{\scriptstyle\eqref{eqn:cola-ww}}}{{\leq}}2\varepsilon.$$

$$v_n:=(\Vert {\rho }_i(x_n^{\ast }){\Vert }_{X_i^{\ast }}{)}_{i\in I}\in {\ell }^q(I)\text{mbox}foralln\in \mathbb{N},$$

$$\|\rho_{i}(y_{j}^{*})\|_{X_{i}^{*}}\leq\frac{1}{j}\sum_{k=1}^{j}\|\rho_{i}(x_{n_{k}}^{*})\|_{X^{*}_{i}}=\psi_{i}\Big{(}\frac{1}{j}\sum_{k=1}^{j}v_{n_{k}}\Big{)},$$

$$L_w(X):=\{A\subseteq X:A\text{mbox}isL\text{mbox}-weaklycompact\}.$$

$$\lim_{N\to\infty}\sup_{f\in B_{L^{\infty}(\nu)}}\Big{|}\Big{\langle}\frac{1}{N}\sum_{k=1}^{N}x^{*}_{n_{k}},i(f)\Big{\rangle}\Big{|}=\lim_{N\to\infty}\Big{\|}\frac{1}{N}\sum_{k=1}^{N}i^{*}(x^{*}_{n_{k}})\Big{\|}_{L^{1}(\nu)}=0.$$

$$L_{w}(X)\supseteq\mathcal{Q}(X):=\big{\{}T(C):\,T:L^{1}[0,1]\to X\mbox{ operator, }C\in wk(L^{1}[0,1])\big{\}}.$$

$$\Big{\{}\sum_{n\in I_{k}}h_{n}(\cdot)x_{n}:\,k\in\mathbb{N}\Big{\}}$$

$$C:=\Big{\{}\sum_{n\in I_{k}}a_{n}x_{n}:\,k\in\mathbb{N},\,(a_{n})_{n\in I_{k}}\in\{-1,1\}^{I_{k}}\Big{\}}$$

$$y_m=\sum _{n\in I_{k_m}}{a}_{n,m}{x}_n$$

$$Z:=\overline{\mathrm{span}}\{r_n(\cdot )x_n:n\in \mathbb{N}\}\subseteq L^1([0,1],X),$$

$$\Big{\langle}g_{k},\sum_{n\in I_{k}}r_{n}(\cdot)x_{n}\Big{\rangle}=\Big{\langle}\sum_{n\in I_{k}}a_{n}e_{n}^{*},\sum_{n\in I_{k}}S(r_{n}(\cdot)x_{n})\Big{\rangle}\\ =\Big{\langle}\sum_{n\in I_{k}}a_{n}e_{n}^{*},\sum_{n\in I_{k}}e_{n}\Big{\rangle}=\sum_{n\in I_{k}}a_{n}=1.$$

$$\Big{\{}\sum_{n\in I_{k}}r_{n}(\cdot)x_{n}:\,k\in\mathbb{N}\Big{\}}$$

$$H:=\{h\in L^1(\nu ,X):h(\omega )\in T(B_Y)\text{mbox}for\mu \text{mbox}-a.e.\omega \in \Omega \}$$

$$\tilde{T}(f):=T\circ f\text{mbox}forallf\in L^2(\nu ,Y).$$

$$L^2([0,1],E{)}^{\ast }=L^2([0,1],X)\text{mbox}and{L}^1([0,1],X{)}^{\ast }=L^{\infty }([0,1],E),$$

Full text

Research supported by projects MTM2014-54182-P and MTM2017-86182-P (AEI/FEDER, UE)

1. Introduction

A subset $C$ of a Banach space $X$ is said to be Banach-Saks if every sequence $(x_{n})_{n}$ in $C$ admits a Cesàro convergent subsequence $(x_{n_{j}})_{j}$, i.e. the sequence of arithmetic means $(\frac{1}{k}\sum_{j=1}^{k}x_{n_{j}})_{k}$ is convergent (in the norm topology) to some element of $X$. A Banach space $X$ is said to have the Banach-Saks property if its closed unit ball $B_{X}$ is a Banach-Saks set. An operator $T:Y\to X$ between Banach spaces is said to be Banach-Saks if so is $T(B_{Y})$. Every Banach-Saks set is relatively weakly compact (see e.g. [28, Proposition 2.3]) and so every space having the Banach-Saks property is reflexive [31], and every Banach-Saks operator is weakly compact. The converse statements are not true in general [2]. Every super-reflexive space (like $L^{p}(\nu)$ for a non-negative measure $\nu$ and $1<p<\infty$) has the Banach-Saks property (see e.g. [12, p. 124]). For any non-negative measure $\nu$, the space $L^{1}(\nu)$ enjoys the weak Banach-Saks property, that is, every weakly compact subset of $L^{1}(\nu)$ is Banach-Saks, a result due to Szlenk [40] (cf. [12, p. 112]). At this point it is convenient to recall the Erdös-Magidor theorem [15] (cf. [28, Corollary 2.6] and [33, Theorem 2.1]) which implies, in particular, that every sequence in a Banach-Saks set admits a subsequence such that all of its subsequences are Cesàro convergent to the same limit:

Theorem 1.1** (Erdös-Magidor).**

Every bounded sequence $(x_{n})_{n}$ in a Banach space $X$ admits a subsequence $(x_{n_{j}})_{j}$ such that

1. (i)

either all subsequences of $(x_{n_{j}})_{j}$ are Cesàro convergent (to the same limit); 2. (ii)

or no subsequence of $(x_{n_{j}})_{j}$ is Cesàro convergent.

As we will see, for a finite measure $\nu$, the weak Banach-Saks property of $L^{1}(\nu)$ yields a somehow similar property for its dual $L^{1}(\nu)^{*}=L^{\infty}(\nu)$ by considering the $w^{*}$-topology and the Mackey topology, namely: every $w^{*}$-null sequence in $L^{\infty}(\nu)$ admits a subsequence such that all of its subsequences are Cesàro convergent to [math] with respect to $\mu(L^{\infty}(\nu),L^{1}(\nu))$. Recall that, for an arbitrary Banach space $X$, the Mackey topology $\mu(X^{*},X)$ is the (locally convex) topology on $X^{*}$ of uniform convergence on all weakly compact subsets of $X$. So, for any finite measure $\nu$, the space $L^{1}(\nu)$ satisfies the following property which is the main object of study of this paper:

Definition 1.2**.**

A Banach space $X$ is said to have property $(\mu^{s})$ if every $w^{*}$-null sequence in $X^{*}$ admits a subsequence such that all of its subsequences are Cesàro convergent to [math] with respect to $\mu(X^{*},X)$.

The paper is organized as follows. In the preliminary Section 2 we point out that property $(\mu^{s})$ is stronger than the so-called property (K) invented by Kwapień in connection with some results of Kalton and Pełczýnski [24]:

Definition 1.3**.**

A Banach space $X$ is said to have property (K) if every $w^{*}$-null sequence in $X^{*}$ admits a convex block subsequence which converges to [math] with respect to $\mu(X^{*},X)$.

Property (K) (and some variants) have been also studied by Frankiewicz and Plebanek [19], Figiel, Johnson and Pełczýnski [17], de Pagter, Dodds and Sukochev [10], Avilés and the author [1]. In Section 2 we also give some basic examples of Banach spaces having property $(\mu^{s})$. For a reflexive space $X$, $(\mu^{s})$ is equivalent to the Banach-Saks property of $X^{*}$ (Proposition 2.2). In particular, any super-reflexive space has $(\mu^{s})$. For a $C(L)$ space (where $L$ is a compact Hausdorff topological space), $(\mu^{s})$ is equivalent to the fact that $C(L)$ is Grothendieck (Proposition 2.3). So, for instance, $\ell_{\infty}$ has property $(\mu^{s})$.

In Section 3 we discuss the role of “strong generation” in the study of property $(\mu^{s})$. To be more precise we need some terminology:

Definition 1.4**.**

Let $X$ be a Banach space and let $\mathcal{H}$ and $\mathcal{G}$ be two families of subsets of $X$. We say that $\mathcal{H}$ is strongly generated by $\mathcal{G}$ if for every $H\in\mathcal{H}$ and every $\varepsilon>0$ there is $G\in\mathcal{G}$ such that $H\subseteq G+\varepsilon B_{X}$. If in addition $\mathcal{G}=\{nG_{0}:n\in\mathbb{N}\}$ for some $G_{0}\subseteq X$, we simply say that $\mathcal{H}$ is strongly generated by $G_{0}$.

We will be mainly interested in the case $\mathcal{H}=wk(X)$, the family of all weakly compact subsets of the Banach space $X$.

Definition 1.5**.**

Let $X$ be a Banach space and let $\mathcal{G}$ be a family of subsets of $X$. We say that $X$ is strongly generated by $\mathcal{G}$ if $wk(X)$ is strongly generated by $\mathcal{G}$. If in addition $\mathcal{G}=\{nT(B_{Y}):n\in\mathbb{N}\}$ for some operator $T:Y\to X$ from a Banach space $Y$, we say that $Y$ strongly generates $X$ or that $T$ strongly generates $X$.

Banach spaces which are strongly generated by a reflexive space (i.e. SWCG spaces) or by a super-reflexive space have been widely studied, see e.g. [16, 25, 29] and the references therein. All SWCG spaces and their subspaces have property (K), see [1, Corollary 2.3]. We show that property $(\mu^{s})$ is enjoyed by every subspace of a Banach space which is strongly generated by an operator with Banach-Saks adjoint (Theorem 3.1). This assumption is satisfied by the so-called strongly super weakly compactly generated spaces (S2WCG) studied recently by Raja [35] and Cheng et al. [8]. In particular, any Banach space which is strongly generated by a super-reflexive space (e.g. $L^{1}(\nu)$ for a finite measure $\nu$) has property $(\mu^{s})$.

We prove that a SWCG space $X$ has property $(\mu^{s})$ if (and only if) every $w^{*}$-null sequence in $X^{*}$ admits a subsequence which is Cesàro convergent to [math] with respect to $\mu(X^{*},X)$. We do not know whether such equivalence holds for arbitrary Banach spaces. The case of SWCG spaces is generalized to Banach spaces which are strongly generated by less than $\mathfrak{p}$ weakly compact sets (Theorem 3.7). Recall that $\mathfrak{p}$ is the least cardinality of a family $\mathcal{M}$ of infinite subsets of $\mathbb{N}$ such that:

• •

$\bigcap\mathcal{N}$ is infinite for every finite subfamily $\mathcal{N}\subseteq\mathcal{M}$.

• •

There is no infinite set $A\subseteq\mathbb{N}$ such that $A\setminus M$ is finite for all $M\in\mathcal{M}$.

In general, $\omega_{1}\leq\mathfrak{p}\leq\mathfrak{c}$. Under CH cardinality less than $\mathfrak{p}$ just means countable, but in other models there are uncountable sets of cardinality less than $\mathfrak{p}$, see e.g. [5] for more information.

In Section 4 we study the stability of property $(\mu^{s})$ under $\ell^{p}$-sums for $1\leq p\leq\infty$. Pełczýnski showed that the $\ell^{1}$-sum of $\mathfrak{c}$ copies of $L^{1}[0,1]$ fails property (K), see [17, Example 4.I] (cf. [19]). In particular, this implies that property $(\mu^{s})$ is not preserved by arbitrary $\ell^{1}$-sums. We prove that $(\mu^{s})$ is preserved by $\ell^{1}$-sums of less than $\mathfrak{p}$ summands (Theorem 4.2), as well as by arbitrary $\ell^{p}$-sums whenever $1<p<\infty$ (Theorem 4.4). On the other hand, in Example 4.6 we point out the existence of a sequence of finite-dimensional spaces whose $\ell^{\infty}$-sum fails property (K), which answers a question left open in [1, Problem 2.19].

In Section 5 we consider a natural weakening of properties $(\mu^{s})$ and (K) to deal with certain families of relatively weakly compact sets. This idea is applied to some Banach lattices and Lebesgue-Bochner spaces.

Definition 1.6**.**

Let $X$ be a Banach space and let $\mathcal{A}$ be a family of subsets of $X$. We say that $X$ has

1. (i)

property $(\mu^{s}_{\mathcal{A}})$ if every $w^{*}$-null sequence in $X^{*}$ admits a subsequence such that all of its subsequences are Cesàro convergent to [math] uniformly on each element of $\mathcal{A}$; 2. (ii)

property (KA*) if every $w^{*}$-null sequence in $X^{*}$ admits a convex block subsequence which converges to [math] uniformly on each element of $\mathcal{A}$.*

For instance, the so-called property ($k$) of Figiel, Johnson and Pełczýnski [17] coincides with property (KA) when $\mathcal{A}$ is the family

[TABLE]

see [10, Lemma 8.1]. Every weakly sequentially complete Banach lattice with weak unit has property ($k$), see [17, Proposition 4.5(b)]. This can also be obtained as a consequence of a result of de Pagter, Dodds and Sukochev (see [10, Theorem 5.3]) stating that every Banach lattice $X$ with order continuous norm and weak unit has property (KA) when $\mathcal{A}$ is the family of all order bounded subsets of $X$. We sharpen those results by proving that, in fact, such Banach lattices have property $(\mu^{s}_{\mathcal{A}})$ when $\mathcal{A}$ is the family described in (1.1) or the family of all $L$-weakly compact sets, respectively (Corollary 5.2 and Theorem 5.1).

Finally, we focus on the Lebesgue-Bochner space $L^{1}(\nu,X)$, where $\nu$ is a finite measure and $X$ is a Banach space. It is known that if $X$ contains a subspace isomorphic to $c_{0}$, then $L^{1}([0,1],X)$ contains a complemented subspace isomorphic to $c_{0}$, see [14]. When applied to the space $\ell^{\infty}$, this shows that properties $(\mu^{s})$ and (K) do not pass from $X$ to $L^{1}([0,1],X)$ in general (cf. [17, Remark 6.5]). In fact, in Theorem 5.6 we prove that $L^{1}([0,1],X)$ fails property (KA), for the family $\mathcal{A}$ of all $\delta\mathcal{S}$-sets of $L^{1}([0,1],X)$, whenever $X$ contains a subspace isomorphic to $c_{0}$.

Definition 1.7**.**

A set $K\subseteq L^{1}(\nu,X)$ is said to be a $\delta\mathcal{S}$-set if it is uniformly integrable and for every $\delta>0$ there exists a weakly compact set $W\subseteq X$ such that $\nu(f^{-1}(W))\geq 1-\delta$ for every $f\in K$.

The collection of all $\delta\mathcal{S}$-sets of $L^{1}(\nu,X)$ will be denoted by $\delta\mathcal{S}(\nu,X)$ or simply $\delta\mathcal{S}$ if no confussion arises. These sets play an important role when studying weak compactness in Lebesgue-Bochner spaces. Any $\delta\mathcal{S}$-set of $L^{1}(\nu,X)$ is relatively weakly compact, while the converse is not true in general. For more information on these sets, see [37] and the references therein. Concerning positive results, we show that $L^{1}(\nu,X)$ has property $(\mu^{s}_{\delta\mathcal{S}})$ whenever $X$ is a subspace of a S2WCG space (Theorem 5.8). In general, the assumption that $X^{*}$ has the Banach-Saks property is not enough to ensure that $L^{1}(\nu,X)$ has property $(\mu^{s}_{\delta\mathcal{S}})$ (Example 5.9).

2. Notation and preliminaries

The symbol $|S|$ stands for the cardinality of a set $S$. All our vector spaces are real. Given a sequence $(f_{n})_{n}$ in a vector space, a convex block subsequence of $(f_{n})_{n}$ is a sequence $(g_{k})_{k}$ of the form

[TABLE]

where $(I_{k})_{k}$ is a sequence of finite subsets of $\mathbb{N}$ with $\max(I_{k})<\min(I_{k+1})$ and $(a_{n})_{n}$ is a sequence of non-negative real numbers such that $\sum_{n\in I_{k}}a_{n}=1$ for all $k\in\mathbb{N}$. An operator is a continuous linear map between Banach spaces. By a subspace of a Banach space we mean a closed linear subspace. Given a Banach space $X$, its norm is denoted by either $\|\cdot\|_{X}$ or simply $\|\cdot\|$, and we write $B_{X}=\{x\in X:\|x\|\leq 1\}$. The topological dual of $X$ is denoted by $X^{*}$ and the adjoint of an operator $T$ is denoted by $T^{*}$. The evaluation of $x^{*}\in X^{*}$ at $x\in X$ is denoted by either $x^{*}(x)$ or $\langle x^{*},x\rangle$. The weak (resp. weak∗) topology on $X$ (resp. $X^{*}$) is denoted by $w$ (resp. $w^{*}$).

Lemma 2.1**.**

Let $X$ be a Banach space and let $\mathcal{A}$ be a family of subsets of $X$. If $X$ has property $(\mu_{\mathcal{A}}^{s})$, then it also has property (KA).

Proof.

Bear in mind that if $(u_{n})_{n}$ is a sequence in a topological vector space which is Cesàro convergent to [math], then it admits a convex block subsequence converging to [math]. Indeed, define

[TABLE]

for every $k\in\mathbb{N}$. Then $(w_{k})_{w}$ is a convex block subsequence of $(u_{n})_{n}$ converging to [math], because $(v_{k})_{k}$ converges to [math] and $v_{k}=\frac{1}{2}(v_{k-1}+w_{k})$ for all $k\geq 2$. ∎

In particular, property $(\mu^{s})$ implies property (K). The converse is not true in general, since any reflexive space has property (K), there are reflexive spaces which fail the Banach-Saks property (see [2]) and, moreover, we have:

Proposition 2.2**.**

Let $X$ be a Banach space. The following statements are equivalent:

1. (i)

$X^{*}$* has the Banach-Saks property;* 2. (ii)

$X$* is reflexive and has property $(\mu^{s})$;* 3. (iii)

$X$* contains no subspace isomorphic to $\ell^{1}$ and has property $(\mu^{s})$.*

Proof.

(i)$\Rightarrow$(ii): Clearly, the Banach-Saks property of $X^{*}$ implies that $X$ has property $(\mu^{s})$. On the other hand, as we mentioned in the introduction, every space with the Banach-Saks property is reflexive.

The implication (ii)$\Rightarrow$(iii) is obvious, while (iii)$\Rightarrow$(ii) follows from Lemma 2.1 and the fact that any Banach space with property (K) and without subspaces isomorphic to $\ell^{1}$ is reflexive, see [1, Theorem 2.1].

Finally, (ii)$\Rightarrow$(i) follows from the fact that, if $X$ is reflexive, then $\mu(X^{*},X)$ agrees with the norm topology of $X^{*}$ and $B_{X^{*}}$ is $w^{*}$-sequentially compact. ∎

Proposition 2.3**.**

Let $L$ be a compact Hausdorff topological space. The following statements are equivalent:

1. (i)

$C(L)$* is Grothendieck (i.e. every $w^{*}$-convergent sequence in $C(L)^{*}$ is weakly convergent);* 2. (ii)

$C(L)$* has property $(\mu^{s})$.*

Proof.

$C(L)^{*}$ is isomorphic (in fact, order isometric) to the $L^{1}$-space of a non-negative measure, so it has the weak Banach-Saks property. The implication (i)$\Rightarrow$(ii) follows at once from this.

(ii)$\Rightarrow$(i): Apply Lemma 2.1 and the fact that $C(L)$ is Grothendieck if (and only if) it has property (K), see [1, Corollary 2.5]. ∎

A subspace $Y$ of a Banach space $X$ is said to be $w^{*}$-extensible in $X$ if every $w^{*}$-null sequence in $Y^{*}$ admits a subsequence which can be extended to a $w^{*}$-null sequence in $X^{*}$. It is easy to check that: (i) any complemented subspace is $w^{*}$-extensible; and (ii) if $B_{X^{*}}$ is $w^{*}$-sequentially compact, then any subspace is $w^{*}$-extensible in $X$ (see [41, Theorem 2.1]). The following is straightforward:

Remark 2.4**.**

Let $X$ be a Banach space and let $Y\subseteq X$ be a subspace which is $w^{*}$-extensible in $X$. Let $\mathcal{A}$ be a family of subsets of $Y$. If $X$ has property $(\mu^{s}_{\mathcal{A}})$, then $Y$ has property $(\mu^{s}_{\mathcal{A}})$ as well.

In general, the statement of Remark 2.4 is not true for arbitrary subspaces. For instance, $\ell^{\infty}$ has property $(\mu^{s})$ (by Proposition 2.3 and the fact that $\ell^{\infty}$ is Grothendieck) but $c_{0}$ does not (since it fails (K), see e.g. [24, Proposition C]).

3. Strong generation and property $(\mu^{s})$

Our first result in this section provides a sufficient condition for property $(\mu^{s})$.

Theorem 3.1**.**

Let $X$ be a Banach space which is strongly generated by an operator with Banach-Saks adjoint. Then every subspace of $X$ has property $(\mu^{s})$.

The proof of Theorem 3.1 requires two lemmas which will be used again later.

Lemma 3.2**.**

Let $X$ be a Banach space. Let $\mathcal{H}$ and $\mathcal{G}$ be two families of subsets of $X$ such that $\mathcal{H}$ is strongly generated by $\mathcal{G}$.

1. (i)

If $(x_{n}^{*})_{n}$ is a bounded sequence in $X^{*}$ converging to [math] uniformly on each element of $\mathcal{G}$, then it also converges to [math] uniformly on each element of $\mathcal{H}$. 2. (ii)

If $X$ has property $(\mu^{s}_{\mathcal{G}})$, then it also has property $(\mu^{s}_{\mathcal{H}})$.

Proof.

(ii) is an immediate consequence of (i) applied to the corresponding sequences of Cesàro means. For the proof of (i), let $c>0$ be a constant such that $\|x_{n}^{*}\|_{X^{*}}\leq c$ for all $n\in\mathbb{N}$. Fix $H\in\mathcal{H}$ and take any $\varepsilon>0$. Pick $G\in\mathcal{G}$ such that $H\subseteq G+\varepsilon B_{X}$ and choose $n_{0}\in\mathbb{N}$ such that $\sup_{x\in G}|x_{n}^{*}(x)|\leq\varepsilon$ for all $n\geq n_{0}$. Then $\sup_{x\in H}|x_{n}^{*}(x)|\leq(1+c)\varepsilon$ for all $n\geq n_{0}$. ∎

Lemma 3.3**.**

Let $X$ and $Y$ be Banach spaces and let $T:Y\to X$ be an operator such that $T^{*}$ is Banach-Saks. Then $X$ has property $(\mu^{s}_{\{T(B_{Y})\}})$.

Proof.

Let $(x_{n}^{*})_{n}$ be a $w^{*}$-null sequence in $X^{*}$. Since $T^{*}:X^{*}\to Y^{*}$ is Banach-Saks and $w^{*}$-$w^{*}$-continuous, there is a subsequence of $(x_{n}^{*})$, not relabeled, such that for every further subsequence $(x_{n_{k}}^{*})_{k}$ we have that $(T^{*}(x^{*}_{n_{k}}))_{k}$ is Cesàro convergent to [math] in norm, that is,

[TABLE]

This shows that $X$ has property $(\mu^{s}_{\{T(B_{Y})\}})$. ∎

Proof of Theorem 3.1.

Since any Banach-Saks operator is weakly compact, the space $X$ is SWCG. In particular, $X$ is weakly compactly generated and so $B_{X^{*}}$ is $w^{*}$-sequentially compact (see e.g. [12, p. 228, Theorem 4]). Therefore, every subspace is $w^{*}$-extensible in $X$ and so it suffices to prove that $X$ has property $(\mu^{s})$ (see Remark 2.4 and the paragraph preceding it). To this end, let $Y$ be a Banach space and let $T:Y\to X$ be an operator such that $T^{*}$ is Banach-Saks and $wk(X)$ is strongly generated by $T(B_{Y})$. By Lemmas 3.3 and 3.2(ii), $X$ has property $(\mu^{s})$. ∎

An operator between Banach spaces $T:Y\to X$ is said to be super weakly compact if the ultrapower $T^{\mathcal{U}}:Y^{\mathcal{U}}\to X^{\mathcal{U}}$ is weakly compact for every free ultrafilter $\mathcal{U}$ on $\mathbb{N}$. This is equivalent to being uniformly convexifying in the sense of Beauzamy [3], see e.g. [21, Theorem 5.1]. A Banach space $X$ is said to be strongly super weakly compactly generated (S2WCG) if it is strongly generated by a super weakly compact operator (see [35]). In general:

strongly generated by a super-reflexive space $\Longrightarrow$ S2WCG $\Longrightarrow$ strongly generated by an operator with Banach-Saks adjoint.

The first implication is clear. The second one holds because an operator is super weakly compact if and only if its adjoint is super weakly compact (see [3, Proposition II.4]) and any super weakly compact operator is Banach-Saks (see [4, Théorème 3]). An example of a S2WCG space which is not strongly generated by a super-reflexive space can be found in [35, Example 3.10]. The converse of the second implication above is neither true in general, even for reflexive spaces. Indeed, there are spaces with the Banach-Saks property which are not super-reflexive (see [31]), while every S2WCG reflexive space is super-reflexive (cf. [35, Theorem 1.9]).

Corollary 3.4**.**

Every subspace of a S2WCG Banach space has property $(\mu^{s})$.

Corollary 3.5**.**

Let $\nu$ be a finite measure. Then every subspace of $L^{1}(\nu)$ has property $(\mu^{s})$.

Remark 3.6**.**

The previous results are formulated in terms of “subspaces” because the corresponding classes of Banach spaces are not hereditary. Indeed, Mercourakis and Stamati (see [29, Theorem 3.9(ii)]) showed that there exist subspaces of $L^{1}[0,1]$ which are not SWCG.

The Erdös-Magidor Theorem 1.1 is also valid for Fréchet spaces and, under certain set theoretic assumptions, for other classes of locally convex spaces, see [33]. Along this way, we have the following:

Theorem 3.7**.**

Let $X$ be a Banach space which is strongly generated by a family $\mathcal{G}\subseteq wk(X)$ with $|\mathcal{G}|<\mathfrak{p}$ (e.g. a SWCG space). Then $X$ has property $(\mu^{s})$ if (and only if) every $w^{*}$-null sequence in $X^{*}$ admits a subsequence which is Cesàro convergent to [math] with respect to $\mu(X^{*},X)$.

To deal with the proof of Theorem 3.7 we need two lemmas. The first one is a standard diagonalization argument.

Lemma 3.8**.**

Let $\{\mathcal{S}_{\alpha}\}_{\alpha<\gamma}$ be a collection of families of infinite subsets of $\mathbb{N}$, where $\gamma$ is an ordinal with $\gamma<\mathfrak{p}$. Suppose that, for each $\alpha<\gamma$, we have:

1. (i)

if $B\subseteq\mathbb{N}$ is infinite and $B\setminus A$ is finite for some $A\in\mathcal{S}_{\alpha}$, then $B\in\mathcal{S}_{\alpha}$; 2. (ii)

every infinite subset of $\mathbb{N}$ contains an element of $\mathcal{S}_{\alpha}$.

Then every infinite subset of $\mathbb{N}$ contains an element of $\bigcap_{\alpha<\gamma}\mathcal{S}_{\alpha}$.

Proof.

Fix an infinite set $B\subseteq\mathbb{N}$. We will first construct by transfinite induction a collection $\{B_{\alpha}:\alpha<\gamma\}$ of subsets of $B$ with the following properties:

1. ($P_{\alpha}$)

$B_{\alpha}\in\mathcal{S}_{\alpha}$ for all $\alpha<\gamma$; 2. ($Q_{\alpha,\beta}$)

$B_{\beta}\setminus B_{\alpha}$ is finite whenever $\alpha\leq\beta<\gamma$.

For $\alpha=0$ we just use (ii) to select any subset $B_{0}$ of $B$ belonging to $\mathcal{S}_{0}$. Suppose now that $1\leq\gamma^{\prime}<\gamma$ and that we have already constructed a collection $\{B_{\alpha}:\alpha<\gamma^{\prime}\}$ of subsets of $B$ such that ($P_{\alpha}$) and ($Q_{\alpha,\beta}$) hold whenever $\alpha\leq\beta<\gamma^{\prime}$. In particular, for any finite set $I\subseteq\gamma^{\prime}$ the intersection $\bigcap_{\alpha\in I}B_{\alpha}$ is infinite. Since $\gamma^{\prime}<\mathfrak{p}$, there is an infinite set $B_{\gamma^{\prime}}\subseteq B$ such that $B_{\gamma^{\prime}}\setminus B_{\alpha}$ is finite for all $\alpha<\gamma^{\prime}$. Property (i) implies that $B_{\gamma^{\prime}}\in\bigcap_{\alpha<\gamma^{\prime}}\mathcal{S}_{\alpha}$. Now property (ii) implies that, by passing to a further subset of $B_{\gamma^{\prime}}$ if necessary, we can assume that ($P_{\gamma^{\prime}}$) holds. Clearly, ($Q_{\alpha,\beta}$) also holds for every $\alpha\leq\beta\leq\gamma^{\prime}$. This finishes the inductive construction.

Since $\gamma<\mathfrak{p}$ and for any finite set $I\subseteq\gamma$ the intersection $\bigcap_{\alpha\in I}B_{\alpha}$ is infinite, there is an infinite set $C\subseteq B$ such that $C\setminus B_{\alpha}$ is finite for every $\alpha<\gamma$. From (i) it follows that $C\in\bigcap_{\alpha<\gamma}\mathcal{S}_{\alpha}$. ∎

The second lemma will also be used in Section 4.

Lemma 3.9**.**

Let $\{E_{i}\}_{i\in I}$ be a family of topological vector spaces with $|I|<\mathfrak{p}$ and let $E:=\prod_{i\in I}E_{i}$ be equipped with the product topology. For each $i\in I$, we denote by $\rho_{i}:E\to E_{i}$ the $i$th-coordinate projection. Let $(u_{n})_{n}$ be a sequence in $E$ satisfying the following condition:

• ($\star$)

for every infinite set $A\subseteq\mathbb{N}$ and every $i\in I$ there is an infinite set $B\subseteq A$ such that the subsequence $(\rho_{i}(u_{n}))_{n\in C}$ is Cesàro convergent to [math] in $E_{i}$ for every infinite set $C\subseteq B$.

Then there is a subsequence of $(u_{n})_{n}$ such that all of its subsequences are Cesàro convergent to [math] in $E$.

Proof.

We will apply Lemma 3.8. For each $i\in I$, let $\mathcal{S}_{i}$ be the family of all infinite sets $A\subseteq\mathbb{N}$ such that for every infinite set $C\subseteq A$ the corresponding subsequence $(\rho_{i}(u_{n}))_{n\in C}$ is Cesàro convergent to [math] in $E_{i}$. It suffices to check that conditions (i) and (ii) of Lemma 3.8 hold for this choice.

Indeed, (ii) follows immediately from ($\star$). On the other hand, fix $i\in I$, $A\in\mathcal{S}_{i}$ and an infinite set $B\subseteq\mathbb{N}$ such that $B\setminus A$ is finite. To check that $B\in\mathcal{S}_{i}$, take any strictly increasing sequence $(n_{k})_{k}$ in $B$. There is $k_{0}\in\mathbb{N}$ such that $n_{k}\in A$ for all $k>k_{0}$, hence $(\rho_{i}(u_{n_{k}}))_{k>k_{0}}$ is Cesàro convergent to [math] in $E_{i}$ and so is $(\rho_{i}(u_{n_{k}}))_{k}$. This shows that $B\in\mathcal{S}_{i}$. Thus, condition (i) of Lemma 3.8 is also satisfied. ∎

Proof of Theorem 3.7.

To prove that $X$ has property $(\mu^{s})$ it suffices to check that it has property $(\mu^{s}_{\mathcal{G}})$ (Lemma 3.2(ii)). For each $G\in\mathcal{G}$, let $R_{G}:X^{*}\to C(G)$ be the operator given by $R_{G}(x^{*}):=x^{*}|_{G}$ (the restriction of $x^{*}$ to $G$).

Let $(x_{n}^{*})_{n}$ be a $w^{*}$-null sequence in $X^{*}$. Fix an infinite set $A\subseteq\mathbb{N}$ and $G\in\mathcal{G}$. Since $(R_{G}(x_{n}^{*}))_{n\in A}$ is bounded, we can apply the Erdös-Magidor Theorem 1.1 to find an infinite set $B\subseteq A$ such that either all subsequences of $(R_{G}(x_{n}^{*}))_{n\in B}$ are Cesàro convergent (to the same limit), or no subsequence of $(R_{G}(x_{n}^{*}))_{n\in B}$ is Cesàro convergent. The assumption on $X$ excludes the second possibility and ensures that all subsequences of $(R_{G}(x_{n}^{*}))_{n\in B}$ are Cesàro convergent to [math].

We can now apply Lemma 3.9 to the family of Banach spaces $\{C(G)\}_{G\in\mathcal{G}}$ and the sequence $(u_{n})_{n}$ in $\prod_{G\in\mathcal{G}}C(G)$ defined by $u_{n}:=(R_{G}(x_{n}^{*}))_{G\in\mathcal{G}}$. So, there is a subsequence of $(x_{n}^{*})_{n}$ such that all of its subsequences are Cesàro convergent to [math] uniformly on each $G\in\mathcal{G}$. This proves that $X$ has property $(\mu^{s}_{\mathcal{G}})$. ∎

4. $\ell^{p}$-sums

The $\ell^{p}$-sum ($1\leq p\leq\infty$) of a family of Banach spaces $\{X_{i}\}_{i\in I}$ is denoted by

[TABLE]

When $p\neq\infty$ we identify the dual of $(\bigoplus_{i\in I}X_{i})_{\ell^{p}}$ with $(\bigoplus_{i\in I}X^{*}_{i})_{\ell^{q}}$, where $q$ is the conjugate exponent of $p$, i.e. $1/p+1/q=1$, and for each $j\in I$ we denote by

[TABLE]

the $j$th-coordinate projections.

Lemma 4.1**.**

Let $\{X_{i}\}_{i\in I}$ be a family of Banach spaces and $X:=(\bigoplus_{i\in I}X_{i})_{\ell^{1}}$. Let $(x_{n}^{*})_{n}$ be a bounded sequence in $X^{*}$ such that for every $i\in I$ the sequence $(\rho_{i}(x_{n}^{*}))_{n}$ is $\mu(X_{i}^{*},X_{i})$-null. Then $(x_{n}^{*})_{n}$ is $\mu(X^{*},X)$-null.

Proof.

It is known that $wk(X)$ is strongly generated by the family $\mathcal{G}$ consisting of all weakly compact subsets of $X$ of the form

[TABLE]

where $J\subseteq I$ is finite and $W_{i}\in wk(X_{i})$ for every $i\in J$ (see e.g. [23, Lemma 7.2(ii)]). Clearly, $(x_{n}^{*})_{n}$ converges to [math] uniformly on each element of $\mathcal{G}$. From Lemma 3.2(i) it follows that $(x_{n}^{*})_{n}$ converges to [math] with respect to $\mu(X^{*},X)$. ∎

Theorem 4.2**.**

Let $\{X_{i}\}_{i\in I}$ be a family of Banach spaces having property $(\mu^{s})$. If $|I|<\mathfrak{p}$, then $(\bigoplus_{i\in I}X_{i})_{\ell^{1}}$ has property $(\mu^{s})$.

Proof.

Write $X:=(\bigoplus_{i\in I}X_{i})_{\ell^{1}}$. Let $(x_{n}^{*})_{n}$ be a $w^{*}$-null sequence in $X^{*}$. Then $(\rho_{i}(x_{n}^{*}))_{n}$ is $w^{*}$-null in $X_{i}^{*}$ for all $i\in I$. Since each $X_{i}$ has property $(\mu^{s})$, we can apply Lemma 3.9 to the family of locally convex spaces $\{(X^{*}_{i},\mu(X_{i}^{*},X_{i}))\}_{i\in I}$ and the sequence $(u_{n})_{n}$ in $\prod_{i\in I}X^{*}_{i}$ defined by $u_{n}:=(\rho_{i}(x_{n}^{*}))_{i\in I}$. Therefore, there is a subsequence of $(x_{n}^{*})_{n}$, not relabeled, such that for every further subsequence $(x^{*}_{n_{k}})_{k}$ and every $i\in I$, the sequence $(\rho_{i}(x^{*}_{n_{k}}))_{k}$ is Cesàro convergent to [math] with respect to $\mu(X_{i}^{*},X_{i})$. Now, apply Lemma 4.1 to the sequence of arithmetic means of any such subsequence $(x^{*}_{n_{k}})_{k}$ to conclude that $(x^{*}_{n_{k}})_{k}$ is Cesàro convergent to [math] with respect to $\mu(X^{*},X)$. ∎

The following lemma isolates an argument used in the proof of Theorem 2.15 in [1], which says that property (K) is preserved by arbitrary $\ell^{p}$-sums for $1<p<\infty$. We will use it to prove the same statement for property $(\mu^{s})$ (Theorem 4.4 below).

Lemma 4.3**.**

Let $\{X_{i}\}_{i\in I}$ be a family of Banach spaces, let $1<p<\infty$ and write $X:=(\bigoplus_{i\in I}X_{i})_{\ell^{p}}$. Let $(y_{j}^{*})_{j}$ be a sequence in $X^{*}$ such that:

1. (i)

for every $i\in I$ the sequence $(\rho_{i}(y^{*}_{j}))_{j}$ is $\mu(X_{i}^{*},X_{i})$-null; 2. (ii)

there is a norm convergent sequence $(\tilde{\xi}_{j})_{j}$ in $\ell^{q}(I)$ such that

[TABLE]

for all $j\in\mathbb{N}$.

Then $(y_{j}^{*})_{j}$ is $\mu(X^{*},X)$-null.

Proof.

Fix any weakly compact set $L\subseteq B_{X}$ and $\varepsilon>0$. Since $(\tilde{\xi}_{j})_{j}$ is norm convergent in $\ell^{q}(I)$, there exist a finite set $I_{0}\subseteq I$ and $j_{0}\in\mathbb{N}$ such that

[TABLE]

where $\psi_{i}\in\ell^{q}(I)^{*}$ denotes the $i$th-coordinate functional. Bearing in mind that

[TABLE]

from (4.1) we get

[TABLE]

The previous inequality and Hölder’s one imply that

[TABLE]

For each $i\in I$, the sequence $(\rho_{i}(y_{j}^{*}))_{j}$ converges to [math] uniformly on the weakly compact set $\pi_{i}(L)\subseteq X_{i}$. Thus, we can find $j_{1}>j_{0}$ such that

[TABLE]

Therefore, for every $j>j_{1}$ and $x\in L\subseteq B_{X}$ we have

[TABLE]

This shows that $(y_{j}^{*})_{j}$ is $\mu(X^{*},X)$-null. ∎

Theorem 4.4**.**

Let $\{X_{i}\}_{i\in I}$ be a family of Banach spaces having property $(\mu^{s})$ and let $1<p<\infty$. Then $(\bigoplus_{i\in I}X_{i})_{\ell^{p}}$ has property $(\mu^{s})$.

Proof.

Write $X:=(\bigoplus_{i\in I}X_{i})_{\ell^{p}}$. Let $(x_{n}^{*})_{n}$ be a $w^{*}$-null sequence in $X^{*}$. Define

[TABLE]

so that $\|v_{n}\|_{\ell^{q}(I)}=\|x_{n}^{*}\|_{X^{*}}$. Since $(v_{n})_{n}$ is bounded and $\ell^{q}(I)$ has the Banach-Saks property, there exist a subsequence of $(x_{n}^{*})_{n}$, not relabeled, such that all subsequences of $(v_{n})_{n}$ are Cesàro convergent in norm (to the same limit).

On the other hand, since every element of $X^{*}=(\bigoplus_{i\in I}X^{*}_{i})_{\ell^{q}}$ is countably supported, we can assume without loss of generality that $I$ is countable. Then, as in the proof of Theorem 4.2, we can find a subsequence of $(x_{n}^{*})_{n}$, not relabeled, such that for every further subsequence $(x_{n_{k}}^{*})_{k}$ and every $i\in I$, the sequence $(\rho_{i}(x^{*}_{n_{k}}))_{k}$ is Cesàro convergent to [math] with respect to $\mu(X_{i}^{*},X_{i})$.

We claim that any subsequence $(x_{n_{k}}^{*})_{k}$ is Cesàro convergent to [math] with respect to $\mu(X^{*},X)$. Indeed, define $y_{j}^{*}:=\frac{1}{j}\sum_{k=1}^{j}x_{n_{k}}^{*}$ for all $j\in\mathbb{N}$. We will show that $(y^{*}_{j})_{j}$ is $\mu(X^{*},X)$-null by checking that it satisfies conditions (i) and (ii) of Lemma 4.3. Obviously, (i) holds. On the other hand, for each $i\in I$ and $j\in\mathbb{N}$ we have

[TABLE]

where $\psi_{i}\in\ell^{q}(I)^{*}$ denotes the $i$th-coordinate functional. Hence, condition (ii) of Lemma 4.3 holds by taking $\tilde{\xi}_{j}:=\frac{1}{j}\sum_{k=1}^{j}v_{n_{k}}$ for all $j\in\mathbb{N}$. ∎

Remark 4.5**.**

The previous result provides examples of separable Banach spaces having property $(\mu^{s})$ which do not embed isomorphically into any SWCG space, like $\ell^{p}(\ell^{1})$ and $\ell^{p}(L^{1}[0,1])$ for $1<p<\infty$, see [25, Corollary 2.29].

The following example answers in the negative a question left open in [1, Problem 2.19]. It also shows that property $(\mu^{s})$ is not preserved by countable $\ell^{\infty}$-sums.

Example 4.6**.**

There is a sequence $(X_{n})_{n}$ of finite-dimensional Banach spaces such that $(\bigoplus_{n\in\mathbb{N}}X_{n})_{\ell^{\infty}}$ fails property (K).

Proof.

Johnson [22] proved the existence of a sequence $(X_{n})_{n}$ of finite-dimensional Banach spaces such that, for every separable Banach space $X$, its dual $X^{*}$ is isomorphic to a complemented subspace of $(\bigoplus_{n\in\mathbb{N}}X_{n})_{\ell^{\infty}}$.

Bearing in mind that property (K) is inherited by complemented subspaces, the fact that the space $(\bigoplus_{n\in\mathbb{N}}X_{n})_{\ell^{\infty}}$ fails property (K) follows from the existence of separable Banach spaces whose dual fails property (K), like $C[0,1]$ and the predual of the James tree space $JT$. Indeed, $C[0,1]^{*}$ is isomorphic to the $\ell^{1}$-sum of $\mathfrak{c}$ many copies of $L^{1}[0,1]$ (see e.g. [38, p. 242, Remark 5]) and so it fails property (K), according to Pełczyński’s example mentioned in the introduction. On the other hand, $JT$ is not reflexive and contains no subspace isomorphic to $\ell^{1}$, hence $JT$ fails property (K), see [1, Theorem 2.1]. ∎

5. Applications

5.1. Banach lattices

Given a Banach lattice $X$, we write

[TABLE]

Recall that a bounded set $A\subseteq X$ is said to be $L$-weakly compact if every disjoint sequence contained in $\bigcup_{x\in A}[-|x|,|x|]$ (the solid hull of $A$) is norm null. Every $L$-weakly compact set is relatively weakly compact, while the converse does not hold in general. $L$-weak compactness and relative weak compactness are equivalent for subsets of the $L^{1}$-space of a non-negative measure. More generally, if $X$ has order continuous norm, then a set $A\subseteq X$ is $L$-weakly compact if and only if it is approximately order bounded, i.e. for every $\varepsilon>0$ there is $x\in X^{+}$ such that $A\subseteq[-x,x]+\varepsilon B_{X}$. For more information on $L$-weakly compact sets, see [30, §3.6].

Theorem 5.1**.**

Let $X$ be a Banach lattice with order continuous norm and weak unit. Then $X$ has property $(\mu^{s}_{L_{w}(X)})$.

Proof.

Such a Banach lattice is order isometric to a Köthe function space over a finite measure space, see e.g. [27, Theorem 1.b.14]. So, we can assume that $X$ is a Köthe function space over a finite measure space, say $(\Omega,\Sigma,\nu)$. Let $i:L^{\infty}(\nu)\to X$ be the inclusion operator. Since $X$ has order continuous norm, $i^{*}(X^{*})\subseteq L^{1}(\nu)$ and so $i^{*}:X^{*}\to L^{1}(\nu)$ is $w^{*}$-$w$-continuous (see e.g. [27, p. 29]).

The order continuity of the norm also ensures that $L_{w}(X)$ is strongly generated by $i(B_{L^{\infty}(\nu)})$ (see e.g. [32, Lemma 2.37(iii)]). Therefore, in order to prove that $X$ has property $(\mu^{s}_{L_{w}(X)})$ we only have to check that $X$ has property $(\mu^{s}_{\{i(B_{L^{\infty}(\nu)})\}})$ (by Lemma 3.2(ii)). To this end, let $(x^{*}_{n})_{n}$ be a $w^{*}$-null sequence in $X^{*}$. Then $(i^{*}(x^{*}_{n}))_{n}$ is weakly null in $L^{1}(\nu)$, which has the weak Banach-Saks property. Hence there is a subsequence of $(x^{*}_{n})_{n}$, not relabeled, such that any further subsequence $(i^{*}(x^{*}_{n_{k}}))_{k}$ is Cesàro convergent to [math] in the norm of $L^{1}(\nu)$. In particular,

[TABLE]

This shows that $X$ has property $(\mu^{s}_{\{i(B_{L^{\infty}(\nu)}\})})$. ∎

If $X$ is a weakly sequentially complete Banach lattice, then

[TABLE]

Indeed, in this case $X$ has order continuous norm (see e.g. [30, Theorems 2.4.2 and 2.5.6]) and any operator $T:L^{1}[0,1]\to X$ is regular (see e.g. [30, Theorem 1.5.11]), so it maps approximately order bounded (i.e. relatively weakly compact) subsets of $L^{1}[0,1]$ to approximately order bounded (i.e. $L$-weakly compact) subsets of $X$. Thus, we get the following:

Corollary 5.2**.**

Let $X$ be a weakly sequentially complete Banach lattice with weak unit. Then $X$ has property $(\mu^{s}_{\mathcal{Q}(X)})$.

A Banach lattice is said to have the positive Schur property (PSP) if every weakly null sequence of positive vectors is norm null. Obviously, this property is satisfied by the $L^{1}$-space of any non-negative measure. The PSP implies weak sequential completeness (see e.g. [30, Theorem 2.5.6]) and so the order continuity of the norm. On the other hand, it is known that the PSP is equivalent to saying that every relatively weakly compact set is $L$-weakly compact (see e.g. [20, Theorem 3.14]). So, Theorem 5.1 implies that every Banach lattice with the PSP and weak unit has property $(\mu^{s})$. In fact, such a Banach lattice is S2WCG, as we show in Corollary 5.4 below. The key is the following lemma (cf. [18, Corollary 5.6]):

Lemma 5.3**.**

Let $X$ be a Köthe function space with order continuous norm over a finite measure space $(\Omega,\Sigma,\nu)$. Then the inclusion operator $i:L^{\infty}(\nu)\to X$ is super weakly compact.

Proof.

Since $X$ has order continuous norm, every order interval of $X$ is weakly compact (see e.g. [30, Theorem 2.4.2]). Hence $i$ is weakly compact and so is $i^{*}:X^{*}\to L^{1}(\nu)$ (see the proof of Theorem 5.1). Every weakly compact operator taking values in the $L^{1}$-space of a finite measure is super weakly compact, see e.g. [3, p. 123, Remarque] (cf. [18, Proposition 5.5]). Therefore, $i^{*}$ is super weakly compact and the same holds for $i$ (see [3, Proposition II.4]). ∎

Corollary 5.4**.**

Let $X$ be a Banach lattice with the positive Schur property and weak unit. Then $X$ is S2WCG.

Proof.

As in Theorem 5.1, we can assume that $X$ is a Köthe function space over a finite measure space $(\Omega,\Sigma,\nu)$. Then $X$ is strongly generated by the inclusion operator $i:L^{\infty}(\nu)\to X$ (bear in mind that every weakly compact subset of $X$ is $L$-weakly compact). On the other hand, $i$ is super weakly compact by Lemma 5.3. ∎

The previous corollary is an improvement of [6, Proposition 5.6], where it was shown that such Banach lattices are SWCG.

Remark 5.5**.**

The conclusion of Theorem 5.1 and Corollaries 5.2 and 5.4 can fail in the absence of weak unit. For instance, let $X$ be the $\ell^{1}$-sum of $\mathfrak{c}$ many copies of $L^{1}[0,1]$. Then $X$ has the PSP and fails property $(\mu_{\mathcal{Q}(X)}^{s})$, since it does not have property (k) (see [17, Example 4.I]).

5.2. Lebesgue-Bochner spaces

The first result of this subsection is based on the proof of Emmanuele’s result [14] on complemented copies of $c_{0}$ in Lebesgue-Bochner spaces, cf. [7, Theorem 4.3.2].

Theorem 5.6**.**

Let $X$ be a Banach space containing a subspace isomorphic to $c_{0}$. Then $L^{1}([0,1],X)$ fails property (KδS).

For the proof of Theorem 5.6 we need a lemma.

Lemma 5.7**.**

Let $X$ be a Banach space containing a $c_{0}$-sequence $(x_{n})_{n}$. Let $(h_{n})_{n}$ be a sequence of $\{-1,1\}$-valued measurable functions on $[0,1]$ and let $(I_{k})_{k}$ be a sequence of finite subsets of $\mathbb{N}$ with $\max(I_{k})<\min(I_{k+1})$ for all $k\in\mathbb{N}$. Then

[TABLE]

is a $\delta\mathcal{S}$-set of $L^{1}([0,1],X)$.

Proof.

It suffices to show that the set

[TABLE]

is relatively weakly compact in $X$. To this end, let $(y_{m})_{m}$ be a sequence in $C$. For each $m\in\mathbb{N}$ we write

[TABLE]

for some $k_{m}\in\mathbb{N}$ and $a_{n,m}\in\{-1,1\}$. By passing to a subsequence, not relabeled, we can assume that one of the following alternatives holds:

• •

There is $k\in\mathbb{N}$ such that $k_{m}=k$ for all $m\in\mathbb{N}$. In this case, $(y_{m})_{m}$ is a bounded sequence in a finite-dimensional subspace of $X$ and, therefore, it admits a norm convergent subsequence.

• •

$k_{m}<k_{m+1}$ for all $m\in\mathbb{N}$. In this case, since $(x_{n})_{n}$ is a $c_{0}$-sequence, the same holds for $(y_{m})_{m}$ and so it is weakly null.

Thus $C$ is relatively weakly compact, as required. ∎

Proof of Theorem 5.6.

We denote by $(e_{n})_{n}$ and $(e^{*}_{n})_{n}$ the usual bases of $c_{0}$ and $\ell^{1}$, respectively. Let $(x_{n})_{n}$ be a $c_{0}$-sequence in $X$ and let $(r_{n})_{n}$ be the sequence of Rademacher functions on $[0,1]$. Then $(r_{n}(\cdot)x_{n})_{n}$ is a $c_{0}$-sequence in $L^{1}([0,1],X)$ which spans a complemented subspace

[TABLE]

see e.g. the proof of Theorem 4.3.2 in [7]. Let $P:L^{1}([0,1],X)\to Z$ be a projection and let $T:Z\to c_{0}$ be the isomorphism satisfying $T(r_{n}(\cdot)x_{n})=e_{n}$ for all $n\in\mathbb{N}$. Consider the operator $S:=T\circ P:L^{1}([0,1],X)\to c_{0}$. Note that $(S^{*}(e_{n}^{*}))_{n}$ is a $w^{*}$-null sequence in $L^{1}([0,1],X)^{*}$.

Claim. $(S^{*}(e_{n}^{*}))_{n}$ does not admit convex block subsequences converging to [math] uniformly on each $\delta\mathcal{S}$-set. Indeed, let $(g_{k})_{k}$ be any convex block subsequence of $(S^{*}(e_{n}^{*}))_{n}$. Write $g_{k}=\sum_{n\in I_{k}}a_{n}S^{*}(e_{n}^{*})$, where $(I_{k})_{k}$ is a sequence of finite subsets of $\mathbb{N}$ with $\max(I_{k})<\min(I_{k+1})$ and $a_{n}\geq 0$ satisfy $\sum_{n\in I_{k}}a_{n}=1$. Then for each $k\in\mathbb{N}$ we have

[TABLE]

Hence $(g_{k})_{k}$ does not converge to [math] uniformly on the $\delta\mathcal{S}$-set

[TABLE]

(Lemma 5.7). This proves that $L^{1}([0,1],X)$ fails property (KδS). ∎

From now on $(\Omega,\Sigma,\nu)$ is a finite measure space. Given a Banach space $X$, the identity operator $i:L^{2}(\nu,X)\to L^{1}(\nu,X)$ strongly generates $L^{1}(\nu,X)$. Indeed, this can be checked as in the case of real-valued functions, bearing in mind that any weakly compact subset of $L^{1}(\nu,X)$ is uniformly integrable (see e.g. [13, p. 104, Theorem 4]). On the other hand, $L^{2}(\nu,X)$ is super-reflexive whenever $X$ is super-reflexive, see [9] (cf. [11, Ch. IV, Corollary 4.5]). In particular, $L^{1}(\nu,X)$ is strongly generated by a super-reflexive space (and so it has property $(\mu^{s})$, by Corollary 3.4) whenever $X$ is super-reflexive.

Theorem 5.8**.**

Let $X$ be a S2WCG Banach space and let $Z\subseteq X$ be a subspace. Then $L^{1}(\nu,Z)$ has property $(\mu^{s}_{\delta\mathcal{S}(\nu,Z)})$.

Proof.

The space $L^{1}(\nu,X)$ is weakly compactly generated because $X$ is (see e.g. [13, p. 252, Corollary 11]). Therefore, as in the proof of Theorem 3.1, it suffices to check that $L^{1}(\nu,X)$ has property $(\mu^{s}_{\delta\mathcal{S}(\nu,X)})$.

Let $Y$ be a Banach space which strongly generates $X$ through a super weakly compact operator $T:Y\to X$. We can assume that $Y$ is reflexive and $T$ is injective, according to Theorem 4.5 and Proposition 4.6 in [34]. Then $\delta\mathcal{S}(\nu,X)$ is strongly generated by

[TABLE]

(see [36, proof of Theorem 2.7]). Let $\tilde{T}:L^{2}(\nu,Y)\to L^{2}(\nu,X)$ be the “composition” operator defined by

[TABLE]

Let $i:L^{2}(\nu,X)\to L^{1}(\nu,X)$ be the identity operator and define $S:=i\circ\tilde{T}$. Since $Y$ is reflexive and $T$ is injective, we have $H\subseteq S(\rho B_{L^{2}(\nu,Y)})$ for $\rho:=(\nu(\Omega))^{1/2}$ (see [26, proof of Theorem 1]), hence $\delta\mathcal{S}(\nu,X)$ is strongly generated by $S(B_{L^{2}(\nu,Y)})$.

Since $T$ is super weakly compact, so is $\tilde{T}$ (see [3, p. 126, Corollaire]), hence $S$ is super weakly compact as well. Therefore, $S^{*}$ is super weakly compact (see [3, Proposition II.4]) and so $S^{*}$ is Banach-Saks (see [4, Théorème 3]). From Lemmas 3.3 and 3.2(ii) we conclude that $L^{1}(\nu,X)$ has property $(\mu^{s}_{\delta\mathcal{S}(\nu,X)})$. ∎

Our final example shows that, in the statement of Theorem 5.8, the S2WCG property cannot be replaced by the Banach-Saks property of the dual. It is known (and not difficult to check) that if $X$ is a reflexive Banach space, then every relatively weakly compact subset of $L^{1}(\nu,X)$ is a $\delta\mathcal{S}$-set, so in this particular setting properties $(\mu^{s})$ and $(\mu^{s}_{\delta\mathcal{S}})$ are equivalent.

Example 5.9**.**

There exists a Banach space $X$ such that $X^{*}$ has the Banach-Saks property but $L^{1}([0,1],X)$ fails property $(\mu^{s})$.

Proof.

Schachermayer [39] constructed an example of a Banach space $E$ having the Banach-Saks property such that $L^{2}([0,1],E)$ does not have it. The failure of the property is witnessed by a uniformly bounded weakly null sequence $(f_{n})_{n}$ in $L^{2}([0,1],E)$ (see [39], proof of Proposition 3).

Set $X:=E^{*}$ and let $i:L^{2}([0,1],X)\to L^{1}([0,1],X)$ be the identity operator. Since $E$ is reflexive, the same holds for $L^{2}([0,1],E)$ and we have

[TABLE]

see e.g. [13, IV.1]. Moreover, $(f_{n})_{n}$ is a $w^{*}$-null sequence in $L^{1}([0,1],X)^{*}$. No subsequence $(f_{n_{k}})_{k}$ is Cesàro convergent to [math] in the norm of $L^{2}([0,1],E)$, so it cannot be Cesàro convergent to [math] uniformly on the weakly compact set $i(B_{L^{2}([0,1],X)})$. This shows that $L^{1}([0,1],X)$ fails property $(\mu^{s})$. ∎

Acknowledgements

This research was supported by projects MTM2014-54182-P and MTM2017-86182-P (AEI/FEDER, UE).