# Weakly null sequences in Banach spaces

Every weakly null sequence in a Banach space, as a subset, is clearly relatively weakly compact. To quantify the elementary fact, we need the following quantities:

$$\delta_{0}((x_{n})_{n}):=\sup_{x^{*}\in B_{X^{*}}}\limsup_{n}|\langle x^{*},x_{n}\rangle |$$ for a bounded sequence $$(x_{n})_{n}$$ of a Banach space $$X$$, where $$B_{X^{*}}$$ is the closed unit ball of $$X^{*}$$.

For a bounded subset $$A$$ of $$X$$ we set: $$\operatorname{wck}_{X}(A)=\sup\{\textrm{d}(\textrm{clust}_{X^{**}}((x_{n})_{n}),X)\colon (x_{n})_{n}\text{ is a sequence in }A\},$$ where $$\textrm{clust}_{X^{**}}((x_{n})_{n})=\cap_{n=1}^{\infty}\overline{\{x_{m}:m>n\}}^{w^*}$$ is the set of all weak$$^{*}$$-cluster points of $$(x_{n})_{n}$$ in $$X^{**}$$. It follows from the Eberlein-Smulyan theorem that $$\operatorname{wck}_{X}(A)=0$$ if and only if $$A$$ is relatively weakly compact.

Let $$A$$ and $$B$$ be non-empty subsets of a Banach space $$X$$, we set $$\textrm{d}(A,B)=\inf\{\|a-b\|\colon a\in A,b\in B\},$$ $$\widehat{\textrm{d}}(A,B)=\sup\{\textrm{d}(a,B)\colon a\in A\}.$$ $$\textrm{d}(A,B)$$ is the ordinary distance between $$A$$ and $$B$$, and $$\widehat{\textrm{d}}(A,B)$$ is the (non-symmetrised) Hausdorff distance from $$A$$ to $$B$$. When $$A$$ is a bounded subset of $$X$$, we set $$\textrm{wk}_{X}(A)=\widehat{\textrm{d}}\big(\overline{A}^{\sigma(X^{**},X^{*})},X\big).$$

It is a direct consequence of the Banach-Alaoglu theorem that $$A$$ is relatively weakly compact if and only if $$\textrm{wk}_{X}(A)=0$$.

Question 1. $$\operatorname{wck}_{X}(\{x_{n}:n=1,2,\cdots\})\leq \delta_{0}((x_{n})_{n})$$ for every bounded sequence $$(x_{n})_{n}$$ of a Banach space $$X$$ ?

Question 2. $$\operatorname{wk}_{X}(\{x_{n}:n=1,2,\cdots\})\leq \delta_{0}((x_{n})_{n})$$ for every bounded sequence $$(x_{n})_{n}$$ of a Banach space $$X$$ ?

Thank you !

• for a bounded sequence $(x_n)$, let $C\subseteq X^{**}$ be the set of its weak$^*$ cluster points. Isn't $\delta_0 = \sup\{ \|\mu\|_{X^{**}} : \mu\in C \}$ and $\textrm{wck}_X = \sup\{ \|\mu+X\|_{X^{**}/X} : \mu\in C\}$ ? If so, the inequality is evident. Jun 9 at 5:41
• I'm a bit confused: I assume by a cluster point of a sequence you mean the limit of a convergent subnet of the sequence? If so, you don't need the Eberlein-Smulyan theorem to see the characterization of weak compactness that you mention. Jun 9 at 6:41
• @OnurOktay In your argument, the second equality seems to be false because in my question, the sequence $(x_{n})_{n}$ is considered as a subset $A$. Jun 9 at 8:07
• @JochenGlueck A cluster point of a sequence does not mean the limit of a convergent subnet of the sequence. I add the definition of the set of all weak*-cluster points of a sequence in my question. Jun 9 at 10:21
• What are the definitions of $d$ and $\hat{d}$? Jun 9 at 16:29

I answer Question 1 by myself and I am sure my proof is correct.

Let $$A=\{x_{n}:n=1,2,\cdots\}$$ and let $$0 be arbitrary. Then there exists a sequence $$(z_{n})_{n}$$ in $$A$$ so that $$\textrm{d}(\textrm{clust}_{X^{**}}((z_{n})_{n}),X)>c$$. Let $$Y=\overline{\textrm{span}}\{x_{n}\colon n=1,2,\ldots\}$$ and $$i_{Y}\colon Y\rightarrow X$$ be the inclusion map. Since $$i_{Y}^{**}\colon Y^{**}\rightarrow X^{**}$$ is an isometric embedding, we get $$\textrm{d}(\textrm{clust}_{Y^{**}}((z_{n})_{n}),Y)\geqslant \textrm{d}(\textrm{clust}_{X^{**}}((z_{n})_{n}),X)>c.$$ Let $$\varepsilon>0$$. Take any $$y^{**}_{0}\in \textrm{clust}_{Y^{**}}((z_{n})_{n})$$ and let $$d=\textrm{d}(y^{**}_{0},Y)$$. By the Hahn-Banach theorem, there exists $$y^{***}_{0}\in S_{Y^{***}}$$ so that $$\langle y^{***}_{0},y^{**}_{0}\rangle=d$$ and $$\langle y^{***}_{0},y\rangle=0$$ for all $$y\in Y$$. We let $$C=B_{Y^{*}}\cap \{y^{***}\in Y^{***}\colon |\langle y^{***},y^{**}_{0}\rangle-d|<\varepsilon\}.$$ By Goldstine's theorem, $$y^{***}_{0}\in \overline{C}^{\sigma(Y^{***},Y^{**})}$$. Since $$\langle y^{***}_{0},y\rangle=0$$ for all $$y\in Y$$, we get $$0\in \overline{C}^{\sigma(Y^{*},Y)}$$. Since $$Y$$ is separable, there exists a weak$$^{*}$$ null sequence $$(f_{m})_{m}$$ in $$C$$. By passing to a subsequence, we may assume that the limit $$\lim\limits_{m}\langle y^{**}_{0},f_{m}\rangle$$ exists, which is denoted by $$a$$. By the definition of $$C$$, $$|a-d|\leqslant \varepsilon$$. Since $$y^{**}_{0}\in \textrm{clust}_{X^{**}}((z_{n})_{n})$$, we get a subsequence $$(y_{n})_{n}$$ of $$(z_{n})_{n}$$ so that $$|\langle y^{**}_{0}-y_{n},f_{m}\rangle|<\frac{1}{n}$$ for $$m=1,2,\ldots,n$$. This implies that $$\lim\limits_{n\to\infty}\langle f_{m},y_{n}\rangle=\langle y^{**}_{0},f_{m}\rangle$$ for each $$m$$ and then $$\lim\limits_{m\to\infty}\lim\limits_{n\to\infty}\langle f_{m},y_{n}\rangle=a$$.

Claim. $$|a|\leq \delta_{0}((x_{n})_{n})$$.

Case 1. the set $$\{y_{n}:n=1,2,\cdots\}$$ is finite.

In this case, there exists a subsequence $$(y_{k_{n}})_{n}$$ and $$y_{0}\in Y$$ so that $$y_{k_{n}}=y_{0}$$ for all $$n$$. Hence $$a=\lim_{m}\lim_{n}\langle f_{m},y_{k_{n}}\rangle=\lim_{m}\langle f_{m},y_{0}\rangle=0.$$ The claim holds trivially.

Case 2. the set $$\{y_{n}:n=1,2,\cdots\}$$ is infinite.

In this case, we get two strictly increasing sequences $$(k_{i})_{i},(l_{i})_{i}$$ of positive integers so that $$y_{k_{i}}=x_{l_{i}}(i=1,2,\cdots)$$. Hence, for each $$m$$, we get $$|\langle y^{**}_{0},f_{m}\rangle|=\lim\limits_{i\to\infty}|\langle f_{m},y_{k_{i}}\rangle|\leq \delta_{0}^{Y}((x_{n})_{n}):=\sup_{y^{*}\in B_{Y^{*}}}\limsup_{n}|\langle y^{*},x_{n}\rangle|.$$ By the Hahn-Banach theorem, we get $$|a|\leq \delta_{0}((x_{n})_{n})$$.

By Claim, we get $$d\leq \delta_{0}((x_{n})_{n})+\varepsilon$$ and so is $$c$$. Letting $$\varepsilon\rightarrow 0$$, we get $$c\leq \delta_{0}((x_{n})_{n})$$. As $$c$$ was arbitrary, the proof is complete.