Are (L)-sets in dual Banach spaces weak* pre-compact or weak* sequentially pre-compact?

Question

Let $X$ be a Banach space. A subset $B$ of the dual $X$ is said to be $(L)$ set if any weakly null sequence $(x_n)\in X$ converges uniformly to zero on $B$.

It is well Known in the theory that Dunford-Pettis sets (i.e sets of $X$ on which any weakly null sequnece $(f_n) \in X'$ converges uniformly to zero) are weaky precompact. I want to know if the dual counterpart of this result still holds, that is, any $(L)$ set $B \in X$ is weak* precompact? or weak* sequentially precompact? I searched where located the proof of the result for Dunford-Pettis sets but I have found nothing. I feel that the result for $(L)$ sets is true but I can not prove it.

Many thanks in advance for your help.

Edit: Although the conclusion in the comment of Daniel Fischer is true, I dont know if for every sequence $\left( f_{n}\right) \subset B$ one can find a subsequence $\left( f_{n_{k}}\right) $ which is weak* de Cauchy, that is, the sequence $\left( f_{n_{k}}\left( x\right) \right) $ is de Cauchy in the scalar field for every $x \in X$. In other words, does the weak* precompacity implies the sequential weak* one? It is well known in the theory that this is not the case for the compacity if the topology of the space is not metrisable; see Sequentially compact space.

Thanks in advances.

Answer 1

This is not a complete solution, but perhaps just a step forward.

If $A\subseteq X^{\ast}$ is an L-set, which is not weak* sequentially compact, then $A$ contains an $\ell^1$ sequence $(f_n)$, which has no weak* convergent subsequence by Rosenthal's theorem. Conversely, if $(f_n)$ is an $\ell^1$ sequence in $X^{\ast}$, which has no weak* convergent subsequence, then $\{f_n:n\in\mathbb{N}\}$ is an L-set that is not weak* sequentially compact.

So, the problem reduces to whether there exists such a sequence in $X^{\ast}$.

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A few notes in addition: Suppose $X^{\ast}$ contains such a sequence. Then,

1. $X$ is not separable, for bounded subsets of $X^{\ast}$ are weak* seq. cpt. for separable Banach spaces.
2. $X$ contains a copy of $\ell^1$, and $X^{\ast}$ contains a copy of $M([0,1])$ - the space of bounded Radon measures on $[0,1]$.
3. The operator $T:X\to\ell^{\infty}$, $Tx = (f_n(x))$ is completely continuous, but $T^{**}:X^{**}\to(\ell^{\infty})^{**}$, $T^{**}u = (u(f_n))$ is not completely continuous. In fact, $T$ is c.c. iff $\{f_n:n\in\mathbb{N}\}$ is an L-set for $X$, and $T^{**}$ is c.c. iff $\{f_n:n\in\mathbb{N}\}$ is a DP-set for $X^{**}$ (so weak* seq. cpt. in $X^{*}$)
4. A Banach space $Y$ has reciprocal Dunford-Pettis property (RDPP) iff every L-set for $Y$ is relatively weakly compact in $Y^{*}$ (thus relatively weak* sequentially compact). Therefore, $X$ does not have RDPP.