Let $X$ be a separable Banach space and $1<p<\infty$. We say that a sequence $(x\_{n})\_{n}$ in $X$ is weakly $p$-convergent to $x\in X$ if the sequence $(x\_{n}-x)\_{n}$ is weakly $p$-summable. A subset $K$ of $X$ is said to be relatively weakly $p$-compact if $K$ is contained in $S(B\_{l\_{p^{\*}}})$ for some operator $S$ from $l\_{p^{\*}}$ into $X(\frac{1}{p}+\frac{1}{p^{\*}}=1)$. My question is: A subset $K$ of $X$ is relatively weakly $p$-compact if and only if every sequence in $K$ admits a weakly $p$-convergent subsequence? Thank you.



A characterization of relatively weakly $p$-compact sets
The answer is no even for separable $X$. Let $X=(\sum\_{n=1}^\infty \ell\_1^n)\_{p^\*}$. $B\_X$ has your $p$-subsequence property (same argument as for the unit ball of $\ell\_{p^\*}$). It is not relatively weakly $p$-compact because $X$ is not isomorphic to a quotient space of $\ell\_{p^\*}$.


