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.

$\begingroup$ You should add that $X$ is separable or replace "$B_{\ell_p^*}$" by "$B_{\ell_p^*(\Gamma)}$ for some $\Gamma$". $\endgroup$ – Bill Johnson Nov 15 '15 at 15:49

$\begingroup$ Is my question is true if $X$ is separable? $\endgroup$ – Dongyang Chen Nov 15 '15 at 16:00
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 $\ell_{p^*}$.

$\begingroup$ If $B_{X}$ is relatively weakly $p$compact, then it seems that $X$ is only isomorphic to a complemented subspace of a quotient of $l_{p^{*}}$. Moreover, I do not understand why $X$ is not isomorphic to a quotient of $l_{p^{*}}$. $\endgroup$ – Dongyang Chen Nov 16 '15 at 15:27

$\begingroup$ If $TB_Y \supset B_X$ then $T$ is surjective and $X$ is isomorphic to quotient of $Y$. What is there not to understand? $\endgroup$ – Bill Johnson Nov 16 '15 at 16:18

$\begingroup$ If $X$ is isomorphic to a quotient of $Y$, then $X^*$ is isomorphic to a subspace of $Y^*$. That is not the case herethe relevant $Y$ is super reflexive and the relevant $X$ is not. $\endgroup$ – Bill Johnson Nov 16 '15 at 16:21

$\begingroup$ Does every sequence in $B_{c_{0}}$ admit a weakly $1$summable subsequence? $\endgroup$ – Dongyang Chen Nov 16 '15 at 22:55

$\begingroup$ I guess you mean weakly 1convergent rather than weakly 1summable. The answer is no; consider the summing basis for $c_0$. $\endgroup$ – Bill Johnson Nov 17 '15 at 22:44