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.
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asked Nov 14, 2015 at 23:42
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$\begingroup$ You should add that $X$ is separable or replace "$B_{\ell_p^*}$" by "$B_{\ell_p^*(\Gamma)}$ for some $\Gamma$". $\endgroup$– Bill JohnsonCommented Nov 15, 2015 at 15:49
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$\begingroup$ Is my question is true if $X$ is separable? $\endgroup$– Dongyang ChenCommented Nov 15, 2015 at 16:00
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1 Answer
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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 $\ell_{p^*}$.
answered Nov 16, 2015 at 13:59
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$\begingroup$ 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 $l_{p^{*}}$. $\endgroup$ Commented Nov 16, 2015 at 15:27
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$\begingroup$ If $TB_Y \supset B_X$ then $T$ is surjective and $X$ is isomorphic to quotient of $Y$. What is there not to understand? $\endgroup$ Commented Nov 16, 2015 at 16:18
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$\begingroup$ If $X$ is isomorphic to a quotient of $Y$, then $X^*$ is isomorphic to a subspace of $Y^*$. That is not the case here--the relevant $Y$ is super reflexive and the relevant $X$ is not. $\endgroup$ Commented Nov 16, 2015 at 16:21
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$\begingroup$ Does every sequence in $B_{c_{0}}$ admit a weakly $1$-summable subsequence? $\endgroup$ Commented Nov 16, 2015 at 22:55
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$\begingroup$ I guess you mean weakly 1-convergent rather than weakly 1-summable. The answer is no; consider the summing basis for $c_0$. $\endgroup$ Commented Nov 17, 2015 at 22:44