A sequence $(x_{n})_{n}$ in a Banach space $X$ is said to be unconditionally $p$summable if $$\sup_{x^{*}\in B_{X^{*}}}\Bigl(\sum_{n=m}^{\infty}\lvert\langle x^{*},x_{n}\rangle\rvert^{p}\Bigr)^{1/p}\rightarrow 0\qquad(m\rightarrow \infty).$$ We say that an operator $T:X\rightarrow Y$ is unconditionally $p$summing if $(Tx_{n})_{n}$ is unconditionally $p$summable in $Y$ whenever $(x_{n})_{n}$ is weakly $p$summable in $X$. We can prove that $T$ is unconditionally $p$summing whenever $T^{**}$ is unconditionally $p$summing. Is the converse true?

$\begingroup$ It is usual for people to add one of the "toplevel" tags like fa.functionalanalysis, or you can use the tag banachspaces. $\endgroup$– Yemon ChoiCommented Sep 20, 2015 at 0:47

2$\begingroup$ Yes! If you want to use MO, please use it properly. $\endgroup$– Bill JohnsonCommented Sep 20, 2015 at 3:39

$\begingroup$ Thank you for your advice, Yemon. I'll use the tag banachsapces later. $\endgroup$– Dongyang ChenCommented Sep 20, 2015 at 16:19
1 Answer
The answer is no. Bourgain and Delbaen constructed a Banach space $X$ that has the Schur property and $X^{**}$ is isomorphically universal for separable Banach space (it is even isomorphic to the the second dual of $C[0,1]$). Every operator from $\ell_p$, $1<p<\infty$, and from $c_0$ into $X$ is thus compact, so that every operator with domain $X$ is unconditionally $p$summing for all $1\le p < \infty$. But the second adjoint of the identity operator on $X$ is not unconditionally $p$summing for any $1\le p < \infty$ because it is an isomorphism on copies of $\ell_p$ for all $p$.
Bourgain, J.; Delbaen, F. A class of special $L_\infty$ spaces. Acta Math. 145 (1980), no. 34, 155–176.

1$\begingroup$ Typo in second sentence, unless Anthony Bourdain has branched out even further... $\endgroup$ Commented Sep 20, 2015 at 20:16

4$\begingroup$ Thanks, Yemon. Jean is better at cooking than Anthony is at math, but only marginally so. :) $\endgroup$ Commented Sep 20, 2015 at 20:46

$\begingroup$ Thanks, Bill. If $X$ is a Banach space such that $X^{*}$ does not contain isomorphic copy of $l_{1}$, is any operator from $c_{0}$ to $X^{*}$ compact? At least, this is true for $X=c_{0},l_{p}(1<p<\infty)$. $\endgroup$ Commented Sep 21, 2015 at 0:14

$\begingroup$ Yes. If $Y$ does not contain an isomorphic copy of $c_0$, then every operator from $c_0$ to $Y$ is compact. But if $X^*$ contains a copy of $c_0$, then $X$ contains a complemented copy of $\ell_1$ (an old result of Pelczynski that is in standard textbooks; probably AlbiacKalton has this). $\endgroup$ Commented Sep 21, 2015 at 0:35

$\begingroup$ "$X^{*}$ does not contain a copy of $l_{1}$" should be corrected to be "$X$ does not contain a copy of $l_{1}$". $\endgroup$ Commented Sep 21, 2015 at 14:00