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 "top-level" tags like fa.functional-analysis, or you can use the tag banach-spaces. $\endgroup$ – Yemon Choi Sep 20 '15 at 0:47
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    $\begingroup$ Yes! If you want to use MO, please use it properly. $\endgroup$ – Bill Johnson Sep 20 '15 at 3:39
  • $\begingroup$ Thank you for your advice, Yemon. I'll use the tag banach-sapces later. $\endgroup$ – Dongyang Chen Sep 20 '15 at 16:19

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. 3-4, 155–176.

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    $\begingroup$ Typo in second sentence, unless Anthony Bourdain has branched out even further... $\endgroup$ – Yemon Choi Sep 20 '15 at 20:16
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    $\begingroup$ Thanks, Yemon. Jean is better at cooking than Anthony is at math, but only marginally so. :) $\endgroup$ – Bill Johnson Sep 20 '15 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 $X=c_{0},l_{p}(1<p<\infty)$. $\endgroup$ – Dongyang Chen Sep 21 '15 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 Albiac-Kalton has this). $\endgroup$ – Bill Johnson Sep 21 '15 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 $l_{1}$". $\endgroup$ – Dongyang Chen Sep 21 '15 at 14:00

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