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Let $1\leq p<\infty$. We say that an operator $T:X\rightarrow Y$ is unconditionally $p$-converging if $T$ takes a weakly $p$-summable sequence to a norm null sequence.

Question: Is every unconditionally $p$-converging operator $T$ from $L_{1}[0,1]$ to every Banach space $Y$ weakly compact?

It is known that this question is false for $p=1$. But I do not know whether it is true for $1<p<\infty$.

Thank you!

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No. Take a projection $P$ from $L_1$ onto a subspace isometrically isomorphic to $\ell_1$ and use the fact that $\ell_1$ has the Schur property.

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  • $\begingroup$ Is there a space $X$ such that my question is true for some $1<p<\infty$, but is false for $p=1$. $\endgroup$
    – Dongyang Chen
    Apr 12, 2016 at 23:56
  • $\begingroup$ I don't know the answer. $\endgroup$
    – Bill Johnson
    Apr 13, 2016 at 1:10

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