Suppose that $T$ is an operator from a Banach space $X$ to a Banach space $Y$. Let $1<p<q<\infty$. If $TS$ is compact for any operator $S:l_{p}\rightarrow X$, is $TR$ compact for any operator $R:l_{q}\rightarrow X$?
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asked Sep 27, 2015 at 14:36
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Yes. If $R:\ell_q \to X$ and $TR$ is not compact then there is a normalized block basis $u_n$ of the unit vector basis for $\ell_q$ s.t. $TRu_n$ is bounded away from zero. Let $V:\ell_p \to \ell_q$ be the bounded linear operator that maps $e_n$ to $u_n$ and set $S=RV$. Then $TS$ is not compact.
answered Sep 27, 2015 at 15:14
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$\begingroup$ By the proof of Pitt's Theorem , if $TR$ is non-compact, we can get a normalized basic sequence $(x_{n})_{n}$ that is equivalent to a block basis $(u_{n})_{n}$ of the unit vector basis of $l_{q}$ such that $TRx_{n}$ is bounded away from zero. I do not know how to get $TRu_{n}$ is bounded away from zero. $\endgroup$ Commented Sep 27, 2015 at 15:28
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$\begingroup$ Let $P$ be a projection from $l_{q}$ onto $\overline{span}\{u_{n}:n=1,2,...\}$ and $A$ be an isomorphism from $\overline{span}\{u_{n}:n=1,2,...\}$ onto $\overline{span}\{x_{n}:n=1,2,...\}$ such that $Au_{n}=x_{n}$ for all $n$. Let $\widetilde{R}=RAP:l_{q}\rightarrow X$. Then $T\widetilde{R}(u_{n})=TRx_{n}$. We can replace $R$ by $\widetilde{R}$. $\endgroup$ Commented Sep 27, 2015 at 16:19