Let $X$ be a Banach space and $T$ be a compact operator from $c_{0}$ into $X^{*}$. Let $\epsilon>0$. Is there a compact operator $S$ from $X$ into $l_{1}$ such that $S^{*}|_{c_{0}}=T$ and $\|S\|\leq (1+\epsilon)\|T\|$?
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asked Oct 3, 2015 at 16:32
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2$\begingroup$ Yes. If $T:Y\to X^*$ is even just weakly compact, then $T^{**}$ maps $Y^{**}$ into $X^*$ and extends $T$. In a first course in functional analysis one should learn that an bounded linear operator $S:V\to U$ is weakly compact if and only if $S^{**}V^{**} \subset U^{**}$. So this question is arguably more appropriate for another site. (I say ``arguably" because such a basic fact is probably hard to find in text books on real analysis.) $\endgroup$– Bill JohnsonCommented Oct 3, 2015 at 18:47
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$\begingroup$ I forget the basic fact mainly because I have never taught functional analysis in my university. In fact, in my question, we can let $S=T^{*}J_{X}$, where $J_{X}:X\rightarrow X^{**}$ is the canonical embedding. Anyway, thanks, Bill. $\endgroup$– Dongyang ChenCommented Oct 3, 2015 at 21:34
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