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Let $X$ be a separable, infinite-dimensional complex Banach space and $Y\subseteq X$ an infinite-dimensional closed subspace. Suppose $K:Y\to X$ is an arbitrary compact operator. I would like to find an infinite-dimensional closed subspace $Z\subseteq Y$ such that the restriction $K|_Z$ has a compact extension $\widetilde{K}$ to all of $X$.

In other words, I would like to find a compact operator $\widetilde{K}:X\to X$ such that $\widetilde{K}z=Kz$ for all $z\in Z$.

Thanks to Lindenstrauss, it is known that if $X^*=L_1(\mu)$ for some measure $\mu$ then $K$ has a compact extension. However, what I need is weaker. It suffices for my purposes to find an infinite-dimensional restriction with a compact extension.

Recall that $X$ is called subprojective just in case every infinite-dimensional closed subspace $Y$ admits a further infinite-dimensional closed subspace $Z\subseteq Y$ which is complemented in $X$. So, if $X$ is subprojective then obviously I get what I want.

What if $X$ is not subprojective? Can we still get some suitable $\widetilde{K}$? Can we always get a $\widetilde{K}$?

Probably this is already known, which is the reason for my question. Thanks guys!

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You can always extend a nuclear operator even to a nuclear operator. Every compact operator is nuclear on some infinite dimensional subspace, so your question has a positive answer.

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  • $\begingroup$ Perfect! Thank you very much. I think this solves a problem I've been working on. If so, I will let you know. $\endgroup$ – Ben W Oct 15 '14 at 18:10

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