TITLE: Extending compact operators
QUESTION [6 upvotes]: 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!

REPLY [6 votes]: 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.