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!



Extending compact operators
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.


