restriction a non compact operator to compact operator

Question

If $T\in\mathcal{B}(X,Y)$ is not compact can the restriction of $T$ to an infinite dimensional subspace of $X$ be compact?

Answer 1

Of course, take any operator $T$ with infinite dimensional kernel, then $T|_{\ker T}=0$ is compact.

More smart example. Take any infinite dimensional Banach spaces $X$ and $Y$. Let $S\in\mathcal{K}(X)$, then restriction of $T:=S\oplus_\infty 1_Y\in\mathcal{B}(X\oplus_\infty Y)$ to $X$ is $S$ which is compact by construction.