restriction a non compact operator to compact operator If  $T\in\mathcal{B}(X,Y)$ is not compact can the restriction of $T$ to an infinite dimensional subspace of $X$ be compact?
 A: 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.
A: What the OP probably has is mind, is the question under which circumstances, for a given operator $T\colon X\to Y$ one can find a subspace of $X_0$ of $X$ which is infinite-dimensional and on which $T$ is compact. 
This is true for operators which are not bounded below on finite-codimensional subspaces of $X$ (in other words, for operators $T$ which are not upper semi-Fredholm). In this case we have Kato's lemma which asserts that for every $\varepsilon > 0$ you will find a basic sequence $(x_n)_{n=1}^\infty$ in $X$ so that $T$ restricted to $X_0:=\overline{{\rm span}}\{x_n\colon n\in \mathbb{N}\}$ is compact and $\|T|_{X_0}\|\leqslant \varepsilon$. 
This fact is necessary to prove that sum of two strictly singular operators is again strictly singular.
