Answer to Question #38357 in Functional Analysis for sima

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

Yes. Look at the following example. Take any infinite dimensional Banach spaces X and Y. Let S∈K(X), then restriction of T:=S⊕∞1Y∈B(X⊕∞Y) to X is S which is compact by construction.