Composition of bounded operator and compact operators

Question

In Hilbert space $H$, is it true that the composition of operators $ST$ and $TS$ of the bounded operator $S$ and the compact operator $T$ are compact?

Answer 1

Yes (and more generally in a Banach space). If $B$ is the closed unit ball, $\overline{TB}$ is compact, so $STB \subseteq S(\overline{TB})$ which is compact (in a Hausdorff space, a continuous image of a compact set is compact). Therefore $ST$ is a compact operator.

$SB \subseteq \|S\| B$, so $TSB \subseteq \|S\| TB \subseteq \|S\| \overline{TB}$ which is compact, so $TS$ is a compact operator.