# Composition of bounded operator and compact operators

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?

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.