I understand that $\mathscr{K}(\mathscr{H})$ is a closed subspace of $\mathscr{B}(\mathscr{H})$. Where $\mathscr{K}$ is the compact operators and $\mathscr{H}$ is the bounded operators.
I need to prove that if $S \in \mathscr{H}$ and $T \in \mathscr{K}$ then ST is compact. I started by defining D(ST) = {u| $u \in D(T), Tu \in D(S)$}.
I started to say that this implies that D(S) is closed due to the compactness of T, but I don't think that is necessarily right since I do not know for sure that D(S) includes all possible Tu. If D(S) is closed, then S itself is closed because it is bounded.
But if D(S) was closed, would closed + bounded = convex?
If everything is wrong, please correct my reasoning and maybe offer a hint as to the correct path for a solution. Much appreciated
To show the question, it is quite straightforward to just use the sequential definition of compact operators.
As for your claims about the domain, if you are given $S,T\in\mathcal{B(H)}$, are you not already assuming the domain to be the whole Hilbert space?
Anyways closed+ bounded = convex is definiely wrong (take any disconnected closed and bounded set for example)