TITLE: Is every Hilbert space a Banach algebra?
QUESTION [10 upvotes]: Let $H$ be a Hilbert space. Could we say that, always there is a multiplication on $H$, that makes it into a Banach algebra? If not, under which conditions does it exist?

REPLY [4 votes]: Another Banach-algebra structure on the Hilbert space $\mathsf{hs}(H)$ of Hilbert-Schmidt operators on a Hilbert space $H$ is just operator multiplication (composition). There is a natural involution on this algebra but it does not make it a C*-algebra.