Is every Hilbert space a Banach algebra?

Question

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?

Answer 1

Let $J $ be an index for the cardinality of an orthonormal basis of $H $. Then $H $ is isometrically isomorphic to $\ell^2 (J) $, so it is enough to discuss the problem on this latter space.

Define the product $fg $ pointwise, i.e. $fg (j):=f (j)g (j) $. The question is whether this product stays in $\ell^2$, and whether the norm is submultiplicative. We have $$ \|fg\|_2^2=\sum_j|f (j)|^2\,|g (j)|^2\leq\|f\|_\infty^2\,\|g\|_2^2\leq\|f\|_2^2\,\|g\|_2^2, $$ so the two norm and the pointwise product make $\ell^2$ a Banach algebra.

Answer 2

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.