# Pointwise multiplication as Hilbert-Schmidt operator on L^2

1. Apr 16, 2010

### schieghoven

Hello,

The map $\psi \mapsto A \psi$ for $\psi \in L^2$, where $A$ is a function and $A \psi$ represents pointwise multiplication, is a bounded map on $L^2$ if A is essentially bounded. However, the map is not necessarily Hilbert-Schmidt, as for example the identity map (corresponding to $A(x) \equiv 1$) demonstrates.

My question is: can I put stronger conditions on A to ensure the map is Hilbert-Schmidt, rather than just bounded?

I hope this is the right group to post this question... there doesn't seem to be a functional analysis group!

Many thanks

Dave