Pointwise multiplication as Hilbert-Schmidt operator on L^2

[schieghoven]

Hello,

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

 

[jvicens]

Hello Dave,

Thank you for your question. You are correct, the map \psi \mapsto A \psi is a bounded map on L^2 if A is essentially bounded. This means that there exists a constant M>0 such that \|A\psi\|_{L^2} \leq M \|\psi\|_{L^2} for all \psi \in L^2 .

To ensure that the map is also Hilbert-Schmidt, we can impose the condition that A is a Hilbert-Schmidt operator. This means that the operator norm of A is finite, i.e. \|A\|_{op} < \infty . In other words, the map \psi \mapsto A\psi is not only bounded, but also has a finite norm as an operator on L^2 .

Another way to ensure that the map is Hilbert-Schmidt is to require that A is a compact operator. This means that the image of any bounded set in L^2 under the map \psi \mapsto A\psi is relatively compact. In other words, the map "compresses" the space L^2, and the resulting image is a relatively compact subset of L^2. This condition is stronger than the Hilbert-Schmidt condition, as all Hilbert-Schmidt operators are compact, but not all compact operators are Hilbert-Schmidt.

I hope this helps answer your question. If you have any further inquiries, please do not hesitate to ask.