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Pointwise multiplication as Hilbert-Schmidt operator on L^2

  1. Apr 16, 2010 #1
    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
     
  2. jcsd
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