I came across the following question: Given a self-adjoint, positive definite Hilbert-Schmidt operator $C$ and a self-adjoint, bounded and positive operator $A$ - both acting on a separable Hilbert space $\mathcal H$ - is then $$ B = CAC^{-1} $$ a bounded operator on $\mathcal H$, too? If not, which requirements on $A$ would guarantee the boundedness of $B$ (except $A$ being the identity)?
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6$\begingroup$ Here is a counterexample: take $H$ generated by two orthogonal sequence $(a_n)$ and $(b_n)$, $A$ the operator $2*Id$ + the operator which switch $a_n$ and $b_n$ for all $n$, and $C$ the operator which acts on $a_n$ by multiplication by $1/n$ and on $b_n$ by multiplication by $1/n^2$. then $(C A C^{-1}) (b_n)=n.b_n$ Generalizing this idea you can construct counterexample for a very large class of $A$. (for example for any operator which have two infinite dimensional disjoint spectral projection...) $\endgroup$– Simon HenryCommented Sep 9, 2014 at 9:02
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$\begingroup$ @SimonHenry Why not leave this as an answer? $\endgroup$– Yemon ChoiCommented Sep 22, 2014 at 1:28
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$\begingroup$ Because the question was "Which requirements on A..." and I have no idea on if and how my counterexample really allow to says something about that ? (and because I think that the answer is "this is true for almost no $A$" and I was hoping that someone would prove this) $\endgroup$– Simon HenryCommented Sep 22, 2014 at 13:27
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This is more a comment than an answer (but I am not entitled) since it is merely a reformulation of your condition albeit in a way which may shed some light. By the spectral theorem, we can assume that $C$ is the operation of multiplication by an $\ell^2$ sequence $(x_n)$ on $\ell^2$. Then the condition is that the operator $A$ also act continuously on the weighted $\ell^2$ space consisting of those sequences $(y_n)$ for which $\sum \left (\frac { y_n}{x_n}\right )^2<\infty$.
