Lately I have to use a lot of functional calculus. A question that keeps popping up and that I don't manage to resolve is the following:



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> Let $A,B$ be self-adjoint (not necessarily bounded) operators such that $\pm A\leq B$. Is it true that $B^{-1} A$ is a bounded operator?
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In case this is false, would the result be implied by the stronger condition $A^2\leq B^2$ and $0\leq A\leq B$?



Does $\pm A \leq B$ imply that $B^{-1} A$ is bounded?
On $R^2$, consider the matrices $B\_N=\pmatrix{N&0\cr 0&1}$, $A\_N=\pmatrix{0&\sqrt{N}\cr \sqrt{N}& 0}$. It is easily checked that $B\_N\pm A\_N$ is positive definite, but $B\_N^{-1}A\_N$ is of order $\sqrt{N}$. You can build infinite dimensional operators using $B\_N$ and $A\_N$ as diagonal blocks.


