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:
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?
In case this is false, would the result be implied by the stronger condition $A^2\leq B^2$ and $0\leq A\leq B$?
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.
The first question has been answered by Michael Renardy, and the answer is no. The second answer should have a positive answer. To avoid any domain issues, let me explain rather why $A^2 \leq B^2$ implies that $A B^{-1}$ has norm $\leq 1$ (for bounded operators $B^{-1} A$ is the adjoint of $A B^{-1}$ so it also has norm $\leq 1$, but I never work with unbounded operators so not sure whether this is true always).
For $\xi$ in the domain of $B$, if $\eta = B \xi$, $$\|A B^{-1} \eta \|^2 = \| A \xi\|^2 = \langle A^2 \xi,\xi\rangle \leq \langle B^2 \xi,\xi\rangle = \|\eta\|^2.$$ This shows that $A B^{-1}$ extends to a norm $\leq 1$ operator on the closure of the image of $B$, which (I guess since you write $B^{-1}$) is assumed to be the whole space.
Another way to write the same proof is, for $A,B \geq 0$, $$ A^2 \leq B^2 \iff B^{-1} A B^{-1} \leq 1 \iff \|A B^{-1}\| \leq 1.$$ The first equivalence is just the fact that the operation $ X \mapsto B X B$ preserves positive operators.
