# Does $\pm A \leq B$ imply that $B^{-1} A$ is bounded?

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 $$0\leq A\leq B$$?

• For bounded positive operators, the condition $A^2 \leq B^2$ is equivalent to saying that $B^{-1} A$ has norm $\leq 1$. So your second question should have a positive answer. (But I am not very used to unbounded operators; what you get for free is that $A B^{-1}$ has dense domain and is bounded of norm $\leq 1$, I am not even sure that $B^{-1} A$ has dense domain). Apr 13, 2021 at 8:13
• @MikaeldelaSalle Thank you for your comment. Is equivalence you mention easy to see? Apr 13, 2021 at 8:20

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.

• Thank you very much for the neat counterexample! Do you by chance also know about the second condition, i.e. whether $A^2 \leq B^2$ might save us? Apr 13, 2021 at 8:15

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.

• That is super neat, thanks! Indeed, I assume that the image of $B$ is the full space (was a bit sloppy there). Apr 13, 2021 at 9:19
• If $ST$ is densely defined, then $(ST)^\ast\supset T^\ast S^\ast$. Thus $B^{-1}A\subset (AB^{-1})^\ast$ is bounded as well. Apr 13, 2021 at 9:42
• @MaoWao Thanks. But is $T^* S^*$ densely defined? Apr 13, 2021 at 13:46
• Not necessarily. But in this case, $D(B^{-1}A)=D(A)$ is dense. Apr 13, 2021 at 13:52