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 $0\leq A\leq B$?

  • $\begingroup$ 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). $\endgroup$ Apr 13, 2021 at 8:13
  • $\begingroup$ @MikaeldelaSalle Thank you for your comment. Is equivalence you mention easy to see? $\endgroup$ Apr 13, 2021 at 8:20
  • $\begingroup$ I added an answer. $\endgroup$ Apr 13, 2021 at 8:29

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$ 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.

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

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