# Inequality between matrix elements of positive self-adjoint operators

We have three positive semi-definite self-adjoint operators $$\hat{A}_-$$, $$\hat{B}$$, $$\hat{A}_+$$ on the Hilbert space $$\mathcal{H}$$. They are unbounded operators and satisfy the following inequality

$$$$\hat{A}_-~\le~\hat{B}~\le~\hat{A}_+,$$$$

$$\hat{A}_-~\le\hat{B}$$ means $$\le\hat{B}-\hat{A}_-$$ is a positive semi-definite self-adjoint operator.

Is it possible to prove the following relation

$$$$|\langle \psi_1|\hat B - \hat A_-|\psi_2\rangle|\le|\langle \psi_1|\hat A_+ - \hat A_-|\psi_2\rangle|$$$$

$$\forall~\psi_1,\psi_2\in\mathcal{H}$$?

• You should try if it works for 2-by-2 matrices. Jun 15, 2021 at 2:32
• @NarutakaOZAWA: I just saw your comment just after typing my answer :) Jun 15, 2021 at 2:38
• @Darth Vader: Well, my comment was much shorter than yours. Jun 15, 2021 at 4:23

This doesn't even hold for bounded operators. I will give a counterexample involving $$2 \times 2$$ matrices. Take $$A_{+}= Id_2$$, the $$2 \times 2$$ identity matrix. $$A_{-}=0$$ and $$B= \begin{pmatrix} 1 &0\\0 &0 \end{pmatrix}$$. Let $$x_1=\begin{pmatrix} 1\\ -1 \end{pmatrix}$$. Let $$x_2=\begin{pmatrix} -2\\ -1 \end{pmatrix}$$.
Then $$|x_2^tBx_1|=2> 1=|x_2^tA_{+}x_1|$$.