Inequality between matrix elements of positive self-adjoint operators

Question

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

\begin{equation} \hat{A}_-~\le~\hat{B}~\le~\hat{A}_+, \end{equation}

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

Is it possible to prove the following relation

\begin{equation} |\langle \psi_1|\hat B - \hat A_-|\psi_2\rangle|\le|\langle \psi_1|\hat A_+ - \hat A_-|\psi_2\rangle| \end{equation}

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

Answer 1

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 $|x_2^tBx_1|=2> 1=|x_2^tA_{+}x_1|$.

Since unbounded operator technically means "not necessarily bounded", this gives a counterexample to your question