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}


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

1 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


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