Let $\mathcal{H}$ be a separable Hilbert space, $A$ be a bounded self-adjoint operator on it, $u$ and $v$ be two vectors in $\mathcal{H}$.
Of course it is false that $$ \left|\langle u,\, A Bv\rangle\right| \leq \|A\| \left|\langle u,\, Bv\rangle\right| $$ where $B$ is any other bounded operator.
However, is the following true?
Let $f$ be Borel bounded and $g$ be Borel bounded and non-negative. If $\mu_{u,v}$ is the spectral measure of the triple $A$, $u$ and $v$, then \begin{align} \left|\langle u,\, f\left(A\right) g\left(A\right) v\rangle\right| &= \left|\int_{\sigma\left(A\right)}f\left(\lambda\right)g\left(\lambda\right)\mathrm{d}\mu_{u,v}\left(\lambda\right)\right| \\ &\leq \int_{\sigma\left(A\right)}\left|f\left(\lambda\right)g\left(\lambda\right)\right|\mathrm{d}\mu_{u,v}\left(\lambda\right)\\ &\leq \|f\|_{\infty}\int_{\sigma\left(A\right)}\left|g\left(\lambda\right)\right|\mathrm{d}\mu_{u,v}\left(\lambda\right)\\ &= \|f\|_{\infty}\int_{\sigma\left(A\right)}g\left(\lambda\right)\mathrm{d}\mu_{u,v}\left(\lambda\right)\\ &= \|f\|_{\infty} \langle u,\, g\left(A\right) v\rangle\end{align}
where the penultimate equality is valid as $g\geq0$. Is this correct?