Let $H$ be a Hilbert space and $T : H \rightarrow H$ be a self-adjoint operator.
Let $\sigma(T)$ be the spectrum of $T$, $\Sigma$ be the Borel sigma algebra of $\sigma(T)$ and $E$ be a resolution of the identity on $\Sigma$ (for $T$).
Suppose that $f$ is a bounded Borel function on $\sigma(T)$ and $A$ is a Borel set of $\sigma(T)$.
Consider the operator $S : H \rightarrow H$ defined by $$ \langle Sx, y \rangle = \int_A f(\lambda)\, dE_{x,y}. $$
I claim that $|\langle Sx, y \rangle| \leq \|f \|_{\infty} \|E(A)x\| \|E(A)y\|$, where $\|f\|_{\infty} = \sup \{ |f(\lambda)|: \lambda \in \sigma(T) \}$.
It is easy to see that
$$ \left|\int_A f(\lambda) \, dE_{x,y} \right| \leq \|f\|_{\infty} |E_{x,y}| (A). $$
But, how can I show that $|E_{x,y}|(A) \leq \|E(A)x\| \|E(A)y\|$?
Any help will be appreciated!
Let $\omega_1, \dots, \omega_n$ be disjoint subsets of $A$ with $\cup_{i=1}^n \omega_i = A$.
Note that
\begin{align} \sum_{i=1}^n |E_{x,y}(\omega_i)| = \sum_{i=1}^n |\langle E(\omega_i)x,y\rangle| &= \sum_{i=1}^n |\langle E(\omega_i)x,E(\omega_i)y\rangle| \\ &\leq \sum_{i=1}^n \|E(\omega_i) x\| \|E(\omega_i)y\|\\ &\leq\left(\sum_{i=1}^n \|E(\omega_i)x\|^2\right)^{\frac{1}{2}} \left(\sum_{i=1}^n \|E(\omega_i)y\|^2\right)^{\frac{1}{2}} \\ &=\left(\left\|\sum_{i=1}^n E(\omega_i)x \right\|^2\right)^{\frac{1}{2}} \left(\left\|\sum_{i=1}^n E(\omega_i)y \right\|^2\right)^{\frac{1}{2}} \\ &=\|E(A)x\| \|E(A)y\|. \end{align}
This implies that $|E_{x,y}|(A) \leq \|E(A)x\| \|E(A) y\|$.