I hope that somebody can help me with the following problem:


Let $A$ be a positive operator on $\mathbf{B}(\mathcal{H})$, ( $\mathcal{H}$ is a Hilbert space) with its spectral measure $E$. Show that for every Borel set $\mathbf{B}$ from the domain of $E(\cdot)$ the following equality holds
$$f(\| AE(\mathbf{B})\|) = \| f(A)E(\mathbf{B})\|, $$
where $f$ is an arbitrary continuous increasing function such that $f(0)=0$. Is it also true when $f(0) \geq 0$?


I have no idea how to solve the main part. The answer for the second part is probably negative, because if I take e.g. $f(x)=x^2+1$, then


$$\| (A^2+I)E(\mathbf{B}) \| \leq \|AE(\mathbf{B})\|^2 +1$$ 
and the equality does not hold for every $A$.



Positive operators - norm equality
The $L^2$ view usually helps, but I don't think it makes things simpler in this case. 


Note that since $E(B)$ is a spectral projection of $A$, you have $f(A)\,E(B)=f(A\,E(B))$ (easy to see since the relation holds for any monomial).


Then the question reduces to whether $\|f(A)\|=f(\|A\|)$ for a positive operator. Since by the spectral mapping theorem $\sigma(f(A))=f(\sigma(A))$, the positivity, monotonicity and continuity of $f$ guarantee that $f$ commutes with $\max$. So
$$\|f(A)\|=\max\{t:\ t\in\sigma(f(A))\}=\max\{t:\ t\in f(\sigma(A))\}=f(\max\{t:\ t\in\sigma(A)\})=f(\|A\|).$$


