Positive operators - norm equality 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 $A$.
 A: 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
$$\|f(A)\|=\max\{t:\ t\in\sigma(f(A))\}=\max\{t:\ t\in f(\sigma(A))\}=f(\max\{t:\ t\in\sigma(A)\})=f(\|A\|).$$
A: Following above steps for an arbitrary $\mathbf{B}$ we get that
$$ f \left(\| M_{\phi}E(\mathbf{B})\| \right)= f \left( \sup_{x \in \mathbf{B}} \ \phi(x) \right) =  \sup_{x \in \mathbf{B}} \ f(\phi(x)) = \| f(M_{\phi})E(\mathbf{B})\|.$$
Since unitary operators preserve the norm the above equality is true for an arbitrary positive operator $A$. Moreover, now it is clear that with $f(0) \geq 0$ the property holds as well. We can even generalize it for a positive $\tau$-measurable operator, where $\tau$ is a faithful normal semi-finite trace on some semi-finite von Neumann algebra. Do you agree with my answer?
