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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 $A$.

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Check the identity for a multiplication operator on some $L^2$. – Michael Jun 16 '11 at 22:22
I am not pretty sure what do you mean. Did you mean to use the spectral theorem and take $M_{\phi}$ a multiplication operator on some $L^2(\mu)$ unitarly equivalent to our operator $A$. For the spectral measure of $M_{\phi}$ ( which is $\textbf{1}_{\phi^{-1}(\cdot)}$) choose $\mathbf{B}$ to be the whole space to get the identity i.e. $E(\mathbf{X})=I$. Now $f(\| M_{\phi} \|) = f(\sup_x \phi(x))$ and $\| f(\phi) \| = \sup_x f(\phi(x))$ and since $f$ is increasing continuous and $f(0)=0$ the equality holds. I think that for any set $\mathbf{B}$ it will be satisfied in a similar way. – Romanov Jun 16 '11 at 23:15
up vote 1 down vote accepted

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 $$\|f(A)\|=\max\{t:\ t\in\sigma(f(A))\}=\max\{t:\ t\in f(\sigma(A))\}=f(\max\{t:\ t\in\sigma(A)\})=f(\|A\|).$$

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Nice alternative, thank you for that! – Romanov Jun 17 '11 at 17:13
(I felt that $||f(A)|| = f(||A||)$ could be seen easily in the $L^2$-picture.) – Michael Jun 17 '11 at 22:40

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?

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You don't really need any unitary operators. Since you are working with a fixed positive operator $A$, you can work on the abelian vN algebra it generates. Also, the question doesn't make sense for unbounded operators, as it deals with the norm (neither side of the equality make sense for an unbounded operator if the set B is unbounded). – Martin Argerami Jun 17 '11 at 17:18
Ok, but if I modify it a little bit i.e. I take $\tau$-measurable positive operator $A$ and von Neumann algebra generated by the spectral projection of $A$. By the bi-commutant theorem the spectral projections of $f(A)$ will be there as well. Thus for any projection from this von Neumann algebra we will get this equality I mean $$ f( \| AE\|) = \| f(A)E \|$$. Here $A$ can be even unbounded, but \|AE\| makes sense since $A$ is $\tau$-measurable. – Romanov Jun 17 '11 at 19:44

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