# Is a bounded Borel function of a normal operator normal

I am playing around with Borel functional calculus to try to understand it, and made the following argument:

Let $T\in B(H)$ (bounded operator on Hilbert space) be normal. Let $f\in C(\sigma(T))$ (continuous function on the spectrum of T). Let $\mathcal{A}=\text{Alg}_{C^*}(I,T)$ be the C*-algebra generated by $T$ and the identity operator I.

Using the continuous functional calculus we can write $T=\Psi(\text{Id}_{\sigma(T)})$ for $\Psi$: $C(\sigma(T))\to \mathcal{A}$ an isometric $*$-isomorphism. Further we can give meaning to $f(T)$ for any $f\in C(\sigma(T))$, consistently with how polynomials would be defined, by setting $f(T)=\Psi(f(\text{Id}_{\sigma(T)}))$.

We can expand this to $BB(\sigma(T))$ (the bounded borel functional calculus). Now if we set let $g \in BB(\sigma(T))$ and define $S=\Psi(g)$ since $\Psi$ is now a $*$-homomophism and $$S^*S = \Psi(g)^*\Psi(g)=\Psi(g^*g)=\Psi(gg^*)=SS^*$$

Since a bounded borel function commutes with its complex conjugate. So S is normal. Ergo any Borel function of a normal operator is normal.

I think the conclusion seems a bit strong, so I wonder if there is a flaw in my reasoning? If so, when is it true?

What I wan't to use the property for is proving that $\sigma(S)=g(\sigma(T))$, so if the above is wrong an argument for that would be a nice addition :)

The functional calculus works as stated. However, the spectral mapping result that you want to prove is not true in general. For example, consider the operator $M$ of multiplication by $x$ on $L^{2}[0,1]$. The spectrum of $M$ is $[0,1]$. If you let $g(x)=x$ for $x \in [0,1/2)\cup(1,2,1]$ and $g(1/2)=50$, then $g(M)$ does not have $50$ in its spectrum. On the other hand, if you consider $M$ on $L^{2}_{\mu}[0,1]$ where $\mu$ is Lebesgue measure with an additional atom at $1/2$, then $g(M)$ does have $50$ in its point spectrum. Look into essential range: http://en.wikipedia.org/wiki/Essential_range , and see if you can formulate something in terms of the spectral measure.
• First of all thanks for the answer, interesting that the result is indeed true. Regarding the second part actually, we have proven that the relevant spectral measure only has the empty set as null set, would that not remove the problem? Else I guess the result should be that $\sigma(S)=\text{EssRan}_g(\sigma(T))$. – Henrik Sep 2 '14 at 17:12
• @Henrik If $M$ has an eigenvalue $\lambda$ (i.e., is in the point spectrum,) then changing $f$ at $\lambda$ definitely changes $f(M)$. However, if $M$ has only continuous spectrum, then changing $f$ at $\lambda$ doesn't affect $f(M)$ at all, and that latter case means that the spectral mapping theorem cannot apply to $f$ because $f(\sigma(T))$ is changed by changing $f$ at $\lambda$, even though $f(M)$ is not changed in that case. That's why you need essential range relative to the spectral measure. – DisintegratingByParts Sep 2 '14 at 18:02
• Just to be sure, I expect you to be right on this, but let me present my argument for why you are not so you can tell me what I get wrong. $f(M)=\int_{\sigma(M)} f(\lambda) dE(\lambda)$ so if I change $f$ on a $E$-nullset $f(M)$ would not change, but $f(\sigma(T))$ would. If there are no $E$-null sets this though can not happen. I.e. the essential range and ordinary range coincide. – Henrik Sep 2 '14 at 18:25
• @Henrik : A normal operator has continuous spectrum and/or point spectrum. If there is no point spectrum then a set with a single point is $E$-null, which means you can change $f$ at any single point without affecting $f(T)$, even though you will change $f(\sigma(T))$ by changing $f$ on a point of the spectrum. If you have only point spectrum, then things are different. – DisintegratingByParts Sep 2 '14 at 18:48