A normal operator is the continuous functional calculus of an adjoint operator

Question

Let $H$ be a hilbert space and suppose $T \in B(H)$ is a normal operator. Is it true that there exist $S \in B(H)$ self-adjoint such that $T=f(S)$ being $f: \sigma(S) \longrightarrow \mathbb{C}$ a continuous function?

At the outset I thought the result was false for the following reason: if $H$ is separable then there exist a normal operator $T \in B(H)$ such that $\sigma(T)=K$, where $$ K=\overline{\{ t+ i sin(1/t): t \in (0,1] \} } $$ Therefore, if such an $S$ exist, we would have that $$ K=\sigma(T)=\sigma(f(S))=f(\sigma(S)) $$ Since $\sigma(S) \subset \mathbb{R}$ (since $S$ is self-adjoint) this tells us that $K$ is the image of a continuous function $f: \sigma(S)\subset \mathbb{R} \longrightarrow \mathbb{C}$. Writing $f$ as $f(t)=f_1(t)+if_2(t)$, the continuity of $f$ implies that $f_1$ and $f_2$ are both continuous, also the definition of $K$ and the above equation tells us that $(0,1] \subset \sigma(S)$. Putting these facts together we would have that the function $g(t)=sin(1/t)$ can be extended to a continuous function $f_2$ defined on a compact subset of $\mathbb{R}$ (namely, $\sigma(S)$) which contains $(0,1]$ (and this isn't possible according to me).

However, in the book An Introduction to Operator Algebras by Kehe Zhu it is claimed (without proof or hint) that the result is true. That being said, these are my two questions:

1. What is wrong with the "counterexample" I provided?
2. How can I prove the claim? Any hint will be appreciated.

In advance thank you very much.

Answer 1

I started to write yesterday a slightly more direct approach, that contains ideas mostly mentioned in the comments.

If we denote the Cantor set by $\def\cC{\mathcal C}\cC$ and by $f:\cC\to\sigma(T)$ a continuous Borel surjection (that exists by the universality of the Cantor set among compact metric spaces), there exists a Borel right-inverse $g:\sigma(T)\to\cC$. We write $$ T=\int_{\sigma(T)}\lambda\,dE(\lambda) $$ via the Spectral Theorem, and we define $$ S=\int_{\sigma(T)}g(\lambda)\,dE(\lambda). $$ The definition makes sense because $g$ is bounded Borel, and $S$ is selfadjoint because $g$ is real-valued. And by functional calculus we have $$f(S)=\int_{\sigma(T)}(f\circ g)(\lambda)\,dE(\lambda)=\int_{\sigma(T)}\lambda\,dE(\lambda)=T. $$

Answer 2

The following proof uses quite advanced machinery - I'm sure there are more elementary proofs, but I can't come up with one.

Let $\mathcal{C}$ be the Cantor set. Since the Cantor set is universal among compact metrizable spaces, there exists a continuous surjection $f: \mathcal{C} \rightarrow \sigma(T)$. This dualizes to an injective $\ast$-homomorphism $f^\ast: C(\sigma(T)) \rightarrow C(\mathcal{C})$. As $T$ is normal, we may let $\iota: C(\sigma(T)) \rightarrow \mathbb{B}(H)$ be the functional calculus map, which is an injective $\ast$-homomorphism. Let $A$ be the von Neumann algebra generated by $\iota(C(\sigma(T)))$, which is the same as the von Neumann algebra generated by $T$ and so in particular abelian. Then $\iota$ can be regarded as a $\ast$-homomorphism from $C(\sigma(T))$ to $A$. As abelian von Neumann algebras are injective in the category of unital abelian $C^\ast$-algebras and $\ast$-homomorphisms (see, for example, Theorem 2.4 in this paper), there exists a $\ast$-homomorphism $g: C(\mathcal{C}) \rightarrow A$ s.t. $g \circ f^\ast = \iota$.

Now, let $x \in C(\mathcal{C})$ be the usual inclusion map $\mathcal{C} \rightarrow \mathbb{R}$. In particular, $x$ is self-adjoint. Hence, $S = g(x)$ is a self-adjoint element of $\mathbb{B}(H)$. We claim that $f(S) = T$. Indeed, $f(S) = f(g(x)) = g(f(x))$ as $g$ is a $\ast$-homomorphism. But $f(x)$, as an element of $C(\mathcal{C})$, is simply $f^\ast(\mathrm{Id}_{\sigma(T)})$. Thus, $f(S) = g(f(x)) = g \circ f^\ast(\mathrm{Id}_{\sigma(T)}) = \iota(\mathrm{Id}_{\sigma(T)}) = T$. This proves the claim.

Answer 3

You've got some great answers for how to prove the claim already, but I wanted to elaborate on your "counterexample" a bit.

The only mistake with your counterexample is the insistence that $\sigma(S) = [0,1]$ with $f_2(t) = \sin(1/t)$. The operator $S$ doesn't have to respect your construction/definition of $K$. It's sufficient to find a Borel map $g: K \rightarrow \sigma(S) \subset \mathbb{R}$ and a continuous map $f: \sigma(S) \rightarrow K$ such that $f \circ g = \text{Id}_K$, which is what the other answers construct for you.

You're right that this would be impossible if we required $\sigma(S) = [0, 1]$! Even if we're allowed to change our attempted mapping from the one you prescribed for $K$, the Hahn-Mazurkiewicz theorem says that $K$ isn't a continuous image of the unit interval. That's because your $K$ is the closed topologist's sine curve which isn't locally connected. So it's impossible to find a self-adjoint $S$ with spectrum $[0,1]$ that does the job.

The other answers show that if you let $\sigma(S)$ be, say, the Cantor set $\mathcal{C}$, then you can definitely make it work. You can continuously map from $\mathcal{C}$ onto pretty much anything, including your topologist's sine curve.

So your counterexample was placing an artificial restriction on what the spectrum of $S$ could be, leading you to a mistaken conclusion. If you only require $\sigma(S) \subset \mathbb{R}$ instead of $\sigma(S) = [0,1]$, it becomes doable.