# Spectral mapping theorem for polynomials in $z,\overline{z}$ and direct construction of the function calculus for a normal operator

Suppose that $A$ is an element in Banach algebra and $p$ is a polynomial. Then we have an equality $p(\sigma(A))=\sigma(p(A))$ where $p(A)$ has an elementary meaning. This theorem (the spectral mapping theorem) is a key point in constructing the (isometric) functional calculus in the context of $C^*$-algebras $f \mapsto f(A)$ for $f \in C(\sigma(A))$ where $A$ is self adjoint. This can be done before we prove theorems such Gelfand-Najmark: in other words even before we know that each commutative unital $C^*$-algebra is of the form $C(X)$ for some compact $X$, we can state that for a self-adjoint element $A$ in some $C^*$-algebra, the $C^*$-algebra generated by $A$ is isomorphic to $C(\sigma(A))$. I wonder if it is possible to give an analogous direct proof for a normal operator, which would involve a similar theorem as spectral mapping theorem but in the context of polynomials in $z$ and $\overline{z}$. This would be enough since every continuos complex valued function on $C(\sigma(A))$ (where $A$ is normal, therefore $\sigma(A)$ need not to be contained in the real line anymore) can be aproximated by such polynomials and if $A$ is normal then $p(A)$ makes sense for such polynomials. Of course I'm interested in a ,,elementary'' proof of such spectral mapping theorem (not using Gelfand Najmark theorem).

• Tried that once, too. All I can say is that it works for self-adjoint operators, but I couldn't manage for normal operators. – user1688 Jan 23 '16 at 11:46
• It's indeed easy for self-adjoint, basically because you can factor a polynomial $p(z)$ in one variable into linear factors. If you google something like "spectral mapping + several variables," you'll find papers that prove the same thing for $p(z,\overline{z})$ (without GN theory), but the proofs are so unenlightening that you will quickly go back to the standard treatment via G-N. – Christian Remling Jan 23 '16 at 14:56

If I interpret your question correctly, then you want an elementary proof of the fact that if $A$ is normal, then $\sigma(p(A,A^*))=f(\sigma(A))$, where $f(z):=p(z,\overline{z})$.
By looking at approximate eigenvectors, it's easy to see that $f(\sigma)\subseteq \sigma(p)$, so the question really is about the reverse inclusion. I might be wrong, but I'm not sure a completely elementary proof is possible. As I mentioned in my comment, this topic has of course been studied, see for example this paper.
Alternatively, the following argument works. It's not exactly what you are asking for; while it does avoid Gelfand-Naimark, we are using the following facts: (1) The spectrum of an operator $A\in B(H)$ agrees with its spectrum in an arbitrary $C^*$-subalgebra of $B(H)$; (2) if $A\in\mathcal A\subseteq B(H)$, with $\mathcal A$ a commutative $C^*$-algebra, then $\sigma(A)=\{\varphi(A)\}$, where $\varphi :\mathcal A\to\mathbb C$ ranges over the complex homomorphisms; (3) $\varphi(A^*)=\overline{\varphi(A)}$ (establishing this is usually the first step in proving GN)
We can now work in the $C^*$-algebra generated by $A$, which of course also contains $p(A,A^*)$. If now $z\in\sigma(p(A,A^*))$, then $z=\varphi(p(A,A^*))=p(\varphi(A),\overline{\varphi(A)})=f(\varphi(A))\in f(\sigma(A))$, as desired.