# spectral mapping theorem

Let $\mathcal{A}$ be a unital $C^{*}$-algebra (http://planetmath.org/CAlgebra). Let $x$ be a normal element in $\mathcal{A}$ and $\sigma(x)$ be its spectrum.

The continuous functional calculus provides a $C^{*}$-isomorphism

$C(\sigma(x))\longrightarrow\mathcal{A}[x]$

$\;\;f\mapsto f(x)$

between the $C^{*}$-algebra $C(\sigma(x))$ of complex valued continuous functions on $\sigma(x)$ and the $C^{*}$-subalgebra $\mathcal{A}[x]\subseteq\mathcal{A}$ generated by $x$ and the identity of $\mathcal{A}$.

Let $x\in\mathcal{A}$ be as above. Let $f\in C(\sigma(x))$. Then

 $\sigma(f(x))=f(\sigma(x)).$

Proof : Since $C(\sigma(x))$ and $\mathcal{A}[x]$ are isomorphic we must have

 $\sigma(f)=\sigma_{\mathcal{A}[x]}(f(x))$

where $\sigma_{\mathcal{A}[x]}(f(x))$ denotes the spectrum of $f(x)$ relative to the subalgebra $\mathcal{A}[x]$.

By the spectral invariance theorem we have $\sigma_{\mathcal{A}[x]}(f(x))=\sigma(f(x))$. Hence

 $\sigma(f)=\sigma(f(x))$

Thus, we only have to prove that $f(\sigma(x))=\sigma(f)$.

$f$ is defined on $\sigma(x)$ so $f(\sigma(x))$ is precisely the image of $f$.

Let $\lambda\in\mathbb{C}$. The function $f-\lambda$ is invertible if and only if $f-\lambda$ has no zeros.

Equivalently, $f-\lambda$ is not invertible if and only if $f-\lambda$ has a zero, i.e. $f(\lambda_{0})=\lambda$ for some $\lambda_{0}$.

The previous statement can be reformulated as: $\lambda\in\sigma(f)$ if and only if $\lambda$ is in the image of $f$.

We conclude that $\sigma(f)=f(\sigma(x))$, and this proves the theorem. $\square$

Title spectral mapping theorem SpectralMappingTheorem 2013-03-22 17:30:08 2013-03-22 17:30:08 asteroid (17536) asteroid (17536) 4 asteroid (17536) Theorem msc 46L05 msc 47A60 msc 46H30