spectral mapping theorem

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

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

$C(\sigma (x))⟶𝒜[x]$

$f\mapsto f(x)$

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

Spectral Mapping Theorem - Let $x\in 𝒜$ be as above. Let $f\in C(\sigma (x))$. Then

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

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

$$\sigma (f)={\sigma }_{𝒜[x]}(f(x))$$

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

By the spectral invariance theorem we have ${\sigma }_{𝒜[x]}(f(x))=\sigma (f(x))$. 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 $\lambda \in ℂ$. 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. $\mathrm{\square }$