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Let
 be a unital  -algebra. Let  be a normal element in
 and  be its spectrum.
The continuous functional calculus provides a  -isomorphism
between the  -algebra
 of complex valued continuous functions on  and the  -subalgebra
![$ \mathcal{A}[x] \subseteq \mathcal{A}$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img13.png "$ \mathcal{A}[x] \subseteq \mathcal{A}$") generated by  and the identity of
 .
Spectral Mapping Theorem - Let
 be as above. Let
 . Then
Proof : Since
 and
![$ \mathcal{A}[x]$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img20.png "$ \mathcal{A}[x]$") are isomorphic we must have
where
![$ \sigma_{\mathcal{A}[x]}(f(x))$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img22.png "$ \sigma_{\mathcal{A}[x]}(f(x))$") denotes the spectrum of  relative to the subalgebra
![$ \mathcal{A}[x]$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img24.png "$ \mathcal{A}[x]$") .
By the spectral invariance theorem we have
![$ \sigma_{\mathcal{A}[x]}(f(x))=\sigma(f(x))$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img25.png "$ \sigma_{\mathcal{A}[x]}(f(x))=\sigma(f(x))$") . Hence
Thus, we only have to prove that
 .
 is defined on  so
 is precisely the image of  .
Let
 . The function
 is invertible if and only if
 has no zeros.
Equivalently,
 is not invertible if and only if
 has a zero, i.e.
 for some  .
The previous statement can be reformulated as:
 if and only if  is in the image of  .
We conclude that
 , and this proves the theorem. 
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![$ C(\sigma(x)) \longrightarrow \mathcal{A}[x]$](http://images.planetmath.org:8080/cache/objects/9891/l2h/img7.png "$ C(\sigma(x)) \longrightarrow \mathcal{A}[x]$"){kind=small}
![$\displaystyle \sigma(f) = \sigma_{\mathcal{A}[x]}(f(x)) $](http://images.planetmath.org:8080/cache/objects/9891/l2h/img21.png "$\displaystyle \sigma(f) = \sigma_{\mathcal{A}[x]}(f(x)) $"){kind=small}
