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%%% Primary Title: $C^*$-algebra homomorphisms are continuous
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%%% Filename: HomomorphismsOfCAlgebrasAreContinuous.tex
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\begin{document}

 {\bf Theorem -} Let $\mathcal{A}, \mathcal{B}$ be \htmladdnormallink{$C^*$-algebras}{http://planetmath.org/encyclopedia/CAlgebra.html} and $f:\mathcal{A} \longrightarrow \mathcal{B}$ a \htmladdnormallink{*-homomorphism}{http://planetmath.org/encyclopedia/CategoryOfCAlgebras3.html}. Then $f$ is \htmladdnormallink{bounded}{http://planetmath.org/encyclopedia/ContinuousLinearMapping.html} and $\|f\| \leq 1$ (where $\|f\|$ is the \htmladdnormallink{norm}{http://planetmath.org/encyclopedia/OperatorNorm.html} of $f$ seen as a \htmladdnormallink{linear operator}{http://planetmath.org/encyclopedia/LinearOperator.html} between the spaces $\mathcal{A}$ and $\mathcal{B}$).

For this reason it is often said that \htmladdnormallink{homomorphisms}{http://planetmath.org/encyclopedia/Automorphism3.html} between $C^*$-algebras are \htmladdnormallink{automatically continuous}{http://planetmath.org/encyclopedia/ContinuousLinearMapping.html}.

{\bf Corollary -} A *-isomorphism between $C^*$-algebras is an \htmladdnormallink{isometric isomorphism}{http://planetmath.org/encyclopedia/IsometricIsomorphism.html}.\\
$\;$

{\bf \emph{\htmladdnormallink{Proof}{http://planetmath.org/encyclopedia/Proof.html} of Theorem :}} Let us first suppose that $\mathcal{A}$ and $\mathcal{B}$ have \htmladdnormallink{identity elements}{http://planetmath.org/encyclopedia/IdentityElement.html}, both denoted by $e$.

We denote by $\sigma(x)$ and $R_{\sigma}(x)$ the \htmladdnormallink{spectrum}{http://planetmath.org/encyclopedia/SpectralRadius.html} and the \htmladdnormallink{spectral radius}{http://planetmath.org/encyclopedia/SpectralRadius.html} of an element $x \in \mathcal{A}$ or $\mathcal{B}$.

Let $a \in \mathcal{A}$ and $\lambda \in \mathbb{C}$. If $a- \lambda e$ is \htmladdnormallink{invertible}{http://planetmath.org/encyclopedia/Invertible3.html} in $\mathcal{A}$, then $f(a- \lambda e)$ is invertible in $\mathcal{B}$. Thus,
\begin{displaymath}
\sigma(f(a)) \subseteq \sigma(a)\,.
\end{displaymath}
Hence $R_{\sigma}(f(a)) \leq R_{\sigma}(a)$ for every $a \in \mathcal{A}$. Therefore, by the result from \htmladdnormallink{this entry}{http://planetmath.org/encyclopedia/NormAndSpectralRadiusInCAlgebras.html},
\begin{displaymath}
\|f(a)\| = \sqrt{R_{\sigma}(f(a)^*f(a))} = \sqrt{R_{\sigma}(f(a^*a))} \leq \sqrt{R_{\sigma}(a^*a)}= \|a\|\,.
\end{displaymath}

We conclude that $f$ is bounded and $\|f\| \leq 1$.

If $\mathcal{A}$ or $\mathcal{B}$ do not have identity elements, we can consider their \htmladdnormallink{minimal unitizations}{http://planetmath.org/encyclopedia/Unitization.html}, and the result follows from the above argument. $\square$

{\bf \emph{Proof of Corollary :}} This follows from the fact that $f^{-1}$ is also a *-homomorphism and therefore $\|f^{-1}(b)\|\leq \|b\|$ for every $b \in \mathcal{B}$. $\square$

\end{document}