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Version 9 |
| {\bf Theorem -} Let $\mathcal{A}, \mathcal{B}$ be \PMlinkname{$C^*$-algebras}{CAlgebra} and $f:\mathcal{A} \longrightarrow \mathcal{B}$ a *-homomorphism. Then $f$ is \PMlinkname{bounded}{ContinuousLinearMapping} and $\|f\| \leq 1$ (where $\|f\|$ is the \PMlinkname{norm}{OperatorNorm} of $f$ seen as a linear operator between the spaces $\mathcal{A}$ and $\mathcal{B}$). |
{\bf Theorem -} Let $\mathcal{A}, \mathcal{B}$ be \PMlinkname{$C^*$-algebras}{CAlgebra} and $f:\mathcal{A} \longrightarrow \mathcal{B}$ a *-homomorphism. Then $f$ is \PMlinkname{bounded}{ContinuousLinearMapping} and $\|f\| \leq 1$ (where $\|f\|$ is the \PMlinkname{norm}{OperatorNorm} of $f$ seen as a linear operator between the spaces $\mathcal{A}$ and $\mathcal{B}$). |
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| For this reason it is often said that homomorphisms between $C^*$-algebras are \PMlinkname{automatically continuous}{ContinuousLinearMapping}. |
For this reason it is often said that homomorphisms between $C^*$-algebras are \PMlinkname{automatically continuous}{ContinuousLinearMapping}. |
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| {\bf Corollary -} A *-isomorphism between $C^*$-algebras is an \PMlinkname{isometric isomorphism}{IsometricIsomorphism}.\\ |
{\bf Corollary -} A *-isomorphism between $C^*$-algebras is an \PMlinkname{isometric isomorphism}{IsometricIsomorphism}.\\ |
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$\;$ |
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| {\bf \emph{Proof of Theorem :}} Let us first suppose that $\mathcal{A}$ and $\mathcal{B}$ have identity elements, both denoted by $e$. |
{\bf \emph{Proof of Theorem :}} Let us first suppose that $\mathcal{A}$ and $\mathcal{B}$ have identity elements, both denoted by $e$. |
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| We denote by $\sigma(x)$ and $R_{\sigma}(x)$ the spectrum and the spectral radius of an element $x \in \mathcal{A}$ or $\mathcal{B}$. |
We denote by $\sigma(x)$ and $R_{\sigma}(x)$ the spectrum and the spectral radius of an element $x \in \mathcal{A}$ or $\mathcal{B}$. |
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| Let $a \in \mathcal{A}$ and $\lambda \in \mathbb{C}$. If $a- \lambda e$ is invertible in $\mathcal{A}$, then $f(a- \lambda e)$ is invertible in $\mathcal{B}$. Thus, |
Let $a \in \mathcal{A}$ and $\lambda \in \mathbb{C}$. If $a- \lambda e$ is invertible in $\mathcal{A}$, then $f(a- \lambda e)$ is invertible in $\mathcal{B}$. Thus, |
| \begin{displaymath} |
\begin{displaymath} |
| \sigma(f(a)) \subseteq \sigma(a)\,. |
\sigma(f(a)) \subseteq \sigma(a)\,. |
| \end{displaymath} |
\end{displaymath} |
| Hence $R_{\sigma}(f(a)) \leq R_{\sigma}(a)$ for every $a \in \mathcal{A}$. Therefore, by the result from \PMlinkname{this entry}{NormAndSpectralRadiusInCAlgebras}, |
Hence $R_{\sigma}(f(a)) \leq R_{\sigma}(a)$ for every $a \in \mathcal{A}$. Therefore, by the result from \PMlinkname{this entry}{NormAndSpectralRadiusInCAlgebras}, |
| \begin{displaymath} |
\begin{displaymath} |
| \|f(a)\| = \sqrt{R_{\sigma}(f(a)^*f(a))} = \sqrt{R_{\sigma}(f(a^*a))} \leq \sqrt{R_{\sigma}(a^*a)}= \|a\|\,. |
\|f(a)\| = \sqrt{R_{\sigma}(f(a)^*f(a))} = \sqrt{R_{\sigma}(f(a^*a))} \leq \sqrt{R_{\sigma}(a^*a)}= \|a\|\,. |
| \end{displaymath} |
\end{displaymath} |
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| We conclude that $f$ is \PMlinkescapetext{bounded} and $\|f\| \leq 1$. |
We conclude that $f$ is \PMlinkescapetext{bounded} and $\|f\| \leq 1$. |
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| If $\mathcal{A}$ or $\mathcal{B}$ do not have identity elements, we can consider their minimal unitizations, and the result follows from the above \PMlinkescapetext{argument}. $\square$ |
If $\mathcal{A}$ or $\mathcal{B}$ do not have identity elements, we can consider their minimal unitizations, and the result follows from the above \PMlinkescapetext{argument}. $\square$ |
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| {\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 $b \in \mathcal{B}$. $\square$ |
{\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 $b \in \mathcal{B}$. $\square$ |
