Theorem - Let $\mathcal{A}, \mathcal{B}$ be $C^*$ -algebras and $f:\mathcal{A} \longrightarrow \mathcal{B}$ a *-homomorphism. Then $f$ is bounded and $\|f\| \leq 1$ (where $\|f\|$ is the norm of $f$ seen as a linear operator between the spaces $\mathcal{A}$ and $\mathcal{B}$ ).

For this reason it is often said that homomorphisms between $C^*$ -algebras are automatically continuous.

Corollary - A *-isomorphism between $C^*$ -algebras is an isometric isomorphism.
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Proof of Theorem : Let us first suppose that $\mathcal{A}$ and $\mathcal{B}$ have identity elements, both denoted by $e$ .

We denote by $\sigma(x)$ and $R_{\sigma}(x)$ the spectrum and the spectral radius of an element $x \in \mathcal{A}$ or $\mathcal{B}$ .

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,

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 minimal unitizations, and the result follows from the above argument. $\square$

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