$C^{*}$-algebra homomorphisms are continuous

Theorem - Let $𝒜,ℬ$ be $C^{*}$-algebras (http://planetmath.org/CAlgebra) and $f:𝒜⟶ℬ$ a *-homomorphism. Then $f$ is bounded (http://planetmath.org/ContinuousLinearMapping) and $\parallel f\parallel \le 1$ (where $\parallel f\parallel$ is the norm (http://planetmath.org/OperatorNorm) of $f$ seen as a linear operator between the spaces $𝒜$ and $ℬ$).

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

Corollary - A *-isomorphism between $C^{*}$-algebras is an isometric isomorphism (http://planetmath.org/IsometricIsomorphism).
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Proof of Theorem : Let us first suppose that $𝒜$ and $ℬ$ 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 $x\in 𝒜$ or $ℬ$.

Let $a\in 𝒜$ and $\lambda \in ℂ$. If $a-\lambda e$ is invertible in $𝒜$, then $f(a-\lambda e)$ is invertible in $ℬ$. Thus,

$$\sigma (f(a))\subseteq \sigma (a).$$

Hence $R_{\sigma }(f(a))\le R_{\sigma }(a)$ for every $a\in 𝒜$. Therefore, by the result from this entry (http://planetmath.org/NormAndSpectralRadiusInCAlgebras),

$$\parallel f(a)\parallel =\sqrt{R_{\sigma }(f{(a)}^{*}f(a))}=\sqrt{R_{\sigma }(f(a^{*}a))}\le \sqrt{R_{\sigma }(a^{*}a)}=\parallel a\parallel .$$

We conclude that $f$ is and $\parallel f\parallel \le 1$.

If $𝒜$ or $ℬ$ do not have identity elements, we can consider their minimal unitizations, and the result follows from the above . $\mathrm{\square }$

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