injective $C^{*}$-algebra homomorphism is isometric

Theorem - Let $𝒜$ and $ℬ$ be $C^{*}$-algebras (http://planetmath.org/CAlgebra) and $\mathrm{\Phi }:𝒜⟶ℬ$ an injective *-homomorphism. Then $\parallel \mathrm{\Phi }(x)\parallel =\parallel x\parallel$ and $\sigma (\mathrm{\Phi }(x))=\sigma (x)$ for every $x\in 𝒜$, where $\sigma (y)$ denotes the spectrum of the element $y$.

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Proof: It suffices to prove the result for unital $C^{*}$-algebras, since the general case follows directly by considering the minimal unitizations of $𝒜$ and $ℬ$. So we assume that $𝒜$ and $ℬ$ are unital and we will denote their identity elements by $e$, being clear from context which one is being used.

Let us first prove the second part of the theorem for normal elements $x\in 𝒜$. It is clear that $\sigma (\mathrm{\Phi }(x))\subseteq \sigma (x)$ since if $x-\lambda e$ invertible for some $\lambda \in 𝒞$, then so is $\mathrm{\Phi }(x)-\lambda e=\mathrm{\Phi }(x-\lambda e)$. Suppose the inclusion is strict, then there is a non-zero function $f\in C(\sigma (x))$ whose restriction to $\sigma (\mathrm{\Phi }(x))$ is zero (here $C(\sigma (x))$ denotes the $C^{*}$-algebra of continuous functions $\sigma (x)⟶ℂ$). Thus we have, by the continuous functional calculus, that $f(x)\ne 0$ and also that

$\mathrm{\Phi }(f(x))=f(\mathrm{\Phi }(x))=0$

by the continuous functional calculus and the result on this entry (http://planetmath.org/CAlgebraHomomorphismsPreserveContinuousFunctionalCalculus). Thus, we conclude that $\mathrm{\Phi }$ is not injective and which is a contradiction. Hence we must have $\sigma (\mathrm{\Phi }(x))=\sigma (x)$.

Let $R_{\sigma }(z)$ denote the spectral radius of the element $z$. From the norm and spectral radius relation in $C^{*}$-algebras (http://planetmath.org/NormAndSpectralRadiusInCAlgebras) we know that, for an arbitrary element $x\in 𝒜$, we have that

${\parallel x\parallel }^2=R_{\sigma }(x^{*}x)$

Since the element $x^{*}x$ is normal, from the preceding paragraph it follows that $R_{\sigma }(x^{*}x)=R_{\sigma }(\mathrm{\Phi }(x^{*}x))$, and hence we conclude that

${\parallel x\parallel }^2=R_{\sigma }(x^{*}x)=R_{\sigma }\left(\mathrm{\Phi }{(x)}^{*}\mathrm{\Phi }(x)\right)={\parallel \mathrm{\Phi }(x)\parallel }^2$

i.e. $\parallel \mathrm{\Phi }(x)\parallel =\parallel x\parallel$.

Since $\mathrm{\Phi }$ is isometric, $\mathrm{\Phi }(𝒜)$ is closed *-subalgebra of $ℬ$, i.e. $\mathrm{\Phi }(𝒜)$ is a $C^{*}$-subalgebra of $ℬ$, and it is isomorphic to $𝒜$. Using the spectral invariance theorem we conclude that $\sigma (x)=\sigma (\mathrm{\Phi }(x))$ for every $x\in 𝒜$. $\mathrm{\square }$