Let $\mathcal{A},\mathcal{B}$ be two unital C*-algebras and consider a *-homomorphism $\pi: \mathcal{A} \rightarrow \mathcal{B}$. I know that in general $\pi$ is contractive, i.e. $\vert\vert \pi(A) \vert\vert \leq \vert\vert A \vert\vert, \forall A\in \mathcal{A}$.
I want to show that under the additional assumption that $\pi(A)>0, \forall A>0$ one has equality, i.e. $\pi$ is an isometry: $\vert\vert \pi(A) \vert\vert = \vert\vert A \vert\vert, \forall A\in \mathcal{A}$
The crucial step in establishing the previous inequality lies in the fact that $\forall A\in\mathcal{A}$
$$r(\pi(AA^*)) \leq r(AA^*),$$
where $r$ denotes the spectral radius, respectively. Since $AA^*$ is positive, I sense that the additional condition enters at this point, but I can't finish the proof.
Am I on the wrong foot? Help is highly appreciated!
Best Answer
Your assumption is that $\pi$ is faithful. Being a $*$-homomorphism, this implies that $\pi$ is injective, because $$ \pi(X)=0\ \implies\ \pi(X^*X)=\pi(X)^*\pi(X)=0\ \implies\ X^*X=0\ \implies\ X=0. $$ Now we need to use that the image of $\pi$ is closed and so it is a C$^*$-algebra (I added a proof at the end).
We showed that $\pi$ is injective, so bijective onto its image. Then $\pi^{-1}$ is a $*$-homomorphism, thus contractive. Then, for any $X\in\mathcal A$, $$ \|X\|=\|\pi^{-1}\pi(X)\|\leq\|\pi(X)\|\leq\|X\|. $$
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Proof that a homomorphic image of a C$^*$-algebra is closed (this is how Kadison-Ringrose prove it in Theorem 4.1.9 of their first volume).
Let $Y\in\overline{\pi(\mathcal A)}$. We want to show that $Y\in\pi(\mathcal A)$.
We have that $Y=\lim\pi(A_n)$ for some sequence $\{A_n\}$ in $\mathcal A$. As $Y^*=\lim\pi(A_n^*)$, the real and imaginary parts of $Y$ are limits of the corresponding real and imaginary parts of the $A_n$. So we may assume, without loss of generality, that $Y$ is selfadjoint.
Also, by choosing a subsequence if necessary, we may assume that $\|\pi(A_{n+1})-\pi(A_n)\|<2^{-n}$ for all $n$. In particular, $\sigma(\pi(A_{n+1}-A_n))\subset[-2^{-n},2^{-n}]$.
Choose continuous functions $f_n$ with the property that $f_n(t)=t$ when $t\in[-2^{-n},2^{-n}]$ and $|f_n(t)|\leq2^{-n}$ for all $n$. Let $$ A=A_1+\sum_nf_n(A_{n+1}-A_n)\in\mathcal A $$ (the series converges because $\|f_n(A_{n+1}-A_n)\|<2^{-n}$ for all $n$).
The key point is that, since $f_n$ is the identity on the spectrum of $\pi(A_{n+1}-A_n)$, it is the identity when evaluated on the operator. Then, as $\pi$ is a continuous $*$-homomorphism, $$ \pi(A)=\lim_m\pi(A_1)+\sum_{n=1}^m\pi(f_n(A_{n+1}-A_n))\\ =\lim_m\pi(A_1)+\sum_{n=1}^mf_n(\pi(A_{n+1}-A_n))\\ =\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1}-A_n)\\ =\lim_m\pi(A_1)+\sum_{n=1}^m\pi(A_{n+1})-\pi(A_n)\\ =\lim_m\pi(A_m)=Y. $$