TITLE: Sufficient condition for a *-homomorphism between C*-algebras being isometric
QUESTION [6 upvotes]: 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!

REPLY [4 votes]: Simpler. Assume $\pi:\mathcal{A}\to\mathcal{B}$ is an injective $\ast$-Algebra Homomorphism. (No assumption of continuity.) It is shown, that $\pi$ is automatically an isometry.
Lemma 1. In a $C^{\ast}$-Algebra $a\leq M\Longleftrightarrow \|a\|\leq M$, if $a\geq 0$.
A simple proof involves observing the spectrum $\sigma(M-a)$ in relation to $\sigma(a)$.
Lemma 2. $\pi(a)\geq 0\Longleftrightarrow a\geq 0$.
Proof. $(\Longleftarrow)$: If $a\geq 0$, then $\pi(a)=\pi(\sqrt{a}^{\ast}\sqrt{a})=\pi(\sqrt{a})^{\ast}\pi(\sqrt{a})\geq0$.
$(\Longrightarrow)$: If $\pi(a)\geq 0$, then $\pi(a)=\sqrt{\pi(a)^{\ast}\pi(a)}=\sqrt{\pi(a^{\ast}a)}$.
Now for $c\in\mathcal{A}$ positive, $\pi(\sqrt{c})^{2}=\pi(\sqrt{c}^{2})=f(c)$. Now $\pi(c), \pi(\sqrt{c})\geq 0$, since $c,\sqrt{c}\geq 0$ (by the first part). Hence $\pi(\sqrt{c})=\sqrt{\pi(c)}$.
Thus $\pi(a)=\sqrt{\pi(a^{\ast}a)}=\pi(\sqrt{a^{\ast}a})$. Since $\pi$ is injective, it follows that $a=\sqrt{a^{\ast}a}\geq 0$.
$\blacksquare$
Corollary. $\pi$ is an isometry.
Proof. Consider $a\in\mathcal{A}$ positive. Set $M:=\|a\|$ and $N:=\|\pi(a)\|$.
(1) It holds that $M-a\geq 0$, thus $M-\pi(a)=\pi(M-a)\geq 0$, thus $\pi(a)\leq M$, thus $\|\pi(a)\|\leq M=\|a\|$.
(2) Similarly $\pi(N-a)=N-\pi(a)\geq 0$, thus $N-a\geq 0$, thus $a\leq N$, thus $\|a\|\leq N=\|\pi(a)\|$.
Hence $\|\pi(a)\|=\|a\|$ for $a$ positive.
For $a\in\mathcal{A}$ a general element: $\|a\|^{2}=\|a^{\ast}a\|=\|\pi(a^{\ast}a)\|=\|\pi(a)^{\ast}\pi(a)\|=\|\pi(a)\|^{2}$. Hence $\|a\|=\|\pi(a)\|$.
$\blacksquare$