Problem: "Let $(A,\|\cdot\|)$ and $(B,\|\cdot\|)$ be unital $C^*$-algebras and let $\phi:A\to B$ be an injective $^*$-homomorphism. Show that $\phi$ is isometric. Hint: Treat the case of self-adjoint elements firstly and use the fact that it suffices to consider the case when $A$ and $B$ are commutative."
I'll collect together the key facts that, I think, I need to use:
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For $C^*$-algebras $(A,\|\cdot\|),(B,\|\cdot\|)$ we call $\phi: A\to B$ a $^*$-homomorphism if (i) $\phi$ is linear, (ii) $\phi(a_1a_2)=\phi(a_1)\phi(a_2)\,\forall a_1,a_2\in A$, and (iii) $\phi(a^*)=\phi(a)^*\,\forall a\in A$.
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For $C^*$-algebras $(A,\|\cdot\|),(B,\|\cdot\|)$ and $\phi: A\to B$ we have that $r(a)=\|a\|\,\forall a\in A: a=a^*$, where $r(a)$ is the spectral radius of $a\in A$.
Attempt: I've made multiple attempts at this problem, and I'm not sure which, if any, are going to yield any fruit. I know that in order to show that this injective $^*$-homomorphism is isometric I need to show that $\|a\|=\|\phi(a)\|,\,\forall a\in A$.
Consider firstly those $a\in A:a=a^*$. Then we know that by the $C^*$-property $\|a\|^2=\|a^*a\|$ for self-adjoint $a\in A$. Then:
$$\|a\|^2=\|a^*a\|=r(a^*a)=r(a^*)r(a)=r(a)^2$$
But I don't see that this gets me anywhere, other than reiterating what I already know. Is there a connection between $r(a)$ and $\phi(a)$ that I can make use of? It seems to me, that in some sense, if the properties above for $\phi$ held for $r$ I might be able to get somewhere.
Alternatively, since we have an injective homomorphism we know that $\phi$ maps the identity element in $A$ to the identity element in $B$. Then, consider:
$$1=\|e_B\|=\|\phi(e_A)\|=\|\phi(a^*a)\|=\|\phi(a^*)\phi(a)\|=\|\phi(a)^*\phi(a)\|$$
And then $\phi(a)^*=b^*$ for some $b\in B:b=b^*$. Then we have that,
$$\|\phi(a)^*\phi(a)\|=\|b^*b\|=\|b\|^2$$
This, again, doesn't tell me anything that I already know.
Can anybody direct me on how best to proceed? In particular, in accordance with the hint, what exactly am I being told when hinted that "it suffices to consider the case when $A$ and $B$ are commutative?
Best Answer
The easiest way to show that $\phi$ is isometric goes as follows: Using that $\lVert a \rVert$ equals its spectral radius for self-adjoint $a$, one sees that the norm on $A$ is uniquely determined. Now, define a norm $$ \rho(a) := \lVert \phi(a)\rVert \qquad (a \in A). $$ Then the norm $\rho$ makes $A$ into a C*-algebra. Therefore, $\rho = \lVert \cdot \lVert$. It follows that $\phi$ is isometric.