A lecturer once gave a very elementary proof that $*$-homomorphisms between C*-algebras are always norm-decreasing. It is well-known that this holds for a $*$-homomorphism between a Banach algebra and a C*-algebra, but all the proofs I find involve the spectral radius and so.
If I remember it well, the proof he gave used the C*-algebra structure in the domain, and (as always) had something to do with a geometric series.
Does anyone knows how to do so?
Let $f\colon A\to B$ be a *-homomorphism. Let us note that $f$ cannot enlarge spectra of self-adjoint elements in $A$, that is, for all $y\in A$ self-adjoint we have $\mbox{sp}(f(y))\setminus \{0\} \subseteq \mbox{sp}(y)\setminus \{0\}$. By the spectral radius formula, we have $\|y\|=r(y)$. Now, let $x\in A$. It follows that
$\|f(x)\|^2 = \|f(x^*x)\| = r(f(x^*x))\leqslant r(x^*x)=\|x\|^2$. $\square$
Another strategy (involving geometric series) is to notice that $f$ is positive and $\|f\|=\|f(1)\|=1$ (in the unital case). In this case, you can tweak the proof given by julien in Why is every positive linear map between $C^*$-algebras bounded?
