Ytukyg

Consider $mathcal{A}, mathcal{B}$ two unital $C^*$-algebras and $varphi: mathcal{A} rightarrow mathcal{B}$. Let $A in mathcal{A}$ be a selfadjoint element and consider $f in C(sigma(A), mathbb{C})$. I need to prove the following:

$(i)$ $sigma(f(A)) = f(sigma(A)) = {f(lambda) : lambda in sigma(A)}$;

$(ii)$ $varphi(f(A)) = f(varphi(A))$.

I know that as $A$ is selfadjoint, $sigma(A) subset [-vertvert Avertvert, vertvert A vertvert]$. I also know that if $mathcal{P}$ is a polynomial, then $sigma(mathcal{P}(A)) = mathcal{P}(sigma(A))$. I was thinking about approximating any function $f in C(sigma(A), mathbb{C})$ by polynomials, but I'm not sure how the spectrum behaves.

Also, for $(ii)$, I know that as $varphi$ is a $*$-homomorphism, then $sigma(varphi(A)) subset sigma(A)$.

In fact, if $lambda in sigma(varphi(A))$, we have that $(lambda1 - varphi(A))$ is not invertible. But using the fact that $varphi$ is a $*$-homomorphism, we can write $lambda1 - varphi(A) = varphi(lambda1 - A)$ and it follows that $lambda1 - A$ is not invertible. So we get that $lambda in sigma(A)$.

I don't know how to proceed in any of the cases. Could someone help me?

The first part, use the fact that the Gelfand transform is an isomorphism between $C^*(A)$ and $C(sigma(A))$, hence preserves spectra, and that $sigma_{C^*(A)}(f(A))=sigma_{mathcal A}(f(A))$.

For the second part, approximate $f$ by polynomials, which commute with a $*$-homomorphism.