I am reading book "A short course on Spectral Theory", written by William Averson and I got some stucks.
Lemma 2.4.4 (page 54) claimed that:
Let $A$ be a normal operator on a Hilbert space $H$ and assume that the $*-$ algebra generated by $A$ and the identity has a cyclic vector. Then $A$ is diagonalizable.
In order to proof it, the author stated that
Recalling that the functional calculus for normal operators provides a $*-$ homomorphism $f \in C(X) \rightarrow f(A) \in B(H)$.
In here, $X$ is the spectrum of $A$.
If we denote $ \varphi : f \in C(X) \rightarrow f(A) \in B(H)$, what is the adjoint $\varphi ^*$ of $\varphi$?
Furthermore, how to prove that $\varphi$ is a $*-$ homomorphism?
Let $A$ and $B$ be two (Banach) *-algebras. Then a map $\varphi:A\to B$ is called a *-homomorphism if:
The functional calculus for normal operators consists in a *-homomorphism $\varphi: C(X)\to B(H)$ between the $*$-algebras $C(X)$ and $B(H)$.
Here, $C(X)$ is a *-algebra with pointwise function sum and multiplication, and the * is given by complex conjugation, whereas in $B(H)$ the operations are operator sum and composition, and the * is given by the adjoint operator.
Therefore, if we denote $\varphi(f)$ as $f(A)$, we get that $\varphi(f)^*=f(A)^*=\bar{f}(A)$. In particular, $f(A)$ is self-adjoint iff $f$ is a real valued function.
To prove that there exist a $*$-homomorphism $\varphi: C(X)\to B(H)$, you construct it as follows. You first set $\varphi(1\mapsto x)=I$, $\varphi(x\mapsto x)=A$, $\varphi(x\mapsto \bar{x})=A^*$, and then extend it first to all polynomials $p(x,x^*)$ by linearity and multiplicativity, and then to all continuous functions in $C(X)$ by density. It is not hard to check that the resulting extended map is still a $*$-homomorphism, in fact an isometric one. Its range is the sub-*-algebra of $B(H)$ generated by $A$ (i.e. the smallest *-algebra containing $A$).