Bounded function of compact normal operator on Hilbert space is normal Let $H$ be a Hilbert space and consider a compact normal linear operator $A:H \to H$. Moreover, let $f$ be a bounded function on the spectrum $\sigma(A)$ of $A$ and consider the operator $f(A)$ in the sense of functional calculus.
I want to show that $f(A)$ is also a normal operator. I know that a bounded linear operator $B:H \to H$ is normal if and only if 
$$ ||Bx||=||B^\star x|| \quad \forall x \in H,$$
which might be useful in this context.
 A: For a compact normal operator $A\ne 0$, there are eigenvalues $\lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_n\mathcal{H}$ with eigenvalue $\lambda_n$ such that
$$
                  A = \sum_{n} \lambda_n P_n, \;\; I=\sum_{n}P_n\\
                P_n P_m = 0,\;\; n\ne m, \\
                  P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=\sum_{n}f(\lambda_n)P_n$ is normal because $f(A)^*=\sum_n \overline{f(\lambda_n)}P_n$ commutes with $f(A)$. In fact,
$$
       f(A)^*f(A)=\sum_{n}|f(\lambda_n)|^2P_n=\sum_{n}|\overline{f(\lambda_n)}|^2P_n=f(A)f(A)^*.
$$
Alternatively,
\begin{align}
      \|f(A)x\|^2 &= \|\sum_n \lambda_n P_n x\|^2 \\
   & =\sum_n |\lambda_n|^2\|P_nx\|^2 \\
   & =\sum_n |\overline{\lambda_n}|^2\|P_nx\|^2  = \|f(A)^*x\|^2
\end{align}
A: For any $f \in B(\sigma(A))$  we have $f(A) \in C^*(A)$.
Since $f \mapsto f(A)$ is a $*$-homomorphism, we also have $f(A)^* = \overline{f}(A) \in C^*(A)$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $ C^*(A)$ is a commutative $C^*$-algebra.
A: The functional calculus is a $*$-homomorphism, and since the algebra $B(\sigma(A))$ of bounded functions on $\sigma(A)$ is commutative, we have 
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$
