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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.

For a compact normal operator $Ane 0$, there are eigenvalues $lambda_n$ and orthogonal projections $P_n$ onto the corresponding eigenspaces $P_nmathcalH$ with eigenvalue $lambda_n$ such that
$$
A = sum_n lambda_n P_n, ;; I=sum_nP_n\
P_n P_m = 0,;; nne m, \
P_n^2 = P_n = P_n^*.
$$
Suppose $f$ is a bounded function on the spectrum of $A$.
Then $f(A)=sum_nf(lambda_n)P_n$ is normal because $f(A)^*=sum_n overlinef(lambda_n)P_n$ commutes with $f(A)$. In fact,
$$
f(A)^*f(A)=sum_n|f(lambda_n)|^2P_n=sum_n|overlinef(lambda_n)|^2P_n=f(A)f(A)^*.
$$
Alternatively,
beginalign
|f(A)x|^2 &= |sum_n lambda_n P_n x|^2 \
& =sum_n |lambda_n|^2|P_nx|^2 \
& =sum_n |overlinelambda_n|^2|P_nx|^2 = |f(A)^*x|^2
endalign

The functional calculus is a $*$-homomorphism, and since the algebra $B(sigma(A))$ of bounded functions on $sigma(A)$ is commutative, we have
$$f(A)f^*(A)=(ff^*)(A)=(f^*f)(A)=f^*(A)f(A).$$

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 $f(A)^* = overlinef(A) in C^*(A)$.

Therefore $$f(A)f(A)^* = f(A)^*f(A)$$

since $ C^*(A)$ is a commutative $C^*$-algebra.