# 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.

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}
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 $$f(A)^* = \overline{f}(A) \in C^*(A)$$.
Therefore $$f(A)f(A)^* = f(A)^*f(A)$$
since $$C^*(A)$$ is a commutative $$C^*$$-algebra.