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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.

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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}

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

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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.

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