How to prove $f(A^*) =f(A)^*$

Question

I'm studying Conway's functional calculus.

If $mathcal{X}$ is banach space and $A in mathcal{B}(mathcal{X})$, and $f in mathtt{Hol}(A)$ , show that $f(A)^* = f(A^*)$

Firstly, I tried it for polynomials, and it is trivial. Similarly, we can prove it at rational function.
Then, I want to apply Runge's theorem, which indicates that there is $f_n$(rational) uniformly converges to $f$.
So, $f_n(A^*) to f(A^*)$ is clear, but $f_n(A)^* to f(A)^*$ is not clear. Is it right approach?
Thanks a lot.

Fred · Accepted Answer

For $ lambda in rho(A)$ we have
$$R_{lambda}(A)^*=R_{lambda}(A^*).$$
Thus
begin{align}f(A)^*=left( frac{1}{2 pi i} int_{gamma}f(lambda)R_{lambda}(A) d lambda right)^* &=frac{1}{2 pi i} int_{gamma}f(lambda)R_{lambda}(A)^* d lambda \ &=frac{1}{2 pi i} int_{gamma}f(lambda)R_{lambda}(A^*) d lambda \ &=f(A^*).end{align}

How to prove $f(A^) =f(A)^$

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