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.

Answer 1

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}