Showing that the Holomorphic Functional Calculus preserves adjoints. Let $T\in B(X)$ for some complex Banach space $X$. For any holomorphic $f$ on $\Omega\supset \sigma(T)$ I'd like to show that $f(T^*)=f(T)^*$, where $f(T)$ is defined via the Holomorphic Functional Calculus. My idea was as follows: We know, via basic facts of the adjoint that
$$f(T^*)=\frac{1}{2\pi i}\int_\Gamma f(z)(zI-T^*)^{-1}dz=\frac{1}{2\pi i}\int_\Gamma \left(f(z)(zI-T)^{-1}\right)^*dz.$$
We also know, by definition of the Bochner integral that
$$f(T)=\frac{1}{2\pi i}\lim_{n\to\infty}\int_\Gamma s_ndz$$
where $s_n$ are simple functions converging to $f(z)(zI-T)^{-1}$. Now I had always held the vague belief that $T\mapsto T^*$ was strong operator continuous, so, as the adjoint distributes over finite sums, and the integral of simple functions is defined as a finite sum, we would simply have
$$f(T)^*=\left(\frac{1}{2\pi i}\lim_{n\to\infty}\int_\Gamma s_ndz\right)^*=\frac{1}{2\pi i}\lim_{n\to\infty}\int_\Gamma s_n^*dz=f(T^*).$$
I thought I should first verify my vague belief before patting myself on the back, and unfortunately I discovered that adjoint map is not in fact strongly continuous, and so my proof fails. Does anyone have a better idea  how to approach it? I'd really appreciate it.
 A: I believe this follows from the fact that integration and application of a continuous linear operator can be interchanged.
Note that we have $(A^*)^{-1} = (A^{-1})^*$.
\begin{eqnarray}
(f(T^*) \phi) x &=& {1 \over 2\pi i} (\int_\Gamma f(z) (z-T^*)^{-1} dz \,\phi)x \\
&=& {1 \over 2\pi i} (\int_\Gamma f(z) (z-T^*)^{-1} \phi \, dz)x \\
&=& {1 \over 2\pi i} \int_\Gamma f(z) ((z-T^*)^{-1} \phi) x\, dz \\
&=& {1 \over 2\pi i} \int_\Gamma f(z) (((z-T)^{-1} )^*\phi) x\, dz \\
&=& {1 \over 2\pi i} \int_\Gamma f(z) (\phi((z-T)^{-1} x) \, dz \\
&=& {1 \over 2\pi i} \int_\Gamma\phi( f(z)(z-T)^{-1} x) \, dz \\
&=& \phi({1 \over 2\pi i} \int_\Gamma f(z)(z-T)^{-1} x \, dz )\\
&=& \phi({1 \over 2\pi i} \int_\Gamma f(z)(z-T)^{-1} dz \, x)\\
&=& \phi(f(T))x \\
&=& ((f(T)^*) \phi)x
\end{eqnarray}
Since this holds for all $x \in X, \phi \in X^*$ we have
$f(T^*) = f(T)^*$.
A: Suppose
$$
                   f(z) = \sum_{n=0}^{\infty}a_n z^{n}
$$
converges in a region where $|z| < r$ and suppose $\sigma(T)\subset B_r(0)$. Then the functional calculus will give
$$
                   f(T) = \sum_{n=0}^{\infty}a_nT^n.
$$
From this it follows that
$$                   f(T)^*=\sum_{n=0}^{\infty}\overline{a_n}(T^*)^n \ne f(T^*).
$$
If you define $f^*(z)=\overline{f(\overline{z})}$, then $f(T)^*=f^*(T^*)$.
Using $f^*(z)=\overline{f(\overline{z})}$ as a definition of $f^*$, and using the identity $\sigma(T^*)=\overline{\sigma(T)}$ gives
\begin{align}
   f(T)^*&=\left( \frac{1}{2\pi i}\oint_{C}f(\lambda)(\lambda I-T)^{-1}d\lambda \right)^* \\
  &= -\frac{1}{2\pi i}\oint_{C}\overline{f(\lambda)}(\overline{\lambda}I-T^*)^{-1}d\overline{\lambda} \\
  &= -\frac{1}{2\pi i}\oint_{C}f^*(\overline{\lambda})(\overline{\lambda}I-T^*)^{-1}d\overline{\lambda} \\
  &= \frac{1}{2\pi i}\oint_{\overline{C}}f^*(\mu)(\mu I-T^*)^{-1}d\mu = f^*(T^*).
\end{align}
The negative disappears in the last line because of contour orientation.
