Showing that the Holomorphic Functional Calculus preserves adjoints.

Question

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.

Answer 1

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 $f(T^*) = f(T)^*$.

Answer 2

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.