Is $f(a)$ normal in $A\ $?

Question

Fact $:$ Let $K$ be a compact subset of $\mathbb C$ and $\Omega$ be an open set containing $K.$ Then there exists a cycle $\Gamma : = \sum\limits_{j=1}^{n} n_{j} \gamma_{j}$ in $\Omega \setminus K$ such that $\text {ind}_{\Gamma} (z) = 1$ for all $z \in K$ and $\text {ind}_{\Gamma} (z) = 0$ for all $z \in \Omega^{\complement}.$

With the help of the above fact a question has been proposed in our lecture note. Here it is $:$

Let $A$ be a unital Banach algebra endowed with an involution $\ast$ and $a \in A$ be a normal element (i.e. $aa^{\ast} = a^{\ast}a$). Let $\text {Hol} (a)$ denote the set of all functions which are holomorphic on an open neighborhood of the spectrum $\sigma (a)$ of $a$ and $f \in \text {Hol} (a)$ be such that it is defined and holomorphic on $\Omega,$ where $\Omega$ is an open set containing a compact set $K.$ Let $\Gamma$ be as above. Now define $$f(a) : = \frac {1} {2 \pi i} \int_{\Gamma} f(z) (z - a)^{-1}\ dz = \sum\limits_{j = 1}^{n} \frac {n_j} {2 \pi i} \int_{\gamma_{j}} f(z) (z - a)^{-1}\ dz.$$ Then find $f(a)^{*}$ and also analyze whether $f(a)$ is normal or not.

I think that $$f(a)^{\ast} = - \frac {1} {2 \pi i} \int_{\Gamma} \overline {f(z)} (\overline {z} - a^{\ast} )^{-1}\ dz.$$ Although I am not quite sure about that. Also I am perplexed as to how do I compute $f(a) f(a)^{\ast}$ or $f(a)^{\ast} f(a)\ $? Any suggestion in this regard would be greatly appreciated.

Thanks for your time.

Answer 1

I will try to give you a hint rather than a rigorous proof.

If $f$ is a polynomial, then, doubtless, $f(a)^*=\overline{f}(a^*)$, where $\overline{f}$ is the polynomial obtained from $f$ by conjugating its coefficients. Moreover, since $a$ and $a^*$ commute, we have that $f(a)$ and $f(a)^*=\overline{f}(a^*)$ commute. For arbitrary “nice enough” function $f$ the things should be the same.

If you assume that $||x^*||=||x||$ for any $x\in A$, then you can prove $f(a)^*=\overline{f}(a^*)$ as follows. Note that $f(a)$ is defined as a certain Bochner’s integral, see this wiki page. For simple functions you, obviously, can swap the integration and the involution $*$, since such integral is given by a finite sum. For an arbitray function (not necessarily simple) you can swap them as well, using the very definition of Bochner’s integral and the identity $||x^*||=||x||$. Namely, you should use the fact that, if $y_n\to y$ as $n\to\infty$, then $y_n^*\to y^*$. But after swapping the integral and the involution, we have to “conjugate the measure” as well as the function that we integrate. You forgot to “conjugate the measure”. So, your expression for $f(a)^*$ differs from the correct one by $dz\mapsto d\overline{z}$. After this little change your formula turns into $\overline{f}(a^*)$.

To see that $f(a)$ is normal, you should rewrite $f(a)f(a)^*=f(a)\overline{f}(a^*)$ as a double integral and then change the order of factors corresponding to $a$ and $a^*$, using that $a$ is normal.

Answer 2

Assume that $A$ is a $B^\ast$-algebra (that is, the involution satisfies $\|aa^\ast\|= \|a^2\|$; in particular $a \to a^\ast$ is continuous). Let $r_z$ denote the resolvent, that is $r_z(a):=(ze-a)^{-1}$ $(z \notin \sigma(a))$. From the holomorphic functional calculus we first need the following fact:

If $a,b \in A$, $f \in Hol(a)$, $g \in Hol(b)$ then $ab=ba$ implies $f(a)g(b)=g(b)f(a)$: First, $ab=ba$ implies $br_z(a) = r_z(a)b$. Thus $b$ can be passed through the integral defining $f(a)$, leading to $bf(a)=f(a)b$. Now the same conclusion for the commuting elements $f(a),b$ leads to $f(a)g(b)=g(b)f(a)$.

Now let $a$ be normal, $f \in Hol(a)$, $\sigma(a) \subseteq D_f=\Omega_f$, and $\Gamma_f$ an admissible cycle for $a$ in $\Omega_f$. Set $g(z):= \overline{f(\overline z})$ for $z \in\Omega_g:=\{\overline{z}:z \in \Omega_f\}$. Now $g$ is holomorphic and $\sigma(a^\ast) \subseteq \Omega_g$. Thus $g \in Hol(a^\ast)$. Now consider the conjugate cycle $\Gamma_g$ defined as $\Gamma_g:=-\overline{\Gamma_f}$, that is each path $\gamma:[0,1] \to \Omega_f$ in $\Gamma_f$ is replaced by the path $t \mapsto \overline{\gamma(1-t)}$. Note here that conjugation changes the rotational direction of a closed path. Thus $\Gamma_g$ is an admissible cycle for $a^\ast$ in $\Omega_g$. Now $$ f(a)^\ast=\left(\frac{1}{2\pi i} \int_{\Gamma_f} f(z)r_z(a)dz\right)^\ast = -\frac{1}{2\pi i} \int_{\Gamma_f} \overline{f(z)}r_z(a)^\ast dz $$ $$ = -\frac{1}{2\pi i} \int_{\Gamma_f} \overline{f(z)}r_{\overline z}(a^\ast) dz = \frac{1}{2\pi i} \int_{\Gamma_g} \overline{f(\overline z)}r_{z}(a^\ast) dz = \frac{1}{2\pi i} \int_{\Gamma_g} g(z)r_{z}(a^\ast) dz =g(a^\ast). $$ Thus $f(a)^\ast f(a) = g(a^\ast) f(a) =f(a) g(a^\ast) = f(a)f(a)^\ast $.