# Proposed proof of operator theory result

Hi I am interested in checking my proposed solution to the following problem in Operator Theory: Please give me hints as to how to improve the proposed proof rather than the full correct solution.

Question: Let $\mathcal{A}$ be a unital Banach Algebra. Consider $\{a_{n}\}_{n} \subset \mathcal{A}$ where $a_{n} \to a \in \mathcal{A}$. Suppose that $\sigma(a_{n})$, $\sigma(a) \subset \Omega$ where $\Omega \subset \mathbb{C}$ is an open set. Show that for any $f \in \text{hol}(\Omega)$ it follows that $f(a_{n}) \to f(a)$.

Proposed proof:

Choose $\Gamma$ such that winding number $\text{ind}_{\Gamma}(\lambda) = 1$ for all $\lambda \in \sigma(a_{n}) \cup \sigma(a)$ and $\{ z : \text{ind}_{\Gamma}(z) \neq 0 \} \subset \Omega$.

In the following I use that $f(z)\big((z-a_{n})^{-1}-(z-a)^{-1}\big)$ is continuous on compact set $\Gamma$ and therefore bounded and also I use the second resolvent identity which states that $(z-a_{n})^{-1}-(z-a)^{-1} = (z-a_{n})^{-1}(a_{n}-a)(z-a)^{-1}$.

\begin{align} 0 \leq \|f(a_{n})- f(a)\| &= \bigg\| \frac{1}{2 \pi i} \int_{\Gamma}f(z)\big((z-a_{n})^{-1}-(z-a)^{-1}\big)dz\bigg\| \\& = \bigg\|\frac{1}{2 \pi i} \int_{\Gamma}f(z)(z-a_{n})^{-1}(a_{n}-a)(z-a)^{-1}dz\bigg\| \\& \leq \frac{1}{2 \pi }\bigg(\sup_{z \in \Gamma} \bigg\| f(z)(z-a_{n})^{-1}(a_{n}-a)(z-a)^{-1}\bigg\| \bigg\| \Gamma \bigg\|\bigg) \\& \leq \frac{1}{2 \pi }\bigg(\sup_{z \in \Gamma} \bigg\| f(z)(z-a_{n})^{-1}\bigg\| \bigg\|(a_{n}-a)\bigg\| \bigg\|(z-a)^{-1}\bigg\| \bigg\| \Gamma \bigg\|\bigg) \to 0 \end{align} Therefore $f(a_{n}) \to f(a)$ in $\mathcal{A}$. $\square$

Thanks for any assistance.

• Seems OK, apart from the fact that you have a pure imaginary number greater or equal to a real number. You want to delete the $i$. – uniquesolution Sep 9 '15 at 13:38
• @uniquesolution Thanks for your response. Oh you mean drop the $i$ since $|\frac{1}{2 \pi i}| = \frac{1}{2 \pi}$? – user103184 Sep 9 '15 at 16:54
• - Yes, precisely. – uniquesolution Sep 10 '15 at 11:50