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Again, I'm struggling with a proof.

Cauchy integral theorem for convex sets (Preliminary lemma)

Let $\Omega\subseteq \mathbb{C}$ open and convex, $ p\in\Omega,\ f$ : $\Omega\rightarrow \mathbb{C}$ continuous, $f\in H(\Omega\backslash \{p\})$ ($f$ analytic on $\Omega \backslash \{p\}$. Then there is a $F\in H(\Omega)$ (F analytic on $\Omega$) with $F'(z)=f(z)$ for all $ z\in\Omega$.

It also folows that $\displaystyle \int\limits_{\gamma}f(z)\mathrm{d}z=0$ for every piecewise continuously differentiable and closed path $\gamma\in\Omega$.

Cauchy integral formula

Let $\Omega\subseteq \mathbb{C}$ be open and convex, $f\in H(\Omega)$ ($f$ analytic on $\Omega$) and $\gamma$ a piecewise continuously differentiable and closed path in $\Omega$. Then, we have: $$ f(z)\cdot \mathrm{ind}_{\gamma}(z)=\frac{1}{2\pi i}\int\frac{f(\xi)}{\xi-z}\mathrm{d}\xi\ , (z\in\Omega\backslash \gamma^{*}) $$ where $\mathrm{ind}_\gamma$ is the winding number and $\gamma^* = \gamma([a,b])$ for $\gamma:[a,b]\rightarrow\mathbb{C}$.


Fix $z\in\Omega\backslash \gamma^{*}$ and define $g:\Omega\rightarrow \mathbb{C}$ by: $$ g(\xi):=\left\{\begin{array}{l} \frac{f(\xi)-f(z)}{\xi-z},\ \xi\in\Omega\backslash \{z\}\\ f'(z),\ \xi=z \end{array}\right. $$ Then we have $g$ continuous on $\Omega$ and $g \in H(\Omega\backslash \{z\})$.

$\color{red}{\text{(1) Why is g continuous on } \Omega \text{ and analytical on } \Omega\backslash \{z\}}$

From the Cauchy integral theorem for convex sets we have $\displaystyle \frac{1}{2\pi i}\int\limits_{\gamma}g(\xi)\mathrm{d}\xi=0$, so:

$\color{red}{\text{(2) Where does that } \frac{1}{2\pi i} \text{ come from? There is no factor in the mentioned Cauchy integral theorem}}$

$$ \frac{1}{2\pi i}\int_{\gamma}\frac{f(\xi)-f(z)}{\xi-z}\mathrm{d}\xi\ =0\ (z\in\Omega\backslash \gamma^{*}) $$

This completes the proof.

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up vote 1 down vote accepted

(1) $g$ is analytic on $\Omega\backslash \{z\}$ because both the function $\xi\mapsto f(\xi)-f(z)$ and the function $\xi \mapsto \frac{1}{\xi-z}$ are analytic there, hence the product is analytic. $g$ is continuous on $\Omega\backslash \{z\}$ since it is analytic there, the only point in question is at $z$, but this follows since $f$ is differentiable there.

(2) The factor doesn't matter, and is a convenience so that you end up with $f(z)\cdot \mathrm{ind}_{\gamma}(z)$, but it will work out anyway, even if you don't multiply it.

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For (1), notice that for $\xi\in\Omega\setminus\{z\}$, $g(\xi)$ is a quotient of analytic functions for which the denominator is non-vanishing. Therefore $g$ is analytic and so also continuous on $\Omega\setminus\{z\}$. To see that $g$ is continuous at $z$, notice that \begin{equation} g(z)=f'(z)=\lim_{\xi\to z} \frac{f(\xi)-f(z)}{\xi-z}=\lim_{\xi\to z} g(\xi). \end{equation} Thus $g$ is continuous at $z$ and so also on all of $\Omega$.

For (2), notice that $g$ satisfies the hypotheses the Cauchy integral theorem for convex sets, which yields $\int_\gamma g(\xi) d\xi=0$. Multiplying by $\frac{1}{2\pi i}$ gives the desired conclusion.

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$\,(1)\,$ It is analytic in $\,\Omega- \{z\}\,$ because it is the quotient of two analytical functions and the denominator never vanishes there, and it is continue in $\,\Omega\,$ because$$\lim_{\xi\to z}\frac{f(\xi)-f(z)}{\xi - z}=f'(z)$$

$\,(2)\,$ Yes, there is. Check here

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