# Showing that a function is analytic

Let $f: \partial D(0,1) \mapsto \mathbb{R}$ be continuous. Define $g: D(0,1) \mapsto \mathbb{C}$ by $g(z)= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(e^{i\theta}) \frac{e^{i\theta}+z}{e^{i\theta}-z} d\theta$ . Show that $g$ is analytic. I don't know how to do this. Maybe I can use the definition or the Cauchy Riemann equations but there must be a more elegant way of proving that a function defined by a integral is analytic. Thx.

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## 2 Answers

Use Morera's theorem. First of all, show that $g$ is continuous on $D(0,1)$. Next, let $\gamma$ be any piecewise $C^1$ closed curve contained in $D(0,1)$ (it is enough to take $\gamma$ any triangle or any rectangle). Then, by Fubini's theorem $$\int_\gamma g(z)\,dz=\frac{1}{2\,\pi}\int_{-\pi}^{\pi}f(e^{i\theta})\Bigl(\int_\gamma \frac{e^{i\theta}+z}{e^{i\theta}-z}\,dz\Bigr)\,d\theta=0.$$

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You have $$g(z) = {1\over 2\pi i}\int_{S^1} {f(w)(w+z)\,dw\over w(w-z)}$$ As long as $z$ does not lie in $S^1$, you can differentiate under the integral sign and determine the function's derivative.

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