Why is this function holomorphic in $K(a,\rho)$?

Question

If $\phi:\partial K(a,\rho)\to \mathbb{R}$ is a Lebesgue-integrable function, why is

$$z\mapsto \frac{1}{2\pi}\int_{0}^{2\pi}\frac{e^{i\theta}+\frac{z-a}{\rho}}{e^{i\theta}-\frac{z-a}{\rho}}\phi(a+e^{i\theta})d\theta$$

holomorphic in $K(a,\rho)$?

Answer 1

You combine Morera's theorem and Fubini's theorem. For simplicity I will assume $a=0$, $\rho=1$.

Step 1. Show that the function $$z\mapsto f(z)=\int_{0}^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}\,\phi(e^{i\theta})\,d\theta$$ is continuous.

Step2. Given a piecewise $C^1$ closed curve $\gamma$ contained in $K(0,1)$, then $$ \int_\gamma f(z)\,dz=\int_\gamma\Bigl(\int_{0}^{2\pi}\frac{e^{i\theta}+z}{e^{i\theta}-z}\,\phi(e^{i\theta})\,d\theta\Bigr)\,dz=\int_{0}^{2\pi}\Bigl(\int_\gamma\frac{e^{i\theta}+z}{e^{i\theta}-z}\,\phi(e^{i\theta})\,dz\Bigr)\,d\theta=0. $$ I have used Fubini's theorem to interchange the order of integration and Cauchy's theorem to conclude that the integral is equal to $0$.

Step 3. Apply Morera's theorem.