# Show that $f(z)=\frac{1}{2\pi}\int\limits_0^{2\pi}f\left(\frac{e^{i\theta}+z}{1+\overline{z}e^{i\theta}}\right)d\theta$

Let $$f$$ be analytic on domain $$\Omega$$ which contains the closed unit disk $$\overline{\mathbb{D}}$$. Show that
(a) $$f(0)=\frac{1}{2\pi}\int_0^{2\pi}f(e^{i\theta})d\theta$$
(b) Use part (a) to show that whenever $$z\in\mathbb{D}$$, $$f(z)=\frac{1}{2\pi}\int_0^{2\pi}f\left(\frac{e^{i\theta}+z}{1+\overline{z}e^{i\theta}}\right)d\theta$$ Hint: Consider the conformal self maps of the unit disk

For part (a) I used the Cauchy integral formula on the unit disk and then substituted the parameterization $$z(\theta)=e^{i\theta}$$ where $$0<\theta<2\pi$$

But for part (b) I don't see how it is related with part (a) and how the automorphisms on the unit disk comes into play.
Anyway I know that $$\Phi_{\alpha}:\mathbb{D}\rightarrow\mathbb{D}$$ defined as $$\Phi_\alpha(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$$ is an automorphism when $$|\alpha|<1$$

Appreciate your help

## 1 Answer

Consider the function $$g_z(w) = \frac{z+w}{1+\overline{z}w}$$ for $$|z|<1$$ and $$|w|\leq 1$$. When $$|w| = 1$$, observe that $$|g_z(w)| = \left| \frac{z+w}{1+\overline{z}w} \right| = \left| \frac{z+w}{1+\overline{z}w} \right| \cdot \left| \frac{1}{\overline{w}} \right| = \left| \frac{z+w}{\overline{w} + \overline{z}} \right| = \left| \frac{z+w}{\overline{z+w}} \right| = 1.$$ When $$|w| < 1$$ observe that, for example, $$|g_z(0)| = |z| < 1$$. By the maximum principle, $$|g_z(w)| < 1$$ in this case. Therefore, we see that $$f \circ g_z$$ is analytic on a domain containing the closed unit disc.

So by part (a), we have $$(f\circ g_z)(0) = \frac{1}{2 \pi} \int_{0}^{2 \pi} (f \circ g_z)(e^{i \theta}) d\theta,$$ and the result follows.

• Thank you!! It works – Charith Jeewantha Mar 31 at 11:37
• @Eliot can you explain a bit more on how $|g_z(w)|<1$ implies $f(g_z)$ is analytic on a domain (that is an open set) containing the closed unit disc? – Charith Jeewantha Mar 31 at 12:34
• No problem. It is given in the problem that $f$ is analytic in a domain which contains the closed unit disc. When $|w| <1$, $g_z(w)$ takes values in the unit disc, and when $|w|=1$, $g_z(w)$ takes values on the boundary of the unit disc. So $g_z(w)$ is analytic on a domain containing the unit disc. Therefore, $f \circ g_z$ is analytic on a domain containing the unit disc. – Elliot Herrington Mar 31 at 21:51