Is the function $V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it})$ analytic on the unit disc $|z|<1$?

Question

Let $\mathbb {T}$ be the unit circle in the complex plane and let $\mathbb{D}$ denote the open unit disc. Let $\mu$ be a complex Borel measure on $\mathbb {T}$ . Is the function $V$ defined as $$V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it})\;\; \;\;z\in \mathbb{D} $$ analytic on $\mathbb{D}$? If yes, then How?

Answer 1

Let $$F(z,t)= \sum_{n=-\infty}^{\infty} z^{|n|}e^{-i|n|t} =\sum_{n=0}^{\infty} z^{n}e^{-int}+\sum_{n=1}^{\infty} z^{n}e^{-int} \\= \frac{1}{1-ze^{-it}}+ \frac{ze^{-it}}{1-ze^{-it}} =\frac {e^{it}+z}{e^{it}-z}$$ $z\mapsto F(z,t)$ is analytic on $\mathbb D$. Since $\mathbb T $ is compact, as $\mu$ is a Borel measure and hence finite on very compact sets in particular $\mu(\Bbb T)<\infty$. For fixed $z\in \Bbb D$ the map $t\mapsto F(z,t)$ is bounded on $\Bbb{T}$ and thus $V (z)\in L^1(\Bbb T)$

$$V (z)=\int_{ \mathbb {T} }\frac {e^{it}+z}{e^{it}-z}d\mu (e^{it}) = \sum_{n=-\infty}^{\infty} z^{|n|}\int_{ \mathbb {T} }e^{-i|n|t}d\mu (e^{it}) \\= \mu(\mathbb T) + \sum_{n=1}^{\infty} \left(2\int_{ \mathbb {T} }e^{-int}d\mu (e^{it})\right)z^{n} \;\; \;\;z\in \mathbb{D}$$

that is

$$V (z)=\mu(\mathbb T) + \sum_{n=1}^{\infty} a_nz^{n} \;\; \;\;z\in \mathbb{D}$$ with $$a_n =\left(2\int_{ \mathbb {T} }e^{-int}d\mu (e^{it})\right) \qquad\text{and}\qquad \mu(\mathbb T) =\int_{ \mathbb {T} }d\mu (e^{it}) $$

Answer 2

$\bullet\;$ First show, for each fixed $t$, the function $$ \frac {e^{it}+z}{e^{it}-z} $$ is analytic on the open unit disk.

$\bullet\;$ Then find what is needed to verify this calculation: if $\gamma$ is a closed contour contained in the open unit disk, then $$ \int_\gamma V (z)\;dz= \int_{ \mathbb {T} }\int_\gamma \frac {e^{it}+z}{e^{it}-z}\;dz\;d\mu (e^{it}) = \int_{\mathbb T} 0\;d\mu(e^{it}) = 0. $$

$\bullet\;$ Conclude that $V(z)$ is analytic.