I am studying Banach Spaces of Analytic Functions by Hoffman. In Chapter 5 Page 61, the textbook claims that
If $u \in L^1 (\mathbb T)$ then the function $F: \mathbb D \to \mathbb C$ \begin{align*} F(z) = \exp \left[ \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^{i\theta}+z}{e^{i\theta} - z} u(e^{i\theta})\right] d\theta \end{align*} is analytic.
I proved this directly by computing it. I was wondering if I could analyticity of $z\mapsto \frac{e^{i\theta} +z}{e^{i\theta}-z}$, $z\in \mathbb D$ to conclude that $F$ is analytic. Can I possibly use some theorem to conclude this easily?
Here's my approach anyways: Let $(h_n)$ be sequence in $\mathbb C$ of nonzero terms converging to $0$ and $z \in \mathbb D$ Then $\lvert e^{i\theta } - z - h_n \rvert \to \lvert e^{i\theta} - z \rvert > 0$. Thus there exists $M > 0$ and $N \in \mathbb N$ such that $\lvert e^{i\theta } - z - h_n \rvert > M$ for all $n \ge N$. Let $g(z)=\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{e^{i\theta}+z}{e^{i\theta} - z} u(e^{i\theta})$Consider the following for $n\ge N$:
\begin{align*} \left\lvert \frac{g(z+h_n) - g(z)}{h_n} \right\rvert &\le \frac{1}{2\pi} \int_{-\pi}^{\pi} \left\lvert \frac{2h_n e^{i\theta}}{(e^{i\theta}-z-h_n)(e^{i\theta}-z)} \right\rvert \lvert u(e^{i\theta}) \rvert d\theta \\ &\le \frac{2h_n}{M \inf_{e^{i\theta} \in \mathbb T}{|e^{i\theta}-z|}} \lVert u \rVert_1 \end{align*} This shows that $g$ is analytic and hence $e^g$ is analytic.
This is a general result, namely given $F(z,t), z \in \Omega, t \in [a,b]$ ($\Omega$ some open plane set) that is jointly continuous in $(z,t)$ and analytic in $z$ for every fixed $t \in [a,b]$, and a (piecewise) smooth simple curve (doesn't need to be closed) $\gamma$ parametrized by $\zeta(t)$ that is disjoint of $\Omega$, then $f(z)=\int_\gamma F(z,t)u(t)d\zeta(t)$ is analytic in $\Omega$ for any $u$ integrable on $[a,b]$ since $\gamma$ disjoint from $\Omega$ means that for every $z_0 \in \Omega$ there is a small closed ball around it in $\Omega$ and at a positive distance from $\gamma$ so for any triangle $T$ in that ball, one can switch integrals in $\int_Tf(z)dz=\int_T(\int_\gamma F(z,t)u(t)d\zeta(t))dz=\int_\gamma u(t)d\zeta(t)\int_T F(z,t)dz=0$ and $f$ is clearly continuous so Morera implies $f$ differentiable on the small ball, hence in $\Omega$ as it is covered by such.
One can replace $u(t)d\zeta(t)$ by some finite measure $d\mu$ (not necessarily absolutely continuous with respect to $dt$) and even just assume only some joint uniform integrability condition for $|F(z,t)|$ (plus of course analyticity in $z$) so Fubini still applies and the integrals can be switched as can limits in $z$ with the integral to ensure continuity plus Morera.
Also one can weaken the conditions on $\gamma$ to allow it to be an infinite curve parametrized by $t \in (a,b)$ some open interval that can be infinite at either or both ends, but then one has to make sure Fubini applies, so rather than state general conditions, it is better to look at such cases by hand as sometimes one can have just conditional convergence of the integral in $t$ but still can switch the integrals by some Cauchy process or the like etc
Here of course one uses that $ F(z,t)=\frac{e^{it}+z}{e^{it} - z}$ is as nice as it goes for $z$ away from the unit circle, so the integral defines analytic functions in both the unit disc and outside of it.