Is this Function Holomorphic (Riesz-Herglotz Representation)?

Question

Let $g(z)$ be a holomorphic function on the closed unit disc $\overline {\mathbb D}$

Is the function $ g_1(w) = 1/2\pi \int_0^{2\pi} \operatorname{Re}(g(e^{i\theta}))(e^{i\theta} + w)/(e^{i\theta} - w) \, d\theta $ holomorphic on the open unit disc $ {\mathbb D}$ ?

From the context (see below) I presume that it is but I don't see how to prove this.

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Context

This is part of a proof here on p.29 :
https://su.diva-portal.org/smash/get/diva2:1069992/FULLTEXT01.pdf
to establish the Riesz-Herglotz representation of a holomorphic function on ${\mathbb D}$ with a non-negative real part.

I follow the proof up to establishing an expression for $w \in \mathbb D,$ $$ g(w) = \frac 1 {2\pi} \int_0^{2\pi} g(e^{i\theta}) \operatorname{Re} \left( \frac{e^{i\theta} + w}{e^{i\theta} - w}\right) \, d\theta $$

I can also see that $\operatorname{Re}(g(w)) = \operatorname{Re}(g_1(w))$ so if $g_1$ is holomorphic one can then say that $g, g_1$ differ only by a complex constant.

Then I could continue to follow the proof. Help would be appreciated.

Answer 1

Theorem 7 in Rudin's RCA (it is actually Theorem 10.7 in my Indian edition of the book) immediately tells you that $g_1$ is holomorphic in $D$. Here are the details: let $g_2(w)=\frac 1 {2\pi }\int_0^{2\pi }\frac 1 {e^{i\theta }-w} d\mu (\theta )$ where $d\mu (\theta ) =Re (g(e^{i\theta )})e^{i\theta } \, d\theta$ and $g_3(w)=\frac 1 {2\pi }\int_0^{2\pi }\frac 1 {e^{i\theta }-w} d\nu (\theta )$ where $d\nu (\theta ) =Re (g(e^{i\theta )}) \, d\theta$. Then $g_1(w)=g_2(w)+wg_3(w)$. $g_2$ and $g_3$ are both holomorphic by the theorem in Rudin's book.

Answer 2

The answer goes along the lines of the answer to the question Analytic functions defined by integrals.

In our case, we begin by making two easy observations:

(1). The function $\theta \mapsto Re[g(e^{i \theta})]$ is bounded for $\theta \in [0,2\pi)$. This is due to the assumption that $g$ is holomorphic in $\overline{\mathbb{D}}$.

(2). The function $w \mapsto (e^{i \theta} + w)/(e^{i \theta} - w)$ is holomorphic in $\mathbb{D}$ for any fixed $\theta \in [0,2\pi)$.

From (1) and (2), we quickly infer the following:

(3). Given a compact set $U \subseteq \mathbb{D}$, there exists a constant $C \in \mathbb{R}$, such that for any $w \in U$, it holds that $$\int_0^{2\pi}\left|Re[g(e^{i \theta})]\frac{e^{i \theta} + w}{e^{i \theta} - w}\right|\mathrm{d}\theta \leq C.$$

Morera's theorem now gives that $g_1$ is holomorphic in $\mathbb{D}$. Indeed, if $T \subseteq \mathbb{D}$ is any triangle, it holds that \begin{align} \oint_Tg_1(w)\mathrm{d}w & = \frac{1}{2\pi}\oint_T\int_0^{2\pi}Re[g(e^{i \theta})]\frac{e^{i \theta} + w}{e^{i \theta} - w}\mathrm{d}\theta\:\mathrm{d}w \\ & = \frac{1}{2\pi}\int_0^{2\pi}Re[g(e^{i \theta})]\underbrace{\oint_T\frac{e^{i \theta} + w}{e^{i \theta} - w}\mathrm{d}w}_{= 0\:\mathrm{due~to~ (2)}}\:\mathrm{d}\theta \\ & = 0. \end{align} Here, it was valid to change the order of integration as (3) allows us to apply Fubini's theorem and change the order of integration.