Good day, I am having some problems with this question. Our supervisor's notes gave an answer, but she doesn't know how to get to it (it was set by an old lecturer).
Given a continuous function $f: S(0,1)\to R$, construct a holomorphic function $g:B(0,1) \to C$ which extends to a holomorphic $h:\overline{B(0,1)} \to C$ such that the real part of $h$ is equal to $f$ for all $x \in S(0,1)$.
$S(0,1)$ is circle of radius $1$ about $0$. $B(0,1)$ and $B(0,1)$ are the open and closed balls of radius $1$ about $0$, respectively.
The solution in the notes was $\displaystyle h(z) = \int_0^1 \frac{f(e^{2\pi i \theta})}{1-ze^{-2\pi i \theta}} \, d\theta $. I would like for someone to show how this was obtained, or if it was even correct. Much thanks.
(I tried to see if it has properties the question wants, but I could not simplify or evaluate it with what I know).
The only values of $z$ for which you need to evaluate $h(z)$ are those on the boundary. As for "properties the question wants", there seem to be two of those:
To show $h$ is holomorphic, I'd try using Morera's theorem:
\begin{align} \int\limits_\text{some closed curve} \!\!\!\!\!\!\!\!\! h(z)\,dz & = \int\limits_\text{curve} \int_0^1 \frac{f(e^{2\pi i \theta})}{1-ze^{-2\pi i \theta}} \, d\theta \, dz \\[10pt] & = \int_0^1 \int\limits_\text{curve} \frac{f(e^{2\pi i \theta})}{1-ze^{-2\pi i \theta}} \, dz\,d\theta & & \text{Is this step valid? See below.} \\[10pt] & = \int_0^1 0 \,d\theta & & \ldots\,\text{and is this step valid?} \\[10pt] & = 0. \end{align}
The interchange of order of integration is valid if the double integral of the absolute value of the function is finite. That's Fubini's theorem. And that holds if the function being integrated is bounded and continuous.
For now, I'm leaving this as an incomplete answer, not addressing the second bullet point above.