I'm trying to work on this exercise from Conway's Complex Analysis textbook (Exercise 3, Chapter 9, Section 1). Here $G$ denotes a symmetric region, $G_+$ the points of $G$ above the real axis and $G_0$ the points of $G$ on the real axis.
Let $f\colon G_+ \cup G_0 \to \mathbb{C}_\infty$ be a continuous function such that $f$ is meromorphic on $G_+$. Also suppose that $f(x) \in \mathbb{R}$ if $x \in G_0$. Show that there is a meromorphic function $g\colon G \to \mathbb{C}_\infty$ such that $g(z) = f(z)$ for $z \in G_+ \cup G_0$.
I read about using an adapted version of the Morera's theorem to prove this on Wikipedia but I have no idea what they were referring to.
I thought about trying to use the standard Morera's theorem on triangles that passed through or had a pole in its interior but had no ideia how to work on that.
Can anyone give me some hints on what to do here? Thanks in advance!
Best Answer
The idea of the reflection principle is that you define
$$ g(z) = \begin{cases} f(z) & z \in G_+ \\ f(z) & z \in G_0 \\ \overline{f(\bar z)} & z \in G_- \end{cases} $$
Now you need to prove several claims:
I'll sketch the proof of the last claim in that list; the others are easier to show.
Choose an open, simply connected set $G_*$ with $G_0 \subseteq G_* \subseteq G$, but small enough so that $G_*$ doesn't contain any poles. This is possible because $f$ takes finite values on $G_0$ and $f$ is continuous, so the poles can't have a limit point in $G_0$.
Now use Morera's theorem. Choose a triangle in $G_*$. If the triangle is fully contained in $G_+$ or $G_-$ then there's no problem here; the path integral around the boundary is $0$ by Cauchy's theorem. If the triangle crosses the boundary $Im(z)=0$ then break the triangle into two smaller triangles, one in $G_+ \cup G_0$ and the other in $G_- \cup G_0$, and compute the two path integrals separately.
Say $T_-$ is the triangle intersected with $G_- \cup G_0$. Then $\int_{T_- - iy} g(z)dz = 0$ for all $y>0$ (assuming $y$ is small enough to keep this adjusted triangle inside our domain), and we have $$0 = \lim_{y \to 0 ^+} \int_{T_- - iy}g(z)dz = \lim_{y \to 0 ^+} \int_{T_-} g(z+iy)dz = \int_{T_-} f(z)dz$$ where the last step relied on $f$ being uniformly continuous on the relevant region. We have uniform continuity because $f$ is continuous on a closed, bounded set that contains the relevant region for this limit of integrals.
Use the same argument to conclude $\int_{T_+} g(z)dz = 0$. Then combining the two parts of the path, we get the conclusion we need to prove $g$ is holomorphic on $G_*$ by Morera's theorem.