Understanding this wording of the Schwarz Reflection Principle I'm trying to understand what the following is trying to say:

If a function $f(z)$ is continuous on a domain $S$ that lies in the region $Im(z) \geq 0$ and meets the real axis in an interval, and if $f$ is real on that interval and analytic as a function of $z$ when $Im(z) > 0$, then $f$ can be extended to an analytic function on the region $S \cup \overline{S}$, where $\overline{S} = \{z : \overline{z} \in S\}$.

I've tried writing this differently to understand what it's saying, but I'm still struggling at some parts. What I have so far is:


*

*"$f(z)$ is continuous on a domain $S$ that lies in the region $Im(z) \geq 0$":


$f$ is continuous on $S \subset Im(z)$


*

*"and meets the real axis in an interval":


$f(z) \cap \mathbb{R} \neq \varnothing$, for $z \in I$, where $I$ is some interval


*

*"$f$ is real on that interval":


$f(z) \in \mathbb{R}$, for all $z \in I$ (this, and/or the previous part is probably wrong since this part makes the previous part pointless)


*

*"and analytic as a function of $z$ when $Im(z) > 0$":


$f(z)$ is analytic for all $Im(z) > 0$


*

*"then $f$ can be extended to an analytic function on the region $S \cup \overline{S}$":


then $f : S \rightarrow Y$ can be extended to some function $F:S \cup \overline{S} \rightarrow Y$ s.t. $F|_{S} = f$.
How can I clean this up?
 A: I'm going to deal more with the intuition of the Reflection Principle, as I feel the wording is easier to understand after that.
Consider the following picture:

Let $S$ be the red region in both cases.
Now, if we have some function $f$ defined on $S$ such that:

*

*$f$ is continuous on $S$


*$f(x)\in\mathbb R$ where $x$ is real, and $x\in S$.


*$f(z)$ is analytic when $f(z)$ isn't on the real axis.
Then we can apply the Schwarz reflection principle.
After doing this,$f$ will be extend to a function $\hat{f}$.
Whenever $\hat{f}$ is on $S$ (the red), we have that $\hat{f} = f$.
But when $\hat{f}$ is on $\overline{S}$ (the pink), we get that $\hat{f}$ is it's own thing (thus extending $f$).  Of course, $\hat{f}$ is analytic on $S\cup\overline{S}$ (the red and the pink).
A: Let's go through this piece by piece as you have.

$f(z)$ is continuous on a domain $S$ that lies in the region $Im(z) \geq 0$

This means $S\subseteq\{z\in\mathbb{C}:Im(z)\geq 0\}$ and $f:S\to\mathbb{C}$ is a continuous function.

and meets the real axis in an interval

This is referring to $S$, not $f$.  So $S\cap\mathbb{R}$ is an interval $I\subseteq\mathbb{R}$.

$f$ is real on that interval

This means $f(z)\in\mathbb{R}$ for all $z\in I$ (where as above, $I=S\cap\mathbb{R}$).

and analytic as a function of $z$ when $Im(z) > 0$

This means that the restriction of $f$ to the set $U=\{z\in S:Im(z)>0\}$ is an analytic function $U\to\mathbb{C}$.

then $f$ can be extended to an analytic function on the region $S \cup \overline{S}$

This means basically what you said, but you left out the key fact that $F$ is analytic.  To be precise, it means there exists an analytic function $F:S\cup\overline{S}\to\mathbb{C}$ such that $F|_S=f$.
A: As usual in complex analysis, for understanding this theorem you need to look at its proof, which is in two parts :


*

*Let $U$ be a simply connected open contaning an interval of the real line, and such that $U^+ = U \cap \{Im(z) > 0\}$ and $ U^-= U \cap \{Im(z) < 0\}$ are simply connected, and let $f(z)$ be holomorphic on $U^+ \cup U^-$ and continuous on $U$.
Consider $F(z) = \int_{z_0}^z f(u)du$. Since $U^+$ is simply connected, $f(z)$ has a holomorphic anti-derivative there, i.e. $F(z)$ is holomoprhic on $U^+$. Clearly the same is true for $U^-$, and note that $f(z) = F'(z)$ on $U^+ \cap U^-$ and is continuous on $U$. A little work shows that $f(z) = F'(z)$ on $U$ so that $F(z)$ is holomorphic on $U$, and by the holomorphic  $\implies$ analytic theorem, its derivative $f(z)$ is holomorphic on $U$.

*Let $f(z)$ be  holomorphic on $U^+$ (simply connected) and real on $U \cap \{Im(z) = 0\}$, then  $g(z) =\begin{cases}f(z) \text{ if } Re(z) > 0\\ \overline{f(\overline{z})} \text{ otherwise}\end{cases}$ is holomorphic on $U^+ \cup U^-$ (with $U^-$ the complex conjugate of $U^+$) and continuous on $U$, so by the preceding theorem it is holomorphic on $U$.
