TITLE: Understanding the Schwarz reflection principle
QUESTION [9 upvotes]: I am currently reading Stein and Shakarchi's Complex Analysis, and I think there is something I am not quite understanding about the Schwarz reflection principle. Here is my problem:
Suppose $f$ is a holomorphic function on $\Omega^+$ (an open subset of the upper complex plane) that extends continuously to $I$ (a subset of $\mathbb{R}$). Let $\Omega^-$ be the reflection of $\Omega^+$ across the real axis.
Take $F(z) = f(z)$ if $z \in \Omega^+$ and $F(z) = f(\overline{z})$ is $z \in \Omega^-$. We can extend $F$ continuously to $I$. Why isn't the function $F$ holomorphic on $\Omega^+ \cup I \cup \Omega^-$?
I think there's some detail of a proof that I overlooked. My intuition tells me that $F$ isn't holomorphic for the same reason that a function defined on $\mathbb{R}^+$ isn't necessarily differentiable at zero if you extend it to be an even function.

REPLY [3 votes]: You left out the condition that $f$ takes real values on $I.$ Pretend that $0 \in I,$ it changes nothing. Once we succeed in extending to a holomorphic $G$ on both sides of the real axis, this says that the power series of $G$ around $0$ has all real coefficients, if for no better reason than that all derivatives of $G$ at $0$ are real. In turn, this says that $$ G( \bar{z}) = \overline{G(z)}   $$