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.
Best Answer
$f(\bar{z})$ is not holomorphic (unless $f$ is constant), as $d(f(\bar{z}))=f'(\bar{z}) d\bar{z}$ is not a multiple of $dz$ (but rather of $d\bar{z}$). Perhaps more intuitively: holomorphic = conformal and orientation-preserving; $f(\bar{z})$ is conformal, but changes the orientation (due to the reflection $z\mapsto\bar{z}$). Hence your function $F$ is not holomorphic on $\Omega^-$.
On the other hand, $\overline{f(\bar{z})}$ is holomorphic, as there are two reflections. If $f$ is real on $I$ then by gluing $f(z)$ on $\Omega^+$ with $\overline{f(\bar{z})}$ on $\Omega^-$ you get a function continuous on $\Omega^+\cup I\cup\Omega^-$ and holomorphic on $\Omega^+\cup\Omega^-$. It is then holomorphic on $\Omega^+\cup I\cup\Omega^-$ e.g. by Morera theorem.