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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.

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up vote 4 down vote accepted

$f(\bar{z})$ 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.

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Thank you! This was exactly my problem! I had assumed that $f(\overline{z})$ was holomorphic because it was a pretty "nice and simple" function. I just started studying complex analysis, so I had the definition of holomorphic memorized without much intution, but I understand the idea a lot better now! –  Alan C May 27 '11 at 23:49
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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 $$ G( \bar{z}) = \overline{G(z)} $$

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Thanks! That was the statement and construction given by the book, and while I could understand it, I didn't understand why my construction didn't work. –  Alan C May 27 '11 at 23:51
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