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.


$f(\bar{z})$ 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.

  • $\begingroup$ 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! $\endgroup$ – Alan C May 27 '11 at 23:49
  • $\begingroup$ @Alan C: It's not the definition, but it's been said that the parts of math where all the manipulations Cauchy and Euler and co. did are valid is exactly made up of holomorphic functions of one variable. $\endgroup$ – Gunnar Þór Magnússon Jan 15 '16 at 5:56

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 $$ G( \bar{z}) = \overline{G(z)} $$

  • $\begingroup$ 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. $\endgroup$ – Alan C May 27 '11 at 23:51

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