Schwarz reflection principle, understanding the conjugated function:

Question

Given a symmetric region $\Omega$, say, symmetric w.r.t. the real line, and f(z) defined and analytic only on $\Omega^{+}$, we can analytically continue the function to $\Omega^{-}$ with the analytic continuation $\bar{ f{(\bar{z}})}$.

But how can I see that $\bar{ f{(\bar{z}})}$, restricted to $\Omega^{+}$ is exactly f(z) itself, which is a key point in the definition of analytic continuation?

In $\Omega^{+}$, the inputs are z (whereas the inputs for $\Omega^{-}$ are $\bar{z}$), but then I still have $\bar{ f{(z})}$, which is not exactly f(z).

The restriction to the upper region should be exactly f(z). Where am I going wrong?

Thanks,

Answer 1

Let's organize our ideas. The map $z \mapsto \overline{f(\overline{z})}$ need not be an analytic extension of $f$. Instead, if we know that $f: \Omega^+ \to \Bbb C$, then we can define $F: \Omega \to \Bbb C$ by putting: $$F(z) = \begin{cases} f(z), \text{ if } z \in \Omega^+ \\ \overline{f(\overline{z})}, \text{ if } z \in \Omega^-\end{cases}$$This map is well defined if $f(\Bbb R)\subset \Bbb R$. One can prove that $F$ is analytic in $\Omega$, and clearly $F\big|_{\Omega^+} = f$. That's the extension you want.

Answer 2

Assuming $f:\Omega^+\to\Omega^+$, if $z\in\Omega^-$, $f$ is not defined. We use analytic continuation to create a function $F$ which is defined over $\Omega$.

If $z\in\Omega^-$, $\overline{z}\in\Omega^+$, and so $f(\overline{z})\in\Omega^+$.

As $\Omega$ is symmetric, we want $F(\Omega^-)\to\Omega^-$ and we get the functional equation $F(z)=\overline{F(\overline{z})}$.