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Suppose $f$ is a analytic function defined on $\bar{D}(0;1)$ and has real value on the boundary. I'm trying to show $f$ can be extended to entire plane by $$g(z) = \begin{cases}f(z) &, \lvert z\rvert \leqslant 1\\ \frac{1}{\overline{f(\overline{z}^{-1})}}, &, \lvert z\rvert > 1\end{cases} $$ I tried to use $\gamma(z)=e^{iz}$, which sends real axis to unit circle, for Schwarz Reflection Principle. But I did not get the result. May I get a help?

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  • $\begingroup$ You may find guidance in this question, especially by looking on @DanielFischer answer. $\endgroup$ Commented Aug 3, 2015 at 16:02
  • $\begingroup$ I checked that question, but I couldn't know how to get $\frac{1}{\overline{f(\overline{z}^{-1})}}$ $\endgroup$ Commented Aug 3, 2015 at 16:24

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You can write various forms of the reflection principle, it seems to me that the formula that you are writing is correct for a map from unit disk to C which maps a unit circle to a unit circle (and not the real axis). In the case of a map from unit disk which maps the boundary to the real axis the formula is $$ g(z) = \begin{cases} f(z) \quad |z|\leq 1, \\ \overline{f(\frac{1}{\bar{z}})} \quad |z|>1 \end{cases} $$ which is what TrialAndError has.

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  • $\begingroup$ Note this differs from the linked question, as we have the boundary real in this case, but they had the boundary tending to |f| = 1. $\endgroup$ Commented Dec 7, 2020 at 22:56
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Assuming that $f$ is real on the boundary, then, for $|z|=1$, \begin{align} \sum_{n=0}^{\infty}f_{n}z^{n} & = \overline{\sum_{n=0}^{\infty}f_{n}z^{n}} \\ & = \sum_{n=0}^{\infty}\overline{f_{n}z^{n}} \\ & = \sum_{n=0}^{\infty}\overline{f_n}\frac{1}{z^{n}}. \end{align}

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  • $\begingroup$ And therefore $f$ is the real constant $f_0.$ $\endgroup$
    – zhw.
    Commented Jul 20, 2017 at 6:37

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