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Let $f\colon D\to \mathbb{C}$, where $D\subseteq \mathbb{C}$, be such that $f(x)$ is real for all $x \in D\cap\mathbb{R}$. Let $\overline{a}$ denote the complex conjugate of $a\in\mathbb{C}$.

My question. Does $f$ satisfy $\overline{f(x)}= f(\overline{x})$ for all $x\in D$? If not, under which conditions on $ f $ does this hold true?

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Of course not, since you can change any value of $f$ to let this equlity fail.

If $f(z)$ is entire, it holds. Let $g(z)=\overline{f(z)}-f(\bar{z})$, it's easy to verify that $g$ is entire. However $g(x)=0$ for all $x\in\mathbb R$, which has limit points. Therefore $g\equiv 0$.

Edit: Just as Chinnapparaj R said in the comment, since $D$ is open, $D\cap \mathbb R$ is also open and then the zero set of $g$ has a limit point.

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  • $\begingroup$ Sorry, there was a typo in the domain of $f$. I guess that now $f$ analytic in $D$ is sufficient for the equality to hold true. $\endgroup$ – Ludwig May 22 at 0:36
  • $\begingroup$ $g $ is not zero for all $x \in \Bbb R$. Since $D\cap \Bbb R \neq \varnothing$, $D$ contains a non empty intervel of $\Bbb R$ and hence has a limit points. So by identity theorem, $g$ vanishes $\endgroup$ – Chinnapparaj R May 22 at 0:37

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