# $f(x)\in \mathbb{R}$, $x\in D\cap\mathbb{R}$, implies $\overline{f(x)}= f(\overline{x})$, $x\in D$?

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?

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.
• 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. – Ludwig May 22 at 0:36
• $g$ 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 – Chinnapparaj R May 22 at 0:37