Is $f(x) \in \mathbb{R}$ if $x \in \mathbb{R}$? Let $f:\Omega\subset\mathbb{C}\rightarrow\mathbb{C}$ be holomorphic with $\Omega$ open. Assume there is an infinite sequence $(x_n)_n$ in $\Omega \cap \mathbb{R}$ that converges to $x_0 \in \Omega \cap \mathbb{R}$ and assume that for every $n$, $f(x_n) \in \mathbb{R}$. Let $D_r$ be the open disk of radius $r&gt;0$ and centre $x_0$ such that $D_r \subseteq \Omega$. Is it true that $f(x) \in \mathbb{R}$ for every $x \in D_r \cap \mathbb{R}$?
I tried proving this by constructing a holomorphic function $g$ such that $g(z)$ is real whenever $z$ is real and $g(x_n)=f(x_n)$. Then by the identity theorem it follows that $g=f$. Unfortunately, I was unable to find such $g$.
Any help would be welcome. Thank you in advance.
Hint: $g(z) = \frac12\big( f(z) + \overline{f(\overline z)} \big)$. (This is the same construction as in a proof of the Schwarz reflection principle.)
First of all, $D_r$ seems not to be part of the question or problem?
Secondly: your assumptions is that there exists a particular sequence $(x_n)$ converging to a particular point $x$ and you want to show a general statement that would be &quot;$f(x)$ is real whenever $x$ is real&quot;. There is a clear lack of assumptions to show that.
This is the only thing that can be shown with your assumptions:
Let $f : \Omega \to \mathbb{C}$ be continuous. If $x \in \Omega$ is the limit of the sequence $(x_n) \in (\Omega)^{\mathbb{N}}$ and if for every $n$, $f(x_n) \in \mathbb{R}$, then $f(x)\in \mathbb{R}$. This is because $f$ is continuous at $x$ and because $\mathbb{R}$ is a closed subset of $\mathbb{C}$.