I have the following problem: Let $G$ be a domain in the upper half plane such that $\partial G\cap\mathbb{R}$ is an open set of $\mathbb{R}$, i.e. a disjoint union of open intervals. Suppose that $f:G\cup(\partial G\cap\mathbb{R})\to\mathbb{C}$ is a continuous and injective function, holomorphic on $G$ and such that $f(z)\in\mathbb{R}$ for all $z\in\mathbb{R}\cap\partial G$. My question is this: Is it possible that $f$ assumes a real value on $G$? I believe no, but I cannot prove this.
I was thinking of possible ways to apply the maximum modulus principle for harmonic functions on the real or imaginary part, but I have no information on the rest of the boundary of $G$. Moreover I don't see how to take advantage of the injectivity of $f$;
Another idea that might help would be to extend $f$ to a holomorphic function defined on the domain $G\cup(\mathbb{R}\cap \partial G)\cup G^*$ (this is readily seen to be a domain) (where $^*$ denotes image under complex conjugation) by the Riemann-Schwarz reflection principle. Then again, I don't see why this new setting would help me rule out the possibility of $f$ assuming real values on interior points (note that $f_1(z)=\bar{f}(\bar{z})$ where $f_1$ is the extention)
Yes, $f$ can assume a real value on $G$. Suppose that $G=\left\{re^{i\theta}\,\middle|\,r\in(0,\infty)\wedge\theta\in\left(0,\frac{2\pi}3\right)\right\}$ and that $f(z)=z^2$. Then $\delta G\cap\mathbb R=(0,\infty)$, which is an open subset of $\mathbb R$, and $f$ is injective. But $i\in G$ and $f(i)=-1\in\mathbb R$.