# Intersection $f(G\cap \mathbf{R}) \subset \mathbf{R} \Leftrightarrow f(\bar{z}) = \overline{f(z)}$

$G\subset \mathbf{C} , G= \{z\in \mathbf{C}| \overline{z}\in G\} = \overline{G}$, then for $f\in \mathcal{O}(G)$ it holds that:$$f(G\cap \mathbf{R}) \subset \mathbf{R} \Leftrightarrow \forall z\in G : f(\overline{z}) = \overline{f(z)}$$

This is an example in the script of our professor, however there is no proof for it and it hasn't been shown during the lectures either.

Proof : $"\Rightarrow " :$ Let $f(G\cap \mathbf{R}) \subset \mathbf{R}$, so all points on the real axis are mapped to the real axis. That means $f$ does not have any imaginary part, because then this would not be true anymore. Since f does not have any imaginary part. it follows also that : $f(\overline{z}) = \overline{f(z)}$

$"\Leftarrow" :$ Let $f(\overline{z}) = \overline{f(z)}$, so f can not have any imaginary part, otherwise this would not be true anymore. But if f doesn't have any imaginary part, then the map of the real axis is always on the real axis. So $f(G\cap \mathbf{R}) \subset \mathbf{R}$

What I mean with not having any imaginary part is that it is of the form $f(z) = z ; f(z)=2z; f(z) = 2z+z^{3}$ etc.

I am not sure if this is the right thought, and how to express this better.

Help is greatly appreciated.

-
Thanks for your attention. I tried to show it in my edited post. – PumaDAce Dec 6 '11 at 16:07
Yes, G is a domain. :) – PumaDAce Dec 6 '11 at 16:21
Ah, okay. I think it's good to mention that sort of thing, just for reassurance. – Dylan Moreland Dec 6 '11 at 16:25
I don't follow your $"\Rightarrow"$. If I write $f = u(x, y) + iv(x, y)$, then we find that $v(x, 0) = 0$ for all $x \in G \cap \mathbf{R}$. I don't see how to finish off from there -- maybe there are facts about harmonic functions that we could use? – Dylan Moreland Dec 6 '11 at 16:49

($\Rightarrow$) I think it should be straightforward to check that the function $\overline{f(\bar z)}$ is holomorphic on $G$. Then $f(z) - \overline{f(\bar{z})}$ is a holomorphic function on $G$ which vanishes along an interval on the real axis.

($\Leftarrow$) I think you have the right idea, but we should write down some equations. Try to use the fact that $z = \bar z$ if and only if $z \in \mathbf R$; if you take $z \in \mathbf R \cap G$ and conjugate $f(z)$, what happens?

The problem is very related to the Schwarz reflection principle.

-

:) :) :) Dylan Moreland :) :) :)

$$"\Rightarrow: "$$ If $f(G\cap\mathbf{R})\subset \mathbf{R}$, then for every $z_{0} \in \mathbf{R}$ we can write $f(z)= \sum_{k=0}^{\infty} a_{k}(z-z_{0})^{k}$, $a_{k}\in \mathbf{R}$. From this it follows immediately that $\overline{f(\overline{z})}=f(z)$ in a neighborhood of the real axis. Because of the identity theorem these two holomorphic functions must be the same everywhere.$$"\Leftarrow":$$ If $z\in \mathbf{R}$, $\overline{z}=z$. Then $f(\overline{z})=\overline{f(z)}= f(z)$ $\forall z\in \mathbf{R}$, so $f(z) \in \mathbf{R}$ , which means that $f(G\cap \mathbf{R})\subset \mathbf{R}$

Is this ok? Thank you for your help.

-
No problem! I agree with your "$\Leftarrow$". For the other direction, it isn't clear to me how you get $a_k \in \mathbf R$. I think the holomorphicity should be obtainable without thinking about power series--let me try to work that out. The identity theorem is definitely what you want to use. – Dylan Moreland Dec 6 '11 at 19:06