Real valued analytic function defined on a connected set is constant

Question

Let $G$ be a connected set and $f : G \rightarrow \mathbb{C}$ a real valued analytic function. Prove that $f$ is constant.

My idea to prove the result is to prove a subset $A \neq \varnothing$ of the connected set $G$ is both open and closed. So $G=A$ Take $f(w) = a$ $$A = \{z\colon z \in G, f(z) = a\}$$ Now I want to show that $A$ is infinite. How to do it? After that it is easy to prove $A=G$.

If you don't have the open mapping principle yet, I think the easiest way to prove it is using the Cauchy-Riemann equations to conclude that the partial derivatives are $0$. — Daniel Fischer, Jul 07 '13 at 12:22
You should also assume that $G$ is an open subset of $\Bbb C$ otherwise it is clearly false. — Siméon, Jan 01 '14 at 18:00
@HopelessFool: Take $f(z)=z$ for all $z \in \Bbb C$ and $G = \Bbb R$ which is a connected subset of $\Bbb C$. — Siméon, Jan 02 '14 at 08:57
@Siméon Thank you for pointing it out. Very nice. Anything else related to this question to learn? — Supriyo, Jan 02 '14 at 13:42

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

5

This isn 't an answer to your question, but rather an answer to your problem.

Let $f$ be an holomorphic function on a connected set $G$.

There exist functions $u,v$ such that $f=u+iv$.

Now using Daniel Fischer's hint, since $f$ is holomorphic, $u_x=v_y$ and $u_y=-v_x$. By hypothesis $\text{im}(f)\subseteq \Bbb R$, therefore $v=\textbf 0$ and it follows that $u_x=\textbf 0= v_x$.

Finally use the $f'=u_x+iv_x$ and the fact that $G$ is connected to conclude. (This is used in the last step).

edited Apr 13 '17 at 12:20

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answered Jul 07 '13 at 12:28

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@Dutta Anytime.${}$ – Git Gud Jul 07 '13 at 14:54

Real valued analytic function defined on a connected set is constant

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