3

Let $U\subset \mathbb{C}$ be a nonempty connected open set such that for every $z\in U$, $\overline z\in U$.

Let $f$ be analytic on $U$. Suppose $f(x)\in\mathbb R$ for every $x\in U\cap\mathbb R$. Prove that $f(\overline{z})=\overline{f(z)}$ for any $z \in U$.

By definition, I know that $f$ analytic on $U$ means that for every $z_0 \in U$, there exists $r>0$ and a sequence of complex numbers $\left(a_n\right)_{n=0}^\infty$ such that $f(z)=\sum_{n=0}^\infty a_n\left(z-z_0\right)^n$ on the disc $D(z_0,r)\subseteq U$. I see that $f$ takes points without imaginary components to other points without imaginary components. But I don't see how this implies a symmetry that $f(\overline{z})=\overline{f(z)}$ for $z \in \mathbb C \setminus \mathbb R$. It seems like a very strong conclusion and I'm not sure how I would prove it.

I have also shown in the preceding question that $U\cap\mathbb R$ contains an open interval, however I am not sure if that detail is meant to be helpful to this question.

EthanAlvaree
  • 3,412
  • 2
    Earlier today, you had a question about $\overline{f(\overline{z})}$. That can help you a lot with this one. – Daniel Fischer Feb 11 '15 at 14:30
  • Yes, you're right. Here is a link to my other question: http://math.stackexchange.com/questions/1143343/if-f-is-analytic-prove-that-overlinef-overlinez-is-also-analytic – EthanAlvaree Feb 11 '15 at 14:35
  • 2
    Now, here you have $V = U$. Can you see the connection? – Daniel Fischer Feb 11 '15 at 14:36
  • Great! I think I've about got it now. I posted a proof attempt in a comment to Jonas's answer. Does that do it? – EthanAlvaree Feb 11 '15 at 16:28
  • Yes, it does. You had previously shown that $U\cap \mathbb{R}$ contains a non-empty open interval, call it $I$. This interval $I$ is the subset of $U$ which has an accumulation point [all of its points are accumulation points] in $U$. By hypothesis, we have $g(z) = f(z)$ for all $z\in I$. Then the identity theorem tells us that $g \equiv f$ on all of $U$. – Daniel Fischer Feb 11 '15 at 16:42

1 Answers1

4

$f(z)$ and $g(z)=\overline{f(\overline z)}$ are analytic functions on a connected open set that are equal on a set containing limit points in the domain, that interval on the real line you showed exists. Hence $f=g$ by the identity theorem.

Jonas Meyer
  • 53,602
  • I think I understand! We will apply the uniqueness/identity theorem: "Suppose both $f$ and $g$ are analytic on $U$, and there is $S\subseteq U$ with an accumulation point in $U$ such that $f(z) = g(z)$ for $z ∈ S$. Then $f ≡ g$ in $U$." In this example, we are given that $f$ is analytic on $U$, and choose $g$ as $g(z)=\overline{f\left(\overline{z}\right)}$; a previous question proves $g$ is also analytic. Now for every $x \in U ∩ \mathbb{R}$, we have $x=\overline{x}$, so $f(x)=f(\overline{x})$. Also, since $f(x) \in \mathbb{R}$, this showns $f(x)=\overline{f(\overline{x})}$ on $U∩\mathbb{R}$. – EthanAlvaree Feb 11 '15 at 16:23
  • (cont.) Therefore, the uniqueness theorem gives that $f(x)=\overline{f(\overline{x})}$ on $U$. The only details I think I've missed are (1) the part of the theorem about "an accumulation point in $U$" which I haven't mentioned anything about... And (2) I also haven't used the open interval (subset of $U \cap \mathbb{R}$) which I showed exists in a previous question. Did I miss something in my proof? – EthanAlvaree Feb 11 '15 at 16:26
  • Yes, you seem to have missed why $f(x) = \overline{f(\overline x)}$ for $x\in U\cap \mathbb R$ implies $f(z) = \overline{f(\overline z)}$ for $z\in S$ for some set $S$ with an accumulation point in $U$. Take $S=U\cap \mathbb R$; you can show that this set has an accumulation point in $U$, using the fact that it contains an interval. – Jonas Meyer Feb 11 '15 at 17:03