Theorem 1.1 Part 2 Lang's Complex Analysis proof explanation

Question

Page 294, Part 2 chapter 1 in Langs "Complex Analysis."

Let $U^+$ be an open set in the upper half plane, and suppose $\delta U^+$ contains an interval $I$ of real numbers. Let $U^-$ be the reflection of $U^+$ across the real axis (i.e. the set of $\bar{z}$ with $z \in U^+$) Let $U = U^+ \cup I \cup U^-$

$Theorem 1.1:$ Suppose that $U$ is open.

$i)$ If $f$ is a function on $U$, analytic on $U^+$ and $U^-$, and continuous on $I$, then $f$ is analytic on $U$.

$ii)$ If $f$ is a function on $U^+ \cup I$, analytic on $U^+$ and continuous on $I$, and $f$ is real valued on $I$, then $f$ has a unique analytic continuation $F$ on $U$, and satisfies $F(z)=\overline{f({\bar{z}})}$.

---

Now the beginning of the proof says is that it is clear that BECAUSE $f$ is real valued on $I$ that $F$ is continuous on $I$. Is this just true because $\overline{f({\bar{z}})}=f(z)$ because $z$ is on the x-axis so $\bar{z} = z$ and $f$ is real valued so $\bar{f}=f$? General insight appreciated! Y'all are the best!!

Answer 1

Here's an answer using the pasting lemma. If you haven't seen it before, it is perhaps easiest considered as a generalization the following basic theorem:

• $f : (a,b) \to \mathbb{R}$ is continuous at $c \in (a,b)$ if and only if $f(c) = \lim_{x\to c^+} f(x)$ and $f(c) = \lim_{x\to c^-} f(x)$.

---

You are given that $F|_{U^+\cup I} = f$ is continuous, and from this, we can also conclude $F|_{U^-\cup I} = \overline{f(\bar{\cdot})}$ is continuous.

You have shown that if $f(x)$ is real for all $x \in I$, then $f(x) = \overline{f(\bar{x})}$ for all $x \in I$. That is, the two function $F|_{U^+\cup I}$ and $F|_{U^-\cup I}$ agree on the intersection of $U^+\cup I$ and $U^-\cup I$ , each of which is closed in $U$, so by pasting, $F$ is continuous.