1
$\begingroup$

I have issues finishing the proof of an exercise concerned with an application of the Schwarz reflection principle.

Let $f: \mathbb{C} \to \mathbb{C}$ be a holomorphic function and $(a,b) \subset \mathbb{R}$. a real nonempty interval such that $f$ only takes real values on $(a,b)$. Show that $f$ only takes real values on all of $\mathbb{R}$.

My idea is to solve this using the following two theorems.

$\text{Theorem 1 (Schwarz reflection principle):}$ Suppose $G \subset H^{+}$ where $H^{+}$ denotes the upper half plane. Suppose that $K \subset \partial G$ where $K$ denotes an interval of the real axis. Let $f: G \to \mathbb{C}$ be holomorphic extendable to a continous function $f: \bar{G} \to \mathbb{C}$ where $\bar{G}$ denotes the topological closure of $G$. Suppose that $f(K) \subset \mathbb{R}$. Let $\tau$ denote the complex conjugation such that $\tau(G)$ is the reflection of $G$ across the real axis. Define

$$ F: G \cup K \cup \tau(G), \ F(z)=\begin{cases} f(z) & , \ z \in G \\ \overline{f(\bar{z})} & , \ z \in \tau(G) \\ f(z)=\overline{f(\bar{z})} & , \ z \in K \end{cases} $$ Then $F$ is holomorphic.

and

$\text{Theorem 2 (Uniqueness theorem):}$ Let $D \subset \mathbb{C}$ be a domain (an open, connected set). Let $J$ be a subset of $D$ having an accumulation point $a \in D$. Let $h_1,h_2: D \to \mathbb{C}$ be holomorphic. If $h_1=h_2$ on $J$ then $h_1=h_2$ on $D$.

Consider the upper half plane $H^{+}$ and $f|_{H^{+}} : H^{+} \to \mathbb{C}$. Then $f|_{H^{+}}$ is holomorphic because $f$ is holomorphic. This function can be continously extended to a function $f|_{\bar{H^{+}}}: \bar{H^{+}} \to \mathbb{C}$. Then I may define $K:=(a,b) \subset \partial H^{+}$. Let $H^{-}$ denote the lower half plane. By the Schwarz reflection principle there is a holomorphic function $F: H^{+} \cup K \cup H^{-} \to \mathbb{C}$ which is a holomorphic extension of $f|_{H^{+}}$.

I wanted to proceed as follows:

1) Show that $F$ extends to an entire function $\tilde{F}$, such that $\tilde{F}$ only takes real values on all of $\mathbb{R}$.
2) Use the uniqueness theorem to show $\tilde{F}=f$.

My issue is that I do not see how to prove 1).

$\endgroup$
6
  • $\begingroup$ Showing that $F$ has a continuous extension to the entire plane requires/is equivalent to showing that $f\lvert_{\mathbb{R}}$ is real. Forget about the reflection principle [only for this, not generally ;)] and just consider $g \colon z \mapsto \overline{f(\bar{z})}$. $\endgroup$ Jun 9, 2020 at 15:53
  • $\begingroup$ take the Taylor series centered at the middle of $(a,b)$ say - any point inside the interval will do - and show that it has real coefficients as they are (up to positive constants) successive derivatives which can be taken on the interval $(a,b)$ only; the result follows since the Taylor series has infinite radius of convergence at any point $\endgroup$
    – Conrad
    Jun 9, 2020 at 16:00
  • $\begingroup$ @DanielFischer I was about to post the same suggestion; at least, I think it's the same. Define $g \colon z \mapsto \overline{f(\bar{z})}$ on the reflection of the domain of $f$ in the real axis. Then $g$ is analytic on this reflected domain. (Very easy, e.g. straight from the definition of analyticity as local expandability as a power series). By the identity theorem, if $f$ and $g$ are equal on an interval of the real axis, then they are equal everywhere. This is much easier than using the reflection principle. Is that more or less the same as what you were saying? $\endgroup$ Jun 9, 2020 at 16:07
  • $\begingroup$ @CalumGilhooley With the additional simplification that since $f$ is assumed to be entire, we may shorter say that we define $g$ on $\mathbb{C}$ in that way. $\endgroup$ Jun 9, 2020 at 16:10
  • $\begingroup$ @Conrad Turn that into an answer please. $\endgroup$
    – zhw.
    Jun 9, 2020 at 22:05

3 Answers 3

2
$\begingroup$

Let $S:=(a,b)\times \Bbb R$ and $S^+=(a,b)\times (0,\infty)$ and $S^-=(a,b)\times (-\infty,0)$. Now, go through the the proof of Schwarz's reflection principal, for example, see page 211 of Conway's Complex analysis.

Now, $f\big|S^+$ is holomorphic and $f\big|(a,b)\times \{0\}$ is real valued continuous. So, the proof actually gives $$g:z\longmapsto \begin{cases}\overline {f(\overline z)} &\text{ if }z\in S^-\\ f(z) & \text{ if }z\in S^+\cup \big((a,b)\times \{0\}\big)\end{cases}$$ is a homlomorphic extension of $f\big|S^+$ on $S$. But, $f\big| S$ already is an extension, so indentity theorem gives $f(z)=g(z)=\overline{f(\overline z)}$ for all $z\in S^-$.

Now, the map $\widetilde f:H^-\ni z\mapsto \overline{f(\overline z)}$ is also holomorphic, and coincide with $f$ on $S^-$. So, $\widetilde f=f\big|H^-$, by identity theorem. So, $f(z)=\overline{f(\overline z)}$ for all $z\in H^-$. Hence, by continuity of $f$ we have $f\big|\Bbb R\subseteq\Bbb R$.

$\endgroup$
2
$\begingroup$

As per my comment, a direct solution to the problem follows noting that if $f$ is a holomorphic function and there is a real interval $(a,b)$ for which the restriction of $f$ to it is real, all the derivatives of $f$ are real on $(a,b)$ - this is clear since we can take $f'(c)=\lim_{h \to 0} \frac{f(c+h)-f(c)}{h}$ with $h$ small real so $f'$ is real on $(a,b)$ and we can use induction on the order of the derivative (or simply noting that $f$ restricted to $(a,b)$ is real analytic and its derivatives as a real analytic function of $1$ variable are the same as the ones as a complex function by analytic continuation).

But then picking $c \in (a,b)$ (for example $c =\frac{a+b}{2}$) the Taylor series of $f$ at $c$ is

$\sum \frac{f^{(n)}(c)}{n!}(x-c)^n$ which is obviously real for any real $x$ for which the series converges; since $f$ is entire, the Taylor series sums to $f(x)$ for all real $x$ so we are done!

$\endgroup$
0
$\begingroup$

Another proof: Let $u = \text{Re } f, v= \text{Im }f.$ Then $v$ is a real analytic function on $\mathbb R$ that vanishes in $(a,b).$ By the identity principle for real analytic functions, $v=0$ on $\mathbb R.$ Thus $f=u$ on $\mathbb R$ as desired.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.