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).
Best Answer
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!