Proving existence of equal holomorph periodic function composition from strip onto annulus

Question

The question:

Let $a,b \in \mathbb{R} \cup \{ + \infty , -\infty \} , a < b$.

$\theta : S(a,b) \rightarrow \mathbb{C}$ by $\theta (z) := e^{2 \pi i z}$.

Let $f \in \mathcal{O}(S(a,b))$ with $\forall z\in S(a,b): f(z+1)=f(z)$.

Show there exists exactly one $ F \in \mathcal{O}(K_{e^{-2 \pi b}, e^{-2 \pi a}}(0))$ such that

$\forall z\in S(a,b) : f(z) = (F \circ \theta)(z) $.

(Here K is the annulus and $S(a,b) = \{ z \in \mathbb{C}: a < Im(z) < b \}$)

I know that $\theta$ is a holomorph mapping from $S(a,b)$ onto $K_{e^{-2 \pi b},e^{-2 \pi b}}(0)$. I'm not at all sure how to abuse the periodicity/holomorphicity of $f$ to find $F$. Does anyone have any pointers?

Answer 1

First we prove that there exists a function $F: K_{e^{-2\pi b}, e^{-2\pi i a}}(0) \rightarrow \mathbb{C}$, such that

$$ f(z) = F (\Theta (z)) $$

for all $z\in S(a,b)$. We have that $Im(S(a,b))= K_{e^{-2\pi b}, e^{-2\pi i a}}(0)$ (why?). Thus, the only thing we need to check is that

$$ \Theta(z_1)=\Theta(z_2) \ \Rightarrow \ f(z_1)=f(z_2). $$

However, this follows from the 1-periodicity of $f$ and the fact

$$ \{ z\in \mathbb{C} \ : \ e^z=1 \} = 2\pi i \mathbb{Z}.$$

Using the chain rule we compute

$$ \Theta '(z)=2\pi i e^{2\pi i z} $$

for all $z\in \mathbb{C}$. Thus we have $\Theta ' (z) \neq 0$ for all $z\in \mathbb{C}$ and hence (by the inverse function theorem) we have that $\Theta$ is locally biholomorphic. Therefore, we can write (locally)

$$ F(z)= f\circ \Theta^{-1}(z).$$

As $\Theta^{-1}$ is holomorphic by the inverse function theorem we have that $F$ is holomorphic. Furthermore, the same argument implies that $F$ is unique. Namely, suppose we have $\tilde{F}$ such that

$$ f= \tilde{F} \circ \Theta,$$

then we get

$$ \tilde{F} = f\circ \Theta^{-1} = F. $$

Answer 2

Let $U$ be the annulus such that $V=\mathbb{Z}+ \frac{\log U}{2i\pi}$ is the strip where $f$ is holomorphic.

• Let $L_1(z)=\frac{\log z}{2i \pi}$, choosing the branch of $\log$ holomorphic on $\mathbb{C}\setminus (-\infty,0]$.

  Since $f$ is holormorphic on $V$, then $f(L_1(z))$ is holomorphic on $U \setminus (-\infty,0]$.

• Let $L_2(z)=\frac{\log z}{2i \pi}$, choosing the branch of $\log$ holomorphic on $\mathbb{C} \setminus [0,\infty)$.

  Since $f$ is holormorphic on $V$, then $f(L_2(z))$ is holomorphic on $U \setminus [0,\infty)$.

• Using that $f(z) = f(z+1)$ and $L_2(z)-L_1(z) \in \mathbb{Z}$ we get $F(z) = f(L_1(z)) = f(L_2(z))$ so that $F$ is holomorphic on $U \setminus \{0\} = U$ and $f(z) = F(e^{2i \pi z})$. The unicity of $F$ follows the same argument.