Aufgaben:Problem 1
Contents
Part a)
Prove that, if \(f\) is an elliptic function, then \(f\) is constant if and only if \(f\) has no poles.
Solution part a)
Definition (elliptic function): A function \(f\) is called elliptic if it has the following two properties:
(a) \(f\) is doubly periodic.
(b) \(f\) is meromorphic (its only singularities in the finite plane are poles).
(Taken from Apostol, 1.4)
Liouville's theorem: If \(f\) is holomorph on \( \mathbb{C} \) and bounded, then \(f\) is constant.
To prove: Let \(f\) be an elliptic function: \(f\) is constant \(\Leftrightarrow\) \(f\) has no poles.
\( "\Rightarrow" \) Let \(f\) be constant on \( \mathbb{C} \) \( \rightarrow f\) has no poles.
\( "\Leftarrow" \) Let \(P\) be the fundamental region to \(f\), where \(f\) is elliptic and has no poles.
\( f(\mathbb{C}) = f(P) \) is compact and thus bounded. Because \(f\) is both holomorph and bounded on \(\mathbb{C}\), Liouville's theorem tells us that \(f\) has to be a constant function. \(\square\)
Part b)
\( \vdash: \) Let \( Q = \left[ 0, 1\right] \times \left[ 0, 1 \right] \subset \mathbb{C} \) be the unit square, and let \( f \) be a holomorphic function on a neighborhood of \( Q \). Suppose further that \( f( z + i) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, 1 \right] \) and \( f( z + 1) - f(z) \in \mathbb{R}_{\geq 0} \) for any \( z \in \left[ 0, i \right] \). Then \(f\) is constant.
Proof
At first, we prove a little Lemma:
Pierre's Lemma: If \( f = u + iv \) is a holomorphic function then \( u, v \) are meromorphic.
\( \color{red}{So \: beautiful! \: See \: discussion} \)
Proof of Lemma:
If \( a, b \in \mathbb{R} \smallsetminus \{ 0 \} \), then \( a \) and \( b i \) are linearly independent. ( \( \mathbb{C} \) is a two-dimensional \( \mathbb{R} \)-vectorspace with basis \( 1 \) and \( i \). ) If \( v \) is zero then \( f = u \) is constant (see critic in discussion), if \( u = 0 \) then \( v \) is meromorphic anyway and yay. \( \square \)
Now we can start with the main proof. We write \( z = x + iy \) and \( f(x + iy) = u(x,y) + i v(x,y) \) where \( u, v \) are real valued functions. By assumption we get:
$$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
Therefore $$\forall z\in [0,i]: v(z+1)=v(z)$$ $$\forall z\in[0,1] : v(z+i)=v(z)$$
And we see that v is doubly-periodic on \(Q\). Pierre's Lemma implies that \(u\) and \(v\) are meromorphic. Because \(f\) has no poles on \(Q\), \(u\) and \(v\) don't have either. Therefore, \(u\) and \(v\) are holomorphic on \(Q\).
We thus see that \( v \) can be doubly-periodic, analytically continued on \(\mathbb{C}\), so that it's an elliptic function with respect to the lattice \( \Omega = \{ m + in : m, n \in \mathbb{Z} \} \).
We've shown in part a) that a elliptic, holomorphic function has to be constant. Thus \( v \) is constant.
From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
