Aufgaben:Problem 1
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 1
We divide \( \partial Q = Q_1 \cup Q_2 \cup Q_3 \cup Q_4 \) where \( Q_{1, ..., 4} \) are the edges of the square. We imagine \( Q_1 \) to be the interval \( [0, 1] \), and proceed to numerate the edges in positive orientation.
Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = 0 \) by Cauchy's theorem.
Then:
$$ \int_{\partial Q } f(z) \, dz = \left( \int_{Q_1 = [0, 1]} + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) f(z) \, dz $$
By using the parametrizations $$\gamma_1 : [0, 1] \rightarrow Q_2, t \mapsto 1 + it $$ $$ \gamma_2 : [0, 1] \rightarrow \tilde{Q_3}, t \mapsto i + t $$ $$ \gamma_3 : [0, 1] \rightarrow \tilde{Q_4}, t \mapsto it $$
(where \( \tilde{Q_3} \) and \( \tilde{Q_4} \) are the edges with negative orientation), we get:
$$ \begin{align} \int_{\partial Q } f(z) \, dz &= \int_{[0, 1]} f(t) \, dt + \int_{[0, 1]} f(1 + it) \dot{\gamma_1} \, dt - \int_{[0,1]} f(t + i) \dot{\gamma_2} \, dt - \int_{[0, 1]} f(it) \dot{\gamma_3} \, dt \\ &= \int_{[0, 1]} f(t) - f(t + i) \, dt + i \cdot \int_{[0, 1]} f(1 + it) - f(it) \, dt \end{align} $$
And by assumption we have \( f(t) - f(t + i) \in \mathbb{R}_{\leq 0}\) and \( f(1 + it) - f(it) \in \mathbb{R}_{\geq 0} \). But \( 1 \) and \( i \) are linearly independent in the \( \mathbb{R} \) - vector space \( \mathbb{C} \). Since the integrands are real we get \( \forall t \in [0,1] : f(t) - f(t + i) = 0 \), \( f(1 + it) - f(it) = 0 \).
We now can doubly-periodically and holomorphically continue \( f \) on \( \mathbb{C} \) to an elliptic function with the fundamental pair of periods \( \left( 1, i \right) \) by using the following argument:
The continuation is obviously continuous, and by the identity principle we have that \( f(z) \equiv f(z + i ) \) on a neighborhood of each \( z \in [0, 1] \) and \( f(z) \equiv f(z + 1) \) for each \( z \in [0, i] \) (note: the identity principle holds since every point \( z \in \partial Q \) is a limit point of the coincidence set of \( f ( z + i ) = f (z) \), respectively \( f( z + 1) = f(z) \)).
We then see easily that the limit \( f'(z) = \lim_{\Delta z \rightarrow 0} \frac{f(z + \Delta z ) - f(z)}{\Delta z} = \lim_{\Delta z \rightarrow 0} \frac{f(z + i + \Delta z ) - f(z + i)}{\Delta z} = f'(z + i) \) for \( z \in [0,1] \) and of course also \( f(z) = f(z + 1) \) for \( z \in [0, i] \) exists. Thus the continuation has an existing derivative of first order at all points in \( \mathbb{C} \) which is, by its periodicity on the boundary, also continuous, ergo the continuation of \( f \) is an entire function. But since \( f \) is entire and elliptic, it follows from part a) that \( f \equiv const \). \( \square \)
Proof 2
You have to be careful here! You can't use Liouville's theorem without proof. If you want to state this proof in the exam you have to learn a proof for Liouville. For all I've seen this is a rather long and complicated proof. If you know a short and easy one, I guess we would all be glad if you stated it here. A.
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:
$$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$
$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$
Since \( f \) is holomorphic, \( v \) is a harmonic function. We can harmonically and doubly-periodically continue \( v \) to \( v : \mathbb{C} \rightarrow \mathbb{R} \). Furthermore \( v(Q) = v(\mathbb{C}) \) and since \( Q \) is compact we have \( \forall z \in \mathbb{C} \left| v (z) \right| \leq M \) for some \( M \in \mathbb{R} \). We now apply Liouville's theorem for harmonic functions and get \( v \equiv const \).
From the Cauchy-Riemann equations it follows immediately that also \( u \) has to be constant. \( \square \)
Liouville's theorem for harmonic functions: Let \( \lambda : \mathbb{C} \rightarrow \mathbb{R} \) be a harmonic function and \( \forall z \in \mathbb{C} \left| \lambda (z) \right| \leq M \) for some \( M \in \mathbb{R} \). Then \( \lambda \) is constant.
