Difference between revisions of "Aufgaben:Problem 1"
(→Part b)) |
|||
| Line 31: | Line 31: | ||
===Proof=== | ===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: | |
| − | + | $$ \forall y \in \left[ 0, 1 \right] : v(1, y) = v(0, y) $$ | |
| − | \ | + | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$ |
| − | + | 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. | |
| − | + | ''Proof.'' | |
| − | + | In ''Complex Analysis'' by Th. Gamelin, Chapter 3, Section 4, we have the following property of a harmonic function \( \lambda \): | |
| − | + | Let \( B_r(z_0) \) be the ball of radius \( r \) around \( z_0 \in \mathbb{C} \). Then: | |
| − | + | ||
| − | + | ||
| − | + | $$ \lambda (z_0) = \int_{0}^{2\pi} \lambda ( z_0 + re^{i\theta} ) \frac{d\theta}{2\pi} $$ | |
| − | + | (Still working on that...) | |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
Revision as of 02:43, 1 January 2015
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
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.
Proof.
In Complex Analysis by Th. Gamelin, Chapter 3, Section 4, we have the following property of a harmonic function \( \lambda \):
Let \( B_r(z_0) \) be the ball of radius \( r \) around \( z_0 \in \mathbb{C} \). Then:
$$ \lambda (z_0) = \int_{0}^{2\pi} \lambda ( z_0 + re^{i\theta} ) \frac{d\theta}{2\pi} $$
(Still working on that...)
