Difference between revisions of "Aufgaben:Problem 1"
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(Another idea for a proof) |
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$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$ | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$ | ||
− | Since \( f \) is | + | 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. | |
− | + | ''The proof is much too complex to state here. I asked the main assistant by mail if it would be okay to state that without proof. More informations to follow.'' | |
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− | |||
− | + | ===Proof 1=== | |
− | + | ||
− | + | ||
− | ===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: | 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: | ||
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$$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$ | $$ \forall x \in \left[ 0, 1 \right] : v(x, 0) = v(x, 1) $$ | ||
− | Since \( f \) is holomorphic | + | Since \( f \) is a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = \int_{\partial Q } u + iv \, dz = 0 \). 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. |
− | + | Then: | |
− | + | $$ \int_{\partial Q } u + iv \, dz = \int_{\partial Q } u(z) \, dz + i \int_{\partial Q } v(z) \, dz = \int_{\partial Q } u(z) \, dz + i \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) v(x,y) \, dz $$ | |
− | + | But since the function values of \( v \) on \( Q_1 \) are the same as on \( Q_3 \) and are the same on \( Q_2 \) as on \( Q_4 \) we get | |
+ | |||
+ | $$ \int_{\partial Q } v(z) \, dz = 0 $$ | ||
+ | |||
+ | because of the converse orientation of the \( Q_1 \) and \( Q_3 \) part, respectively the \( Q_2 \) and \( Q_4 \) part. | ||
+ | |||
+ | Thus we have \( \int_{\partial Q } u(z) \, dz = 0 \). We write, since \( \forall y \in \left[ 0, 1 \right] : u(1, y) = u(0, y) + \alpha (y) \) and \( \forall x \in \left[ 0, 1 \right] : u(x, 1) = u(x, 0) + \beta (x) \) where \( \alpha, \beta \in \mathbb{R}_{\geq 0} \): | ||
+ | |||
+ | $$ \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) u \, dz = \int_{[0, 1]} u(t,0) \, dt - \int_{[0, 1]} u(t, 1) \, dt + \int_{[0, 1]} u(1, t) \, dt - \int_{[0, 1]} u(0,t) \, dt = \int_{[0, 1]} - \alpha (t) + \beta (t) \, dt = 0 $$ | ||
+ | |||
+ | (...) more to follow |
Revision as of 01:07, 2 January 2015
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 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.
The proof is much too complex to state here. I asked the main assistant by mail if it would be okay to state that without proof. More informations to follow.
Proof 1
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 a holomorphic function the integral \( \int_{\partial Q } f(z) \, dz = \int_{\partial Q } u + iv \, dz = 0 \). 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.
Then:
$$ \int_{\partial Q } u + iv \, dz = \int_{\partial Q } u(z) \, dz + i \int_{\partial Q } v(z) \, dz = \int_{\partial Q } u(z) \, dz + i \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) v(x,y) \, dz $$
But since the function values of \( v \) on \( Q_1 \) are the same as on \( Q_3 \) and are the same on \( Q_2 \) as on \( Q_4 \) we get
$$ \int_{\partial Q } v(z) \, dz = 0 $$
because of the converse orientation of the \( Q_1 \) and \( Q_3 \) part, respectively the \( Q_2 \) and \( Q_4 \) part.
Thus we have \( \int_{\partial Q } u(z) \, dz = 0 \). We write, since \( \forall y \in \left[ 0, 1 \right] : u(1, y) = u(0, y) + \alpha (y) \) and \( \forall x \in \left[ 0, 1 \right] : u(x, 1) = u(x, 0) + \beta (x) \) where \( \alpha, \beta \in \mathbb{R}_{\geq 0} \):
$$ \left( \int_{Q_1 } + \int_{Q_2 } + \int_{Q_3 } + \int_{Q_4 } \right) u \, dz = \int_{[0, 1]} u(t,0) \, dt - \int_{[0, 1]} u(t, 1) \, dt + \int_{[0, 1]} u(1, t) \, dt - \int_{[0, 1]} u(0,t) \, dt = \int_{[0, 1]} - \alpha (t) + \beta (t) \, dt = 0 $$
(...) more to follow