(Adding solutions and problem descriptions) |
|||
| Line 14: | Line 14: | ||
=== Clinton, 2014 === | === Clinton, 2014 === | ||
| + | Suppose for contradiction that <math>S</math> is bounded; that is, <math>\exists R \forall z</math> such that <math>|z| \geq R \implies u(z) \neq 0</math>. Let <math>f = u + iv</math> analytic, where <math>v</math> is the global analytic conjugate for <math>u</math>. We will show that <math>g = e^{f(z)}</math> is constant, thus <math>u</math> is constant. | ||
| − | + | Let <math>z_0 = R+0i</math>. Then as <math>|z_0|=R \geq R</math>, <math>u(z_0) \neq 0</math>. Without loss of generality (as we could multiply <math>u,f</math> by <math>-1</math>) let <math>u(z_0) < 0</math>. Consider a point <math>p</math> with <math>|p|\geq R</math>. Let <math>\gamma_p</math> be the path along <math>C_{|p|}</math> clockwise to the origin, followed by the path along the real axis from <math>|p|+0i</math> to <math>|R|</math>. This path lies outside <math>B_R(0)</math>, and thus <math>u(z) \neq 0</math> on this path; so by the contrapositive to the intermediate value theorem, as <math>u</math> is harmonic and thus continuous, <math>u(z) < 0</math> at <math>p</math>. Thus <math>u(z) < 0 \forall z \in \mathbb C - B_R(0)</math>. | |
| − | < | + | Consider now the analytic function <math>g = e^{u+iv} = e^ue^{iv}</math>. For <math>z \in \mathbb C -B_R(0)</math>, <math>|g(z)| = |e^u||e^{iv}| = |e^u| < 1</math> as <math>u(z)<0</math>. On <math>\overline{B_R(0)}</math>, as <math>g</math> is continuous on a compact set, it achieves a maximum <math>M</math>. Thus <math>g</math> is a bounded entire function, so by Liouville, <math>g</math> is constant. |
| + | |||
| + | <math>0 \equiv g' \equiv f' e^f</math>. As <math>e^f</math> is nonzero, <math>f' \equiv 0</math> and thus <math>u' \equiv 0</math>; so <math>u</math> is constant, a contradiction. | ||
| + | |||
| + | Thus no nonconstant <math>u</math> with bounded zero set exists. | ||
== Problem 3 == | == Problem 3 == | ||
Revision as of 07:05, 7 August 2014
Post solutions for mock qual #2 here. Please indicate authorship!
Contents
Problem 1
Problem 2
Suppose $ u: \mathbb C \to \mathbb R $ is a non-constant harmonic function. Show that the zero set $ S = \{z \in \mathbb C | u(z) = 0\} $ is unbounded as a subset of $ \mathbb C $.
Clinton, 2014
Suppose for contradiction that $ S $ is bounded; that is, $ \exists R \forall z $ such that $ |z| \geq R \implies u(z) \neq 0 $. Let $ f = u + iv $ analytic, where $ v $ is the global analytic conjugate for $ u $. We will show that $ g = e^{f(z)} $ is constant, thus $ u $ is constant.
Let $ z_0 = R+0i $. Then as $ |z_0|=R \geq R $, $ u(z_0) \neq 0 $. Without loss of generality (as we could multiply $ u,f $ by $ -1 $) let $ u(z_0) < 0 $. Consider a point $ p $ with $ |p|\geq R $. Let $ \gamma_p $ be the path along $ C_{|p|} $ clockwise to the origin, followed by the path along the real axis from $ |p|+0i $ to $ |R| $. This path lies outside $ B_R(0) $, and thus $ u(z) \neq 0 $ on this path; so by the contrapositive to the intermediate value theorem, as $ u $ is harmonic and thus continuous, $ u(z) < 0 $ at $ p $. Thus $ u(z) < 0 \forall z \in \mathbb C - B_R(0) $.
Consider now the analytic function $ g = e^{u+iv} = e^ue^{iv} $. For $ z \in \mathbb C -B_R(0) $, $ |g(z)| = |e^u||e^{iv}| = |e^u| < 1 $ as $ u(z)<0 $. On $ \overline{B_R(0)} $, as $ g $ is continuous on a compact set, it achieves a maximum $ M $. Thus $ g $ is a bounded entire function, so by Liouville, $ g $ is constant.
$ 0 \equiv g' \equiv f' e^f $. As $ e^f $ is nonzero, $ f' \equiv 0 $ and thus $ u' \equiv 0 $; so $ u $ is constant, a contradiction.
Thus no nonconstant $ u $ with bounded zero set exists.
