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.
