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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. | 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 - | + | 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 - D_R(0)</math>. |
| − | Consider now the analytic function <math>g = e^{u+iv} = e^ue^{iv}</math>. For <math>z \in \mathbb C - | + | Consider now the analytic function <math>g = e^{u+iv} = e^ue^{iv}</math>. For <math>z \in \mathbb C -D_R(0)</math>, <math>|g(z)| = |e^u||e^{iv}| = |e^u| < 1</math> as <math>u(z)<0</math>. On <math>\overline{D_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. | <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. | ||
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== Problem 6 == | == Problem 6 == | ||
| + | Suppose <math>f</math> is analytic on <math>D_1(0)</math> and <math>|f(z)| < 1</math> for all <math>z</math> in the unit disk. Prove that if <math>f(0) = a \neq 0</math>, then <math>f</math> has no zeroes on <math>D_{|a|}(0)</math>. | ||
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| + | === Clinton, 2014 === | ||
| + | Let <math>\phi_a := \dfrac{z-a}{1-\bar a z}</math>, the automorphism of the unit disk <math>D = D_1(0)</math> taking <math>a \to 0</math> and <math>0 \to -a</math>. Let <math>g \equiv \phi_a \circ f</math>; then as <math>f: D \to D</math>, <math>\phi_a: D \to D</math>, we have <math>g: D \to D</math> with <math>g(0) = \phi_a(a) = 0</math>. So by the Schwarz lemma, <math>g(z) \leq z</math>. | ||
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| + | For contradiction suppose there exists <math>b \in D_{|a|}(0)</math> such that <math>f(b) = 0</math>. Then | ||
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| + | <math>|a| = |-a| = |\phi_a(0)| = |g(b)| \leq |b| < |a|</math>, a contradiction. Therefore no such zeroes exist. | ||
== Problem 7 == | == Problem 7 == | ||
Revision as of 06:34, 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 - D_R(0) $.
Consider now the analytic function $ g = e^{u+iv} = e^ue^{iv} $. For $ z \in \mathbb C -D_R(0) $, $ |g(z)| = |e^u||e^{iv}| = |e^u| < 1 $ as $ u(z)<0 $. On $ \overline{D_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.
Problem 3
Problem 4
Problem 5
Problem 6
Suppose $ f $ is analytic on $ D_1(0) $ and $ |f(z)| < 1 $ for all $ z $ in the unit disk. Prove that if $ f(0) = a \neq 0 $, then $ f $ has no zeroes on $ D_{|a|}(0) $.
Clinton, 2014
Let $ \phi_a := \dfrac{z-a}{1-\bar a z} $, the automorphism of the unit disk $ D = D_1(0) $ taking $ a \to 0 $ and $ 0 \to -a $. Let $ g \equiv \phi_a \circ f $; then as $ f: D \to D $, $ \phi_a: D \to D $, we have $ g: D \to D $ with $ g(0) = \phi_a(a) = 0 $. So by the Schwarz lemma, $ g(z) \leq z $.
For contradiction suppose there exists $ b \in D_{|a|}(0) $ such that $ f(b) = 0 $. Then
$ |a| = |-a| = |\phi_a(0)| = |g(b)| \leq |b| < |a| $, a contradiction. Therefore no such zeroes exist.
