Spring 2016 - Problem 12

Solution.

1. Observe that the conformal mapping $\func{\phi}{\H}{\D}$, $\phi\p{z} = \frac{z - i}{z + i}$ maps $\cl{\R} \to \D$, so it suffices to find a positive harmonic function on $\H$ which is unbounded as $\abs{z} \to \infty$. The function $u\p{z} = \log\,\abs{z + i}$ works: if $\Im{z} > 0$, then

   $$\abs{z + i}^2 = \abs{\Re{z}}^2 + \abs{i\Im{z} + i}^2 \geq \abs{\Im{z} + 1}^2 > 1,$$

   so $u\p{z} > 0$ on $\H$. Furthermore, it is clear that $u$ is unbounded as $\abs{z} \to \infty$, and so

   $$v\p{z} = u\p{\phi^{-1}\p{z}} = \log\,\abs{i\frac{1 + z}{1 - z} + i} = \log\,\abs{\frac{1 + z}{1 - z} + 1}$$

   works.

2. For $r > 0$, let $u_r\p{z} = u\p{rz}$, which is a harmonic function on a neighborhood of $\cl{\D}$. We have the Poisson integral for $u_r$:

   $$u_r\p{te^{i\theta}} = \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} u_r\p{e^{i\theta}} \,\diff\theta = \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} u\p{re^{i\theta}} \,\diff\theta.$$

   Observe that the integrand $g_r$ is bounded:

   $$\abs{\frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} u_r\p{e^{i\theta}}} \leq \frac{M}{1 - t^2} \eqqcolon A.$$

   Hence, $A - g_r \geq 0$, so by Fatou's lemma,

   $$\begin{gathered} \frac{1}{2\pi} \int_0^{2\pi} \liminf_{r\to1}\,\p{A - \frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} u\p{re^{i\theta}}} \,\diff\theta \leq \liminf_{r\to1}\frac{1}{2\pi} \int_0^{2\pi} A - \frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} u\p{re^{i\theta}} \,\diff\theta \\ \implies \limsup_{r\to1}\,u_r\p{te^{i\theta}} \leq \frac{1}{2\pi} \int_0^{2\pi} \frac{1 - \abs{t}^2}{\abs{e^{i\theta} - te^{i\theta}}^2} \p{\limsup_{r\to1} u\p{re^{i\theta}}} \,\diff\theta. \end{gathered}$$

   By construction, the left-hand side is $u\p{te^{i\theta}}$ and the right-hand side is non-negative, and so this becomes

   $$u\p{te^{i\theta}} \leq 0,$$

   so $u\p{z} \leq 0$ on all of $\D$.