Spring 2013 - Problem 12

Solution.

1. We have the Schwarz integral formula for the upper-half plane:

   $$f\p{z} = \frac{1}{\pi i} \int_\R \frac{\Re{f\p{x}}}{x - z} \,\diff{x} + C$$

   for $z \in \H$ and some real constant $C$. Replacing $f$ with $-if$, we see

   $$-if\p{z} = \frac{1}{\pi i} \int_\R \frac{\Im{f\p{x}}}{x - z} \,\diff{x} + C \implies f\p{z + i\epsilon} = \frac{1}{\pi} \int_\R \frac{\Im{f\p{x + i\epsilon}}}{x - z} \,\diff{x} + iC.$$

   We will show in (2) that $C = 0$.

2. The map $h \mapsto \int_\R hg_\epsilon \,\diff{x}$ defines a linear function on $C_0\p{\R}$, i.e., the closure of compactly supported continuous functions on $\R$. Thus, we may view $\set{g_\epsilon}_\epsilon \subseteq \p{C_0\p{\R}}^*$, the space of Radon measures. We wish to apply Banach-Alaoglu, so we need to show that $\norm{g_\epsilon}_{\p{C_0\p{\R}}^*}$ is uniformly bounded.

   Write $z = x + iy$. For $\epsilon > 0$, consider the contour $\gamma$, which is an upper semicircle centered at $x + i\epsilon$ of radius $R$. Notice that if $0 < \epsilon < y < R$, then $z$ is in the interior of $\gamma$ and $\conj{z} + 2i\epsilon$ is outside. Then by the Cauchy integral formula,

   $$\begin{aligned} f\p{z} &= \frac{1}{2\pi i} \int_\gamma \p{\frac{1}{\zeta - z} - \frac{1}{\zeta - \conj{z} - 2i\epsilon}} f\p{\zeta} \,\diff\zeta \\ &= \frac{1}{\pi} \int_\gamma \frac{y - \epsilon}{\p{\zeta - z}\p{\zeta - \conj{z} - 2i\epsilon}} f\p{\zeta} \,\diff\zeta \\ &= \frac{1}{\pi} \int_{-R}^R \frac{y - \epsilon}{t^2 + \p{y - \epsilon}^2} f\p{x + i\epsilon + t} \,\diff{t} + \frac{i}{\pi} \int_0^\pi \frac{y - \epsilon}{R^2e^{2i\theta} + \p{y - \epsilon}^2} f\p{x + i\epsilon + Re^{i\theta}} Re^{i\theta} \,\diff\theta. \end{aligned}$$

   In the second integral, $\abs{f\p{x + i\epsilon + Re^{i\theta}}} \leq \frac{1}{\epsilon + R\sin\theta}$, and so the integrand tends to $0$ as $R \to \infty$. Thus, by dominated convergence, we may send $R \to \infty$ to get

   $$f\p{z} = \frac{1}{\pi} \int_\R \frac{y - \epsilon}{\p{t - x}^2 + \p{y - \epsilon}^2} f\p{t + i\epsilon} \,\diff{t}$$

   via a translation change of variables. Taking imaginary parts and multiplying both sides by $y$, we see

   $$\begin{gathered} \int_\R \frac{y - \epsilon}{\p{t - x}^2 + \p{y - \epsilon}^2} g_\epsilon\p{t} \,\diff{t} = \Im{f\p{z}} \leq \frac{1}{\Im{z}} \\ \implies \int_\R \frac{y\p{y - \epsilon}}{\p{t - x}^2 + \p{y - \epsilon}^2} g_\epsilon\p{t} \,\diff{t} \leq 1. \end{gathered}$$

   Sending $y \to \infty$ and applying dominated convergence, we see $\int_\R g_\epsilon\p{t} \,\diff{t} \leq 1$. Thus, $\set{g_\epsilon}_\epsilon$ is uniformly bounded, so by Banach-Alaoglu and taking $\epsilon = \frac{1}{n}$, we get a subsequence $g_{n_k}$ which converges weakly to a Radon measure $\mu$. In particular, $\frac{1}{x - z} \in C_0\p{\R}$, so by weak convergence,

   $$\int_\R \frac{g_{n_k}\p{x}}{x - z} \,\diff{x} \xrightarrow{k\to\infty} \int_\R \frac{\diff\mu\p{x}}{x - z},$$

   which gives the representation

   $$f\p{z} = \int_\R \frac{\diff\mu\p{x}}{x - z} + iC.$$

   Notice that if $z = iy$, then $\abs{\frac{y}{x - iy}} = \frac{y}{\sqrt{x^2 + y^2}}$ is bounded. Since $\mu$ is a finite measure, we see that $\frac{1}{x - z} \in L^1\p{\R, \mu}$, is dominated by $1$, so by dominated convergence,

   $$\lim_{n\to\infty} \int_\R \frac{y}{x - iy} \,\diff\mu\p{x} = -\frac{1}{i} \mu\p{\R} = i\mu\p{\R}.$$

   If $C \neq 0$, then $yf\p{iy}$ would be unbounded, which implies that $C = 0$. Thus,

   $$i = \lim_{n\to\infty} yf\p{iy} = i\mu\p{\R} \implies \mu\p{\R} = 1,$$

   so $\mu\p{\R}$ is a probability measure.