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Spring 2012 - Problem 7

harmonic functions, Harnack's inequality

Let $\set{u_n\p{z}}_n$ be a sequence of real-valued harmonic functions on $\D \coloneqq \set{z \in \C \mid \abs{z} < 1}$ that obey

u_1\p{z} \geq u_2\p{z} \geq u_3\p{z} \geq \cdots \geq 0 \quad\text{for all } z \in \D.

Prove that $z \mapsto \inf_n u_n\p{z}$ is a harmonic function on $\D$ .

Solution.

Let $0 < r < R < 1$ so that $\cl{B\p{0, r}} \subseteq \cl{B\p{0, R}} \subseteq \D$ . Since $\set{u_n}_n$ is a decreasing sequence, we $u_n - u_m$ is a non-negative harmonic function for $n \leq m$ . Thus, by Harnack's inequality, we have for any $z \in \cl{B\p{0, r}}$ that

\abs{u_n\p{z} - u_m\p{z}} = u_n\p{z} - u_m\p{z} \leq \frac{R + r}{R - r}\p{u_n\p{0} - u_m\p{0}} = \frac{R + r}{R - r}\abs{u_n\p{0} - u_m\p{0}}.

Since $\set{u_n\p{0}}_n$ is a decreasing sequence bounded below (by $0$ ), it is convergent, hence Cauchy. Thus, because $r, R$ are independent of $n, m$ , it follows that $\set{u_n}_n$ is uniformly Cauchy on compact sets, so it converges locally uniformly to $u = \inf_n u_n$ . Since each $u_n$ was continuous, locally uniform convergence implies that $u$ is also continuous, so we may integrate.

Let $B\p{z, R}$ be such that $\cl{B\p{z, R}} \subseteq \D$ . Then because $u_n$ is harmonic,

u_n\p{z} = \frac{1}{2\pi} \int_0^{2\pi} u_n\p{z + Re^{i\theta}} \,\diff\theta.

By uniform convergence, we may take $n \to \infty$ and interchange the limit and integral to get

u\p{z} = \frac{1}{2\pi} \int_0^{2\pi} u\p{z + Re^{i\theta}} \,\diff\theta,

so $u$ satisfies the mean value property, hence harmonic. We reproduce the proof of this fact below:

Let

\rho\p{z} = C\exp\p{-\frac{1}{1 - \abs{z}^2}} \chi_{\D\p{z}}

with $C$ such that $\norm{\rho}_{L^1} = 1$ . Then $\rho$ is a smooth function supported in $\D$ . For $\epsilon > 0$ , set $\rho_\epsilon\p{z} = \frac{1}{\epsilon}\rho\p{\frac{z}{\epsilon}}$ so that $\rho_\epsilon$ is supported on $B\p{0, \epsilon}$ and set $u_\epsilon = \rho_\epsilon * u$ .

Let $z \in \D$ and $\epsilon$ small enough so that $\cl{B\p{z, \epsilon}} \subseteq \D$ . Then

\begin{aligned} u_\epsilon\p{z} &= \int_{B\p{0,\epsilon}} \rho_\epsilon\p{w} u\p{z - w} \,\diff{w} \\ &= \int_0^\epsilon \int_0^{2\pi} r\rho_\epsilon\p{re^{i\theta}} u\p{z - re^{i\theta}} \,\diff\theta \,\diff{r} \\ &= \int_0^\epsilon r\rho_\epsilon\p{r} \int_0^{2\pi} u\p{z - re^{i\theta}} \,\diff\theta \,\diff{r} \\ &= \int_0^\epsilon 2\pi r\rho_\epsilon\p{r} u\p{z} \,\diff{r} \\ &= u\p{z} \int_0^\epsilon \int_0^{2\pi} r\rho_\epsilon\p{re^{i\theta}} \,\diff\theta \,\diff{r} \\ &= u\p{z} \norm{\rho_\epsilon}_{L^1} \\ &= u\p{z}. \end{aligned}

Thus, $u_\epsilon = u$ whenever $\epsilon$ is small enough, and because $u_\epsilon$ is a convolution with a smooth function against a continuous function, it follows that $u$ is also smooth. To show that $u$ is harmonic, we have by the divergence theorem

\begin{aligned} \int_{B\p{z,R}} \Delta u\p{w} \,\diff{m} &= \int_{\partial B\p{z,R}} \nabla u\p{z + Re^{i\theta}} \cdot \p{\cos\theta, \sin\theta} \,\diff\theta \\ &= \int_{\partial B\p{z,R}} \pder{u}{x}\p{z + Re^{i\theta}} \cos\theta + \pder{u}{x}\p{z + Re^{i\theta}}\sin\theta \,\diff\theta \\ &= \int_{\partial B\p{z,R}} \pder{u}{r}\p{z + Re^{i\theta}} \,\diff\theta \\ &= \pder{}{r} \int_{\partial B\p{z,R}} u\p{z + Re^{i\theta}} \,\diff\theta \\ &= 0, \end{aligned}

since $u$ satisfies the mean value property. Thus, because $u$ is smooth, $\Delta u$ is continuous, and so

\Delta u\p{z} = \lim_{R\to0} \frac{1}{m\p{B\p{z,R}}} \int_{B\p{z,R}} \Delta u\p{w} \,\diff{w} = 0,

so $u$ is harmonic.