Suppose $u$ is a positive harmonic function in $U$, and $u(re^{i\theta}) \to 0$ as $r \to 1$, for every $e^{i\theta} \ne 1$. Prove that there is a constant $c$ such that
$$u(re^{i\theta}) = cP_r(\theta).$$
Notes:
- $U$ is the open unit disc.
- $P$ is the Poisson kernel: $P_r(\theta) = \frac{1-r^2}{1-2r\cos\theta+r^2}$.
- This is self-study. Not homework.
My thoughts:
Since $u$ is a positive harmonic function on $U$, it's the Poisson integral of a positive finite measure $\mu$ on $T$ (the unit circle). Write $\mu = fd\sigma + \mu_s$ (the Lebesgue decomposition of $\mu$). Since $u$ has $0$ radial limits a.e., $f = 0$ a.e. and $\mu \perp \sigma$. I guessed that $c = \|\mu\|$ and tried to evaluate $u(re^{i\theta}) – cP_r(\theta)$ but I got stuck.
Is there a better way? Thanks.
Best Answer
You are not using the full power of the assumption that the radial limits are zero for all points except one (not merely a.e.). Here are some facts which you might already know:
Lemma 1. There is a universal constant $c>0$ such that for all $0<\epsilon<1$ and all $t\in [-\epsilon,\epsilon]$ we have $P_{1-\epsilon}(t)\ge c\epsilon^{-1}$.
Sketch of proof. Since $t\mapsto P_r(t)$ is symmetric and decreasing in both directions, it suffices to consider $t=\epsilon$. Look at the denominator of the kernel:
$$1-2(1-\epsilon)\cos \epsilon+(1-\epsilon)^2 = 2(1-\epsilon)(1-\cos \epsilon) +\epsilon^2 = O(\epsilon^2) $$ Since the numerator is $1-r^2\ge \epsilon$, we conclude that the function $\epsilon\,P_{1-\epsilon}(\epsilon)$ is bounded by a positive constant from below on $(0,1]$. $\Box$
Lemma 2. Suppose $u$ is a positive harmonic function on $U$ and $\lim_{r\to 1} u(re^{i\theta})=0$ for some $\theta\in [0,2\pi)$. Then $\lim_{\epsilon\to 0}\epsilon^{-1}\mu(N_\epsilon(e^{i\theta}))=0$, where $N_\epsilon$ stands for the $\epsilon$-neighborhood (really, a circular arc with midpoint at $e^{i\theta}$).
Proof. Let $r=1-\epsilon$. Using Lemma 1, obtain
$$ u(re^{i\theta})\ge \int_{-\epsilon}^{\epsilon} u(re^{i(\theta-t)})P_r(t)\,dt \ge c\epsilon^{-1}\mu(N_\epsilon(e^{i\theta})) $$ and the conclusion follows. $\Box$
The assumptions in your question imply that $\lim_{\epsilon\to 0}\epsilon^{-1}\mu(N_\epsilon(\xi))=0$ for all $\xi\in T\setminus \{1\}$. This implies (by a covering argument) that $\mu(T\setminus \{1\})=0$, and therefore $\mu$ is a positive point mass at $1$.