Let $\{U_n\}_{n\geq 1}$ be a sequence of positive harmonic functions on a domain $\Omega$ and let $z_0\in \Omega$. Suppose that $\lim_{n\longrightarrow \infty}U_n(z_0)=\infty$. How does one show that $U_n$ converges uniformly to infinity on every compact subset of $\Omega$?
The hint provided is to use the following: Let $U$ be a positive harmonic function on a domain $\Omega$ and let $\overline{D(a,R)}\subset\Omega$. Then $U(a+re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{R^2-r^2}{R^2+r^2-2Rrcos(\theta-t)}U(a+Re^{i\theta})dt$ for $0\leq r<R.$
Hints: you can easily reduce this to the case when $\Omega$ is a disk. Let $f_n(z)=e^{-(u_n+iv_n)}$ where $v_n$ is a harmonic conjugate of $u_n$. Then $\{f_n\}$ is uniformly bounded and hence it is a normal family. If a subsequence converges uniformly on compact subsets to $f$ then $f$ is holomorphic. An application of Rouche's Theorem tells you that $f$ is either identically $0$ or it has no zeros. [If $f(z_1)=0$ but $f$ not identically $0$ consider a small circle around $z_1$ such that $f$ has no zeros on this circle; then show that $f_n$ and $f$ have the same number of zeros inside the circle. Note that $f_n$ has no zeros]. The hypothesis tells you that $f(z_0)=0$. Hence $f \equiv 0$. This implies that $u_n(z) \to \infty$.