Show that the pointwise limit of monotone increasing harmonic functions is harmonic Let $\Omega \subseteq \mathbb{R}^n$ be open and consider a sequence $\{f_k\}_{k \in \mathbb{N}}$, $f \in C^2(\Omega)$ of harmonic functions in $\Omega$ such that $0 \le f_k \le f_{k+1}$ and for which $\displaystyle f(x) \equiv \sup_{k \in \mathbb {N}} f_k(x) < \infty$, each $x \in \Omega$. Prove that $f$ is harmonic in $\Omega$. 
 A: One way to answer this problem is to use Harnack's Convergence Theorem, which was stated in my PDE course as follows:
Let $f_k \in C^2(\Omega)$, $\Delta f_k = 0$, and $f_k \to f$ uniformly on compact subsets of $\Omega$. Then $f \in C^\infty(\Omega)$ and $\Delta f = 0$.
So, it suffices to prove that $\{f_k\}$ is uniformly Cauchy (and thus uniformly convergent) on an arbitrary compact $K \subseteq \Omega$. But then, we can reduce things even further and just show that, for each $x_0 \in K$, there is some ball $B(x_0, r_{x_0}) \subseteq \Omega$ on which $\{f_k\}$ is uniformly Cauchy. Just use a quick $N, \epsilon$ argument to show that that this second statement implies the first (relying heavily on the fact that $K$ is compact). 
To prove that $\{f_k\}$ is uniformly Cauchy on such a ball, we can use another result of Harnack, called Harnack's inequality. In my course, the inequality was stated in the following form:
Suppose $a \in \mathbb{R}^n$ and $R > 0$, with $f \ge 0$ a harmonic function in $B(a, 4R)$. Then there is some constant $C> 0$ such that $\displaystyle \sup_{\overline{B(a, R)}}f \le C\inf_{\overline{B(a, R)}}f$.
So now, for $x_0 \in K$, pick $r_{x_0} >0 $ small enough so that $B(x_0 , 4r_{x_0}) \subseteq \Omega$. Let $\epsilon > 0$ be given, and let $C$ be the constant associated to $B(x_0, r_{x_0})$ in the statement of Harnack's Inequality. Because $\{f_k(x_0)\}$ is a convergent sequence, it is Cauchy, meaning there is some $N_{x_0}$ such that $m \ge n \ge N_{x_0}$ implies $(f_m - f_n)(x_0) = |(f_m - f_n)(x_0)| < \frac{\epsilon}{C}$. Then, whenever $m \ge n \ge N_{x_0}$ and $y \in B(x_0 , 4r_{x_0})$, Harnack's inequality gives:
$$\begin{align} |(f_m - f_n)(y)| = (f_m - f_n)(y) \le \displaystyle \sup_{\overline{B(x_0, r_{x_0})}}(f_m - f_n) \le C\inf_{\overline{B(x_0, r_{x_0})}}(f_m - f_n) \le C(f_m - f_n)(x_0) < \epsilon. \end{align}$$
This completes the proof.
A: By Lebesgue's monotone convergence theorem, the function $f$ satisfies the mean value property, and thus harmonic.
