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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 $\Omega$.

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Do you know Harnack's inequality? It implies that any family of positive harmonic functions that is uniformly bounded at one point of the domain is in fact uniformly bounded on every compact set, and equicontinuous there. – user53153 Dec 29 '12 at 1:22
Here is another way: Let $D(a,r)$ be a disk contained in $\Omega$. Since $f_k(a)\le f(a)$, it follows (why?) that $\int_D f_k \le \pi r^2 f(a)<\infty$. Hence the limit function $f$ is integrable on $D$. Next, prove a version of Koebe's (?) theorem for integrable functions. It goes like this: let $\phi$ be a standard bump function (spherically symmetric). Then $f*\phi=f$ by the mean value property. Hence $f$ is actually $C^\infty$ smooth. – user53153 Dec 29 '12 at 2:21
@PavelM: Yes, we did study the Harnack inequality and I see how to use it now. Thank you! – JZShapiro Dec 29 '12 at 4:04
Sure. Consider posting an answer, since you're the one practicing these problems. – user53153 Dec 29 '12 at 4:06
up vote 3 down vote accepted

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.

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Good job. // The formulation of Harnack's inequality is missing the crucial "$f>0$" part. // In this subject one eventually one gets so much used to the equivalence of "converges uniformly on every compact set" and "every point has a neighborhood where convergence is uniform", that the same term "locally uniform convergence" is used for both without thinking or saying anything. (But it usually happens after passing the quals, I think) – user53153 Dec 29 '12 at 19:43
@PavelM: Oh yes I forgot that assumption and have just added it. – JZShapiro Dec 29 '12 at 20:25

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