There is a tedious, simple but hopefully fruitful exercise I always wanted to do. It is to review all the different proofs of the Harnack inequality and regularity of solutions to elliptic equations that I know, *but only * for the Laplace equation. First, because it is a good way to really get your hands on some of the ideas of several deep theorems (like those of De Giorgi-Nash-Moser and Krylov-Safonov) in the simplest possible setting. Second, because looking at all the different proofs it is possible to trace the evolution of analysis and PDEs through the last century (and a bit before that) and appreciate the level maturity reached in several fields: potential theory, singular integrals, calculus of variations, fully non linear elliptic PDE and free boundary problems. The `simple’ and `elementary’ Laplace equation lies at the intersection of all these fields, so every new breakthrough reflected on our understanding of this equation, each new proof emphasizing a different approach or point of view. Each of the proofs that I will discuss are based on one of the following:

- The mean value property (the proof you learn in your typical complex variables or introductory PDE course).
- The Poisson Kernel for the ball (the proof from potential theory).
- The Calderón-Zygmund theorem (ok not exactly a `Harnack inequality’, but it should be on this list anyway) which uses the machinery of singular integrals.
- The De Giorgi-Nash-Moser theorem, which follows the variational point of view and it is best suited for quasilinear equations or equations in divergence form.
- The Aleksandrov-Bakelman-Pucci estimate and the Krylov-Safonov’s `Harnack’s inequality’, which follows the comparison principle point of view and it is best suited for fully non linear equations or equations in non-divergence form.

So I am going to review each theorem and its proof but only for Laplace’s equation: . To start off easy, I am going to do first the proof via the mean value property.

** First proof: mean value property **

The mean value property says basically this

*Let be a function in the unit ball of . If and is a sphere contained in and centered at , then equals the average of on *

It is not hard to prove with some calculus, one basically looks at the function `Average of on the sphere of radius centered at ‘= and shows that , and since by continuity , the theorem follows. To show one sees (by say, a change of variables) that and this last integral is zero thanks to Stokes’ theorem and the fact that . Moreove, integrating the result with respect to the radius of the sphere one gets the same statement where instead of average over a *sphere* we have an average over a *ball*.

With this, one may prove easily Harnack’s inequality for harmonic functions, which I will state formally for the first time

Theorem 1For any nonnegative harmonic function in we have the inequality

*Proof*. Let , then the ball of radius centered at (call it ) is completely contained in , thus by the mean value property

but is also contained in and since is nonnegative we have , again by the mean value property. This finishes the proof.

That is for today, in the next post I will explain some of the consequences of this theorem and maybe move on to the proof with potential theory methods.

(Note: this post was made using Luca Trevisan’s Latex to WordPress program, which is very useful although I am still getting used to using it. It allows you to prepare your post in a latex editor and then translate it into HTML code which WordPress can read, I strongly recommend it)

Hey Nestor, would you define a few things for me?

1. f(r)

2. that appears in the second paragraph of the first proof

and , I assume means the sphere of radius r centered at x.

Thanks

Hi K,

1) is the average of the the function u on the sphere of radius r, in formulas

here stands for the surface measure of the sphere.

2) is the outer normal derivative of on , which is indeed the sphere of radius .

Hope that clears it up!

Oh, I’m sorry. I just noticed the tick mark (quotations) defining the f. Before, I read it as , which made no sense. Thanks!