Theorem 1 Let with , such that , where , if ; for all , if . If satisfies
We will show a basic argument why Harnack inequality implies Holder continuity. Assume is a weak solution to a uniform elliptic equation
in . Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution we will have
where is a constant independent of .
First one will know , see this note.
If is a weak solution, so are and . And they are strictly positive in due to the strong maximum principle, otherwise will be a constant. By Harnack inequality, we will have
Adding these two inequalities up we will have
Denote , and choose such that , by interation we know
Then for any , choose such that , we will have
So Holder continuity follows.
— 1. Definition of Newtonian Potential —
Definition 1 Let be a bounded open set, we define the Newtonian potential of is the function on by
where is the fundamental solution of Laplace’s equation.
Theorem Suppose is a subsolution of , . Then for all , ,
Theorem Let , satisfies
If there exists such that
If as , then in .
Proof: Without loss of generality assume on , or we can add a large constant on to make it positive on boundary. , there exists such that
Since , we have
Also on since is non-negative on boundary. Then maximum principle tells in , where . Then we can first take and then take to get in .
Remark The condition is essential, otherwise one can take plus a large constant as a counterexample.
I am now reviewing Perron’s method and it is a fun to go over a similar argument like this. The materials come from the lecture by Prof. Li and T.Aubin’s book.
Theorem On a compact Riemannian manifold without boundary , the equation
with and on , has a solution .