Theorem 1Let with , such that , where , if ; for all , if . If satisfiesthen .

# Category: PDE

# Harnack inequality implies Holder continuity

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

hence

Denote , and choose such that , by interation we know

Then for any , choose such that , we will have

So Holder continuity follows.

# Newtonian Potential

**— 1. Definition of Newtonian Potential —**

Definition 1Let be a bounded open set, we define the Newtonian potential of is the function on bywhere is the fundamental solution of Laplace’s equation.

# Regularity of scalar elliptic equation by Moser iteration

TheoremSuppose is a subsolution of , . Then for all , ,

# Behavior of harmonic functions comparing with a Green’s function

TheoremLet , satisfiesIf 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 .

RemarkThe condition is essential, otherwise one can take plus a large constant as a counterexample.

# A Perron-type method

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.

TheoremOn a compact Riemannian manifold without boundary , the equationwith and on , has a solution .