# Conservative, solenoidal vector fields and Hardy space

Theorem 1 Let ${E \in L^p ({\mathbb R}^n; {\mathbb R}^n), B \in L^q ({\mathbb R}^n ; {\mathbb R}^n)}$ with ${1 < p < \infty, \frac{1}{p} + \frac{1}{q} = 1}$, such that ${B = \nabla \phi}$, where ${\phi \in L^{\frac{nq}{n-q}}}$, if ${q < n}$ ; ${\phi \in L_{loc}^s}$ for all ${s < \infty}$, if ${q \ge n}$. If ${E}$ satisfies

$\displaystyle div E = 0 \quad \text{ in } D'({\mathbb R}^n),$

then ${E \cdot B \in \mathcal{H}^1}$.

# Harnack inequality implies Holder continuity

We will show a basic argument why Harnack inequality implies Holder continuity. Assume ${u \in H^1(B_2)}$ is a weak solution to a uniform elliptic equation

$\displaystyle -\partial_i(a^{ij} \partial_j u) = 0,$

in ${B_2}$. Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution ${u}$ we will have

$\displaystyle \sup_{B_1} u \le C \inf_{B_1} u,$

where ${C}$ is a constant independent of ${u}$.

First one will know ${u \in L^\infty(B_1)}$, see this note.

If ${u}$ is a weak solution, so are ${\sup_{B_2} u - u}$ and ${u - \inf_{B_2} u}$. And they are strictly positive in ${B_1}$ due to the strong maximum principle, otherwise ${u}$ will be a constant. By Harnack inequality, we will have

$\displaystyle \sup_{B_2} u - \inf_{B_1} u \le C(\sup_{B_2} u - \sup_{B_1} u),$

$\displaystyle \sup_{B_1} u - \inf_{B_2} u \le C(\inf_{B_1} u - \inf_{B_2} u).$

Adding these two inequalities up we will have

$\displaystyle \text{osc}_{B_2} u + osc_{B_1} u \le C(osc_{B_2} u - osc_{B_1}),$

hence

$\displaystyle osc_{B_1} u \le \frac{C-1}{C+1} osc_{B_2} u.$

Denote ${\theta = \frac{C-1}{C+1} < 1}$, and choose ${\alpha > 0}$ such that ${\theta = 2^{- \alpha}}$, by interation we know

$\displaystyle osc_{B_{2^{-k}}} u \le \theta^k osc_{B_1} u \le 2 \theta^k \|u\|_{L^\infty(B_1)}.$

Then for any ${0 < r < 1}$, choose ${k \in {\mathbb N}}$ such that ${2^{-k-1} < r \le 2^{-k}}$, we will have

$\displaystyle osc_{B_r} u \le osc_{B_{2^{-k}}} u \le 2^{-\alpha k + 1} \|u\|_{L^\infty(B_1)} = 2^{1+\alpha} \cdot 2^{- \alpha ( k +1)} \|u\|_{L^\infty(B_1)}< 2^{1+\alpha} \|u\|_{L^\infty(B_1)} r^\alpha.$

So Holder continuity follows.

# Newtonian Potential

— 1. Definition of Newtonian Potential —

Definition 1 Let ${\Omega}$ be a bounded open set, we define the Newtonian potential of ${f}$ is the function ${N_f}$ on ${{\mathbb R}^n}$ by

$\displaystyle N_f(x) = \int_\Omega \Gamma(x-y) f(y)\, dy,$

where ${\Gamma}$ is the fundamental solution of Laplace’s equation.

# Regularity of scalar elliptic equation by Moser iteration

Theorem Suppose ${v \in W^{1,2}(B_R)}$ is a subsolution of ${-\partial_i (a^{ij}(x) \partial_j u) =0}$, ${\lambda I \le (a^{ij}(x)) \le \Lambda I}$. Then for all ${p>0}$, ${0<\theta <1}$,

$\displaystyle \sup_{B_{\theta R}} v \le C(n, \lambda, \Lambda, p) (1- \theta)^{- \frac{n}{p}} \left( \frac{1}{|B_R|} \int_{B_R} (v^+)^p \right)^{\frac{1}{p}}.$

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

Theorem Let ${E \subset \subset B_1}$, ${u \in C^0 (\bar{B_1} \setminus E) \cap C^2 (B_1 \setminus E)}$ satisfies

$\displaystyle - \Delta u =0, \quad \text{in } B_1 \setminus E, \text{ with } u \ge 0 \text{ on } \partial B_1,$

If there exists ${G \in C^2(\bar{B_1} \setminus E)}$ such that

$\displaystyle - \Delta G =0, \quad \text{in } B_1 \setminus E \text{ and } \lim_{dist(x,E) \rightarrow 0} G(x) = +\infty.$

If ${\frac{u^-}{G(x)} \rightarrow 0}$ as ${dist(x,E) \rightarrow 0}$, then ${u \ge 0}$ in ${(B_1 \setminus E)}$.

Proof: Without loss of generality assume ${G \ge 0}$ on ${\partial B_1}$, or we can add a large constant on ${G}$ to make it positive on boundary. ${\forall \varepsilon >0}$, there exists ${0 < \delta(\epsilon) < \epsilon}$ such that

$\displaystyle \frac{u^-}{G(x)} \le \varepsilon \text{ whenever } 0 < dist(x,E) \le \delta.$

Since ${\frac{-u}{G} \le \frac{\max(0, -u)}{G} = \frac{u^-}{G} \le \varepsilon}$, we have

$\displaystyle u(x) \ge - \varepsilon G(x) \text{ for } 0 < dist(x,E) \le \delta.$

Also ${u \ge - \varepsilon G(x)}$ on ${\partial B_1}$ since ${G}$ is non-negative on boundary. Then maximum principle tells ${ u \ge - \varepsilon G}$ in ${B_1 \setminus E_\delta}$, where ${E_\delta = \{x\in B_1: dist(x,E) \le \delta \}}$. Then we can first take ${\delta \rightarrow 0^+}$ and then take ${\varepsilon \rightarrow 0^+}$ to get ${u \ge 0}$ in ${B_1 \setminus E}$. $\Box$

Remark The condition ${\frac{u^-}{G(x)} \rightarrow 0}$ is essential, otherwise one can take ${-G}$ 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.

Theorem On a compact Riemannian manifold without boundary ${(M,g)}$, the equation

$\displaystyle -\Delta_g u + a(x)u = -u^p$

with ${u>0, a(x) \in \mathcal{C}^\infty(M)}$ and ${a(x)<0}$ on ${M}$, ${1 has a solution ${u \in \mathcal{C}^\infty(M)}$.