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

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# 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 ${p>0}$, ${0<\theta <1}$,

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

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# 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.