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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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.