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

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

Remark 1 Without loss of generality one can assume ${v \ge 0}$. Since if ${v \in W^{1,2}}$ is a subsolution, ${v^+ \in W^{1,2}}$ is also a subsolution. Since the result is scaling invariant, one can assume ${R = 1}$. Further more one can assume ${\theta = \frac{1}{2}}$, if we have ${\sup_{B_{1/2}} v \le C(n, \lambda, \Lambda, p) (\int_{B_1} v^p )^{\frac{1}{p}}}$, for any ${0 < \theta <1}$, any ${\varepsilon >0}$, we can take a point ${x \in B_\theta}$ such that ${v(x)^p \ge (\sup_\theta u)^p - \varepsilon}$. Then

$\displaystyle (\sup_\theta u)^p \le \varepsilon + (\sup_{B_{\frac{1- \theta}{2}}(x)} v)^p \le \varepsilon + \frac{C}{|B_{1-\theta}|} \int_{B_{1-\theta} (x)} v^p \le \varepsilon + \frac{C}{(1-\theta)^n} \int_{B_1} v^p.$

Let ${\varepsilon \rightarrow 0}$, we can derive the inequality for any ${\theta \in (0,1)}$.

Lemma Let ${\varphi(t)}$ be a nonnegative bounded function on ${(0,1)}$, if there exist ${0< \theta <1}$, ${A,B, \alpha}$ nonnegative constant such that

$\displaystyle \varphi(t) \le \theta \varphi(s) + \frac{A}{(s-t)^\alpha} +B$

for all ${0 < t . Then

$\displaystyle \varphi(t) \le C(\theta, \alpha) \left[ \frac{A}{(s-t)^\alpha} +B \right].$

Proof: Set ${t_0=t, t_i = t + (1 -\tau^i)(s-t)}$, for some ${0<\tau<1}$ to be chosen later. Then ${t_{i+1} - t_i = (\tau^i - \tau^{i+1}) (s-t) = \tau^i (1- \tau) (s-t)}$. So

$\displaystyle \varphi (t_i) \le \theta \varphi(t_{i+1}) + A(t_{i+1} - t_i)^{-\alpha} +B = \theta \varphi(t_{i+1}) + \frac{A}{(1 - \tau)^\alpha} (s-t)^{- \alpha} \tau^{-i \alpha} +B.$

After iterating we will have

$\displaystyle \varphi(t_o) \le \theta^k \varphi(t_k) + \frac{A}{(1 - \tau)^\alpha} (s-t)^{- \alpha} \cdot \sum_{i=0}^{k-1} \left( \frac{\theta}{\tau^\alpha} \right)^i +\sum_{i=0}^{k-1} \theta^i B.$

Now we can choose ${\tau}$ such that ${\tau^\alpha > \theta}$. Let ${k \rightarrow \infty}$,

$\displaystyle \varphi(t_0) \le C(\theta, \alpha) [A(s-t)^{- \alpha} +B].$

$\Box$

Remark 2 We only need to prove for case ${p \ge 2}$, and use the above lemma to recover the cases when ${0. That is because for any ${0 < q< p}$, ${0 < \theta R < R <1}$,

$\displaystyle \sup_{B_{\theta R}} v \le C (1- \theta)^{- \frac{n}{p}} \left( \frac{1}{|B_R|} \int_{B_R} v^p \right)^{\frac{1}{p}} \le C (1- \theta)^{- \frac{n}{p}} \sup_{B_R} v^{\frac{p-q}{p}} \cdot \left( \frac{1}{|B_R|} \int_{B_R} v^q \right)^{\frac{1}{p}}$

$\displaystyle \le \frac{1}{2} \sup_{B_R} v + \frac{C}{[(1- \theta)R]^\frac{n}{q}} \left(\int_{B_R} v^q \right)^{\frac{1}{q}} \le \frac{1}{2} \sup_{B_R} v + \frac{C}{[(1- \theta)R]^\frac{n}{q}} \left(\int_{B_1} v^q \right)^{\frac{1}{q}}.$

Set ${t = \theta R, s = R}$ and ${\varphi(t) = \sup_{B_t} v}$. Apply the above lemma we will have

$\displaystyle \varphi(t) \le \frac{C}{(s-t)^{\frac{n}{q}}} \left(\int_{B_1} v^q \right)^{\frac{1}{q}}.$

Proof of the theorem: By the above remarks, it suffices to prove the case when ${v \le 0, p \ge 2}$. Since

$\displaystyle \int_{B_R} a^{ij} \partial_i v \partial_j \varphi \le 0$

for all ${\varphi \ge 0 \in W^{1,2}_0 (B_R)}$. Take ${\varphi = \xi^2 v^{p-1}}$, where ${\xi }$ is a cut off function. Then we have

$\displaystyle \int_{B_R} a^{ij} \partial_i v \partial_j (\xi^2 v^{p-1}) \le 0,$

which implies

$\displaystyle (p-1) \int_{B_R} a^{ij} \partial_i v \partial_j v \cdot v^{p-2} \xi^2 \le -2 \int_{B_R} a^{ij} \partial_i v \partial_j \xi \cdot v^{p-1}.$

By ellipticity, we have

$\displaystyle (p-1) \int_{B_R} |\nabla v|^2 \cdot v^{p-2} \xi^2 \le C(\lambda , \Lambda) \int_{B_R} (|\nabla v| \cdot v^{\frac{p-2}{2}} \xi) \cdot (|\nabla \xi| v^{\frac{p}{2}}).$

By Holder inequality we will have

$\displaystyle \int_{B_R} |\nabla v|^2 \cdot v^{p-2} \xi^2 \le \frac{C}{(p-1)^2} \int_{B_R} |\nabla \xi|^2 v^p \le C \int_{B_R} |\nabla \xi|^2 v^p.$

Since ${|\nabla v^{\frac{p}{2}}|^2 = | v^{\frac{p}{2} -1} \cdot \nabla v|^2 = v^{p-2} \cdot |\nabla v^2|}$, we can derive ${|\nabla (\xi v^{\frac{p}{2}})|^2 = |\nabla \xi \cdot v^{\frac{p}{2}} + \xi \cdot \nabla v^{\frac{p}{2}}|^2 \le 2|\nabla \xi|^2 \cdot v^p + 2\xi^2 \cdot |\nabla v|^2 v^{p-2}}$, which implies

$\displaystyle \int_{B_R} |\nabla (\xi v^{\frac{p}{2}})|^2 \le 2 \int_{B_R} |\nabla \xi|^2 v^p + 2 \int_{B_R} \xi^2 \cdot |\nabla v|^2 v^{p-2} \le C \int_{B_R} |\nabla \xi|^2 v^p.$

By Sobolev inequality, we have

$\displaystyle \left( \int_{B_R} |\xi v^\frac{p}{2}|^{\frac{2n}{n-2}} \right) ^{\frac{n-2}{n}} \le C \int_{B_R} |\nabla \xi|^2 v^p.$

Now let ${R_k = R(\theta + \frac{1-\theta}{2^k})}$, so ${R_0 = R}$ and ${R_k \rightarrow \theta R}$ as ${k \rightarrow \infty}$. Take ${\xi_k \in C_c^\infty (B_{R_k})}$, ${0 \le \xi_k \le 1}$, and ${\xi_k \equiv 1}$ on ${B_{R_{k+1}}}$. Then ${|\nabla \xi_k| \le \frac{C}{R_k - R_{k+1}} \le \frac{C 2^{k+1}}{(1-\theta)R}}$. Therefore

$\displaystyle \left( \int_{B_{R_{k+1}}} v^{\frac{np}{n-2}} \right)^{\frac{n-2}{n}} \le \frac{C4^k}{(1-\theta)^2R^2} \int_{B_{R_k}} v^p.$

Take ${p_k = p (\frac{n}{n-2})^k}$, Then

$\displaystyle \left( \int_{B_{R_{k+1}}} v^{p_{k+1}} \right)^{\frac{n-2}{n}} \le C4^k \int_{B_{R_k}} v^{p_k}.$

Therefore

$\displaystyle \|v\|_{L^{p_{k+1}} (B_{R_{k+1}})} \le (C4^k)^\frac{1}{p_k} \|v\|_{L^{p_k}(B_{R_k})}.$

After iteration, we will have

$\displaystyle \|v\|_{L^{p_{k+1}} (B_{R_{k+1}})} \le C^{\sum_{i=1}^k \frac{1}{p_i}} \cdot 4^{\sum_{i=1}^k \frac{i}{p_i}} \cdot \|v\|_{L^p (B_R)}.$

Since ${\sum_{i=1}^k \frac{i}{p_i} \le C \sum_{i=1}^k (\frac{n-2}{n})^i \cdot i < \infty,}$ we can take ${k \rightarrow \infty}$ and get

$\displaystyle \|v\|_{L^\infty (B_{\theta R})} \le C \|v\|_{L^p (B_R)}.$

$\Box$