# Scaled Harnack inequality $\sup_{B_r} v \le c\,(1-r)^{-p}\, \inf_{B_r} v$

Where can I find a proof of the following scaled version of Harnack inequality?

Let $$v$$ be a non-negative solution of $${L}u = 0$$ in $$B_1$$, with $$L$$ a uniformly elliptic operator. Then, for $$r<1$$, there exist constants $$c$$ and $$p$$ such that $$\sup_{B_r} v \le c\,(1-r)^{-p}\, \inf_{B_r} v.$$

• It follows from applying the usual Harnack inequality to a sequence of balls. For example, applying Harnack in $B_1$ gives $u|_{B_{1/2}} \leq Cu(0)$. Applying it in balls of radius $1/2$ centered at points on $\partial B_{1/2}$ gives $u|_{B_{3/4}} \leq C^2u(0)$. Continuing we get $u|_{B_{1-2^{-k}}} \leq C^ku(0)$, giving polynomial growth near the boundary with power $p \sim \log C / \log 2$. Commented Aug 29, 2019 at 20:56
• @ConnorMooney How does the last inequality in your remark give that polynomial growth $(1-r)^{-p}$? Could you add more details on this in an answer, please?
– Riku
Commented Aug 30, 2019 at 9:52
• Sure, please see my answer below. Commented Aug 30, 2019 at 15:07

The usual Harnack inequality says that $$C^{-1}u(x) \leq \inf_{B_{\rho/2}(x)}u \leq \sup_{B_{\rho/2}(x)} u \leq Cu(x)$$ for any $$B_{\rho}(x) \subset B_1$$ and $$C$$ universal depending on the ellipticity constants, etc. of $$L$$. Applying this with $$x = 0$$ and $$\rho = 1$$ gives $$C^{-1}u(0) \leq \inf_{B_{1-2^{-1}}}u \leq \sup_{B_{1-2^{-1}}} u \leq Cu(0).$$ Applying it again for all $$x \in \partial B_{1/2}$$ and $$\rho = 1/2$$, and using the previous inequality, gives $$C^{-2}u(0) \leq \inf_{B_{1-2^{-2}}}u \leq \sup_{B_{1-2^{-2}}}u \leq C^2u(0).$$ Proceeding inductively with $$x \in \partial B_{1-2^{-k}}$$ and $$\rho = 2^{-k}$$ gives $$C^{-k}u(0) \leq \inf_{B_{1-2^{-k}}}u \leq \sup_{B_{1-2^{-k}}}u \leq C^ku(0),$$ so in particular $$\sup_{B_{1-2^{-k}}}u \leq C^{2k} \inf_{B_{1-2^{-k}}}u.$$ For $$r \in (0,\,1)$$ choose $$k \geq 1$$ such that $$1-2^{1-k} \leq r \leq 1-2^{-k}$$, and let $$p = 2\frac{\log C}{\log 2}$$. Since $$B_r \subset B_{1-2^{-k}}$$ the previous inequality gives $$\sup_{B_r} u \leq C^{2k} \inf_{B_r}u = 2^p(2^{1-k})^{-p} \inf_{B_r}u.$$ Since $$1-r \leq 2^{1-k}$$ the right side is bounded above by $$2^p(1-r)^{-p}\inf_{B_r}u$$, completing the proof.