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 $\sup_{B_r} v \le c\,(1-r)^{-p}\, \inf_{B_r} v.$

  • $\begingroup$ 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 $p \sim \log C / \log 2$. $\endgroup$ Commented Aug 29, 2019 at 20:56
  • $\begingroup$ @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? $\endgroup$
    – Riku
    Commented Aug 30, 2019 at 9:52
  • $\begingroup$ Sure, please see my answer below. $\endgroup$ Commented Aug 30, 2019 at 15:07

1 Answer 1


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.


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