## Harnack inequality under scaling

Thm: Suppose ${\Omega}$ is domain in ${\mathbb{R}^n}$. ${u}$ is a harmonic function in ${\Omega}$, ${u\geq 0}$. For any subdomain ${\Omega'\subset\subset \Omega}$, there exists a constant ${C}$ such that

$\displaystyle \sup_{\Omega'}u\leq C(n,\Omega,\Omega')\inf_{\Omega'}u$

To prove this Harnack inequality, there is an intermediate step

$\displaystyle \sup_{B_R}u\leq 3^n\inf_{B_R}u$

whenever ${B_{4R}\subset\Omega}$.

Here the constant is independent of ${R}$, actually the constant ${3^n}$ is not so important. We can use the above Thm to give another proof.

Suppose ${v(x)=u(Rx)}$ for ${x\in B_1}$. Let ${\Omega=B_2}$ and ${\Omega'=B_1}$, applying the thm

$\displaystyle \sup_{B_1}v\leq C(n)\inf_{B_1}v$

convert back to ${u}$

$\displaystyle \sup_{B_R}u\leq C(n)\inf_{B_R}u$