Question

State and prove harnacks theorem

Answer 1

By Poisson's formula

{\displaystyle f(x)={\frac {1}{\omega _{n-1}}}\int _{|y-x_{0}|=R}{\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\cdot f(y)\,dy,} {\displaystyle f(x)={\frac {1}{\omega _{n-1}}}\int _{|y-x_{0}|=R}{\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\cdot f(y)\,dy,}

where ωn − 1 is the area of the unit sphere in Rn and r = |x − x0|.

Since

{\displaystyle R-r\leq |x-y|\leq R+r,} {\displaystyle R-r\leq |x-y|\leq R+r,}

the kernel in the integrand satisfies

{\displaystyle {\frac {R-r}{R(R+r)^{n-1}}}\leq {\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\leq {\frac {R+r}{R(R-r)^{n-1}}}.} {\displaystyle {\frac {R-r}{R(R+r)^{n-1}}}\leq {\frac {R^{2}-r^{2}}{R|x-y|^{n}}}\leq {\frac {R+r}{R(R-r)^{n-1}}}.}

Harnack's inequality follows by substituting this inequality in the above integral and using the fact that the average of a harmonic function over a sphere equals its value at the center of the sphere:

{\displaystyle f(x_{0})={\frac {1}{R^{n-1}\omega _{n-1}}}\int _{|y-x_{0}|=R}f(y)\,dy.} {\displaystyle f(x_{0})={\frac {1}{R^{n-1}\omega _{n-1}}}\int _{|y-x_{0}|=R}f(y)\,dy.}