An exercise unsolved.
The complete question is: Let $u(x)$ be a harmonic function defined on the unit ball, with boundary condition: $u(x) = g , x\in\partial\Omega$. Let function $G$ be the Green function on unit ball. Then try to use the equation: $$u(x) = -\int_{\partial\Omega}u(y)\frac{\partial G}{\partial\boldsymbol{n}_y}(x,y)\,dS_y , x\in\Omega$$ to prove Harnack's inequality again.
Harnack's inequality: If $u(x)\in C^2(\Omega)$ is a non-negative harmonic function defined on $\Omega\subset\mathbb{R}^3$, then for any $\Omega'\subset\subset\Omega$, there exists constant $C = C(\Omega, \Omega')>0$ such that $$\frac{1}{C}u(y)\le u(x)\le Cu(y) , \forall x,y\in \bar{\Omega}'$$
I have computed the specific formula of $u(x)$: $$u(x) = \frac{1 – |x|^2}{4\pi}\int_{\partial B_1}\frac{g(y)}{|x-y|^3}\,dS_y$$ And I don't not how to continue.
Best Answer
I will continue with what I understand to be your assumption, that $\Omega$ is specialised to the unit ball in three dimensions. Then given $\Omega' \subset\subset \Omega$ there is some $r<1$ such that any $x\in \overline{\Omega'}$ must satisfy $\lvert x \rvert \le r$. Consequently for $x,y\in \overline{\Omega'}$, \begin{eqnarray*} u(x)&=&\frac{1-\lvert x \rvert^{2}}{4\pi} \int_{\partial B_{1}} \frac{g(z)}{\lvert x - z \rvert^{3}} d S_{z}\\ &\le& \frac{1}{4\pi} \int_{\partial B_{1}} \frac{g(z)}{\left(1-r\right)^{3}} d S_{z} = \frac{\left(1+r\right)^{3}}{4\pi \left(1-r\right)^{3}} \int_{\partial B_{1}} \frac{g(z)}{\left(1+r\right)^{3}} d S_{z}\\ &\le& \frac{\left(1+r\right)^{3}}{4\pi \left(1-r\right)^{3}} \int_{\partial B_{1}} \frac{g(z)}{\lvert y-z\rvert^{3}} d S_{z} = \frac{\left(1+r\right)^{2}}{ \left(1-r\right)^{4}}\, \frac{1-r^{2}}{4\pi} \int_{\partial B_{1}} \frac{g(z)}{\lvert y-z\rvert^{3}} d S_{z}\\ &\le& \frac{\left(1+r\right)^{2}}{ \left(1-r\right)^{4}}\, \frac{1-\lvert y \rvert^{2}}{4\pi} \int_{\partial B_{1}} \frac{g(z)}{\lvert y-z\rvert^{3}} d S_{z} = \frac{\left(1+r\right)^{2}}{ \left(1-r\right)^{4}} u(y). \end{eqnarray*}