PDE Evans, 2nd edition, Exercise 2.7
Use Poisson's formula for the ball to prove
$$r^{n-2} \frac{r-|x|}{(r+|x|)^{n-1}}u(0) \le u(x) \le r^{n-2} \frac{r+|x|}{(r-|x|)^{n-1}}u(0)$$
whenever $u \in C^2(B^0(0,r)) \cap C(\partial B(0,r))$ is positive and harmonic in $B^0(0,r) \subset \mathbb{R}^n$. This is an explicit form of Harnack's inequality.
My proof attempt is in the answer, but I would like to know if it's correct.
Best Answer
Triangle inequality and reverse triangle inequality assert that $$r-|x|\le|x-y|\le r+|x|$$ for all $x \in B^0(0,r)$ and $y \in \partial B(0,r)$ (which means $|y|=r$).
Now we appeal to Poisson's formula: $$u(x) = \frac{r^2-|x|^2}{n \alpha(n)r}\int_{\partial B(0,r)} \frac{g(y)}{|x-y|^n} dS(y)$$
Thus, \begin{align} u(x) &\le \frac{r^2-|x|^2}{n \alpha(n)r}\int_{\partial B(0,r)} \frac{g(y)}{(r-|x|)^n} dS(y) \\ &= r^{n-2} \frac{r+|x|}{(r-|x|)^{n-1}} \frac{1}{n \alpha(n)r^{n-1}} \int_{\partial B(0,r)} g(y) \, dS(y) \\ &= r^{n-2} \frac{r+|x|}{(r-|x|)^{n-1}} u(0) \end{align} and \begin{align} u(x) &\ge \frac{r^2-|x|^2}{n \alpha(n)r}\int_{\partial B(0,r)} \frac{g(y)}{(r+|x|)^n} dS(y) \\ &= r^{n-2} \frac{r-|x|}{(r+|x|)^{n-1}} \frac{1}{n \alpha(n)r^{n-1}} \int_{\partial B(0,r)} g(y) \, dS(y) \\ &= r^{n-2} \frac{r-|x|}{(r+|x|)^{n-1}} u(0) \end{align} Therefore, Harnack's inequality has been established, as desired.