Help with proof of Poisson's formula for the half-space

Question

Here follows a theorem from L. Evans PDE and a short part of its proof

Theorem

Assume $g\in C(\mathbb{R}^{n-1})\cap L^\infty(\mathbb{R}^{n-1})$, and define $u$ by

$$ u(x)=\frac{r^2-|x|^2}{n\alpha(n)}\int_{\partial\mathbb{R_+^n}}\frac{g(y)}{|x-y|^n}\,dy\qquad x\in\mathbb{R_+^n}. $$ Then (i) $u\in C^\infty(B^0(0,r))$,

(ii) $\Delta u=0$ in $B^0(0,r)$, and

(iii) $\lim_{x\rightarrow x^0,x\in B^0(0,r)}u(x)=g(x^0)$ for each point $x^0\in\partial B(0,r)$

The proof of a similar theorem then states that after a direct calculation follows $$1=\int_{\partial\mathbb{R_+^n}}\frac{r^2-|x|^2}{n\alpha(n)}\cdot\frac{1}{|x-y|^n}dy.$$

This is the part I don't get, I simply don't know how to show this. I can see we can take the first part out of the integral, but due to my lack of calculus skills, I don't know how to even begin with solving this integral.

Answer 1

You seem to be confusing the Poisson kernel for the half-space (Sec 2.2, Thm 14 in Evans) and the Poisson kernel for the ball (Sec 2.2, Thm 15 in Evans). The theorem you have "quoted" appears to be some kind of average of Thm 14 and Thm 15, so it doesn't make sense. For example, what is $r$ in the this context? In the Poisson kernel for the ball, it is the radius, but for the Poisson kernel for the half-space there shouldn't be an $r$.

The Poisson kernel in the half-space (using Evans' notation) is $$K(x,y) = \frac {2x_n} {n \alpha (n)} \frac 1 {\vert x-y \vert ^n} $$ not the kernel you have written. Moreover, Evans proves in Thm 14 that $u\in C^\infty(\mathbb R^n_+) \cap L^\infty(\mathbb R^n_+)$, $\Delta u =0$ in $\mathbb R^n_+$, and $\lim_{\stackrel{x\to x_0}{x\in\mathbb R^n_+}} u(x)=g(x_0)$ for each $x_0\in \mathbb R^{n-1}$, not what you have written.

A proof that $\int_{\partial \mathbb R^n_+} K(x,y) \, dy = 1$ has already been given here.

Finally, as a small note, the kernel you have written is almost the Poisson kernel for the ball, except you are missing a factor of $\frac1r$ and the integral would be over $\partial B_r$.