This question stems from a step of a proof in this paper: http://www.cmap.polytechnique.fr/~ammari/papers/04AKS.pdf
The question itself I didn't feel to be "research level" as it is only an asymptotic estimate, so I'm posting this here.
The step occurs near the bottom of page 8:
"Since $\displaystyle-\frac{1}{4\pi}\frac{t}{(|x-x^\prime |^2+t^2)^{3/2}}$ is the Poisson kernel for the half space and $f(\cdot,h\zeta^\prime)$ has compact support in $\Omega$,
$-\frac{1}{4\pi}\int_\Omega \frac{h^2 (\zeta-\zeta^\prime)}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime =-\frac{h}{2} f(x,0) \frac{\zeta - \zeta^\prime}{ |\zeta-\zeta^\prime |} +O(h^2)\text{"}$
$x,x^\prime \in \Omega\subset\mathbb{R}^2$ where $\zeta,\zeta^\prime \in \mathbb{R}$. So as a big picture, we are looking at functions of $(x,h\zeta)$ and $(x^\prime , h\zeta^\prime)$ which are in $\mathbb R^3$. As $h$ goes to zero, the third component of both go to the boundary (i.e. 0). The Poisson theorem about $x$ going to the boundary that I mention below requires the third component of both to equal 0.
In Evans' PDE book chapter 2.2 theorem 14, we have that if $\zeta^\prime=0$ and (ignoring one of the $h$ in front) then we would have
$-\frac{1}{4\pi}\int_\Omega \frac{h (\zeta-\zeta^\prime)}{(|x-x^\prime|^2 + h^2|\zeta-\zeta^\prime |^2 )^{3/2}} f(x^\prime , h\zeta^\prime) \,dx^\prime =-f(x,0)$ as $h\to 0$. I'm confused on (a) how they get the $O$ estimate and (b) how the fraction $\frac{\zeta - \zeta^\prime}{ |\zeta-\zeta^\prime |}$ pops out. I think it has to do with $\zeta^\prime$ possibly not being on $\partial\mathbb{R}^{n}_+$ which is what is required in Evan's theorem, but I'm stuck there.
For (a): Let $\displaystyle u(x)= \frac{2 x_n}{n\alpha(n)}\int_{\partial\mathbb{R}^n_+}\frac{g(y)}{|x-y|^n}\, dy$ with $x\in\mathbb{R}^n_+$ and $y\in\partial\mathbb{R}^n_+$. For convenience we write: $u(x)=\int_{\partial\mathbb{R}^n_+}K(x,y)g(y)\, dy$ with $K(x,y)$ defined accordingly.
Following the steps in Evan's proof: Let $\epsilon>0$ be given and $\delta>0$ such that $|g(y)-g(x_0)|<\epsilon$ if $|y-x_0|<\delta$ and $y\in\partial\mathbb{R}^n_+$. we can find that for $x_0$ on the boundary, if $|x-x_0|<\delta/2$ So as $x\to x_0\in\partial\mathbb{R}^n_+$, $|u(x)-g(x_0)|=\left |\int_{\partial\mathbb{R}^n_+}K(x,y)[g(y)-g(x_0)]\, dy \right| \leq \int_{\partial\mathbb{R}^n_+ \cap B(x_0,\delta)}K(x,y)|g(y)-g(x_0)|\, dy +\int_{\partial\mathbb{R}^n_+ \setminus B(x_0,\delta)}K(x,y)|g(y)-g(x_0)|\, dy$ $=I+J$
In the book it is shown $I\leq\epsilon$ and that $J=O(x_n)$. So what I'm wondering is this: does this mean I can write:
$u(x) = g(x_0) + O(x_n)$ as $x_n \to 0$?