Questions about the proof of Poisson's formula for half-space in Page 38 of Evans' book I have questions about the proof of Theorem 14 (poisson's formula for half-space) in Page 38.
Let $K(x,y)$ be the Poisson's kernel for $\mathbb R^n_+$:
$$K(x,y)=\frac{2x_n}{na(n)}\frac{1}{|x-y|^n}dy\quad (x\in\mathbb R^n_+,\; y\in \partial \mathbb R_+^n).$$
Let
$$u(x)=\frac{2x_n}{na(n)}\int_{\partial \mathbb R_+^n}\frac{g(y)}{|x-y|^n}dy\quad  (x\in \mathbb R_+^n).$$
The author claims that since $x\mapsto K(x,y)$ is smooth for $x\neq y$, we easily verify as well $u\in C^\infty(R_+^n)$, with
$$\Delta u(x)=\int_{\partial\mathbb R_+^n}\Delta_xK(x,y)g(y)dy=0\quad (x\in \mathbb R_+^n).$$ Suppose $g\in C(R^{n-1})\cap L^\infty(R^{n-1})$. $g$ is bounded. 
My questions are 

  
*
  
*How to prove $u\in C^\infty(\mathbb R_+^n)$?
  
*How to prove that we can take the laplacian operator into the integration when calculating
  $$\Delta u(x)=\Delta_x\int_{\partial \mathbb R_+^n}K(x,y)g(y)dy\;?$$
  

Thanks!
 A: This should prove that you can differentiate under the integral:
$$\partial_i u(x) = \lim_{h \to 0}\frac{u(x+e_i h)-u(x)}{h} = \lim_{h \to 0}\frac{\int_{\partial R_+^n} K(x+e_i h,y)g(y)dy -\int_{\partial R_+^n} K(x,y)g(y)dy}{h} = \lim_{h \to 0} \int_{\partial R_+^n} \frac{K(x+e_i h,y) -K(x,y)}{h}g(y)dy  = \int_{\partial R_+^n} \partial_{i,x}K(x,y)g(y)dy$$
(Passing the last limit inside uses dominated convergence). Smoothness of $u$ now follows from smoothness of $K$ and $g$.
EDIT: On the dominated convergence.
Since $K$ is smooth we can apply the mean value theorem to estimate $\frac{K(x+e_i h,y) -K(x,y)}{h}$, for $|h|< \frac{x_n}{2}$ (just making sure that we do not hit $\partial R_+^n$). Thus for bounded values of $y$ say $|y-x'|< 1$ ($x'=(x_1,\ldots,x_{n-1})$) we may bound the integrand by a constant. Outside this ball we see that the derivative of the kernel wrt. $x_i$ is less than the function, which is constructed from the partial derivative by replacing $x$ with $(x',0)$. Outside $|y-x'|<1$ we now have a function that decreases like $|y-x'|^{-(n+1)}$, whereas the radial volume element goes like $r^{n-1}dr$. Hence the outside integral remains finite, and we have constructed a $L^1$ bound, and we may now use the dominated convergence to pass the limit inside.
