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 $\mathbb R^n_+$: $$K(x,y)=\frac{2x_n}{na(n)}\frac{1}{|x-y|^n}\quad (x\in\mathbb R^n_+,\; y\in \partial \mathbb R_+^n).$$
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 question is
How to prove that we can take the laplacian operator into the integration when calculating $$\Delta u(x)=\Delta_x\int_{\partial \mathbb R_+^n}K(x,y)g(y)dy\;?$$
My attempt
The first thing is to find a first order partial differential (this is enough) $$ \frac{u(x+h_me_i)-u(x)}{h_m}=\int_{\partial \mathbb R_+^n}\frac{k(x+h_me_i,y)-k(x,y)}{h_m}g(y)dy $$ In order to use LDCT(Lebesgue's Dominated Convergence Theorem), we need to take function $f_m(y)=\frac{k(x+h_me_i,y)-k(x,y)}{h_m}g(y) $ and find $L_1$ function $h(y)$ such that $|f_m(y)|<h(y)$, but it is not easy to find $h(y)$
$$\frac{k(x+h_me_i,y)-k(x,y)}{h_m}g(y)=\frac{2x_n}{h_mna(n)}(\frac{1}{|x+h_me_i-y|^n}-\frac{1}{|x-y|^n})g(y)$$
Also, by MVT $$\frac{k(x+h_me_i,y)-k(x,y)}{h_m}g(y)=k_{x_i}(x+h_m'e_i)g(y)=\frac{-2x_n}{a(n)}\frac{1}{|x+h_m'e_i-y|^{n+2}}(x_i+h_m'-y_i)g(y)$$
Using Dominated Convergence Theorem to move derivatives inside the integral is the right idea. When doing so, keep in mind that once we fix $x\in \mathbb R_+^n$, there is no singularity in $|x - y|^{-n}$ or any of it's derivatives because $y\in \partial \mathbb R_+^n$. In what follows, I'll give the argument for moving the first-order derivatives inside the integral. The argument for moving second-order derivatives inside the integral is similar, so I'll leave it to you.
Fix $x\in \mathbb R_+^n$. For any $h$ satisfying $0< |h|< x_n/2$ and any $i\in \{1, \ldots, n -1\}$ we have \begin{equation*} \begin{split} \frac{u(x + he_i) - u(x)}{h} & = \frac{2x_n}{n\alpha(n)}\int_{\mathbb R^{n - 1}}\frac 1 h\left(\frac{1}{|x + he_i - y|^n} - \frac{1}{|x - y|^n}\right)g(y)\; \mathrm d y. \end{split} \end{equation*} As $h\to 0$, the integrand on the right-hand side converges for all $y\in \mathbb R^{n - 1}$ to \begin{equation*} g(y)\frac{\partial}{\partial x_i}|x- y|^{-n} = -ng(y)\frac{x_i - y_i}{|x - y|^{n+ 2}} \in L^1(\mathbb R^{n - 1}; \mathrm d y), \end{equation*} where the integrability follows from the assumption that $g$ is bounded and the fact that $(x_i - y_i)|x - y|^{-(n + 2)}$ is smooth in $y$ (again, there is no singularity since $x\in \mathbb R_+^n$ is fixed and $y\in \partial \mathbb R_+^n$) and decays as $|y|^{-(n + 1)}$ as $|y|\to \infty$. Now, for all $0< |h|< x_n/2$ we have \begin{equation*} \begin{split} \left|\frac 1 h\left(\frac{1}{|x + he_i - y|^n} - \frac{1}{|x - y|^n}\right)\right| & = \left|\frac 1 h\int_0^1\frac{\mathrm d}{\mathrm d t}|x - y + the_i|^{-n}\; \mathrm d t\right|\\ & \leq C_n\int_0^1|x - y + the_i|^{-(n + 1)}\; \mathrm d t\\ & \leq C_n(|x - y| - |h|)^{-(n + 1)}\\ & \leq C_n|x - y|^{-(n + 1)} \in L^1(\mathbb R^{n - 1}\; \mathrm d y). \end{split} \end{equation*} The Dominated Convergence Theorem now guarantees that for $i \in \{1, \ldots, n - 1\}$, \begin{equation*} \frac{\partial u}{\partial x_i}(x) = \frac{2x_n}{n\alpha(n)}\int_{\mathbb R^{n - 1}}g(y)\frac{\partial}{\partial x_i}(|x- y|^{-n})\; \mathrm d y = \int_{\mathbb R^{n -1}}g(y)\frac{\partial K}{\partial x_i}(x,y)\; \mathrm d y. \end{equation*} The partial derivative of $u$ with respect to $x_n$ is computed in a similar manner. For every $0< |h|< x_n/2$ we have \begin{equation*} \frac{u(x + he_n) - u(x)}{h} = \frac 2{n\alpha(n)}\int_{\mathbb R^{n - 1}}\frac{g(y)}{h}\left(\frac{x_n + h}{|x + he_n - y|^n} - \frac{x_n}{|x - y|^n}\right)\; \mathrm d y, \end{equation*} and, for each $y\in \mathbb R^{n - 1}$, the integrand converges as $h\to 0$ to \begin{equation*} g(y)\frac{\partial}{\partial x_n}\left(\frac{x_n}{|x- y|^n}\right) = g(y)\left(\frac 1{|x- y|^n} - \frac{nx_n^2}{|x- y|^{n + 2}}\right) \in L^1(\mathbb R^{n - 1}; \mathrm d y). \end{equation*} Moreover, for any $0< |h|< x_n/2$ we have \begin{equation*} \begin{split} \left|\frac{g(y)}{h}\left(\frac{x_n + h}{|x + he_n - y|^n} - \frac{x_n}{|x- y|^n}\right)\right| & = |g(y)|\left|\frac 1 h\int_0^1\frac{\mathrm d}{\mathrm d t}\left(\frac{x_n + h}{|x -y + the_n|^n}\right)\; \mathrm d t\right|\\ & \leq |g(y)|\int_0^1\left(\frac 1{|x -y + the_n|^n} + n\frac{x_n+ th}{|x - y + the_n|^{n + 1}}\right)\; \mathrm d t\\ & \leq C_n|g(y)|\int_0^1\frac 1{|x + the_n - y|^n}\; \mathrm d t\\ & \leq C_n|g(y)|(|x- y| - |h|)^{-n}\\ & \leq C_n|g(y)||x- y|^{-n} \in L^1(\mathbb R^{n - 1}; \mathrm d y). \end{split} \end{equation*} The Dominated Convergence Theorem now ensures that \begin{equation*} \frac{\partial u}{\partial x_n}(x) = \frac 2{n\alpha(n)}\int_{\mathbb R^{n - 1}}g(y)\frac{\partial}{\partial x_n}\left(\frac{x_n}{|x- y|^n}\right)\; \mathrm d y = \int_{\mathbb R^{n - 1}}g(y)\frac{\partial K}{\partial x_n}(x,y)\; \mathrm d y. \end{equation*}