On poisson Equation:Interchanging $\int$ and $\lim$ 
Let $\Phi(x)=\frac{1}{|x|^{(n-2)}},\, (x\neq0\in\mathbb R^n, n\geq3)$, $\,f\in C^2(\mathbb R^n)$ with compact support. Then $$\lim_{h\rightarrow0}\int_{\mathbb R^n}\Phi(y)[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy= \int_{\mathbb R^n}\Phi(y)\lim_{h\rightarrow0}[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy$$

Why are limit and integral interchangeable? $\Phi$ has no definition on $0$, I'm not sure about how to deal with it.
 A: $\Phi(x)=\frac{1}{|x|^{(n-2)}},\, (x\neq0\in\mathbb R^n, n\geq3)$, $\,f\in C^2(\mathbb R^n)$ with compact support. Then
 $$\lim_{h\rightarrow0}\int_{\mathbb R^n}\Phi(y)[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy= \int_{\mathbb R^n}\Phi(y)\lim_{h\rightarrow0}[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy$$
There is $R>0$ such that, $Supp [f(x+he_i-\cdot)-f(x-\cdot)]\subset B_R(0).$ and 
$$\left|\frac{f(x+he_i-y)-f(x-y)}{h}\right| = \left|\int_0^1 \nabla f(x-y+the_i)\cdot e_i dt \right|\le \|\nabla f\|_\infty $$
Therefore 
$$ \Phi(y)\left|\frac{f(x+he_i-y)-f(x-y)}{h}\right| \le \Phi(y) \|\nabla f\|_\infty\in L^1(B_R(0))$$
and 
$$\Phi(y)\lim_{h\rightarrow0}\frac{f(x+he_i-y)-f(x-y)}{h} = \Phi(y)\nabla f(x-y)\cdot e_i\in L^1(B_R(0))$$
Indeed by polar coordinate 
$$\int_{B_R(0)}\Phi(y)dy = c_n\int_0^R r^{n-1}r^{-n+2}dr <\infty.$$
Form convergence dominated theorem, it follows that 
\begin{split}&&\lim_{h\rightarrow0}\int_{\mathbb R^n}\Phi(y)[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy\\
&=&\lim_{h\rightarrow0}\int_{B_R(0)}\Phi(y)[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy\\
&=& \int_{B_R(0)}\Phi(y)\lim_{h\rightarrow0}[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy\\
&=& \int_{\mathbb R^n}\Phi(y)\lim_{h\rightarrow0}[\frac{f(x+he_i-y)-f(x-y)}{h}]\,dy.
\end{split}
