exercise prove that if $f\in L^1(\mathbb R^3)\cap L^{\infty}(\mathbb R^3)$ and said $ G_f(x)=\int dy \frac{f(y)}{4\pi|x-y|}$ we have$\frac{\partial}{\partial x_i} G_f(x)=\int dy \frac{\partial}{\partial x_i} \frac{f(y)}{4\pi|x-y|}$
my solution I have used this following theorem
theorem: let $\Omega \subset \mathbb R^n$$(X,\mu)$ a measure space let $z_0\in \Omega, g\in L^1(X,\mu)$ and suppose that for a given vector $v\in \mathbb R^n$ exist $\frac{\partial}{\partial v}f(z,x) \forall z\in (z_0-\delta,z_0+\delta) $ a e $\mu $ and suppose $|\frac{\partial}{\partial v}f(z,x)|\leq g(x)\forall z\in (z_0-\delta,z_0+\delta) $ a e $\mu $ we have that given $F(z)=\int_X f(z,x)d\mu(x)$ we have $\frac{\partial}{\partial v}F(z)=\int_X \frac{\partial}{\partial v}f(z,x)d\mu(x)$
back story of problem in class we have solved this problem using another approach , I asked to m professor and he said he was not possible to use this theorem but I think es in this way
step 1 fixed $x_0$ we have that $\frac{\partial}{\partial x_i} \frac{f(y)}{4\pi|x-y|}=\frac{y_1-x_1}{4\pi|x-y|^3}f(y)$ this function $\forall x \in (x_0-\delta,x_0+\delta)$fixed is derivable a e $y$ because I have to exlude the points where y=x
step 2 case $x_0>0$ I can alwas say that \delta is sufficiently small suh that $x_0-\delta>0$ so I can say taht $|\frac{y_1-x_1}{4\pi|x-y|^3}f(y)|$<\begin{equation} \begin{cases} \frac{\left\lVert f(y) \right \rVert_{\infty}} {|x_0-\delta-y|^2 } \quad y\in B_{x_0-\delta}(1)\\ |f(y)| \quad \text{otherwise} \end{cases}\,.\end{equation}
and similar cases for $x_0>0$ and $x_0=0$ let me know how can this be improved