Differentiation of the Newtonian potential of a function (Sandro Salsa's PDE book)

Question

I am studying the PDE book Partial Differential Equations in Action, From Modelling to Theory, by Sandro Salsa. I would like to ask two questions about the differentiation of the Newtonian potential of a function $f$, $$ u(x)=\int_{\mathbb{R}^3} \Phi(x-y) f(y)\,\mathrm{d}y=(\Phi\ast f)(x), $$ where $$\Phi(x)=\frac{1}{4\pi}\frac{1}{|x|}$$ is the fundamental solution of the Poisson equation $-\Delta u=\delta_0$.

1. In a footnote in page 130, it is stated the following: if $g\in C^1(\mathbb{R}^3)$ and $|g(x)|\leq M/|x|^{2+\epsilon}$, then $$ \frac{\partial}{\partial x_j}\int_{\mathbb{R}^3}\frac{1}{|x-y|}g(y)\,\mathrm{d}y=\int_{\mathbb{R}^3}\frac{1}{|x-y|}\frac{\partial g}{\partial y_j}(y)\,\mathrm{d}y. $$ Usually, results on differentiation under the integral sign requiere bounds on the derivative, so that the Dominated Convergence Theorem can be applied. I do not know how to proceed in this case.

2. In Theorem 3.9, it is said that if $f\in C_c^2(\mathbb{R}^3)$ (i.e., $C^2$ with compact support), then its Newtonian potential $u=\Phi\ast f$ is $C^2(\mathbb{R}^3)$ and $-\Delta u=f$. The fact that $f$ is $0$ outside a ball and $1/|x|$ is locally integrable allows differentiating under the integral sign. But, in Remark 3.6, it is said that the theorem holds when $f\in C^1(\mathbb{R}^3)$ and $|f(x)|\leq M/|x|^{2+\epsilon}$. Any idea or reference on why this statement is true?

Answer 1

1. Some conditions should be put on derivatives of $g$ as the following example shows. Let $g(y)=\frac{\sin e^{y_1}}{(1+|y|^4)}$, then the integral in the rhs diverges. For the statement to hold it would be enough to demand additionally $|\nabla g(x)|\leq M/|x|^{2+\epsilon}$.
2. It is enough to prove the theorem for $f\in C^1(\Omega)$ in a bounded domain $\Omega$. This is done in Glibarg D., Trudinger N. "Elliptic Partial Differential Equations of Second Order", Lemma 4.2, under a weaker assumption on $f$, namely Holder continuity. Actually it is enough to demand that $f$ should satisfy the Dini condition: $$ \int_{|x-y|\le1}\frac{1}{|x-y|^3}|f(y)-f(x)|\,\mathrm{d}y<\infty. $$