# Lemma from Gilbarg/Trudinger

This is from Gilbarg/Trudinger "Elliptic Partial Differential Equations of Second order" Lemma 4.1 (p.54 of the third edition) The lemma states that for a bounded integrable function $f$, the newtonian potential is $C^1(\mathbb{R}^n)$.

Basically, I don't really get how this inequality happens : $\int_{|x-y|\leq 2\epsilon} \left(|D_i\Gamma| + \frac{2}{\epsilon}|\Gamma| \right)dy \leq (\frac{2n\epsilon}{n-2})$ .

Here, $\Gamma$ is the fundamental solution to Laplace: That is , $\Gamma(|x-y|) = \frac{1}{n(2-n)\omega_n}|x-y|^{2-n}$ where $\omega_n$ is the unit ball volume. I know that we can bound the derivatives by $|D^{\beta}\Gamma(x-y)| \leq C|x-y|^{2-n-|\beta|}$ where we are using the generalized derivative notation here, but I don't see how to apply this to the problem. In fact, for very small values of $y$ near $x$, both $|\Gamma|$ and $|D_i \Gamma|$ are unbounded, so how can this integral be small?

This is because the growth of $$r^{1-n}$$ as $$r\to0$$ is just not enough to induce blow up. More specifically $$\int_{r<\varepsilon}\frac{\mathrm{d}y}{r^{n-1}} = C\int_0^{\varepsilon}\frac{r^{n-1}\mathrm{d}r}{r^{n-1}} =C\varepsilon,$$ where we used polar coordinates.