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?
Thank you for your time.
Best Answer
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.