Show: If $u \in W^{1,p}(B_4(0))$ then $\int_{B_4(0)} \frac{|\nabla u|^p}{|x-a|^{n-1}} dx < \infty$ for a.e. $a \in B_1(0)$.

Question

I'm currently working through the article "Topology and Sobolev Spaces" by Brezis and Li and as basis for the proof of an important result the following fact is used:

If $u \in W^{1,p}(B_4(0))$ then $\int_{B_4(0)} \frac{\mid\nabla u(x) \mid^p}{\mid x-a \mid^{n-1}} dx < \infty$ for a.e. $a \in B_1(0)$. (p. 330, remark 1.1)

How does one show that / Why is that evident?

I'd appreciate any help!

Answer 1

The function $$F(a) = \int_{B_4(0)} \frac{\mid\nabla u(x) \mid^p}{\mid x-a \mid^{n-1}} dx$$ is the convolution of $|\nabla u|^p$ (restricted to the ball) with the Riesz potential $I(y) = 1/|y|^{n-1}$. Moreover, since only the values of $x,a$ with $|x-a|\le 8$ appear in the integral, we can just as well convolve with the restriction of the $I$, namely $$J(y) = \begin{cases}1/|y|^{n-1},\quad & |y|\le 8, \\ 0,\quad & |y|>8\end{cases}$$ With this setup, it remains to observe that

1. Both $|\nabla u|^p$ and $J$ are in $L^1$.
2. The convolution of $L^1$ functions in again in $L^1$.
3. A function that is in $L^1$ is finite almost everywhere.

More generally, the above shows that the Riesz potential of an integrable function with bounded support is finite almost everywhere.