# It is true that $\int_{\Omega} |u|^s |\nabla u|^p dx =+\infty$ for some $u\in W_0^{1, p}(\Omega)$?

During my math class, the professor said what follows.

Let $$\Omega$$ be an open bounded domain in $$\mathbb{R}^N$$ and $$s\ge 1, p>1$$. Thus it would be $$\int_{\Omega} |u|^s |\nabla u|^p dx =+\infty \quad\mbox{ for some u\in W_0^{1, p}(\Omega)}.$$

Actually, I did not understand why. When I asked the professor, he answered that it is because the function $$f(u)=|u|^s$$ is not bounded.

Here's an example, denote with $$r(x_1,..,x_n)= \sqrt{\sum_i x_i^2}$$ the radius function. Then you have for example if $$n=6, p=2, s>2$$ that $$u(x_1,...,x_n) = \frac1{\sqrt{r(x_1,...,x_n)}}$$
is in $$W^{1,p}( B_1(0))$$, but $$\int |u|^s \ \|\nabla u\|^p = const \cdot \int_0^1 r^{n-1}r^{-s/2} \cdot r^{-5p/2}dr = const \cdot \int r^{-s/2}dr$$
is infinite for $$s>2$$. Its easy to modify this example to find counter-examples for any $$p$$, provided $$n$$ and $$s$$ are large enough. It is not clear to me what the set of triples $$(n,p,s)$$ looks like for which one can find a $$u$$ that makes the integral diverge.
• s.harp, thank you for the answer. Could you please clarify the details of the last sequence of computations? And why it is in $W^{1,2}(B_1(0))$? Thank you in advance! Feb 6, 2022 at 17:42
• Note that $\|\nabla f(r)\| =\frac12 \|f'(r) \frac1r\|$, whence $\|\nabla 1/\sqrt r \| = \frac14 \|r^{-5/2}\|$. Thats where the last term comes from. The $r^{n-1}$ term comes from spherical coordinates (volume element). The middle term is the $|u|^s$. To check that $u\in W^{1,2}$ you just need to see that $\int_0^1 r^{n-1} r^{-5\cdot p /2} dr <\infty$ and $\int_0^1 r^{n-1} r^{-p/2} dr$ for $p=2$, $n=6$, because thats what comes out when you integrate over the unit ball in spherical coordinates. Feb 6, 2022 at 18:35