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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.

But it remains not clear for me. Could someone please help me to understand that fact?

Thank you in advance!

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1 Answer 1

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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.

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  • $\begingroup$ 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! $\endgroup$
    – C. Bishop
    Feb 6, 2022 at 17:42
  • $\begingroup$ 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. $\endgroup$
    – s.harp
    Feb 6, 2022 at 18:35

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