I was searching some counterexample for I was searching some counterexample for Rellich-Kondrachov Compactness Theorem (You can see: PDE, Evans, chapter 5), for the case $q=p^*$ and I found this exemple in Math over flow.

But I don't understood the final lines of answer proposed. For example,why

$$\varepsilon^{-\frac{n}{p^*}} u\bigg(\frac{x}{\varepsilon}\bigg)$$

is bounded in $W^{1,p}$, and why this function has a constant non-zero norm in $L^{p^∗}$, and why has no convergent subsequences there for $\varepsilon\to 0$, since it converges a.e. to zero.

Can someone a hint? Thanks.

  • $\begingroup$ If you can find a counterexample, then it isn't a theorem. $\endgroup$ Jun 15, 2015 at 16:15

1 Answer 1


Let me give you a hint:

First, let us define $$v_\epsilon:= \epsilon^{-\frac{n}{p^*}}u\left(\frac{x}{\epsilon}\right) $$ Can you compute $\|v_\epsilon\|_{L^p}$ and $\|\nabla v_\epsilon \|_{L^p}$ in term of $u$? Try to write it down explicitly. Then you will know why it is bounded in $W^{1,p}$

Secondly, the fact that $v_\epsilon$ has no convergent subsequence in $L^{p^*}$ is because it has a constant non-zero norm in $L^{p^*}$ and $v_\epsilon\to 0$ a.e. To see why, suppose $v_\epsilon\to v$ for some $v$ in $L^{p^*}$, then we know that $\|v\|_{L^{p^*}}>0$ and hence $v$ is not 0 a.e., however, this will contradict to the fact that $v_\epsilon\to 0$ a.e.

Moreover, in this book, page 324, exercise 11.15, a easier counterexample is provided. You may check it out. The idea is same but I think the calculation is easier.

  • $\begingroup$ I understood, thanks you so much. $\endgroup$
    – Irddo
    Jun 15, 2015 at 21:39
  • $\begingroup$ (in case the link expires, the linked book is "A First Course in Sobolev Spaces (Graduate Studies in Mathematics) New ed. Edition") $\endgroup$ Sep 23, 2022 at 7:04

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