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.
Best Answer
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.