Question

Let $U$ be an open bounded subset of $\mathbb{R}^n$ with $C^{1}$ boundary. Let $1 \leq p < n$ and $p^{\ast} = pn/(n-p)$. Then the Sobolev space $W^{1,p}(U)$ is contained $L^{p^{\ast}}(U)$ and there is a constant $C$, depending only on $p$, $n$, and $U$, such that $$ ||u||_{L^{p^{*}}(U)} \leq C ||u||_{W^{1,p}(U)} $$ for every $u \in W^{1,p}(U)$ (cf. Theorem 2 in Section 5.6.1 of Partial Differential Equations by Evans).

The Rellich-Kondrachov Compactness Theorem says that $W^{1,p}(U)$ is compactly embedded into $L^{q}(U)$ for every $1 \leq q < p^{*}$. This means two things:

(i) There is a constant $C$, depending only on $p$, $n$, and $U$, such that $$ \displaystyle{ ||u||_{L^q(U)} \leq C||u||_{W^{1,p}(U)} } $$ for every $u \in W^{1,p}(U)$.

(ii) Every bounded sequence $(u_k)$ in $W^{1,p}(U)$ has a subsequence $(u_{k_j})$ that converges in $L^q(U)$.

Is there a standard counterexample that shows we cannot take $q=p^{\ast}$ in the Rellich-Kondrachov Compactness Theorem? In other words, I am asking for a sequence $(u_k)$ that is bounded in the $W^{1,p}(U)$ norm but has no convergent subsequence in ${L^{p^{\ast}}(U)}$. Note that such a sequence would have a subsequence that converges in $L^q(U)$ for every $1 \leq q < p^{\ast}$ but diverges in ${L^{p^{*}}(U)}$.

Thanks.

Answer 1

Yes, there is a standard way. Take any nonzero $u\in W^{1,p}(U)$, with support in a ball, say w.l.o.g $\operatorname{supp}(u)\subset B(0,r)\subset U$, and consider $$ u_\epsilon(x):= u\big(\frac{x}{\epsilon}\big)$$ Under this action by dilatations, the $L^q$ norm of $u$ and the $L^{p}$ norm of $\nabla u$ rescale with the same powers exactly for $q=p^*$: $$ \|u_ \epsilon\| _ q = \epsilon^{n/q} \|u\|_ q$$ $$ \| \nabla u_ \epsilon\|_p = \epsilon^{\frac{n-p}{p}} \|\nabla u\| _ p$$ This means that the normalized family , for all $0 < \epsilon \le 1$,

$$ \epsilon ^ { - \frac {n} {p ^ *} } u \Big( \frac{x} {\epsilon} \Big) \ , \ \ \ 0 < \epsilon \le r $$

is bounded in $W^{1,p}$, and has a constant non-zero norm in $ L^{p*} $, and of course has no convergent subsequences there for $\epsilon \to 0$, since it converges a.e. to zero. Note also that it converges to $0 $ in $L^q$ for all $ q < p^*$, as it has to be.

Answer 2

This question has cropped up in a work of mine.When the p* norm is computed without using the gradient value of the sequence,the Gagliardo-Nirenberg default condition' 1-n/p + n/p* = 0 appears to impose a restriction on the sequence index epsilon,since the sequence index disappears.This has theeffect' that the W^(1,p) bound condition the sequence has to compulsorily verify is ignored and this sounds illogical,specially when the Rellich-Kondrachov theorem tacitly makes use of such a bound in its proof.On the other hand,since the gradient values of the sequence have to be used in integration to verify the W^(1,p) bound and keeping in mind that there is a formula that expresses a C^1 function of compact support in terms of its gradient(this by integration by parts of the fundamental solution of the Laplacian),it is possible to obtain estimates for the p* norm in terms of the L^infinity norm of the defining test function u and what is more,such an estimate involves positive powers of the sequence index epsilon even though the Gagliardo-Nirenberg default condition is used,showing that the sequence actually converges to zero in p* norm! The actual meaning therefore is that the p* norm computations,one without using the gradient of the sequence and the other using the gradient and its implied value in integration,are different and if forcibly compared,leads to the contradiction that u vanishes identically.We have worked out the proof that relies on nontrivial facts such as strong bounds for the Hardy-Littlewood maximal function. It is to be noted further that the condition q < p* is only a sufficient condition to establish the Rellich-Kondrachov theorem by interpolation and there is nothing to support that the theorem can not be established by a procedure that does not require interpolation. In summary,the counter-example seems vacuous,showing that compactness at p* may still be an open problem!