# Compact embeddings of Sobolev spaces: a counterexample showing the Rellich-Kondrachov theorem is sharp

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.

• Isn't the name Kondrashev? Commented Mar 8, 2012 at 17:49
• The Russian is В.И.Кондрашов. Therefore, according with the BGN/PCGN romanization of Russian, the English transliteration should be "Kondrashov". Nevertheless, for some reason, "Rellich-Kondrachov theorem" gets more Google results than "Rellich-Kondrashov theorem" (small numbers, anyway) Commented May 25, 2012 at 17:54
• @PietroMajer that's because before we used the French-based latinization (ш= French ch =English sh). Commented Oct 17, 2020 at 10:58

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 $0 < \epsilon \le 1$,

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

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.

• In this answer, it is not clear to see why the sequence $\{u_\varepsilon\}_\varepsilon$ has no convergent subsequence in $L^{p^\star}$. Basically, there are two types of lacking the compactness property: Unbounded domains and critical exponents. In both cases, it is more convenient to dilate a "good" function plus a scaling of this function if working in bounded domains. Commented Jan 7, 2015 at 12:56
• @QA Ngô In fact, $u_\epsilon\to0$ a.e., in $L^{p^*}$, and in $W^{1,p}$, as $\epsilon\to0$. The claim is on the normalized family: check the last lines. Commented Jan 7, 2015 at 15:32
• You can find this in Adams, Sobolev Spaces. Commented Nov 4, 2015 at 14:53
• Apologies for coming back to very old material, but do you have any comment on the the answers below which claim that there is something amiss in your answer? Commented Apr 7, 2021 at 23:11
• Actually no comment Commented Apr 22, 2021 at 9:37

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 1-n/p + n/p* = 0 appears to impose a restriction on the sequence index epsilon,since the sequence index disappears.This has the `effect' 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!

• In respect of the Question raised by Ànonimous',if only one can,and indeed one should inevitably,give-up using the àpriori given,generic'Lebesgue norm and use instead the gradient estimated(from a reproducing kernel formula) Lebesgue norm,the dilemma is resolved for good.Indeed by a well known finite measure theoretic result of A.Villani,Rellich-Kondrachov compactness(RK) seen for q<p* is transferred to RK at p* by Cauchy Continuity.Thus any measure of non-compactness at p* has to be zero.The inf.metric on W^(1,p) on the bounded set is incomplete as well in respect of a uniform gauge. Commented Oct 16, 2020 at 15:17