# 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.

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Isn't the name Kondrashev? –  timur Mar 8 '12 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) –  Pietro Majer May 25 '12 at 17:54

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.

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