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 $$ ||u||_{L^q(U)} \leq C||u||_{W^{1,p}(U)} $$ 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.