The <Rellich-Kondrachov compactness theorem> says that if $U\subset\mathbb R^n$ is a bounded Lipschitz domain, then
$$
W^{1,p}(U)\Subset L^q(U)
$$
for $1\leq q<p^*=np/(n-p)$ when $p<n$. When $p=n$, the embedding is compact into every finite $L^q$, and when $p>n$ it is compact into $C^0(\overline U)$, hence into every $L^q$.

The boundedness of the domain is essential. Choose a nonzero $\phi\in C_c^\infty(0,1)$ and set
$$
u_j(x)=\phi(x-2j)
\qquad(x>0).
$$
Translation invariance gives $\|u_j\|_{W^{1,1}(\mathbb R_+)}=\|\phi\|_{W^{1,1}}$, so after a fixed rescaling these functions lie in the unit ball. Their supports are pairwise disjoint and
$$
\|u_j-u_k\|_{L^1(\mathbb R_+)}=2\|\phi\|_{L^1}
\qquad(j\ne k).
$$
No subsequence is <Cauchy sequence>[Cauchy] in $L^1$, so the unit ball is not compact. This is the standard <failure of Rellich compactness on an unbounded domain>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 / 2 / f / Solution

Translations of one nonzero compactly supported smooth function have equal $W^{1,p}$ norms. On an unbounded domain they can have pairwise disjoint supports, so their mutual $L^p$ distances stay fixed and no subsequence converges in $L^p$.


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 Incoming links: Failure of Rellich compactness on an unbounded domain