The critical Sobolev embedding is not compact

Given: the Axiom of Choice, $n\ge 2$, $\mathrm{\Omega }=B(0,1)\subset {\mathbb{R}}^n$, $1\le p<n$, $p^{\ast }=\frac{np}{n-p}$, and $u_k(x):=k^{(n-p)\mathrm{/}p}\max \{0,1-k\mathrm{\mid }x\mathrm{\mid }\}$ for $k\ge 1$.

Scaling. For measurable nonnegative $h$ and $k>0$, $\int h(kx)k^{n}dx=\int h(y)dy$. (A linear map $T$ of ${\mathbb{R}}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with ${\lambda }_n(T[E])=\mathrm{\mid }\det T\mathrm{\mid }{\lambda }_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not, Euclidean balls have positive finite Lebesgue measure)

Membership in $W_0^{1,p}(\mathrm{\Omega })$. Each $u_k$ is Lipschitz on $\bar{\mathrm{\Omega }}$ and vanishes on $\mathrm{\partial }\mathrm{\Omega }$. Its Sobolev trace is therefore zero, because the trace agrees with boundary values for continuous Sobolev functions; the trace-kernel theorem then gives $u_k\in W_0^{1,p}(\mathrm{\Omega })$. To establish the missing premise for that chain rule, on the ball put $r_{\epsilon }(x)=\sqrt{\mathrm{\mid }x{\mathrm{\mid }}^2+{\epsilon }^2}$. These smooth functions converge uniformly to $\mathrm{\mid }x\mathrm{\mid }$, and ${\mathrm{\partial }}_i{r}_{\epsilon }=x_i\mathrm{/}{r}_{\epsilon }$ converges almost everywhere to $x_i\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$, with modulus at most $1$. Dominated convergence passes their weak test identities to the limit, proving $\mathrm{\mid }x\mathrm{\mid }\in W^{1,p}(\mathrm{\Omega })$ with that weak gradient for finite $p$. The scalar truncation chain rule then gives the cone gradient and membership. (Dominated convergence, Classical derivatives agree with weak derivatives) (Chain rule for globally Lipschitz scalar maps of Sobolev functions, Positive, negative, and truncated Sobolev functions, The trace agrees with classical restriction for continuous Sobolev functions, The kernel of the trace is the closure of the test functions, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms)

Almost-everywhere subsequences. An $L^{p^{\ast }}$-convergent sequence has a subsequence converging almost everywhere to a representative of its limit ($1<p^{\ast }<\mathrm{\infty }$). (Assuming Countable Choice, $L^p$-convergent sequences have almost-everywhere convergent subsequences, The space $L^p(\mu )$ as the quotient by null functions)