Subcritical compactness for $W_0^{1,p}$ on arbitrary bounded open sets

Given: the Axiom of Choice, a bounded open set $\mathrm{\Omega }\subseteq {\mathbb{R}}^n$, $1\le p<n$, $p^{\ast }=np\mathrm{/}(n-p)$, a target exponent $1\le q<p^{\ast }$, and a sequence $(u_j)$ with $M:={\sup }_j\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W_0^{1,p}(\mathrm{\Omega })}<\mathrm{\infty }$.

$L^p$-compactness. Some subsequence $(u_{j_k})$ converges in $L^p(\mathrm{\Omega })$; the extraction uses Countable and Dependent Choice. (Compactness of $W_0^{1,p}(\mathrm{\Omega })\hookrightarrow L^p(\mathrm{\Omega })$ on bounded open sets, The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$), The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain)

Sobolev inequality for $W_0^{1,p}$ of an arbitrary bounded open set. $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^{p^{\ast }}(\mathrm{\Omega })}\le C\mathrm{\parallel }Dv{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$ for all $v\in W_0^{1,p}(\mathrm{\Omega })$, with $C=C(n,p)$; the zero extensions of the $u_j$ therefore satisfy $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{L^{p^{\ast }}(\mathrm{\Omega })}\le CM$. (The Sobolev inequality for zero-boundary Sobolev closures on open sets, Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, The Sobolev conjugate exponent and the scaling identity)

For $p=1$, supply the endpoint separately: the zero extension $v$ of a zero-boundary class belongs to $W^{1,1}({\mathbb{R}}^n)$. Choose smooth compactly supported $v_j\to v$ in $W^{1,1}$ by Compactly supported smooth functions are dense in W^{k,p}(R^n). The endpoint The p=1 Gagliardo-Nirenberg-Sobolev inequality on differences makes $(v_j)$ Cauchy in $L^{n\mathrm{/}(n-1)}$; completeness and the almost-everywhere subsequence theorem identify this limit with $v$, since $v_j\to v$ in $L^1$ as well. Passing to the limit in the endpoint inequality gives $\mathrm{\parallel }v{\mathrm{\parallel }}_{n\mathrm{/}(n-1)}\le C\mathrm{\parallel }Dv{\mathrm{\parallel }}_1$, and restriction supplies the bound asserted in [F2]. (Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$, Assuming Countable Choice, $L^p$-convergent sequences have almost-everywhere convergent subsequences)

Lyapunov interpolation and H"older. For $p\le b<p^{\ast }$ and $1\mathrm{/}b=\theta \mathrm{/}p+(1-\theta )\mathrm{/}{p}^{\ast }$, $\mathrm{\parallel }g{\mathrm{\parallel }}_b\le \mathrm{\parallel }g{\mathrm{\parallel }}_p^{\theta }\mathrm{\parallel }g{\mathrm{\parallel }}_{p^{\ast }}^{1-\theta }$; for $b\le p$, $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^b(\mathrm{\Omega })}\le \mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}b-1\mathrm{/}p}\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$. (Lyapunov interpolation inequality for $L^p$ norms, Holder's inequality for integrals, including the endpoint cases, The space $L^p(\mu )$ as the quotient by null functions)

Under Countable Choice, each $L^q(\mathrm{\Omega })$, $1\le q<\mathrm{\infty }$, is complete. (Riesz-Fischer completeness of $L^p$ for $1\le p\le \mathrm{\infty }$)