The critical Sobolev embedding into every finite $L^q$

The case $q\le n$. By [F1] with $a=q$ and $b=n$, $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^q(\mathrm{\Omega })}\le \mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}q-1\mathrm{/}n}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^n(\mathrm{\Omega })}\le \mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}q-1\mathrm{/}n}\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,n}(\mathrm{\Omega })}$.

The case $q>n$. Put $s:=\frac{nq}{n+q}$. Then $n\mathrm{/}2<s<n$ and $s^{\ast }=\frac{ns}{n-s}=q$ by [F3]. By the domain hypothesis, $\mathrm{\Omega }$ is a $W^{1,s}$-extension domain, so the subcritical embedding [F2] with $p=s$ and $r=q=s^{\ast }$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^q(\mathrm{\Omega })}\le C(n,s,q,\mathrm{\Omega })\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,s}(\mathrm{\Omega })}$. Holder [F1] applied to each component with $a=s<b=n$ gives $\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^s}\le \mathrm{\mid }\mathrm{\Omega }{\mathrm{\mid }}^{1\mathrm{/}s-1\mathrm{/}n}\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^n}$ for $\mathrm{\mid }\alpha \mathrm{\mid }\le 1$; summing over the finitely many multi-indices yields $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,s}(\mathrm{\Omega })}\le C(\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid },n,s)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,n}(\mathrm{\Omega })}$. Combining the two estimates proves the case $q>n$; since $s=nq\mathrm{/}(n+q)$, the resulting constant depends only on $n,q,\mathrm{\Omega }$.

Conclusion. Steps 1.1 and 1.2 cover $q\le n$ and $q>n$, so every finite $q$ is covered with a constant depending only on $n,q,\mathrm{\Omega }$. To see that these constants cannot remain bounded as $q\to \mathrm{\infty }$, choose $a\in \mathrm{\Omega }$ and $R>0$ with $\overline{B(a,2R)}\subset \mathrm{\Omega }$, and choose $\eta$ as in [F4]. Define $w(x)=\eta (x)\log \log (1+R\mathrm{/}\mathrm{\mid }x-a\mathrm{\mid })$ for $x\mathrm{\ne }a$, assigning any finite value at $a$. If $r=\mathrm{\mid }x-a\mathrm{\mid }\le R\mathrm{/}2$, then $\mathrm{\mid }D\log \log (1+R\mathrm{/}r)\mathrm{\mid }=\frac{R}{r(r+R)\log (1+R\mathrm{/}r)}\le \frac{1}{r\log (R\mathrm{/}r)}\mathrm{.}$ Thus polar coordinates [F4] give ${\int }_{B(a,R\mathrm{/}2)}\mathrm{\mid }Dw{\mathrm{\mid }}^{n}dx\le C{\int }_0^{R\mathrm{/}2}\frac{dr}{r(\log (R\mathrm{/}r){)}^n}<\mathrm{\infty }$ for $n\ge 2$; also $w\in L^n$ near $a$ because $\log \log (1+R\mathrm{/}r)=O(r^{-1\mathrm{/}2})$ and the polar $L^n$ integral of $r^{-1\mathrm{/}2}$ converges for $n\ge 2$, and on the rest of its compact support $w$ is smooth with bounded derivatives. For every coordinate line with nonzero transverse displacement from $a$, $w$ is smooth and compactly supported on that line. Apply the fundamental theorem (If $f:[a,b]\to {\mathbb{R}}^m$ is differentiable with integrable $f^{\mathrm{\prime }}$ then ${\int }_a^b{f}^{\mathrm{\prime }}=f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz) to $w$ times a test function. The transverse singleton of excluded lines is null since $n\ge 2$, and $w,Dw\in L^n\subseteq L^1$ on their bounded support. Fubini (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability) therefore integrates the section identities to the weak derivative identity, proving $w\in W^{1,n}(\mathrm{\Omega })$. It is essentially unbounded on every neighbourhood of $a$: for every $M>0$ it exceeds $M$ on a punctured ball about $a$, so $E_M$ has positive measure. If $q_j\to \mathrm{\infty }$ and admissible constants satisfied $C(q_j)\le C_{\ast }$, then for each $M>0$, the set $E_M=\{x:\mathrm{\mid }w(x)\mathrm{\mid }>M\}$ has positive measure and $C_{\ast }\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,n}(\mathrm{\Omega })}\ge \mathrm{\parallel }w{\mathrm{\parallel }}_{L^{q_j}(\mathrm{\Omega })}\ge M\mathrm{\mid }{E}_M{\mathrm{\mid }}^{1\mathrm{/}{q}_j}\mathrm{.}$ Letting $j\to \mathrm{\infty }$ gives $C_{\ast }\mathrm{\parallel }w{\mathrm{\parallel }}_{W^{1,n}}\ge M$ for every $M$, a contradiction. Hence the best constants necessarily diverge as $q\to \mathrm{\infty }$, and no $L^{\mathrm{\infty }}$ endpoint is asserted.