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On a bounded Lipschitz domain $\Omega \subset {\mathbb{R}}^d$, the Rellich-Kondrachov compactness theorem states that $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le p<\infty$ andwith $1/q=0$ for $q=\infty$. Thus $q<p^{\ast }$ when $p<d$, every finite $q$ is allowed when $p=d$, and the embedding into continuous functions with their uniform norm is compact when $p>d$.

For the main proof, a Sobolev extension operator puts a bounded sequence into a common compactly supported region of ${\mathbb{R}}^d$. The Sobolev fundamental theorem of calculus on lines gives the uniform translation estimate $\Vert u(\cdot +h)-u{\Vert }_p\le |h|\Vert \nabla u{\Vert }_p$. Convolution with a mollifier therefore approximates the sequence uniformly in Lp space. At any fixed smoothing scale, its derivatives and supremum are uniformly bounded, so the Arzelà-Ascoli theorem gives a convergent subsequence. A diagonal choice and the uniform approximation give strong Lp space convergence. The finite measure of $\Omega$ handles $q<p$, while Lp interpolation inequality with the bounded Sobolev embeddings upgrades this to the larger subcritical finite Lp spaces; for $p>d$, Morrey's inequality gives uniform equicontinuity directly and the Arzelà-Ascoli theorem gives uniform convergence. Boundedness of the domain is essential.

The Sobolev embedding theorem $H^2(\Omega )\hookrightarrow L^{\infty }(\Omega )$ in dimension three, combined with the preceding estimate, givesOne direct justification of this embedding is a bounded Sobolev extension operator into $H^2({\mathbb{R}}^3)$, followed by Fourier inversion and the Cauchy-Schwarz inequality: ${\int }_{{\mathbb{R}}^3}(1+|\xi {|}^2{)}^{-2}d\xi <\infty$ bounds the inverse Fourier integral by the $H^2$ norm.

For radial forcing, the preceding part makes $u=U(r)$ and $f=F(r)$. Set $a=1/2$, $b=2$ and $H(r)={\int }_a^r{s}^{2}F(s)ds$. The radial equation gives $r^2{U}^{\prime }(r)=C_0+H(r)$. Both zero boundary values implyMoreover $\Vert H{\Vert }_{\infty }\le ({\int }_a^b{s}^{2}ds{)}^{1/2}({\int }_a^b{s}^2|F{|}^{2}ds{)}^{1/2}\le C\Vert f{\Vert }_2$. Since $r\ge a$, this proves the radial Poisson gradient estimate:

For a nonradial counterexample choose $x_0$ in the shell and a nonconstant $\phi \in C_c^{\infty }(B(0,1))$. For sufficiently small $\epsilon$, set $u_{\epsilon }(x)={\epsilon }^{1/2}\phi ((x-x_0)/\epsilon )$ and $f_{\epsilon }=\Delta u_{\epsilon }$. These are smooth with support away from the boundary. Scaling givesThus the failure of L2-to-Linfinity Poisson gradient bounds in three dimensions is explicit, despite the valid $H^2$ and value-supremum estimates.

Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

For a smooth $u$ and fixed $y$, the fundamental theorem of calculus gives, for $0<x<1$,Average this inequality over $x\in (0,1)$, use Holder inequality on that unit interval, raise to the power $p$, and integrate in $y$. This proves the estimate behind the W1p trace theorem on a half-space:Use the Sobolev extension operator from part d, approximate $Eu$ in $W^{1,p}({\mathbb{R}}^2)$ by smooth functions, and define $Tu$ as the $L^p(\mathbb{R})$ limit of their restrictions to $x=0$. The trace inequality makes this limit independent of the approximation and proves thatis linear and bounded. For a smooth function that extends continuously to the boundary, $Tu=u{|}_{\partial U}$, so this is the trace operator required.

For $U=(0,\infty )\times {\mathbb{R}}^{n-1}$, the formula $Eu(x_1,x^{\prime })=u(|x_1|,x^{\prime })$ defines a bounded Sobolev extension operator. Its weak normal derivative changes sign across the boundary, while its tangential weak derivatives are reflected unchanged.