Rellich Kondrachov and Sobolev Compactness — Examples

By [F2] each translate satisfies $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^p}$ and ${\mathrm{\partial }}_j{u}_k={\tau }_{k{e}_1}({\mathrm{\partial }}_j\psi )$ with $\mathrm{\parallel }{\mathrm{\partial }}_j{u}_k{\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }{\mathrm{\partial }}_j\psi {\mathrm{\parallel }}_{L^p}$; [F1] identifies these classical derivatives with the weak derivatives, so by [F4] $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{W^{1,p}({\mathbb{R}}^n)}=\mathrm{\parallel }\psi {\mathrm{\parallel }}_{W^{1,p}({\mathbb{R}}^n)}$ for every $k$. Hence ${\sup }_k\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{W^{1,p}}<\mathrm{\infty }$.

By [F3] fix $R$ with $K\subseteq B(0,R)$; if $\mathrm{\mid }k-l\mathrm{\mid }>2R$ then $\mathrm{supp}{u}_k$ and $\mathrm{supp}{u}_l$ are disjoint, so $\mathrm{\parallel }{u}_k-u_l{\mathrm{\parallel }}_{L^p}^p={\int }_{\mathrm{supp}{u}_k}\mathrm{\mid }{u}_k{\mathrm{\mid }}^p+{\int }_{\mathrm{supp}{u}_l}\mathrm{\mid }{u}_l{\mathrm{\mid }}^p=2\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^p}^p>0$.

Let $(u_{k_j})$ be any subsequence. Its indices tend to infinity, so for every tail there are two indices in it whose difference exceeds $2R$. Step 2.1 makes their distance the fixed positive value $2^{1\mathrm{/}p}\mathrm{\parallel }\psi {\mathrm{\parallel }}_p$, so this subsequence is not Cauchy and cannot converge. Hence no subsequence converges in $L^p$, and the refuted claim is false. Countable Choice is inherited through the smooth weak-derivative and Sobolev interfaces.

By [F1] and [F2], $\mathcal{F}$ is bounded, ${\sup }_k\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p}<\mathrm{\infty }$, and ${\sup }_k\mathrm{\parallel }{\tau }_h{u}_k-u_k{\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }{\tau }_h\phi -\phi {\mathrm{\parallel }}_{L^p}\to 0$ as $\mathrm{\mid }h\mathrm{\mid }\to 0$: the family is uniformly translation continuous.

Fix $R>0$ and choose a radius $A>0$ with $\mathrm{supp}\phi \subseteq B(0,A)$ and then an integer $k>R+A$; then up to a null set the support of $u_k$ lies outside $B(0,R)$, so ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R}\mathrm{\mid }{u}_k{\mathrm{\mid }}^{p}dx=\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^p}^p=\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p}^p>0$, and this value is independent of $R$; hence no $R$ makes the tails uniformly small, and the tightness condition of [F4] fails.

By [F3] the family is not relatively compact, so — although it is bounded and uniformly translation continuous — it violates the conclusion of the criterion [F4]; the two remaining hypotheses do not force compactness, and the refuted claim is false. Countable and Dependent Choice are used only through the criterion [F4] and its negated instance; the Sobolev and translation suppliers also use Countable Choice.

By [F1] applied to $\mathrm{\mid }\phi {\mathrm{\mid }}^p$ and to $\mathrm{\mid }D\phi {\mathrm{\mid }}^p$, $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{L^p}=j^{-n\mathrm{/}p}\cdot j^{n\mathrm{/}p}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p}=1$ and $\mathrm{\parallel }D{f}_j{\mathrm{\parallel }}_{L^p}=j^{-1}\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}$, and the same scaling holds for every derivative component. The classical derivatives are the weak derivatives by Classical derivatives agree with weak derivatives, so $(f_j)$ is bounded in $L^p$ and in $W^{1,p}({\mathbb{R}}^n)$; moreover ${\int }_{B(0,R)}\mathrm{\mid }{f}_j{\mathrm{\mid }}^{p}dx={\int }_{B(0,R\mathrm{/}j)}\mathrm{\mid }\phi (y){\mathrm{\mid }}^{p}dy\to 0$ as $j\to \mathrm{\infty }$ for each fixed $R$ by [F4], since $\mathrm{\mid }\phi {\mathrm{\mid }}^p{\mathbf{1}}_{B(0,R\mathrm{/}j)}\le \mathrm{\mid }\phi {\mathrm{\mid }}^p$ and the indicators tend to zero except at the null point $0$.

For fixed $h$ the segment bound [F2] applied to $\phi$ at $y=(x-h)\mathrm{/}j$ and $y^{\mathrm{\prime }}=x\mathrm{/}j$ gives $\mathrm{\mid }{f}_j(x-h)-f_j(x)\mathrm{\mid }\le j^{-n\mathrm{/}p}\mathrm{\mid }h\mathrm{\mid }{j}^{-1}{\int }_0^1\mathrm{\mid }\mathrm{\nabla }\phi ((x-th)\mathrm{/}j)\mathrm{\mid }dt$; taking $L^p$ norms and applying [F3] together with the translation and scaling identities of [F1] yields $\mathrm{\parallel }{\tau }_h{f}_j-f_j{\mathrm{\parallel }}_{L^p}\le \mathrm{\mid }h\mathrm{\mid }{j}^{-1}\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}\le \mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}$, a bound independent of $j$ that tends to $0$ with $\mathrm{\mid }h\mathrm{\mid }$; hence the family is uniformly translation continuous.

The family is not tight: by [F1], ${\int }_{\mathrm{\mid }x\mathrm{\mid }>R}\mathrm{\mid }{f}_j{\mathrm{\mid }}^{p}dx={\int }_{\mathrm{\mid }y\mathrm{\mid }>R\mathrm{/}j}\mathrm{\mid }\phi (y){\mathrm{\mid }}^{p}dy\to {\int }_{{\mathbb{R}}^n}\mathrm{\mid }\phi {\mathrm{\mid }}^p=1$ for every fixed $R$ by [F4], so no $R$ makes the tails uniformly small; and it is not relatively compact, because if a subsequence converged in $L^p({\mathbb{R}}^n)$ to some $g$, then $\mathrm{\parallel }g{\mathrm{\parallel }}_{L^p}=1$ by continuity of the norm, while step 1.1 forces $g=0$ almost everywhere on each ball $B(0,R)$ and hence on all of ${\mathbb{R}}^n$, a contradiction. So boundedness and uniform translation continuity alone do not give relative compactness on ${\mathbb{R}}^n$. Countable Choice is inherited through the scaling, Sobolev and completed-product interfaces.

By [F1] and [F4], $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{L^p}^p=c_p^p{\int }_0^{2\pi }\mathrm{\mid }\sin (jx){\mathrm{\mid }}^{p}dx=1$, and all supports lie in the fixed bounded set $[0,2\pi ]$, so the family is bounded and tight.

Put $h_j:=\pi \mathrm{/}j$. For $x\in (h_j,2\pi )$ both $x$ and $x-h_j$ lie in $(0,2\pi )$, so [F2] gives $f_j(x-h_j)-f_j(x)=-2c_p\sin (jx)$ and hence $\mathrm{\mid }{f}_j(x-h_j)-f_j(x){\mathrm{\mid }}^p=2^p{c}_p^p\mathrm{\mid }\sin (jx){\mathrm{\mid }}^p$; integrating over $(h_j,2\pi )$ and using [F1] with the trivial bound ${\int }_0^{h_j}\mathrm{\mid }\sin (jx){\mathrm{\mid }}^{p}dx\le h_j$ gives $\mathrm{\parallel }{\tau }_{h_j}{f}_j-f_j{\mathrm{\parallel }}_{L^p}^p\ge 2^p{c}_p^p(c_p^{-p}-h_j)=2^p(1-c_p^p{h}_j)\to 2^p$.

Hence ${\mathrm{lim\hspace{0.17em}inf}}_j\mathrm{\parallel }{\tau }_{h_j}{f}_j-f_j{\mathrm{\parallel }}_{L^p}\ge 2>0$ although $h_j=\pi \mathrm{/}j\to 0$, so the family is not uniformly translation continuous; by [F3] it is not relatively compact in $L^p(\mathbb{R})$, and the refuted claim is false. Countable Choice is used by the scaling interface [F1] and the necessity lemma [F3].

By [F1], $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^{p^{\ast }}}^{p^{\ast }}=k^{(n-p)p^{\ast }\mathrm{/}p}{k}^{-n}\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^{p^{\ast }}}^{p^{\ast }}=\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^{p^{\ast }}}^{p^{\ast }}>0$ because $(n-p)p^{\ast }\mathrm{/}p=n$, and $\mathrm{\parallel }D{u}_k{\mathrm{\parallel }}_{L^p}=k^{(n-p)\mathrm{/}p+1}{k}^{-n\mathrm{/}p}\mathrm{\parallel }D{u}_1{\mathrm{\parallel }}_{L^p}=\mathrm{\parallel }D{u}_1{\mathrm{\parallel }}_{L^p}$, while $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{L^p}=k^{(n-p)\mathrm{/}p}{k}^{-n\mathrm{/}p}\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^p}=k^{-1}\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^p}\to 0$; by [F2] all $u_k$ lie in $W_0^{1,p}(\mathrm{\Omega })$, so the sequence is bounded in $W_0^{1,p}(\mathrm{\Omega })$ and converges to $0$ almost everywhere and in $L^p(\mathrm{\Omega })$.

Suppose a subsequence converged in $L^{p^{\ast }}(\mathrm{\Omega })$ to some $w$. Then $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^{p^{\ast }}}=\mathrm{\parallel }{u}_1{\mathrm{\parallel }}_{L^{p^{\ast }}}>0$ by continuity of the norm, while [F3] provides a further subsequence converging almost everywhere to a representative of $w$; since $u_k(x)\to 0$ for every $x\mathrm{\ne }0$, that representative vanishes almost everywhere, forcing $w=0$ and contradicting the positive norm. Hence no subsequence converges in $L^{p^{\ast }}(\mathrm{\Omega })$ and the refuted compactness claim is false; the companion subcritical statement Subcritical compactness for $W_0^{1,p}$ on arbitrary bounded open sets shows that the strict inequality $q<p^{\ast }$ cannot be relaxed. The Axiom of Choice is inherited through [F2].

By [F1], $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\le (k+1{)}^{-1}\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p}+\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p}$ is bounded, and $\mathrm{\parallel }{u}_k{\mathrm{\parallel }}_{\mathrm{\infty }}=(k+1{)}^{-\alpha }\mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$, so the representatives converge uniformly to $0$ on $\mathrm{\Omega }$.

For $x=0$ and $y=e_1\mathrm{/}(k+1)$ in $\bar{\mathrm{\Omega }}$, one has $u_k(x)=(k+1{)}^{-\alpha }$ and $u_k(y)=(k+1{)}^{-\alpha }\phi (e_1)=0$ because the bump support is inside the unit ball. Therefore $[u_k{]}_{C^{0,\alpha }(\bar{\mathrm{\Omega }})}\ge (k+1{)}^{-\alpha }\mathrm{/}\mathrm{\mid }{e}_1\mathrm{/}(k+1){\mathrm{\mid }}^{\alpha }=1$ for every $k$.

If a subsequence converged in $C^{0,\alpha }(\bar{\mathrm{\Omega }})$ to some $w$, then it would converge uniformly, hence $w=0$ by step 1.1, and by [F2] the seminorms would converge: $\mathrm{\mid }[u_{k_j}{]}_{C^{0,\alpha }}-[w{]}_{C^{0,\alpha }}\mathrm{\mid }\le [u_{k_j}-w{]}_{C^{0,\alpha }}\le \mathrm{\parallel }{u}_{k_j}-w{\mathrm{\parallel }}_{C^{0,\alpha }}\to 0$, forcing the seminorms of step 1.2 to tend to $0$ — a contradiction, since they are at least $1$. Hence the endpoint exponent in Morrey--Rellich compactness for $p>n$ cannot yield compactness, and the refuted claim is false.

For all sufficiently large $j$, the support of $\psi (j\cdot )$ on $\mathrm{\Omega }$ lies inside the flat patch, where $\mathrm{\Omega }$ is the upper half-space; the restriction of the ambient smooth function is therefore smooth up to the boundary and belongs to $W^{1,p}(\mathrm{\Omega })$. By [F3] and the change of variables over that half-space, $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}=j^{(n-p)\mathrm{/}p}{j}^{-n\mathrm{/}p}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^p({\mathbb{R}}_{+}^n)}=j^{-1}\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^p({\mathbb{R}}_{+}^n)}$ and $\mathrm{\parallel }D{u}_j{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}=j^{(n-p)\mathrm{/}p}{j}^{1-n\mathrm{/}p}\mathrm{\parallel }D\psi {\mathrm{\parallel }}_{L^p({\mathbb{R}}_{+}^n)}=\mathrm{\parallel }D\psi {\mathrm{\parallel }}_{L^p({\mathbb{R}}_{+}^n)}$, so ${\sup }_j\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}<\mathrm{\infty }$ (discarding the finitely many initial indices if needed).

By [F1] and [F2], $T{u}_j$ is the restriction of $u_j$ to $\mathrm{\partial }\mathrm{\Omega }$, which equals $j^{(n-p)\mathrm{/}p}g(j{x}^{\mathrm{\prime }})$ on the flat patch and $0$ elsewhere. By [F2] and [F3] with $m=n-1$, $\mathrm{\parallel }T{u}_j{\mathrm{\parallel }}_{L^{q_{\mathrm{\partial }}}(\mathrm{\partial }\mathrm{\Omega })}^{q_{\mathrm{\partial }}}=j^{(n-p)q_{\mathrm{\partial }}\mathrm{/}p}{j}^{-(n-1)}\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{q_{\mathrm{\partial }}}}^{q_{\mathrm{\partial }}}=\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{q_{\mathrm{\partial }}}}^{q_{\mathrm{\partial }}}>0$, because $\frac{(n-p)q_{\mathrm{\partial }}}{p}=n-1$; and $T{u}_j(x^{\mathrm{\prime }})\to 0$ for every $x^{\mathrm{\prime }}\mathrm{\ne }0$ on the patch, since $g$ is compactly supported, while off the patch $T{u}_j=0$ for all large $j$.

Suppose a subsequence of $(T{u}_j)$ converged strongly in $L^{q_{\mathrm{\partial }}}(\mathrm{\partial }\mathrm{\Omega })$ to some $v$. Then $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^{q_{\mathrm{\partial }}}}=\mathrm{\parallel }g{\mathrm{\parallel }}_{L^{q_{\mathrm{\partial }}}}>0$ by continuity of the norm [F5], while by [F4] a further subsequence converges almost everywhere to a representative of $v$; since the traces converge to $0$ at every boundary point except the single point $0$, which has surface measure zero, that representative vanishes almost everywhere, so $\mathrm{\parallel }v\mathrm{\parallel }=0$ by [F5], a contradiction. Hence the bounded $W^{1,p}$-sequence $(u_j)$ has no subsequence whose traces converge strongly at the critical boundary exponent, and the refuted compactness claim is false. The Axiom of Choice is inherited through the trace interface [F1].

By [F1] and [F2] every pair $x,y\in [0,1]$ satisfies $\mathrm{\mid }{u}_j^{\ast }(x)-u_j^{\ast }(y)\mathrm{\mid }\le M\mathrm{\mid }x-y{\mathrm{\mid }}^{1-1\mathrm{/}p}$ (with exponent $1$ when $p=\mathrm{\infty }$), a modulus independent of $j$; by [F3] also $\mathrm{\parallel }{u}_j^{\ast }{\mathrm{\parallel }}_{\mathrm{\infty }}\le 2M$. Hence $\{u_j^{\ast }\}$ is uniformly bounded and equicontinuous, and for $1<p\le \mathrm{\infty }$ the exponent $1-1\mathrm{/}p$ is positive, so the modulus tends to $0$ with $\mathrm{\mid }x-y\mathrm{\mid }$.