Smooth Approximation and Sobolev Extension — Examples

For every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$, integration by parts on the two half-lines gives ${\int }_{\mathbb{R}}\mathrm{\mid }x\mathrm{\mid }{\phi }^{\mathrm{\prime }}(x)dx={\int }_{-\mathrm{\infty }}^0(-x){\phi }^{\mathrm{\prime }}(x)dx+{\int }_0^{\mathrm{\infty }}x{\phi }^{\mathrm{\prime }}(x)dx={\int }_{-\mathrm{\infty }}^0\phi (x)dx-{\int }_0^{\mathrm{\infty }}\phi (x)dx=-{\int }_{\mathbb{R}}\mathrm{sgn}(x)\phi (x)dx\mathrm{.}$ The boundary terms vanish at infinity by compact support and at $0$ because $\mathrm{\mid }0\mathrm{\mid }=0$. Thus $u^{\mathrm{\prime }}=\mathrm{sgn}$ weakly; both $u$ and $u^{\mathrm{\prime }}$ are bounded on every bounded open interval $J$, so $u{\mathrm{\mid }}_J\in W^{1,\mathrm{\infty }}(J)$ and $u\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,\mathrm{\infty }}(\mathbb{R})$. Globally $u\mathrm{\notin }{L}^{\mathrm{\infty }}(\mathbb{R})$.

Fix $R>0$ with $\mathrm{supp}\rho \subseteq [-R,R]$ and set $\tilde{\rho }(x)=R\rho (Rx)$. This is a nonnegative even unit-mass kernel supported in $[-1,1]$, with ${\tilde{\rho }}_{R\epsilon }={\rho }_{\epsilon }$. For any fixed $\epsilon >0$ and bounded open interval $I$, enlarge $I$ to a bounded open interval $J$ with $\mathrm{dist}(\bar{I},\mathbb{R}\setminus J)>R\epsilon$. Apply [F4] to $u{\mathrm{\mid }}_J\in W^{1,\mathrm{\infty }}(J)$ and $\tilde{\rho }$ at scale $R\epsilon$: on $I$ its convolution agrees with the globally defined $u_{\epsilon }$, since the kernel only samples $J$. Thus $u_{\epsilon }$ is smooth on $I$ and $u_{\epsilon }^{\mathrm{\prime }}={\rho }_{\epsilon }\ast \mathrm{sgn}$ there. Since $I$ was arbitrary, both assertions hold on $\mathbb{R}$.

For finite $p$, fix a bounded open interval $J$ with $\bar{I}\subset J$. Step 1.1 gives $u{\mathrm{\mid }}_J\in W^{1,p}(J)$, since $J$ has finite length. Apply [F5] to this restriction and the unit-support kernel $\tilde{\rho }$ at scale $R\epsilon$. For all sufficiently small $\epsilon$, the interior mollification agrees with $u_{\epsilon }$ on $I$, so $\mathrm{\parallel }{u}_{\epsilon }-u{\mathrm{\parallel }}_{W^{1,p}(I)}\to 0$.

Evenness at the corner: because $\rho$ is even, so is ${\rho }_{\epsilon }$, and the change of variable $y\mapsto -y$ gives $u_{\epsilon }^{\mathrm{\prime }}(0)={\int }_{\mathbb{R}}{\rho }_{\epsilon }(-y)\mathrm{sgn}(y)dy={\int }_{\mathbb{R}}{\rho }_{\epsilon }(y)\mathrm{sgn}(y)dy=0,$ the last integral vanishing because $y\mapsto {\rho }_{\epsilon }(y)\mathrm{sgn}(y)$ is odd and integrable.

The derivative $u_{\epsilon }^{\mathrm{\prime }}$ is continuous on $\mathbb{R}$ with $u_{\epsilon }^{\mathrm{\prime }}(0)=0$ by step 3.2, so there is $\delta >0$ with $\mathrm{\mid }{u}_{\epsilon }^{\mathrm{\prime }}(x)\mathrm{\mid }<1\mathrm{/}2$ for $\mathrm{\mid }x\mathrm{\mid }<\delta$; on $(0,\delta )$ one therefore has $\mathrm{sgn}=1$ and $\mathrm{\mid }{u}_{\epsilon }^{\mathrm{\prime }}(x)-\mathrm{sgn}(x)\mathrm{\mid }=1-u_{\epsilon }^{\mathrm{\prime }}(x)>1\mathrm{/}2$. If the interval $I$ contains $0$ in its interior, then either $(0,\delta )\cap I$ or $(-\delta ,0)\cap I$ has positive length, so the essential supremum defining $\mathrm{\parallel }{u}_{\epsilon }^{\mathrm{\prime }}-\mathrm{sgn}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I)}$ is at least $1\mathrm{/}2$; by the norm convention of [F7], $\mathrm{\parallel }{u}_{\epsilon }-u{\mathrm{\parallel }}_{W^{1,\mathrm{\infty }}(I)}\ge \mathrm{\parallel }{u}_{\epsilon }^{\mathrm{\prime }}-\mathrm{sgn}{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(I)}\ge 1\mathrm{/}2$ for every $\epsilon >0$, and $u_{\epsilon }\to ̸u$ in $W^{1,\mathrm{\infty }}(I)$.

Thus the mollifications of $\mathrm{\mid }x\mathrm{\mid }$ converge in $W^{1,p}(I)$ for every finite $p$ but never in $W^{1,\mathrm{\infty }}(I)$ across the corner; the failure is not an artefact of this sequence, by the continuity argument of [F6] for arbitrary smooth approximants.

The hypotheses of [L1] with $k=1$ hold: $u\in W^{1,p}(\mathrm{\Omega };\mathbb{K})$ vanishes almost everywhere outside the compact set $K_0\subset \mathrm{\Omega }$, and $1\le p\le \mathrm{\infty }$ with the same scalar field.

Applying [L1] with $k=1$: the zero extension $E_{0}u$ lies in $W^{1,p}({\mathbb{R}}^n;\mathbb{K})$, its first weak derivatives are $D_i(E_{0}u)=E_0(D_{i}u)$ almost everywhere, and the norms agree, so $\mathrm{\parallel }{E}_{0}u{\mathrm{\parallel }}_{W^{1,p}({\mathbb{R}}^n)}=\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}$ by [L2].

Scope. The conclusion is an extension statement for the class of $u$; it uses only compact essential support and $W^{1,p}$ regularity and yields no membership in $W_0^{1,\mathrm{\infty }}(\mathrm{\Omega })$, which by [L3] would require approximating $u$ by test functions in the $W^{1,\mathrm{\infty }}$ norm.

The constant $u\equiv 1$ satisfies $u\in L^p((0,1))$ for every $1\le p\le \mathrm{\infty }$, since ${\int }_0^1\mathrm{\mid }1{\mathrm{\mid }}^p=1<\mathrm{\infty }$ and $\mathrm{\mid }1\mathrm{\mid }$ is bounded. For every $\phi \in C_c^{\mathrm{\infty }}((0,1))$ the functions $\phi$ and the constant $0$ give ${\int }_0^{1}u{\phi }^{\mathrm{\prime }}={\int }_0^1{\phi }^{\mathrm{\prime }}=\phi (1)-\phi (0)=0=-{\int }_0^{1}0\cdot \phi ,$ because $\phi$ is supported in $(0,1)$. By [F1] the zero class is the weak derivative of $u$, so $u\in W^{1,p}((0,1))$ for every $p$ by [F7].

The zero extension $H={\chi }_{(0,1)}$ is measurable with $\mathrm{\mid }H\mathrm{\mid }\le 1$, so $H\in L^p(\mathbb{R})$ for every $p$ by [F4]. For $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R})$ the distributional derivative acts by $⟨\mathrm{\partial }{T}_H,\phi ⟩=-{\int }_{\mathbb{R}}H{\phi }^{\mathrm{\prime }}=-{\int }_0^1{\phi }^{\mathrm{\prime }}=-(\phi (1)-\phi (0))=⟨{\delta }_0-{\delta }_1,\phi ⟩,$ so $\mathrm{\partial }{T}_H={\delta }_0-{\delta }_1$ in ${\mathcal{D}}^{\mathrm{\prime }}(\mathbb{R})$.

Fix the cut-offs ${\psi }_m$ of [F5]. They satisfy ${\psi }_m(0)=1$, ${\psi }_m(1)=0$ for $m\ge 1$, and $⟨{\delta }_0-{\delta }_1,{\psi }_m⟩={\psi }_m(0)-{\psi }_m(1)=1-0=1$ for every $m\ge 1$.

Suppose, for contradiction, that $\mathrm{\partial }{T}_H$ were a regular distribution $T_v$ with $v\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$, that is, that $H$ had a weak derivative in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$. Testing the identity $T_v={\delta }_0-{\delta }_1$ against each ${\psi }_m$ of step 1.3 gives ${\int }_{\mathbb{R}}v{\psi }_m=⟨{\delta }_0-{\delta }_1,{\psi }_m⟩=1$ for every $m\ge 1$.

On the other hand $v{\psi }_m\to 0$ almost everywhere and $\mathrm{\mid }v{\psi }_m\mathrm{\mid }\le \mathrm{\mid }v\mathrm{\mid }{\chi }_{(-1,1)}$, which is integrable because $v$ is locally integrable; [F6] therefore gives ${\int }_{\mathbb{R}}v{\psi }_m\to 0$. This contradicts step 2.1. Hence $\mathrm{\partial }{T}_H$ is not a regular distribution, and by [F2] the function $H$ has no weak first derivative in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1(\mathbb{R})$.

If $H$ belonged to $W^{1,p}(\mathbb{R})$ for some $1\le p\le \mathrm{\infty }$, then by [F7] its weak first derivative would be an $L^p$ class, hence in particular a weak $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ derivative, contradicting step 3.1. Therefore $H\mathrm{\notin }{W}^{1,p}(\mathbb{R})$ for every $p$, while $H{\mathrm{\mid }}_{\mathrm{\Omega }}=u$ and $u\in W^{1,p}(\mathrm{\Omega })$ by step 1.1. The map $u\mapsto {\chi }_{(0,1)}u$ therefore fails to send $W^{1,p}(\mathrm{\Omega })$ into $W^{1,p}(\mathbb{R})$ for each exponent, so it is not an extension operator in the sense of Sobolev extension domains and extension operators.

The set $\mathrm{\Omega }$ is open, because it is the intersection of the open disc $B(0,1)$ with the open set $\{y\mathrm{\ne }0\}\cup \{x<0\}$, and it is bounded; the branch $u$ is of class $C^{\mathrm{\infty }}$ on $\mathrm{\Omega }$ with $0<u<2\pi$ and $\mathrm{\nabla }u(x,y)=(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}),\mathrm{\mid }\mathrm{\nabla }u(x,y)\mathrm{\mid }=\frac{1}{r},r=\sqrt{x^2+y^2}\mathrm{.}$

Endpoint estimate. Let $g\in C^1([0,\epsilon ])$ and let $1\le p<\mathrm{\infty }$. For every $s\in (0,\epsilon )$ the fundamental theorem gives $g(0)=g(s)-{\int }_0^s{g}^{\mathrm{\prime }}(t)dt$, so $\epsilon \mathrm{\mid }g(0)\mathrm{\mid }\le {\int }_0^{\epsilon }\mathrm{\mid }g\mathrm{\mid }+\epsilon {\int }_0^{\epsilon }\mathrm{\mid }{g}^{\mathrm{\prime }}\mathrm{\mid }$, and [F5] applied to the two summands yields $\mathrm{\mid }g(0){\mathrm{\mid }}^p\le C(\epsilon ,p)({\int }_0^{\epsilon }\mathrm{\mid }g{\mathrm{\mid }}^p+{\int }_0^{\epsilon }\mathrm{\mid }{g}^{\mathrm{\prime }}{\mathrm{\mid }}^p)$ with $C(\epsilon ,p)=2^{p-1}\max \{{\epsilon }^{-1},{\epsilon }^{p-1}\}$; the same estimate holds on $(-\epsilon ,0)$ for the endpoint $0$.

Integrability. Since $\mathrm{\mid }u\mathrm{\mid }\le 2\pi$ and $\mathrm{\Omega }\subseteq B(0,1)$ has finite area, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }u{\mathrm{\mid }}^p<\mathrm{\infty }$; by [F3], [F7] and $\mathrm{\mid }\mathrm{\nabla }u\mathrm{\mid }=1\mathrm{/}r$ of step 1.1, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^p={\int }_{\mathrm{\Omega }}{r}^{-p}\le {\int }_0^{1}2\pi r^{1-p}dr=\frac{2\pi }{2-p}<\mathrm{\infty },$ because $p<2$ makes the one-dimensional integral converge at $r=0$.

Membership. The function $u$ is $C^{\mathrm{\infty }}$ on the open set $\mathrm{\Omega }$, so by [F1] its classical partial derivatives of step 1.1 are its weak derivatives; by step 2.1 they lie in $L^p(\mathrm{\Omega })$ together with $u$, and the membership criterion of [F2] gives $u\in W^{1,p}(\mathrm{\Omega })$ for every $1\le p<2$.

Upper strip. Suppose ${\phi }_j\in C^{\mathrm{\infty }}({\mathbb{R}}^2)$ satisfy $\mathrm{\parallel }{\phi }_j{\mathrm{\mid }}_{\mathrm{\Omega }}-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\to 0$. Fix $0<\epsilon <1\mathrm{/}4$ and put $S_{+}=(1\mathrm{/}4,3\mathrm{/}4)\times (0,\epsilon )\subset \mathrm{\Omega }$. The function $u=\arctan (y\mathrm{/}x)$ extends $C^1$ to the closed strip, with $u(x,0)=0$. For each $j$, apply the endpoint estimate of step 1.2 to $g_j(x,y)={\phi }_j(x,y)-u(x,y)$ on each vertical section of $S_{+}$ and integrate in $x$ using [F4]. Since $\epsilon$ is fixed, its constant is fixed, and ${\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^{p}dx\le C(\epsilon ,p){\int }_{S_{+}}(\mathrm{\mid }{\phi }_j-u{\mathrm{\mid }}^p+\mathrm{\mid }{\mathrm{\partial }}_y{\phi }_j-{\mathrm{\partial }}_{y}u{\mathrm{\mid }}^p)\le C(\epsilon ,p)\mathrm{\parallel }{\phi }_j-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}^p⟶0.$

Lower strip. On $S_{-}=(1\mathrm{/}4,3\mathrm{/}4)\times (-\epsilon ,0)$ the function $u=2\pi -\arctan (\mathrm{\mid }y\mathrm{\mid }\mathrm{/}x)$ extends $C^1$ to the closed strip with $u(x,0)=2\pi$. Applying step 1.2 to $g_j={\phi }_j-u$ on each vertical section and integrating in $x$ gives, for the same fixed $\epsilon$, ${\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^{p}dx\le C(\epsilon ,p){\int }_{S_{-}}(\mathrm{\mid }{\phi }_j-u{\mathrm{\mid }}^p+\mathrm{\mid }{\mathrm{\partial }}_y{\phi }_j-{\mathrm{\partial }}_{y}u{\mathrm{\mid }}^p)\le C(\epsilon ,p)\mathrm{\parallel }{\phi }_j-u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}^p⟶0.$

Contradiction. For every $j$ the elementary inequality $\mathrm{\mid }2\pi {\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^p+\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^p)$ holds pointwise on $(1\mathrm{/}4,3\mathrm{/}4)$, so $\frac{1}{2}(2\pi {)}^p={\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }2\pi {\mathrm{\mid }}^{p}dx\le 2^{p-1}({\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0){\mathrm{\mid }}^{p}dx+{\int }_{1\mathrm{/}4}^{3\mathrm{/}4}\mathrm{\mid }{\phi }_j(x,0)-2\pi {\mathrm{\mid }}^{p}dx)\underset{j\to \mathrm{\infty }}{\overset{}{\to }}0$ by steps 4.1 and 4.2, which is impossible since $(2\pi {)}^p\mathrm{/}2>0$.

Therefore no sequence of globally smooth functions converges to $u$ in $W^{1,p}(\mathrm{\Omega })$ for any $1\le p<2$, so ambient smooth restrictions are not dense on this bounded open set. By contrast [F6] gives that density for every bounded $C^k$ domain, so the conclusion cannot be extended to arbitrary open sets. Indeed $\mathrm{\Omega }$ is not a bounded $C^1$ domain in the graph sense: near a slit point $(x_0,0)$ with $0<x_0<1$, its complement is only a line segment and has empty interior, so $\mathrm{\Omega }$ is dense on both sides of that segment. A one-sided graph domain has a nonempty open complementary side in every sufficiently small chart neighbourhood, which rules out such a chart here.

On $\mathrm{\Omega }$ the branch $u$ is real-valued with $0<u<1$, the function $u$ is $C^{\mathrm{\infty }}$, and by [F5] $\mathrm{\mid }\mathrm{\nabla }u(x,y)\mathrm{\mid }=\frac{1}{2\pi \sqrt{x^2+y^2}}=\frac{1}{2\pi r}$ at every point of $\mathrm{\Omega }$; $\mathrm{\Omega }$ is open and bounded. It is path connected: on every circle $\mathrm{\mid }z\mathrm{\mid }=r<1$, the removed cusp occupies an arc around the positive real axis, while either complementary arc from a point of $\mathrm{\Omega }$ to $(-r,0)$ stays in $\mathrm{\Omega }$; the negative real segment then joins $(-r,0)$ to $(-1\mathrm{/}2,0)$.

Integrability. Since $\mathrm{\mid }u\mathrm{\mid }\le 1$ and $\mathrm{\Omega }\subseteq B(0,1)$ has finite area, ${\int }_{\mathrm{\Omega }}\mathrm{\mid }u{\mathrm{\mid }}^{3\mathrm{/}2}<\mathrm{\infty }$; and [F2] gives ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^{3\mathrm{/}2}dxdy\le (2\pi {)}^{-3\mathrm{/}2}{\int }_0^1{r}^{-3\mathrm{/}2}2\pi rdr=(2\pi {)}^{-1\mathrm{/}2}{\int }_0^1{r}^{-1\mathrm{/}2}dr=2(2\pi {)}^{-1\mathrm{/}2}<\mathrm{\infty }\mathrm{.}$

Consequently $u\in W^{1,3\mathrm{/}2}(\mathrm{\Omega })$: the $C^{\mathrm{\infty }}$ representative of step 1.1 is ACL with classical derivatives ${\mathrm{\partial }}_{x}u,{\mathrm{\partial }}_{y}u$, which are measurable and, by step 2.1, lie in $L^{3\mathrm{/}2}(\mathrm{\Omega })$, and $u\in L^{3\mathrm{/}2}(\mathrm{\Omega })$; the implication of [F1] applies.

Suppose $U\in W^{1,3\mathrm{/}2}({\mathbb{R}}^2)$ satisfies $U{\mathrm{\mid }}_{\mathrm{\Omega }}=u$ almost everywhere, and take its ACL representative $U^{\ast }$ from [F1]. For almost every $x\in (0,1\mathrm{/}2)$: the vertical section $U^{\ast }(x,\cdot )$ is absolutely continuous on the compact interval $[-\sqrt{1-x^2},\sqrt{1-x^2}]$; since $U^{\ast }=u$ almost everywhere on $\mathrm{\Omega }$ while $u$ is continuous on each of the two open pieces of the section, $U^{\ast }$ agrees on each piece with the continuous function $y\mapsto u(x,y)$, so the values at the two ends of the gap are $U^{\ast }(x,x^2)=\frac{\arctan x}{2\pi },U^{\ast }(x,-x^2)=1-\frac{\arctan x}{2\pi }\mathrm{.}$