Sobolev Traces and Zero Boundary Values — Examples

The operator $T$ is well defined, linear and bounded. By [F1] the representative $u^{\ast }$ and hence the pair $(u^{\ast }(a),u^{\ast }(b))$ depend only on the class, so $T$ is well defined on classes; if $u,v$ are classes with representatives $u^{\ast },v^{\ast }$, then $u^{\ast }+v^{\ast }$ and $\lambda u^{\ast }$ are the absolutely continuous representatives of $u+v$ and $\lambda u$ by [F1]'s uniqueness and the linearity of the fundamental-theorem identity, so $T$ is linear. With $\epsilon :=(b-a)\mathrm{/}2$, [F2] gives for $1<p<\mathrm{\infty }$ the bound $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}(\frac{2}{b-a}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p}^p+(\frac{b-a}{2}{)}^{p-1}\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^p}^p)$ and the same bound at $b$, hence $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{{\mathbb{K}}^2}\le C(b-a,p)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(I)}$; for $p=1$, $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le \frac{2}{b-a}\mathrm{\parallel }u{\mathrm{\parallel }}_{L^1}+\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^1}$ and the same at $b$.

Endpoint vanishing implies membership in $W_0^{1,p}(I)$. Assume $u^{\ast }(a)=u^{\ast }(b)=0$. For $0<\delta <(b-a)\mathrm{/}4$ choose ${\chi }_{\delta }\in C_c^{\mathrm{\infty }}(I)$ with $0\le {\chi }_{\delta }\le 1$, equal to one on $[a+\delta ,b-\delta ]$, vanishing on $(a,a+\delta \mathrm{/}2)\cup (b-\delta \mathrm{/}2,b)$, and $\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le C\mathrm{/}\delta$. By [F5], $({\chi }_{\delta }u{)}^{\mathrm{\prime }}={\chi }_{\delta }{u}^{\mathrm{\prime }}+{\chi }_{\delta }^{\mathrm{\prime }}u$. The terms containing $1-{\chi }_{\delta }$ tend to zero in $L^p$ by dominated convergence. Since $u^{\ast }(t)={\int }_a^t{u}^{\mathrm{\prime }}$, [F8] gives $\mathrm{\mid }{u}^{\ast }(t){\mathrm{\mid }}^p\le (t-a{)}^{p-1}{\int }_a^t\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$ (also at $p=1$), hence ${\int }_a^{a+\delta }\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}u{\mathrm{\mid }}^p\le C^p{p}^{-1}{\int }_a^{a+\delta }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p\to 0$; the right endpoint is identical. Thus ${\chi }_{\delta }u\to u$ in $W^{1,p}$. For each fixed $\delta$, [F6] extends ${\chi }_{\delta }u$ by zero to $V_{\delta }\in W^{1,p}(\mathbb{R})$. Choose ${\zeta }_{\delta }\in C_c^{\mathrm{\infty }}(I)$ equal to one near its compact support, and use [F7] to choose ${\psi }_{\delta }\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with $\mathrm{\parallel }{\psi }_{\delta }-V_{\delta }{\mathrm{\parallel }}_{W^{1,p}(\mathbb{R})}<\delta \mathrm{/}(1+C_{{\zeta }_{\delta }})$. Then [F5] gives $\mathrm{\parallel }{\zeta }_{\delta }{\psi }_{\delta }-{\chi }_{\delta }u{\mathrm{\parallel }}_{W^{1,p}(I)}<\delta$, and ${\zeta }_{\delta }{\psi }_{\delta }\in C_c^{\mathrm{\infty }}(I)$. Combining these approximations proves membership in the closure [F4].

Membership in $W_0^{1,p}(I)$ implies $Tu=0$. Every $\phi \in C_c^{\mathrm{\infty }}(I)$ vanishes on a neighbourhood of $a$ and of $b$, so its absolutely continuous representative vanishes at both endpoints and $T\phi =(0,0)$. By step 1.1 the map $T$ is bounded, hence continuous; if $u\in W_0^{1,p}(I)$ and ${\phi }_m\in C_c^{\mathrm{\infty }}(I)$ with ${\phi }_m\to u$ by [F4], then $Tu={\lim }_{m}T{\phi }_m=0$.

Conclusion and the two worked functions. Steps 2.1 and 1.2 prove the equivalence $u\in W_0^{1,p}(I)\iff Tu=(0,0)$, and step 1.1 gives well-definedness, linearity and boundedness of $T$; with [F1] this is the assertion that the trace of $u$ is the pair of endpoint values of $u^{\ast }$ and depends only on the class. On $I=(0,1)$: for $u(x)=x(1-x)$ the absolutely continuous representative is $u^{\ast }=u$ with $u^{\ast }(0)=u^{\ast }(1)=0$, so $Tu=(0,0)$ and $u\in W_0^{1,p}(0,1)$ by step 1.2; for $u\equiv 1$ the representative is $u^{\ast }\equiv 1$ with $Tu=(1,1)\mathrm{\ne }(0,0)$, so $u\mathrm{\notin }{W}_0^{1,p}(0,1)$ by step 2.1.

The trace is the classical restriction, with a controlled norm. The affine function is smooth on $\bar{\mathrm{\Omega }}$; its gradient is the constant $c$, so ${\int }_{\mathrm{\Omega }}\mathrm{\mid }u{\mathrm{\mid }}^{p}dx\le 2^{p-1}(\mathrm{\mid }c{\mathrm{\mid }}^p{\int }_{\mathrm{\Omega }}\mathrm{\mid }x{\mathrm{\mid }}^{p}dx+\mathrm{\mid }d{\mathrm{\mid }}^p\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid })\le C(R,n,p)(\mathrm{\mid }c\mathrm{\mid }R+\mathrm{\mid }d\mathrm{\mid }{)}^p$ and ${\int }_{\mathrm{\Omega }}\mathrm{\mid }{\mathrm{\partial }}_{i}u{\mathrm{\mid }}^{p}dx=\mathrm{\mid }{c}_i{\mathrm{\mid }}^p\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\le C(R,n,p)(\mathrm{\mid }c\mathrm{\mid }R+\mathrm{\mid }d\mathrm{\mid }{)}^p$ by [F4]. Hence $u\in W^{1,p}(\mathrm{\Omega })$ with $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}(\mathrm{\Omega })}\le C^{\mathrm{\prime }}(R,n,p)(\mathrm{\mid }c\mathrm{\mid }R+\mathrm{\mid }d\mathrm{\mid })$, and $u\in C(\bar{\mathrm{\Omega }})$, so $Tu=(c\cdot x+d){\mathrm{\mid }}_{\mathrm{\partial }\mathrm{\Omega }}$ by [F1].

Fractional membership and the boundary norm. For $1<p<\mathrm{\infty }$ put $\theta :=1-1\mathrm{/}p$. By [F2] and step 1.1, $Tu\in W^{\theta ,p}(\mathrm{\partial }\mathrm{\Omega })$ with $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{W^{\theta ,p}(\mathrm{\partial }\mathrm{\Omega })}\le C(R,n,p)(\mathrm{\mid }c\mathrm{\mid }R+\mathrm{\mid }d\mathrm{\mid })$; at $p=1$ the same computation gives only $Tu\in L^1(\mathrm{\partial }\mathrm{\Omega })$, and no space $W^{0,1}$ is introduced. For the $L^p$ value, parametrise the sphere by $x=R\omega$; by [F3] and the definition of the surface integral, $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{L^p(\mathrm{\partial }\mathrm{\Omega })}^p={\int }_{\mathrm{\partial }\mathrm{\Omega }}\mathrm{\mid }c\cdot x+d{\mathrm{\mid }}^{p}dS=R^{n-1}{\int }_{S^{n-1}}\mathrm{\mid }c\cdot R\omega +d{\mathrm{\mid }}^{p}d\sigma (\omega )$, which is the classical sphere integral.

Conclusion. Steps 1.1 and 2.1 prove that the affine class lies in $W^{1,p}(\mathrm{\Omega })$ with trace its classical restriction, that the trace lies in the fractional boundary space for $1<p<\mathrm{\infty }$ with the stated bound, and that its boundary $L^p$ norm is the displayed surface integral.

Two different boundary restrictions, one class. The set $\mathrm{\partial }\mathrm{\Omega }$ has measure zero by [F2], so $f$ and $\tilde{f}$ agree off a null set; their restrictions to $\mathrm{\Omega }$ are both identically zero, so [F1] identifies their common $L^p(\mathrm{\Omega })$ class with the zero element of $W^{k,p}(\mathrm{\Omega })$ for every $k$ and $p$. Their classical restrictions to $\mathrm{\partial }\mathrm{\Omega }$ are $0$ and $1$ respectively, which differ at every point of $\mathrm{\partial }\mathrm{\Omega }$. Since $f$ is continuous on $\bar{\mathrm{\Omega }}$, $Tf=0$ by [F3]; and $T\tilde{f}=Tf$ because $\tilde{f}$ is a representative of the class of $f$ and $T$ is defined on classes. Hence the boundary values of a representative carry information invisible to $T$.

An $L^2$ class with no finite boundary values. On the unit ball $B(0,1)\subset {\mathbb{R}}^n$ put $g(x):=(1-\mathrm{\mid }x\mathrm{\mid }{)}^{-1\mathrm{/}4}$. Then ${\int }_B{g}^{2}dx={\int }_B(1-\mathrm{\mid }x\mathrm{\mid }{)}^{-1\mathrm{/}2}dx={\int }_0^1(1-r{)}^{-1\mathrm{/}2}{r}^{n-1}\sigma (S^{n-1})dr\le \sigma (S^{n-1}){\int }_0^1(1-r{)}^{-1\mathrm{/}2}dr=2\sigma (S^{n-1})<\mathrm{\infty }$ by [F4], so $g\in L^2(B)$. On the other hand $g(x)\to \mathrm{\infty }$ as $\mathrm{\mid }x\mathrm{\mid }\to 1$, so $g$ is unbounded on every neighbourhood of $\mathrm{\partial }B$: no finite boundary values can be assigned from pointwise evaluation.

Conclusion. Step 1.1 shows that two functions with the same interior class can have different classical boundary restrictions, while both have zero trace; step 1.2 shows that a single $L^p$ class need not be bounded near the boundary, so pointwise boundary evaluation is not a well-defined operation on $L^p(\mathrm{\Omega })$ classes. For $W^{1,p}$ classes the Sobolev trace supplies boundary data independent of representatives; this does not extend pointwise evaluation to all $L^p$ classes.

The full seminorm of the line jump. Let $g={\mathbf{1}}_{(0,1)}$ on $\mathbb{R}$ and put $q=p\theta >0$. Symmetry and Tonelli give $[g{]}_{\theta ,p}^p=2{\int }_0^1({\int }_{-\mathrm{\infty }}^0(x-y{)}^{-1-q}dy+{\int }_1^{\mathrm{\infty }}(y-x{)}^{-1-q}dy)dx=(2\mathrm{/}q){\int }_0^1(x^{-q}+(1-x{)}^{-q})dx=(4\mathrm{/}q){\int }_0^1{x}^{-q}dx$. Thus it is finite exactly when $q<1$, and infinite when $q\ge 1$; at $q=1$ the logarithmic divergence is explicit. Since $q=p-1$, the threshold is exactly $p=2$.

Transfer to the boundary and conclusion. Because $\mathrm{\partial }\mathrm{\Omega }$ has finite surface measure and the straight chart has bounded density, $g\in L^p(\mathrm{\partial }\mathrm{\Omega })$ is a bounded function on a finite-measure boundary, hence an $L^p$ class by [F4]; choose a subordinate cutoff equal to one near the closed segment. In that atlas its representation is exactly the line jump from step 1.1, while all other localised representations are bounded Lipschitz multiples and coordinate transforms of it. The multiplier and atlas estimates of Chart independence of the fractional boundary norm therefore make the norm finite when $p<2$ and infinite when $p\ge 2$, independently of the atlas. By [F3] the trace range equals the boundary space, so for $p\ge 2$ there is no $u\in W^{1,p}(\mathrm{\Omega })$ with $Tu=g$. For $1<p<2$, step 1.1 gives $p\theta <1$ and $[g{]}_{\theta ,p}<+\mathrm{\infty }$, so $g$ is in the trace range in that exponent range: the range genuinely depends on $p$.

Endpoint vanishing implies zero extension in $W^{1,p}(\mathbb{R})$. Assume $u^{\ast }(0)=u^{\ast }(1)=0$ and let $U$ be the extension of $u^{\ast }$ by zero outside $(0,1)$. Then $U$ is continuous on $\mathbb{R}$ and absolutely continuous on every compact interval: on a compact interval meeting $(0,1)$ any finite family of disjoint subintervals has $U$-increments equal to the corresponding $u^{\ast }$-increments after intersecting with $(0,1)$, with the increments over pieces crossing an endpoint bounded by $\mathrm{\mid }{u}^{\ast }\mathrm{\mid }$ evaluated there, so the absolute-continuity modulus of $u^{\ast }$ controls $U$. Hence $U$ is differentiable a.e. with $U^{\mathrm{\prime }}=u^{\mathrm{\prime }}$ a.e. on $(0,1)$ and $U^{\mathrm{\prime }}=0$ a.e. outside, so $U^{\mathrm{\prime }}\in L^p(\mathbb{R})$ with $\mathrm{\parallel }{U}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^p(\mathbb{R})}=\mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^p(I)}$; also $U\in L^p(\mathbb{R})$ with the same norm as $u$. By the ACL characterization [F3] applied on $\mathbb{R}$, $U\in W^{1,p}(\mathbb{R};\mathbb{K})$.

Zero extension in $W^{1,p}(\mathbb{R})$ implies endpoint vanishing. Conversely, let $U\in W^{1,p}(\mathbb{R})$ be the zero extension of $u$ and let $\tilde{U}$ be its unique continuous locally absolutely continuous representative [F2]. Since $U=0$ a.e. on $(-1,0)$ and on $(1,2)$, continuity of $\tilde{U}$ forces $\tilde{U}=0$ on $(-1,0]$ and on $[1,2)$: a continuous function vanishing a.e. on an interval vanishes identically there. On $(0,1)$ the classes of $U$ and of $u$ coincide, so $\tilde{U}=u^{\ast }$ a.e. on $(0,1)$, and as both are continuous they agree identically there; taking the limits at the endpoints gives $u^{\ast }(0)=\tilde{U}(0)=0$ and $u^{\ast }(1)=\tilde{U}(1)=0$.

The equivalence chain. By [F1] the conditions (i), (ii) and (iii) are mutually equivalent: $Tu=0$ means exactly $u^{\ast }(0)=u^{\ast }(1)=0$, and this is exactly the criterion for membership in the closure space $W_0^{1,p}(I)$ of [F4]. Steps 1.1 and 1.2 add (ii)$\iff$(iv), so all four conditions are equivalent.

The two worked functions. For $u(x)=x(1-x)$ one has $u^{\ast }=u$ (a polynomial is its own absolutely continuous representative), $u^{\ast }(0)=u^{\ast }(1)=0$ and hence all four conditions hold by step 2.1; explicitly the zero extension is continuous, is absolutely continuous on $\mathbb{R}$ with classical derivative $1-2x$ on $(0,1)$ and $0$ outside, so its weak derivative is the zero extension of $u^{\mathrm{\prime }}$, in agreement with step 1.1. For $u\equiv 1$ one has $u^{\ast }\equiv 1$, so $Tu=(1,1)\mathrm{\ne }(0,0)$ and $u^{\ast }(0)=u^{\ast }(1)=1\mathrm{\ne }0$; by [F1] the conditions (i), (ii), (iii) fail, and (iv) fails as well: if the zero extension belonged to $W^{1,p}(\mathbb{R})$, step 1.2 applied to it would force $u^{\ast }(0)=u^{\ast }(1)=0$, contradicting $u^{\ast }\equiv 1$.

The inside norm. The function $u_{\delta }$ is the restriction to ${\mathrm{\Omega }}_{\alpha }$ of the $C^{\mathrm{\infty }}({\mathbb{R}}^2)$ function $(x,y)\mapsto \theta (y\mathrm{/}\delta )$, whose classical derivatives are its weak derivatives on ${\mathrm{\Omega }}_{\alpha }$ by [F1], with ${\mathrm{\partial }}_x{u}_{\delta }=0$ and ${\mathrm{\partial }}_y{u}_{\delta }={\delta }^{-1}{\theta }^{\mathrm{\prime }}(y\mathrm{/}\delta )$ by [F1]. Its support in ${\mathrm{\Omega }}_{\alpha }$ lies in the strip $0<y<2\delta$, whose area is ${\int }_0^{2\delta }2y^{\alpha }dy=2^{\alpha +2}{\delta }^{\alpha +1}\mathrm{/}(\alpha +1)$ by [F2]. Since $\mathrm{\mid }{u}_{\delta }\mathrm{\mid }\le 1$ and $\mathrm{\mid }{\mathrm{\partial }}_y{u}_{\delta }\mathrm{\mid }\le {\delta }^{-1}\mathrm{\parallel }{\theta }^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$, this gives $\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{L^p}^p\le C{\delta }^{\alpha +1}$ and $\mathrm{\parallel }{\mathrm{\partial }}_y{u}_{\delta }{\mathrm{\parallel }}_{L^p}^p\le C{\delta }^{\alpha +1-p}$, hence $u_{\delta }\in W^{1,p}({\mathrm{\Omega }}_{\alpha })$ and $\mathrm{\parallel }{u}_{\delta }{\mathrm{\parallel }}_{W^{1,p}({\mathrm{\Omega }}_{\alpha })}^p\le C^{\mathrm{\prime }}{\delta }^{\alpha +1-p}$ for $0<\delta <\frac{1}{2}$ and a constant independent of $\delta$.