Endpoint trace commutes with Sobolev truncation on an interval

(Well-definedness of $T$) By [F1] the class $u$ has exactly one continuous locally absolutely continuous representative, it extends uniquely to an absolutely continuous function $u^{\ast }$ on $[a,b]$, and its endpoint values are determined by the class; hence $T$ is well defined, and [F1] also gives $u^{\ast }(x)=u^{\ast }(a)+{\int }_a^x{u}^{\mathrm{\prime }}$ for every $x\in [a,b]$.

(Norm estimate) With $\epsilon =(b-a)\mathrm{/}2$, [F2] bounds $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }$ and $\mathrm{\mid }{u}^{\ast }(b)\mathrm{\mid }$ by $C_p(a,b)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}}$, hence $\mathrm{\parallel }Tu{\mathrm{\parallel }}_{{\mathbb{R}}^2}\le \sqrt{2}{C}_p(a,b)\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{1,p}}$ for every $u$.

(Linearity) For $u,v\in W^{1,p}(I)$ the class $u+v$ has weak derivative $u^{\mathrm{\prime }}+v^{\mathrm{\prime }}$ by [F4], and $u^{\ast }+v^{\ast }$ is continuous and absolutely continuous; by the uniqueness clause of [F1] it is the representative of $u+v$, so $(u+v{)}^{\ast }=u^{\ast }+v^{\ast }$ and $T(u+v)=Tu+Tv$; for $c\in \mathbb{R}$ the class $cu$ has weak derivative $c{u}^{\mathrm{\prime }}$ and representative $c{u}^{\ast }$ by [F1] and [F4], so $T(cu)=cTu$; hence $T$ is linear.

(Truncation membership) The constant function $k$ lies in $W^{1,p}(I)$ with weak derivative $0$ by [F5], so $w:=u-k$ lies in $W^{1,p}(I)$ with weak derivative $u^{\mathrm{\prime }}$ by [F4]; by [F6] the classes $w^{+}=\max \{w,0\}$ and $w^{-}=\max \{-w,0\}$ lie in $W^{1,p}(I)$ with $D{w}^{+}=1_{\{w>0\}}{u}^{\mathrm{\prime }}$ and $D{w}^{-}=-1_{\{w<0\}}{u}^{\mathrm{\prime }}$ a.e.; in particular $(u-k{)}^{+}\in W^{1,p}(I)$.

(Endpoint values of the truncated class) Apply [F1] to the class $w^{+}$ of step 1.4: it has a unique continuous representative $w^{\ast }$ on $[a,b]$, absolutely continuous, with $w^{\ast }(x)=w^{\ast }(a)+{\int }_a^{x}D{w}^{+}$. The continuous function $x\mapsto (u^{\ast }(x)-k{)}^{+}$ on $[a,b]$ is a representative of $w^{+}$ because $u^{\ast }-k$ represents $u-k$ and the positive part is well defined on a.e. classes [F6]; two continuous representatives of one class agree on the dense interval $(a,b)$, hence by continuity on all of $[a,b]$. Therefore $T(w^{+})=((u^{\ast }(a)-k{)}^{+},(u^{\ast }(b)-k{)}^{+})=((Tu-k{)}^{+})$ componentwise.

($W_0^{1,p}\subseteq \ker T$) Each $\phi \in C_c^{\mathrm{\infty }}(I)$, extended by zero, is its own absolutely continuous representative with $\phi (a)=\phi (b)=0$, so $T\phi =0$ by [F1]; by step 1.2 the map $T$ is bounded, hence continuous, and by [F3] the space $W_0^{1,p}(I)$ is the closure of $C_c^{\mathrm{\infty }}(I)$, so $T$ vanishes on all of $W_0^{1,p}(I)$.

(Endpoint decay when $Tu=0$) Assume $Tu=0$, so $u^{\ast }(a)=u^{\ast }(b)=0$. For $x\in (a,b)$, [F1] gives $u^{\ast }(x)=u^{\ast }(a)+{\int }_a^x{u}^{\mathrm{\prime }}={\int }_a^x{u}^{\mathrm{\prime }}$ and, symmetrically, $u^{\ast }(x)=u^{\ast }(b)-{\int }_x^b{u}^{\mathrm{\prime }}=-{\int }_x^b{u}^{\mathrm{\prime }}$; hence $\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le {\int }_a^x\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$ and $\mathrm{\mid }{u}^{\ast }(x)\mathrm{\mid }\le {\int }_x^b\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$. For $p>1$ [F8] turns this into $\mathrm{\mid }{u}^{\ast }(x){\mathrm{\mid }}^p\le (x-a{)}^{p-1}{\int }_a^x\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$ and $\mathrm{\mid }{u}^{\ast }(x){\mathrm{\mid }}^p\le (b-x{)}^{p-1}{\int }_x^b\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$, with the same inequalities for $p=1$ since $(x-a{)}^0=(b-x{)}^0=1$.

(Cutoffs and convergence to $u$) Assume $Tu=0$ and fix $0<\delta <(b-a)\mathrm{/}4$. Let $b_0$ be the bump of [F7] and put ${\eta }_{\delta }(x):=1-b_0((x-a)\mathrm{/}\delta )$, ${\rho }_{\delta }(x):=1-b_0((b-x)\mathrm{/}\delta )$, ${\chi }_{\delta }:={\eta }_{\delta }{\rho }_{\delta }$. Then $0\le {\chi }_{\delta }\le 1$, ${\chi }_{\delta }=1$ on $[a+2\delta ,b-2\delta ]$, ${\chi }_{\delta }=0$ on $[a,a+\delta ]\cup [b-\delta ,b]$, so $\mathrm{supp}{\chi }_{\delta }\subseteq [a+\delta ,b-\delta ]\subset I$, and the chain and product rules give $\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}\mathrm{\mid }\le 2M\mathrm{/}\delta$ with $M:=\mathrm{\parallel }{b}_0^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}<\mathrm{\infty }$ [F7]. By [F10] ${\chi }_{\delta }u\in W^{1,p}(I)$ with $D({\chi }_{\delta }u)={\chi }_{\delta }^{\mathrm{\prime }}u+{\chi }_{\delta }{u}^{\mathrm{\prime }}$; moreover ${\chi }_{\delta }\to 1$ pointwise on $I$, so dominated convergence [F9] gives $\mathrm{\parallel }(1-{\chi }_{\delta })u{\mathrm{\parallel }}_p\to 0$ and $\mathrm{\parallel }(1-{\chi }_{\delta })u^{\mathrm{\prime }}{\mathrm{\parallel }}_p\to 0$. For the remaining term, [F8] and step 2.3 give pointwise a.e. $\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p\le (x-a{)}^{p-1}{\int }_a^x\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$ on $(a,a+2\delta )$, so Tonelli [F9] yields ${\int }_a^{a+2\delta }\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p\le \frac{(2\delta {)}^p}{p}{\int }_a^{a+2\delta }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$, and reflecting at $b$ the same bound holds on $(b-2\delta ,b)$; consequently ${\int }_I\mathrm{\mid }{\chi }_{\delta }^{\mathrm{\prime }}u{\mathrm{\mid }}^p\le (2M\mathrm{/}\delta {)}^p{\int }_{(a,a+2\delta )\cup (b-2\delta ,b)}\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p\le \frac{(4M{)}^p}{p}{\int }_{(a,a+2\delta )\cup (b-2\delta ,b)}\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p\to 0$ because the endpoint regions shrink to null sets and $\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$ is integrable [F9]. Hence $D({\chi }_{\delta }u)\to u^{\mathrm{\prime }}$ in $L^p$ and ${\chi }_{\delta }u\to u$ in $L^p$, that is, ${\chi }_{\delta }u\to u$ in $W^{1,p}(I)$.

(Each ${\chi }_{\delta }u$ lies in $W_0^{1,p}$) Fix $\delta$ as in step 3.1. Since ${\chi }_{\delta }u$ vanishes a.e. outside the compact set $[a+\delta ,b-\delta ]\subset I$, its zero extension $E_0({\chi }_{\delta }u)$ lies in $W^{1,p}(\mathbb{R})$ with equal norm by [F11], and by the density corollary there are ${\phi }_j\in C_c^{\mathrm{\infty }}(\mathbb{R})$ with ${\phi }_j\to E_0({\chi }_{\delta }u)$ in $W^{1,p}(\mathbb{R})$. By [F13] choose $\psi \in C_c^{\mathrm{\infty }}(I)$ with $\psi =1$ on a neighbourhood of $[a+\delta ,b-\delta ]$; then $\psi {\phi }_j\in C_c^{\mathrm{\infty }}(I)$ and $\psi {\phi }_j-{\chi }_{\delta }u$ is the restriction to $I$ of $\psi ({\phi }_j-E_0({\chi }_{\delta }u))$, so the multiplication and restriction bounds of [F12] give $\mathrm{\parallel }\psi {\phi }_j-{\chi }_{\delta }u{\mathrm{\parallel }}_{W^{1,p}(I)}\le C_{\psi }\mathrm{\parallel }{\phi }_j-E_0({\chi }_{\delta }u){\mathrm{\parallel }}_{W^{1,p}(\mathbb{R})}\to 0$. Thus ${\chi }_{\delta }u$ lies in the closure of $C_c^{\mathrm{\infty }}(I)$, which is $W_0^{1,p}(I)$ by [F3].

((i) concluded: $\ker T=W_0^{1,p}$) Steps 1.1-1.3 show that $T$ is well defined, linear and bounded (with the estimate of step 1.2); step 2.2 gives $W_0^{1,p}(I)\subseteq \ker T$. Conversely, if $Tu=0$, then step 3.1 exhibits ${\chi }_{\delta }u\to u$ in $W^{1,p}(I)$ with ${\chi }_{\delta }u\in W_0^{1,p}(I)$ for each $\delta$ by step 4.1; since $W_0^{1,p}(I)$ is closed in $W^{1,p}(I)$ [F3], $u\in W_0^{1,p}(I)$. Hence $\ker T=W_0^{1,p}(I)$.

(The truncation iff) Fix $k\in \mathbb{R}$. By step 2.1, $T((u-k{)}^{+})=((Tu-k{)}^{+})$; by step 5.1, $(u-k{)}^{+}\in W_0^{1,p}(I)$ exactly when this trace pair vanishes, that is, exactly when $(u^{\ast }(a)-k{)}^{+}=0$ and $(u^{\ast }(b)-k{)}^{+}=0$, equivalently $u^{\ast }(a)\le k$ and $u^{\ast }(b)\le k$, equivalently $Tu\le (k,k)$ componentwise; membership $(u-k{)}^{+}\in W^{1,p}(I)$ is step 1.4.

(The particular identities and discharge) Taking $k=0$ in step 6.1 gives $T(u^{+})=(Tu{)}^{+}$ and $u^{+}\in W_0^{1,p}(I)$ exactly when $Tu\le (0,0)$. Applying step 6.1 to the class $-u$, whose representative is $-u^{\ast }$ and whose trace pair is $-Tu$, gives $T((-u{)}^{+})=(-Tu{)}^{+}=(Tu{)}^{-}$, that is $T(u^{-})=(Tu{)}^{-}$; since $\mathrm{\mid }u\mathrm{\mid }=u^{+}+u^{-}$, linearity of $T$ from step 1.3 gives $T\mathrm{\mid }u\mathrm{\mid }=T(u^{+})+T(u^{-})=(Tu{)}^{+}+(Tu{)}^{-}=\mathrm{\mid }Tu\mathrm{\mid }$ componentwise. Steps 5.1, 6.1 and the present step prove all the assertions; the Axiom of Choice was used only through the representative interfaces [F1], the endpoint estimate [F2] and the truncation, extension, density and multiplication interfaces [F6, F11, F12], which assume it or Countable Choice [A1].