Zero trace, zero boundary values and zero extension agree on an interval

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 $u^{\ast }\equiv 1$.