The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction

The smooth case: pointwise bound. Let $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ and extend it by zero to ${\mathbb{R}}^n$. Write $x=x_{\perp }+x_{n}e$ with $x_{\perp }\perp e$. For each fixed $x_{\perp }$, the profile $g(s):=\phi (x_{\perp }+se)$ is smooth and supported in $(a,b)$, since $\mathrm{\Omega }\subset \{a<y\cdot e<b\}$. If $x=x_{\perp }+x_{n}e\in \mathrm{\Omega }$, then $a<x_n<b$ and $g(a)=0$. Applying [F2] componentwise (in $\mathbb{R}$ or ${\mathbb{R}}^2$ according to the scalar field) gives $\phi (x)=g(x_n)-g(a)={\int }_a^{x_n}D\phi (x_{\perp }+se)\cdot eds$. Hence $\mathrm{\mid }\phi (x)\mathrm{\mid }\le {\int }_a^{x_n}\mathrm{\mid }D\phi (x_{\perp }+se)\mathrm{\mid }ds$. Holder [F3], followed by $x_n-a\le b-a$ and enlargement of the integration interval, yields $\mathrm{\mid }\phi (x){\mathrm{\mid }}^p\le (b-a{)}^{p-1}{\int }_a^b\mathrm{\mid }D\phi (x_{\perp }+se){\mathrm{\mid }}^{p}ds$.

The smooth case: integration. If $n=1$, then $\mathrm{\mid }\mathrm{\Omega }\mathrm{\mid }\le b-a$, so integrating step 1.1 gives ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\phi {\mathrm{\mid }}^p\le (b-a{)}^p{\int }_a^b\mathrm{\mid }D\phi (se){\mathrm{\mid }}^{p}ds=(b-a{)}^p{\int }_{\mathrm{\Omega }}\mathrm{\mid }D\phi {\mathrm{\mid }}^p$ by the change of variable $y=se$ and the support of $D\phi$ in $\mathrm{\Omega }$. Now suppose $n\ge 2$. Choose orthonormal coordinates with last vector $e$ and write $x=x_{\perp }+x_{n}e$; by [F5] integration on ${\mathbb{R}}^n$ is integration over $(x_{\perp },x_n)\in {\mathbb{R}}^{n-1}\times \mathbb{R}$. For each $x_{\perp }$, the slice $S_{x_{\perp }}:=\{x_n:(x_{\perp },x_n)\in \mathrm{\Omega }\}$ is a measurable subset of $(a,b)$, so $\mathrm{\mid }{S}_{x_{\perp }}\mathrm{\mid }\le b-a$. Integrating step 1.1 over $\mathrm{\Omega }$ and applying Tonelli [F4] gives ${\int }_{\mathrm{\Omega }}\mathrm{\mid }\phi {\mathrm{\mid }}^{p}dx\le (b-a{)}^{p-1}{\int }_{{\mathbb{R}}^{n-1}}{\int }_{S_{x_{\perp }}}{\int }_a^b\mathrm{\mid }D\phi (x_{\perp }+se){\mathrm{\mid }}^{p}dsd{x}_{n}d{x}_{\perp }\le (b-a{)}^p{\int }_{{\mathbb{R}}^{n-1}}{\int }_a^b\mathrm{\mid }D\phi (x_{\perp }+se){\mathrm{\mid }}^{p}dsd{x}_{\perp }$. The last integral equals ${\int }_{{\mathbb{R}}^n}\mathrm{\mid }D\phi {\mathrm{\mid }}^{p}dx={\int }_{\mathrm{\Omega }}\mathrm{\mid }D\phi {\mathrm{\mid }}^p$, since $\phi$ and its gradient vanish outside $\mathrm{\Omega }$ and $\mathrm{\Omega }$ lies in the slab. Therefore $\mathrm{\parallel }\phi {\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}^p\le (b-a{)}^p\mathrm{\parallel }D\phi {\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}^p$.

The general class. By [F1] there are ${\phi }_k\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };\mathbb{K})$ with ${\phi }_k\to u$ in $W^{1,p}(\mathrm{\Omega };\mathbb{K})$; by step 2.1, $\mathrm{\parallel }{\phi }_k{\mathrm{\parallel }}_{L^p}\le (b-a)\mathrm{\parallel }D{\phi }_k{\mathrm{\parallel }}_{L^p}$ for every $k$. Both sides are continuous in the $W^{1,p}$ norm: $\mathrm{\parallel }{\phi }_k{\mathrm{\parallel }}_p\to \mathrm{\parallel }u{\mathrm{\parallel }}_p$ and $\mathrm{\parallel }D{\phi }_k{\mathrm{\parallel }}_p\to \mathrm{\parallel }Du{\mathrm{\parallel }}_p$ by [F6]. Passing to the limit gives $\mathrm{\parallel }u{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}\le (b-a)\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$, so the asserted inequality holds with $C(p)=1$.