Take $u\in W_0^{1,p}(U)$ and extend it by zero outside $U$. The <zero extension of W01> belongs to $W^{1,p}(\mathbb R^n)$ and the <Sobolev inequality> gives
$$
\|u\|_{L^{p^*}(U)}\leq C_S\|Du\|_{L^p(U)}.
$$
Because $U$ has finite measure, the <Holder inequality> gives
$$
\|u\|_{L^p(U)}
\leq |U|^{1/p-1/p^*}\|u\|_{L^{p^*}(U)}
\leq C\|Du\|_{L^p(U)}.
$$
Thus
$$
\|u\|_{W^{1,p}(U)}
\leq C'\|Du\|_{L^p(U)}.
$$
The reverse estimate follows directly from $\|Du\|_{L^p}\leq\|u\|_{W^{1,p}}$. After adjusting constants,
$$
\boxed{C\|Du\|_{L^p(U)}\leq\|u\|_{W^{1,p}(U)}\leq C'\|Du\|_{L^p(U)}}.
$$
This is the <Poincare inequality> on $W_0^{1,p}(U)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 105 / 1 / c / ii / Solution

{title2=$W_0^{1,p}(U)\hookrightarrow W^{1,p}(\mathbb R^n)$}

By the definition of $W_0^{1,p}(U)$ as the closure of $C_c^\infty(U)$, extending a function by zero outside $U$ gives a bounded map into $W^{1,p}(\mathbb R^n)$ and does not create a boundary distribution.


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 Incoming links: Zero extension of W01