The Sobolev inequality for zero-boundary Sobolev closures on open sets

Zero extension. Let $E_{0}u$ be the extension of $u$ by zero. By [F2], $E_{0}u\in W^{1,p}({\mathbb{R}}^n;\mathbb{K})$, its weak gradient is the zero extension of $Du$, and $\mathrm{\parallel }{E}_{0}u{\mathrm{\parallel }}_{L^{p^{\ast }}({\mathbb{R}}^n)}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^{p^{\ast }}(\mathrm{\Omega })}$, $\mathrm{\parallel }D(E_{0}u){\mathrm{\parallel }}_{L^p({\mathbb{R}}^n)}=\mathrm{\parallel }Du{\mathrm{\parallel }}_{L^p(\mathrm{\Omega })}$ componentwise.

For $1<p<n$, apply [F3] to $E_{0}u$ and use step 1.1. For $p=1$, choose ${\phi }_j\in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ converging to $u$ in $W^{1,1}(\mathrm{\Omega })$ by [F1]; their zero extensions converge to $E_{0}u$ in $W^{1,1}({\mathbb{R}}^n)$ by [F2]. The endpoint estimate [F3] applied to differences shows that these extensions are Cauchy in $L^{n\mathrm{/}(n-1)}$. By [F4] their limit in that space exists; an almost-everywhere subsequence, followed by an $L^1$ almost-everywhere subsequence, identifies it with $E_{0}u$. Passing to the limit in the endpoint estimate gives $\mathrm{\parallel }{E}_{0}u{\mathrm{\parallel }}_{n\mathrm{/}(n-1)}\le C(n)\mathrm{\parallel }D(E_{0}u){\mathrm{\parallel }}_1$. Step 1.1 transfers both cases to $\mathrm{\Omega }$, proving the assertion.