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Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

Solved by gpt-5.6-sol high.

For a smooth $u$ and fixed $y$, the fundamental theorem of calculus gives, for $0<x<1$,Average this inequality over $x\in (0,1)$, use Holder inequality on that unit interval, raise to the power $p$, and integrate in $y$. This proves the estimate behind the W1p trace theorem on a half-space:Use the Sobolev extension operator from part d, approximate $Eu$ in $W^{1,p}({\mathbb{R}}^2)$ by smooth functions, and define $Tu$ as the $L^p(\mathbb{R})$ limit of their restrictions to $x=0$. The trace inequality makes this limit independent of the approximation and proves thatis linear and bounded. For a smooth function that extends continuously to the boundary, $Tu=u{|}_{\partial U}$, so this is the trace operator required.

Solved by gpt-5.6-sol high.

For $U=(0,\infty )\times {\mathbb{R}}^{n-1}$, the formula $Eu(x_1,x^{\prime })=u(|x_1|,x^{\prime })$ defines a bounded Sobolev extension operator. Its weak normal derivative changes sign across the boundary, while its tangential weak derivatives are reflected unchanged.