If the <Poincare inequality with a boundary trace> failed, after normalization there would be $u_k\in H^1(U)$ with
$$
\|u_k\|_{L^2(U)}=1,
\qquad
\|Du_k\|_{L^2(U)}+\|u_k\|_{L^2(\partial U)}\longrightarrow0.
$$
As in part 1(c)(i), a subsequence converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to a <constant function> $u$. The <Sobolev trace theorem> is a bounded linear map, so the traces converge weakly while their norms tend to zero; hence the trace of $u$ is zero. A constant with zero trace is zero, contradicting $\|u\|_2=1$. Consequently
$$
\boxed{\|u\|_{L^2(U)}\leq C_2\left(\|Du\|_{L^2(U)}+\|u\|_{L^2(\partial U)}\right).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 105 / 1 / c / ii / Solution

{c}

On a bounded connected Lipschitz domain $U$,
$$
\|u\|_{L^2(U)}\leq C\left(\|Du\|_{L^2(U)}+\|\operatorname{Tr}u\|_{L^2(\partial U)}\right)
$$
for every $u\in H^1(U)$. If this failed, the <Rellich-Kondrachov compactness theorem> would produce a normalized strong $L^2$ limit with zero <gradient>, hence a <constant function>; continuity of the <Sobolev trace theorem> would force that constant to vanish, contradicting its unit <norm>.

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