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

Suppose the claimed Poincare-Wirtinger inequality were false. There would be $u_k\in H^1(U)$ such that, after settingwe have $(v_k{)}_U=0$, $\Vert v_k{\Vert }_2=1$, and $\Vert D{v}_k{\Vert }_2\to 0$. The sequence is bounded in $H^1(U)$, so part 1(b)(ii) supplies a subsequence converging strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $v$.

The weak gradient of $v$ is zero. Because $U$ is connected, $v$ is a constant function; its mean is zero, so $v=0$. Strong convergence would then give $\Vert v_k{\Vert }_2\to 0$, contradicting $\Vert v_k{\Vert }_2=1$. Therefore

If the Poincare inequality with a boundary trace failed, after normalization there would be $u_k\in H^1(U)$ withAs 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 $\Vert u{\Vert }_2=1$. Consequently