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

Write $u_Q=|Q{|}^{-1}{\int }_{Q}u$. Since $u-u_Q$ has arithmetic mean zero,and the pairwise-difference identity givesFirst take $u$ smooth. Join $x$ to $y$ by changing one coordinate at a time and apply the Cauchy-Schwarz inequality:For a one-dimensional slice $v$, the fundamental theorem of calculus yieldsIntegrating the $i$th summand over $x,y\in Q$ therefore gives at most $L^{n+2}\Vert D_{i}u{\Vert }_{L^2(Q)}^2$. HenceThe density of smooth functions in a Sobolev space extends the estimate to every $u\in H^1({\mathbb{R}}^n)$. Thus

With the standard Sobolev space inner product, $u_i\rightharpoonup u$ weakly in $H^1(U)$ meansfor every $v\in H^1(U)$. Equivalently, every bounded linear functional on $H^1(U)$ takes convergent values on the sequence.

The Sobolev space $H^1(U)$ is a separable Hilbert space. A bounded sequence therefore has a weakly convergent subsequence by the weak subsequence of a bounded Hilbert-space sequence; write $u_{i_j}\rightharpoonup u$ in $H^1(U)$. Since $U$ is bounded with smooth boundary, the Rellich-Kondrachov compactness theorem says that $H^1(U)\hookrightarrow L^2(U)$ is compact. Passing to a further subsequence gives

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