Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 105 / 2

The functionalis bounded by the Cauchy-Schwarz inequality. Hence $H=\ker I$ is a closed vector subspace of the Hilbert space $H^1(U)$. Every closed subspace of a Hilbert space is complete in the inherited norm, so $H$ is a Hilbert space with the standard $H^1$ inner product.

If the asserted Poincare inequality failed, after rescaling there would be $u_j\in H$ withThe sequence is bounded in $H^1(U)$. By the Rellich-Kondrachov compactness theorem, a subsequence converges strongly in $L^2(U)$ to some $u$ and weakly in $H^1(U)$. Its weak gradient is zero, so connectedness of $U$ makes $u$ almost everywhere constant. Continuity of the integral under $L^2$ convergence gives ${\int }_{U}u=0$, hence $u=0$. Strong convergence would then imply $\Vert u_j{\Vert }_2\to 0$, contradicting the normalization. Therefore the required constant exists.

The assumed inequality is equivalent toand equality holds at $w$. For $v\in H$ and $t\in \mathbb{R}$, expand $J(w+tv)\ge 0$ and use $J(w)=0$:Since this holds for both signs of arbitrarily small $t$, the linear coefficient vanishes:Every $v\in H^1(U)$ is a mean-zero function plus a constant. The same identity holds for constants because ${\int }_{U}w=0$, so it holds for all $v\in H^1(U)$. This is precisely the weak formulation ofwhere the Neumann boundary condition is the natural boundary condition encoded by the weak formulation.