The restricted <Sobolev trace theorem>[trace map] is continuous, and $V$ is its kernel, so $V$ is a <closed vector subspace> of the <Hilbert space> $H^1(U)$. It is therefore complete in the $H^1$ norm. Part a shows that the gradient norm
$$
\lVert v\rVert_V=\lVert\nabla v\rVert_{L^2(U)}
$$
is equivalent to that norm, so it is complete as well. It comes from the <inner product>
$$
(u,v)_V=\int_U\nabla u\mathbin\cdot\nabla v\,dx.
$$
Consequently $(V,\lVert\cdot\rVert_V)$ is a <Hilbert space>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 105 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-105/2/d/solution