Suppose the claimed <Poincare inequality with a partial Dirichlet boundary> fails. There are $v_k\in V$ with
$$
\lVert v_k\rVert_{L^2(U)}>k\lVert\nabla v_k\rVert_{L^2(U)}.
$$
After the normalization $w_k=v_k/\lVert v_k\rVert_{L^2(U)}$,
$$
\lVert w_k\rVert_{L^2(U)}=1,
\qquad
\lVert\nabla w_k\rVert_{L^2(U)}<\frac1k.
$$
Thus $(w_k)$ is bounded in the <Sobolev space> $H^1(U)$. The <Rellich-Kondrashov compactness theorem for H01>[Rellich-Kondrashov compactness theorem] and the corresponding compact embedding for a bounded $C^1$ domain give a subsequence that converges strongly in $L^2(U)$ and weakly in $H^1(U)$ to some $w$. The <Sobolev space with a partial Dirichlet condition> $V$ is a <closed vector subspace>, hence weakly closed, so $w\in V$. Moreover $\nabla w=0$, and connectedness of $U$ makes $w$ a <constant function>. Its <Sobolev trace theorem>[trace] vanishes on the positive-measure set $\Gamma_1$, so that constant is zero. This contradicts
$$
\lVert w\rVert_{L^2(U)}
=\lim_k\lVert w_k\rVert_{L^2(U)}=1.
$$
Therefore some $C_P$ satisfies
$$
\lVert v\rVert_{L^2(U)}\leq C_P\lVert\nabla v\rVert_{L^2(U)}.
$$
Since the reverse bound $\lVert\nabla v\rVert_2\leq\lVert v\rVert_{H^1}$ is immediate,
$$
\lVert\nabla v\rVert_2
\leq\lVert v\rVert_{H^1}
\leq\sqrt{1+C_P^2}\,\lVert\nabla v\rVert_2.
$$
The gradient seminorm is a norm on $V$ because equality to zero would make $v$ a constant whose trace on $\Gamma_1$ is zero.

Solved by gpt-5.6-sol high.

Solution

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

{title2=$V$}

For a boundary portion $\Gamma_D\subseteq\partial U$, the space
$$
V=\{v\in H^1(U):\operatorname{Tr}v|_{\Gamma_D}=0\}
$$
is the kernel of the restricted <Sobolev trace theorem>[trace map], hence a <closed vector subspace> of $H^1(U)$.

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 Incoming links: Sobolev space with a partial Dirichlet condition