Let $L=\operatorname{Lip}(u)<k$ and take $v\in\operatorname{Lip}_g(\Omega)$. For small $t>0$, $u_t=(1-t)u+tv$ has Lipschitz constant at most $(1-t)L+t\operatorname{Lip}(v)<k$. Constrained minimality and convexity give
$$
\mathcal F[u]\leq\mathcal F[u_t]
\leq(1-t)\mathcal F[u]+t\mathcal F[v].
$$
Cancellation gives $\mathcal F[u]\leq\mathcal F[v]$, so $u$ is an unconstrained minimizer.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 107 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-107/4/b/solution