The proximal optimality condition gives
$$
\frac{x_k-x_{k+1}}t\in\partial f(x_{k+1}).
$$
The <subgradient inequality> therefore yields, for every $u$,
$$
\boxed{f(u)\geq f(x_{k+1})+\frac1t(x_k-x_{k+1})^T(u-x_{k+1}).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 339 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-339/1/b/i/solution