Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 326 / 3 / 2 / b / Solution

The indicator functional of a constraint set restricts the conjugate supremum to $[-1,1]$. Complete the square:Its maximizer is $x=p$ when $|p|\le 1$, and the nearer endpoint $x=\mathrm{sign}(p)$ otherwise. The convex conjugate of a constrained quadratic is thereforeThis is the unit-threshold Huber loss; both branches and their first derivatives agree at the transition points. The original PDF constrains $|x|\le 1$, including both signs; the damaged TeX loses that absolute value.