Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 326 / 3 / iv / Solution

For the convex conjugate, use a dual scalar $q$ to avoid confusing it with the fixed centre $z$. Differentiating $qx-\lambda (x-z{)}^2/2$ gives $q-\lambda (x-z)=0$, so the unique maximizing point is $x=z+q/\lambda$. Substitution yieldsThe positivity of $\lambda$ makes the maximized quadratic strictly concave and the convex conjugate finite for every real $q$.