Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 9 / 1 / a / Solution

The statement requires $K\ne \varnothing$; an empty set is closed and convex but contains no minimizer. Assume this necessary condition and put $d={\inf }_{g\in K}\Vert g-f{\Vert }_p$. Since $K$ is closed and $f\notin K$, $d>0$. Choose a minimizing sequence $g_j\in K$ with $\Vert g_j-f{\Vert }_p\to d$.

The midpoint $(g_j+g_k)/2$ lies in the convex set $K$. Applying the stated Clarkson inequality to $g_j-f$ and $g_k-f$ givesThe right side tends to zero, so $g_j$ is a Cauchy sequence. By completeness of Lp spaces, proved in Question 2(a), it converges to some $h$. Closedness puts $h$ in $K$, and continuity of the norm gives $\boxed{\Vert h-f{\Vert }_p=d}$. This is the closest point theorem for a closed convex subset of Lp. The same inequality applied to two minimizers forces their difference to be zero, so the minimizer is also unique.