Solution

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

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

{title2=$h=\operatorname*{argmin}_{g\in K}\|g-f\|_p$}

For $2\le p<\infty$, a nonempty <closed> <convex set> $K$ in an <Lp space> has exactly one nearest point to every $f$. A <minimizing sequence> $g_j$ is a <Cauchy sequence>: the <Clarkson inequality> bounds $\|g_j-g_k\|_p^p$ by $2^{p-1}(\|g_j-f\|_p^p+\|g_k-f\|_p^p)-2^pd^p$, where $d$ is the infimum of the distances. <Completeness of Lp spaces> and the <closed set> property give a limit in $K$. Applying the same inequality to two minimizers proves uniqueness. Nonemptiness is essential.

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 Incoming links: Closest point theorem for a closed convex subset of Lp