Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 69 / 4 / a

Let $u\in \mathcal{U}$, $r=f-u$, $M=\Vert r{\Vert }_{\infty }$, and $E=\{x\in [0,1]:|r(x)|=M\}$. The Kolmogorov criterion for uniform approximation saysEquivalently, no available perturbation $v$ has strictly the same sign as the nonzero residual at every extremal point. This statement concerns a candidate $u$ in the linear subspace; it does not assume every subspace has an attained best approximation. If $M=0$, $u=f$ is already exact and the criterion is automatic.

To see sufficiency, for any competitor $u+v$, the criterion supplies an $x\in E$ with $r(x)v(x)\le 0$. Then $|f(x)-u(x)-v(x)|\ge |r(x)|=M$, so the competitor cannot have smaller supremum norm error.

For necessity, suppose $M>0$ and some $v\in \mathcal{U}$ has $r(x)v(x)>0$ on $E$. Since $E$ is compact, this product has a positive minimum there. By continuity, a neighborhood $O$ of $E$ has $rv\ge \eta >0$. For sufficiently small $\epsilon >0$,On the compact complement of $O$, $|r|$ has a strict gap below $M$, and taking $\epsilon \Vert v{\Vert }_{\infty }$ smaller than this gap preserves it. Hence $u+\epsilon v$ improves the uniform error everywhere, contradicting best uniform approximation.