 Solution

Let $u^{\ast }\in \mathcal{U}$, $e=f-u^{\ast }$, $d=\Vert e{\Vert }_{\infty }$, and $E=\{x:|e(x)|=d\}$. The real Kolmogorov criterion for uniform approximation says $u^{\ast }$ is best if and only ifEquivalently, no direction $v$ has the error's sign strictly at every extremal point. If $d=0$, the criterion is automatic.

For sufficiency, a point with $e(x)v(x)\le 0$ gives $|e(x)-v(x)|\ge d$, so $u^{\ast }+v$ cannot improve the norm. For necessity, suppose $e(x)v(x)>0$ on $E$. Compactness gives a positive lower bound there and on a neighbourhood of $E$. On that neighbourhood, $(e-\tau v{)}^2=e^2-2\tau ev+{\tau }^2{v}^2<d^2$ for sufficiently small positive $\tau$. Off the neighbourhood, $|e|$ has a strict gap below $d$, and small $\tau$ preserves that gap. Thus $u^{\ast }+\tau v$ has smaller error, a contradiction. This also explains why the criterion involves all active extrema.