Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 61 / 6 / a / Solution

We work in the usual real-valued setting of the Chebyshev alternation theorem. For $f\in C[-1,1]$ and an algebraic polynomial $p$ of degree at most $n$, the theorem says that $p$ is a best uniform approximation if and only if its error has $n+2$ ordered points with alternating maximal values:If $f=p$, the zero-error case is included directly.

Existence follows, for example, by taking a minimizing sequence: its supremum norms are bounded, all norms on the finite-dimensional polynomial space are equivalent, and a convergent coefficient subsequence attains the infimum. To prove uniqueness of best uniform polynomial approximation, let $p,q$ both attain the minimum error $E$. Their average $r=(p+q)/2$ has error at most $E$ by the triangle inequality and therefore exactly $E$ by minimality.

If $E=0$, both polynomials equal $f$ and are equal. Otherwise apply the Chebyshev alternation theorem to $r$. At every alternating extremal point, $f-r$ is either $E$ or $-E$. But it is the average of $f-p$ and $f-q$, each lying in $[-E,E]$. An average attains an endpoint of this interval only when both entries equal that endpoint. Hence $p(x_j)=q(x_j)$ at all $n+2$ points. The difference is a degree-at-most-$n$ polynomial with more than $n$ distinct zeros, so it is identically zero. The best approximating polynomial is unique.