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

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.

The original PDF has indices $3^k$; the exponent is lost in the TeX transcription. This lacunary indexing is essential to the positive lacunary Chebyshev series argument below.

Let $K$ be the least nonnegative integer with $3^K>n$, so $K=0$ for $n=0$. DefineFor $K=0$ the sum is empty and means the zero polynomial. Each included Chebyshev polynomial has degree at most $n$. Also $|T_j(x)|\le 1$ on the interval, so summability of the positive coefficients gives uniform convergence by the Weierstrass M-test, andChoose $L=3^K$ and the $L+1$ points $x_j=\cos (j\pi /L)$, $0\le j\le L$. For every omitted index $k\ge K$, the integer $3^{k-K}$ is odd. Consequently,Every term of the tail has the same sign at a given point, and thereforeThis shows both that the error norm is exactly $A_K$ and that it alternates at $L+1\ge n+2$ distinct points. The points are in decreasing order; reversing their order still gives alternation. The Chebyshev alternation theorem proves that this partial sum is the unique best uniform approximation. HenceIn particular, $p_0=0$ and $E_0(f_0)={\sum }_{k=0}^{\infty }{a}_k$. The equal signs of all tail terms at the same extrema are the reason positivity and the odd integer frequency ratios are useful.

An explicit construction gives Bernstein's lethargy theorem in the requested inequality form. SetStrict decrease makes every coefficient positive. Telescoping and the limit assumption giveThusdefines a continuous function by the Weierstrass M-test and the uniform limit theorem. Apply the positive lacunary Chebyshev series calculation. At $n=0$,For $n\ge 1$, choose $K\ge 1$ so that $3^{K-1}\le n<3^K$. Its error isTherefore the function satisfiesThis explicit Chebyshev construction for Bernstein lethargy also has $E_n(f_{ϵ})\to 0$. It shows that continuity imposes no universal speed of convergence of best polynomial approximation, even though convergence itself follows from the Weierstrass approximation theorem.