Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 4

For real-valued $f\in C[-1,1]$, there is a unique best uniform approximation $p_{\ast }\in {\mathcal{P}}_n$, where ${\mathcal{P}}_n$ means degree at most $n$. If $E=\Vert f-p_{\ast }{\Vert }_{\infty }>0$, the Chebyshev alternation theorem characterizes it by the existence of $n+2$ ordered pointsand a sign $\epsilon \in \{-1,1\}$ such thatConversely, a polynomial in ${\mathcal{P}}_n$ whose error has this alternation is the best one. If the best error is zero, $f$ itself is the exact polynomial approximant. The real-valued qualification matters: this sign-alternation formulation is not the corresponding criterion for arbitrary complex-valued approximation.

The obstruction to a strict improvement is concrete. If $q$ had smaller error at every alternation point, $q-p_{\ast }$ would have alternating strict signs there. The intermediate value theorem would give at least $n+1$ distinct zeros, impossible for a nonzero polynomial of degree at most $n$.

Since $|T_j(x)|\le 1$ on $[-1,1]$ and the coefficients are summable, the Weierstrass M-test gives uniform convergence of this positive lacunary Chebyshev series and makes its sum continuous.

Let $r$ be the largest integer with $3^r\le n$, and set $r=-1$ when $n=0$. The candidate and its tail size areThe empty sum for $n=0$ is zero. The candidate has degree at most $n$. The triangle inequality bounds its error by the proposed tail sum, and equality holds at $x=1$, since every Chebyshev polynomial has $T_j(1)=1$ and all tail coefficients are positive. It remains to prove that another polynomial cannot improve this error.

Put $L=3^{r+1}$, the first omitted degree, so $L>n$. At the points $y_j=\cos (j\pi /L)$ for $j=0,\dots ,L$, every omitted term hasbecause $3^{k-r-1}$ is odd. Uniform convergence therefore givesThese $L+1$ distinct points run in decreasing order; reversing their order still gives alternating signs. Since $L+1\ge n+2$, select $n+2$ consecutive ones and apply the Chebyshev alternation theorem. It proves both optimality and uniqueness of the displayed candidate. Thus the best polynomial and the error stay unchanged between successive powers of three; this exact truncation property depends on the synchronized tail signs, not on a general rule that truncating a Chebyshev expansion is optimal.