Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 3

For a real continuous $2\pi$-periodic function $f$, the trigonometric Chebyshev alternation theorem says that $t^{\ast }\in {\mathcal{T}}_n$ is its best uniform approximation if and only if there are $2n+2$ distinct cyclically ordered points in one period,and a sign $\epsilon \in \{1,-1\}$ such thatThe best approximant exists and is unique. Alternation is cyclic: the last and first displayed extrema also have opposite signs across the closing interval. If the error is zero, $f=t^{\ast }$ and the characterization holds trivially.

The sufficiency can be seen from zero counting. A strictly improving trigonometric polynomial would have a difference from $t^{\ast }$ with the alternating nonzero signs at these points, forcing at least $2n+2$ zeros on the circle. A nonzero member of ${\mathcal{T}}_n$ has at most $2n$ zeros: multiplying its Laurent expression by $e^{inx}$ gives an ordinary polynomial of degree at most $2n$ in $e^{ix}$. This contradiction is the alternation mechanism used in the next part.

The coefficient sum is finite, so the Weierstrass M-test makes the cosine series uniformly convergent and $f$ continuous. Put $Q=5^{m+1}$ and $A_m={\sum }_{k=m+1}^{\infty }{a}_k$. The indicated partial sum belongs to ${\mathcal{T}}_n$ because its largest frequency is $5^m\le n$. Its tail satisfiesand equality holds at $x=0$. Thus its supremum norm error is exactly $A_m$.

At the $2Q$ points $x_j=j\pi /Q$, $0\le j<2Q$, every omitted frequency is an odd integer multiple of $Q$. Indeed for $k\ge m+1$,Uniform convergence permits summing these values, givingThese are cyclically alternating extrema. Since $n<Q$, there are $2Q\ge 2n+2$ of them, so the trigonometric Chebyshev alternation theorem proves that the partial sum is the unique best approximant. Consequently the positive lacunary trigonometric series hasThe positivity of every tail coefficient makes the same tail attain its full coefficient sum with alternating signs; a triangle bound alone would not establish optimality.