 Solution

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.