Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 69 / 6 / a / Solution

For $f\in C(\mathbb{T})$, define $E_j(f)={\inf }_{T\in {\mathcal{T}}_j}\Vert f-T{\Vert }_{\infty }$, where ${\mathcal{T}}_j$ is the space of degree-at-most-$j$ trigonometric polynomials. The first inverse theorem for trigonometric approximation states that a universal constant $C$ satisfiesFor perspective, take best approximants at dyadic degrees and telescope their differences. Each difference between consecutive approximants has supremum norm at most twice the earlier error, and the Bernstein inequality for trigonometric polynomials multiplies that bound by its degree when differentiating. The resulting dyadic sum is bounded by the displayed sum because $E_j$ decreases with $j$; the remaining approximation error is also bounded by an average of earlier errors.

Now suppose $E_j(f)\le K{j}^{-\alpha }$ for large $j$, absorbing finitely many earlier terms into a constant. If $0<\alpha <1$, comparison with an integral gives ${\sum }_{j=1}^n{j}^{-\alpha }=O(n^{1-\alpha })$. If $\alpha =1$, the same comparison gives ${\sum }_{j=1}^n{j}^{-1}=O(1+\ln n)$. Substitution proves the requested two rates:The logarithm is the endpoint contribution of the harmonic sum and cannot generally be removed, as the following Weierstrass function shows.