Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 67 / 3 / Solution

Use the periodic modulus of continuitySubdividing a displacement $t$ into at most $\lceil |t|/\delta \rceil$ steps of size at most $\delta$ proves the chaining inequalityFor $|t|\le \delta$, the sharper bound $\omega (f,\delta )$ holds directly.

Set $\delta =n^{-\alpha }$, where $0<\alpha <1$ and $n\ge 1$. Positivity and unit normalized mass of the Fejér kernel giveThe portion $|t|\le \delta$ is at most $\omega (f,\delta )$. For $0<|t|\le \pi$, $\sin (|t|/2)\ge |t|/\pi$ and $|\sin (nt/2)|\le 1$, soApply this bound and the chaining inequality to the two tails:Let $\beta =1-\alpha >0$. Calculus gives ${\sup }_{u\ge 1}{u}^{-\beta }\log u=1/(e\beta )$, attained at $u=e^{1/\beta }$, and $u^{-\beta }\le 1$. Thus a possible explicit constant for the fractional-scale Fejér approximation bound isThe constant depends only on $\alpha$, not on $f$ or $n$. Since a continuous periodic function is uniformly continuous, the right-hand side tends to zero. The logarithmic tail estimate is what permits every $\alpha <1$; a crude global-oscillation bound would not prove the whole requested range.