Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 318 / 1

Averaging the integral representations of the Fourier partial sums givesThe finite trigonometric sum isSince $D_k(u)=\sin ((k+1/2)u)/(2\sin (u/2))$, the Fejér kernel is thereforeand

The displayed square shows that $F_n\ge 0$. Every Fourier partial sum preserves the constant function, so ${\sigma }_n(1)=1$. Substituting $f=1$ in the integral formula givesNonnegativity then yields

Because the Fejér kernel has normalized integral one,For $f\in \mathrm{Lip}\alpha$ this impliesOn $0\le t\le \pi$, the inequalities $\sin (t/2)\ge t/\pi$ and $|\sin (nt/2)|\le n|\sin (t/2)|$ giveSplitting the integral at $1/n$ givesFor $0<\alpha <1$, both terms are $O(n^{-\alpha })$. For $\alpha =1$, the second is $O((\log n)/n)$. Uniformly in $x$,where the constants absorb the Lipschitz continuity constant $M$.

At the cusp of $f_0(t)=|t|$, positivity and evenness of the Fejér kernel giveSince $\sin (t/2)\le t/2$,Summing the supplied lower bound over $I_k$, $1\le k\le n^{\prime }=\lfloor n/2\rfloor$, yieldsThe harmonic series satisfies ${\sum }_{k=1}^{n^{\prime }}{k}^{-1}\ge c^{\prime }\log n$, so

The modulus of continuity of the periodic function $f_0$ obeys $\omega (f_0,1/n)=1/n$. If a universal Jackson-type estimateheld for all continuous periodic $f$, it would give $O(1/n)$ for $f_0$, contradicting the lower bound. Thus