At the cusp of $f_0(t)=|t|$, positivity and evenness of the <Fejér kernel> give
$$
\sigma_n(f_0,0)-f_0(0)
=\frac2\pi\int_0^\pi F_n(t)t\,dt.
$$
Since $\sin(t/2)\leq t/2$,
$$
\sigma_n(f_0,0)
\geq\frac4{\pi n}\int_0^\pi
\frac{\sin^2(nt/2)}{t}\,dt.
$$
Summing the supplied lower bound over $I_k$, $1\leq k\leq n'=\lfloor n/2\rfloor$, yields
$$
\sigma_n(f_0,0)
\geq\frac{c}{n}\sum_{k=1}^{n'}\frac1k.
$$
The <harmonic series> satisfies $\sum_{k=1}^{n'}k^{-1}\geq c'\log n$, so
$$
\boxed{|\sigma_n(f_0,0)-f_0(0)|
\geq c_1'\frac{\log n}{n}}.
$$

The <modulus of continuity> of the periodic function $f_0$ obeys $\omega(f_0,1/n)=1/n$. If a universal <Jackson-type estimate>
$$
\|\sigma_n(f)-f\|_\infty\leq C\omega(f,1/n)
$$
held for all continuous periodic $f$, it would give $O(1/n)$ for $f_0$, contradicting the lower bound. Thus
$$
\boxed{\text{no such universal Jackson-type estimate holds for Fejér sums}.}
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-318/1/c/solution