Because the <Fejér kernel> has normalized integral one,
$$
\sigma_n(f,x)-f(x)
=\frac1\pi\int_{\mathbb T}F_n(t)
[f(x-t)-f(x)]\,dt.
$$
For $f\in\operatorname{Lip}\alpha$ this implies
$$
|\sigma_n(f,x)-f(x)|
\leq\frac{2M}{\pi}\int_0^\pi F_n(t)t^\alpha\,dt.
$$
On $0\leq t\leq\pi$, the inequalities $\sin(t/2)\geq t/\pi$ and $|\sin(nt/2)|\leq n|\sin(t/2)|$ give
$$
F_n(t)\leq\min\left(\frac n2,\frac{\pi^2}{2nt^2}\right).
$$
Splitting the integral at $1/n$ gives
$$
\int_0^\pi F_n(t)t^\alpha\,dt
\leq\frac n2\int_0^{1/n}t^\alpha\,dt
+\frac{\pi^2}{2n}\int_{1/n}^{\pi}t^{\alpha-2}\,dt.
$$
For $0\lt\alpha\lt1$, both terms are $O(n^{-\alpha})$. For $\alpha=1$, the second is $O((\log n)/n)$. Uniformly in $x$,
$$
\boxed{
\|\sigma_n(f)-f\|_\infty\leq
\begin{cases}
c_\alpha n^{-\alpha},&0\lt\alpha\lt1,\\
c_1(\log n)/n,&\alpha=1,
\end{cases}}
$$
where the constants absorb the <Lipschitz continuity> constant $M$.

Solution

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

 Solution

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