= Solution
Periodicity and the <Hölder condition> give $|f(x-t)-f(x)|\le M|t|^\alpha$ for $|t|\le\pi$. The positive normalized <Fejér kernel> therefore yields
$$
\|\sigma_{n-1}f-f\|_\infty\le\frac{2M}{\pi}\int_0^\pi t^\alpha F_{n-1}(t)\,dt.
$$
The exponential-sum representation of the kernel bounds $F_{n-1}(t)\le n/2$. Also $\sin(t/2)\ge t/\pi$ for $0\le t\le\pi$, so $F_{n-1}(t)\le\pi^2/(2nt^2)$. Splitting at $1/n$ and using $0<\alpha<1$ gives
$$
\begin{aligned}
\int_0^\pi t^\alpha F_{n-1}(t)\,dt&\le\frac n2\int_0^{1/n}t^\alpha\,dt+\frac{\pi^2}{2n}\int_{1/n}^\pi t^{\alpha-2}\,dt\\
&\le\left(\frac1{2(\alpha+1)}+\frac{\pi^2}{2(1-\alpha)}\right)n^{-\alpha}.
\end{aligned}
$$
Hence the <Hölder error bound for Fejér summation> is
$$
\boxed{\|\sigma_{n-1}f-f\|_\infty\le M\left(\frac1{\pi(\alpha+1)}+\frac\pi{1-\alpha}\right)n^{-\alpha}.}
$$
The constant depends on the <Hölder exponent> and $M$, but not on $n$ or $f$ beyond its <Hölder seminorm> bound.
Back to article page
