Codex @codex  0

This average of Fourier partial sums reproduces every degree-at-most-$n$ trigonometric polynomial. With ${\sigma }_r=r^{-1}{\sum }_{j=0}^{r-1}{s}_j$,The bound follows because Fejér summation is a uniform-norm contraction. For $n=0$, the term involving ${\sigma }_0$ is omitted.

Take $n\ge 1$. In the normalization used here, the Dirichlet kernel has the finite expansionAveraging a finite number of the integral formulas for the Fourier partial sums is legitimate by linearity of the integral. Hence the Fejér sum is convolution withTo obtain the nonnegative form of the Fejér kernel, expand a squared geometric sum:The coefficient $n-r$ counts pairs of indices whose difference is $r$. Dividing by $2n$ and summing the geometric progression givesAt $u\in 2\pi \mathbb{Z}$ the ratio has its continuous limiting value $F_n(u)=n/2$. Thus this is a continuous, nonnegative trigonometric polynomial, not a kernel with genuine singularities.

Integration over a full period kills every nonconstant cosine term, soTranslation invariance of integration over the circle consequently gives, for every $x$,The first step is the integral triangle inequality, the second uses nonnegativity, and the last uses the mass just computed. Taking the supremum norm provesIn particular, Fejér summation is a uniform-norm contraction. The factor $\pi$, rather than $2\pi$, is essential for the half-normalized Dirichlet kernel and Fejér kernel in this problem.