= Solution
For a continuous $2\pi$-periodic <function>, define the <Fourier coefficients> and <Fourier partial sum> by
$$
\widehat f(k)=\frac1{2\pi}\int_{-\pi}^{\pi}f(t)e^{-ikt}\,dt,\qquad
s_n(f;x)=\sum_{|k|\leq n}\widehat f(k)e^{ikx}.
$$
The <Fejér sum> is the arithmetic average of the first $n$ <Fourier partial sums>:
$$
\boxed{\sigma_n(f)=\frac1n\sum_{j=0}^{n-1}s_j(f)
=\sum_{|k|<n}\left(1-\frac{|k|}{n}\right)\widehat f(k)e^{ikx}.}
$$
Thus $s_n$ has degree at most $n$, whereas $\sigma_n$ has degree at most $n-1$. These indexing conventions matter in the identity in (b). Throughout, $m$ is a positive integer; when $n=0$, any term $n\sigma_n$ is interpreted as zero without introducing $\sigma_0$.
For later estimates, the <Fejér kernel> in the PDF's normalization is
$$
F_n(t)=\frac1{2n}\left|\sum_{j=0}^{n-1}e^{ijt}\right|^2
=\frac1{2n}\frac{\sin^2(nt/2)}{\sin^2(t/2)},\qquad
\sigma_n(f;x)=\frac1\pi\int_{-\pi}^{\pi}f(x-t)F_n(t)\,dt.
$$
The apparent singularities are removable. The squared modulus makes $F_n$ nonnegative, and integrating its finite exponential expansion leaves only the $n$ diagonal terms, so $\pi^{-1}\int F_n=1$. In particular <Fejér summation is a uniform-norm contraction>:
$$
\|\sigma_nf\|_\infty\leq\frac1\pi\int_{-\pi}^{\pi}F_n(t)\|f\|_\infty\,dt=\|f\|_\infty.
$$
Back to article page
