= Solution
The indexing of the <Fejér sums> gives
$$
(n+m)\sigma_{n+m}(f)=\sum_{j=0}^{n+m-1}s_j(f),\qquad
n\sigma_n(f)=\sum_{j=0}^{n-1}s_j(f).
$$
Subtracting removes precisely the initial $n$ <Fourier partial sums>. Thus
$$
\boxed{v_{n,m}(f)=\frac{(n+m)\sigma_{n+m}(f)-n\sigma_n(f)}m}.
$$
For $n=0$, omit the second term, so that no undefined $\sigma_0$ is needed. Apply the triangle inequality and the <uniform-norm contraction of Fejér summation> estimate to get
$$
\|v_{n,m}(f)\|_\infty
\le\frac{n+m}{m}\|\sigma_{n+m}(f)\|_\infty
+\frac n m\|\sigma_n(f)\|_\infty
\le\left(1+\frac{2n}{m}\right)\|f\|_\infty.
$$
Hence the <operator norm> of the <de la Vallée Poussin sum> is at most \b[$1+2n/m$].
Back to article page
