= Solution
The difference between the two cumulative sums of <Fourier partial sums> is
$$
(n+m)\sigma_{n+m}(f)-n\sigma_n(f)=\sum_{j=n}^{n+m-1}s_j(f).
$$
Hence the <de la Vallée Poussin sum> satisfies
$$
\boxed{v_{n,m}(f)=\frac{n+m}{m}\sigma_{n+m}(f)-\frac nm\sigma_n(f).}
$$
Using the <uniform-norm contraction of Fejér summation> on each term and the <triangle inequality> in the <supremum norm>,
$$
\begin{aligned}
\|v_{n,m}(f)\|_\infty
&\leq\frac{n+m}{m}\|\sigma_{n+m}(f)\|_\infty+\frac nm\|\sigma_n(f)\|_\infty\\
&\leq\left(1+\frac{2n}{m}\right)\|f\|_\infty.
\end{aligned}
$$
This proves the requested <operator norm> bound. When $n=0$ the identity reduces to $v_{0,m}=\sigma_m$ and the bound is one.
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