I'm reading the proof of the sewing lemma and i'm not able to proof a quite straightforward result that is implicetly used in the paper.
Given an interval $\left[s,t\right]$ and a partition $\mathcal{P}$ of such interval we define the mesh size as $|\mathcal{P}|=\sup_\limits{[u,v] \in \mathcal{P}} \left|u-v \right|$.
Given $\beta>1$ we have that $$ \sum_\limits{[u,v] \in \mathcal{P}} |u-v|^\beta \le \left|t-s\right| \ \ \left|\mathcal{P}\right|^{\beta-1} $$ The result is used to prove that an operator is a Riemann type limit. As far as I know this result holds only if $\beta>1$, so i guess that $\sum_\limits{[u,v] \in \mathcal{P}} |u-v|^\beta$ is not bounded by $|\mathcal{P}|^\beta$. Is this correct?
For the closed interval $[u,v]\in \mathcal{P}$ and $\beta > 1$ we have $|u-v| \leqslant |\mathcal{P}|$ and, hence, $|u-v|^{\beta-1} \leqslant |\mathcal{P}|^{\beta-1}$.
Thus,
$$\sum_\limits{[u,v] \in \mathcal{P}} |u-v|^\beta =\sum_\limits{[u,v] \in \mathcal{P}} |u-v|^{\beta-1} |u-v|\leqslant |\mathcal{P}|^{\beta-1}\sum_\limits{[u,v] \in \mathcal{P}} |u-v| = |\mathcal{P}|^{\beta-1}|t-s|$$
The last equality follows because, by definition of a partition, the subintervals are non-overlapping and cover the interval $[s,t]$. Whence,
$$\sum_\limits{[u,v] \in \mathcal{P}} |u-v|= t-s = |s-t|$$