I am studying the book 'A course on Rough Paths' by Friz and Hairer at the moment and I have a problem with one conclusion they're making on page 11 (also described below).
Their claim is the following:
Let $X:[0,T]^2\rightarrow V$, $T>0$, $V$ Banach space satisfy the following:
$X(s,t) \leq C |t-s|^a$, $a>1$.
Then we have for a partition $P$ of $[0,T]$:
$\sum_{[s,t]\in P} |X(s,t)| \leq CT|P|^{a-1}$, where $|P|$ denotes the length of the largest element of $P$.
My naive approach was the following:
$\sum_{[s,t]\in P} |X(s,t)| \leq C\sum_{[s,t]\in P}|t-s|^a \leq C\sum_{[s,t]\in P} |P|^a$,
but now I would have to estimate the number of summands by $T/p$, with $p$ being the length of the smallest element of $P$.
Am I just not seeing the right approach?
Thank you in advance!
Best,
Luke
I don't see any problem with what you propose. First of all by definition $$p = |P|.$$ Then it should be clear that the number of summands is smaller or equal then $T/|P|.$ Thus you get exactly the estimate that you need.