I am wondering if what I show below is allowed, and why it is allowed.
Assume you have an interval $[0,T]$. For each $n$ you divide the interval in $2^n$ n parts. In each interval you have different functions $o(\Delta t)$. We have the summation
$\sum\limits_{i=1}^{2^n}o(\Delta t)=2^no(\Delta t)=T\frac{o(\Delta t)}{\frac{T}{2^n}}=T\frac{o(\Delta t)}{\Delta t}$.
So this goes to zero as n goes to infinity.
I have seen this argumentation in a book. Is it allowed or do we need more constraints on the $o(\Delta t)$ functions? The problems I have are:
Let n be given, the $o(\Delta t)$ functions may be different, so can we just say they are $2^n$ times an $o(\Delta t)$ function?
When $n$ increases, we get new $o(\Delta t)$-functions. Does this affect the result? For the limit to go to zero as $n$ increases we need it to be the same $o(\Delta t)$-function?
So does the result hold, or are there counterexamples where they don't hold? Do we need more conditions on the functions for it to hold?
The argument is on page 50(printed page) or page 104-105(scroll-meny page) on the link here:
https://www.nb.no/items/URN:NBN:no-nb_digibok_2011011705004?page=103
UPDATE:
I think the statement is false. Assume for each $i,n$ we have functions $o_{i,n}(\Delta t)$. They are defined by $o_{i,i}(T/2^i)=1$, and zero for every other combination of $i$ and $n$ and $\Delta t$. Then these functions are $o(\Delta t)$ because they are identically zero when $\Delta t$ gets small enough. So:
$\sum\limits_{i=1}^{2^n}o_{i,n}(T/2^n)\ge o_{n,n}(T/2^n)=1.$
Is this argument correct? If so, is there a reason why it would work in the link I provided?
We have by definition that on each interval
$$o_i(\Delta t)= \omega_i(\Delta t)\cdot \Delta t=\omega_i(\Delta t)\cdot \frac{T}{2^n}$$
with $\omega_i(\Delta t)\to 0$ as $\Delta t\to 0$ therefore
$$\sum_{i=1}^{2^n}o_i(\Delta t)=\Delta t \cdot \sum_{i=1}^{2^n}\omega_i(\Delta t)=2^no(\Delta t)$$
indeed
$$\Delta t \cdot \sum_{i=1}^{2^n}\omega_i(\Delta t)=2^n\cdot\frac{\sum_{i=1}^{2^n}\omega_i(\Delta t)}{2^n}\cdot \Delta t=2^n\cdot\omega(\Delta t)\cdot \Delta t$$
and
$$\omega(\Delta t)=\frac{\sum_{i=1}^{2^n}\omega_i(\Delta t)}{2^n} \to 0$$
indeed since for any $\epsilon >0$ eventually for some $N\ge n$ we have that $\forall i, \;|\omega_i(\Delta t)|<\epsilon$
$$\left|\frac{\sum_{i=1}^{2^n}\omega_i(\Delta t)}{2^n}\right|\le \frac{\sum_{i=1}^{2^n}|\omega_i(\Delta t)|}{2^n}\le \frac{2^n\cdot \epsilon}{2^n}=\epsilon$$