Consider a strictly positive sequence $\{u_n\}$ of real numbers (i.e. $\forall n, u_n >0$). Suppose that $\displaystyle \lim_{n \to \infty} u_n = 0$ and that $\{u_n\}$ is not decreasing.
Consider the sequence $\{l_n\}$ defined by : $\forall n \in \mathbb N, l_n := \min\left\{i-n,\text{ such that } u_i < u_n,\forall i> n\right\}$.
This sequence counts the number of iterations needed before a term of the sequence $\{u_n\}$, smaller than the current term, appears. In a previous post, I wondered whether such a sequence could be said to be bounded (when $\{u_n\}$ is not decreasing), but the answer is no: I've been given a counter-example. (See the post).
I'm now wondering what kind of conditions need to be added to the sequence $\{u_n\}$ for the sequence $\{l_n\}$ to become bounded.
At first, I thought of a "Cauchy sequence" condition, but I have the impression that the counterexample $\{u_n\}$ given in the previous post is indeed a Cauchy sequence, yet the associated sequence $\{l_n\}$ is not bounded.
I'm really having trouble visualizing the kind of condition that would imply boundedness on $\{l_n\}$, except when the $\{u_n\}$ sequence is decreasing. (In fact, if $\{u_n\}$ is a decreasing sequence, it is clear that $\forall n \in \mathbb N, l_n = 1$, so in this case, the sequence is bounded). Do you have any clues?
Given that $u_n>0$ converges to $0$ but isn't a decreasing sequence,
each of the following is equivalent:
"$l_n$ is bounded"
"there exists $K$ such that $\inf B_{j+1}<\inf B_j$ for all $j$, where $B_j$ denotes $\{u_{jK},u_{jK+1},\cdots,u_{(j+1)K-1}\}$"
"there exists $K$ such that $c_{j+1}<c_j$ for all $j$, where $\displaystyle c_j$ denotes $\inf_{n\leq jK}u_n$"
"the gap between any pair of consecutive elements of $T$ is bounded, where $T$ denotes the set of integers $\tau$ such that $u_i>u_\tau$ whenever $i<\tau$"
Not too hard to prove.