Showing that this particular sequence of functions defined by recursion is point-wise bounded

Question

Let $(s_n)_n$ be a summable sequence. Consider the following sequence of functions $(z_n)_{n\geq 0}$, $z_n:[0,+\infty)\rightarrow\mathbb{R}$, defined recursively as \begin{align*} z_0(x)&\equiv 0\\ z_1(x)&=x\\ z_{n+1}(x)&=\sqrt{z_{n}(x)}\bigg(\sqrt{z_{n-1}(x)}+s_n\bigg) \end{align*} It is sufficient for my purposes to show that such sequence is pointwise bounded. Since it is pointwise non-negative (by induction) with an upper bound I would conclude. Notice that I don't need a uniform bound (which in fact does not exist in my opinion), instead I need to show that \begin{equation*} \forall \,x\geq 0\quad \exists\quad L=L(x)>0\quad\text{s.t.} \quad\lim_{n\to+\infty} z_n(x)<L. \end{equation*} Obviously every map of the sequence is non-decreasing, as one can check by induction. First I tried some numerical calculation in order to explore its behavior (I still didn't know at the time if it was bounded or not) and it seems that it admits a pointwise limit to an unbounded function. Here the graph. In this graph on the x-axis we have the integer value $n$ of $z_n$ while on the y-axis are plotted the values of the $z_n(x)$ for several values of $x$ (in this case the values are $1,\dots,10$). As you can see each "orbit" seems to approach a limit value. The sequence $(s_n)_n$ is actually defined $s_n=2^{-\frac{2}{3}\beta(n+1)}$ for $\beta>0$ (in the graph $\beta=0.7$). I want to prove the result for this specific sequence (and for every value of $\beta>0$) but I'm confident that it holds for every summable sequence $s_n$. Of course the value of the limit depends on $s_n$, and in particular for small values of $\beta$ the convergence is slow, but I still observed it numerically (in this case the limit has a big value). I tried different approaches, e.g. evaluating the difference \begin{align*} |z_{n+1}(x)-z_{n}(x)|=\left|\sqrt{z_{n}(x)}\bigg(\sqrt{z_{n-1}(x)}+s_n\bigg)-\sqrt{z_{n-1}(x)}\bigg(\sqrt{z_{n-2}(x)}+s_{n-1}\bigg)\right| \end{align*} but I can't do any progress in this direction. I also tried to take the logarithm and see if some tricks are possible \begin{equation*} \log z_{n+1}=\frac{1}{2} \log z_n(x)+\log(\sqrt{z_{n-1}(x)}+s_n) \end{equation*} but my attempts were unsuccessful. Also computed the ratio, which for $n$ sufficiently large, neglecting $s_n$, we can approximate as \begin{equation*} \frac{z_{n+1}(x)}{z_n(x)}=\frac{\sqrt{z_{n}(x)}\bigg(\sqrt{z_{n-1}(x)}+s_n\bigg)}{\sqrt{z_{n-1}(x)}\bigg(\sqrt{z_{n-2}(x)}+s_{n-1}\bigg)}\approx \sqrt\frac{z_{n}(x)}{z_{n-2}(x)} \end{equation*} but nothing comes to my mind at this point. I need to formalize this result in order to prove the existence of so called self-similar solutions for a dyadic model of turbulence, which is the topic of my master thesis. Thanks to anyone who can give any suggestions or ideas.

Answer 1

The main Idea is to bound $$z_{n+1}(x)\le \max\{x,1\}\prod_{i=1}^n(1+s_i)$$ by induction, and then prove that $\prod_{i=1}^n(1+s_i)$ converges iff $s_n$ are summable.

(Here I'm supposing everything is positive. Otherwise, work with $|z_n|$ and $|s_n|$)