I have a recurrence of the form
$u_0=0, u_1=50, u_n=-a_{n-1}+u_{n-1}+50$, where $a_{n-1}$ is a probabilistic amount which I can not describe in a simple formula. But I can set bounds for the original recurrence in the following way:
$u_n\le-0.001 u_{n-1}^2+u_{n-1}+50$, and
$u_n\ge-0.001 u_{n-1}^2+0.9 u_{n-1}+45$
Plotting the two functions which correspond to the right side of each inequality; i.e.
$f_1(x)=-0.001 x^2+x+50$ and
$f_2(x)=-0.001 x^2+0.9 x+45$
I get the attached figure, which shows $f_1$ in blue and $f_2$ in red, and function $f_3(x)=x$ in black.
Now, my question is:
Say that $f_1$ intersects with $f_3$ in $(x_1,y_1)$, and $f_2$ intersects with $f_3$ in $(x_2,y_2)$
Is it correct to say that (based on Banach fixed point theorem) $u_n$ is convergent, and its limit $l$ satisfies $y_2 \le l \le y_1$ ?
Update:
recently, I was able to find another lower bound of the form $u_n \ge f_3(u_{n-1}) = 0.001 u_{n-1}^2+0.005 u_{n-1}+50$
Now upper and lower bounds intersect like the 
Does that change my problem?
Without additional restrictions imposed on the sequence $\{a_n\}$, the sequence $\{u_n\}$ is not necessarily convergent. For instance, a sequence $0, 50, 90, 125, 150, 170, 171, 170, 171,\dots $ satisfies $f_2(u_{n-1})\le u_n \le f_1(u_{n-1})$ for each $n\ge 1$, but does not converge.
Let’s investigate the behavior of the sequence $\{u_n\}$. Condition $f_3(u_n)\le f_1(u_n)$ implies $0\le u_n\le 497.5$. We have $$[y_2, y_1]=[50\sqrt{19}-50,100\sqrt{5}]~\simeq [168, 224].$$ If $u_n\le y_1$ then $f_1(u_n)<y_1$ so $u_m<y_1$ for all $m\ge n$. If $u_n<y_2$ then $u_{n+1}>u_n$, so $u_{m+1}>u_{m}$ for all $m\ge n$ or $u_m\ge y_2$ for some $m\ge n$. In both cases $u_m<y_1$ for all $m\ge n$. In the first case, moreover, $\{u_m:m\ge n\}$ is a monotonic bounded sequence, which, therefore, has a limit $u\le y_2$. If $u<y_2$ then $f_2(u)>u$ so, by continuity of the function $f_2$, there exists $n$ such that $u_{n+1}\ge f_2(u_n)>u$, a contradiction. So $u=y_2$.