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 $y_2 \le l \le y_1$ ?


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 attached shape Does that change my problem?

  • $\begingroup$ Welcome to Math.SE! I have tried to improve the readability of your question by improving the $\rm \LaTeX$ code of the question. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. I've edited away the $\times$ sign since it can be confused with the variable $x$ in your question. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Apr 27 '19 at 15:51
  • $\begingroup$ @gnusupporter-8964民主女神-地下教會 Thank you very much. $\endgroup$ – Angie Apr 27 '19 at 16:10

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 $u=y_2$.

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  • $\begingroup$ Thank you very much, and sorry for my weak mathematics background, 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 bound and lower bound intersect in one point, does that change anything? $\endgroup$ – Angie May 2 '19 at 8:12
  • $\begingroup$ @Angie Now the situation looks even worse because the graph suggests that the domain $D$ between the upper and lower bound is bigger. And if $D$ contains a square $S$ with two opposite vertices (s,s) and $t$ placed at the line $x=y$ then we can reach in a sequence $s\le u_n\le t$ for some $n$ then we can continue it with arbitrary values between $s$ and $t$. But if our case even this simple observation is not needed because the sequence from the answer still satisfies the given conditions but does not converge. $\endgroup$ – Alex Ravsky May 2 '19 at 20:31
  • $\begingroup$ Thanks a lot, I choose the second lower boundary which intersects with the upper one, just to say that $u_n$ is surely bounded. but can I describe the behavior of $u_n$ relative to the intersection points with y=x? In order to understand the direction of $u-n$ I computed $u_{n+1}-u_n$ where $f_2(u_n)-u_n \le u_{n+1}-u_n \le f_1(u_n)-u_n$ this amount will change its sign outside the area bounded by intersection points, so, is it correct to say that for some n $y_2 \le u_n \le y_1$? $\endgroup$ – Angie May 3 '19 at 6:49
  • $\begingroup$ @ Almost correct. I updated my answer. $\endgroup$ – Alex Ravsky May 4 '19 at 7:24
  • $\begingroup$ I couldn't figure out why $\{u_m : m \ge n\}$ is monotonic , It is a key point, but not clear for me. $\endgroup$ – Angie May 4 '19 at 10:11

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