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Let $a_1=\frac 2 3 , \ a_{n+1}=a_n-a^2_n$ for $n\ge 1$, $a_n$ is monotonically decreasing and bounded: $0\le a_{n+1} \le 1$ and $\displaystyle\lim_{n\to\infty} a_n=0$.

Is there a uniformly continuous function $f:\mathbb R \to \mathbb R$ such that $a_{n+1}=f(a_n), \ \forall n\in \mathbb N$ ?

I don't know how to start here but I'm pretty sure there is such function since $a_n$ is converging then from the definition of uniform continuity, we'll have $\forall \epsilon >0 :\exists \delta >0: \forall n,m\in \mathbb N: |a_n-a_m|<\delta\Rightarrow |f(a_n)-f(a_m)|<\epsilon$

$a_{n+1}=f(a_n)=f(a_{n-1}-a_{n-1}^2)$ so I need to find a function that will "push the index" by 1, like $f(x)=x+1$ and is uniformly continuous, but I have no idea how to find such function.


marked as duplicate by Guy Fsone, tomasz, The Phenotype, Cameron Williams, Namaste calculus Jan 28 '18 at 1:42

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  • $\begingroup$ The function $f(x) = x - x^2 + C$ looks like an interesting candidate of a general function that satisfies the relation $a_{n+1} = f(a_n)$. $\endgroup$ – DanZimm Jun 16 '14 at 12:11

Isn't $f(x)=x-x^2$ what you are looking for?

Edit: This function is uniformly continuous on $[0,1]$ as a continuous function on the compact interval $[0,1]$. As noticed by DanZimm, this function is unfortunately not uniformly continuous on the unbounded set $\mathbb{R}$. But as proposed by Daw and Ihf you may just define piecewise your function outside of the interval of interest to solve this little problem. That is considering for $a<b$, $a,b \in \mathbb{R}$ the function

$g(x) = \left\{ \begin{array}{l l} f(a) & \text{if } x \leq a\\ f(x) & \text{if } a \leq x \leq b \\ f(b) & \text{if } x \geq b \end{array} \right.$

In particular you said that $0 \leq a_n \leq 1$ in your question so it makes sens to choose $a=0$ and $b=1$ which leads to the answers of the other posts. It is now clear that the function $g$ is continuous on the interval $[a-1,b+1]$ and constant everywhere else thus uniformly continuous on $\mathbb{R}$.

  • $\begingroup$ This is not uniformly continuous on $\mathbb{R}$. $\endgroup$ – DanZimm Jun 16 '14 at 12:13
  • $\begingroup$ True but it is on $[0,1]$ as continuous function on a compact interval. $\endgroup$ – Surb Jun 16 '14 at 12:14
  • $\begingroup$ @DanZimm, your question is if there is such a function. I believe this is the only candidate... $\endgroup$ – vonbrand Jun 16 '14 at 12:16
  • $\begingroup$ I think vonbrand is correct. But I also think that proving that it is the only candidate could be little harder... $\endgroup$ – Surb Jun 16 '14 at 12:18
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    $\begingroup$ @GinKin Well, you can see it like this. Yes in the answers of Ihf and daw, they took $a=0$ and $b=1$ but note that every $a\leq 0$ and $b\geq 1$ will fulfill the assumptions. You just "cut" the function somewhere else. I wanted to emphase the way you can construct the function to get uniforme continuity starting from only continuity. $\endgroup$ – Surb Jun 16 '14 at 12:41

What about $$ f(x) = \max(x-x^2,\ 0)? $$ It is globally Lipschitz continuous, hence uniformly continuous.


How about $f(x)=x-x^2$ for $x\in [0,1]$ and $0$ otherwise?

  • $\begingroup$ That's the same function proposed by @daw. $\endgroup$ – lhf Jun 16 '14 at 12:20
  • $\begingroup$ What is left is showing the function is bounded above on [0,1] and since $a_n$ is also bounded on [0,1] we'll be done ? $\endgroup$ – GinKin Jun 16 '14 at 12:25
  • $\begingroup$ @lhf you were ~ 15s faster :) $\endgroup$ – daw Jun 16 '14 at 12:26

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