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Let $f:[0,1]\to[0,1]$ be a smooth, convex (downward) function satisfying $$ f(0)=f(1)=1,\quad \lim_{x\to 0}f'(x)=-\infty,\quad \lim_{x\to 1}f'(x)=+\infty. $$

I am confident to be able to argue that $f$ has exactly two fixed points in $[0,1]$ (one of them being $1$, of course.)

I would like to show that for any starting value $x\in (0,1)$, the sequence of function iterates $f(x), f(f(x)),\ldots$ converges to the fixed point which is not $1$.

I know from the convexity of $f$ that there exist $0<x_-<x_+<1$ such that $f'(x_\pm)=\pm1$ and that $f$ on the interval $(x_-,x_+)$ is non-expansive.

I was thinking to try and argue that for any starting value the iterates $f^i(x)$ would eventually lie in $(x_-,x_+)$ and to then apply Banach's fixed point theorem.

My questions are:

  • Is it clear that the fixed point lies in the interval $(x_-,x_+)$? (I doubt it)
  • In order to apply Banach's fixed-point theorem, would I have to show that $f((x_-,x_+))\subset (x_-,x_+)$?
  • Is there a different approach that would guarantee convergence of the function iterates without checking additional conditions?

Thank you.


Thanks to the efforts of richard and froggie it now seems that convergence of the iterates cannot be guaranteed under the conditions specified above.

I would therefore like to add the following assumptions: ($p$ denotes the fixed point which is not $1$)

  • $-1<f'(p)<1$.
  • If $c=\min_x f(x)$, then $-1<f'(c)<1$.

I think that with these additional assumptions it should be possible to prove convergence of the function iterates from every starting point.

share|cite|improve this question
Do functions with these properties exist? – user108903 Dec 16 '12 at 15:30
Yes, the maybe easiest example would be $f(x)=2-\sqrt{x}-\sqrt{1-x}$. – Eckhard Dec 16 '12 at 15:35
Oops, I misread your question. Thanks for the example. – user108903 Dec 16 '12 at 15:41
Do you assume that $f$ is differentiable on $(0,1)$? – 23rd Dec 16 '12 at 15:52
Yes. In fact, I'm happy to assume the existence of as many derivatives as necessary. I added smoothness to the original question. – Eckhard Dec 16 '12 at 15:57
up vote 1 down vote accepted

Since $f(1)=1$ and $\lim_{x\to 1}f'(x)=+\infty$, it is easy to see that there exists $a\in(0,1)$, such that $f(a)<a$. Define $g(x)=f(x)-x$ on $[0,1]$. Since $g(0)=1>0$ and $g(a)<0$, there exists $p\in[0,a]$, such that $g(p)=0$, i.e. $f(p)=p$. Since $g(p)=g(1)=0$ and $g$ is convex on $[p,1]$, either $g\equiv 0$ on $[p,1]$ or $g(x)<0$ on $(p,1)$. The former case cannot happen, because $\lim_{x\to 1}g'(x)=+\infty$. Therefore, $f$ has a unique fixed point $p$ in $[0,1)$.

Unfortunately, it could happen that $f'(p)<-1$. In this situation, the iteration of $f$ cannot converge to $p$.

When $-1<f'(p)<1$, note that the iteration of $f$ on $(p-\delta,p+\delta)$ converges to $p$ for some $\delta>0$. Then we can define $I=(l,r)$ to be the maximal interval containing $p$ such that the iteration of $f$ on $I$ converges to $p$. By definition, $f(I)\subset I$. Since $I$ is maximal, $f(l),f(r)\notin I$, i.e. $f(l),f(r)\in\{l,r\}$. Then there are two cases: $l=0$ and $r=1$ or $f(l)=r$ and $f(r)=l$. For the latter case, by the maximality of $I$, we can conclude that $f'(r)< 0$, and hence $f'(p)<0$. Moreover, due to $f(I)\subset I$, we know that $f'(l)f'(r)\ge 1$.

share|cite|improve this answer
Thank you for your answer, Richard. Would the condition $-1<f'(p)<1$ be sufficient for the iteration of $f$ to converge to $p$ for any starting value? In the notation of my original question this would mean $x_-<p<x_+$. – Eckhard Dec 16 '12 at 16:32
@Eckhard: I have updated my answer for $-1<f'(p)<1$. I avoid to use your notations $x_\pm$, because the point $x$ for $f'(x)=\pm 1$ may not be unique. – 23rd Dec 16 '12 at 17:20
Thanks again. I don't quite see yet why the maximal interval $I$ must be open, but otherwise the argument seems compelling. – Eckhard Dec 16 '12 at 17:28
@Eckhard: It is because if $f^n(x)\to p$, then there exists a neighborhood of $x$ converging to $p$ under the iteration of $f$. – 23rd Dec 16 '12 at 17:33
@Eckhard: I agree with you. Moreover, I think only $f'(c)\ge -1$ is sufficient to exclude the situation $f(l)=r,f(r)=l$. The reason is as follows. Letting $f(x)=c$, then $f'(x)=0$ and $f$ is decreasing on $[0,x]$. Therefore, $f'(x)>f'(r)\Rightarrow r\le x\Rightarrow l=f(r)\ge f(x)=c\Rightarrow f'(c)\le f'(l)<-1$. – 23rd Dec 17 '12 at 8:15

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