Convergence of fixed-point iteration for convex function 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.

Edit:
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.
 A: 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 $f'(l)f'(r)\ge 1$.
