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
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
. - If $c=\min_x f(x)$, then $-1
.
I think that with these additional assumptions it should be possible to prove convergence of the function iterates from every starting point.