Let $f:[-1,1]\to\mathbb{R}$ be a $C^2-$function with $f(0)=0$, $f^{\prime}(0)=1$ and $f^{\prime}(x)>1$ for all $x>0$. Let $a_1\in (0,1)$ and define recursively $a_{n+1}$ by equation $f(a_{n+1})=a_n$. Prove that $a_n\to 0$ but $\sum a_n = \infty$.

I was doing this exercise but I am kind of confused by the definition of the sequence, we should have the injection for this and I cannot say anything about the function around $-1$. If I consider $a_n \in (0,1)$ it works well and I can prove that $a_n\to 0$ , with the respective conditions, but I also know that can exist negative points $a_i$ near $0$ for some $i$ so it becomes a problem. And what about $\sum a_n = \infty$? I am out of ideas for this.

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    $\begingroup$ How do you know that $a_n\in[-1,1]$? If it isn't, what is $f(a_n)$? $\endgroup$ – Don Thousand Feb 23 at 16:25
  • $\begingroup$ From the given conditions, $f(x)>x$ for $x>0$, hence $a_n$ should be divergent (or at least sooner or later leave $[-1,1]$. $\endgroup$ – Hagen von Eitzen Feb 23 at 16:26
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    $\begingroup$ This has been asked (and answered) recently: math.stackexchange.com/q/4035021/42969. $\endgroup$ – Martin R Feb 23 at 16:26
  • $\begingroup$ @HagenvonEitzen: Note that the (somewhat unusual) recursion is $f(a_{n+1})=a_n$, so $(a_n)$ is decreasing. $\endgroup$ – Martin R Feb 23 at 16:27
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    $\begingroup$ If $a_n \in (0,1)$ is not given then the equation $f(a_{n+1})=a_n$ may have multiple solutions for $a_{n+1}$ and the sequence is not well-defined. $\endgroup$ – Martin R Feb 23 at 16:36

As Martin R says, it’s quite necessary to assume $0 < a_n < 1$, and then $a_n$ decreases to zero. Behavior of the series $a_n$ is (of course) trickier.

We know by Taylor that $a_n=f(a_{n+1})=a_{n+1}+\int_0^{a_{n+1}}{tf’’(a_{n+1}(1-t))dt}=a_{n+1}+a_{n+1}^2(c +o(1))$ where $c=f’’(0)/2$. Thus $a_{n+1}^{-1}-a_n^{-1}=a_{n+1}^{-1}(1-(1+(c+o(1))a_{n+1})^{-1})=\frac{(c+o(1))a_{n+1}}{a_{n+1}(1+(c+o(1))a_{n+1})}=c+o(1)$ (so $c \geq 0$). It follows that $0<a_n^{-1}=O(n)$ so $a_n \geq C/n$ for some positive constant $C$.


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