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Let $f\in C([0,1],[0,1])$ be such that: $$\forall x\in [0,1], \; \exists k\in \mathbb N, \; f^{\circ k}(x)=0.$$

Is it true that $f$ is nilpotent (i.e., that there is some $k$ such that $f^{\circ k}=0$)?

Here $f^{\circ k}$ denotes the $k$th iterate of $f$.

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    $\begingroup$ why can't you just have shrinking triangles to $0$? $\endgroup$ Sep 14 at 17:32
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    $\begingroup$ What is true is that necessarily $f(0)=0$, by Sarkovski's theorem, because otherwise there would be other periodic orbits which obviously will not visit $x=0$. $\endgroup$ Sep 14 at 19:47
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    $\begingroup$ I am confused. @ChristianRemling, isn't $f \circ f = 0$ in your example? I am probably tired and not thinking right... $\endgroup$
    – Malkoun
    Sep 14 at 19:48
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    $\begingroup$ @Christian: I don't understand your counterexample: doesn't it satisfy precisely $f\circ f=0$? $\endgroup$ Sep 14 at 19:52
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    $\begingroup$ After replacing $f$ by $f^k$ for some $k$, we can assume that $f(0)=f(1)=0$. Now $f$ cannot have any fixed points apart from $x=0$, so we can apply the intermediate value theorem to $f(x)-x$ to see that $f(x)\leq x$ for all $x$, with equality only when $x=0$. The sets $Z_m=(f^m)^{-1}\{0\}$ are closed and their union is $[0,1]$, so some $Z_m$ must have nonempty interior by the Baire Category Theorem. I am not sure how much that helps. $\endgroup$ Sep 14 at 20:29

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Yes, this implies that $f$ is nilpotent.

As explained in my comment, $f(0)=0$ because otherwise Sarkovski's theorem would give us other periodic orbits which, of course, won't visit $x=0$. We also know that $f(x)<x$ for $x>0$.

Decompose the open set $\{x: f(x)>0\}=\bigcup I_n$ into its connected components. Clearly, since each set $[0,a]$ is invariant, the zeros of $f$ must accumulate at $0$. On the other hand, I claim that the $I_n$ do not accumulate at $x=0$.

Indeed, if they did, we could start out with any $I_0$ and then $f(\overline{I_0})=[0,b_0]$ for some $b_0>0$. By assumption, $I_1\subseteq [0,b_0]$ for some $I_1$. Let $K_1=\{x\in \overline{I_0}: f(x)\in \overline{I_1}\}$. Since $0\notin\overline{I_n}$ for all $n$, the orbit of any $x\in K_1$ will not yet have reached zero after one iteration.

Continue in this style: $f(\overline{I_1})=[0,b_1]\supseteq I_2$ for some $I_2$. Let $K_2 =\{x\in K_1: f^2(x)\in \overline{I_2}\}$. The compact sets $K_n$ are nested, so $\bigcap K_n\not=\emptyset$, but if $x\in K_n$, then $f^k(x)\not= 0$ for $k\le n$, so this point never reaches zero.

It follows that $f=0$ on $[0,d]$ for some $d>0$, but then everything is clear because now $f(x)\le x-\delta$ for some fixed $\delta>0$ for $x\ge d$, and each iteration brings us closer to the set $[0,d]$ by at least $\delta$.

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    $\begingroup$ I am having trouble understanding this argument (and perhaps others are too given the lack of upvotes). Why must $f(\bar{I_0})$ contain $0$, and why must there be some $I_1 \subseteq [0, b_0]$? I think I at least understand the last paragraph, and in the first paragraph we can avoid appealing to Sharkovski's theorem using Neil Strickland's idea in the comments to replace $f$ by $f^k$ as necessary. $\endgroup$ Sep 15 at 3:09
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    $\begingroup$ @QiaochuYuan: I think $\exists I_1: I_1 \subseteq [0, b_0]$ is because we are assuming (for a contradiction) that the $I_n$ do accumulate at $0$. $\endgroup$
    – Ville Salo
    Sep 15 at 4:38
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    $\begingroup$ $f(\overline{I_0}) \ni 0$ is because the endpoints of $\overline{I_0}$ map to $0$ or the connected component $I_0$ would be larger. $\endgroup$
    – Ville Salo
    Sep 15 at 4:40
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    $\begingroup$ @QiaochuYuan: Re first paragraph, actually we don't need to replace $f$ with $f^k$, either: by the intermediate value theorem $f$ has a fixed point, and clearly $f$ cannot have any fixed point different from $0$. $\endgroup$ Sep 15 at 6:39
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    $\begingroup$ @JochenGlueck Otherwise for some $a>0$ you get the restriction of $f$ as $g:[0,a]\to [0,a]$ locally nilpotent with $g^{-1}(\{0\})=\{0\}$, and this is clearly absurd. $\endgroup$
    – YCor
    Sep 15 at 8:03

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