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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 $f$.

YCor
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Dattier
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    why can't you just have shrinking triangles to $0$? – mathworker21 Sep 14 '22 at 17:32
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    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 $x=0$. – Christian Remling Sep 14 '22 at 19:47
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    I am confused. @ChristianRemling, isn't $f \circ f = 0$ in your example? I am probably tired and not thinking right... – Malkoun Sep 14 '22 at 19:48
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    @Christian: I don't understand your counterexample: doesn't it satisfy precisely $f\circ f=0$? – Nicolast Sep 14 '22 at 19:52
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    @Malkoun: Yes, I misread the desired statement as $f\circ f=f$. – Christian Remling Sep 14 '22 at 20:06
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    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. – Neil Strickland Sep 14 '22 at 20:29
  • @NeilStrickland: I had similar thoughts, but I'm not sure we're getting much mileage out of an interval $(a,b)$ with $a>0$ and $f=0$ on $(a,b)$. The question might be if necessarily $f=0$ on $[0,d]$ for some $d>0$. (That would imply the claim immediately because we get closer to this set by a fixed amount during each iteration.) – Christian Remling Sep 14 '22 at 20:33
  • Lazy comment: I'm pretty sure that this is somewhat related: https://mathoverflow.net/q/34059/167834 – Alessandro Della Corte Sep 14 '22 at 21:05
  • If $f$ is increasing on an interval $[0,d]$ for some $d > 0$, then due to $f(x) \le x$ for all $x \in [0,1]$ it follows that $f^n(x) \le f^n(d)$ for all $n \ge 0$ and all $x\in [0,d]$. So there exists $k$ such that $f^k = 0$ on $[0,d]$, and thus one can conclude that $f$ is nilpotent by the argument in @ChristianRemling's comment. So when trying to construct a counterexample, one apparently needs quite ugly behaviour of $f$ close to $0$. – Jochen Glueck Sep 14 '22 at 22:21
  • @AlessandroDellaCorte The problem you link to uses the Baire Category Theorem in its solution, and to me that is the only resemblance/relevance. One might as well say that this is related to the Open Mapping Theorem for linear maps between Banach spaces, because that uses BCT... – Yemon Choi Sep 14 '22 at 22:33
  • Dédicace à Hybridex..... – Dattier Nov 14 '22 at 16:14

1 Answers1

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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 $\delta$.

YCor
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Christian Remling
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    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. – Qiaochu Yuan Sep 15 '22 at 03:09
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    @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 $0$. – Ville Salo Sep 15 '22 at 04:38
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    $f(\overline{I_0}) \ni 0$ is because the endpoints of $\overline{I_0}$ map to $0$ or the connected component $I_0$ would be larger. – Ville Salo Sep 15 '22 at 04:40
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    @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 $0$. – Jochen Glueck Sep 15 '22 at 06:39
  • Hmm, why does the invariance of $[0,a]$ for each $a$ imply that the zeros of $f$ accumulate at $0$? – Jochen Glueck Sep 15 '22 at 06:47
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    @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. – YCor Sep 15 '22 at 08:03
  • @YCor: Ah, I see. Thanks. – Jochen Glueck Sep 15 '22 at 08:43
  • @QiaochuYuan: I guess your concerns have been addressed already in the other comments, but here it is one more time anyway: $I_0=(a,b)$ is by definition a connected component of ${ f>0}$, so $f(a)=f(b)=0$. And then $I_1\subseteq [0,b_0]$ because I want to show that $f=0$ on some $[0,d]$, so I assume, to obtain a contradiction, that there are (arbitrarily small) $I_n$ arbitrarily close to zero. – Christian Remling Sep 15 '22 at 13:39