13
$\begingroup$

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 $f$.

$\endgroup$
11
  • 1
    $\begingroup$ why can't you just have shrinking triangles to $0$? $\endgroup$ Sep 14, 2022 at 17:32
  • 2
    $\begingroup$ 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, 2022 at 19:47
  • 2
    $\begingroup$ 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, 2022 at 19:48
  • 1
    $\begingroup$ @Christian: I don't understand your counterexample: doesn't it satisfy precisely $f\circ f=0$? $\endgroup$
    – Nicolast
    Sep 14, 2022 at 19:52
  • 2
    $\begingroup$ 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, 2022 at 20:29

1 Answer 1

7
$\begingroup$

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 $\delta$.

$\endgroup$
8
  • 1
    $\begingroup$ 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, 2022 at 3:09
  • 2
    $\begingroup$ @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, 2022 at 4:38
  • 2
    $\begingroup$ $f(\overline{I_0}) \ni 0$ 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, 2022 at 4:40
  • 2
    $\begingroup$ @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, 2022 at 6:39
  • 1
    $\begingroup$ @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, 2022 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.