# Does locally nilpotent imply nilpotent for continuous self-maps of intervals?

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 $$f$$.

• why can't you just have shrinking triangles to $0$? Sep 14 at 17:32
• 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 $x=0$. Sep 14 at 19:47
• I am confused. @ChristianRemling, isn't $f \circ f = 0$ in your example? I am probably tired and not thinking right... Sep 14 at 19:48
• @Christian: I don't understand your counterexample: doesn't it satisfy precisely $f\circ f=0$? Sep 14 at 19:52
• 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. Sep 14 at 20:29

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 $$f(x) for $$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 $$\delta$$.

• 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. Sep 15 at 3:09
• @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 $0$. Sep 15 at 4:38
• $f(\overline{I_0}) \ni 0$ is because the endpoints of $\overline{I_0}$ map to $0$ or the connected component $I_0$ would be larger. Sep 15 at 4:40
• @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 $0$. Sep 15 at 6:39
• @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 at 8:03