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Hello fellow mathematicians,
Consider a continuous function $f:[0,1]\rightarrow [0,1]$ such that $f(0)=0$, and for every $x\in [0,1]$ there exists $k=k(x)>0$ such that the $k$th iteration of $f$ at $x$ is equal to $0$, i.e. $f^{(k)}(x)=0$. I want to show that there exists $n>0$ such that $f^{(n)}(x)=0$ for all $x\in [0,1]$.
I defined the sets $A_k=\{x\in [0,1] \mid f^{(k)}(x)=0\}$ and noticed that they are increasing and cover $[0,1]$ but it doesn't seem to help much.
Any ideas?
As already suggested in the comments, we want to follow one of the proofs for the problem where at each point some derivative is $0$.
Let $\Omega$ be the set of points $x\in [0,1]$ for which there is an interval $(a,b)\ni x$ and $k>0$ such that $f^{(k)}(y)=0$ for every $y\in (a,b)\cap [0,1]$. Following one of the proofs from the post we can use Baire's theorem to deduce that $\Omega$ is dense in $[0,1]$. The next step is show that $[0,1]\setminus \Omega$ has no isolated points.
If $x$ is such that there exist $a,b,n_1,n_2$ such that $f^{(n_1)}(a,x)=\{0\}$ and $f^{(n_2)}(x,b)=\{0\}$, then $f^{(n)}(a,b)=\{0\}$ by continuity, where $n=\max\{n_1,n_2\}$. However, it could be that $x$ is an isolated point and $(a,x)\cup (x,b)$ can be covered by infinitely many intervals, where at each of them some iteration of $f$ is $0$. At this point we take cases.
First case: There are $n_1,n_2$, overlapping $(a,b),(c,d)$ and $x\in (a,b)\cup (c,d)$ such that $f^{(n_1)}(a,b)=\{0\}$, $f^{(n_2)}(c,d)=\{0\}$ and $f^{(m)}(x)\neq 0$, where $m=\min \{n_1,n_2\}$. In this case, let $r=|n_1-n_2|$, and note that we can find $\varepsilon>0$ such that $f^{(r)}(0,\varepsilon)=\{0\}$ by looking at the last point $y$ where $f^{(m)}(y)=0$ and using continuity. Now for every $x\in [0,1]$ we can find $k>0$ and $(a,b)\ni x$ such that $f^{(k)}(y)<\varepsilon$ for every $y\in (a,b)\cap [0,1]$. By compactness we can get a finite subcover $[0,1]\subset\cup_{i=1}^N (a_i,b_i)$, $N<\infty$, and $k_i>0$ such that $f^{(k_i)}(x)<\varepsilon$ for every $x\in (a_i,b_i)\cap [0,1]$. Then $f^{(k_i+r)}(x)=0$ for every $x\in (a_i,b_i)\cap [0,1]$. Then we just take the maximum.
Second case: The following is true for every $n_1,n_2$ and overlapping $(a,b),(c,d)$. If $f^{(n_1)}(a,b)=\{0\}$ and $f^{(n_2)}(c,d)=\{0\}$, then $f^{(m)}(x)=0$ for every $x\in (a,b)\cup (c,d)$, where $m=\min \{n_1,n_2\}$. In this case we are done regarding isolated points by the argument above.
