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?

  • $\begingroup$ Refer to Primer of Real Functions by R P Boas for this and many other applications of Baire Category Theorem. $\endgroup$ May 25 at 23:13
  • $\begingroup$ @geetha290krm Thanks! Do you have a precise reference? $\endgroup$
    – Michael
    May 26 at 9:00
  • $\begingroup$ @Nick That's the obvious thing to do but it is not clear to me how to show it is open $\endgroup$
    – Michael
    May 26 at 9:01
  • $\begingroup$ @Nick The $A_k$s are closed, not necessarily open, precisely because of continuity and because $A_k = \left(f^{(k)}\right)^{-1}(\{0\})$ with $\{0\}$ closed. Showing that one of them is open (in $[0,1]$) would solve the question using connectedness though. $\endgroup$
    – Bruno B
    May 26 at 9:36
  • 2
    $\begingroup$ I have a proof now. Thanks everyone $\endgroup$
    – Michael
    May 28 at 20:19

1 Answer 1


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.


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