# If for each point there is an iteration of $f$ which is equal to $0$ then some iteration of $f$ is identically $0$

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?

• Refer to Primer of Real Functions by R P Boas for this and many other applications of Baire Category Theorem. May 25 at 23:13
• @geetha290krm Thanks! Do you have a precise reference? May 26 at 9:00
• @Nick That's the obvious thing to do but it is not clear to me how to show it is open May 26 at 9:01
• @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. May 26 at 9:36
• I have a proof now. Thanks everyone May 28 at 20:19

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.