Let $f:[0,1]\to [0,1]$ be a continuous function with a unique fixed point $x_{0}$

Assume that $\forall x\in [0,1], \exists n\in \mathbb{N}$ such that $f^{n}(x)=x_{0}$. Does this implies that there is a uniform $n$ with this property?

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    $\begingroup$ Does there even exist a non-constant such $f$? $\endgroup$ – user7530 Dec 24 '14 at 1:57
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    $\begingroup$ @user7530 yes for example $\begin{cases} -x+1 &0<x<1/2\\1/2 & 1/2<x<1 \end{cases}$ $\endgroup$ – Ali Taghavi Dec 24 '14 at 4:57
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    $\begingroup$ Period three implies chaos is possibly relevant. Though, I'm not sure how to incorporate the assumption that $f$ has only one fixed point. Can we make a function with one fixed point and a period of size three? If so, the theorem implies that there is a period of every order, hence your conjecture is false. $\endgroup$ – 6005 Dec 30 '14 at 9:09
  • $\begingroup$ A better explanation of the theorem is here. $\endgroup$ – 6005 Dec 30 '14 at 9:10
  • $\begingroup$ In fact, I believe it suffices to find a function with a unique fixed point and a period of size that is not a power of two. $\endgroup$ – 6005 Dec 30 '14 at 9:11

We can separate the proof into a few lemmas, each giving slightly stronger conditions on $f$. We will use Cantor's intersection theorem a few times. Note that I use the convention, in this post, that the interval $(2,1)$ is that same as $(1,2)$. As an appetizer, let's take:

Lemma 1: If $z<x_0$, then $f(z)>z$ and if $z>x_0$, then $f(z)<z$.

This follows immediately from $f(z)-z$ having but one zero and from the clear fact that $f(0)>0$ and $f(1)<1$.

Lemma 2: For any open set containing $x_0$, there must be some $c\in S$ other than $x_0$ such that $f(c)=x_0$.

We prove this by contradiction. In particular, suppose that there were some $\delta>0$ such that for any $z$, we'd have that $0<|z-x_0|<\delta$ implies $f(z)\neq x_0$. It is clear that there must be some such $z$ such that $|f(z)-x_0|\geq \delta$, as otherwise the set of such $z$ would have itself as its image, and no point therein could ever iterate to $x_0$. Further, notice that since if $z<x_0$, then $f(z)>z$, it follows that $f(z)\geq x_0+\delta$, which is the smallest value greater than $z$ which is sufficiently far from $x_0$. A similar condition holds if $z>x_0$, so we assume WLOG that $z<x_0$. Let $z$ be the greatest such number, which, since the set of such $z$ is non-empty and compact, must exist by continuity.

However, it then follows that there is some $z'$ and $n$ with $x_0<z'<x_0+\delta$ and $f^n(z')=z$, since if no such $z'$ existed, then every iterate of every $z'$ in $(x_0,x_0+\delta)$ would remain in the set $(z,x_0+\delta)\setminus\{x_0\}$ and would therefore never iterate to $x_0$. This means we have a $z'$ such that $f^{n+1}(z')\geq x_0+\delta > z'$. However, since $f^{n+1}$ is eventually constant as well, it follows that it would have to hold that $f^{n+1}(z')<z'$, since $z'>x_0$. This is a contradiction, proving the lemma.

Lemma 3: For any set of the form $(k,x_0)$, there is a $c\in(k,x_0)$ such that $f^2(c)=x_0$.

This is simple given lemma $2$; assuming WLOG that $k<x_0$ we can clearly, the image of $(k,x_0)$ under $f$ must be of the form $(a,b)$ for $a<b$ where $k<a\leq x_0$, where the lower bound comes from lemma $1$ and the upper bound comes from continuity. Clearly, if $b>x_0$, since otherwise we would have $f((k,x_0))\subset (k,x_0)$. If $a<x_0$, then the lemma follows immediately from the intermediate value theorem. So, the only difficult case is where $a=x_0$ - but then $f((k,x_0))=(x_0,b)$ and since $(k,x_0)\cup (x_0,b)$ is a punctured neighborhood of $x_0$, it follows that there is a $c$ in it such that $f(c)=x_0$, by lemma $2$. This suffices to prove lemma $3$.

Lemma 4: There exists a $k$ such that $f([x_0,k])=\{x_0\}$.

To prove this, consider the partition $P$ of the set $f^{-2}([0,1]\setminus\{x_0\})$ into connected components. We can impose a relation on sets in $P$ as follows: $$S_1\lesssim S_2 \Leftrightarrow \overline{S_1}\subseteq f(S_2)$$ that is, $S_1\lesssim S_2$ means that the closure of $S_1$ is in the image of $S_2$. Suppose we had an infinite descending chain $$S_1\gtrsim S_2 \gtrsim S_3 \gtrsim S_4 \gtrsim \ldots.$$ It follows that there must be some compact subset $S'_1$ of $S_1$ such that $f(S'_1)=\overline{S_2}$ - for instance, if we let $\overline{S_2}=[a,b]$ (as $S_2$ is connected and its closure is therefore a closed interval), then we must have $x,y\in S_1$ with $f(x)=a$ and $f(y)=b$, and setting $S'_1=[x,y]\cap f^{-1}(\overline{S_2})$ yields a compact subset of $S_1$ with the desired image, since by $S_1$ being connected we know if $x,y\in S_1$ so $[x,y]\subseteq S_1$ and intersecting it with the appropriate preimage ensures $f(S'_1)\subseteq \overline{S_2}$ and the intermediate value theorem ensures $f(S'_1)\supseteq \overline{S_2}$. We can, moreover, find a compact set $S'_i\subset S_i$ with $f(S_i')=\overline{S_{i+1}}$ by a similar construction. From here, we can write: $$S'_1\supset S'_1\cap f^{-1}(S'_2)\supset S'_1\cap f^{-1}(S'_2)\cap f^{-2}(S'_3)\supset S'_1\cap f^{-1}(S'_2)\cap f^{-2}(S'_3)\cap f^{-3}(S'_4)\supset\ldots$$ which is an infinite descending chain of non-empty compact sets (non-empty since $f(S'_1)=\overline{S_2}\supset S'_2$ and therefore $S'_1\cap f^{-1}(S'_2)$ is non-empty - and an analogous argument may be presented for all terms) and by Cantor's intersection theorem, we'd have that it contained an element $z$. This element would have $f^{n}(z)\in S_{n+1}$ and as the $S$ were disjoint from from $\{x_0\}$, that element could never iterate to $x_0$.

However, if we suppose lemma $4$ was false, we could construct an infinite descending chain, because that, alongside lemma $3$, would imply that there were infinitely many $S\in P$ contained in any interval $(x_0,k)$, implying that for any $S\in P$ there existed a $S'\in P$ with $S'\lesssim S$, which easily allows us to construct an infinite descending chain.

Lemma 5: There is an open set $S$ containing $x_0$ such that $f^2(S)=\{x_0\}$.

The proof of this follows from lemma $4$ telling us that a non-empty open set $(x_0,k)$ exists and then essentially reusing the proof of lemma $3$ to get the requisite symmetry, since lemma $3$ essentially states that if one side of the fixed point does not immediately go to the fixed point, it goes to the other side.

The proof from here is trivial; let $K=[0,1]\setminus S$. We know that $$\bigcap_{n=0}^{\infty} f^{-n}(K)=\emptyset$$ and since each $f^{-1}(K)$ is compact, by Cantor's intersection theorem, we have that there is some $f^{-n}(K)$ which is empty. Then, it follows that $f^{n+2}(x)=x_0$ for any $x$.

Remark: The above proof relies heavily on the topological properties of $[0,1]$. Even in a simple space like $S_1\cong \mathbb R / \mathbb Z$ we have counterexamples; for instance, the map, where $0\leq x < 1$: $$f(x+n)=\begin{cases}n && \text{if }x<\frac{1}2\\ n+2x-1 &&\text{if }x\geq \frac{1}2 \end{cases}$$ taken mod $\mathbb Z$ is well-defined, continuous, and has every point eventually iterating to $\mathbb Z$ (i.e. the equivalence class of $0$), but fails to have a uniform $n$, since $1-\frac{1}{2^{n+1}}$ takes $n$ steps to reach $0$. We can modify the above to work over the unit disk as, in polar coordinates: $$g(x,\theta)=\begin{cases}0 &&\text{if }0\leq \theta \leq \pi \\ (\alpha(t)\beta(\theta),2\theta-\pi)&&\text{if }\pi \leq \theta \leq 2\pi\end{cases}$$ where $\alpha$ and $\beta$ are a continuous functions with $\alpha(0)=\alpha(1)=0$ and for $0<t<1$, $0<\alpha(t)\leq 1$ and $\beta(\pi)=\beta(2\pi)=0$ and for $\pi<\theta<2\pi$ and $0<\beta(\pi)\leq 1$. By projecting onto a plane, we can extend the above to work for a closed ball of any dimension - and, moreover, by a similar construction, we can prove that any topology with an open subset homeomorphic to an open ball does not satisfy the condition. This eliminates a very wide class of topologies - so $[0,1]$ seems to be very special in the fact that it satisfies this.

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  • $\begingroup$ I think this works. Note that lemma 2 relies on the fact that $f^{n+1}$ must also have exactly one fixed point. $\endgroup$ – user7530 Dec 26 '14 at 19:43
  • $\begingroup$ I'm unsure about lemma 4. The sets $S_i$ are either open or half-open, not compact. If you take their closures when computing their intersection, couldn't the intersection point $z$ be outside of any of the $S_i$ themselves? $\endgroup$ – user7530 Dec 26 '14 at 19:51
  • $\begingroup$ @user7530 I have changed that part somewhat to reflect your comment. $\endgroup$ – Milo Brandt Dec 26 '14 at 20:04
  • $\begingroup$ "there must be some compact subset such that..." why is this true? I still don't follow. $\endgroup$ – user7530 Dec 26 '14 at 20:06
  • $\begingroup$ We know $\overline{S_2}\subseteq f(S_1)$. Since $\overline{S_2}$ is a closed interval $[a,b]$, it follows that if we choose some $x$ and $y$ in $S_1$ such that $f(x)=a$ and $f(y)=b$, then $S'_1=[x,y]\cap f^{-1}(\overline{S_2})$ satisfies the condition (since the preimage is closed and, since $S_1$ is an interval containing $x$ and $y$, it contains $[x,y]$). $\endgroup$ – Milo Brandt Dec 26 '14 at 20:10

This is not a complete answer, but shows that for some $n$ (and hence for all sufficiently large $n$), there is a non-empty open subset $U$ of $[0,1]$ such that $f^n(x) = x_0$ for every $x \in U$.

Let $A_n = \{x \in [0,1] \vert f^n(x) = x_0\}$. Since $f^n$ is continuous, each $A_n$ is closed. Since $x_0$ is a fixed point, $A_n \subseteq A_{n+1}$. Each $x \in [0,1]$ belongs to some $A_n$, so $\bigcup A_n = [0,1]$. Since $[0,1]$ is a compact metric space, in particular complete, by the Baire category theorem, some $A_n$ has a non-empty interior. Define that to be $U$.

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