Prove that the iterate $f^{n}$ is a constant function.

Question

Let $A \subset \mathbb{R}$ be a finite set with $|A| = n$ and let $f : A \to A$ satisfy the strict contraction condition $|f(x) - f(y)| < |x - y|$ for all $x \neq y$ in $A$. Prove that $f$ is not injective and that the iterate $f^{n}$ is a constant function.

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My idea

I was able to show that $f$ is not injective, but not that the iterate is a constant function. So, I let the $n$ elements of $A$ be $x_1 < x_2 < \dots<x_n$ and I noted $y_i=f(x_i)$ for every $i \in \{ 1, \dots n \}$. I suppose that $f$ is injective, and because $f: A \to A$ and A is finite with $n$ elements, we can say that $f$ is bijective. If we note $k$ be the smallest difference in module between any $2$ elements of $A$, we can let $x_i$ and $x_j$ be those elements, then we get that $|y_i-y_j|<|x_i-x_j|=k$ and $y_i, y_j$ are also elements from $A$, which is a contradiction of our suppose so we get that $f$ is not injective. Now I don't know what I should to do to show that the iterate is constant. Hope one of you can help me find a solution without orbits (I've seen these kind of problem use orbits).

Answer 1

Claim I Let $A \subset \mathbb{R}$ be a finite set with $|A| > 1$ and let $f : A \to A$ satisfies the strict contraction condition $|f(x) - f(y)| < |x - y|$ for all $x \neq y$ in $A$. Then $|f(A)|< |A|$. In particular $f$ is not injective.

Proof. Let $x,y\in A$, $x\neq y$ be such that

$$|x-y|=m:=\min_{\substack{w,z\in A\\ w\neq z}} |w-z|.$$

Then $|f(x)- f(y)|< |x-y|=m$, that implies $f(x)=f(y)$. So $f$ is not injective. Since $A$ is a finite set this implies that $f$ is not surjective. So $|f(A)|< |A|.$

Claim II. If $|A|=n\geq 2$ we have $|f^{n-1}(A)|=1$.

Proof. Note that $f^{k+1}(A)=f(f^k(A))\subset f^k(A)$, and in particular we can consider the restriction

$$f\colon f^k(A) \rightarrow f^k(A).$$

If there is $k\leq n-2$ such that $|f^k(A)|=1$ then $f^{n-1}(A)\subset f^k(A)$ contains only a point and we are done. So suppose that $|f^k(A)|> 1$ for every $k\leq n-2$. By Claim I

$$|f^{k+1}(A)| < |f^k(A)|,$$

Since $|A|=n$ it follows by induction that $|f^{k+1}(A)|\leq n-(k+1)$, so in particular $|f^{n-1}(A)|\leq 1$. Since this set is non-empty it follows that $|f^{n-1}(A)|=1$.

Claim III. for $n\geq 2$ we have $f^{n-1}$ is constant.

Proof. The image of $f^{n-1}$ is $f^{n-1}(A)$, that contains just a point, so $f^{n-1}$ is constant.

Claim IV. for $n\geq 1$ we have $f^{n}$ is constant.

Proof. For $n=1$ this is obvious. For $n\geq 2$ we have that $f^{n-1}$ is constant, so $f^n$ is constant.

Answer 2

This is trivial if $n = 1$. If $n > 1$, then you have shown that $f$ is not injective, and hence it is not surjective (do you see why?). Hence, there is some $a \in A$ that is not in the range of $f$ and we can consider the restriction $g: A' \to A'$ of $f$ to $A' = A \setminus \{a\}$. By induction, $g^{n-1}$ is a constant function: say $g^{n-1}(x) = b$ for some $b$ and any $x \in A'$. But that implies that $f^{n-1} [A] = \{f^{n-1}(a), b\}$ and then, we can only have that $f^n[A] = \{b\}$. To see this, note that $f(a) \in A'$, because it is in the range of $f$ and hence is not equal to $a$. And then $f^n(a) = f^{n-1}(f(a)) = b$, because $f^{n-1}(x) = g^{n-1}(x) = b$ for every $x \in A'$.

Answer 3

Put $x_1<x_2<\cdots<x_n$. For the injectivity, since the set is finite an injective function will also be surjective and $|x_n-x_1|$ is the biggest distance realized in the set. Since $f$ is surjective we have $f(x_i)=x_1$ and $f(x_j)=x_n$ for some $i$ and $j$ and we have that $|f(x_i)-f(x_j)|=|x_n-x_1|>|x_i-x_j|$ which is a contradiction.

Since $A$ is finite there is a shortest (nonzero) distance realized and iterating $f$ strictly reduces distances so eventually the distance between two iterates will be less then the minimal distance so it must be zero. This means that for some $k$ the iterate $f^{(k)}$ is constant. The worst case that can happen is that we started with two elements that realize the biggest distance ($x_n$ and $x_1$) and that the closest elements in the sets are $x_1$ and $x_2$ (or equivalently $x_{n-1}$ and $x_n$). It will take at most $n$ steps so $f^{(n)}$ will be constant.

Answer 4

By assumption $f(A) \subset A$, so $f(f(A)) \subset f(A)$, and inductively $f^{k+1}(A) \subset f^k(A)$ for each $k$. So the sets $f^k(A)$ are nested in each other.

In addition the condition $|f(x) - f(y)| < |x - y|$ for $x \neq y$ will still hold when $f$ is restricted to any $f^k(A)$. So the assumptions of the problem hold on $f^k(A)$ for each $k$. Since you have shown that in this situation $f$ is not injective whenever the set it's acting on has at least two elements, the cardinality of $f^k(A)$ drops by at least one on each iteration until the set no longer has at least $2$ elements. In other words, $f^{k+1}(A) \subsetneq f^k(A)$ with each iteration until $f^k(A)$ has one element, after which subsequent $f^k(A)$ are this same element.

Since the original $A$ has $n$ elements, this can take at most $n-1$ iterations. Thus $f^k(A)$ has one element for all $k \geq n -1$. Call this element $a$. Since $n > n-1$, $f^n$ has the constant value $a$ on $A$.

Answer 5

Consider the point $x$ which maximizes the function $\operatorname{min}_{y\in A}\{d(x,y)\}$. After one iteration of $f$, this point should be vacant and at each later iteration should remain vacant. To give a little more insight, the point $x$ has to be vacant after the first step on, because any point that maps to $x$ basically increases the distance to other points. Hence, either all points are mapped there which finishes the proof or it remains vacant. Thus, at each iteration the number of occupied slots decreases. This works even if we take $n$ different contraction and apply them in any way possible.