Let $f:[-1,1]\longrightarrow[-1,1]$ be a continuous function satisfying the following condition : $$\tag{1}\forall x\in[-1,1]\!\setminus\!\{0\},\quad|f(x)|<|x|.$$ Is that condition sufficient to assert that the sequence $\left(f^n\right)_{n\in\mathbb N}$ pointwisely converges to the constant function $0$ ? ($f^n$ means $f\circ\dots\circ f$, $n$ times). Because, my idea to prove the pointwise convergence is to consider the quantity $$\|f\| = \sup_{x\in[-1,1]\setminus\{0\}}\frac{|f(x)|}{|x|}$$ so by $\rm(1)$, $\|f\|\le1$. We can show that $\|f^n\|\le\|f\|^n$, so if $\|f\|<1$ then $\|f^n\|\to0$ hence $\left(f^n\right)_{n\in\mathbb N}$ pointwisely converges to $0$, otherwise if $||f\|=1$ ... I don't know. Can someone help ?
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- $\begingroup$ +1. Interesting question. I think the answer is yes if the function is required to be continuous, but no if it is not required to be continuous. $\endgroup$Adam Rubinson– Adam Rubinson2025-09-09 21:38:18 +00:00Commented Sep 9 at 21:38
- $\begingroup$ Use the same method as in Fixed point convergence for contractive mappings taking $x^\star=0$. $\endgroup$Daniel Smania– Daniel Smania2025-09-10 01:44:38 +00:00Commented Sep 10 at 1:44
- 1$\begingroup$ Let $M(x) = \max \{ \lvert f(t)\rvert : \lvert t\rvert \leqslant x\}$ for $0 \leqslant x \leqslant 1$. Then let $\mu_0 = 1$, $\mu_{n+1} = M(\mu_n)$. By the same reasoning used in the answers, $(\mu_n)$ is either strictly decreasing and converging to $0$, or there is a $k$ with $\mu_n = 0$ for all $n \geqslant k$. Thus the convergence is even uniform. $\endgroup$Dermot Craddock– Dermot Craddock2025-09-10 09:25:22 +00:00Commented Sep 10 at 9:25
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2 Answers
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2 Note that $f(0)=0$ and if $f_n(x)=f^{(n)}(x)$ we have $$|f_n(x)|=|f(f_{n-1}(x))| < |f_{n-1}(x)|$$ unless $f_{n-1}(x)=0$ .
Fix $x \ne 0$ and consider $y_n=|f_n(x)|$ which is either strictly decreasing or eventually $0$ so in any case it has a limit $y$.
Assume $y >0$ and then note that the limit points of the sequence $f_n(x)$ are $\pm y$. Say wlog $z_k=f_{n_k}(x) \to y$ so $f(z_k) \to f(y)$. But $f(z_k)=f_{n_{k}+1}(x)$ are in the sequence $f_n(x)$ so their limit points are still $\pm y$ hence $f(y)=\pm y$ and that clearly contradicts the hypothesis since we assumed $y>0$
So for all $x$ we have indeed $f_n(x) \to 0$
- $\begingroup$ It looks like our answers are similar. $\endgroup$Adam Rubinson– Adam Rubinson2025-09-09 23:21:06 +00:00Commented Sep 9 at 23:21
- $\begingroup$ @Adam I think that is the natural way to do this so no wonder $\endgroup$Conrad– Conrad2025-09-09 23:21:59 +00:00Commented Sep 9 at 23:21
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0 Let $x\in [-1,1].$ Define $f^0(x):=x.$ Since $\vert f(t)\vert < \vert t\vert\ \forall\ t,$ let $n\in\mathbb{N}$ and put $t=f^{n-1}(x),$ giving $\vert f^n(x)\vert < \vert f^{n-1}(x)\vert.$ Since this is true for all $n\in\mathbb{N},$ we see that the sequence $ \left(\vert f^n(x)\vert\right)_{n=0}^{\infty}$ is strictly decreasing.
Suppose, by way of contradiction, that $\displaystyle\lim_{n\to\infty}\vert f^n(x)\vert\neq 0.$ Then by the monotone convergence theorem there exists $\gamma>0$ such that $\displaystyle\lim_{n\to\infty}\vert f^n(x)\vert = \gamma^+.$
We have a sequence, $t_n:=f^n(x)$ such that, $\displaystyle\lim_{n\to\infty} \vert t_n\vert = \gamma$ and $\displaystyle\lim_{n\to\infty} \vert f(t_n)\vert = \gamma.$ This implies the existence of a subsequence $(t_{k_n})_{n=1}^{\infty}$ such that $\left(\displaystyle\lim_{n\to\infty} t_{k_n} = -\gamma\quad\text{ or } \quad\displaystyle\lim_{n\to\infty} t_{k_n} = \gamma\right)\quad$ and $\quad\left(\displaystyle\lim_{n\to\infty} f(t_{k_n}) = -\gamma\quad\text{ or }\quad\displaystyle\lim_{n\to\infty} f(t_{k_n}) = \gamma\right).$
One of the four cases must be true. Suppose, for example, that $\displaystyle\lim_{n\to\infty} t_{k_n} = \gamma\quad$ and $\quad\displaystyle\lim_{n\to\infty} f(t_{k_n}) = -\gamma.$ since $f$ is continuous, we must have that $f(\gamma)=-\gamma.$ Similar statements can be made for the other three cases. The point is that, in any of the four cases, we certainly have $\vert f(\gamma)\vert = \vert \gamma\vert.$
This contradicts the fact that $\vert f(x)\vert < \vert x \vert$ for all $x\in [-1,1].$ Thus $\displaystyle\lim_{n\to\infty}\vert f^n(x)\vert= 0,$ (that is, no such $\gamma$ exists), and so $\displaystyle\lim_{n\to\infty} f^n(x)=0.$ Since $x$ was arbitrary, the result follows.