Uniform convergence to $0$ of the $n$-th iterate of $f$

Question

Let $a>0$ and $f:[-a,a]\to[-a,a]$, continuous, such that $$\forall x\in[-a,a]-\{0\}:\left\vert f(x)\right\vert <\left\vert x \right\vert.$$

Denote $f^n$ the $n$-th composition of $f$. We have to show that $f_n$ converges uniformly to the zero function.

If $x_0\ne 0$ and $\forall n\in\mathbb{N}^* f_n(x_0)\ne 0$ then the sequence $(|f_n(x_0)|)_{n\in\Bbb{N}}$ is decreasing and bounded below so it converges.

If we have $f_{n+1}(x_0)=f(f_n(x_0))$ the we can conclude that the $limit$ is zero ( $l\neq0$ and $\vert l\vert<\vert l\vert$) but I need the convergence of $(f_n(x_0))_{n\in\Bbb{N}}$ with I don't how can I prove this.

Any ideas on how can I continue?

Answer 1

I work by contradiction. Suppose $f_n$ does not converge uniformly to $0$. This is equivalent on saying that $$ \exists \varepsilon >0 : \exists \{ x_n \}_{n \geq 1} \subseteq [-a,a] : f_n(x_n) > \varepsilon $$

Now, for all $n$, you have $|x_n| > |f(x_n)| > \dots > |f_n (x_n)| > \varepsilon $. So the sequence $\{ x_n \}_{n \geq 1}$ infact is contained in $[-a, -\varepsilon] \cup [ \varepsilon,a] $. Consider the continuous function $$[-a, -\varepsilon] \cup [ \varepsilon,a] \longrightarrow \mathbb{R} \ \ \ \ x \mapsto \frac{|f(x)|}{|x|}$$ and call $q = \max \{ \frac{|f(x)|}{|x|} : x \in [-a, -\varepsilon] \cup [ \varepsilon,a]\} $ which exists since $[-a, -\varepsilon] \cup [ \varepsilon,a]$ is compact. Clearly $q < 1$.

You can prove by induction on $k$ that for all $n \geq 1$ and for al $k \in \{1, \dots, n \}$ the following holds: $$ |f_k(x_n)| \leq q^k |x_n| $$

Finally, you can get the contradiction $$\varepsilon < |f_n(x_n)| \leq q^n |x_n| \leq q^n a \to 0$$

Answer 2

Take som $\varepsilon > 0$. Let's show that $$\sup_{x\in[-a,a]\setminus(-\varepsilon,\varepsilon)}\frac{|f(x)|}{|x|}=q_{\varepsilon}<1$$ Assuming the contradiction, we get sequance $x_n\in[-a,a]$ for which \begin{equation} \frac{|f(x_n)|}{|x_n|}\to 1 \end{equation}

Without loss of generality it can be assumed that $x_n$ is convergent (otherwise the correponding subsequance can be selected). Suppose $x_n\to c\in[-a,a]\setminus(-\varepsilon,\varepsilon)$. By the above limit $|f(c)|=|c|$, which is a contradiction, since $c\in[-a,a]\setminus{\{0\}}$.

Now, by continuity $f(0)=0$, hence on the interval $x\in[-\varepsilon,\varepsilon]$ $$|f^n(x)|\le |x|\le \varepsilon$$ for any natural $n$.

On the interval $x\in[-a,a]\setminus(-\varepsilon,\varepsilon)$ $$|f^n(x)|\leq q_{\varepsilon}|f^{n-1}(x)|\le\dots\le q_{\varepsilon}^n|x|\le q_{\varepsilon}^n a,$$ which can be less than $\varepsilon$ by choosing $n$ sufficiently big. So, for sufficiently big $n$ $$|f^n(x)|\le \varepsilon, \quad x\in[-a,a].$$

Since $\varepsilon$ was an arbitrary, therefore $f^n$ converges uniformely to 0.