# $\|f(x)\|<\|x\|$ and $x_{n+1}=f(x_n)$ with $f$ being continuous imply that $x_n\to 0$

Let $f:{\bf R}^d\to{\bf R}^d$ be a continuous function such that $\|f(x)\|<\|x\|$ for every point $x\neq 0$. Fix a point $x_1\in{\bf R}^d$, and define recursively $x_{n+1}=f(x_n)$ for $n\geq 1$. Show that the sequence $(x_n)_{n=1}^\infty$ converges to $0$.

[Some thoughts]

• The restriction of $f$ on a closed ball $B:=B(0,R)$ of radius $R$ is a continuous function from $B$ to $B$, which implies by the Brouwer fixed point theorem that one must have $f(0)=0$ since $f$ cannot have other fixed points.
• The sequence $(x_n)$ is bounded.
• If the limit of $(x_n)$ exists, it must be $0$ by the recursive relation and continuity of $f$.

Directly showing $(x_n)$ is Cauchy seems impossible. I have also tried to argue by contradiction that there exists convergent subsequence $x_{n_k}\to x\neq 0$, which seems giving not much information.

This is an exercise in real analysis. I would be also interested in seeing how this could be solved in the view of dynamical systems.

It is clearly a bounded sequence. Also $\|x_n\|$ decreases to a limit $L$ say. It has a convergent subsequence $(x_{n_i})$, by the Bolzano-Weierstrass theorem. If $x_{n_i}\to y$ then $\|y\|=L$ and if $y\ne0$, $x_{n_i+1}\to f(y)$ and as then $\|f(y)\|<L$ we get a contradiction. So $y=0$, $L=0$ and $x_n\to0$.

• that is smooth indeed Jun 26, 2017 at 19:15
• Sorry I don't follow your reasoning. As I understand, $(x_{n_i})$ and $(x_{n_i+1})$ are two different subsequences. Without knowing a priori the limit of $(x_n)$ exists, where does the contradiction come from?
– user9464
Jun 26, 2017 at 19:27
• @Jack $x_{n_i+1}=f(x_{n_i})\to f(y)$. So $\|x_n\|$ has a subsequence tending to $\|f(y)\|<L$. Jun 26, 2017 at 19:30
• Your argument assumes that $\|x_n\|$ has a limit in the beginning. But how do you prove that the limit exists in the first place?
– user9464
Jun 26, 2017 at 19:33
• @Jack A bounded monotone sequence converges! Jun 26, 2017 at 19:34

I would like to rephrase Lord Shark the Unknown's smart argument here.

Since the sequence $(\|x_n\|)$ is bounded and monotone, it has a limit $L$. (Note at this point that one cannot argue that $(x_n)$ has a limit.) It suffices to show that $L=0$. Suppose $L>0$. By the boundedness of $(x_n)$, we have a subsequence $x_{n_k}$ such that $$x_{n_k}\to x,\ k\to\infty\tag{1}$$ By the continuity of $f$ and the recursive relation, one must also have $$x_{n_k+1}\to f(x),\ k\to\infty.\tag{2}$$ (1) and (2) imply that $$\|x_{n_k}\|\to\|x\|,\quad \|x_{n_k+1}\|\to\|f(x)\|,\ k\to\infty$$ But $\|x_n\|\to L$ as $n\to\infty$. Thus one must have $$\|x\|=\|f(x)\|=L>0$$ which is a contradiction.

I was confused by his answer but eventually find out how his argument remedies my partial work in the problem. I assumed by contradiction that $x_{n_k}\to x\neq 0$, which implies that $x_{n_k+1}\to f(x)$. I was not able to get a contradiction because I thought one would need know a priori that the limit of $(x_n)$ exists. But it is enough to notice that the limit of $(\|x_n\|)$ exists.

Set

$$L= \inf \{ C\in \mathbb{R} \ \vert \ \forall x\in B(0,\Vert x_1 \Vert): \Vert f(x) \Vert \leq C \Vert x \Vert \}.$$

Using the continuity of $f$ and $\Vert \cdot \Vert$ and the fact $\Vert f(x) \Vert < \Vert x \Vert$ we get $0\leq L<1$. Thus, we have

$$\Vert x_{n+1}\Vert \leq L^n \cdot \Vert x_1 \Vert \rightarrow 0$$

for $n\rightarrow \infty$. Thus, $\lim_{n\rightarrow \infty} x_n = 0$.

Added: As $\Vert f(x) \Vert < \Vert x \Vert$ for all $x\in B(0, \Vert x_1 \Vert)$ we get $L\leq 1$. Assume that $L=1$, then for every $n\in \mathbb{N}$ exists $x_n\in B(0, \Vert x_1 \Vert)$ such that

$$\Vert f(x_n) \Vert \geq \left(1- \frac{1}{n} \right) \Vert x_n \Vert.$$

By compactness we may w.l.o.g. assume that $x_n \rightarrow x\in B(0, \Vert x_1 \Vert)$. However, then we have by the continuity of $f$ and $\Vert \cdot \Vert$ that

$$\Vert f(x) \Vert \geq \Vert x \Vert,$$

• Thanks for your answer. Would you elaborate how you "use the continuity of $f$ and ..." to get $0\leq L<1$?