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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.

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3 Answers 3

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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 $x_n\to0$.

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  • $\begingroup$ that is smooth indeed $\endgroup$
    – Simonsays
    Commented Jun 26, 2017 at 19:15
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    $\begingroup$ 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? $\endgroup$
    – user9464
    Commented Jun 26, 2017 at 19:27
  • $\begingroup$ @Jack $x_{n_i+1}=f(x_{n_i})\to f(y)$. So $\|x_n\|$ has a subsequence tending to $\|f(y)\|<L$. $\endgroup$ Commented Jun 26, 2017 at 19:30
  • $\begingroup$ 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? $\endgroup$
    – user9464
    Commented Jun 26, 2017 at 19:33
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    $\begingroup$ @Jack A bounded monotone sequence converges! $\endgroup$ Commented Jun 26, 2017 at 19:34
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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.

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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

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

which is a contradiction.

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    $\begingroup$ Thanks for your answer. Would you elaborate how you "use the continuity of $f$ and ..." to get $0\leq L<1$? $\endgroup$
    – user9464
    Commented Jun 26, 2017 at 19:58
  • $\begingroup$ @Jack Sorry for being so sketchy. I added some details. Please let me know if something is unclear. $\endgroup$ Commented Jun 26, 2017 at 20:09

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