1
$\begingroup$

At the moment, I am studying for an exam and I came across the following exercise:

Consider the map $f:[0,1] \to \mathbb{R}$, $f(x)=1-\arctan(x)$. Prove the following statements:

a) $f$ has a unique fixed point

b) For each $x \in [0,1]$, the sequence defined by $$x_0=x,\phantom{aa}x_{n+1}=f(x_n)$$ converges to the fixed point of $f$

I have no problems with a) but I have trouble with b): If I am right, $f$ is not contractive, so we cannot apply the Banach fixed point theorem and that is my problem. (I only know that there is some intervall $[a,b] \subseteq [0,1]$ such that $[a,b]$ contains the fixed point of $f$ and such that the restriction $f|_{[a,b]}$ satisfies the conditions of the Banach theorem but I am not sure whether this might help.) So, how to solve part b)?

$\endgroup$
  • $\begingroup$ Cobweb plots are helpful to visualize the dynamics of $x_{n+1}=f(x_n)$, see, for example, en.wikipedia.org/wiki/Cobweb_plot $\endgroup$ – Michael Jun 25 '16 at 21:27
  • $\begingroup$ Can you find a value $f_{min} = \min_{x \in [0,1]} f(x)$? Then $x_k\geq f_{min}$ for all $k>0$. So you can consider the function restricted to the interval $[f_{min}, 1]$. $\endgroup$ – Michael Jun 25 '16 at 21:34
  • $\begingroup$ The problem is at $0$, because $f'(0)=-1$. Away from $0$ you have $|f'|<1$. Can you show that $f([0,1])=[a,b]$ where $a>0$, and that $f([a,b]) \subset [a,b]$? Then you can apply the Banach fixed point theorem on $[a,b]$ instead of $[0,1]$. It will be useful here to note that $f$ is decreasing, so $f([a,b])=[f(b),f(a)]$. $\endgroup$ – Ian Jun 25 '16 at 21:37
  • $\begingroup$ @Ian: I already know that we can restrict $f$ to a suitable intervall $[a,b] \subseteq [0,1]$ and that we can apply the Banach theorem to $[a,b]$. However, I have no idea how this should help with b). So, if we choose $x_0 \in [a,b]$ and we apply the iteration to $x_0$, it is clear that the resulting sequence converges to the fixed point. But I am not sure how to handle the case that $x_0 \in [0,1]\setminus [a,b]$ $\endgroup$ – russoo Jun 25 '16 at 21:46
  • 1
    $\begingroup$ A more useful example of essentially the same trick: show that if $x_0>0$ and $a>0$ then the sequence $x_{k+1}=\frac{x_k+a/x_k}{2}$ converges. $\endgroup$ – Ian Jun 25 '16 at 23:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.