Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function and there exists a $A \in [0,1[$ so that $|f'(t)| \leq A$ for every $t \in \mathbb{R}$. Then $f$ has a unique fixed point.

How does one prove the existence of this fixed point? I succeeded in proving its unicity, however that's not much worth if you can't verify its existence.

And what about it if $|f'(t)|<1$ for every $t \in \mathbb{R}$? Does $f$ then still have a fixed point?

  • $\begingroup$ ...so that $|f'(t)|$ what, now? $\endgroup$ – Chris Jun 9 '17 at 23:31
  • $\begingroup$ It's been edited! $\endgroup$ – simp Jun 9 '17 at 23:33
  • $\begingroup$ Without loss of generality $f(0) \geq 0$, can $f(t) - t > 0$ for all $t \geq 0$? $\endgroup$ – Contravariant Jun 9 '17 at 23:44
  • $\begingroup$ Show that the sequence $x_{n+1} = f(x_n)$ is Cauchy. $\endgroup$ – copper.hat Jun 9 '17 at 23:44
  • $\begingroup$ Well, it seems to me like you split up for the case $f(0) > 0$ or $f(0) < 0$, and then examine $f(x) - x$ or $f(x) + x$ respectively, to find a zero. $\endgroup$ – Chris Jun 9 '17 at 23:44

For the first, uniqueness is straightforward, all you need to do is show existence.

Suppose $f(0) >0$. Note that for $t >0$ $f(t) = f(0) + f'(\xi) t \le f(0)+At$, and hence $f(t) -t \le f(0) + (A-1)t$. Since $A-1 <0$, there is some $t'$ such that $f(t')-t' < 0$ and the intermediate value theorem shows that there is some $t^*$ such that $f(t^*) = t^*$.

If $f(0) <0$ apply the previous reasoning to $-f$.

For the second part, try $f(x) = ({1 \over 2}+{1 \over \pi} \arctan x) x +{1 \over \pi}$, then $f'(x) = { \arctan x \over \pi }+{ x\over \pi(1+x^2)} + {1 \over 2} $ and $f(x) > x$ for all $x$.

Note that $f''(x) = {2 \over \pi (1+x^2)^2} >0$ and $\lim_{x \to \infty} f'(x) = 1$, hence $f'(x) \in (0,1)$ for all $x$.

  • $\begingroup$ @Chris: Thanks for catching that. Unfortunately $|f'(x)| < 1$ but it is no longer as obvious. $\endgroup$ – copper.hat Jun 10 '17 at 0:07

Since for any $x, y\in \mathbb R$, $$|f(x)-f(y)|=|f'(\xi)||x-y|\le A|x-y|,$$ now we can apply Banach Fixed Point Theorem to conclude that $f$ has a unique fixed point..

  • $\begingroup$ What if $A = 1$? what about $f(x) = x + c$, $c \ne 0$? $\endgroup$ – Robert Lewis Jun 10 '17 at 0:17
  • $\begingroup$ @RobertLewis: Hi! Did you see that the Med. changed ownership. End of an era. $\endgroup$ – copper.hat Jun 10 '17 at 0:18
  • $\begingroup$ @copper.hat: Yeah, it's gone. I was at the closing party on 30 November or was it 1 December? In any event, the last day was 30 November as I recall. thebolditalic.com/… $\endgroup$ – Robert Lewis Jun 10 '17 at 0:22
  • $\begingroup$ @RobertLewis: It is sad. At least Moe's across the street is still there, but not the same without the Med. :-(. $\endgroup$ – copper.hat Jun 10 '17 at 0:24
  • $\begingroup$ @copper.hat: Nope. End of an era! $\endgroup$ – Robert Lewis Jun 10 '17 at 0:28

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