Question is to check if :

$f : [0,\infty]\rightarrow [0,\infty]$ which is continuous and bounded has a fixed point.

I have first of all considered boundedness.

So, $f(x)$ should not have $x$ as multiple i mean $f(x)$ should not have a polynomial factor.

I took $f(x)=\frac{1}{g(x)}$ and $g(x)$ should not have positive root (assuming it to be polynomial)

for fixed point, $\frac{1}{g(x)}=x$, we have $x.g(x)=1$

for simple $g(x)$ i have ended up with concluding that $f(x)$ have fixed point.

I do not have any clear idea how to porve this.. So, I have written what are all i have thought could be helpful...

please give some hint to crack this.

Thank you..

  • 7
    $\begingroup$ If $f$ is bounded by $M$, consider $f\lvert_{[0,M]} \colon [0,M] \to [0,M]$. $\endgroup$ – Daniel Fischer Oct 7 '13 at 13:13
  • 1
    $\begingroup$ Oh. I got it... any continuous function from a convex compact set to itself has a fixed point.. i.e., $f\lvert_{[0,M]}$ has a fixed point and so is $f$... Thank you.. :) $\endgroup$ – user87543 Oct 7 '13 at 13:16
  • $\begingroup$ Assume $f(x)-x>0$ for all $x\in [0,\infty[$. $\endgroup$ – Michael Hoppe Oct 7 '13 at 13:20
  • 4
    $\begingroup$ Boundedness assumption is redundant if the notion of continuity at infinity is appropriately defined. Indeed, $[0, \infty]$ is homeomorphic to $[0, 1]$ (via the mapping $x \mapsto \frac{x}{x+1}$ or something like this) so you only need to prove that a continuous function $g:[0,1]\to[0,1]$ has a fixed point. $\endgroup$ – Sangchul Lee Oct 7 '13 at 13:29
  • $\begingroup$ @sos440 : I am not very much aware of technical details here but, yours argument does makes a lots of sense to me.... $\endgroup$ – user87543 Oct 7 '13 at 13:31

For fixed point $x_1$, we have $f(x_1) = x_1$ (that is what fixed point means). Let's say $0$ is not a fixed point. Then $f(0) \gt 0$ (it can't be less, by definition of $f$, and we assume it's not equal). so there are values of $x$ for which $f(x) > x$; that is, the function $g(x) = f(x) - x$ is positive.

Now, $f(x)$ is bounded, so there is some $K$ such that $f(x) < K$ for all $x$. Especially, $f(K) < K$, so $g(K) < 0$.

Now, the function $g(x)$ is continuous, and it is positive at some points (for instance, at $0$), and it is negative at some points (for instance, at $K$), so it has to be $0$ at some points (specifically between $0$ and $K$). That's your fixed point.


Your Answer

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