12
$\begingroup$

Let $f$ be continuous and decreasing everywhere on $\mathbb{R}$. Show that:

1) $f$ has a unique fixed point

2) $f\circ f$ has either an infinite number of fixed points or an odd number of fixed points.

The first part is easy and I am sure it is available on this website. The basic idea is to use the fact that a decreasing function satisfies $\lim_{x \to -\infty}f(x) = A\text{ or }\infty$ and $\lim_{x \to \infty}f(x) = B\text{ or }-\infty$ and in each case apply the intermediate value theorem on $g(x) = f(x) - x$. The uniqueness of the fixed point is also easy to understand as $f(a) - a = 0 = f(b) - b$ would imply $b - a = f(b) - f(a)$. If $b \neq a$ then this goes against the decreasing nature of $f$.

It is the second part of the problem which is bit troublesome. Let $h(x) = f(f(x))$ let $c$ be the unique fixed point of $f$ so that $f(c) = c$. This means that $f(f(c)) = f(c) = c$ so that $c$ is also a fixed point of $h = f \circ f$. But counting the number of fixed points of $h$ seems tricky. Any hints are welcome!

$\endgroup$
8
$\begingroup$

Note that if $d$ is a fixed point of $f\circ f$ but not of $f$, then so is $f(d)$. So you can make pairs of fixed points.

$\endgroup$
6
$\begingroup$

Hint: Suppose $h$ only has finitely many fixed points. Call the set of fixed points $A$. Where does $f$ map $A$? (And what does uniqueness of the fixed point of $f$ then tell you?)

Added: Since the problem appears to be solved, here's how to complete the solution: define $$\begin{align}A_<&=\{x\in A\mid x<f(x)\},\\A_=&=\{x\in A\mid x=f(x)\},\\A_>&=\{x\in A\mid x>f(x)\}.\end{align}$$ Note that these three sets are disjoint and $A=A_<\cup A_=\cup A_>$. Note that $A_=$ consists of a single element $c$, the fixed point of $f$. Also note that $f$ maps $A_<$ bijectively onto $A_>$.

So $|A_<|=|A_>|=:k$ and $|A_=|=1$. We conclude that $|A|=2k+1$, which is an odd number.

$\endgroup$
  • 1
    $\begingroup$ although i have accepted the other answer (because I was able to use his hint more easily compared to your hint) your idea about the bijection between the sets $A_{<}$ and $A_{>}$ is nice. +1 $\endgroup$ – Paramanand Singh Mar 23 '14 at 12:14
3
$\begingroup$

Hint:

  • Let $d = f(f(d))$.
  • If $d$ is a fix-point of $f$, then there is only one such $d$.
  • Otherwise $d \neq f(d)$; let $\hat{d} = f(d)$, then $$f(f(\hat{d})) = f(f(f(d))) = f(d) = \hat{d},$$ that is, $\hat{d}$ is another fix-point of $(f\circ f)$.

I hope this helps $\ddot\smile$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.