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!


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.


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.

  • 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


  • 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 $\ddot\smile$


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