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!
Best Answer
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.