Let $f : R → R$ be a continuous decreasing function. Prove that the system
$$x = f(y),$$
$$y = f(z),$$
$$z = f(x)$$
has a unique solution.
Here using the intermediate value theorem, it can clearly be seen that when
$\displaystyle{\lim_{x \to -\infty}}f(x)-x$ = $\infty $
& $\displaystyle{\lim_{x \to \infty}}f(x)-x$ = –$\infty $ &
we get some $x_0$ such that $f(x_0) = x_0$
But how or why does that imply that $x_0 = x = y = z$
Best Answer
$h(x) = f(f(f(x)))$ is strictly decreasing with $\lim_{x \to -\infty}h(x)-x = \infty$ and $\lim_{x \to \infty}h(x)-x = -\infty$, so that $h$ has a unique fixed point $a$.
Then $(x, y, z) = (f(f(a)), f(a), a)$ satisfies $$ \tag{*} x=f(y) \, , \, y=f(z) \, , \, z=f(x) \, . $$
If $(x, y, z)$ is any solution of $(*)$ then $h(x) = x$, $h(y) = y$ and $h(z) = z$ implies $x=y=z= a$ because the fixed point of $h$ is unique.