Monotonicity of $(f \circ f)$ and $(f \circ f \circ f)$ for strictly monotonic $f$

Question

Let $f$ be a strictly monotonic function.

What can we say about $(f \circ f)$ and $(f \circ f \circ f)$,

1. when $f$ is strictly increasing,
2. when $f$ is strictly decreasing?

Please provide a proof.

Answer 1

If f is strictly increasing then whenever a< b, then f(a)< f(b). Then, since f(a)< f(b), fof(a)= f(f(a))< f(f(b))= fof(b).

Answer 2

This is more a long comment than a definite answer:

You cannot conclude in general if you do not impose constraints on the domain of definition of $f$, for instance take $f(x)=\pm(x^2-1)$ whether you want an increasing or decreasing function on $\mathbb R^+$.

$f(x)=x^2-1$ is $\nearrow$ on $\mathbb R^+$ but $f\circ f$ presents two monotonicities and $f\circ f\circ f$ has three on that same interval. (i.e. has resp. 2 and 3 intervals where it is strictly monotonic).

Nevertheless, there is a case when you can conclude easily, this is when $f$ is strictly monotonic and continuous on whole $\mathbb R$ because now you get a bijective mapping from $\mathbb R\to\mathbb R$ and no matter how many times you compose the function, $f^{\circ[n]}$ will stay a continuous bijection and thus will be strictly monotonic.

The case when $f$ is not necessarily continuous but still monotonic on whole $\mathbb R$ can be dealt with the numerous hints that were provided to you in the comment section.