Question regarding the existence of a general solution to the functional equation $f \circ f = af$

Question

I would like to know how to determine all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $f\circ f =a f$ (where $a$ is any real number)

Is this problem completely solved (without or with additional assumptions on $f$ )

A friend sent me the problem with $a=3$ , if we assume $f$ bijective then $f $ is linear $3x$

Thanks for sharing any relevant information

Answer 1

Well, clearly $f(x) = 3x$ for all $x$ in the image of $f$. In particular, the image of $f$ is a set $S$ with $3S \subseteq S$. There are many such sets: all of $\mathbb R$, the positive reals, the set $\{27, 81, 243, 729, \dots\}$, the rationals, and many more.

(In general, we can take $S$ to be any union of the sets $\{y, 3y, 9y, 27y, \dots\}$ for one or more real numbers $y$.)

If $S$ is such a set, then we can pick an arbitrary $f : \mathbb R \setminus S \to S$ and extend it to all of $\mathbb R$ by $f(x) = 3x$ for all $x \in S$.

The picture does not change much if $3$ is replaced by $a$.

Answer 2

The trivial solution: $f=0$ also holds (if $f$ not bijective).