Finding $f \colon R_+ \to R_+$ such that $sf = f \circ g$, for $g$ strictly increasing Suppose that $g:R_+\mapsto R_+$ is a strictly increasing and smooth function and let $s$ be a positive real number. I am looking for a function $f:R_+\mapsto R_+$ that satisfies
$$sf(x)=f(g(x))$$
for all $x\geq0$. This problem arises if I want to characterize the extremals of an optimal control problem.
I can find $f$ when $g$ is linear, i.e., $g(x)=ax+b$ with $a>0$. More specifically, I get
$$f(x)=C\left[1-\frac{(1-a)x}b\right]^{\ln s/\ln a}$$
provided that $a\neq1$ and $b\neq0$. Here, $C$ is an arbitrary real number.
If $a=1$ and $b\neq0$, then I get
$$f(x)=Cs^{x/b},$$
and if $a\neq1$ and $b=0$, then I have
$$f(x)=Cx^{\ln s/\ln a}.$$
Of course, I do not know whether these solutions are the only ones, but I would conjecture that they are the only smooth ones.
Do you know of any method which would help me solve the functional equation when $g$ is non-linear? Can you think of other examples (i.e., other functions $g$), for which the functional equation can be solved?
 A: Edit: I didn't quite understand the notation, therefore my old answer is mostly outdated. Still, despite a clarification, the function $f$ that you gave may still be undefined whenever $1-\frac{(1-a)x}{b} < 0$.
Other than that, here is what you could try: use a Taylor expansion to derive certain properties of $f$. Here is an example:
$$f(g(x)) = f(y_0) + [g(x) - y_0]f'(y_0) + ...$$
Plugging this into the functional equation yields, assuming $f$ is smooth enough around $x=0$ for example (you can take any other value for $x$, but that's the easiest one for the next formulas):
$$s[f(0) + xf'(0) + ...] = f(y_0) + [g(x) - y_0]f'(y_0) + ...$$
It now comes down to finding a good choice for $y_0$. For example, taking $y_0 = g(0)$, assuming $g$ is smooth enough around $x=0$ and using yet another Taylor expansion to simplify $g(x) - y_0$ yields
$$s[f(0) + xf'(0) + ...] = f(g(0)) + xg'(0)f'(g(0)) + ...$$
so that after comparing the coefficients you get the conditions
$$sf(0) = f(g(0))$$
and
$$sf'(0) = g'(0)f'(g(0))$$
for example for the first two orders. If you take the case where $g(x) = ax+b$, all these conditions can be summarized as
$$sf^{(n)}(0) = a^nf^{(n)}(b)$$
where $f^{(n)}$ denotes the $n$-th derivative of $f$ for $n \geq 0$, where $f^{(0)} = f$. So, any function that satisfies this condition for every $n\geq 0$ will satisfy your functional equation. For $s = a = 1$, this can be every $b$-periodic function, already showing that a solution is certainly not unique, even among smooth solutions, which makes searching for a specific one that much more complicated.
From there, you can for example assume that $f$ has a Fourier transform $\widehat{f}$ which can be transformed back to $f$. If you assume all of that, then if I'm not mistaken you can rewrite the condition as
$$\int^{\infty}_{-\infty}(s-a^ne^{2\pi i b y})y^n\widehat{f}(y) dy = 0,$$
(using the fact that $\widehat{f^{(n)}}(y) = (2\pi i y)^n \widehat{f}(y)$). This integral formula then has to hold for every $n$. This still gives you a system of infinitely many equations, but at least here you have $f$ (or rather, $\widehat{f}$) appearing only once. Note that this holds when $a = s = 1$ and $f$ is a $b$-periodic function.
For polynomial $g$, using a similar path should lead you to somewhat similar formulas, with more factors however.

Another example that a solution may not be unique or may be trivial, depending on $g$ and $s$: If $g(x) = x$, then for $s=1$ any function satisfies the equation. If $s \neq 1$, only $f(x) \equiv 0$ satisfies it.
