Do there exist a (non-trivial) globally Lipschitz continuous function $g:\mathbf{R}\to\mathbf{R}$ and a non-decreasing function $f:\mathbf{R}\_+\to\mathbf{R}\_+$ such that the identity
\begin{equation}
g(\alpha f(\alpha)s) = f(\alpha) g(s)
\end{equation}
holds for all $\alpha>0$ and $s\in\mathbf{R}$?


Remark:
I am pretty sure that for $f$ polynomial this cannot hold (as then $g$ would have to be less than $1$-homogeneous, and thus not globally Lipschitz).



Existence of functions satisfying a homogeneity condition
Since $f$ is non-decreasing, the function $\alpha\mapsto \alpha f(\alpha)$ is strictly increasing and hence invertible with strictly increasing inverse. Therefore there is a non-decreasing function $k$ such that $k(\alpha f(\alpha)) = f(\alpha)$.




---


*Edit*: As Iosif pointed out in a comment, the notion of the inverse function of $\varphi: \alpha \mapsto \alpha f(\alpha)$ needs to be more carefully described, since $f$ is not assumed continuous. One can define $h(\beta) = \sup \{ \alpha: \varphi(\alpha) \leq \beta \}$ and see that $h$ is a left inverse of $\varphi$. But in general $h$ may fail to be injective. This leaves a gap for the argument below. Readers should refer to Iosif's answer instead, which shows that there must exist a sequence of points $s\_k \searrow 0$ along which $g(s\_k) = C s\_k^{\gamma}$ for $\gamma\in (0,1)$, and thus falsifying Lipschitz condition.




---


So you are looking for equivalently mappings that satisfy
$$ g(\beta s) = k(\beta) g(s)$$
Applying to $s = 1$ you find
$$ g(\beta) = k(\beta) g(1) , \qquad g(\alpha\beta) = k(\alpha\beta) g(1)$$
and applying to $s = \beta$ you find
$$ g(\alpha\beta) = k(\alpha) g(\beta) = k(\alpha)k(\beta) g(1) $$
which implies that $k$ is a group homomorphism of the multiplicative group $\mathbf{R}\_+$. Monotonicity of $k$ then implies that $k(\alpha) = \alpha^\gamma$ for some fixed $\gamma \geq 0$.


The requirement that $k(\alpha f(\alpha)) = f(\alpha)$ implies that
$$ \beta / k(\beta) $$
is strictly increasing, and hence $\gamma < 1$.


In conclusion, there must exist some $\gamma \in [0,1)$ such that $g(\beta s) = \beta^\gamma g(s)$ for all $\beta\in \mathbb{R}\_+$, $s\in \mathbb{R}$.


Except for the case $\gamma = 0$ and $g$ is constant, and the case where $g\equiv 0$ where the identity holds trivially, all other cases have a non-Lipschitz point at $0$.


