I am reading the paper
https://arxiv.org/pdf/1005.0138.pdf
about generalized entropies. They have given two asymptotic scaling relations for the function $g(x)$:
$\lim_{x \mapsto 0}\frac{g(xz)}{g(x)} = z^c, \lim_{x \mapsto 0}\frac{g(x^{1+a})}{x^{ac}g(x)} = (1+a)^d$.
Here, $c$ and $d$ are parameters that will determine the form of the generalized entropy. For the special case $c=d=1$, the classical Shannon entropy is recovered. Above scalings must hold for all values of $z$ and $a$.
Now the authors propose that $g$ has to be in the form
$g(x) = r A^{-d}e^A\Gamma(d+1,A-c \ln x) -rcx$ (*)
with constants $r,A$ and the upper incomplete gamma function $\Gamma(a,b) = \int_b^\infty t^{a-1}e^{-t}dt$. I know that $g(x)$ (the function given in * with incomplete gamma function) satisfy above scaling relations, but can I deduce (*) only from the scaling relations? Are there even many other functions $g(x)$ that satisfy the scaling relations for all $c,d,z,a$?
P.S.: I got an idea how to prove the reverse (from asymptotic to function). Note that due to Kolmogorov-Kinchin axioms it must hold $g(0) = 0$. This allows us to use L'Hospitals rule:
$\lim_{x \mapsto 0}\frac{g(x^{1+a})}{x^{ac}g(x)} = \lim_{x \mapsto 0}\frac{(1+a)x^a g'(x^{1+a})}{acx^{ac-1}g(x)+x^{ac}g'(x)} = \lim_{x \mapsto 0}\frac{(1+a)g'(x^{1+a})}{x^{a(c-1)}g'(x)} = (1+a)^d$
Thus
$\lim_{x \mapsto 0}\frac{g'(x^{1+a})}{x^{a(c-1)}g'(x)} = (1+a)^{d-1}$.
This is a functional equation for $g'(x)$ at least if we do not bother about the limit to zero. It can be solved by expressions like
$g'(x) = (\ln x)^{d-1}x^{c-1}$.
Finally, integration with substitution leads to the desired result. Is this enough to do?