In the Renormalization Group Theory, a key step is the derivation of the so called scaling equation, which in general is in the form of a functional equation of the kind: $$ g(\mu(\lambda)x,\nu(\lambda)y) = \lambda g(x,y) $$ where x, y are real numbers and $\lambda>0$, moreover $\mu(\lambda=1)=\nu(\lambda=1)=1$.
We can assume g differentiable both on x and y and $\mu(\lambda),\nu(\lambda)$ analytic function of $\lambda$.
The (hopefully unique) solution of the functional equation should correspond to $\mu(\lambda)$ and $\nu(\lambda)$ being of the form of power law with two independent exponents.
Is it necessary to add other conditions on the functions $g,\mu$ and $\nu$ to prove this result and its uniqueness? Or is it possible to weaken the hypotheses?
Edit:
Assuming $\mu(\lambda)$ being an invertible function (this is an additional hypothesis), setting $y=0$ we get $$ g(\lambda x,0) = \mu^{-1}(\lambda) g(x,0) $$ It easy to get $$ g(\lambda_1 \lambda_2 x,0) = \mu^{-1}(\lambda_1)\mu^{-1}(\lambda_2) g(x,0)= \mu^{-1}(\lambda_1 \lambda_2) g(x,0) $$ from which follows (at least for differentiable $ \mu^{-1}(\lambda$): $$ \mu^{-1}(\lambda) = \lambda^s$$ and then $ \mu(\lambda) = \lambda^{\frac{1}{s}}$.
My question is if it is necessary to add an additional hypothesis like that of an invertible $\mu(\lambda)$, if there are alternatives to the invertibility or if is possible to proof the result even under weaker conditions.
This is not an answer, of course, but a response to your last comment question, for lack of ease in making extended comments ... Basically I don't know, but there are stringent constraints among g,ν,μ ...
Starting from your $$ g(\mu(\lambda)x,\nu(\lambda) y)=\lambda ~ g(x,y), $$ apply the group property directly, so $$ g(\mu(\lambda_1)x,\nu(\lambda_1) y)=\lambda_1 ~ g(x,y) \Longrightarrow \\ g(\mu(\lambda_1)\mu(\lambda_2)x,\nu(\lambda_1)\nu(\lambda_2) y)=\lambda_1 ~ g(\mu(\lambda_2)x,\nu(\lambda_2)y) = \lambda_1 \lambda_2~ g(x,y) =g(\mu(\lambda_1 \lambda_2)x,\nu(\lambda_1\lambda_2 ) y). $$
Of course $\mu(\lambda_1)\mu(\lambda_2)=\mu(\lambda_1 \lambda_2)$ and $\nu(\lambda_1)\nu(\lambda_2)=\nu(\lambda_1 \lambda_2)$ satisfy it, but there are entire trajectories $g(x,y)=g(\tilde x, \tilde y)$ that do too, yielding potentially different μ,ν. Another way to put it is that g is not invertible, but has level curves on the x-y plane, that is a "symmetry" of g.
I suspect one might have to experiment with toy examples. Take $g=y-x^2$, so $\mu=\lambda^{1/2}$, and $\nu=\lambda$. But then for small $\epsilon$, $y\to y+\epsilon$ and $x\to \sqrt{x^2+\epsilon}$ is such a symmetry...
In any case, μ, ν are related to and are properties of the Wegner function, as follows from differentiating the top expression w.r.t. λ and then setting it equal to 1, $$ x\mu'(1) \partial_1 g(x,y) +y\nu'(1) \partial_2 g(x,y)=g(x,y), $$
where the primes denote derivatives w.r.t. λ at the fixed point, and 1 and 2 w.r.t. the first and second argument, respectively, of g at the fixed point.