This is question on the existence of a decomposability of a function $f(v)$ into multiplicative factors $\lambda(v)$ and $g(v)$. The question is non-trivial since the functions $\lambda(v)$ and $g(v)$ are parametrized with same parameters, say $w$ and $c$ and have a specific functional form; therefore I'm interested in the following question:
Do there exist parameters $w$ and $c$ such that $f$ can be written as $f(v) = \lambda(v,w,c) \cdot g(v,w,c)$ (where the functional forms of $\lambda$ and $g$ are given)?
Note that the parameters has to be same for both the factors. The exact forms of the functions $\lambda$ and $g$ are messy, so I wanted to keep that out in this question. One line of approach is to start with some $w^0, c^0$, and construct the sequence in the following way:
\begin{align*} w^0, c^0 &\rightarrow \lambda(v,w^0, c^0) \\ \lambda(v,w^0, c^0)^{-1} f(v) &\rightarrow g(v,w^1, c^1) \\ w^1, c^1 &\rightarrow \lambda(v,w^1, c^1) \\ \mbox{check if } &||\lambda(v,w^1, c^1) \cdot g(v,w^1, c^1) - f(v)|| < \epsilon \\ \mbox{if not: continue} \end{align*}
What would be the sufficient conditions to ensure that this sequence will converge for every $v$? You may ask in the second step above: `what is the guarantee that I will get such a $w^1,c^1$ from the arbitrary function $\lambda(v,w^0, c^0)^{-1} f(v)$?' The answer is that I have proved that finding such $w^1,c^1$ is guaranteed (due to some other assumptions of the problem). So, you are assured to get such a $g$ function. What I'm still looking for is to show that this sequence converges.
I'll be thankful if you can point me to relevant material that I can read up to solve this problem.