I am looking for an extended version of the implicit/inverse function theorem that would show uniqueness of a strictly increasing implicit function, even when the derivative condition is violated (e.g. $= 0$).
Let's consider this simple example to illustrate:
We have the relation: $f(x, y) = (x-1)^2 + y^2 = 0$.
This relation does not define a function because for every x we have two possibilities $y=x-1$ or $y=1-x$. Using the implicit function theorem, one can show that there exists a function $y(x)$ around the points where $\partial f/\partial y\neq 0$, i.e. everywhere except at $(x_0, y_0) = (1, 0)$. And that this $(x_0, y_0)$ I cannot apply the theorem.However, I am trying to show the uniquess of a strictly increasing function $y(x)$. In this example, with my closed form solution it is straightforward that it exists, and it is: $y(x) = x-1$. I would like to show it using something close to the implicit function theorem, or differential equations. But I cannot because everything breaks at $y=0$. And I don't see how to include the monotonicity $dy/dx > 0$ constraint into the standard proof.
I am asking this question because I am facing a more general problem where I have:
$$f(x, y) = f_1(x) + f_2(y) = 0 \quad \forall (x, y) \in [0,1]^2$$
(and $f_1, f_2$ are continuously differentiable).
I know the existence of a strictly increasing solution $y(x)$, starting with $y(0)=0$ and I need to prove its uniqueness. I showed that, provided that $\partial f/\partial y = 0$ at most at a finite set of $K \geq 0$ points (called $y_k$), then there exists an unique strictly increasing solution (while there may exists more non strictly increasing mappings).
Remark: the idea is that there is still a unique increasing solution even if $\partial f/\partial y = 0$, as long as $f_2(y)$ is not flat.
To show it I am showing that under our condition, $f_2(y)$ is piecewise monotone on the support. And thus, piecewise inversible.
In short, by existence of the solution, I know there will also be, at most, a finite set of $K$ points $x_k$ at which $df/dx = 0$, which will correspond to the $y_k$ points by strict monotonicity of the solution.
Then I just run piecewise inversion in between each subsegment of the support: for $x \in [x_k, x_{k+1}]$, $y(x) = f_2^{-1}(f_1(x))$ and will go from $y_k$ to $y_{k+1}$, etc.
This proof is not too hard in dimension 2, however I am trying to extend the problem to higher dimension now, in which case I think an approach using implicit/inverse function theorem or more generally properties of ODEs could be more useful. But for that I would need to prove uniqueness in dimension 2 using one of these methods and I have not been able to do it because of the $df/dy \neq 0$ condition that always appear (or the determinant $\neq 0$ in the case of systems). But with strict monotonicity, as I showed you in my example, I know this condition can be bypassed. I just don't know how or if it is possible to prove it.
