# Univalent Mapping - Uniqueness of Fixed Point on the Positive Orthant

I am facing a n-dimensional fixed-point problem descending from a game theoretic problem given by the set of equations

$$\forall j \in n: R_{j} (\vec{x}_{-j}) - x_{j} = W \left( A_{j} \exp \left(-\sum_{k \neq j}^{n} c_{j,k} x_{k}\right)\right) - x_{j} =0$$

where $$W$$ is the Lambert W function, $$\vec{x}$$ is the vector of players' choices with $$0 < x_{j}^{-} \leq x_{j} \leq x_{j}^{+} < +\infty$$ $$\forall j$$. The remaining parameters are given by $$A_{j} > 0$$ $$\forall j$$ and $$c_{j,k} \in (0,1)$$ $$\forall j$$ $$\forall k \neq j$$.

I have shown concavity on the original problem and, given the bounds $$\vec{x}_{j}^{-}$$, $$\vec{x}_{j}^{+}$$, the Nash-Debreu-Theorem ensures existence of at least one fixed-point / equilibrium vector $$\vec{x}^{*}$$.

What I am struggling with is showing uniqueness of the fixed-point of this set of equations. I have tried using the contraction mapping (specifically Edelstein's Theorem) and univalent mapping approach (Gale & Nikaido (1965), Rosen (1965)) but could not find more than a sufficient condition for uniqueness so far. Numerically uniqueness seems to hold. I have produced a large sample of these equilibrium problems and solved each from 1000 different starting vectors and each problem exhibited a single solution on $$\mathbb{R}_{\geq0}^{n}$$.

For the contraction mapping, since there is, to my knowledge, no identity for $$W(z_1) - W(z_2)$$, I had work with an upper bound on the integral representation of the difference which essentially gave me the diagonal dominance property of the $$n \times n$$ matrix $$\mathbf{C}$$ with elements $$c_{j,j}=1$$ and $$c_{j,k} \in (0,1)$$ $$\forall j$$ $$\forall k \neq j$$. Unfortunately I cannot make such a parametric restriction ex-ante which is why I believe the contraction mapping argument will not be helpful in proving this.

For the univalent mapping, Gale & Nikaido (1965) and Rosen (1965) guarantee uniqueness of the fixed-point if $$\mathbf{H}^{*} = \frac{1}{2} (\mathbf{H}+ \mathbf{H}^{T})$$ is negative definite where $$\mathbf{H}$$ is the $$n \times n$$ Jacobian matrix of the n-dimensional root finding problem above. Note that the elements of $$\mathbf{H}$$ are given by $$h_{j,j}=-1$$ and $$h_{j,k} \in (-1,0)$$ since the derivative of the Lambert W $$\frac{d W}{d z} \leq 1$$ on $$z \in \mathbb{R}_{\geq0}^{}$$.

I would prove negative definiteness and thereby uniqueness of the fixed-point if I could show that

$$\vec{z}^T \mathbf{H}^{*} \vec{z} < 0 \quad \forall \vec{z} \in \mathbb{R}_{\geq0}^{n} \setminus \vec{0}$$

which is equivalent to having all negative eigenvalues. However it is easy to find counterexamples with parameter combinations where $$\mathbf{H}^{*}$$ has positive eigenvalues for $$n > 2$$.

My question is the following: The univalent mapping above, as I understand it, could give me uniqueness on the entirety of $$\mathbb{R}_{}^{n}$$ but it does not hold generally in my case. My suspicion is the multiplicity of fixed-points depends on allowing negative values for $$x_{j}$$. Is there a property, let's call it constrained definiteness, I could apply to show univalence and therefore uniqueness just on $$\mathbb{R}_{\geq 0}^{n}$$ i.e. under the constraint $$x_{j} \geq 0$$ $$\forall j$$?