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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 $x_{j} \geq 0$ $\forall j$?

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