# Distributed GNE-Seeking under Partial Information Based on Preconditioned Proximal-Point Algorithms
## 3 Iterative Algorithms with Global Information
### Algorithm Development  
**Assumption 4**. : \(\mathbb{F}(\mathbf{x})\) _is a \(\mu\)-strongly monotone function, meaning that \(\langle\mathbb{F}(\mathbf{x})-\mathbb{F}(\mathbf{y}),\mathbf{x}-\mathbf{y}\rangle\geq\mu|\mathbf{ x}-\mathbf{y}|_{2}^{2},\forall\mathbf{x},\mathbf{y}\in\Omega\), and a \(\theta_{0}\)-Lipschitz continuous function, implying that \(|\mathbb{F}(\mathbf{x})-\mathbb{F}(\mathbf{y})|_{2}\leq\theta_{0}|\mathbf{x}-\mathbf{y}|_{2}, \forall\mathbf{x},\mathbf{y}\in\Omega\)._  
The strong monotonicity of \(\mathbb{F}\) implies that a v-GNE exists and is unique. We can write the variational problem for the original game problem as follows:  
\[ \begin{split}\min\limits_{x\in\mathbb{R}^{n}}& \langle\mathbb{F}(\mathbf{x}^{*}),\mathbf{x}\rangle+g(\mathbf{x}),\\ \text{s.t.}&\sum\limits_{i=1}^{N}A_{i}x_{i}\leq\sum \limits_{i=1}^{N}b_{i}.\end{split}  \tag{9}\]  
The objective function is defined by the following Lagrangian function:  
\[\mathcal{L}\big{(}\mathbf{x},\lambda_{g}\big{)}:=\langle\mathbb{F}(\mathbf{x}^{*}),\bm {x}\rangle+g(\mathbf{x})+\lambda_{g}^{\text{T}}\Bigg{(}\sum\limits_{i=1}^{N}(A_{i} x_{i}-b_{i})\Bigg{)}. \tag{10}\]  
Let \(\lambda_{g}\in\mathbb{R}^{m}\) be a global Lagrangian multiplier. To design distributed optimization algorithms, we impose the consistency constraints \(\lambda_{i}=\lambda_{j},\forall(i,j)\in\mathcal{E}\), i.e., \((L\otimes I_{m})\mathbf{\lambda}=0\), where \(I_{m}\) is an identity matrix of order m. Define \(\mathbb{G}(\mathbf{x})=\text{col}(\partial g_{i}(x_{i}))_{i\in\mathcal{I}}\in \mathbb{R}^{n}\), \(\mathbf{\lambda}=\text{col}(\lambda_{1},\cdots,\lambda_{N})=\text{col}(\lambda_{i}) \in\mathbb{R}^{m}\), \(\mathbf{A}=[A_{1},\cdots,A_{N}]\in\mathbb{R}^{m\times n}\), and \(\mathbf{b}=\text{col}(b_{i})\in\mathbb{R}^{m}\). Then,  
\[\mathcal{L}_{1}(\mathbf{x},\mathbf{\lambda}):=\langle\mathbb{F}(\mathbf{x}^{*}),\mathbf{x} \rangle+g(\mathbf{x})+\mathbf{\lambda}^{\text{T}}(\mathbf{A}\mathbf{x}-\mathbf{b}). \tag{11}\]Its corresponding saddle point problem would be  
\[ \begin{array}{ll}\max_{\lambda}\min_{\mathbf{x}}&\langle\mathbb{F}(\mathbf{x}^{*}), \mathbf{x}\rangle+g(\mathbf{x})+\lambda^{\mathrm{T}}(\mathbf{Ax}-\mathbf{b}),\\ &\text{s.t.}&(L\otimes I_{m})\mathbf{\lambda}=0.\end{array}  \tag{12}\]  
Let \(\mathbf{L}=(L\otimes I_{m})\), \(\mathbf{W}=(W\otimes I_{m})\) and \(\mathbf{D}=(D\otimes I_{m})\), the Lagrangian function that defines this saddle point problem is \(\mathcal{L}_{2}(\mathbf{x},\mathbf{\lambda},\mathbf{z})=\langle\mathbb{F}(\mathbf{x}^{*}), \mathbf{x}\rangle+g(\mathbf{x})+\lambda^{\mathrm{T}}(\mathbf{Ax}-\mathbf{b})+\mathbf{z}^{\mathrm{ T}}\mathbf{L}\mathbf{\lambda}\). The optimal conditions for this problem are obtained by sequentially finding the partial derivatives of the variables \(\mathbf{x}\), \(\mathbf{y}\), and \(\mathbf{z}\) as follows:  
\[\left\{ \begin{array}{ll}0\in\mathbb{F}(\mathbf{x})+\mathbb{G}(\mathbf{x})+\mathbf{A}^{ \mathrm{T}}\mathbf{\lambda}\\ 0=\mathbf{L}\mathbf{\lambda}\\ 0\in\mathbf{b}+\mathbf{L}\mathbf{\lambda}+\mathbf{N}_{\mathbb{R}^{Nm}_{\geq 0}}( \mathbf{\lambda})-\mathbf{Ax}-\mathbf{L}\mathbf{z}.\end{array} \right. \tag{13}\]  
It follows from [PERSON]'s dual theorem that the variational problem has an optimal solution \(\mathbf{x}^{*}=\left(x_{i}^{*},\mathbf{x}_{-i}^{*}\right)\) only when \(\lambda_{g}^{*}\in\mathbb{R}^{m}\) satisfies the KKT conditions (13). Lemma 1: _If Assumptions 1-3 hold, then (13) implies (7)._  
Let \(v=\text{col}(\mathbf{x},\mathbf{z},\mathbf{\lambda})\) and consider the following operator:  
\[\mathcal{A}\left[ \begin{array}{c}\mathbf{x}\\ \mathbf{z}\\ \mathbf{\lambda}\end{array} \right]=\left[ \begin{array}{c}\mathbb{F}(\mathbf{x})\\ \mathbf{0}\\ \mathbf{b}\end{array} \right]+\left[ \begin{array}{c}\mathbb{G}(\mathbf{x})\\ \mathbf{0}\\ N_{\mathbb{R}^{Nm}_{\geq 0}}(\mathbf{\lambda})\end{array} \right]+\left[ \begin{array}{c} \mathbf{A}^{\mathrm{T}}\mathbf{\lambda}\\ \mathbf{L}\mathbf{\lambda}\\ -\mathbf{Ax}-\mathbf{L}\mathbf{z}\end{array} \right].