## 4 Game Analysis  
This section is dedicated for the analysis of the proposed game \(\mathcal{G}\). We first demonstrate the existence of the Nash equilibrium. **Theorem 2**. : _The game \(\mathcal{G}=\left(\mathcal{N},\left\{\mathcal{P}_{n}\right\}_{n=1}^{N},\left\{ u_{n}\right\}_{n=1}^{N}\right)\) possesses a Nash equilibrium._  
Proof. : For each player \(n\in\mathcal{N}\), notice the following:  
* In both proposed instances, the actions set of every player is closed and bounded, and hence it is compact. Moreover, both actions' sets are convex. * Theorem 1 shows that \(\|\mathbf{f}_{n}\|^{2}\) is a strictly convex function on \(\mathbf{p}_{n}\), which implies that the utility function of the \(n\)th player is strictly concave on its own actions. * The utility function is a continuous function on the power vector \(\mathbf{p}\). As a result, using Theorem (1) in [26], it follows in a straightforward manner that \(\mathcal{G}=\left(\mathcal{N},\left\{\mathcal{P}_{n}\right\},\left\{u_{n} \right\}\right)\) has a Nash equilibrium point. Due to the fact that each player's utility function is strictly concave in the player's own action \(\mathbf{p}_{n}\), there is only one strategy that maximizes the utility of that player (given a particular strategy profile of the rest of the players). Thus, if every player plays its best response (iteratively), they would achieve a unique Nash equilibrium. For a typical cell \(n\), let the experienced intercell interference at the user equipment (UE) be \(I_{n}^{d}\) and the intercell interference at the base station (BS) be \(I_{n}^{u}\). The best response \(p_{n}^{d}\) and \(p_{n}^{u}\) of cell \(n\) against the actions of the other cells is a crucial quantity for finding the Nash equilibrium. Obviously, the best response should maximize the utility function \(-\|\mathbf{f}\|^{2}=-\|\mathbf{A}\mathbf{p}_{n}-\mathbf{b}\|^{2}\) subject to \(\mathbf{I}\leq\mathbf{C}\mathbf{p}_{n}\leq\mathbf{u}\), where * for problem **P1**, \[\textbf{C}=\left[ \begin{array}{cc}1&0\\ 0&1\end{array} \right],\textbf{1}=\left[ \begin{array}{c}0\\ 0\end{array} \right]\quad\text{and }\textbf{u}=\left[ \begin{array}{c}p_{max}^{d} \\ p_{max}^{u}\end{array} \right],\] (9)
* for problem **P2**, \[\textbf{C}=\left[ \begin{array}{cc}1&0\\ 0&1\\ 1&1\end{array} \right],\textbf{1}=\left[ \begin{array}{c}0\\ 0\end{array} \right]\quad\text{and }\textbf{u}=\left[ \begin{array}{c}p_{max}^{d}+p_{max}^{u} \\ p_{max}^{d}+p_{max}^{u}\\ p_{max}^{d}+p_{max}^{u}\end{array} \right].\] (10)  
For computational efficiency reasons, it is beneficial to transform **A** to a triangular matrix using QR factorization; \(\textbf{A}=\textbf{Q}\textbf{R}\), where **Q** is an orthogonal matrix and **R** is an upper triangular matrix (Since **A** is nonsingular, the diagonal elements of **R** are nonzero). Since **Q** is orthogonal, the objective function can equivalently be written as: \(-\|\textbf{R}\textbf{p}_{n}-\textbf{\tilde{b}}\|^{2}\), where \(\textbf{\tilde{b}}=\textbf{Q}^{T}\textbf{b}\). Now, let \(\textbf{y}=\textbf{R}\textbf{p}_{n}-\textbf{\tilde{b}}\), which implies \(\textbf{p}_{n}=\textbf{R}^{-1}\left(\textbf{y}+\textbf{\tilde{b}}\right)\). Thus, the above problem can equivalently be written as:  
\[\underset{\textbf{y}}{\text{minimize}}\|\textbf{y}\|^{2}\ \ \text{subject to:}\ \ \textbf{\tilde{1}}\leq\textbf{M}\textbf{y}\leq\tilde{\textbf{u}}, \tag{11}\]  
where \(\textbf{M}=\textbf{C}\textbf{R}^{-1}\), \(\textbf{\tilde{1}}=\textbf{1}-\textbf{M}\textbf{\tilde{b}}\) and \(\tilde{\textbf{u}}=\textbf{u}-\textbf{M}\textbf{\tilde{b}}\). This problem is called the least distance problem (LDP). The constraint in (11) can also be written as \(\left( \begin{array}{c}-\tilde{\textbf{u}}\\ \textbf{\tilde{1}}\end{array} \right)\leq\left( \begin{array}{c}-\textbf{M} \\ \textbf{M}\end{array} \right)\textbf{y}\), or equivalently in short \(\textbf{\tilde{1}}\leq\textbf{\tilde{M}}\textbf{y}\). For solving the problem in (11), we restate a useful theorem in ([24], Chap.