How can I build a fixed point argument for Nash Equilibrium in $\mathbb{L}^2$ Banach space?

Question

Consider the sets $Y_1,Y_2\subset\mathbb{L}^2(\Omega,\mathbb{F}_{s},\mathbb{P})$. I am looking for some $y_1$ measurable with respect to $Y_1$ and $y_2$ measurable with respect to $Y_2$ s.t. the pair $(y^*_1,y^*_2)$ constitutes a Nash Equilibrium. In particular, the pair $(y^*_1,y^*_2)\in \mathcal{Y}= \Pi_{i=1}^2Y_i$ is the cartesian product of the strategic choices of evry agent.

I present a setup similar to Vayanos and Wang of $2011$ model, where the set of strategic actions of every agent is in the intercept of their demand functions. Consider that two agents, indexed by $i=\{1,2\}$, have the following utility functions $$g_1(y_1)=-d_1(y_1)p(y_1,y_2)+\mathbb{E}[d_1(y_1)\bar{x}+\bar{z}_1|Y_1]-\gamma_1\mathbb{V}ar[d_1(y_1)\bar{x}+\bar{z}_1|Y_1]$$ and $$g_2(y_2)=-d_2(y_2)p(y_1,y_2)+\mathbb{E}[d_2(y_2)\bar{x}+\bar{z}_2|Y_2]-\gamma_2\mathbb{V}ar[d_2(y_2)\bar{x}+\bar{z}_2|Y_2]$$

where $(\bar{x},\bar{z}_i)$ are i.d and $Y_i$ measurable and also it holds that $\bar{x}\sim N(x,\sigma_{x}^2)$, $\bar{z}_i\sim N(0,\sigma_{z_i}^2)$, $\gamma_i\in (0,+\infty)$, $\mathbb{C}ov(\bar{x},\bar{z_i})\neq 0$, $\mathbb{C}ov(\bar{z_1},\bar{z_2})\neq 0$ and $g_i:Y_i\subset\mathbb{L}^2\rightarrow \mathbb{R}$ where $i\in\{1,2\}$. For the demand functions hold that $$d_i(y_i)=\frac{y_i-p(y_1,y_2)}{2\gamma_i\mathbb{V}ar[\bar{x}|Y_i]}$$ where $y_i$ is measurable with respect to $Y_i$ and $$p(y_1,y_2)=w_1y_1+w_2y_2+w_3y$$ s.t $w_1+w_2+w_3=1$

Every agent needs to solve the following problem $$y^*_i=\max_{y_i}g_i(y_i;y_{-i})$$ where $y_{-i}$ is the strategic choice of agent $-i$ (in our case the agents are two so equilevantely $y^*_1=\max_{y_1}g_i(y_1;y_{2})$) So, I will obtain two solutions for $y_1$ and $y_2$ and by solving the system of them, I result to the pair of $(y_1^*,y_2^*)$

$\textbf{The solutions of the agents' problems}$

By solving the maximization problem of agent-$1$, we obtain: \begin{equation}y_1(y_2)=\frac{1}{1+\omega_1}P_1+\frac{\omega_1\omega_2}{(1-\omega_1)(1+\omega_1)}y_2+\frac{\omega_1\omega_3}{(1-\omega_1)(1+\omega_1)}y\end{equation} where $P_1=\mathbb{E}[\bar{x}|Y_1]-2\gamma_1\mathbb{C}ov(\bar{x},\bar{z}_1|Y_1)$

Similarly, by solving the maximization problem of agent-$2$ \begin{equation}y_2(y_1)=\frac{1}{1+\omega_2}P_2+\frac{\omega_1\omega_2}{(1-\omega_2)(1+\omega_2)}y_1+\frac{\omega_2\omega_3}{(1-\omega_2)(1+\omega_2)}y\end{equation} where $P_2=\mathbb{E}[\bar{x}|Y_2]-2\gamma_2\mathbb{C}ov(\bar{x},\bar{z}_2|Y_2)$

By solving the system of $(y_1,y_2)$ we obtain that $$y_1^*=\frac{(1-\omega_1)(1-\omega_2^2)}{1-\omega_1^2-\omega_2^2}P_1+\frac{\omega_1(1-\omega_1)\omega_2}{1-\omega_1^2-\omega_2^2}P_2+\frac{\omega_1\omega_3}{1-\omega_1^2-\omega_2^2}y$$

and

$$y_2^*=\frac{\omega_1(1-\omega_2)\omega_2}{1-\omega_1^2-\omega_2^2}P_1+\frac{(1-\omega_1^2)(1-\omega_2)}{1-\omega_1^2-\omega_2^2}P_2+\frac{\omega_2\omega_3}{1-\omega_1^2-\omega_2^2}y$$

$\textbf{My question}$

My question is, how can I build the Banach fixed point argument for the aforementioned problem where the set of strategic choices is in the intercept of the demand functions?

Answer 1

Well, jsut to help your post and may someone else come and make clear what you need to check. Every agent needs to solve the following problem $$y^*_i=\max_{y_i}g_i(y_i)$$ So you will obtain a two solutions for $y_1$ and $y_2$ and by solving the system of them, you take the pair of $(y_1^*,y_2^*)$. The question is why sould this be a Nash Equilibrium and is this the unique equilibrium pair of strategies? The answer is deinitely yes for the second part of my-your question, and it is implied by the lienarity of the derivative of $g_i(y_i)$ with respect to $y_i$. So we talk about a unique solution. Why this is a N.E. or how the Banach fixed point theorem is going to be applied here, it is something else that I can not tell, because I do not know. I hope someone else could have a full answer, but in first place, this is a part of your question to be answered.