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I'm trying to formulate a game so that at Nash equilibrium I achieve supply equales demand. Then I ran into this problem.I asked this question in mathexchange as well, not sure where it belongs to.

For all $i,$ $v_{i}\left(x_{i}\right)$ is concave in $x_{i}$. The value function of user $i.$ For all $k$ $c_{k}\left(y_{k}\right)$ is convex in $y_{k}$ and non negative. The cost function of producer $k.$ So that $\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)$ is concave.

Let $x=\left(x_{1},\,\cdots x_{N}\right)$ and $y=\left(y_{1},\cdots y_{M}\right).$

Let price per unit be

$$p=a_{0}+a_{1}\sum_{i}x_{i}+a_{2}\sum_{k}y_{k}.$$

And the objective, which I have verified is an exact potential function of the game, is

$$\Phi\left(x,\, y\right)=\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)-p\sum_{i}x_{i}+p\sum_{k}y_{k}.$$

Show that under some assumptions $\exists\, a_{0},\, a_{1},\, a_{2}$ such that if,

$$\left(x^{},\, y^{}\right)=\underset{y_{k}\in (x^{'}, y^{'})=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x,\, }\Phi\left(x, y\right)$$

then $\sum_{i\in I}x_{i}^{*}=\sum_{k\in K}y_{k}^{*}$I}x_{i}^{'}=\sum_{k\in K}y_{k}^{'}$. I think this statement is equivalent to showing$\exists\, a_{0},\, a_{1},\, a_{2}$such that if, $$\left(x^{},\, y^{}\right)=\underset{y_{k}\in left(x^{"},y^{"}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x,\, y\right) y\right).$$ and if $$\left(x^{'},\, left(x^{'}, y^{'}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i},\,\sum_{i\in I}x_{i}=\sum_{k\in K}y_{k}}{\text{arg max }}\Phi\left(x,\, y\right)$$ then$\left(x^{},\, y^{\left(x^{'}, y^{'}\right)=\left(x^{"}, y^{"} \right)=\left(x^{'},\, y^{'}\right)$\right)$

For the case when $a_{1}=a_{2}=0$ I have proved that $a_0$ is the Lagrange multiplier but I Can't solve for this general case.

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# Optimization with parameters

I'm trying to formulate a game so that at Nash equilibrium I achieve supply equales demand. Then I ran into this problem. I asked this question in mathexchange as well, not sure where it belongs to.

For all $i,$ $v_{i}\left(x_{i}\right)$ is concave in $x_{i}$. The value function of user $i.$ For all $k$ $c_{k}\left(y_{k}\right)$ is convex in $y_{k}$ and non negative. The cost function of producer $k.$ So that $\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)$ is concave.

Let $x=\left(x_{1},\,\cdots x_{N}\right)$ and $y=\left(y_{1},\cdots y_{M}\right).$

Let price per unit be

$$p=a_{0}+a_{1}\sum_{i}x_{i}+a_{2}\sum_{k}y_{k}.$$

And the objective, which I have verified is an exact potential function of the game, is

$$\Phi\left(x,\, y\right)=\sum_{i}v_{i}\left(x_{i}\right)-\sum_{k}c_{k}\left(y_{k}\right)-p\sum_{i}x_{i}+p\sum_{k}y_{k}.$$

Show that under some assumptions $\exists\, a_{0},\, a_{1},\, a_{2}$ such that if $$\left(x^{},\, y^{}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x,\, y\right)$$ then $\sum_{i\in I}x_{i}^{*}=\sum_{k\in K}y_{k}^{*}$.

I think this statement is equivalent to showing $\exists\, a_{0},\, a_{1},\, a_{2}$ such that if $$\left(x^{},\, y^{}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i}}{\text{arg max }}\Phi\left(x,\, y\right)$$ and if $$\left(x^{'},\, y^{'}\right)=\underset{y_{k}\in Y_{k},\, x_{i}\in X_{i},\,\sum_{i\in I}x_{i}=\sum_{k\in K}y_{k}}{\text{arg max }}\Phi\left(x,\, y\right)$$ then $\left(x^{},\, y^{}\right)=\left(x^{'},\, y^{'}\right)$

For the case when $a_{1}=a_{2}=0$ I have proved that $a_0$ is the Lagrange multiplier but I Can't solve for this general case.