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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.

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.

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