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J. B. Rosen proved that in concave games of n players (which assumes that Cartesian product of strategy profiles is convex) if the game satisfies the condition of diagonally strictly concave then there is a unique Nash equilibrium (see here).

Suppose that all users follow gradient descent algorithm (in discrete time space). Suppose that the game satisfies the diagonally strictly concave condition.

Does the strategy profile converge approximately to unique Nash equilibrium (up to some constant depends on the maximal step size)?

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  • $\begingroup$ What does "diagonally strictly concave" mean? $\endgroup$ Jul 2, 2015 at 17:32
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    $\begingroup$ Definition of diagonally strictly concave: Suppose we have $n$ players with strategies that lies in a convex set $S$. Their payoff functions $(u_1,..., u_n)$ are diagonally strictly if for every $x^1, x^2 ∈ S$, we have $(x^1- x^2 )^T \nabla u(x^2) + (x^2 − x^1)^T \nabla u(x^1) > 0$. $\endgroup$
    – Doron
    Jul 4, 2015 at 19:00

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the paper written by J. B. Rosen , has specified the maximum step size required for the convergence to equilibrium points which is a function of the cost functions.

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  • $\begingroup$ Wow! I missed that. Thank you so much behrad! $\endgroup$
    – Doron
    Jul 4, 2015 at 18:49

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