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I have taken an optimization course but It seems that I don't have its basics so any reference would be appreciated,

In the publication , it is stated

any limit point satisfies the Nash equilibrium conditions.

After reading I knew that Nash equilibrium is related to game theory where one of 2 participants can't gain by change of strategy as long as as the other participant is on the same strategy.

I also found that limit point is the stationary point and in other documents that it is the result of the limits operation, where a point doesn't exist in a set but tends to get close to it.

My problem is that I can't relate both of them.

Definition of Nash point in the paper in page 6 for objective function,

\begin{equation*} \begin{aligned} & \underset{x \in \mathcal{X} }{\text{minimize}} & & F(x_1, ...,x_s) = f(x_1,...,x_s) + \sum_{i=1}^s (r_i(x_i)) \\ & \text{subject to} & & F(x_i) \geqslant 0 , \quad where \quad i = 1, ...,s \end{aligned} \end{equation*}

is

\begin{equation*} F(x^*_1,x_{i-1}^*,x_i^*,x_{i+1}^*,...,x_s^*) \leqslant F(x^*_1,x_{i-1}^*,x_i,x_{i+1}^*,...,x_s^*), \forall x_i \in \mathcal{X_i^*} \\ \mathcal{X_i^*} = \mathcal{X_i}(x_1^*,...,x_{i-1}^*,x_{i+1}^*,...,x_s^*) \end{equation*}

where $x_i^*$ is a Nash point.

Mour_Ka
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    Unfortunately, it doesn't seem that your article can be viewed without a subscription. Regardless, it's not good Math.SE practice to require readers to rely on reading external articles to get the necessary context for a question or answer. Please consider providing more details surrounding the claim (specifically, what the limit point is of). (I'm aware that the abstract itself has this claim, but the article claims to explain and prove it.) – Michael Grant May 28 '17 at 12:55
  • @MichaelGrant, Sorry for the confusion, I edited the link for access to the article. I hope if its possible to elaborate the relation between the stationary point and Nash point. That in the case that the limit point is the stationary points. – Mour_Ka May 28 '17 at 13:49

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