# Nash equilibrium for two players game.

Consider a game for two players, say "Player A" and "Player B". The two sets of strategies are denoted by $A$ and $B$, available to the players. Consider a symmetric situation where the players have no means to cooperate and do not share any information about their strategies. The payoff functions are: $$\Phi^A (a,b)=-\frac{1}{1+2e^{-|a-b|/2}}, \ \ \Phi^B (a,b)=-\frac{1}{1+2e^{-|b-a|/2}}$$ and so $\Phi^A (a,b)=\Phi^B (a,b)$. Let $A=[10,+\infty)$ and $B=[0,1]$. Calculate the Nash equilibrium and determine if it is also a Pareto optimum.

This is what I have done so far. If $(\bar a,\bar b)$ is a Nash equilibrium, then $$\bar a= \operatorname{argmax}_{a\in A}\left\{ -\frac{1}{1+2e^{-|a-\bar b|/2}}\right\}$$ $$\bar b= \operatorname{argmax}_{b\in B}\left\{ -\frac{1}{1+2e^{-|\bar a- b|/2}}\right\}$$ Intuitively, I think that: $\bar a=10$, $\bar b=1$ is a Nash equilibrium. But I can not prove this. Also, I do not know how to study Pareto optimum.

Your Nash equilibrium is correct. Both payoffs are greater the smaller the difference between $a$ and $b$, so if this is made as small as possible by choosing the proximate boundaries of the intervals, neither player can improve on their strategy.
This also answers the question about Pareto optimality. Since $a=10$, $b=1$ is the global maximum, there's no way to improve on it, and hence in particular no way to improve on it for one player at the cost of the other.