> Seem neither Von Neumann nor Morgenstern were impressed by Nash's use of Kakutani fixed-points to come up with equilibrium solutions, they see it as impractical and difficult to apply, which it is because it assumes common knowledge of the each player's expected utility.
Von Neumann would have thought no such thing! Nash used a fixed-point theorem for an existence proof, which von Neumann himself had done in earlier work.
"Assumes common knowledge of each player's expected utility" doesn't make sense in this context.
> Since then Nash's theory has been shown to be not very robust with uncertainty.
It's not clear which kind of uncertainty you're referring to here, but it doesn't matter. Nash's theorem is a mathematical theorem, and the proof is sound. It can't become less true over time.
> Nash's theorem is a mathematical theorem, and the proof is sound. It can't become less true over time.
Nash proved the existence of the solution concept he defined. It is true that at least one Nash Equilibrium exists for all games (with every information structure). It is more subjective as to whether 'Nash Equilibria' are particularly useful or interesting to specific classes of games.
Indeed, VN-M had already proposed a different solution concept that they proved existed in a narrower range of games (best response equilibria in 2 player 0 sum games).
In games with asymmetric information structures, stricter equilibrium concepts than Nash Equilibrium are often used, because there are typically a large number of Nash Equilibria for any game.
For example, Bayesian Nash Equilibrium and Perfect Bayesian Equilibrium restrict agents to forming beliefs in a 'Bayesian' manner, whilst the latter also restricts their actions to also be BNE in subgames off the equilibrium path of sequential games.
Von Neumann would have thought no such thing! Nash used a fixed-point theorem for an existence proof, which von Neumann himself had done in earlier work.
"Assumes common knowledge of each player's expected utility" doesn't make sense in this context.
> Since then Nash's theory has been shown to be not very robust with uncertainty.
It's not clear which kind of uncertainty you're referring to here, but it doesn't matter. Nash's theorem is a mathematical theorem, and the proof is sound. It can't become less true over time.