# If $a$ and $b$ are two pure rationalizable strategies, can $0.5a+0.5b$ fail to be a rationalizable strategy?

It is easy to see that if $\sigma$ is mixed rationalizable strategy, then a pure strategy $a$ such that $\sigma(a)>0$ is also rationalizable, but it seems not vice versa.

For a two-player finite game, let $P_1$ be the set of all pure rationalizable strategies for player 1, then the set of her rationalizable strategies is simply $\Delta(P_1)$, which can be seen as a corollary of equivalence of iterated strict dominance and rationalizability for two-player finite games.

What is an example for a finite game with more than two players such that the set of rationalizable strategies is strictly smaller than the mixed extension of the set of pure rationalizable strategies?

If only pure strategies are allowed, only $a$ and $b$ are rationalizable because they are best responses to $L$ and $R$ respectively while $c$ is not a best response to anything. But if you look at the mixed extension of the game then $c$ strictly dominates 0.5$a$+0.5$b$, hence 0.5$a$+0.5$b$ is not rationalizable. As a result, in the mixed extension all $a$, $b$ and $c$ are rationalizable.