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

Question

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?

Answer 1

Yes. Consider a two player game described by the following matrix

\begin{array}{|c|c|c|} \hline & L & R \\ \hline a & 3,0 & 0,0 \\ \hline b & 0,0 & 3,0 \\ \hline c & 2,0 & 2,0 \\ \hline \end{array}

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.

If you insist on an example with three players I can add a third one to this. This third player has exactly one possible strategy, so it does not really matter what his payoffs are and what he does.
If you think that is cheating, the third player can have multiple strategies that do not at all influence the payoff of the first two players.