Consider a static, complete information game with 2 players.
Strategy sets are $S_1=\{U,D\},S_2=\{l,m,r\}$.
Payoffs are irrelevant to this question as I am trying to get the concept of rationalizability correct.
Suppose I want to verify whether $m$ is a rationalizable strategy for player 2.
Then, I want to ask the following question:
$\exists \sigma_1=(q^*,1-q^*)\in\Delta(S_1)$ such that for $\sigma^*_2=(0,1,0),$ $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,\sigma_2)$ for all $\sigma_2\in\Delta(S_2)?$
Now, suppose I have payoff matrix such that I could find $(q^*,1-q^*)$ such that it satisfies both:
(1) $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,(1,0,0))$
(2) $u_2(\sigma_1,\sigma^*_2)\geq u_2(\sigma_1,(0,0,1))$.
This means, I could find a valid range of $q^*$ such that for player 2, choosing $m$ provides a weakly better payoff for her compared to the degenerate (i.e. pure) strategies of $l$ or $r$.
My question is:
If I could find such $q^*$ that satisfies both (1),(2), then I do not have to check for any other strategy profiles in $\Delta(S_2)$, that is any convex combo of $(1,0,0)$ and $(0,0,1)$? My intuition is that for any $\alpha\in[0,1]$, I could simple have:
(1)' $\alpha u_2(\sigma_1,\sigma^*_2)\geq \alpha u_2(\sigma_1,(1,0,0))$
(2)' $(1-\alpha)u_2(\sigma_1,\sigma^*_2)\geq (1-\alpha)u_2(\sigma_1,(0,0,1))$.
and show that (1)'+(2)'implies the degenerate strategy $m$ for player 2 is a best response to some belief, $\sigma_1\in\Delta(S_1)$. Hence, the bottom line is (1),(2) is sufficient, and I do not have to check the convex combo of the two other pure strategies.
