Base Case of Loomis Induction
The base case of the induction in Direct Induction Proof Of Loomis Theorem: when $|I| + |J| = 2$ the matrices are $1 \times 1$ and the Loomis values reduce to a single ratio.
Statement
Let $I$ and $J$ be finite nonempty index types with $|I| + |J| = 2$, so $|I| = |J| = 1$ with single elements $i_0 \in I$ and $j_0 \in J$. Let $A, B \colon I \times J \to \mathbb{R}$ with $B_{i_0, j_0} > 0$. Then $$ \lambda_0 \;=\; \mu_0 \;=\; \frac{A_{i_0, j_0}}{B_{i_0, j_0}}. $$
Proof
Both simplices are singletons: $\Delta(I) = \{e_{i_0}\}$ and $\Delta(J) = \{e_{j_0}\}$, where $e_{i_0}$, $e_{j_0}$ are the unique unit mass strategies. For the unique $x = e_{i_0}$ and $j = j_0$ we have $(xA)_{j_0} = A_{i_0, j_0}$ and $(xB)_{j_0} = B_{i_0, j_0}$, so $$ \lambda_\mathrm{aux}(e_{i_0}) = \frac{A_{i_0, j_0}}{B_{i_0, j_0}}. $$ Taking the supremum over the singleton $\Delta(I)$ gives $\lambda_0 = A_{i_0, j_0} / B_{i_0, j_0}$. The argument for $\mu_0$ is identical. $\square$
References
- [MFoGT, Section 2.8, Exercise 1] Laraki, Renault, and Sorin, Mathematical Foundations of Game Theory. Base case of the Loomis induction.