Weak Duality For Loomis Values
The "easy" direction of the Loomis theorem: any maxmin Loomis ratio is bounded by any minmax Loomis ratio. Used by the direct Loomis induction proof to set up the sandwich $\lambda_0 \le \mu_0$.
Statement
Fix $A, B \colon I \times J \to \mathbb{R}$ with $B$ entrywise positive. With $\lambda_0$ and $\mu_0$ defined in Existence of Loomis Optimisers, $$ \lambda_0 \;\le\; \mu_0. $$
Proof
Let $x_0 \in \Delta(I)$ and $y_0 \in \Delta(J)$ be the optimisers supplied by Existence of Loomis Optimisers, so that for every $j \in J$ and every $i \in I$ $$ (x_0 A)_j \ge \lambda_0\,(x_0 B)_j, \qquad (A y_0)_i \le \mu_0\,(B y_0)_i. $$ Weight the first family by $y_{0,j} \ge 0$ and sum over $j$: $$ \sum_j y_{0,j} (x_0 A)_j \;\ge\; \lambda_0 \sum_j y_{0,j} (x_0 B)_j, $$ i.e. $x_0 A y_0 \ge \lambda_0 \cdot (x_0 B y_0)$. Symmetrically, weight the second family by $x_{0,i} \ge 0$ and sum over $i$: $$ x_0 A y_0 \;\le\; \mu_0 \cdot (x_0 B y_0). $$ Combining, $$ \lambda_0 \cdot (x_0 B y_0) \;\le\; x_0 A y_0 \;\le\; \mu_0 \cdot (x_0 B y_0). $$ By Positive Aggregates xB and By the quantity $x_0 B y_0$ is strictly positive, so dividing gives $\lambda_0 \le \mu_0$. $\square$
Use
This is the first half of the Loomis sandwich. The reverse direction $\mu_0 \le \lambda_0$ is established by ruling out the strict inequality $\lambda_0 < \mu_0$ via the column-drop and row-drop steps of the direct Loomis induction.
References
- [MFoGT, Section 2.5] Laraki, Renault, and Sorin, Mathematical Foundations of Game Theory. Weak duality for the Loomis ratios.