# MiniMax Theorem

Consider the compact sets $X \in \mathbb{R}^n$, $Y \in \mathbb{R}^m$, $A \in \mathbb{R}^n$, $M \in \mathbb{R}^{n \times m}$.

For fixed $(\bar{a},\bar{B}) \in A \times M$, by the MiniMax Theorem we have that

$$\max_{x \in X} \min_{y \in Y} x^\top(\bar{a} + \bar{B} y) \ = \ \min_{y \in Y} \max_{x \in X} x^\top(\bar{a} + \bar{B} y).$$

Say if the following equality holds true, or find a counterexample.

$$\max_{x \in X} \min_{y \in Y} \max_{(a,B) \in A \times M} x^\top(a + B y) \ = \ \min_{y \in Y} \max_{x \in X} \max_{(a,B) \in A \times M} x^\top(a + B y)$$

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so $a_{max}$ and $b_{max}$ can be some functions of $x$ and $y$. Whatever function we have, if $a_{max}$ and $b_{max}$ are still in $A \times B$ then the equality holds. If we can not guarrantee that they are still in $A \times B$ then not necessarily it holds but in this case $min_{x\in X} max_{y\in Y}\geq max_{y\in Y} min_{x\in X}$. Now you wanna know if this happens or not. Right? – Seyhmus Güngören Aug 31 '12 at 16:23
No. To be precise, $(a_{\max},B_{\max})$ are not in $A \times M$, but they are functions whose of the kind $a_{\max}: X \times Y \rightarrow A$. Of course, by definition, $a_{\max}(x,y) \in A$ for all $(x,y) \in X \times Y$. Analogously for $B_{\max}$. But this does not directly imply that $\max_{x \in X} \min_{y \in Y} x^\top ( a_{\max}(x,y) + B_{\max}(x,y)y) = \min_{y \in Y} \max_{x \in X} x^\top ( a_{\max}(x,y) + B_{\max}(x,y)y)$. Otherwise, the question would be trivial. – Adam Aug 31 '12 at 16:30
Because $A \times B$ is also compact and any member of this set will satisfy the third equation. When we maximize $x^T(a+By)$, should we first maximize $a$ while $b$ is constant and vice versa or do they jointly maximize the equation? – Seyhmus Güngören Aug 31 '12 at 16:36
In my notation, $\max_{(a,B) \in A \times M}$ means "joint maximization". – Adam Aug 31 '12 at 16:42
In what context does this problem arise? – copper.hat Aug 31 '12 at 18:18