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Let $C \subset \mathbb{R}^n$, $D \subset \mathbb{R}^m$ be compact sets, and let $\phi : \mathbb{R}^n \times \mathbb{R}^m \rightarrow \mathbb{R}$ be continuous. Define the functions $g : \mathbb{R}^n \rightarrow \mathbb{R}$ and $h : \mathbb{R}^m \rightarrow \mathbb{R}$ by

$g(x) := \max\{\phi(x, y) \mid y \in D\}, \quad h(y) := \min\{\phi(x, y) \mid x \in C\}$

Show that $\max\{\min\{\phi(x, y) \mid x \in C\} \mid y \in D\} = \max\{h(y) \mid y \in D\} \leq \min\{g(x) \mid x \in C\}$.

I have already proved this. Now I tried it with an example:

$\phi(x,y)=xy+y^2,C=D=[-1,1]$

Then $g(x)=\max\{x+1,1-x\}$ because the maximum is attained at $y=\pm1$ and $h(y)=\min\{y+y^2,-y+y^2\}$ because the minimum is attained at $x=\pm1$.

But then the minimum of $g$ in $x$ is $0$ and the maximum of $h$ in $y$ is $2$. I must have made some error! Could you tell me where?

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2 Answers 2

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No contradiction: $$ \max_{y\in[-1,1]} h(y)=0 $$ Please see the graph of $h(y)$:enter image description here

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Let $y^*(x)$ be such that $g(x)=\phi(x,y^*(x))=\max_y\phi(x,y)$, a point where the maximum is achieved. Then, for a given $x$, $$ \phi(x,y)\le \phi(x,y^*(x))=g(x)\quad\text{for all }y. $$ Taking the minimum over $x$, $$ \min_x \phi(x,y) \le \min_x g(x)\quad\text{for all }y. $$ Note that the RHS does not depend on $y$, hence $$ \max_y(\min_x \phi(x,y)) \le \min_x g(x). $$

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  • $\begingroup$ Thank you very much, however, as stated in my post, I have already proved this. ;) My question was about the example I had and where the error was because it didn't verify the theorem. $\endgroup$ Commented Apr 15 at 14:58

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