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Suppose there is a real valued function $f(x,y)$ where $x,y$ are vector variables $x\in\mathbb{R}^n$ $y\in\mathbb{R}^m$. Moreover $f$ is strictly/concave in $x$ and strictly/convex in $y$. $f$ is also continuous. The problem is $$\max_x\min_y f(x,y)=\min_y\max_x f(x,y).$$

Please explain to me how to analyse this problem. From When $\min \max = \max \min$? I see that it is not generally true. But I guess the convexity information given above may force equality.

Thanks a lot

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triomphe
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  • If $f$ (or its partial functions) is not continuous, the $\sf min$ and $\sf max$ may not exist (convexity/concavity is not enough to guarantee continuity). Perhaps you mean $\sf inf$ and $\sf sup$. – Ewan Delanoy Oct 22 '13 at 11:17
  • @EwanDelanoy Thanks. I added continuity. – triomphe Oct 23 '13 at 01:13
  • I guess you also need the the domain of definition to be compact (think of $f(x,y)=xy$ : there is no $\min$ or $\max$ except when $x$ or $y$ is zero). – Ewan Delanoy Oct 23 '13 at 10:40
  • @EwanDelanoy The domain I took to be $\mathbb{R}^n\times \mathbb{R}^m$. So will it work then? – triomphe Oct 23 '13 at 19:40

1 Answers1

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According to the definitions that you made, first the $\min$ and $\max$ should exist. For this you will need a compact space from where you draw $x$ and $y$. In addition to this, if your objective function $f$ is convex in $y$ and concave $x$ then According to Von Neumann's minimax theorem

$$\max_x\min_y f(x,y)=\min_y\max_x f(x,y).$$

holds. This also implies that the topology that you have accepts a saddle point. http://en.wikipedia.org/wiki/Saddle_point If you have such a topology then the order of minimization or maximization doesnt matter. If this does not hold, then you have

$$\max_x\min_y f(x,y)\leq \min_y\max_x f(x,y)$$

Von Neumann's minimax theorem gives us a sufficient condition such that we have a saddle point

Seyhmus Güngören
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