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Show that $f(x_1^*,...x_n^*)=\max\{f(x_1,...,x_n):(x_1,...,x_n)\in\Omega\}$ if and only if $-f(x_1^*,...x_n^*)=\min\{-f(x_1,...,x_n):(x_1,...,x_n)\in\Omega\}$

I am not exactly sure how to approach this problem -- it is very general, so I can't assume anything about the shape of $f$. It seems obvious that flipping the $\max$ problem with a negative turns it into a $\min$ problem. Thoughts?

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2 Answers 2

Follow the chain of equivalences here:

$$\begin{align} & f(x_1^*,...,x_n^*)=\max\{f(x_1,...,x_n):(x_1,...,x_n)\in\Omega\} \\ \iff & (x_1^*,...,x_n^*) \in \Omega \text{ and } \forall (x_1,...,x_n)\in\Omega, f(x_1^*,...x_n^*) \geq f(x_1,...,x_n) \\ \iff & (x_1^*,...,x_n^*) \in \Omega \text{ and } \forall (x_1,...,x_n)\in\Omega, -f(x_1^*,...,x_n^*) \leq -f(x_1,...,x_n) \\ \iff & -f(x_1^*,...,x_n^*)=\min\{-f(x_1,...,x_n):(x_1,...,x_n)\in\Omega\}. \end{align}$$

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I came across this question when it was recently bumped to the front page, so I wrapped your chain of equivalences in an align environment. Hope that's OK. –  Rahul May 16 '12 at 7:03
    
Great! Thanks very much. –  copper.hat May 16 '12 at 7:17

Why can't you just multiply $$\forall \mathbf y \in \Omega,\qquad\max(f(\mathbf x): \mathbf x \in \Omega)\ge f(\mathbf y)$$ with $-1$?

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One cannot recommend the use of a same symbol in the maximum on the LHS and as an argument in the RHS. –  Did May 16 '12 at 5:50
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“Bureaucrat Conrad, you are technically correct — the best kind of correct.” (Futurama, 2acv11: How Hermes Requisitioned His Groove Back) –  fabee May 17 '12 at 8:45

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