TITLE: maximization and minimization
QUESTION [1 upvotes]: 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?

REPLY [2 votes]: 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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