# Show that f ( <msubsup> x 1 &#x2217;<!-- ∗ --> </msubsup> , . .

Paul Duran 2022-05-09 Answered
Show that $f\left({x}_{1}^{\ast },...{x}_{n}^{\ast }\right)=max\left\{f\left({x}_{1},...,{x}_{n}\right):\left({x}_{1},...,{x}_{n}\right)\in \mathrm{\Omega }\right\}$ if and only if $-f\left({x}_{1}^{\ast },...{x}_{n}^{\ast }\right)=min\left\{-f\left({x}_{1},...,{x}_{n}\right):\left({x}_{1},...,{x}_{n}\right)\in \mathrm{\Omega }\right\}$
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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