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Paul Duran 2022-05-09 Answered
Show that f ( x 1 , . . . x n ) = max { f ( x 1 , . . . , x n ) : ( x 1 , . . . , x n ) Ω } if and only if f ( x 1 , . . . x n ) = min { f ( x 1 , . . . , x n ) : ( x 1 , . . . , x n ) Ω }
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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Answers (1)

allstylekvsvi
Answered 2022-05-10 Author has 16 answers
Follow the chain of equivalences here:
f ( x 1 , . . . , x n ) = max { f ( x 1 , . . . , x n ) : ( x 1 , . . . , x n ) Ω } ( x 1 , . . . , x n ) Ω  and  ( x 1 , . . . , x n ) Ω , f ( x 1 , . . . x n ) f ( x 1 , . . . , x n ) ( x 1 , . . . , x n ) Ω  and  ( x 1 , . . . , x n ) Ω , f ( x 1 , . . . , x n ) f ( x 1 , . . . , x n ) f ( x 1 , . . . , x n ) = min { f ( x 1 , . . . , x n ) : ( x 1 , . . . , x n ) Ω } .
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