For some strictly convex functions f 1 </msub> ( x 1 </msub>

kromo8hdcd 2022-05-09 Answered
For some strictly convex functions f 1 ( x 1 ) , f 2 ( x 2 ) , . . . , f n ( x n ), it seems intuitive that for a given sum of x 1 + x 2 + . . . + x n = X, where x 1 , x 2 ,..., x n 0
k { 1 , 2 , , n } , f k ( X ) = max { i = 1 n f i ( x i ) : i = 1 n x i = X  and  i , x i 0 }
Since these functions are convex, the sum of these functions is greatest when the sum of arguments is equal to the argument of just one function. This seems intuitive, but I can't think of how to argue this in formal maths. Any suggestions?
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Answers (1)

exorteygrdh
Answered 2022-05-10 Author has 9 answers
You can use the following theorem: if the maximum of a convex function on a convex set is attained, then it is attained at an extreme point.

In your case, the convex set is a simplex, which is a compact set. The function F ( x ) = i = 1 n f i ( x i ) is convex, and thus continuous. Therefore, the maximum is attained by Weirstrass. The constraints define a convex polytope, whose extreme points are its vertices, which are ( X , 0 , , 0 ) ,   ( 0 , X , 0 , , 0 ) , , ( 0 , , 0 , X ). The result you wish follows from the fact that the maximum is attained at one of those vertices.
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