Superadditive function property

  • Thread starter jetoso
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  • #1
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Homework Statement


Show that a superadditive function has the following property:
For any superadditive function g on XxY (cartesian product):
f(x) = min { y' : y' = argmin g(x,y) }
is nonincreasing in x.


Homework Equations


if g(x,y) is a superadditive on XxY, x in X, y in Y, x1 >= x2, y1 >= y2, then it satisfies the inequality:

g(x1,y1) + g(x2,y2) >= g(x1,y2) + g(x2,y1)


The Attempt at a Solution


Let f(x1) = y', and suppose there is an x2 <= x1 such that f(x2) = y' then, g(x2,y1) - g(x2,y1) <= g(x2,y2) - g(x1,y2).

I am trying to find a contradiction, so that f is increasing in x for x2.
 

Answers and Replies

  • #2
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Nevermind

I think I solved this problem.
Let x2 >= x1, and take y2 >= f(x1) = y'. Then, from the definition of f, we have:
g(x1,f(x1)) - f(x1,y2) <= 0.

Since g is superadditive is satisfies:
g(x2,y2) + g(x1,f(x1)) >= g(x2,f(x1)) + g(x1,y2)

From the first inequality above we get:
g(x2,f(x1)) <= [g(x1,f(x1))-g(x1,y2)] + g(x2,y2) <= 0 + g(x2,y2) = g(x2,y2)

So, we have that:
g(x2,f(x1)) <= g(x2,y2)
for all y2 >= f(x1). Thus, f(x2)<=f(x1) which is nonincreasing in x.
 

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