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We know that for $b \geq a$, and some $s \geq 0$, a concave function $f$ satisfies:

$f(a+s) - f(a) \geq f(b+s) - f(b)$.

This is not a frequent definition of concavity, but can be found, e.g., here.

My question is, what is the name of this property? Is it easy to prove without Taylor Expansions?

independentvariable
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1 Answers1

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The property you are referring to simply states that the directional derivative of a 1-dimensional concave function is non-increasing. You should be able to find a proof for that inequality here. The link proves it for convex functions, for concave functions simply reverse the sign of the inequality. I was unable to find a standard name for the inequality though, and yes it is easy to prove.

batwing
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