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Math Help - Proving a real-valued function cannot be strictly concave

  1. #1
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    Proving a real-valued function cannot be strictly concave

    Hi,

    I've been doing some work that involves the following issue:

    Let f be a real-valued function defined over the positive quadrant of the plane. Suppose the directional derivative of f at (1,1) in the direction <-1,-1> is negative, and the directional derivative of f at (1,1) in the direction <-1,0> is positive. Can the function f be strictly globally concave?

    My wild guess is that f cannot have a strictly concave shape because there seems to be a "kink" of some sort at (1,1), but I don't know how to confirm my intuition. Is there an easy way to prove or disprove this? Any help would be much appreciated!
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  2. #2
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    Re: Proving a real-valued function cannot be strictly concave

    Hey Kreader12.

    I understand your intuition and I think the best way would be to show that a line exists in the form ta + (1-t)b has points for t in [0,1] where it is outside the plane (but a and b are on the deformed plane).

    One thing though I would consider is the Hessian matrix and use that to determine concavity. I think there are results that link the Hessian with concavity (or convexity) but you would need to look them up.
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