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!

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.