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Define function $F$ as $F(x,y,x,t)= (xy-zt)^2$ where $x,y,z,t \geq 0$.

Question: Is this function Convex?

Thanks!

Willie Wong
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Star
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  • Have you tried using the definition of convexivity? http://en.wikipedia.org/wiki/Convex_function#Definition – Paul Feb 15 '13 at 14:55
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    If you have done a large proof along these lines, then including information on it and where you thought you made your mistake would be helpful. There is no reason for us to start from step zero here if we can simply verify some of what you have done. –  Feb 15 '13 at 16:23
  • @Star: I agree with GodricSeer. If you could provide some of the details (or an outline) of your proof, and where in the proof you need help, we will be able to help you better. – Paul Feb 15 '13 at 16:44

1 Answers1

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No, this function is not convex.

Consider the points $(x, y, z, t) = (3, 1, 0, 0), (1, 3, 0, 0)$. Their midpoint is $(2, 2, 0, 0)$. If $F$ is convex, then $F(2, 2, 0, 0) \leq \frac{1}{2}F(3, 1, 0, 0) + \frac{1}{2}F(1, 3, 0, 0)$. However, $16 > (\frac{1}{2} \cdot 9 + \frac{1}{2} \cdot 9)$.

Geoff Oxberry
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