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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!

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Have you tried using the definition of convexivity? en.wikipedia.org/wiki/Convex_function#Definition – Paul Feb 15 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. – Godric Seer Feb 15 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 at 16:44

migrated from scicomp.stackexchange.com Feb 15 at 17:51

1 Answer

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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, $16 > (\frac{1}{2} \cdot 9 + \frac{1}{2} \cdot 9)$.

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