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I want to check does this set is concave

$$\{ (x,y,z) : 0 < z < xy \}$$

  • My solution: it isn't concave as For example: $A_{1} (1; 1; 0.1)$ and $A_{2} (-1; -1; 0.1)$. Both of this points in the set. But their linear combination with $\lambda = 0.5$ isn't in the set $A_{3} (0; 0; 0.1) =>$ set isn't concave

  • I'm trying to use general definition of concave set i.e. $$0 < z_{1} + (1-\alpha)z_{2} < (\alpha x_{1} + (1-\alpha)x_{2})(\alpha y_{1} + (1-\alpha)y_{2}) <=> \\ 0 < z_{1} + (1-\alpha)z_{2} < \alpha^{2} x_{1} y_{1} + \alpha(1-\alpha)x_{1}y_{2} + \alpha(1-\alpha)y_{1}x_{2} + (1-\alpha)^{2}x_{2}y_{2} <=> \\ 0 < z_{1} + (1-\alpha)z_{2} < \alpha (\alpha x_{1} y_{1} + y_{1} x_{2} - \alpha y_{1} x_{2}) + (1-\alpha)(\alpha x_{1} y_{2} + x_{2} y_{2} - \alpha x_{2} y_{2})$$ $=>$ $\begin{cases} \alpha x_{1} y_{1} + (1-\alpha)x_{1}y_{2} > z_{1} \\ \alpha y_{1} x_{2} + (1-\alpha)x_{2} y_{2} > z_{2} \end{cases} <=>$ $\begin{cases} x_{1} (\alpha y_{1} + (1-\alpha)y_{2} > z_{1} \\ x_{2} (\alpha y_{1} + (1-\alpha)y_{2}) > z_{2} \end{cases}$

  • So, I don't know what to do next? I feel I'm pretty close to prove that it isn't concave set. But could you help me to solve it a little bit?

Rodrigo de Azevedo
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David
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  • It looks like you already have a valid counterexample to show the set is concave (i.e. not convex). However the general definition you gave is incomplete, for example what is $\alpha$? – 79037662 Oct 08 '19 at 14:46
  • You basically need two steps: (1) state the definition of convex set (which has "for all" somewhere in it), then (2) show this set does not satisfy the definition, using those points you found. – 79037662 Oct 08 '19 at 14:50
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    How do you define a concave set? – copper.hat Oct 08 '19 at 15:29
  • @copper.hat "Concave" probably means "not convex", but its use is certainly far from universal. https://en.wikipedia.org/wiki/Convex_set#Non-convex_set – 79037662 Oct 08 '19 at 20:05
  • To use the term concave as the negation of convex is very misleading. Concave function is not a negation of convex function. In optics, concave lens is not a negation of convex lens etc. – A.Γ. Oct 09 '19 at 07:00

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