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  • I have a set: $$\{ (x, y, z) : z < x^{2} + y^{2} + 4xy \}$$

I want to prove that this set is convex or not. - I used general definition of convex set i.e.: $$z_{1}\alpha + (1-\alpha)z_{2} < (x_{1}\alpha + (1-\alpha)x_{2} + y \alpha + (1-\alpha)y_{2})^{2} + 2(...)(...)$$ But it looks too complex and I don't come with any solution.

  • After that, I used another approach "It isn't convex set. For example: $A_{1} (0.1; 0.1; 0)$ and $A_{2} (-0.4 ; 0.1; 0)$. Both of this points in the set. But their linear combination with $\lambda = 0.5$ isn't in the set $A_{3} (-0.15; 0.1; 0)$". So, I choose two points and proved that their linear combination isn't in the set, thus it isn't convex set.
  • But, I feel that it isn't a good idea, because I spend too much time for choosing "correct combination". Is there is a way to prove that it is/isn't convex set via general definition? I feel that something can be done with the first method.
Rodrigo de Azevedo
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David
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  • in this case you can immediately tell that the set is not convex even when $y$ and $z$ are fixed – LinAlg Oct 08 '19 at 12:07
  • @LinAlg I didn't catch the idea. Could you clarify please? – David Oct 08 '19 at 12:08
  • These proofs become easier when you know that $x^2$ is convex, and that convex functions must be on the smaller side of an inequality sign. The $x^2$ term is on the wrong side of the inequality sign – LinAlg Oct 08 '19 at 12:14
  • The point is that for any fixed $y$ and $z$, the inequality will be true when $|x|$ is large enough. Take $y$ and $z$ such that the inequality is false for some $x_0$, and you can find $x_1$ and $x_2$ with $x_1 < x_0 < x_2$ such that $(x_1, y,z)$ and $(x_2, y, z)$ are in your set but $(x_0, y, z)$ is not. – Robert Israel Oct 08 '19 at 12:39
  • After a while you will recognise the general form of such problems. – copper.hat Oct 08 '19 at 14:07
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    @LinAlg $x^2$ may be convex, but $4xy$ makes the RHS indefinite. – Rodrigo de Azevedo Oct 09 '19 at 06:59

1 Answers1

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Check some 'standard' points, such as the origin and notice that $(x,y,z)=(0,0,0)$ is not in the set.

Now see if there are points in the set on 'either side' such that the origin is on the line joining the points.

For example, take $(1,0,0), (-1,0,0)$ which are in the set and a convex combination ${ 1 \over 2 } ((1,0,0)+(-1,0,0))$ is not.

copper.hat
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