0

I'm trying to prove or disprove the convexity of $f(x,y)=x^2y^2$.

This is part of a larger function but I think I proved that the rest of the function is convex using Hessian's. The other term in the function is $a(x) = x^4+x^2-2x+5$. The last term is $b(y) = y^2$.

If I plot the function $f(x,y)=x^2y^2$ on Wolfram Alpha it looks pretty convex but I need to prove it using algebra. $f(x,y)$'s Hessian $H$, its minor determinants or its eigenvalues didn't help me to prove that $H$ is positive semi-definite. I can't find two points that disprove the convexity inequality either:

$$\lambda f(x_1,y_1) + (1-\lambda)f(x_2,y_2) \ge f(\lambda (x_1,y_1) + (1-\lambda)(x_2,y_2))$$

Any help would be greatly appreciated.

Thanks a lot and have a great day, Maxime

EDIT: I should have looked more closely at the plot and the Hessian.

Maxime Leclerc
  • 9
  • 3
    The Hessian of $f(x,y)$ is indefinite. The function is clearly not convex, because it is zero on the axes and positive otherwise. Any segment connecting two points on the axes will disprove convexity. – A.Γ. Feb 12 '16 at 20:09
  • Thanks A.G. I'm new at this but you're absolutely right, the points (1,0), (0,1) and (1/2,1/2) disprove convexity. Now I just have to adapt these points for the original function f+a+b. – Maxime Leclerc Feb 12 '16 at 20:25
  • @MaximeLeclerc Proving by definition seems cumbersome here. I would go via the Hessian. The determinant has a negative term that can be made dominating if we set $y=x^2$. So the Hessian is indefinite somewhere far away along the parabola, hence the function is not convex. – A.Γ. Feb 12 '16 at 20:58
  • @AnneBauval That question is a duplicate of this one as the one you linked is from this year while the other is from 2016 even though the linked question has answers – Тyma Gaidash May 14 '23 at 12:14
  • @TymaGaidash I know, but imo the main point is that this link is useful, both for the readers and to avoid duplicate answers. – Anne Bauval May 14 '23 at 12:16

0 Answers0