# Why is Rosenbrock's Banana Function not convex over a positive semi-definite hessian region?

This was a two part question;

Part 1)

"Given f(x,y) = 36(y-x^2)^2+(x-1)^2, show that the hessian is positive semi-definite in the region y <= x^2 + 1/72."

I solved this I think by subbing y <= x^2 + 1/72 into f(x,y) to get;

f(x, x^2 + 1/72) <= x^2 - 2x + 145/144 (I cut out all the algebra, this was the final result)

then solving for the hessian (which was [2,0;0,0] and I think because there are no variables, its just an equality instead of inequality now), giving eigenvalues 2 and 0. This indicates positive semi-definiteness, which should indicate convexity of some form I believe. I'd appreciate some comment on this 'proof',but I do think it shows that the region is positive semi-definite. But the second part asks;

Part 2)

" Explain why the function is not convex in this region? " and I have no idea. To my knowledge, positive semi-definite means convexity. I would really really appreciate some form of explanation here. Thank you very much!

• Positive semi-definiteness does not mean convexity, it means that the issue is decided by higher order terms. For example, $x^4-y^4$ is positive semidefinite at the origin, but obviously not convex. Only positive definiteness is decisive, because the second order terms dominate. – Conifold Apr 27 at 7:30
• @Conifold Positive-semidefinite Hessian in a convex set is equivalent to convexity for $C^2$ functions. – A.Γ. Apr 27 at 7:37
• @A.Γ. thanks, I didn't know it only applied to second quadratic functions only. But yes, positive semi-definite Hessian should imply 'soft' convexity on this region. – Patrick Christie Apr 27 at 7:39
• 1. Parameterizing the boundary and replacing the function with inequality makes no sense in estimation the Hessian. You have to calculate the Hessian and find out if it is pos.semidef, in the region. 2. The region is not convex. – A.Γ. Apr 27 at 7:41
• @A.Γ. Hm, ok. I did it by calculating the hessian then substituting y = x^2+1/72 and have found that it is still positive semi-definite with eigenvalues 0 and 72+288x^2. I still have no idea why it's not convex on the region though. Sorry if I'm missing something in your reply. – Patrick Christie Apr 27 at 7:50

The Hessian of your function $$f$$ is the matrix of its second partial derivatives: $$H(x,y):=\left[\matrix{f_{xx}&f_{xy}\cr f_{xy}&f_{yy}\cr}\right]=\left[\matrix{2+432x^2-144y&\quad-144x\cr -144x&\quad72\cr}\right]\ .$$ The determinant of this matrix is $$\det H(x,y)=144(1+72x^2-72y)\ .$$ This determinant is positive in the region $$\Omega: \ y< x^2+{1\over72}$$. As $$f_{yy}>0$$ everywhere this allows to conclude that the quadratic form associated to $$H$$ is positive definite in all points of $$\Omega$$. This implies that $$f$$ is convex in every convex subregion of $$\Omega$$. It is however not convex in the full region $$\Omega$$ since $$\Omega$$ is not convex. E.g., the segment connecting the two points $$(\pm1,1)\in\Omega$$ contains the point $$(0,1)\notin\Omega$$.