# Convexity of a certain sublevel set

Consider the polynomial of degree $4$ in variable $r$

$$r^4 + (x^2 + y^2)\ r^2 - 2 x y\ r + x^2 y^2$$

The discriminant of this polynomial in $r$ is the following expression (obtained using Mathematica)

$$\Delta := 16 x^2 y^2 \left( -x^6 + x^8 - 27 x^2 y^2 + 33 x^4 y^2 - 4 x^6 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 - 4 x^2 y^6 + y^8 \right)$$

Consider the subset $D$ of $\mathbb{R}^2$ defined by the polynomial inequality $\frac{\Delta}{16 x^2 y^2} \le 0$

$$D : -x^6 + x^8 - 27 x^2 y^2 - 4 x^6 y^2 - 4 x^2 y^6 + 33 x^4 y^2 + 33 x^2 y^4 + 6 x^4 y^4 - y^6 + y^8 \le 0$$

I would like to prove that $D$ is convex. Any help would be appreciated. Thanks!

You can parametrize the zero locus of $\Delta$ (other than the origin) in the first quadrant by $$(x,y) = \left(\frac{(3{-}t)\sqrt{(3{+}t)(1{-}t)}}{8}, \frac{(3{+}t)\sqrt{(3{-}t)(1{+}t)}}{8}\right) \qquad\qquad -1\le t\le 1$$ and then compute that the curvature is $$\frac{(x'y''-y'x'')}{((x')^2+(y')^2)^{3/2}} = \frac{2^{3/2}(3{+}t^2)}{\bigl(3{-}t^2\bigr)^{5/2}}.$$ Since this is always positive, it follows that the curve is strictly convex (since it is symmetric under $x$ and $y$ reflections.

• That solves the problem. Thank you, very helpful! Aug 1, 2014 at 21:59
• You're welcome. Actually, I realized that you can get the whole curve (not just the first quadrant) with a trig substitution: $$(x,y) = \left(\frac12\cos t\ (1{+}\cos^2t)\sqrt{2{-}\cos^2t},\ \ \frac12\sin t\ (2{-}\cos^2t)\sqrt{1{+}\cos^2t}\right).$$ This is a nonsingular parametrization, and, of course, it still gives the curvature as positive everywhere. Aug 2, 2014 at 0:47

This is more of an extended comment than an answer, but here goes.

It might be easier to consider the problem locally and then argue how global convexity follows. Local considerations don't tell if $D$ is connected. Let $f(x,y)=\Delta(x,y)/16x^2y^2$.

Let $z=(x,y)\in\partial D$ and let $\nu(z)$ be the unit outer normal at $z$. Take any vector $a$ orthogonal to $\nu(z)$. We would like to show that $Q(a)=\langle a,\frac{d}{dt}\nu(z+ta)|_{t=0}\rangle\geq0$. (Draw a figure to see why this means convexity near $z$. Or compare with the definition of the second fundamental form.)

The unit outer normal at $z$ is $\nabla f(z)/|\nabla f(z)|$. Using $\frac{d}{dt}\nabla f(z+ta)|_{t=0}=D^2f(z)a$ we get $$\frac{d}{dt}\nu(z+ta)|_{t=0}=|\nabla f|^{-3}(|\nabla f|^2D^2 fa-\langle\nabla f,D^2f a\rangle\nabla f).$$ Thus $$Q(a) = |\nabla\Delta|^{-3}a^T A a,$$ where $A=A(z)$ is the matrix $$A=\langle\nabla f,\nabla f\rangle D^2 f-(D^2 f)(\nabla f)(\nabla f)^T.$$ It remains to check that $A(z)$ is positive definite for all $z\in\partial D$. This seems tedious, but might well be doable.

• That expression has the same sign as the curvature of $f=0$ at a given point $z$, is it right? The problem now reduces to showing that this polynomial $g(x,y)$ is positive on the level set $f=0$. I think it is doable in this case. Very helpful, thank you! Aug 1, 2014 at 21:54
• @orangeskid That expression is the curvature of $f=0$ at $z$ in direction $a$. Now of course you are in two dimensions, so there is only one tangential direction. Pick $a$ to be a unit vector, and the expression gives you the curvature. (I should have written that $Q(a)$ should be positive definite on tangential vectors, not necessarily all.) Anyhow, this approach works in great generality. Aug 1, 2014 at 22:07