# Proving that a semialgebraic set is convex

Let $$C := \left\{ (x, y, z) \in \mathbb{R}^3: \, x^4 + y^2z + z^2 \le 1, y^2 \le z, z \ge 0 \right\}$$ Prove that $$C$$ is convex.

I started the proof using the standard definition of convex set, that is: $$A$$ is convex if for $$x_1, x_2 \in C$$ and $$\lambda \in [0, 1]$$ $$\lambda x_1 + (1 - \lambda)x_2 \in A$$.

That's the beginning of my proof: let $$c_1 = (x_1, y_1, z_1), c_2 = (x_2, y_2, z_2) \in C$$ and let $$\lambda \in [0, 1]$$ $$\lambda c_1 + (1 - \lambda) c_2 = \big(\lambda x_1 + (1 - \lambda)x_2, \lambda y_1 + (1 - \lambda)y_2, \lambda z_1 + (1 - \lambda)z_2 \big).$$

It's quite easy to show that $$\lambda z_1 + (1 - \lambda)z_2 \ge 0$$. However showing the first and second conditions seems to be tought and after some manipulation I don't see how should it be done.

I would appreciate any tips or hints.

• I can see a co-ordinate transform into $(x^4,y^2z,z^2)$, could you maybe try that out? Oct 21, 2019 at 17:29
• @ShreyasPimpalgaonkar can you add some details, please? Oct 21, 2019 at 19:19

The second derivative of the function defining the first inequality is $$\nabla^2 f_1 = \pmatrix{ 12 x^2 & 0 & 0 \\ 0 & 2z & 2y \\ 0 & 2y & 2}$$ with determinant $$\det( \nabla^2 f_1) = 12 x^2 \cdot 4 \cdot (z-y^2).$$ Hence on $$\{ (x,y,z): \ x\ne 0,\ z>0, \ z>y^2\}$$ all leading minors are positive. So $$\nabla^2 f_1$$ is positive definite on this set, and semi-positive definite on its closure, which is the convex set $$\{(x,y,z): \ z \ge y^2\}$$, and $$f_1$$ is convex on this set. This implies that $$C$$ is the intersection of the convex sets $$\{(x,y,z): \ z \ge y^2, \ x^4 + y^2z + z^2 \le 1\}$$ and $$\{(x,y,z): \ z\ge0\}$$.
• A matrix with two negative eigenvalues can have a positive determinant. To prove that the Hessian is positive semidefinite, one needs more than the determinant. One needs to show that all $2^3 - 1 = 7$ principal minors are nonnegative. All three diagonal entries are nonnegative. The determinant is nonnegative. It remains to be proven that the three determinants of $2 \times 2$ submatrices are nonnegative as well. Oct 23, 2019 at 10:58