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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.

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

1 Answer 1

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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 $\{(x,y,z): \ z\ge0\}$.

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  • $\begingroup$ 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. $\endgroup$ Oct 23, 2019 at 10:58
  • $\begingroup$ @RodrigodeAzevedo thanks! I modified the answer. $\endgroup$
    – daw
    Oct 23, 2019 at 11:14

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