Is $f(x,y,z)= x^4+3x^3y+z^4$ convex?

Question

Let $$f(x,y,z)= x^4+3x^3y+z^4$$ Use the Hessian matrix to determine if $f(x,y,z)$ is convex on $\mathbb R^3$.

I have the following Hessian matrix

$$H_f= \left( {\begin{array}{cc} 12x^2+18xy & 9x^2 & 0 \\ 9x^2 & 0 & 0 \\ 0 & 0 & 12x^2 \\ \end{array} } \right)$$

Then I want to calculate the eigenvalues to determine if it is positive semidefinite:

$$p(\lambda) = \det(H_f - I\lambda)= \left| {\begin{array}{cc} 12x^2+18xy -\lambda & 9x^2 & 0 \\ 9x^2 & -\lambda & 0 \\ 0 & 0 & 12x^2 -\lambda \\ \end{array} } \right|$$

Which leads to quite a messy calculation for the determinant:

$$(12x^2+18xy-\lambda)(-\lambda(12x^2-\lambda))-9x^2(9x^2(12x^2-\lambda))=0$$

Is there an easier way to go about this that I'm missing? Any advice about where to go from here would be great!

Answer 1

I don't think it's convex: Consider the section of the surface where $y=1$ and $z=0$. Here you have $f(x, y=1, z=0) = x^4 + 3 x^3$ which is not convex, hence $f(x, y, z)$ is not convex.

For example in the origin $(x=0, y=0, z=0)$ you have that the tangent plane is $ \nabla^T f (x=0, y=0, z=0) \begin{bmatrix} x \\ y \\ z\end{bmatrix} + f (x=0, y=0, z=0) = 0$, but $f (x=-1, y=1, z=0) = -2$. Hence you have a point where the function goes below a tangent plane, that is sufficient condition to state that the function is not convex.

Answer 2

Let $x=1$, $y=-1$, then the upper left component of the Hessian is $-6$, that is negative, hence the Hessian can’t be positive-definite.

Answer 3

The Hessian is positive semidefinite if and only if all of its principal minors are nonnegative. The principal minor produced by the $2 \times 2$ submatrix in the northwest corner is

$$\det \begin{bmatrix} 12x^2+18xy & 9x^2\\ 9x^2 & 0\end{bmatrix} = -81 x^4$$

which is nonpositive. Hence, the Hessian cannot be positive semidefinite for all $(x,y,z) \in \mathbb R^3$.