Convex and concave functions of three variables

Question

How can I test the function for three variables if it is convex or concave

Consider $$f(x,y,z)=xz-xyz$$ whete x,y and z $\in (- \infty,\infty) $

A function ƒ is convex or concave if the Hessian matrix of ƒ $H(f)$ is positive definite or negative definite respectively for all values of x, y and z.

then we get the Hessian matrix of ƒ as following,

$H(f)=\left( \begin{array}{ccc} 0 & -z & 1-y \\ -z & 0 & -x \\ 1-y & -x & 0 \\ \end{array} \right)$

since the determinate of $H(f)$ $$D(H(f))=2 x z - 2 x y z$$

since all critical points of $f(x,y,z)$ are $\{(0, 1, z), (0, y, 0), (x, 1, 0)\}$

In this test I can not determine whether the function is positive definite or negative definite.

Is there another test or correction of the method?

Can I use this theory

where $f:\mathbb{R}^3\rightarrow\mathbb{R} $

Suppose we want to compute the characteristic polynomial of the matrix $H(f)$

$H(f)-\lambda I=\left( \begin{array}{ccc} -\lambda & -z & 1-y \\ -z & -\lambda & -x \\ 1-y & -x & -\lambda \\ \end{array} \right)$

We now compute the determinant of $H(f)-\lambda I$

$$-\lambda ^3+\lambda +\lambda x^2-2 x y z+2 x z+\lambda y^2-2 \lambda y+\lambda z^2$$

From the 2nd derivative test. The Hessian matrix of $f$ which has negative determinant

Is $f$ neither concave nor convex?

Answer 1

You have already found the hessian of $f$, which is well defined everywhere.

$$H=\begin{bmatrix} 0 & -z & 1-y \\ -z & 0 & -x \\ 1-y & -x & 0 \end{bmatrix}$$

To show that $f$ is convex on a convex domain, it is sufficient to show that $H$ is positive semidefinite on that domain's interior, or, equivalently, that the eigenvalues of $H$ are all nonegative on the interior. We can find the eigenvalues, but this requires a lot of messy algebra, and all we really need is their signs. Because $H$ is symmetric, we can instead use the positive semidefinite case of Sylvester's criterion, which states that $H$ is positive semidefinite iff all the principle minors of $H$ have nonnegative determinants. That is,

$$\det\left({\begin{bmatrix} 0 & -z & 1-y \\ -z & 0 & -x \\ 1-y & -x & 0 \end{bmatrix}}\right)\ge 0$$

$$\det\left({\begin{bmatrix} 0 & -x \\ -x & 0 \end{bmatrix}}\right)\ge 0$$

$$\det\left({\begin{bmatrix} 0 & 1-y \\ 1-y & 0 \end{bmatrix}}\right)\ge 0$$

$$\det\left({\begin{bmatrix} 0 & -z \\ -z & 0 \end{bmatrix}}\right)\ge 0$$

Evaluating determinants:

$$-2x(y-1)z\ge 0$$

$$-x^2\ge 0$$

$$-(y-1)^2\ge 0$$

$$-z^2\ge 0$$

From this, we see that $H$ is positive semidefinite only at the point $(x, y, z)=(0, 1, 0)$. Since this set has no interior, $f$ is not convex anywhere for the typical definition of a convex function of several variables.

To find where $f$ is concave, we determine where $-f$ is convex, which in turn implies $-H$ is positive semidefinite. This only reverses the first inequality, and we obtain the same result, that $f$ is not concave anywhere.

Answer 2

While I appreciate Kajelad's thoroughness, there's a much simpler way to look at this if you are simply trying to confirm whether the function is convex or concave or neither. Again, the Hessian is $$H=\begin{bmatrix} 0 & -z & 1-y \\ -z & 0 & -x \\ 1-y & -x & 0 \end{bmatrix}$$ The diagonal has all zero values. That means the sum of the eigenvalues, which is equal to the trace of the matrix, is also zero. So we have only two choices:

1. All of the eigenvalues are zero---i.e., the entire matrix is zero.
2. There is at least one positive eigenvalue and one negative eigenvalue.

Option 1 is true only when $x=z=0$ and $y=1$. Everywhere else, we have option 2. Therefore, the Hessian is indefinite almost everywhere, and the function is neither convex nor concave.