I am trying to figure out if the following polynomials are convex or concave on the region $[0, 1]^3$ for every variable, but I wasn't sure how to go about it for a region rather than the entire function. I plotted them for various values of $x,y,z$ and they appeared concave, but nothing is true until proven.
I have other functions similar to these two, so if you could explain the method that would be very appreciated! $$ f(x,y,z) = 1+y+2z-2yz+xyz $$ $$ g(x,y,z) = 1+2x+2y-z-2xy-yz+xyz $$
The first function is indefinite. For $z=1$ the Hessian over $(x,y)$ is $0.5 \begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$ and has eigenvalues 0.5 and -0.5. The eigenvalues indicate directions of convexity and concavity. For example, an eigenvector corresponding to -0.5 is $(1,-1)$, so the function is concave on lines $\{(x,y,1) + t (1,-1,0) : t \in \mathbb{R}\}$ for any $x,y \in \mathbb{R}$. Similarly, the function is convex lines $\{(x,y,1) + t (1,1,0) : t \in \mathbb{R}\}$ for any $x,y \in \mathbb{R}$.
The second function is also indefinite. For $z=1$ the Hessian over $(x,y)$ is the same as for the first function.
For higher order polynomials you need to check the eigenvalues for all potential values of $(x,y,z)$. If you cannot determine the eigenvalues, you could sample a few values, and if you suspect the function is not indefinite, you may be able to prove that using the positivstellensatz.