Is xy concave or convex or neither in strict positive orthant.

Question

I understand that $f(x,y)=xy$ has a saddle point at $(0,0)$ and neither concave nor convex over the entire $\mathbb{R}^2$. But is that true as well when it's restricted to $\mathbb{R}^2_{++}$ (strictly positive orthant)?

Thanks.

Answer 1

It is neither, you can find the hessian matrix,

$H = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}$ which has two eigenvalues $\lambda_1+\lambda_2 = 0$ and $\lambda_1 \times \lambda_2 = -1 $, i.e., $\lambda_1 = 1$ $\lambda_2 = -1$.

And since convexity has iff relation with H being positive semi-definite (i.e., all eigenvalues greater than or equal to zero) , we can say that the $xy$ is neither convex nor concave.

Answer 2

Consider the values of $f$ at $(1,3),(2,2),(3,1)$ and also at $(1,1),(2,2),(3,3)$.