check convexity wrt generalized inequality

Question

For $f(X) = X^2$, where $X \in \mathbb{R^{3 \times 3}}$. Check if $f$ is convex wrt the cone $\mathbb{R_+^{3 \times 3}}$.

Is the only way to check if this statement false is to find counterexample? (which I felt hard)

Answer 1

Well, given that neither of us seem to have a better way of understanding this question, let's just assume that my interpretation in the comments is correct. What this essentially means is, as you stated, we are trying to determine convexity in each component. So, let's pick on the top-left components. If $X = (x_{ij})_{i,j=1}^3$, then $$(X^2)_{ij} = x_{11}^2 + x_{12}x_{21} + x_{13}x_{31}.$$ Let's consider this as a function $f$ of $5$ variables: $$f(a,b,c,d,e) = a^2 + bc + de.$$ To test convexity, let's compute the Hessian: $$H = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \end{pmatrix}.$$ The function $f$ is convex if and only if $H$ is positive-semidefinite. Let $x = (0, 1, -1, 0, 0)^\top$. Then $$x^\top H x = -2 < 0,$$ implying $H$ is not positive-semidefinite, and hence $f$ is not convex. We could also have proven $f$ is non-convex with an example; this would be a little easier to manage than an entire matrix. Thus the squaring map is not convex with respect to the given cone.