Question

Consider a real symmetric matrix $A\in\mathbb{R}^{n \times n}$. The associated quadratic form $x^T A x$ is a convex function on all of $\mathbb{R}^n$ iff $A$ is positive semidefinite, i.e., if $x^T A x \geq 0$ for all $x \in \mathbb{R}^n$.

Now suppose we have a convex subset $\Phi$ of $\mathbb{R}^n$ such that $x \in \Phi$ implies $x^T A x \geq 0$. Is $x^T A x$ a convex function on $\Phi$ (even if $A$ is not positive definite)? Of course, the answer in general is "no," but we can still ask about the most inclusive conditions under which convexity holds for a given $A$ and $\Phi$. In particular I'm interested in the question:

Suppose we have a quadratic form $Q:\mathbb{R}^{n \times n} \rightarrow \mathbb{R}$. What is the weakest condition on $Q$ that guarantees it will be convex when restricted to the set of positive semidefinite matrices?

Answer 1

$x^2-y^2$ is positive on $[2,3]\times [-1,1]$ but not convex there. This creates problems for any convex sets not containing the origin. You are, probably, after something else not so obviously false. Why don't you just tell us what it is?

Edit: Even then it is false: just take $B_{11}B_{22}$. By the way, for a pure quadratic form, convexity on an open set and convexity on the entire space are the same thing.

Answer 2

The answer to the edited question is no. Let $Q\colon\Phi\to\mathbb{R}$ be the quadratic form $Q(B)=B_{11}B_{22}$. Clearly $Q\gt0$ on the set $\Phi$ of positive definite matrices. Equally clearly, this function is concave on the subset $\{B\in\Phi\colon B_{11}+B_{22}=1\}$.