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?**