Convexity of abs of quadratic form

Question

We know that $f(x) = x^T Q x$ is convex iff $Q \geq 0$. How about $g(x) = |x^T Q x|$ where $Q$ is an arbitrary symmetric matrix?

Answer 1

Let $Q=\operatorname{diag}(1,-1)$.

Then $g(x) = |x_1^2 -x_2^2|$, and if we let $\phi(t) = g((t,1)) = |t^2-1|$ we see that $\phi(-1) = \phi(1) = 0$ but $\phi(0) = 1$, hence $g$ is not convex.