I have the following function $g(r,a,b) = r^2 - (x-a)^2 - (y-b)^2$ to compute the difference between an estimated and a measured radius. $g$ is defined on $F = \mathbb{R^+_0} \times \mathbb{R} \times \mathbb{R}$ (which is non-empty and convex).
For optimization I need to proof that $g$ is convex (and then, that $g^2$ is, too) on $F$.
How can I do so?
I tried two things:
Computing the Hessian. It is a diagonal matrix with entries $2,-2,-2$, hence it is indefinit and I can't gain more information (wanted to use $H_f \text{ pos. semidefinite} \iff f$ convex)
Using $f(y) \ge f(x) + \nabla f^T(y-x)\ \forall x,y \in F \iff f$ convex on $F$. This lead to a complicated inequality that I could not solve.
As @max_zorn (thank you!) pointed out, we can use the relation
$H_f$ pos. semidefinite $\iff f$ convex
Hence the following holds, too:
$H_f$ not pos. semidefinite $\implies f$ not convex
$H_g$ is indefinite (see my question) and hence, $g$ not convex.