Suppose $H:\mathbb{R}^{2n}\to\mathbb{R}$ satisfies the following:
- $H\geq0$ everywhere and $H(x)=0\iff x=0$;
- $H\in\mathcal{C}^2(\mathbb{R}^{2n}\smallsetminus\{0\})$;
- $H$ is positively homogeneous of degree 2 (i.e. for all $x\in\mathbb{R}^{2n}\smallsetminus\{0\},\lambda>0$ we have $H(\lambda x)=\lambda^2H(x)$);
- The Hessian of $H$ is everywhere positive definite, except at most in the origin;
- The level sets of $H$ are all compact.
I know the gradient $\nabla H$ is therefore injective. But can it be proven (in general, for in some cases it is of course possible) it is surjective as well? And if not, can you provide a counterexample? Does this statement's truth depend on the dimension (i.e. on $n$)?
As said in the comments, $H(x)$ must be:
$$H(x)=\sum_{i\leq j}a_{ij}x_ix_j,$$
for $a_{ij}$ real coefficients, by positive homogeneity. Then its Hessian is precisely the matrix $A=(a_{ij})$, which is therefore invertible. But the gradient is:
$$\nabla H(x)=x\cdot A\cdot(e_1,\dotsc,e_{2n})^T,$$
so the equation $\nabla H(x)=y$ as solution $x=y\cdot A^{-1}$.
Actually, having only positive homogeneity those coefficients may depend on the half-line where $x$ lies. Still, the above should yield surjectivity.
Update
Here is how the comments continue in chat: 1, 2, 3, 4, 5.