Let $F: \mathbb{R}^N \to \mathbb{R}$ be a convex function. Let $\nabla^2 F(x)$ be the Hessian matrix.
Are there specific conditions on $F$ (or $\nabla F$) that guarantees $\nabla^2 F(x)$ to be invertible on the domain of $F$ (e.g. all of $\mathbb{R}^N$)?
Context: Newton's method requires inverse of $\nabla^2 F(x)$. Under what condition on $F$ do we have guarantee that $\nabla^2 F(x)$ can be inverted?
As mlc noted, strong convexity of $F$ is a sufficient condition for the invertibility of its Hessian. Strong convexity means that there exists $c>0$ such that the function $F(x)-c\|x\|^2$ is convex. Under this assumption, all eigenvalues of the Hessian are $\ge 2c$, so they are nonzero.
Strong convexity is not necessary; e.g., $F(x)=\sqrt{\|x\|^2+1}$ fails it but nonetheless has invertible Hessian. But a necessary and sufficient condition would be tautological: the Hessian invertible when it's invertible.