I understand that the positive definiteness of the Hessian matrix is a sufficient condition for a function to be strictly convex, i.e. $$f''(x)\succ0, \forall x\in\Omega\implies f(x)\text{ is strictly convex in }\Omega.$$
A counterexample of the inverse statement is $f(x) = x^4$, where $x\in\mathbb{R}$. The second derivative is $12x^2$, and it equals $0$ when $x=0$. So the second derivative is not positive definite in $\mathbb{R}$.
I wonder under what regularity condition (e.g. on $\Omega$ and/or $f(x)$) does strictly convexity of $f(x)$ implies positive definite Hessian matrix? Or is there any probability statement, like Hessian matrix is positive definite with probability 1?