Suppose $f: \mathbb{R}^n \to \mathbb{R}$ is $C^2$ and strictly convex. Consider the set of points in the domain for which the Hessian matrix of $f$ fails to be positive definite:
$$S = \big \{x \in \mathbb{R}^n : \textrm{det }H_f(x)= 0 \big\}.$$
How large can $S$ be? Is there a characterization of strict convexity if '$S$ is not too large,' suitably construed, and the Hessian is globally positive semi-definite?
If $n=1$, I believe that it's necessary and sufficient that $\mathbb{R} \setminus S$ be dense, but this intuitively seem insufficient for $n \ge 2$ (for example, if $S$ contains a line segment).
This is a partial answer that (partially) addresses the one dimensional case, but I’ll fill in more details if I can figure out more details.
A good way to look at this in the one dimensional case is to use the fact that strict monotonicity of the derivative is equivalent to strict convexity, and use facts about derivatives of strictly monotonic functions. This should allow you to show that set of points with zero hessian is small.
In terms of “smallness”, I don’t think that you can do much better than the complement of the zero set of the hessian being dense. For example, you can’t get countability or it being measure 0, because of cantor set type counterexamples. Specifically, consider a function with a second derivative that is zero on some the cantor set, and positive elsewhere, for example defined here
Integrate this twice to get your convex function with uncountable many points where the hessian is 0. Swap the cantor set out for a “fat” cantor set which has positive measure to obtain a counterexample to the measure 0 statement.