Let $K$ be a non-empty convex open subset of $\mathbb{R}^d$, for some $d\in \mathbb{N}$.
Suppose that $f$ is a continuously twice-differentiable function defined on $K$, and that for each $x \in K$ the second differential $\mathrm{d}^2f(x)$ is positive semidefinite, i.e., $\forall h \in \mathbb{R}^d, \mathrm{d}^2f(x)(h)(h)\ge0$. Then, it can be proved that $f$ is convex.
On the other hand, I know that a sort of partial converse holds. In fact, suppose that $g$ is a convex function defined on $K$. In light of Alexandrov Theorem, we know that $g$ is twice differentiable almost everywhere and, being $g$ convex, at the points $x \in K$ where $g$ is twice differentiable we have that $\mathrm{d}^2 g(x)$ is positive semi-definite.
Then, I'm left wondering if an actual converse of Alexandrov Theorem holds, when properly stated. For example:
- The a.e. pointwise version. Is it true that if $h$ is a function defined on $K$, then being $h$ twice differentiable almost everywhere with positive semidefinite second differential (where defined) is equivalent to $h$ being convex? (If not, is it enough to add the continuity of $h$?)
- The weak derivative version. Is it true that if $h$ is a function defined on $K$, the $h$ admitting weak second derivatives almost everywhere with positive semidefinite Hessian is equivalent to $h$ being convex?
References, and other results in this spirit, are welcome.
Best Answer
As pointed out in this answer, the univariate function $x \mapsto \min((1-x)^2, (1+x)^2)$ is a counterexample for the a.e. pointwise version, even with the continuity assumption.
As pointed out by geetha in a previous (deleted) comment, the univariate function $x \mapsto |x|$ is convex. However, the second derivative of this function is (up to a multiplicative constant) the derivative of the $\delta$ distribution, so it can't be represented by an $L^1$ function (neither by a measure, if we wanted to admit also measures as weak second derivatives). So, even the weak derivative version is false.