Strictly convex function: how often can its second derivative be zero? It's a basic fact that a twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ is strictly convex if its derivative is positive everywhere.
The converse is not true: consider, e.g., $f(x) = x^4$, which is strictly convex, with $f ''(0)=0$.
Is there a partial converse, however? 
Is it true, e.g., that a strictly convex twice-differentiable function from $\mathbb{R}$ to $\mathbb{R}$ can have zero second derivative at at most one point? 
Thanks for your help!
 A: We have a partial converse. It's not quite as strong as allowing only finitely many or only isolated zeros of the second derivative, but it's strong enough.
For a function $f \colon I \to \mathbb{R}$, where $I \subset \mathbb{R}$ is an open interval, possibly $I = \mathbb{R}$, convexity can be formulated as
$$\bigl(\forall u,v,w \in I\bigr)\left(u < v < w \Rightarrow \frac{f(v)-f(u)}{v-u} \leqslant \frac{f(w)-f(v)}{w-v}\right)$$
and strict convexity with the strict inequality.
For a differentiable function $f$, these conditions can be seen to be equivalent to


*

*$f$ is convex if and only if the derivative $f'$ is non-decreasing,

*$f$ is strictly convex if and only if $f'$ is strictly increasing.


For a twice continuously differentiable function $f$, the above can be easily seen to be equivalent to


*

*$f$ is convex if and only if $f'' \geqslant 0$ everywhere,

*$f$ is strictly convex if and only if $f'' \geqslant 0$ everywhere and $f''$ does not vanish on any non-empty open interval $J \subset I$.

