calculus - Strictly increasing, strictly convex function: is the second derivative positive?

Consider a twice continuously differentiable function $f \colon \mathbb{R} \to \mathbb{R}$. While $f''(x)>0\ \forall x$ implies strict convexity of $f$, the converse is not true (e.g. $f(x)=x^4$, strictly convex but $f''(0)=0$).

I was wondering whether the additional requirement of $f$ being strictly increasing can ensure $f''(x)>0$. It would at least rule out the example above.

From how I understand this answer, the requirement is sufficient for (once) differentiable functions, but for twice differentiable ones it says:

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

Is it correct that this is satisfied under strict increasingness and hence for $f$ as above, strict increasingness and strict convexity together imply $f''(x)>0\ \forall x$?

Answer

$\newcommand{\Reals}{\mathbf{R}}$Let $h:\Reals \to \Reals$ be a continuous, non-negative function that does not vanish on any open interval, and whose integral over $\Reals$ exists. The function
$$g(x) = \int_{0}^{x} h(t)\, dt$$
is strictly increasing and bounded, say $|g| < M$ for some $M > 0$, and the function
$$f(x) = Mx + \int_{0}^{x} g(t)\, dt = Mx + \int_{0}^{x} (x - t) h(t)\, dt$$
is strictly increasing and strictly convex. By construction, $f'' = h$; the function $h$, however, can vanish at infinitely many points (the integers, the terms of a convergent sequence, a Cantor set...).