Relationship between $f''$ and "$f$ is strongly convex". Suppose $f$ is strongly convex and twice differentiable on some interval $I$:
(1) $\forall x,y \in I: \forall t \in (0,1) \implies f(tx + (1-t)y)<tf(x)+(1-t)f(y)$
(2) $\forall x \in I \implies f''(x) \in \mathbb{R}$
My question is what exactly can we say about $f''$ on $I$? For example obviously, $f'' \geq 0$, but there's more to it for the values of $x$ when $f''(x) = 0$. For example, $f(x) = x$ is definitely not strongly convex on any interval, but $f'' \geq 0: \forall x \in \mathbb{R}$.
I think a sufficient condition would be if $f''(z) = 0 \implies \forall \delta > 0: \exists y_1,y_2 \in (z-\delta,z+\delta):y_1<z<y_2$ and $f''(y_1) > 0 < f''(y_2)$. Or in other words, $f$ is not linear at $z$. This, combined with $f'' \geq 0$ on $I$ should ensure strong convexity.
I would like to clarify my understanding by knowing a near-miss example(s). For example, what if $f$ is a function such that is convex on $\mathbb{R}$ but, say, $f''(0) = 0$ and $\forall \delta > 0: \exists x_1, x_2 \in (-\delta, \delta): x_1 < 0 < x_2:f''(x_1)=0=f''(x_2)?$ In my head I see something that looks kind of like $f''(x) = \sin(\frac{1}{x}) + 1$, but I would like an example where I can express $f$ in elementary functions if possible?
 A: Let $I \subset \mathbb{R}$ be a non-empty interval.

Proposition. Let $f : I \to \mathbb{R}$ be a twice differentiable function. Then $f$ is strictly convex on $I$ if and only if $f'' > 0$ outside a set of empty interior.

Proof:
I do not assume that $f''$ is non-negative on $I$. But Darboux proved that derivative functions satisfy the conclusion of the intermediate value theorem. Since $f''$ is positive outside a set of empty interior, we deduce that $f''$ is non-negative on $I$.
Thus, the function $f$ is convex on $I$. So if $f$ is not strictly convex on $I$, there exists a non-empty interval $J \subset I$ such that $f_{\restriction J}$ is affine. Hence, the function $f''$ is identically zero on $J$, which contradicts our hypothesis. Reciprocally, if $f'' = 0$ on $J$ then $f_{\restriction J}$ is affine and so $f$ is not strictly convex on $I$.
Set $g(x) = x^4 \sin^2(\frac{1}{x})$. Remark that $g$ is non-negative and continuous on $\mathbb{R}$. Consider $f: \mathbb{R} \to \mathbb{R}$ the map defined by
$$\forall x \in \mathbb{R},\ f(x) = \int_0^x \Big(\int_0^y g(s)\, ds\Big) dy. $$
The function $f$ is twice differentiable and $f'' = g$. Thus, $f$ is strictly convex on $\mathbb{R}$ according to the above proposition.
