$C_0$ functions with positive $f'''$ I am looking for functions $f \in C_0(\mathbb{R})$ such that $f'''$ is positive everywhere. So what I tried was solving ODEs like $y'''(x)=e^{-x^2}$ and so on, but I noticed that all these $y$ are in general not in $C_0$. Does anybody have an idea whether there is such a function $f$ with this property?
If you have any question about this problem, please let me know.
 A: Suppose $f'''>0$ everywhere. Then $f'$ is a convex function, and therefore
$$ f'(y) \geqslant f'(x)+f''(x)(y-x), $$
for any $x,y$. Since $f \not\equiv 0$, and $f$ is not linear because the third derivative is positive, there is an $x$ so that $f''(x) \neq 0$. Integrating the inequality with respect to $y$, we find that
$$ f(z+a)-f(a) \geqslant [f'(x)-xf''(x)]z+\frac{1}{2}f''(x)[z^2-2az], $$
for any $x,a,b$. Letting $ z \to \infty$, the left-hand side tends to the bounded $-f(a)$, and the right-hand side is dominated by the $z^2$ term, so we conclude that at any point where $f''(x) \neq 0$, we must have $f''(x)<0$.
Hence $f(x)$ is concave. But this can't happen: since we have the inequality
$$ f(w) \leqslant f(z)+f'(z)(w-z), $$
and $f' \not\equiv 0$, we are forced to conclude that $f(w)$ is bounded away from zero by a linear function as $w$ becomes large (if $f'(z)<0$) or large negative (if $f'(z)>0$).
Therefore there can be no functions that decay at both ends of the real axis and have a positive third derivative.
