"Insanely increasing" $C^\infty$ function with upper bound Let $C^\infty$ denote the collection of functions $f:\mathbb{R}\to\mathbb{R}$ such that for every positive integer $n$, the $n$-th derivative of $f$ exists. For $f\in C^\infty$ we set


*

*$f^{(0)} = f$, and

*$f^{(n+1)} = \big(f^{(n)}\big)'$ for all non-negative integers $n$.


Is there $f\in C^\infty$ with the following properties?


*

*for all $x\in (-\infty, 0]$ we have $f(x)=0$.

*for all non-negative integers $n$ and $x\in (0,\infty)$ we have $f^{(n+1)}(x) > f^{(n)}(x)$

*there is a function $h:\mathbb{R}\to\mathbb{R}$ such that for all non-negative integers $n$ and all $x\in\mathbb{R}$ we have $f^{(n)}(x)\leq h(x)$.


(Additional question for curiosity, answering it is not needed for acceptance of answer: can $h$ be chosen to be continous? Or even $h\in C^\infty$?)
 A: By Willie Wong's comment and answer, $f^{(n)}>0$ on $(0,\infty)$ for all $n=0,1,\dots$. Hence, $f^{(n)}\ge0$ on $\mathbb R$. So, by Bernstein's theorem on completely monotone functions (used "right-to-left"; see the "footnote" for details), $f(x)=\int_0^\infty e^{tx}\mu(dt)$ for some nonzero nonnegative measure $\mu$ on $[0,\infty)$ and all real $x\le1$. So, $f>0$ on $(-\infty,1]$, which contradicts condition 1 in the OP. Thus, there exists no $f$ satisfying conditions 1 and 2. (Condition 3 is not needed for the non-existence.)
"Footnote": The condition $f^{(n)}\ge0$ on $\mathbb R$ for all $n=0,1,\dots$ implies $(-1)^n g^{(n)}\ge0$ on $[0,\infty)$ for all $n=0,1,\dots$, where the function $g\colon[0,\infty)\to\mathbb R$ is defined by reading $f$ "right-to-left": 
$g(y):=f(1-y)$ for $y\in[0,\infty)$. So, $g$ is completely monotone and hence, by Bernstein's theorem, $g(y)=\int_0^\infty e^{-ty}\nu(dt)$ for some nonnegative measure $\nu$ and all real $y\ge0$. Moreover, the measure $\nu$ is nonzero, because otherwise we would have $g=0$ (on $[0,\infty)$), that is, $f=0$ on $(-\infty,1]$, which would contradict the condition that $f^{(n)}>0$ on $(0,\infty)$ for all $n=0,1,\dots$. 
So, $f(x)=g(1-x)=\int_0^\infty e^{-t+tx}\nu(dt)=\int_0^\infty e^{tx}\mu(dt)$ for all real $x\le1$, where $\mu(dt):=e^{-t}\nu(dt)$, and the measure $\mu$ is nonnegative and nonzero. 
A: Combining my comments with that of Terry Tao's:


*

*First we show that $f^{(n)} > 0$ on $(0,\infty)$ for all $n \geq 1$. The argument is given for $f'$, but, extends easily to all $n \geq 1$.
Start by noticing that $f'(a) = 0 \implies f''(a) > 0$ so that $f'$ changes sign at most once, and that if it is not everywhere positive on $(0,\infty)$ there must be an initial interval $(0,\epsilon)$ on which $f' < 0$. Assume WLOG that $\epsilon < \frac12$. On $(0,\epsilon)$, we have that $f(x) \geq x \inf_{y\in (0,x)} f'(y)$ by the mean value theorem. This is incompatible with $f'(y) > f(y)$. Hence we conclude that $f'$ is always positive on $(0,\infty)$. 

*Step 1 implies that $f^{(n)}$ is increasing for all $n \geq 1$. Therefore if $h$ is a function as in condition (3) of the question, so is the increasing function 
$$ \tilde{h}(y) = \inf_{[y,\infty)} h $$
This function is locally bounded, and implies (by Taylor's inequality) that $f$ is real analytic.

*Real analytic functions can't be vanishing on $(-\infty,0)$ and be non-trivial.
