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?


1. for all $x\in (-\infty, 0]$ we have $f(x)=0$.
2. for all non-negative integers $n$ and $x\in (0,\infty)$ we have $f^{(n+1)}(x) > f^{(n)}(x)$
3. 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 $h\in C^\infty$?)



"Insanely increasing" $C^\infty$ function with upper bound
Combining my comments with that of Terry Tao's:


1. 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 $(0,\infty)$.
2. 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.
3. Real analytic functions can't be vanishing on $(-\infty,0)$ and be non-trivial.


