For which $t\ge0$ does there exist a differentiable function $f$ with $f(0)=0$, $f'(x)>f(x)$ for all $x>0$ and with $f'(0)=t$?
This question was inspired by (and is a variation of) the following question and its answer.
Let $f:\mathbb R \to \mathbb R$ be a differentiable function such that $f(0)=0$ and $f'(x)>f(x)$ for all $x>0$.
It could be shown then, as in the answer referred to above,
that $f(x)>0$ for all $x>0$.
The solution goes as follows (with some details added).
Let $g(x)=e^{-x} f(x)$. Because $g'(x)=e^{-x}\Bigl(f'(x)-f(x\Bigr)>0$ we have that $g(x)$ is strictly increasing on the interval $(0,\infty)$. Since $g(0)=0$ we have $g(x)>0$ for all $x\in(0,\infty)$, hence $f(x)=e^x g(x)>0$ whenever $x\in(0,\infty)$.
On close inspection this argument works even if we add to the assumptions the requirement that $f'(0)<0$. This seems contradictory, since then on one hand $f(x)>0$ for all $x>0$, but on the other hand $f(x)<0$ for all $x\in(0,\varepsilon)$, for some small enough $\varepsilon>0$. This contradiction only shows that no such $f$ exist that satisfy all these conditions (including $f'(0)<0$). Thus the following:
Question. For which $t\ge0$ does there exist an $f$ with $f'(0)=t$, and with $f(0)=0$, $f'(x)>f(x)$ for all $x>0$?
Easily $t≥1$ works, e.g. $f(x)=e^{tx}−1$. I wonder if any $t<1$ might also work, in particular if $t=0$ works.
I feel the solution may have something to do with the notion of blow up as understood in this answer, but I don't seem to be able to figure the details. Tried to get a modification of $e^{-\frac1{x^2}}$ but it only works on a bounded interval of positive reals, not on all of $(0,\infty)$. Thank you for your help.
(I included tag integration only because I thought the answer may have also something to do with the average value of $f$ on any interval $[a,b]$, defined as $\frac1{b-a}\int_a^b f(x) dx$. In particular the average value of $f'(x)$ on $[0,b]$ is $\frac1b\int_0^b f'(x) dx=f(b)$.)
