A non-trivial, non-negative, function bounded below by its derivative with $f(0)=0$?

Question

I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question:

Does there exist a continuously differentiable function, $f: [0,1] \rightarrow [0, \infty)$, such that the following hold?

(i) $f(0) = 0$

(ii) $f'(x) \le c f(x)$ for all $x$ ($c$ is a fixed constant)

(iii) $f \not\equiv 0$

This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance.

Answer 1

Let $g(x) = e^{-cx}f(x)$ then $g(x) \geq 0$ for all $x \in [0, 1]$. Now $g'(x) = e^{-cx}\{f'(x) - cf(x)\} \leq 0$. So $g(x)$ is decreasing and $g(0) = 0$ therefore $g(x) \leq 0$ for all $x \in [0, 1]$. It now follows that $g(x) = 0$ for all $x \in [0, 1]$ and hence $f(x) = e^{cx}g(x) = 0$ for all $x \in [0, 1]$.

Its an old popular question and has also been solved on this website earlier but I am not able to find the reference. I had posted a bit difficult solution on my blog post (see problem 2 there).

Answer 2

Let $m = a x ^ n$ be a nonconstant monomial. Then $m'(x) = n a x ^{n-1} > c a x^n= c m(x)$ for some $1 > x > 0 $. Thus, the same can be said for (nonconstant) polynomials (just take $inf$ of $x_m$). Since no polynomials can work, no functions can either by Stone-Weirstrass.