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

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.

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).

• Thank you! The way I came across it, I didn't feel like it was a unique question. Nonetheless, I couldn't actually find it as it was hard to search. Very clever. – mlg4080 Feb 11 '14 at 6:25
• @mlg4080: well searching math questions with math symbols is normally very difficult. Even I am not able to find the link on math.stackexchange.com where this was solved. Hence I posted the solution here. – Paramanand Singh Feb 11 '14 at 6:28

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.

• I'm a little hesitant to make your jump via Stone-Weierstrass. I feel like in that same vein of thought, we could say every continuous function on $[0,1]$ is differentiable everywhere in [0,1]. (Since every polynomial is.) But indeed, the Weierstrass function is itself a counter-example to this claim. Is there some subtle difference I am misunderstanding in your density of polynomials argument? – mlg4080 Feb 11 '14 at 5:52
• Great objection. I didn't think that one through very well... – Frederick Feb 11 '14 at 5:56
• I think that perhaps this can be patched. Since $f$ is assumed to be differentiable, we can find a sequence of polynomials approximating the derivative math.stackexchange.com/questions/496993/… – Frederick Feb 11 '14 at 6:01
• Despite the hasty conclusion, I appreciate the new direction to think about this problem. (Like I said, it's just for fun.) I think your second comment is correct, via Stone-Weierstrass we can find a sequence of polynomials that converges uniformly to $f^\prime(x)$, since $f^\prime$ is continuous, and hence their anti-derivatives will converge uniformly to $f(x)$. I don't think the convergence to $f(x)$ is as trivial as it sounds (for $\mathbb{R}$ it may be, for $\mathbb{R}^n$ it's not quite), but it is indeed true. I still fail to see how this will patch it up. – mlg4080 Feb 11 '14 at 6:12