Derivative proof (hard, at least for me)

[JG89]

Homework Statement

Let f(x) be defined and differentiable on the entire x-axis. Show that if f(0) = 0 and everywhere |f'(x)| <= |f(x)|, then f(x) = 0 identically.

Homework Equations

The Attempt at a Solution

I didn't really know how to get started, but as I worked through the problem, I proved some things that may be helpful.

1) if f(0) = 0, then we have |f'(0)| <= |f(0)| = 0, implying that f'(0) = 0 as well.

2) The derivative is continuous at x = 0. Since f is continuous, particularly at the point x = 0, we have |f(x) - f(0)| < epsilon whenever |x| < delta. The first inequality turns into |f(x)| < epsilon, and we know that |f'(x)| = |f'(x) - f'(0)| <= |f(x)| < epsilon whenever |x| < delta, thus the derivative is continuous at x = 0.

3) By the MVT, there are fixed values a and b such that a < 0 < b and $$f'(0) = \frac{f(b) - f(a)}{b-a}$$, implying that f(b) = f(a).

I was thinking if I could prove that the derivative is constant, then the rest is trivial. I also tried by assuming a contradiction, that for some c, f(c) > 0, but I couldn't derive a contradiction.

Please only give very small hints, I want to work through most of the question on my own.

Edit: Maybe I should assume for two distinct points that the different of their derivatives is not equal to 0 and try to derive a contradiction. I will be back!

 

[Dick]

Here's a small but useful hint. First prove f(x)=0 for x in [0,c] where c<1.