(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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.

2. Relevant equations

3. 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 [tex] f'(0) = \frac{f(b) - f(a)}{b-a} [/tex], 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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Derivative proof (hard, at least for me)

**Physics Forums | Science Articles, Homework Help, Discussion**