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**JG89** Could someone please check this proof for me?

Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and $\displaystyle |f'(x)| \le |f(x)| $ then f(x) = 0 for x in (0,1).

Proof:

Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that $\displaystyle \frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*) $. We know that $\displaystyle f'(c*) \le f(c*) $. So $\displaystyle \frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*) $ since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?