1. ## Another derivative proof

Could someone please check this proof for me?

Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and $|f'(x)| \le |f(x)|$ 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 $\frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*)$. We know that $f'(c*) \le f(c*)$. So $\frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*)$ 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?

2. Originally Posted by JG89
Could someone please check this proof for me?

Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and $|f'(x)| \le |f(x)|$ 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 $\frac{f(c) - f(0)}{c-0} = \frac{f(c)}{c} = f'(c*)$. We know that $f'(c*) \le f(c*)$. So $\frac{f(c)}{c} \le f(c*) \Leftrightarrow f(c) \le cf(c*) < f(c*)$ 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?
There is an perfect solution!
Assume $|f(x)|\leq M\forall x\in (0,1)$
$\forall x\in (0,1),|f(x)|=|f'(x_1)||x|\leq |f(x_1)||x|=$ $|f'(x_2)||x||x_1|=...\leq|f(x_n)||x||x_1|...|x_n|\ leq M|x|^n$
Let $n\rightarrow 0$,we get $f(x)=0$

3. Thanks ynj!

How did you know to write down those inequalities though? Was it by trial-and-error?

4. Originally Posted by JG89
Thanks ynj!

How did you know to write down those inequalities though? Was it by trial-and-error?
Because I have seen this problem in my book before.