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Math Help - Another derivative proof

  1. #1
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    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?
    Last edited by JG89; September 4th 2009 at 01:42 PM.
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  2. #2
    ynj
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    Quote Originally Posted by JG89 View Post
    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
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  3. #3
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    Thanks ynj!

    How did you know to write down those inequalities though? Was it by trial-and-error?
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  4. #4
    ynj
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    Quote Originally Posted by JG89 View Post
    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.
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