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Relating derivative to original function

  1. Aug 7, 2008 #1
    1. The problem statement, all variables and given/known data
    Let [a, b] be an interval in R, and f:[a, b] -> R be differentiable on [a, b]. Assume f(a) = 0, and that there is a number M such that
    [tex]\left|f'(x)\right| \leq M \left|f(x)\right|[/tex]
    for all x in [a, b]. Prove that f is identically 0 on [a, b].

    2. Relevant equations
    Mean Value Theorem?


    3. The attempt at a solution
    I've tried this problem multiple times and I keep hitting a wall. I've tried to approach it through contradiction, letting c be some point in [a, b] such that f(c) > 0, but it always seems that we could find an M sufficiently large such that this is true.
     
  2. jcsd
  3. Aug 7, 2008 #2
    I found some hints in Rudin and I think I've almost got it but I can't justify one step:

    Fix x0 in [a,b], let M0 = sup|f(x)| and M1 = sup|f'(x)| for a < x < x0. (Should be less than or equal to).

    Then for any such x,
    [tex]\left|f(x)\right| \leq M_1 (x_0 - a) \leq A (x_0 - a) M_0[/tex]

    The middle step I can justify with the MVT and the next one with the given relation between f and its derivative. Now how do I show that M0 must be 0. I believe it has something to do with how we choose x0 (we can choose it to be very small). Thanks.
     
  4. Aug 7, 2008 #3

    morphism

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    Should that A be an M?

    Anyway, suppose for a contradiction that M0>0. We may also assume that M>0 (why?). By definition, we have

    [tex]M_0 \leq M (x_0 - a) M_0.[/tex]

    What does this tell us?
     
  5. Aug 8, 2008 #4
    Yes, that should be an M, thank you.

    I can see why M > 0. If M < 0 we would have

    [tex]0 \leq \left| f'(x) \right| \leq M \left| f(x) \right| \leq 0[/tex]

    and we are done.

    Now if M0 > 0, then we must have

    [tex] M (x_0 - a) \geq 1[/tex]

    for all x in (a, x0), but this fails if we choose x to be a + 1/(2M).

    [tex] M (a + \frac{1}{2M} - a) = M(\frac{1}{2M}) = \frac{1}{2} \geq 1[/tex].

    Is this the idea? Thanks for the help.
     
  6. Aug 8, 2008 #5

    morphism

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    Yup, that's the idea. (That last [itex]\geq[/itex] should really be a [itex]<[/itex] though!)
     
  7. Aug 8, 2008 #6
    Right, I meant that as a contradiction. Thanks again for the help.
     
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