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Homework Help: Question about a proof

  1. Nov 7, 2009 #1


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    1. The problem statement, all variables and given/known data

    Show that, if f(0)=0 and |f'(x)| ≤ M |f(x)| for 0 ≤ X ≤ L, show that on 0 ≤ X ≤ L that f(x) ≡ 0.

    2. Relevant equations

    3. The attempt at a solution

    I'm having a hard getting this one started (as of right now, this seems a little over my head). At first, it seems like I need to find some polynomial where f(0)=0. Of course there are a lot of candidates for f(0) = 0.

    Now, I know that |f(x)| = |∫0 to x of f'| ≤ M ∫0 to x of |f|, but can't seem to do anything else with it that progresses me further.

    Any help is truly appreciated.
  2. jcsd
  3. Nov 7, 2009 #2
    You'd be done if you could show that f'(x) = C, for any constant C right? for then f(0) would imply that f(x)=0.
  4. Nov 7, 2009 #3


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    Gold Member

    Nope, he would need to show that [itex]f'(x)=0[/itex] in order to prove that [itex]f(x)=0[/itex]. At the moment, I can't think of any way to do that however.
  5. Nov 7, 2009 #4
    right, blah, that's what I meant.

    Mean value theorem would be my guess. I might try to work though it.
  6. Nov 7, 2009 #5
    Use the MVT multiple times. You should be able to construct a sequence that goes to 0, and "squeezes" f(x) to 0.
  7. Nov 7, 2009 #6
    Yeah. So here is Rudin's Hint for doing this problem:

    If you fix x_o < L, then using mean value you know that

    f'(c)(x - a)=f'(x)- f(a), for some c < x_o. and x in 0<x<L If S is the supremum of f(x), and S' is the supremum of f'(x), you've shown

    f(x)<S'(x - a) < SM(x - a), x<x_o

    From there, I think you can go here:

    f(x)<SM(x_o - a). Now, this can be done for any choice of x, so why not take the supremum of f(x) on the interval (0, x_o). Then you obtain S<SM(x_o - a).
  8. Nov 7, 2009 #7
    I thought I had a proof but I tried to do it in my head and made a stupid mistake.

    It is interesting to note that |f'(x)| ≤ M |f(x)| implies that |f'(x)| ≤ (M + N)|f(x)| with 0 ≤ N. If we let U = M + N for arbitrary N then we can work with a constant that's arbitrarily large instead of M. The trick would be to get it under something on the GT side.
    Last edited: Nov 7, 2009
  9. Nov 7, 2009 #8
    updated my last message
  10. Nov 8, 2009 #9


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    Thx for the help guys. I'll work with your ideas, see what I can come up with and post it back here.
  11. Nov 8, 2009 #10


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    And here we go:


    Suppose M > 0. Then, take a point x_0 and fix x_0 < L.

    Now, let S be the supremum of f and S' be the supremum of f' for a ≤ x ≤ x_0.

    For any such x, f(x) - F(a) = = f'(c)(x-a) by Mean Value Theorem where x < c < a.

    So, |f(x)| ≤ S'(x-a) ≤ S'(x_0 - a) ≤ SM (x_0 - a).

    Hence, S = 0 if M(x_0 - a) < L. Therefore, f(x) = 0 on 0 ≤ X ≤ L.

    End of proof.

    Edit: Any comments on this will be appreciated.
    Last edited: Nov 8, 2009
  12. Nov 8, 2009 #11


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    Note that M has to be > 0 or else it kind of unravels itself.
  13. Nov 8, 2009 #12
    So, |f(x)| ≤ S'(x-a) ≤ S'(x_0 - a) ≤ SM (x_0 - a).

    Hence, S = 0 if M(x_0 - a) < L. Therefore, f(x) = 0 on 0 ≤ X ≤ L.

    I don't think this is obvious enough to just state. It works if (M*x_o-a) is less than one, but this requires a specific choice of x_o, and perhaps not the one which equals b.
  14. Nov 8, 2009 #13


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    Alright, here's a different approach I took:

    Note: ck's are coefficients in the interval [0, 1/M].

    Take any point, x, in the interval [0, 1/2M].

    By MVT, f(x) = f(x) - f(0) = f'(ck)(x-0) = f'(ck)x for ck < x.

    So, |f(x)| ≤ M |f'(ck)| |x| by the condition on f stated in the question.

    Using the condition on f(ck) again,

    |f(ck)| ≤ M |f(ck1)| |ck| then,

    |f(x)|≤ M2 |f(ck1)| |x| |ck|

    Repeatedly using MVT, you obtain

    |f(x)| ≤ Mk |f(ck)| |x| |c1| |c2| etc.. where the c's are contained within [0, 1/2M].

    So, in general,

    |f(x)| ≤ (1/2)k |f(ck)|

    Since f is continuous on [0, 1/2M] f is bounded and as k approaches infinity f(x) goes to zero.

    Therefore, f(x) = 0.

    Note: You'd use the same argument for [1/2M, 1/M] which is too much for me to type out.
    Last edited: Nov 9, 2009
  15. Nov 8, 2009 #14
    Yeah, I like that one.
  16. Nov 8, 2009 #15


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    Alright, thank you.
    Last edited: Nov 9, 2009
  17. Nov 9, 2009 #16


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    Can I get someone else to double check that proof for me? There's a couple of things I'm not 100% confident about:

    i.) That the intervals, [0,1/2M], [1/2M, 1/M] and [0, 1/M] are correct. Especially for 0 <= X <= L.

    ii.) The part where I state |f(x)| ≤ (1/2)^k |f(c_k)|. Is that really OK?
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