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Riemmann/Darboux-Integral Question

  1. Apr 4, 2008 #1
    1. The problem statement, all variables and given/known data

    (i) Suppose that F is continuous on [a,b] and [tex]\int_a^b FG = 0[/tex] for all continuous functions G on [a,b]. Prove that F = 0.

    (ii) Suppose now that G(a) = G(b) = 0. Does it again follow that F must be identically zero?


    2. Relevant equations
    Let P be any partition of [a,b]

    Upper Darboux
    [tex]U(F,P) = \sum_{k=0}^\n sSupF(x_k)(t_k - t_{k-1})[/tex]
    Lower Darboux
    [tex]L(F,P) = \sum_{k=0}^\n iInfF(x_k)(t_k - t_{k-1})[/tex]

    F is integrable iff
    inf U(F,P) = U(F) = L(F) = sup L(F,P)

    3. The attempt at a solution

    For (i) I argued the following.

    Let G = F.
    Clear H = F*F >= 0 and is continuous.
    Let's say H(y) > 0.
    So I can find an interval (y - c, y + c) because H is continuous, where H(x) > H(y)/2.
    Clearly U(H,P) > H(y)/2*c
    Which means U(H) >= H(y)/2*c > 0.

    Thus F must be identically zero.

    For (ii), I said it still needed to be identically zero, and I'm not certain how the argument needs to be changed. Perhaps if F was only nonzero very near a and b, but because of the continuity of F, wouldn't that still bee a problem?
     
  2. jcsd
  3. Apr 4, 2008 #2

    Dick

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    Science Advisor
    Homework Helper

    How about G(x)=F(x)*(x-a)*(b-x)? There's nothing special about (x-a)*(b-x), it's just a positive continuous function on [a,b] that vanishes only at the endpoints.
     
  4. Apr 4, 2008 #3
    Thanks Dick, works perfectly. =)
     
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