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Absolute continuity, function of partition bound

  1. Nov 20, 2013 #1
    Given [itex][a,b][/itex] a bounded interval, and [itex]f \in L^{p} ([a,b]) 1 < p < \infty[/itex], we define:
    [itex]F(x) = \displaystyle \int_{a}^{x} f(t) dt[/itex], [itex]x \in [a,b][/itex]
    Prove that exists [itex]K \in R[/itex] such that for every partition:
    [itex]a_{0} = x_{0} < x_{1} < ... < x_{n} = b[/itex] :
    [itex]\displaystyle \sum_{i=0}^{n-1} \frac{| F(x_{i+1}) - F(x_{i}) |^{p}}{|x_{i+1}-x_{i} |^{p-1}} \leq K[/itex]

    I know that [itex]F(x)[/itex] is absolutely continous and of bounded variation.

    [itex]\frac{|F(x_{i+1}) - F(x_{i})|}{|x_{i+1} - x_{i}|} = F(\xi)[/itex] for some [itex]\xi \in [x_{i}, x_{i+1}][/itex] (Lagrange Intermediate Value Theorem)
    Then
    [itex]\frac{|F(x_{i+1}) - F(x_{i})|^{p}}{|x_{i+1} - x_{i}|^{p-1}} = f(\xi)^{p} (x_{i+1}-x_{i})[/itex]
    As [itex]f \in L^{p}[/itex] when [itex]x_{i+1} \to x_{i}[/itex] [itex]f(x_{i+1}-x_{i})[/itex] when grows slower in some neighborhood of [itex]x_{i}[/itex] than [itex]\frac{1}{x_{i+1}-x_{i}}[/itex] because that function doesn't isn't [itex]L^{1}[/itex].
    I was trying prove that the closenes of the [itex]x_{n}[/itex] can't make the sum diverges, and that neither the number of terms but I don't know how to approach all the issues at the same time.
    When the partition norm tends to zero we got [itex]\displaystyle \int_{a}^{b} | f (t) |^{p} dt[/itex]
    And when it doesn't I think that [itex](2M)^{p} (b-a)[/itex] where [itex]M[/itex] is the maximum of [itex]F(x)[/itex] with [itex]x \in [a,b][/itex] (which exists because [itex]F(x)[/itex] is continous in that closed set).
    So I guess [itex]K = (2M)^{p} (b-a) + \displaystyle \int_{a}^{b} | f (t) |^{p} dt [/itex] works
    But I'm not sure how to prove it.
     
  2. jcsd
  3. Nov 20, 2013 #2
    I got hung up when you wrote:

    ##\frac {F(x_{i+1} - F(x_i)|}{|x_{i+1} - x_i} = F(\zeta) ##... surely you mean ##F'(\zeta)## which would be ##f(\zeta)##.. From the next line I see you did mean that.

    Sum up your ##f^p(\zeta)(x_{i+1} - x_i).## As the width of the partition goes to 0 you will have the Riemann integral of ##f^p##.

    I think you have to show (or state) that if it is Riemann integrable then it is Lebesgue integrable.
    Can you take it from there?
     
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