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A Step Function Proof

  1. Oct 10, 2007 #1
    1. The problem statement, all variables and given/known data
    Let [tex]f[/tex] be a function that is integrable on [tex][a,b][/tex] and bounded by
    [tex]0[/tex] [tex]\leq f(x)\leq M[/tex] for some [tex]M[/tex]. Prove that [tex]f^{2}[/tex] is also integrable on this interval.

    2. Relevant equations
    We've done many problems with step functions [tex]s(x)[/tex][tex]\leq f(x) \leq t(x)[/tex],

    where [tex]s(x)[/tex] and [tex]t(x)[/tex] are step functions. Then we've defined a

    function's lower integral as the supremum of the set of the values of all [tex]\int_a^b s(x)dx[/tex]

    provided that [tex]s(x)[/tex] [tex]\leq f(x)[/tex] and the function's upper integral as the

    infimum of the set of all [tex]\int_a^bt(x)dx[/tex] provided that

    [tex]f(x)[/tex] [tex]\leq t(x)[/tex] on [tex][a,b][/tex].

    3. The attempt at a solution

    Attached is my work. It's a little sloppy, so I'll explain what I tried here:

    I first tried using the weighted mean value theorem, but all that that does is show that [tex]f^{2}[/tex] is bounded.

    I then drew a line and tried again. However, I've realized that my definition of the step functions is faulty because the function is not monotonic, so the second attempt is wrong. As a result, I've been unable to represent my step functions more explicitly using summation notation and expanding it, so I haven't been able to work with the equations. There must be something I'm missing.

    If the function were continuous, then I'd know that [tex]f^{2}[/tex] is continuous and I'd be done, but unfortunately it is not.

    The only thing I know is that [tex]\int_a^b f(x)dx[/tex] is bounded and exists, and so [tex]\int_a^b f(x)^{2}dx[/tex] is also bounded.

    Again, if I can show that the lower integral and the upper integral are equal, then by our definition I will have proven that the function is integrable.
     

    Attached Files:

  2. jcsd
  3. Oct 10, 2007 #2

    EnumaElish

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    Since f is positive, f2 is monotonic. So s < f < t implies s2 < f2 < t2.
     
  4. Oct 10, 2007 #3
    Unfortunately f is not monotonic, since it is not continuous. For example, f(a) could equal 5, and f(b) = 3, when a<b, provided that 3 and 5 are less than M. Moreover, f^2 is neither continuous nor monotonic, but I think I can assert that inequality.

    Still, it doesn't seem to get me anywhere. Because the function is not monotonic, I cannot write my summation notation explicitly enough that it can be expanded and canceled to show that the difference between the supremum of s and the infimum of t is 0.

    I appreciate the help, though. I spent about an hour today working with a T.A. with summation notation, but we ended up getting nowhere. It's possible that that's necessary, but I'm wondering if there's a simpler way to do it.
     
  5. Oct 11, 2007 #4

    EnumaElish

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    I meant that f^2 is monotonic with respect to f, or "square" operation is a monotone transformation. I did not realize that you couldn't represent f as a step function so I was taking it as a given.

    You need to think what the term "integrable" exactly means.
     
    Last edited: Oct 11, 2007
  6. Oct 11, 2007 #5
    Oh monotone transformation. I'm not familiar with that term, but I think I understand what you're saying to mean that each value of f^2 corresponds to one value of f?

    Anyways, I do agree with that inequality, and with the help of a TA in one of my classes I managed to work out the problem. Thanks again for the help; it's much appreciated.

    It says in the instructions that I should edit the topic using the Thread Tools to say "Solved," but I can't see any option under thread tools to change the name. Can someone tell me through a message how to do this?
     
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