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Can anybody check my proof (analysis)?

  1. Dec 5, 2012 #1
    Can anybody tell me if my proof is correct? If not, can you give me a hint? I attached my answer...

    Thanks in advance
     

    Attached Files:

  2. jcsd
  3. Dec 5, 2012 #2

    pasmith

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    Your proof doesn't work: at one point you conclude that [itex]U(f^2,P) - L(f^2,P) = 0[/itex] which is not generally the case: consider f(x) = x on [a,b].
     
  4. Dec 5, 2012 #3
    Thanks, but if [tex]U(f^2,P) \leq B^2(b-a)[/tex] and [tex]L(f^2,P) \geq B^2(b-a)[/tex] then why aren't they equal?
     
  5. Dec 5, 2012 #4

    pasmith

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    Your second inequality isn't true. You should know that for any bounded function g, [itex]L(g,P) \leq U(g,P) \leq (b-a)\sup_{[a,b]} g[/itex] and f^2 is a bounded function if f is.
     
  6. Dec 5, 2012 #5

    pasmith

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    Here's a hint: [itex]\sup f^2 = (\sup |f|)^2[/itex] and [itex]\inf f^2 = (\inf |f|)^2[/itex] (why?). Therefore, on each interval of the partition,
    [tex]
    \sup f^2 - \inf f^2 = (\sup |f|)^2 - (\inf |f|)^2 \\
    = (\sup |f| + \inf |f|)(\sup |f| - \inf |f|) \\
    \leq 2B(\sup |f| - \inf |f|).
    [/tex]
    Now you just have to show that on each interval of the partition,
    [tex]\sup |f| - \inf |f| \leq \sup f - \inf f.[/tex]
    You'll have to classify intervals into three types: those where [itex]f \geq 0[/itex], those where [itex]f \leq 0[/itex], and those where f changes sign.
     
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