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Oscillation Questions

  1. Feb 17, 2013 #1
    Suppose that [itex]f[/itex] is bounded by [itex]M[/itex]. Prove that [itex]ω(f^2,[a,b])≤2Mω(f,[a,b])[/itex].

    I can show that [itex]ω(f,[a,b])≤2M[/itex] and that [itex]ω(f^2,[a,b])≤M^2[/itex] but this procedure is getting me nowhere. I also have a similar problem that likely calls for the same approach:

    Suppose that [itex]f[/itex] is bounded below by [itex]m[/itex] and that [itex]m[/itex] is a positive number. Prove that [itex]ω(1/f,[a,b])≤ω(f,[a,b])/m^2[/itex].

    This one I think I have right but my instructor is telling me that it's wrong. Since all values are positive, by the nature of [itex]\frac{1}{x}[/itex], [itex]\displaystyle\sup f = \frac{1}{\displaystyle\inf f}[/itex] and [itex]\displaystyle\inf f = \frac{1}{\displaystyle\sup f}[/itex]. We can now analyze the oscilation as follows:

    [itex]ω(1/f,[a,b])=\frac{1}{\displaystyle\inf f}-\frac{1}{\displaystyle\sup f}=\frac{ω(f,[a,b])}{(\displaystyle\inf f)(\displaystyle\sup f)}≤\frac{ω(f,[a,b])}{m^2}[/itex]
     
  2. jcsd
  3. Feb 17, 2013 #2

    haruspex

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    It's probably not the most elegant, but you could try breaking it into separate cases according to the signs of sup f and inf f.
    I think you mean [itex]\displaystyle\sup \frac{1}{f} = \frac{1}{\displaystyle\inf f}[/itex] etc. Other than that, your proof looks fine.
     
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