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Question on limits and little-o notation

  1. Apr 18, 2013 #1
    Hello,

    I was studying the theorem of smoothness/compactness in Fourier theory and at the very last step of the proof one gets the result that [itex]\omega F(\omega)\to 0[/itex] when [itex]x\to \infty[/itex]. The author of the book writes this result in little-o notation as: [tex]\omega F(\omega) = o(|\omega|^{-1})[/tex] which I understand, but then he deduces directly that: [tex]F(\omega)=o(|\omega|^{-2})[/tex]. Can anyone explain this last step? Thanks.
     
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  3. Apr 18, 2013 #2

    Simon Bridge

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    Divide both sides by omega.
     
  4. Apr 19, 2013 #3
    Hi Simon,

    if I am not wrong the little-o notation is just a notation and I assume it cannot always be treated as a ordinary equation. The notation [itex]f(x) = o(g(x))[/itex] should be equivalent to the statement: [tex]\lim_{x\to \infty} \frac{f(x)}{g(x)}=0[/tex]
    If that is true then we have: [tex] \lim_{x\to \infty} \omega F(\omega)=0 [/tex] and dividing numerator and denominator by [itex]\omega[/itex] we get: [tex] \lim_{x \to \infty} \frac{F(\omega)}{\omega^{-1}}=0 [/tex] which according to the definition above should be written as: [tex]F(\omega) = o(\omega^{-1})[/tex] which is not the expected result.
     
  5. Apr 19, 2013 #4

    Simon Bridge

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    f(x)=o(g(x)) means that f grows slower than cg

    if wF(w) grows slower than 1/|w| then F(w) must grow slower than ....
    http://en.wikipedia.org/wiki/Little_O_notation
    http://www.math.caltech.edu/~2010-11/1term/ma001a1/bigolittleo.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  6. Apr 19, 2013 #5

    HallsofIvy

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    This makes no sense. Do you mean [itex]\lim_{\omega\to\infty} \omega F(\omega)= 0[/itex]?

    ??? No, dividing both sides of [itex]\lim_{\omega\to\infty}\omega F(\omega)= 0[/itex] by [itex]\omega[/itex] gives [itex]\lim_{\omega\to\infty}F(\omega)= 0[/itex].

    If you are thinking of [itex]\omega F(\omega)[/itex] as a fraction with denominator 1, "Dividing both numerator and denominator by [itex]\omega[/itex]" gives [tex]\frac{F(\omega)}{\omega}[/tex]. That is, the denominator is [itex]\omega[/itex], not [itex]\omega^{-1}[/itex].

     
  7. Apr 25, 2013 #6
    Hi HallsofIvy,
    thanks for your reply. I haven't had time to come back to this earlier.


    Yes. Sorry, I meant to write: [tex]\lim_{\omega \to \infty}\omega F(\omega)=0[/tex]


    Uhm...I am not sure about this claim. What I wanted to say was:
    [tex]\lim_{\omega\to\infty} \omega F(\omega)=\lim_{\omega\to\infty} \frac{\omega F(\omega)}{1} = \lim_{\omega\to\infty} \frac{\frac{1}{\omega}\omega F(\omega)}{\frac{1}{\omega}1} = \lim_{\omega\to\infty} \frac{F(\omega)}{\frac{1}{\omega}} = \lim_{\omega\to\infty} \frac{F(\omega)}{\omega^{-1}}=0 [/tex]

    As you see, the rightmost equality seems to imply that: [tex]F(\omega) = o(|\omega|^{-1})[/tex] but according to the textbook it should be: [itex]F(\omega) = o(|\omega|^{-2})[/itex]. It is stilll totally unclear how the author arrived at this last conclusion.
     
    Last edited: Apr 25, 2013
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