Question on limits and little-o notation

[mnb96]

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 \omega F(\omega)\to 0 when x\to \infty. The author of the book writes this result in little-o notation as: \omega F(\omega) = o(|\omega|^{-1}) which I understand, but then he deduces directly that: F(\omega)=o(|\omega|^{-2}). Can anyone explain this last step? Thanks.

 

[Simon Bridge]

Divide both sides by omega.

 

[mnb96]

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 f(x) = o(g(x)) should be equivalent to the statement: \lim_{x\to \infty} \frac{f(x)}{g(x)}=0
If that is true then we have: \lim_{x\to \infty} \omega F(\omega)=0 and dividing numerator and denominator by \omega we get: \lim_{x \to \infty} \frac{F(\omega)}{\omega^{-1}}=0 which according to the definition above should be written as: F(\omega) = o(\omega^{-1}) which is not the expected result.

 

[Simon Bridge]

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

 

[HallsofIvy]

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 f(x) = o(g(x)) should be equivalent to the statement: \lim_{x\to \infty} \frac{f(x)}{g(x)}=0
If that is true then we have: \lim_{x\to \infty} \omega F(\omega)=0

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

and dividing numerator and denominator by \omega we get: \lim_{x \to \infty} \frac{F(\omega)}{\omega^{-1}}=0

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

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

which according to the definition above should be written as: F(\omega) = o(\omega^{-1}) which is not the expected result.

 

[mnb96]

Hi HallsofIvy,
thanks for your reply. I haven't had time to come back to this earlier.

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

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

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

Uhm...I am not sure about this claim. What I wanted to say was:
\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

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