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.