Talk:Riemann–Lebesgue lemma/Archive 1

extension
I think you can apply the theorem to functions over the whole real line, not just intervals, provided they're still $$L^1$$ of course. The proof has an extra technical step, but this is a much more useful result. For example, see Reed and Simon (though they prove it quite differently, and they also prove that not only does the Fourier Transform of the $$L^1$$ function exist, but that it is $$C^\infty$$). Lavaka 18:13, 13 September 2006 (UTC)

intuitively
I understood everything in this article except for the statement: "Intuitively, the lemma says that if a function oscillates rapidly around zero, then the integral of this function will be small." How is this related to the Riemann-Lebesgue lemma? Why is a rapidly-oscillation necessarily in L1? And how does the lemma imply a small integral? --Zvika (talk) 14:39, 6 December 2007 (UTC)


 * "Function" refers not to f, but to the integrand $$f(x)e^{izx}$$. Does that help? -- Jitse Niesen (talk) 14:54, 6 December 2007 (UTC)


 * I think I understand what you mean now, but I don't think this is what the theorem says. There can be rapidly oscillating functions whose integral does not tend to zero, if they are not absolutely integrable. As I understand the lemma, the point is not that rapidly oscillating functions have small integrals, but that absolutely summable functions have limited oscillations (which also sounds, to me, like a much more powerful result). In any case, even if you do want to keep the current explanation, I think its relation to the theorem should be made clearer. Cheers, --Zvika (talk) 20:08, 6 December 2007 (UTC)

I've deleted that part. It looks like severe confusion. That's not even remotely what the lemma says. Michael Hardy (talk) 14:26, 26 August 2009 (UTC)


 * ...OK, I think I'm starting to see what Jitse had in mind in this edit, but it was certainly not clearly explained in the article, if indeed it was explained at all. Michael Hardy (talk) 14:29, 26 August 2009 (UTC)

Why "around zero"? Michael Hardy (talk) 14:30, 26 August 2009 (UTC)
 * ...Oh... alright, that didn't mean zero in the domain. Michael Hardy (talk) 14:31, 26 August 2009 (UTC)

So if you have a positive function &fnof; whose integral is finite, and then you look at one that oscillates fast between &fnof; and &minus;&fnof;, the integral of the latter will be close to zero. Michael Hardy (talk) 14:33, 26 August 2009 (UTC)

More general
I think the Riemann–Lebesgue lemma is much more general, it holds for each $$f\in L^1(\mathbb{R}^n)$$. Not only for n=1 like stated in this article. --Jobu0101 (talk) 06:43, 17 August 2011 (UTC)

Proof is incorrect?
The "proof" given here seems to rely on the false claim that we can interchange the limits as t goes to 0 and the limit from the approximation by simpler functions. Unless someone else sees how to prove that this is possible in this case, I will probably alter the proof to be the standard argument of approximating in the L1 norm by step functions. — Preceding unsigned comment added by Janjigc (talk • contribs) 04:00, 31 December 2011 (UTC)

Continuity
Should this article not mention that the Fourier transform of such functions is also continuous? At least in some real analysis books, this is also called the RLL. In either case, it should at least be mentioned I think. — Preceding unsigned comment added by 76.121.187.193 (talk) 10:43, 14 April 2012 (UTC)

Proof seems to be a bit unclear
You have written: -- We'll focus on the one-dimensional case, the proof in higher dimensions is similar. Suppose first that f is a compactly supported smooth function. Then integration by parts in each variable yields


 * $$ \left| \int f(x) e^{-izx}dx\right|=\left|\int \frac{1}{iz} f'(x)e^{-izx}dx\right| \leq \frac{1}{|z|}\int|f'(x)|dx \rightarrow 0 \mbox{ as } z\rightarrow\pm\infty. $$

--

How do you know that if f(x) is L1 function then its derivative is L1 function too? Consider:
 * $$ f(x)=\frac{\sin(e^x)}{1+x^2} $$.

or
 * $$ f(x) =

\begin{cases} x^2e^{-x^2}\sin(1/x^2) & |x| \in (0,1]\\ 0 & \text{otherwise} \end{cases} $$

Concluding: If you just write that you take into account only the situation where f(x) is differentiable and additionally f'(x) is L1 function too,then proof is more clear and universal. — Preceding unsigned comment added by 89.79.154.60 (talk) 19:50, 27 February 2014 (UTC)

Your lemma is false
Consider :
 * $$f(x)=\frac{1_{\mathbb{Q}}2 -1}{1+x^2}$$

It's L1$$\scriptstyle \mathrm (\mathbb R),$$ integrable function, but $$\lim_{|\omega|\to +\infty} \int\limits_{\mathbb{R}} f(x) e^{-i \omega x}dx\neq 0$$ — Preceding unsigned comment added by 89.76.155.25 (talk) 18:05, 24 December 2014 (UTC)
 * What is $${1_{\mathbb{Q}}2}$$? If it's a complex number, I think we're okay. 178.39.163.55 (talk) 11:50, 5 July 2015 (UTC)

Article in "References" section (Self-promotion?)
The Researchgate article included in the references appears to have been added by User:Anilped. User:Anilped claims to be "Prof Anil Pedgaonkar", which also happens to be the alias of that article's author. I've made a brief skim of the article, and it seems inconsequential. But, then again, I've never had cause to use the lemma (yet). Maybe the article is of value, in a way I can't see. Can someone with a background in this area check that his generalizations are actually of sufficient novelty and power that we ought be directing readers there?

(If not, note that we should probably remove the last comment in "Other Versions" too.)

2601:240:C400:D60:902F:C0CF:8076:BA68 (talk) 03:21, 17 July 2017 (UTC)