Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f$ be a real-valued, $L^1$ integrable function on the interval $[a,b]$. Then the Riemann-Lebesgue Lemma tells us that: $$\int_a^bf(x)\sin(2\pi nx)dx\rightarrow0 \text{ as } n\rightarrow\infty.$$

Does this have any asymptotic estimate attached to it? i.e. for sufficiently nice $f$ (say continuously differentiable), do we have the estimate that, say: $$\int_a^bf(x)\sin(2\pi nx)dx= O(1/n)$$ or something similar?

Any kind of reference would also be appreciated!

share|improve this question
    
Please see also my answer here. Eric's excellent answer is of course sufficient but the answer to which I linked is for a similar question. –  Amitesh Datta Jul 19 '11 at 0:36
add comment

1 Answer

up vote 9 down vote accepted

For convenience, lets work on the interval $[-\pi,\pi]$. Riemann Lesbegue just says that the Fourier coefficients of $f$ go to zero for an integrable function. If $f$ is in $C^1(\mathbb{T})$ then we have that $\hat{f^'}(n)=in\hat{f}(n)$. (Here $\mathbb{T}$ refers to $[-\pi,\pi]$ with the endpoints identified.) Since $f^'$ is continuous it will be integrable, so Riemann Lesbegue implies the coefficients are $o(1)$. Consequently the coefficients of $f$ are $o\left(\frac{1}{n}\right)$, and hence $$\int_{-\pi}^\pi f(x)\sin(nx)dx=o\left(\frac{1}{n}\right).$$

For a function $f\in C^k(\mathbb{T})$ we get

$$\int_{-\pi}^\pi f(x)\sin(nx)dx=o\left(\frac{1}{n^k}\right).$$

What if $f\in C^1[-\pi,\pi]$, but $f(-\pi)\neq f(\pi)$?

Since the coefficients of the Sawtooth Function have order $\frac{1}{n}$, we will not have a result as strong as before. (The sawtooth function is $C^{\infty}[-\pi,\pi]$).

We can prove that if $f\in C^1[-\pi,\pi]$, but $f(-\pi)\neq f(\pi)$, then the fourier coefficients will be of order $\frac{1}{n}$.

share|improve this answer
    
Thank you Eric. –  John M Jul 18 '11 at 22:19
    
Hi Eric. I was just double-checking this. If you assume $f \in C^1[-\pi,\pi]$ but not necessarily periodic, it seems the non-vanishing boundary term gives you $O(1/n)$ rather than $o(1/n)$. –  John M Jul 19 '11 at 1:18
2  
@John: Yeah, and for the $C^k$ part you also need that the periodic continuation of $f$ is $C^k$. –  Hendrik Vogt Jul 19 '11 at 10:19
    
@Hendrik, @ John: Sorry I forgot to add in that detail! Thanks Hendrik for the correction, I edited the post to reflect that. –  Eric Naslund Jul 19 '11 at 14:40
1  
A remark, Theorem 4.1 in Katznelson's book tells us that R-L can not be improved tailieuhoctap.files.wordpress.com/2007/01/… –  AD. Dec 15 '11 at 17:59
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.