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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!

Mike Spivey
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John M
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1 Answers1

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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 $\frac{1}{n}$.

Joonas Ilmavirta
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Eric Naslund
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  • 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 01:18
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    @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
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    A remark, Theorem 4.1 in Katznelson's book tells us that R-L can not be improved http://tailieuhoctap.files.wordpress.com/2007/01/katznelson-y-an-introduction-to-harmonic-analysis-cup-2004299s_mcf_.pdf – AD - Stop Putin - Dec 15 '11 at 17:59