I'm wondering if there exists a "Quantitative version of of Riemann Lebesgue Lemma" at least for the following case

$ \int_{1}^{\infty}F(t)e^{-2\pi i wt}dt $

where $F(t)$ is a Piecewise cont. function tending zero exponentially.



1 Answer 1


We use the notation $\hat u(\xi)=\int u(x) e^{-2iπ x\cdot \xi} dx$, meaningful for $u\in$ Schwartz space and extendable (weakly) to the huge space of tempered distributions $\mathscr S'(\mathbb R^n)$.

$\bullet$ The Riemann-Lebesgue lemma says that if $u\in L^1(\mathbb R^n)$, then $\hat u\in C^0_{(0)}$ (continuous functions going to zero at infinity).

$\bullet$ To say that $u$ belongs to $H^s(\mathbb R^n)$ for some $s\in \mathbb R$ means $ (1+\vert\xi\vert^2)^{s/2}\hat u(\xi)\in L^2(\mathbb R^n). $

$\bullet$ In particular if $s>n/2$, $H^s(\mathbb R^n)\subset C^0\cap L^\infty$ and with $k\in \mathbb N$, $s>k+n/2$, $H^s(\mathbb R^n)\subset C^k\cap L^\infty$.

So the decay of the Fourier transform at infinity in $\mathbb R^n$ is linked to the regularity of the function: faster is the decay, more regular is the function. So even if your function $F$ is piecewise continuous with compact support in $(1,+\infty)$, the decay of the Fourier transform will measure the local regularity of that function. Just calculate examples with piecewise affine functions.

  • $\begingroup$ Infact the function I'm working on has peaks at integers. Between two integers it's constant. $\endgroup$ Dec 18, 2013 at 14:13

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