Riemann-Lebesgue lemma and absolute continuity

Question

I read that:

"The well-known Riemann-Lebesgue lemma tells us that the Fourier transform of any absolutely continuous measure must vanish at infinity."

Is that true?

Is something similar known concerning singular continuous measures?

Do we know something about the behavior of the Fourier coefficients at infinity in either cases?

For what it's worth, here's a question I posed. Please look at the answer supplied by @5PM, which provides a much better statement of the conditions that govern the behavior of the coefficients at infinity than supplie din my problem statement. http://math.stackexchange.com/questions/318804/relation-between-function-discontinuities-and-fourier-transform-at-infinity — Ron Gordon, Mar 04 '13 at 18:19

score 3 · Answer 1 · answered Mar 04 '13 at 18:50

For example, here's a way to get a singular continuous measure whose Fourier transform does not go to $0$ at infinity: the distribution of a random variable of the form $X = \sum_{n=1}^\infty X_n/n!$, where $X_n$ are iid, taking values $\pm 1$ each with probability $1/2$.

score 0 · Answer 2 · edited Jun 07 '23 at 15:12

The class of measures for which the Riemann-Lebesgue lemma remains true is the well-known class of Rajchman measures see also This class is defined as follows $$ \mathcal R=\{\mu\in M(\Bbb R^d) :\widehat{\mu}(\xi)\to0, \quad|\xi|\to\infty\}.$$ In particular if $f$ is integrable then $|f|dx$ is a Rajchman measure.

However the statement is not true in general for any finite measure for instance, if $\widehat{\delta_0}(\xi)\equiv1$ although $\delta_0$ is a finite measure.

Riemann-Lebesgue lemma and absolute continuity

2 Answers2