generalized Riemann-Lebesgue lemma

Generalized Riemann-Lebesgue lemmaFernando Sanz Gamiz

Lemma 1.

Let $h\mathrm{:}\mathrm{R}\mathrm{\to }\mathrm{C}$ be a bounded measurable function. If $h$ satisfies the averaging condition

$$\underset{c\to +\mathrm{\infty }}{lim}\frac{1}{c}{\int }_0^{c}h(t)𝑑t=0$$

then

$$\underset{\omega \to \mathrm{\infty }}{lim}{\int }_a^{b}f(t)h(\omega t)𝑑t=0$$

with for any $f\mathrm{\in }{L}^{\mathrm{1}}\mathit{}\mathrm{[}a\mathrm{,}b\mathrm{]}$

Proof.

Obviously we only need to prove the lemma when both $h$ and $f$ are real and .

Let ${\mathrm{𝟏}}_{[a,b]}$ be the indicator function of the interval $[a,b]$. Then

$$\underset{\omega \to \mathrm{\infty }}{lim}{\int }_0^b{\mathrm{𝟏}}_{[a,b]}h(\omega t)𝑑t=\underset{\omega \to \mathrm{\infty }}{lim}\frac{1}{\omega }{\int }_0^{\omega b}h(t)𝑑t=0$$

by the hypothesis. Hence, the lemma is valid for indicators, therefore for step functions.

Now let $C$ be a bound for $h$ and choose $ϵ$ $>0$. As step functions are dense in $L^1$, we can find, for any $f\in L^1[a,b]$, a step function $g$ such that , therefore

$\underset{\omega \to \mathrm{\infty }}{lim}\left|{\displaystyle {\int }_a^b}f(t)h(\omega t)𝑑t\right|$ $⩽$ $\underset{\omega \to \mathrm{\infty }}{lim}{\displaystyle {\int }_a^b}\left|f(t)-g(t)\right|\left|h(\omega t)\right|𝑑t+\underset{\omega \to \mathrm{\infty }}{lim}\left|{\displaystyle {\int }_a^b}g(t)h(\omega t)𝑑t\right|$

$⩽$

because ${lim}_{\omega \to \mathrm{\infty }}\left|{\int }_a^{b}g(t)h(\omega t)𝑑t\right|=0$ by what we have proved for step functions. Since $ϵ$ is arbitrary, we are done.

∎

Title	generalized Riemann-Lebesgue lemma
Canonical name	GeneralizedRiemannLebesgueLemma
Date of creation	2013-03-22 17:06:03
Last modified on	2013-03-22 17:06:03
Owner	fernsanz (8869)
Last modified by	fernsanz (8869)
Numerical id	13
Author	fernsanz (8869)
Entry type	Theorem
Classification	msc 42A16
Related topic	RiemannLebesgueLemma
Related topic	FourierCoefficients
Related topic	Integral2