# generalized Riemann-Lebesgue lemma

Generalized Riemann-Lebesgue lemmaFernando Sanz Gamiz

###### Lemma 1.

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

 $\lim_{c\to+\infty}\frac{1}{c}\int_{0}^{c}h(t)\,dt=0$

then

 $\lim_{\omega\to\infty}\int_{a}^{b}f(t)h(\omega t)\,dt=0$

with $-\infty<\!a for any $f\in L^{1}[a,b]$

###### Proof.

Obviously we only need to prove the lemma when both $h$ and $f$ are real and $0=a.

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

 $\lim_{\omega\to\infty}\int_{0}^{b}\mathbf{1}_{[a,b]}h(\omega t)\,dt=\lim_{% \omega\to\infty}\frac{1}{\omega}\int_{0}^{\omega b}h(t)\,dt=0$

Now let $C$ be a bound for $h$ and choose $\epsilon$ $>0$. 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 $\left\|f-g\right\|_{1}<\epsilon$, therefore

 $\displaystyle\lim_{\omega\to\infty}\left|\int_{a}^{b}f(t)h(\omega t)\,dt\right|$ $\displaystyle\leqslant$ $\displaystyle\lim_{\omega\to\infty}\int_{a}^{b}\left|f(t)-g(t)\right|\left|h(% \omega t)\right|\,dt+\lim_{\omega\to\infty}\left|\int_{a}^{b}g(t)h(\omega t)\,% dt\right|$ $\displaystyle\leqslant$ $\displaystyle\lim_{\omega\to\infty}C\left\|f-g\right\|_{1}

because $\lim_{\omega\to\infty}\left|\int_{a}^{b}g(t)h(\omega t)\,dt\right|=0$ by what we have proved for step functions. Since $\epsilon$ is arbitrary, we are done.

Title generalized Riemann-Lebesgue lemma GeneralizedRiemannLebesgueLemma 2013-03-22 17:06:03 2013-03-22 17:06:03 fernsanz (8869) fernsanz (8869) 13 fernsanz (8869) Theorem msc 42A16 RiemannLebesgueLemma FourierCoefficients Integral2