Lemma 1   Let $h \colon \Real \to \Complex$ be a bounded measurable function. If $h$ satisfies the averaging condition $$\lim_{c \to +\infty} \frac{1}{c} \int_0^c h(t) \,d t=0$$ then $$\lim_{\omega \to \infty} \int_a^b f(t)h(\omega t) \,d t=0$$ with $ -\infty < \! a < b < \! +\infty$ 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 $0=a<b<\infty$ .

Let $\mathbf 1_{[a,b]}$ be the indicator function of the interval $[a,b]$ . Then $$\lim_{\omega \to \infty} \int_0^b \mathbf 1_{[a,b]} h(\omega t) \,d t= \lim_{\omega \to \infty} \frac{1}{\omega} \int_0^{\omega b} h(t) \,d t=0$$ by the hypothesis. Hence, the lemma is valid for indicators, therefore for simple functions.

Now let $C$ be a bound for $h$ and choose $\epsilon$ $>0$ . As simple functions are dense in $L^1$ , we can find, for any $f\in L^1[a,b]$ , a simple function $g$ such that $\norm{f-g}_1<\epsilon$ , therefore

\begin{eqnarray*} \lim_{\omega \to \infty} \abs{\int_a^b f(t)h(\omega t)\,d t} & \leqslant & \lim_{\omega \to \infty} \int_a^b \abs{f(t)-g(t)} \abs{h(\omega t)} \,d t + \lim_{\omega \to \infty} \abs{\int_a^b g(t)h(\omega t) \,d t} \\ & \leqslant & \lim_{\omega \to \infty} C\norm{f-g}_1 < C\epsilon \end{eqnarray*} because $\lim_{\omega \to \infty} \abs{\int_a^b g(t)h(\omega t) \,d t}=0$ by what we have proved for simple functions. Since $\epsilon$ is arbitrary, we are done.