- #1
psie
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- TL;DR
- Roughly, at least in the text I'm currently reading, the Riemann-Lebesgue lemma states that the Fourier transform ##\hat f## of an ##L^1(\mathbb R,\mathcal B(\mathbb R),\lambda)## function ##f## vanishes at infinity. I have trouble understanding a concise proof of this lemma.
In the Riemann-Lebesgue lemma, the author says it suffices to prove $$\hat{f}(\xi)\underset{|\xi|\to\infty}{\to}0$$for step functions on ##\mathbb R## only (step functions are simple functions where the sets of the indicator functions are intervals in ##\mathbb R##). This is because the step functions are dense in ##L^1(\mathbb R,\mathcal B(\mathbb R),\lambda)## and so let ##(\varphi_n)## be a sequence of step functions that approximate ##f## in the ##L^1## norm, i.e. ##\|f-\varphi_n\|_1\to0## as ##n\to\infty##. Then we just have to observe \begin{align*}\sup_{\xi\in\mathbb R}|\hat{f}(\xi)-\hat{\varphi}_n(\xi)|&=\sup_{\xi\in\mathbb R}\left|\int f(x)e^{\mathrm{i}x\xi}\,\mathrm{d}x-\int \varphi_n(x)e^{\mathrm{i}x\xi}\,\mathrm{d}x\right|\\ &\leq\|f-\varphi_n\|_1\end{align*}
That's basically the end of the "proof". The above inequality shows ##\hat\varphi_n\to\hat f## in ##L^\infty## (since almost uniform convergence is equivalent to convergence in ##L^\infty##), but how does the proof of the lemma proceed (assuming we've shown the lemma holds for step functions)?
That's basically the end of the "proof". The above inequality shows ##\hat\varphi_n\to\hat f## in ##L^\infty## (since almost uniform convergence is equivalent to convergence in ##L^\infty##), but how does the proof of the lemma proceed (assuming we've shown the lemma holds for step functions)?
). $$|\hat f(\xi)|\leq |\hat{f}(\xi)-\hat{\varphi}_n(\xi)|+|\hat{\varphi}_n(\xi)|\leq\|\hat{f}-\hat{\varphi}_n\|_\infty+|\hat{\varphi}_n(\xi)|.$$Now pick ##n## so large that ##\|\hat{f}-\hat{\varphi}_n\|_\infty<\epsilon## and then for those ##n##, let ##|\xi|\to \infty##.