# A generalized version of the Riemann-Lebesgue lemma

Let $f \in L^1 [0,2\pi ]$ and let $g$ be bounded and $2\pi$-periodic.

Prove that $$\hat{f}(0)\cdot\hat{g}(0)=\lim_{n\to\infty}\frac{1}{2\pi}\intop_0^{2\pi}f(t)g(nt)dt$$ where $\hat{f}(0)$ denotes $f$'s $0$th Fourier coefficient, that is, $$\hat{f}(0) = \frac{1}{2\pi}\intop_0^{2\pi}f(t)dt$$

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## 1 Answer

For the begining assume that $f\in C^1([0,2\pi])$. Note that $$\int\limits_{[0,2\pi]}f(t)g(nt)d\mu(t)= \frac{1}{n}\int\limits_{[0,2\pi n]}f\left(\frac{\tau}{n}\right)g(\tau)d\mu(\tau)= \frac{1}{n}\sum\limits_{k=0}^{n-1}\int\limits_{[2\pi k,2\pi (k+1)]}f\left(\frac{\tau}{n}\right)g(\tau)d\mu(\tau)$$ $$\frac{1}{n}\sum\limits_{k=0}^{n-1}\int\limits_{[0,2\pi]}f\left(\frac{\xi+2\pi k}{n}\right)g(\xi)d\mu(\xi)= \int\limits_{[0,2\pi]}\left(\frac{1}{n}\sum\limits_{k=0}^{n-1}f\left(\frac{\xi+2\pi k}{n}\right)\right)g(\xi)d\mu(\xi)$$ Since $f\in C^1([0,2\pi])$ $$\lim\limits_{n\to\infty}\frac{1}{n}\sum\limits_{k=0}^{n-1}f\left(\frac{\xi+2\pi k}{n}\right)=\frac{1}{2\pi}\int\limits_{[0,2\pi]}f(\zeta)d\mu(\zeta)=\hat{f}(0)$$ Again, since $f\in C^1([0,2\pi])$ and $g\in L^1([0,2\pi])$ is bounded, dominated convergence theorem gives $$\lim\limits_{n\to\infty}\frac{1}{2\pi}\int\limits_{[0,2\pi]}f(t)g(nt)d\mu(t)= \lim\limits_{n\to\infty}\frac{1}{2\pi}\int\limits_{[0,2\pi]}\left(\frac{1}{n}\sum\limits_{k=0}^{n-1}f\left(\frac{\xi+2\pi k}{n}\right)\right)g(\xi)d\mu(\xi)=$$ $$\frac{1}{2\pi}\int\limits_{[0,2\pi]}\lim\limits_{n\to\infty}\left(\frac{1}{n}\sum\limits_{k=0}^{n-1}f\left(\frac{\xi+2\pi k}{n}\right)\right)g(\xi)d\mu(\xi)= \frac{1}{2\pi}\int\limits_{[0,2\pi]}\hat{f}(0)g(\xi)d\mu(\xi)=\hat{f}(0)\hat{g}(0)$$ Now consider linear functional $$\varphi: L^1([0,2\pi])\to\mathbb{C}:f\mapsto\lim\limits_{n\to\infty}\frac{1}{2\pi}\int\limits_{[0,2\pi]}f(t)g(nt)d\mu(t)$$ The proof given above states that for all $f\in C^1([0,2\pi])$ we have $\varphi(f)=\hat{f}(0)\hat{g}(0)$. Consider $f\in L^1([0,2\pi])$ then $$|\varphi(f)|\leq\frac{\operatorname{ess}\sup|g|}{2\pi}\int\limits_{0,2\pi}|f(t)|d\mu(t)= \frac{\Vert g\Vert_{\infty}}{2\pi}\Vert f\Vert_1$$ Since $f\in L^1([0,2\pi])$ is arbitrary $\varphi\in (L^1([0,2\pi]))^*$. Finally we see that equality $\varphi(f)=\hat{f}(0)\hat{g}(0)$ holds for bounded functional $\varphi$ on the dense subspace $C^1([0,2\pi])$ of $L^1([0,2\pi])$, consequently it holds for any function in $L^1([0,2\pi])$.

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This is a beautiful solution. If you have the time and willingness, I'd appreciate if you'd take a few minutes to explain to me what led you to it. Specifically, how did you come up with using Riemann sums... I understand the solution perfectly, but I seem to be lacking at constructing such solutions myself... –  Shai Deshe Dec 13 '11 at 8:47
I don't know how I came to Reinmann sums. I'm just started to transorm original integral (of course using periodicity), and finally came up to Reinmann sums. This was my luck. –  Norbert Dec 13 '11 at 12:32
Can one not easily generalize this to hold for any periodic function? Eg not spesifically $2\pi$-periodic ones –  N3buchadnezzar Dec 12 '14 at 13:13
@N3buchadnezzar, of course! –  Norbert Dec 12 '14 at 14:10