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, 
$$
\hat{f}(0) = \frac{1}{2\pi}\intop_0^{2\pi}f(t)dt
$$
 A: 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 $L^1([0,2\pi])$.
