Fourier analysis — Proving an equality given $f, g \in L^1[0, 2\pi]$ and $g$ bounded We were given a challenge by our Real Analysis professor and I've been stuck on it for a while now. Here's the problem:

Consider the $2\pi$-periodic functions $f, g \in L^1[0, 2\pi]$. If $g$ is bounded show that 
  $$
\lim_{n \to +\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(t)g(nt)\,dt = 
\frac{1}{2\pi}\int_{0}^{2\pi}f(t)\,dt \cdot
\frac{1}{2\pi}\int_{0}^{2\pi}g(t)\,dt.
$$


I first thought of using the Riemann-Lebesgue theorem, i.e. if 
$f \in L^1[a, b]$ then 
$$\lim_{R \to +\infty}\int_{a}^{b}f(t)\cos{Rt} \, dt = 0 
\;\; 
\text{and} 
\;\; 
\lim_{R \to +\infty}\int_{a}^{b}f(t)\sin{Rt} \, dt = 0$$
but it didn't get me far.  

I do not have enough reputation to comment, so I write here a minor mistake of TrialAndError. The integral $$\int_0^{2\pi}f_N(t)S_{g_\delta}^K(nt)dt$$ may cannot be bounded by $\varepsilon/4$, as there may exists a constant term in $S_{g_\delta}^K(t)$. That is the reason why the limit is not $0$. We can assume though that the limit is equal to $0$, since
$$\lim_{n \rightarrow \infty}\dfrac{1}{2\pi}\int_0^{2\pi}f(t)g(nt)dt=
\dfrac{1}{2\pi}\int_0^{2\pi}f(t)dt\dfrac{1}{2\pi}\int_0^{2\pi}g(t)dt\iff$$
$$\lim_{n \rightarrow \infty}\dfrac{1}{2\pi}\int_0^{2\pi}f(t)\left(g(nt)-\dfrac{1}{2\pi}\int_0^{2\pi}g(x)dx\right)dt=0$$
Now, the molifications $g_\delta$ can be selected such that their constant terms are equal to $0$, and thus the constant terms of $S_{g_\delta}^K$ are equal to $0$.
