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.  

Let $\epsilon > 0$ be given.
Let $M$ be a bound for $|g|$, which is assumed to exist. Define $f_{N}(t)=f(t)\chi_{\{ x : |f(x)| \le N\}}(t)$. Let $g_{\delta}$ be a standard mollification of $g$. Then $g_{\delta}\in\mathcal{C}^{\infty}(\mathbb{R})$ is $2\pi$-periodic, $|g_{\delta}| \le M$, and
$$
      \lim_{\delta\rightarrow 0}\int_{0}^{2\pi}|g(t)-g_{\delta}(t)|dt =0.
$$
Let $S_{g_{\delta}}^{K}$ be the truncated Fourier series for $g_{\delta}$; the Fourier series converges uniformly to $g_{\delta}$ as $K\rightarrow\infty$. Then
\begin{align}
   \int_{0}^{2\pi}f(t)g(nt)dt &= \int_{0}^{2\pi}(f(t)-f_{N}(t))g(nt)dt \tag{1}\\
  & +\int_{0}^{2\pi}f_{N}(t)(g(nt)-g_{\delta}(nt))dt \tag{2}\\
  & +\int_{0}^{2\pi}f_{N}(t)(g_{\delta}(nt)-S_{g_{\delta}}^{K}(nt))dt \tag{3}\\
  & +\int_{0}^{2\pi}f_{N}(t)S_{g_{\delta}}^{K}(nt)dt.\tag{4}
\end{align}
The first term on the right is bounded by
$$
     \int_{0}^{2\pi}|f(t)-f_{N}(t)||g(nt)|dt 
     \le M\int_{0}^{2\pi}|f(t)\chi_{\{x : |f(x)| > N\}}|dt, \tag{1}
$$
which tends to $0$ as $N\rightarrow\infty$. Choose $N$ large enough that the above is strictly bounded by $\frac{\epsilon}{4}$. Then, for this fixed $N$,
\begin{align}
\left|\int_{0}^{2\pi}f_N(t)(g(nt)-g_{\delta}(nt))dt\right|
   & \le N\int_{0}^{2\pi}|g(nt)-g_{\delta}(nt)|dt \\
   & = N\int_{0}^{2\pi/n}|g(nt)-g_{\delta}(nt)|d(nt) \\
   & = N\int_{0}^{2\pi}|g(t)-g_{\delta}(t)|dt \rightarrow 0 \mbox{ as } \delta\rightarrow 0. \tag{2}
\end{align}
Choose $\delta > 0$ small enough that the above is strictly bounded by $\frac{\epsilon}{4}$. And $(3)$ is bounded by
$$
             2\pi N\sup_{0\le t\le 2\pi}|g_{\delta}(t)-S_{g_{\delta}}^{K}(t)|,
$$
which is strictly bounded by $\frac{\epsilon}{4}$ for $K$ large enough because $g_{\delta}\in\mathcal{C}^{\infty}(\mathbb{R})$ is periodic. Finally, $(4)$ can be bounded by $\frac{\epsilon}{4}$ by taking $n$ large enough, which follows from the Riemann-Lebesgue lemma. Hence, it follows that the following holds for all large enough $n$:
$$
         \left|\int_{0}^{2\pi}f(t)g(nt)dt\right| < \epsilon
$$
