A Fourier Analysis Question I am stuck at If $f,g\in C[-\pi,\pi]$,and $f,g$ are $2\pi$ periodic, prove that $$\lim_{n\to\infty}\dfrac{1}{2\pi}\int_{-\pi}^\pi f(t)g(nt)\mathbb dt=\big(\dfrac{1}{2\pi}\int_{-\pi}^\pi f(t)\mathbb dt\big)\big(\dfrac{1}{2\pi}\int_{-\pi}^\pi g(t)\mathbb dt\big)$$


I tried to approximate $g(nt)$ by a trigonometric polynomial $T_n(t)$ but the problem is that $n$ is varying here. Nothing seems to work. I tried to interpret the left side as the constant Fourier coefficient of the function $f(x)g(nx)$ and the right side as the product of constant Fourier coefficients of $f(x)$ and $g(x)$ but because $f$ and $g$ are just continuous and not Lipschitz, I cannot replace $f$ and $g$ by their Fourier sums.

Please give me a hint only to get started on this problem.
Let $a_{k}$ and $b_{k}$ be
$$
      a_{k}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ikx}f(x)dx,\;\;\;
      b_{k}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ikx}g(x)dx.
$$
Then $g(x)=\sum_{k=-\infty}^{\infty}b_{k}e^{ikx}$ converges in $L^{2}$ on any finite interval to the periodic extension of $g$. More explicitly, if $S_{N}=\sum_{n=-N}^{N}b_{k}e^{ikx}$, then, for fixed $n=1,2,3,\cdots$,
$$
         \lim_{N\rightarrow\infty}\int_{0}^{2n\pi}|g(x)-S_{N}(x)|^{2}dx= 0.
$$
By a simple change of variables, for fixed $n=1,2,3,\cdots$, one has
$$
         \lim_{N\rightarrow\infty}\int_{0}^{2\pi}|g(nx)-S_{N}(nx)|^{2}dx = 0.
$$
Then, for $f \in L^{2}$, it follows that, for fixed $n=1,2,3,\cdots$,
$$
         \int_{0}^{2\pi}f(x)g(nx)dx = \lim_{N\rightarrow\infty}\int_{0}^{2\pi}f(x)S_{N}(nx)dx.
$$
The above is easily justified using Cauchy-Schwarz:
$$
     \left|\int_{0}^{2\pi}f(x)g(nx)dx-\int_{0}^{2\pi}g(x)S_{N}(nx)dx\right| \\
      \le \int_{0}^{2\pi}|f(x)||g(nx)-S_{N}(nx)|dx \\
      \le \left[\int_{0}^{2\pi}|f(x)|^{2}\right]^{1/2}
          \left[\int_{0}^{2\pi}|g(nx)-S_{N}(nx)|^{2}\right]^{1/2}\rightarrow 0
    \mbox{ as } N \rightarrow \infty \mbox{ for fixed n. }
$$

Therefore, for any fixed $n=1,2,3,\cdots,$
$$
\begin{align}
      \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx &amp;
            =\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)S_{N}(nx)dx \\
       &amp; = \lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\sum_{k=-N}^{N}b_{k}e^{iknx} dx \\
       &amp; = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{iknx}dx \\
       &amp; = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}a_{-kn} \\
       &amp; = \sum_{k=-\infty}^{\infty}b_{k}a_{-kn}.
\end{align}
$$
Finally, Applying Cauchy-Schwarz to the sum on the right gives
$$
\begin{align}
  \left|\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx-a_{0}b_{0}\right|^{2}
   &amp; = \left|\sum_{|k| \ge 1}b_{k}a_{kn}\right|^{2} \\
   &amp; \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k|\ge 1}|a_{kn}|^{2} \\
   &amp; \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k| \ge n}|a_{k}|^{2}\rightarrow 0 \mbox{ as } n\rightarrow\infty.
\end{align}
$$
The last term on the right tends to $0$ because the sum is convergent.
Hint: write $g(t)$ as the combination of $\sin(kt)$ and $\cos(kt)$, i.e., you need only solve the problem for $g(t)=\sin(kt)$ or $\cos(kt)$ for some $k$.