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.
 A: 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 $n=1,2,3,\cdots,$
$$
\begin{align}
      \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx &
            =\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)S_{N}(nx)dx \\
       & = \lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\sum_{k=-N}^{N}b_{k}e^{iknx} dx \\
       & = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{iknx}dx \\
       & = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}a_{-kn} \\
       & = \sum_{k=-\infty}^{\infty}b_{k}a_{-kn}.
\end{align}
$$
Finally, Applying Cauchy-Schwarz to the sum on the right gives
$$
\begin{align}
  \left|\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx-a_{0}b_{0}\right|^{2}
   & = \left|\sum_{|k| \ge 1}b_{k}a_{kn}\right|^{2} \\
   & \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k|\ge 1}|a_{kn}|^{2} \\
   & \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k| \ge n}|a_{k}|^{2}\rightarrow 0 \mbox{ as } n\rightarrow\infty.
\end{align}
$$
The last term on the right tends to $0$ because the sum is convergent.
A: Use the real form of Fourier series.
The first thing that we examine is the integral $\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f(t)g(nt)dt} $.
If we can determine this integral, then we can work out the limit.
Both f and g are continuous and therefore square integrable.  Note  that since they are both periodic of period 2$\pi$, the convergence properties of the Fourier series of both functions should be useful.  Though their Fourier series converge almost everywhere in $[-\pi, \pi]$, but this property is of little use.  But since they are both square integrable, their Fourier series converge respectively to  $f$ and  $g$  in the L2 norm.
Let the Fourier series of  $f$  be
$T(\theta ) = \frac{1}{2}{a_0} + \sum\limits_{j = 1}^\infty  {\left( {{a_j}\cos (j\theta ) + {b_j}\sin (j\theta )} \right)} $
and  the Fourier series of $g$ be 
$S(\theta ) = \frac{1}{2}{A_0} + \sum\limits_{j = 1}^\infty  {\left( {{A_j}\cos (j\theta ) + {B_j}\sin (j\theta )} \right)} $ 
and
${T_k}(\theta ) = \frac{1}{2}{a_0} + \sum\limits_{j = 1}^k {\left( {{a_j}\cos (j\theta ) + {b_j}\sin (j\theta )} \right)} $ and
${S_k}(\theta ) = \frac{1}{2}{A_0} + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}\cos (j\theta ) + {B_j}\sin (j\theta )} \right)} $
their respective partial sums.
Then convergence in the L2 norm means 
$\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi  {{{\left| {{T_k}(t) - f(t)} \right|}^2}}  = 0$ and $\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi  {{{\left| {{S_k}(t) - g(t)} \right|}^2}}  = 0$.
Let $n$ be a fixed positive integer.
Then we also have that
$\mathop {\lim }\limits_{k \to \infty } \int_{ - n\pi }^{n\pi } {{{\left| {{S_k}(t) - g(t)} \right|}^2}}  = 0$.
Thus by a change of variable we get
$\mathop {\lim }\limits_{k \to \infty } n\int_{ - \pi }^\pi  {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt}  = 0$ and so 
$\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi  {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt}  = 0$.

This means ${S_k}(nt)$ tends to $g(nt)$ in the L2 norm and we have $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt} $ tends to $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)g(nt)dt}$.
We can deduce this statement as follows:
$\left| {\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt}  - \frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)g(nt)dt} } \right|$
$ \le \frac{1}{\pi }\int_{ - \pi }^\pi  {|f(t)||{S_k}(nt) - g(nt)|dt} $
$ \le \frac{1}{\pi }\sqrt {\left( {\int_{ - \pi }^\pi  {|f(t){|^2}dt} } \right)} \sqrt {\int_{ - \pi }^\pi  {|{S_k}(nt) - g(nt){|^2}dt} }$  by Holder’s Inequality.
Since $\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi  {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt}  = 0$, the right hand side of the above inequality tends to 0 and so by the Comparison Test,
 $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt} $ tends to $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)g(nt)dt} $ as $k$ tends to infinity.
We now compute $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt} $.
Since $f(t){S_k}(nt) = \frac{1}{2}{A_0}f(t) + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}f(t)\cos (jnt) + {B_j}f(t)\sin (jnt)} \right)} $ ,
$\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt}  = \frac{1}{2}{A_0}\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)dt}  + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)\cos (jnt)} dt + {B_j}\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)\sin (jnt)dt} } \right)} $
$ = \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $.
It follows that 
$\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t){S_k}(nt)dt}  \to \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^\infty  {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $
as $k$ tends to infinity.
Hence $\frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)g(nt)dt}  = \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^{j = \infty } {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $.
Now we claim that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^\infty  {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)}  = 0$.
Observe that $\left| {\sum\limits_{j = 1}^{j = k} {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} } \right| \le \left| {\sum\limits_{j = 1}^{j = k} {{A_j}{a_{jn}}} } \right| + \left| {\sum\limits_{j = 1}^{j = k} {{B_j}{b_{jn}}} } \right| \le \sqrt {\sum\limits_{j = 1}^{j = k} {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^{j = k} {a_{jn}^2} }  + \sqrt {\sum\limits_{j = 1}^{j = k} {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^{j = k} {b_{jn}^2} } $ 
                                         by the Cauchy Schwarz inequality,
$ \le \sqrt {\sum\limits_{j = 1}^\infty  {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty  {a_{jn}^2} }  + \sqrt {\sum\limits_{j = 1}^\infty  {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty  {b_{jn}^2} } $.

And so $\left| {\sum\limits_{j = 1}^\infty  {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} } \right| \le \sqrt {\sum\limits_{j = 1}^\infty  {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty  {a_{jn}^2} }  + \sqrt {\sum\limits_{j = 1}^\infty  {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty  {b_{jn}^2} } $
$ \le \sqrt {\sum\limits_{j = 1}^\infty  {{A_j}^2} } \sqrt {\sum\limits_{j = n}^\infty  {a_j^2} }  + \sqrt {\sum\limits_{j = 1}^\infty  {{B_j}^2} } \sqrt {\sum\limits_{j = n}^\infty  {b_j^2} } $ .
Note that $\frac{1}{\pi }\int_{ - \pi }^\pi  {|g(t){|^2}dt}  = \frac{1}{2}A_0^2 + \sum\limits_{j = 1}^\infty  {\left( {A_j^2 + B_j^2} \right)} $  and  $\frac{1}{\pi }\int_{ - \pi }^\pi  {|f(t){|^2}dt}  = \frac{1}{2}a_0^2 + \sum\limits_{j = 1}^\infty  {\left( {a_j^2 + b_j^2} \right)} $.
So both $\sum\limits_{j = 1}^\infty  {a_j^2} $ and  $\sum\limits_{j = 1}^\infty  {b_j^2} $ are finite and as a consequence,
$\sum\limits_{j = n}^\infty  {a_j^2}  \to 0{\rm{ }}\quad {\rm{and }}\quad \sum\limits_{j = n}^\infty  {b_j^2}  \to 0{\rm{ }}\quad {\rm{as }}\;n \to \infty $ .
And so it follows from the above inequality that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^\infty  {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)}  = 0$.
Hence $\mathop {\lim }\limits_{n \to \infty } \frac{1}{\pi }\int_{ - \pi }^\pi  {f(t)g(nt)dt}  = \frac{1}{2}{A_0}{a_0} = \frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f(t)dt} \frac{1}{\pi }\int_{ - \pi }^\pi  {g(t)dt} $, 
i.e., $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f(t)g(nt)dt}  = \left( {\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {f(t)dt} } \right)\left( {\frac{1}{{2\pi }}\int_{ - \pi }^\pi  {g(t)dt} } \right)$.
Observe that I only use the fact that both functions are periodic, Lebesgue integrable and square integrable. The continuity of both $f$ and $g$ is used only to deduce both integrablity and square integrability. We may of course replace the condition on continuity by Lebsgue integrability and square integrability.
A: 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 $k$.
A: I would suggest using that continuous functions on compact sets are uniformly continuous, and then using that $f$ is Riemann integrable to split $f$ up over a partition of $[-\pi,\pi]$.
Alternative suggestion: Fejér sums?

Okay, I've thought about this some more, and I think it comes down to one of the following lemmata:

Lemma 1: Let $g$ be as in the question, and let $\chi_A$ be the indicator function of a closed subset in $[-\pi,\pi]$. Then
  $$ \lim_{n \to \infty} \int_{-\pi}^{\pi} \chi_A(t)g(nt) \, dt = \mu(A) \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \, dt \right) $$

(From here, I use the usual increasing sequence of step functions argument for Lebesgue integration) (Why do closed sets suffice? For a continuous function, the level sets are closed - see this question.)
or

Lemma 2: Let $g$ be as in the question, and let $\chi_A$ be the indicator function of an interval in $[-\pi,\pi]$. Then
  $$ \lim_{n \to \infty} \int_{-\pi}^{\pi} \chi_A(t)g(nt) \, dt = \mu(A) \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \, dt \right) $$

(This is obviously sufficient to prove the other one, using the Heine-Borel theorem.)
Then you use that and the properties of Riemann sums to sandwich $f$ and take the limit to get the integral. Lemma 2 looks pretty easy to prove.
