I have an equality which I need to prove This question apppear in the chapter of fourier series.
Suppose $f$ is Riemann integrable function in $[a,b]$, and $g$ is a periodic function in $\Bbb R$. The period of $g$ is $T$, and $g$ is integrable in $[0,T]$.
$$\lim_{\lambda \to \infty} \int_a^b f(x) g(\lambda x) dx = \frac{1}{T} \int_a^b f(x) dx \int_0^T g(x) dx$$
Please prove this equality.
 A: Hint: Since $f$ and $g$ are Riemann integrable, they are bounded.
Write
$$
g(x)=\sum_{k\in\mathbb{Z}}a_ke^{2\pi ikx/T}\tag{1}
$$
Then, since $\|f\|_2^2\le\|f\|_1\|f\|_\infty$ and $\|g\|_2^2\le\|g\|_1\|g\|_\infty$, the following sum makes sense
$$
\begin{align}
\int_a^bf(x)g(\lambda x)\,\mathrm{d}x
&=\sum_{k\in\mathbb{Z}}a_k\int_a^bf(x)\,e^{2\pi ik\lambda  x/T}\,\mathrm{d}x\\
&=\sum_{k\in\mathbb{Z}}a_k\hat{f}(k\lambda)\\
&=a_0\hat{f}(0)+\sum_{k\ne0}a_k\hat{f}(k\lambda)\tag{2}
\end{align}
$$
where
$$
\sum_{k\in\mathbb{Z}}a_k^2=\frac1T\|g\|_2^2\le\frac1T\|g\|_1\|g\|_\infty\tag{3}
$$
and
$$
\sum_{k\in\mathbb{Z}}b_k^2=\sum_{k\in\mathbb{Z}}\hat{f}\left(k\frac{T}{b-a}\right)^2=\frac{\|f\|_2^2}{b-a}\le\frac{\|f\|_1\|f\|_\infty}{b-a}\tag{4}
$$
Since $(4)$ is finite, argue that
$$
\lim_{n\to\infty}\sum_{k\ne0}b_{nk}^2=0\tag{5}
$$
Apply $(5)$ to $(2)$ to show that
$$
\lim_{n\to\infty}\sum_{k\ne0}a_k\hat{f}\left(k\frac{nT}{b-a}\right)=0\tag{6}
$$
$(2)$ and $(6)$ show the desired result for $\lambda=\frac{nT}{b-a}$. This can be extended to all $\lambda\to\infty$ by using the fact that translation is continuous on $L^2$ applied to $h(x)=g(e^x)$.
