Evaluate integral of $2\pi$ periodic function's multiplication

Question

Let f and g be $2\pi$ periodic Riemann integrable funtions.

I want to evaluate $\lim_{n \to \infty} \frac{1}{2\pi} \int_0^{2\pi}f(x)g(nx)dx$

I think that this integral express by using fourier coefficient.

How to evaluate this integral?

Answer 1

Let $g_0 = \frac{1}{2\pi}\int_{0}^{2\pi}g(x)dx$ and $f_0=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)dx$. Then \begin{align} \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx &=\frac{1}{2\pi}\int_{0}^{2\pi}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{2\pi}\sum_{k=1}^{n}\int_{\frac{2\pi}{n}(k-1)}^{\frac{2\pi}{n}k}f(x)(g(nx)-g_0)dx+f_0g_0 \\ &=\frac{1}{2\pi}\sum_{k=1}^{n}\int_{\frac{2\pi}{n}(k-1)}^{\frac{2\pi}{n}k}\left(f(x)-f(\frac{2\pi}{n}k)\right)(g(nx)-g_0)dx+f_0g_0 \end{align} Because $g$ is Riemann integrable, then $g$ is uniform bounded by some constant $M$. Let $I_k$ be the interval $[\frac{2\pi}{n}(k-1),\frac{2\pi}{n}k]$ with length $|I_k|=\frac{2\pi}{n}$. Then, assuming $f$ is real, \begin{align} \left|\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx-f_0g_0\right| &\le \frac{1}{2\pi}\sum_{k=1}^{n}\left(\sup_{x\in I_k}f(x)-\inf_{x\in I_k}f(x)\right)2M\frac{2\pi}{n} \\ & =\frac{M}{\pi}\sum_{k=1}^{n}\left(\sup_{x\in I_k}f(x)-\inf_{x\in I_k}f(x)\right)|I_k| \\ & =\frac{M}{\pi}\left(\overline{S}_{\mathcal{P_n}}(f)-\underline{S}_{\mathcal{P_n}}(f)\right) \end{align} where $\overline{S}_{\mathcal{P}}(f)$, $\underline{S}_{\mathcal{P}}(f)$ are the upper and lower Riemann sums of $f(x)$ over the partition $$ \mathcal{P}=\{ 0,\frac{2\pi}{n},2\frac{2\pi}{n},3\frac{2\pi}{n},\cdots\frac{2\pi}{n}\}. $$ Because $f$ is Riemann integrable on $[0,2\pi]$, the right side tends to $0$ as $n\rightarrow\infty$. Hence, $$ \lim_{n\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx = f_0 g_0. $$