Limit of an integral with a periodic function

Question

Let $f , g$ be continuous functions: $f:[0 , 2\pi]\rightarrow\mathbb{R}$ and $g:\mathbb{R}\rightarrow\mathbb{R}$. Assume $\forall x\in\mathbb{R}:g(x+2\pi)=g(x)$ and $$\int\limits_{0}^{2\pi} \! {g(x)} \, \mathrm{d}x=0.$$ Show that $$\lim\limits_{n\rightarrow\infty} \int\limits_{0}^{2\pi} \! {f(x) g(nx)} \, \mathrm{d}x=0.$$

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I've started evaluating the difference upwards but it seems like no use: $$\left| \int\limits_{0}^{2\pi} \! {f(x) g(nx)} \, \mathrm{d}x-0 \right|=\left| \int\limits_{0}^{2\pi} \! {f(x) g(nx)} \, \mathrm{d}x \right|\leq\int\limits_{0}^{2\pi} \! {\left|f(x) g(nx)\right|} \, \mathrm{d}x=\int\limits_{0}^{2\pi} \! {\left|f(x)\right| \left|g(nx)\right|} \, \mathrm{d}x$$ $$=\left|f(\xi)\right| \int\limits_{0}^{2\pi} \! { \left|g(nx)\right|} \, \mathrm{d}x$$ for some $\xi \in ]0 , 2\pi[$. What's next?

Answer 1

Since $g$ is $2 \pi$ periodic with zero mean, it follows that $t \mapsto g(nt)$ is $\frac{2 \pi}{n}$ periodic with zero mean. Consequently, if $I \subset [0,2 \pi]$ is any interval of length $\frac{2 \pi}{n}$, then $\int_0^{2 \pi} 1_I(t) g(nt) dt = 0$.

Let $f_n(t) = f(\frac{2 \pi}{n}\lfloor \frac{nt}{2 \pi} \rfloor)$, and note that $f_n = \sum_{k=0}^{n-1} f(\frac{k}{n}2 \pi)1_{[\frac{k}{n}2 \pi, \frac{k+1}{n}2 \pi)}$ on $[0,2 \pi)$. It follows from the previous paragraph that $\int_0^{2 \pi} f_n(t) g(nt) dt = 0$.

Since $f$ is continuous, it is uniformly continuous on $[0,2 \pi]$. Let $\epsilon>0$ and let $\delta>0$ be such that if $|x-y| < \delta$, then $|f(x)-f(y)| < \epsilon$. Choose $N$ such that $\frac{2 \pi}{N} < \delta$, and let $n \ge N$. Note that $|f(t)-f_n(t)| < \epsilon$ for all $t$.

Then we have the estimate \begin{eqnarray} | \int_0^{2 \pi} f(t) g(nt) dt| &\le &| \int_0^{2 \pi} (f(t)-f_n(t)) g(nt) dt|+ | \int_0^{2 \pi} f_n(t) g(nt) dt| \\ &\le& \int_0^{2 \pi} |f(t)-f_n(t)| |g(nt)| dt + 0\\ &\le& \epsilon \int_0^{2 \pi} |g(nt)| dt \\ &=& \epsilon \int_0^{2 \pi} |g(t)| dt \end{eqnarray} Hence we have the desired result.

Answer 2

An idea: substitution:

$$u:=nx\Longrightarrow dx=\frac{du}{n}\Longrightarrow \int\limits_0^{2\pi} g(nx)dx=\frac{1}{n}\int\limits_0^{2\pi n} g(u)\,du=0$$

since

$$\int\limits_0^{2n\pi}g(u)\,du=\sum_{k=1}^n\;\int\limits_{2(k-1)\pi}^{2k\pi}g(u)\,du$$

but, for any $\,1\le k\le n\,$:

$$\int\limits_{2(k-1)\pi}^{2k\pi}g(u)\,du=\int\limits_{2(k-1)\pi}^{2k\pi}g(u+2k\pi)\,du=\int\limits_0^{2\pi}g(t)\,dt=0$$

where the second equality follows from the substitution

$$t:=u-2(k-1)\pi\,\;,\;\;dt=du$$