Prove $\lim_{n\to \infty}\int_0^Tg(x)h(nx) dx=\cfrac{1}{T}(\int_0^Tg(x)dx)(\int_0^Th(x)dx)$

If $$h(x),g(x)$$ continuous, and $$h(x)$$ is periodic function with a period $$T$$, then prove: $$\lim_{n\to \infty}\int_0^Tg(x)h(nx) dx=\cfrac{1}{T}\left(\int_0^Tg(x)dx\right)\left(\int_0^Th(x)dx\right)$$

• Hello and welcome to MSE! What are your thoughts on this problem? What have you tried so far? What related theorems do you know, like for example the Riemann-Lebesgue Lemma and how did you proof it? – F. Conrad Dec 19 '18 at 3:07
• Do you mean $g(x)$ and $h(x)$, in which $h(x)$ is periodic, or your equation can not fit your assumption. – Nanayajitzuki Dec 19 '18 at 4:07
• @Nanayajitzuki Oh, yes. Thank you for reminding me. I just found this mistake and I ’m correcting it now. – Yan Peng Dec 19 '18 at 8:29
• @F.Conrad Thanks for your suggestions. I’ll come to know about the related theorems, like the Riemann-Lebesgue Lemma you mentioned, and rethink the problem. – Yan Peng Dec 19 '18 at 8:42

(edited for complement of some critical steps)

I think your assumption actually is If $$g(x),h(x)$$ are continuous, and $$h(x)$$ is periodic function with a period $$T$$.

I will give a hint under this assumption (some details may not be so strict), let

$$\widetilde h(x)=h(x)-\frac1{T} \int_0^{T}{h(t){\text d}t}$$

you will get

$$\int_0^{T}{\widetilde h(x){\text d}x}=\int_0^{T}{h(x){\text d}x}-\int_0^{T}{\left( \frac1{T}\int_0^{T}{h(t){\text d}t}\right){\text d}x}=0$$

so you just need to prove

$$\lim_{n\to \infty}\int_0^{T}{g(x)\widetilde h(nx){\text d}x}=\left(\frac1{T}\int_0^{T}\widetilde h(x){\text d}x\right)\left(\int_0^{T}{g(x){\text d}x}\right)=0$$

let $$t=nx$$ (here I assume $$n$$ is any positive real number), where $$[n]$$ means the integer part of $$n$$ and $$a=T(n-[n])

\begin{aligned} \int_0^{T}{g(x)\widetilde h(nx){\text d}x} & = \frac1{n} \int_0^{[n]T+a}{g(t/n)\widetilde h(t){\text d}t} \\ & = \frac1{n} \sum_{k=0}^{[n]} \int_{kT}^{(k+1)T}{g(t/n)\widetilde h(t){\text d}t} + \frac1{n} \int_{[n]T}^{[n]T+a} {g(t/n)\widetilde h(t){\text d}t} \end{aligned}

the continuity of $$g(x)$$ suggests that $$g(t/n)$$ in the $$k$$th period should almost performance as a constant when $$n\to\infty$$, thus for any $$k$$

$$\int_{kT}^{(k+1)T}{g(t/n)\widetilde h(t){\text d}t} \to g(t_{k}) \int_{kT}^{(k+1)T}{\widetilde h(t){\text d}t} = 0$$

as well notice that $$g(x),h(x)$$ are continuous, which means they are bounded in the period, assume $$|g(x)|\le G$$, $$|\widetilde h(x)|\le H$$

$$\frac1{n} \left| \int_{[n]T}^{[n]T+a} {g(t/n)\widetilde h(t){\text d}t} \right| \le \frac1{n} \int_{[n]T}^{[n]T+a} {|g(t/n)||\widetilde h(t)|{\text d}t} \le \frac{G}{n}\int_0^{a}{|\widetilde h(t)|{\text d}t} \le \frac{aGH}{n} \to 0$$

just for curiosity, in general:

If $$h(x)$$ is Lebesgue measurable and periodic in $$\mathbb R$$, $$I$$ is any interval, $$g(x) \in \mathcal L(I)$$, we have

$$\lim_{|\lambda| \to +\infty}\int_I {g(x)h(\lambda x){\text d}x}=\left(\frac1{T}\int_0^{T}h(x){\text d}x\right)\left(\int_I{g(x){\text d}x}\right)$$

under the meaning of Lebesgue integration.

If the periodic one is $$g$$, the equality does not hold. For instance, take $$g(x)=\sin^2x$$, $$h(x)=x$$, $$T=\pi$$.

When $$h$$ has period $$T$$, use first the substitution $$u=nx$$ to get $$\int_0^Tg(x)\,h(nx)\,dx=\frac1n\int_0^{nT} g(x/n)\,h(x)\,dx=\frac1n\sum_{k=0}^{n-1}\int_{kT}^{(k+1)T} g(x/n)\,h(x)\,dx.$$ Next substitute $$v=x-kT$$, to obtain \begin{align} \int_0^Tg(x)\,h(nx)\,dx&=\frac1n\sum_{k=0}^{n-1}\int_{0}^{T} g(\tfrac{x+kT}n)\,h(x+kT)\,dx=\frac1n\sum_{k=0}^{n-1}\int_{0}^{T} g(\tfrac{x+kT}n)\,h(x)\,dx\\ &=\frac1T\int_{0}^{T} \left(\sum_{k=0}^{n-1} g(\tfrac{x+kT}n)\,\frac Tn\right)\,h(x)\,dx. \end{align} Because $$g$$ is uniformly continuous on $$[0,T]$$, the values $$g(\tfrac{x+kT}n)$$ are very close to $$g(\tfrac kTn)$$. So $$\lim_{n\to\infty}\sum_{k=0}^{n-1} g(\tfrac{x+kT}n)\,\frac Tn =\lim_{n\to\infty}\sum_{k=0}^{n-1} g(\tfrac{kT}n)\,\frac Tn=\int_0^Tg(x)\,dx.$$