Csdrhrt

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

I have no idea about this problem. I don’t know how to use the periodicity.

(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

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

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

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

the continuity of $g(x)$ suggests that $g(t/n)$ in the $k$th period should almost performance as a constant when $ntoinfty$, 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

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

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
$$
int_0^Tg(x),h(nx),dx=frac1nint_0^{nT} g(x/n),h(x),dx=frac1nsum_{k=0}^{n-1}int_{kT}^{(k+1)T} g(x/n),h(x),dx.
$$
Next substitute $v=x-kT$, to obtain
begin{align}
int_0^Tg(x),h(nx),dx&=frac1nsum_{k=0}^{n-1}int_{0}^{T} g(tfrac{x+kT}n),h(x+kT),dx=frac1nsum_{k=0}^{n-1}int_{0}^{T} g(tfrac{x+kT}n),h(x),dx\
&=frac1Tint_{0}^{T} left(sum_{k=0}^{n-1} g(tfrac{x+kT}n),frac Tnright),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
$$
lim_{ntoinfty}sum_{k=0}^{n-1} g(tfrac{x+kT}n),frac Tn
=lim_{ntoinfty}sum_{k=0}^{n-1} g(tfrac{kT}n),frac Tn=int_0^Tg(x),dx.
$$