Playing with Integrals: Limits and Integrals 4

This note wasn't a planned one, but yesterday I tried to solve a very beautiful problem proposed to me by Gabriel Stuart Romon. The problem asked to find the following limit

limnabf(x)sin(nx)dx,f(x)C0[a,b].\lim_{n\to\infty}\int^b_a f(x)|\sin(nx)|\,dx, \forall f(x)\subset C^0 [a,b].

The detailed solution to this problem, written by Gabriel may be found here. However, I want to discuss another problem, which I found on math.stackexchange. The key idea of it helped me to elaborate the solution to Gabriel's problem. I hope you'll find it beautiful and interesting.

Problem. If g(x):RRg(x):\mathbb{R}\rightarrow\mathbb{R} is a continuous periodic function with period 1 and f(x):[0,1]Rf(x):[0,1]\rightarrow\mathbb{R} is a continuous function, prove that limn01f(x)g(nx)dx=(01f(x)dx)(01g(x)dx). \lim_{n\to\infty}\int^1_0 f(x)g(nx)\,dx=\left(\int^1_0 f(x)\,dx\right)\left(\int^1_0 g(x)\,dx\right).

Solution. First of all we'll break our single integral into the sum of integrals over the shorter intervals.

01f(x)g(nx)dx=i=0n1ini+1nf(x)g(nx)dx=\int^1_0 f(x)g(nx)\,dx=\sum^{n-1}_{i=0}\int^{\frac{i+1}{n}}_{\frac{i}{n}} f(x)g(nx)\,dx= =1n(i=0n1ii+1f(un)g(u)du)=\frac{1}{n}\left(\sum^{n-1}_{i=0}\int^{i+1}_i f\left(\frac{u}{n}\right)g(u)\,du\right)

By applying mean-value theorem we'll transform our sum into something similar to a Riemann sum and we'll use the fact that g(x)g(x) is periodic in our favor.

1n(i=0n1ii+1f(un)g(u)du)=1ni=0n1(f(ξi)ii+1g(u)du)=\frac{1}{n}\left(\sum^{n-1}_{i=0}\int^{i+1}_{i} f\left(\frac{u}{n}\right)g(u)\,du\right)=\frac{1}{n}\sum^{n-1}_{i=0}\left(f(\xi_i)\int^{i+1}_{i} g(u)\,du\right)= =1n(i=0n1f(ξi))(01g(u)du)=\frac{1}{n}\left(\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(u)\,du\right)

Because every ξi[in,i+1n]\xi_i\in\left[\dfrac{i}{n},\dfrac{i+1}{n}\right], we have that the limit of our sum is 01f(x)dx\int^1_0 f(x)\, dx, so

limn01f(x)g(nx)dx=limn(1ni=0n1f(ξi))(01g(x)dx)=\lim_{n\to\infty}\int^1_0 f(x)g(nx)\,dx=\lim_{n\to\infty}\left(\frac{1}{n}\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(x)\,dx\right)= =(01f(x)dx)(01g(x)dx).=\left(\int^1_0 f(x)\,dx\right)\left(\int^1_0 g(x)\,dx\right).

Note by Nicolae Sapoval
5 years, 9 months ago

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Hello Nicolae! Great post as always. :)

Can you please explain the step where you take f(ξi) f(\xi_i) out of the integral? I fail to understand why it is a valid step.

Many thanks!

Pranav Arora - 5 years, 9 months ago

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Yep, I apply the Mean Value Theorem, so it allows me to pull out the f(ξi)f(\xi_i).

Nicolae Sapoval - 5 years, 9 months ago

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Thanks for the quick response!

I have never seen mean value theorem for integration, I am sorry if I ask stupid questions. From what I see at wiki, it states that this theorem is applicable if φ \varphi (g(x)g(x) in our case) does not change sign on (a,b)(a,b) (or i/n,(i+1)/n)i/n,(i+1)/n) in our case). How do you know that this condition is followed here?

Pranav Arora - 5 years, 9 months ago

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@Pranav Arora Yes, you're right my friend! thanks for pointing that out. Here is the missing part.

In case if g(x)g(x) isn't positive for all reals (so it changes sign) we may find a constant MM, such that g(x)+M>0g(x)+M> 0. The existence of the constant is caused by the afct that the function g(x)g(x) is integrable. By propostion we proved

limn01f(x)(g(nx)+M)dx=(01(g(nx)+M)dx)(01f(x)dx)\lim_{n\to\infty}\int^1_0f(x)(g(nx)+M)\,dx=\left(\int^1_0 (g(nx)+M)\,dx\right)\left(\int^1_0f(x)\,dx\right)\Leftrightarrow limn01f(x)g(nx)dx=(01g(nx)dx)(01f(x)dx).\Leftrightarrow \lim_{n\to\infty}\int^1_0f(x)g(nx)\,dx=\left(\int^1_0 g(nx)\,dx\right)\left(\int^1_0f(x)\,dx\right).

Comment. Those parts with MM are not affected by the limit and theya re equal on the RHSRHS and the LHSLHS, so the problem is done.

Nicolae Sapoval - 5 years, 9 months ago

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