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

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

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):R→Rg(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 lim⁡n→∞∫01f(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=0n−1∫ini+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=0n−1∫ii+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=0n−1∫ii+1f(un)g(u) du)=1n∑i=0n−1(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=0n−1f(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

lim⁡n→∞∫01f(x)g(nx) dx=lim⁡n→∞(1n∑i=0n−1f(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).

#Calculus #Limits #Integration #Goldbach'sConjurersGroup

Note by Nicolae Sapoval
7 years, 5 months ago

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Comments

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 - 7 years, 5 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 - 7 years, 5 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 - 7 years, 5 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

lim⁡n→∞∫01f(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 ⇔lim⁡n→∞∫01f(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 - 7 years, 5 months ago
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