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

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

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

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

\[\frac{1}{n}\left(\sum^{n-1}_{i=0}\int^{i+1}_{i} f\left(\frac{u}{n}\right)g(u)\,du\right)=\frac{1}{n}\sum^{n-1}_{i=0}\left(f(\xi_i)\int^{i+1}_{i} g(u)\,du\right)=\] \[=\frac{1}{n}\left(\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(u)\,du\right)\]

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

\[\lim_{n\to\infty}\int^1_0 f(x)g(nx)\,dx=\lim_{n\to\infty}\left(\frac{1}{n}\sum^{n-1}_{i=0}f(\xi_i)\right)\left(\int^1_0 g(x)\,dx\right)=\] \[=\left(\int^1_0 f(x)\,dx\right)\left(\int^1_0 g(x)\,dx\right).\]

## Comments

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TopNewestHello Nicolae! Great post as always. :)

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

Many thanks! – Pranav Arora · 3 years, 2 months ago

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Mean Value Theorem, so it allows me to pull out the \(f(\xi_i)\). – Nicolae Sapoval · 3 years, 2 months ago

Yep, I apply theLog in to reply

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 \( \varphi \) (\(g(x)\) in our case) does not change sign on \((a,b)\) (or \(i/n,(i+1)/n) \) in our case). How do you know that this condition is followed here? – Pranav Arora · 3 years, 2 months ago

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

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

Comment.Those parts with \(M\) are not affected by the limit and theya re equal on the \(RHS\) and the \(LHS\), so the problem is done. – Nicolae Sapoval · 3 years, 2 months agoLog in to reply