Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ g(x) $ be a continuous periodic function of period 1 on $\mathbb{R}$. Prove that for any integrable function $f(x)$ on $[0,1]$,
$$ \lim_{n \to \infty}\int_0^{1}f(x)g(nx)dx= \int_0^{1}f(x)dx \int_0^{1}g(x)dx.$$

Any help is appreciated.

share|cite|improve this question

Begin with a transformation $u=nx.$

$$\int_{0}^1 f(x)g(nx)dx = \frac{1}{n} \int_{0}^n f(u/n)g(u)du.$$

Break the integrand up into a sum of intervals.

$$ \frac{1}{n} \int_{0}^n f(u/n)g(u)du=\frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du.$$

Make another variable transformation: $v=u-(j-1).$ Because $g(u)$ is 1 periodic $g(u)=g(u+1)=g(u+j-1).$

$$ \frac{1}{n}\sum_{j=1}^n\int_{j-1}^{j} f(u/n)g(u)du = \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f \left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du $$

Rearrange the terms. $$ \frac{1}{n}\sum_{j=1}^n\int_{0}^{1} f\left(\frac{u-(j-1)}{n} \right)g(u-(j-1))du =\int_{0}^1 \left(\frac{1}{n} \sum_{j=1}^n f \left(\frac{u-(j-1)}{n} \right)\right)g(u)du .$$

We have now produced a Riemann sum which converges to an integral in the limit. We are now allowed, by dominated convergence theorem, to say

$$ \begin{align*} \lim_{n\to \infty} \int_{0}^1 f(x)g(nx)dx &= \int_{0}^1 \left(\int_{0}^1 f(z)dz\right) g(u)du \\ &=\int_{0}^1 f(z)dz\int_{0}^1 g(u)du=\int_{0}^1 f(x)dx\int_{0}^1 g(x)dx . \end{align*}$$

share|cite|improve this answer
Thank you so much. – Sume Aug 10 '11 at 3:28

this is not a good proof, just an idea. First consider f to be simple function, we can calculate that this identity holds. Then by the definition of Lebesgue integration, we can approximate f by simple functions. Because g is continuous(bounded), so there's no problem for the left hand side to converge.

share|cite|improve this answer

Ok. I have seen this problem in the book: Principles of Real Analysis by C.D.Aliprantis and O.Burkinshaw, and since I knew that this book has a solution manual, I went and searched over there and got the solution. The problem given in the book is as follows:

$\textbf{Problem.}$ Let $f:(0,\infty) \to \mathbb{R}$ be a real-valued continuous function such that $f(x+1)=f(x)$ for all $x \geq 0$. If $g:[0,1] \to \mathbb{R}$ is an arbitrary continuous function, then show that $$\lim_{n \to \infty} \ \int\limits_{0}^{1} g(x) \cdot f(nx) \ dx = \Biggl( \ \ \int\limits_{0}^{1} g(x) \ dx \Biggr) \cdot \Biggl( \ \ \int\limits_{0}^{1} f(x) \ dx\Biggr)$$

$\textbf{Solution.}$ Please see the book: Problems in real analysis a workbook with solutions Problem 23.14 Page $\textbf{205}$ for a complete solution.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.