# fractional part and integration

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• January 6th 2010, 08:27 AM
Shanks
fractional part and integration
Let $\{x\}$ be the positive fractional part of real number $x$, for example, $\{\pi\}=0.1415926...$ .
Prove that the set $S=\{\{nv\}:n\in Z^+\}$ is dense in [0,1] if $v$ is positive irrational number.

And for any $f(x)\in C[0,1]$(the class of all real-valued continious function on [0,1]), and any positive irrational number $v$, we have

$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}f(\{nv\} )=\int_0^1f(x)dx$.
I know how to do this problem, but i can't write down the formal solution and some details completely.
I need a complete and elaborate solution, thanks in advance!
• January 6th 2010, 10:47 AM
Laurent
Quote:

Originally Posted by Shanks
I know how to do this problem, but i can't write down the formal solution and some details completely.
I need a complete and elaborate solution, thanks in advance!

You should tell us more precisely which details are lacking in your solution, it would spare some time.
• January 7th 2010, 01:38 AM
Shanks
notice the fact that if $t\in S$, then the fractional part of the multiply of t is also in S. To prove S is dense in [0,1], it is suffice to prove that there is a decreasing subsequence in S such that it converges to 0. Or in another word, 0 is the inf limit of S. This can be easily proved by contradiction!
To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x).
that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh)
• January 8th 2010, 12:33 AM
Laurent
Quote:

Originally Posted by Shanks
To prove the limit is equal to the integration, we need the definition of Riemann Integral and the uniform continuity of f(x).
that is, it is suffice to prove that for any $\delta >0$, there exist N, such that if k > N, then the longest distance of two consecutive( the oder of number, not the index n) among the previous k points is less than $\delta$, Or in another word, the distance of any two consecutive points of the previous k points is less than $\delta$. Although I understand what is actually going on, I can't express by word, can't write down the details here!(Doh)

You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral.

I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others.
• January 8th 2010, 02:52 AM
Shanks
Quote:

Originally Posted by Laurent
You need more than just that: not only need the points of the sequence to be close enough, but they need to be "uniformly distributed" (asymptotically) for the sum to converge to the integral.

I give you a magic trick: first prove the limit when $f(x)=e^{2i\pi nx}$ for some non-zero integer $n$ (direct computation), then for trigonometric polynomials, i.e. linear combinations of the previous functions (direct consequence), and finally for general $f$ (using approximation by a trigonometric polynomial). This is the usual way to procede (not very intuitive, I admit), but there may be others.

I know that this is the way written in G.polya book.
But Here I don't want to using approximation of continious function by trigonometric polynomial. I want to prove it by using the definition of Riemann integral.
• January 16th 2010, 03:03 AM
Shanks
The Problem occurs again in Rudin's analysis book, therefore it is a impressive problem.
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