How Does Weyl's Criterion Apply in Proving Uniform Convergence?

[kittensies]

Homework Statement

f is a continuous periodic function on $$\mathbb{R}$$ with period 1, and $${\xi_n}$$ is a equidistributed sequence on [0,1).

Also given is that $$\int_0^1 f(x)dx=0$$

I need to prove that $$\lim_{n\rightarrow \infty}\frac{1}{N}\sum_{n=1}^N f(x+\xi_n) = 0$$ uniformly in x.

Homework Equations

Weyl's criterion?

The Attempt at a Solution

Let f be a trig polynomial. Then $$\frac{1}{N}\sum_{n=1}^N f(x+\xi_n) = \frac{1}{N}\sum_{n=1}^N e^{2\pi in(x+\xi_n)} \leq \frac{1}{N}|\sum_{n=1}^N e^{2\pi in(x)}||\sum_{n=1}^N e^{2\pi in\xi_n)}|$$

By Weyl's criterion, $$\frac{1}{N}|\sum_{n=1}^N e^{2\pi in\xi_n)}|\rightarrow 0$$ as N goes to $\infty$, and $$|\sum_{n=1}^N e^{2\pi in(x)}|$$ goes to 0 since $$\int_0^1 f(x)dx=0$$.

(This part above I'm not certain about -- can I just show that the two parts of something bigger absolutely converges to show that P converges?)

Then we can find a trig polynomial P such that $$sup |f(x)-P(x)|\leq\epsilon$$
Then $$|\dfrac{1}{N}\sum_{n=1}^N P(x+\xi_n) - \int f(x)dx|=|\dfrac{1}{N}\sum_{n=1}^N P(x+\xi_n)|\leq$$
$$\dfrac{1}{N}|\sum_{n=1}^N f(x+\xi_n) - P(x+\xi_n)| +\dfrac{1}{N}|\sum_{n=1}^N P(x+\xi_n) - P(x)| +\int_0^1 |f(x) - P(x)|\leq 3\epsilon$$

Does this work? I'm not sure if the logic follows.

 

[mighty2000]

Also, how does Weyl's criterion fit into this proof? Thanks for any help you can provide.Your attempt at a solution looks good so far. The use of Weyl's criterion is correct, as it shows that the term $\frac{1}{N}\sum_{n=1}^N e^{2\pi in\xi_n}$ converges to 0 as $N\rightarrow \infty$. This is important because it allows you to bound the term $\frac{1}{N}\sum_{n=1}^N e^{2\pi in(x+\xi_n)}$ by a constant multiple of $\frac{1}{N}$, which goes to 0 as well.

To address your question about showing convergence, you are on the right track. To show that $\frac{1}{N}\sum_{n=1}^N e^{2\pi in\xi_n}$ converges, you can use the fact that it is a sequence of complex numbers and use the triangle inequality to bound it by a geometric series. Then, as you mentioned, you can use the absolute convergence of that series to show that the original sequence converges as well.

One thing to note is that you should probably specify that the trig polynomial $P$ is chosen such that it approximates $f$ uniformly, since you need uniform convergence in order to take the limit inside the integral in your final step. Other than that, your proof looks good.