Sum of a periodic function

**bkarpuz** · May 17th 2011, 08:39 AM

Dear MHF members,

Question. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a continuous periodic function with a period of $2\pi$ . If $\alpha/\pi$ is irrational, then for any $x\in\mathbb{R}$ show that the sum $\frac{1}{N}\sum_{n=1}^{N}f(x+n\alpha)$ converges to $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)\mathrm{d}u$ .

Thanks.
bkarpuz

Exercise 19 in Chapter 3 of W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Science/Engineering/Math, 3rd edition, 1976.

**Tinyboss** · May 17th 2011, 09:20 AM

I assume f satisfies some continuity condition, else it's definitely not true. Hint: show that $\{x+n\alpha\text{ mod }2\pi\}_{n\in\mathbb{N}_+}$ is dense in [0,2pi].

**Drexel28** · May 17th 2011, 11:27 AM

Originally Posted by bkarpuz

Dear MHF members,

Question. Suppose that $f:\mathbb{R}\to\mathbb{R}$ is a continuous periodic function with a period of $2\pi$ . If $\alpha/\pi$ is irrational, then for any $x\in\mathbb{R}$ show that the sum $\frac{1}{N}\sum_{n=1}^{N}f(x+n\alpha)$ converges to $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(u)\mathrm{d}u$ .

Thanks.
bkarpuz

Exercise 19 in Chapter 3 of W. Rudin, Principles of Mathematical Analysis, McGraw-Hill Science/Engineering/Math, 3rd edition, 1976.

Tinyboss gave you an idea how to prove this. This theorem is a specific case of one of the equidistribution theorems

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