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Please just a hint, I'd prefer to solve this on my own, but I can't see what to do. Here is the question:

Given that $c \notin 2\pi\mathbb{Q}$, $f$ is everywhere continuous and $f$ is $2\pi$ periodic, prove that $$\lim_{n\to\infty} \frac{1}{n}\sum_{j=0}^{n-1} f(x-cj) = \frac{1}{2\pi}\int_0^{2\pi} f(t)dt.$$

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    Do the numbers $x-cj$ belong to different subintervals of length $2\pi/n$ of $[0,2\pi]$ for different values of $j$? The left-hand side looks a lot like a Riemann sum. – 2017-01-25
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    Also, what is $x$? – 2017-01-25
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    I should have specified that $c$ is real; $x$ is just a real variable, I believe. I don't think that $x-cj$ must be in that interval, though. – 2017-01-25
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    Since $f$ is periodic I meant mod $2\pi$. – 2017-01-25
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    I'm not actually sure - the only things given are that $x\in S^1$ and $c\notin 2\pi\mathbb{Q}.$ – 2017-01-25

2 Answers 2

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$\frac{1}{2\pi}\int_0^{2\pi} f(t)dt$ is the average value of $f(x)$ over one period.

Since $f(x)$ is periodic with period $2\pi$

$f(x-jc) = f(a)$ with $a\in [0,2\pi]$

And since Since $c\ne 2\pi\mathbb Q$

Every $x - jc$ is equivalent to a different value of $a$

And with a large enough value of $n$ this set we expect to be uniformly distributed over $[0,2\pi]$

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    This is very insightful - thank you! – 2017-01-25
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Hint: It's enough to show

$$\lim_{n\to\infty} \frac{1}{n}\sum_{j=0}^n f(cj) = \frac{1}{2\pi}\int_0^{2\pi} f(t)dt.$$

Prove this first for $f(t) = e^{imt}, m \in \mathbb Z.$ Think about Stone-Weierstrass ...

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    I'm not able to see how the statements are equivalent - is it because $x$ falls on the unit circle? It's also not clear to me how this holds true for $f(t) = e^{imt}$... – 2017-01-25