Suppose $f$ is continuous and periodic on the reals with period 1. Prove that if $x\in[0,1]$ is an irrational number, then
$$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N f(nx)=\int_{0}^1f(t)dt$$
Suggestion: First consider $f(t) = e^{2\pi(ikt)}$ where k is an integer.
I can see that this is a limit of a weighted average, but the suggestion throws me off. I've seen the suggestion in fourier transforms but it's not clicking at the moment. Any help would be welcome.
Using the hint you were given, it is easy to verify that $$ \int_0^1 e^{2\pi i k t} \, \mathrm{d}t = \left\{ \begin{array}{lr} 1 & :\ k = 0 \\ 0 & :\ k \neq 0 \end{array} \right. $$ Similarly, for $k \neq 0$ and irrational $x [0,1]$, using geometric series $$ \begin{align*} \lim_{N \to \infty} \frac{1}{N} \left| \sum_{n=1}^N e^{2 \pi i k n x} \right| &= \lim_{N \to \infty} \frac{1}{N} \left| e^{2 \pi i k x} \frac{e^{2 \pi i k N x} - 1}{e^{2 \pi i k x} - 1} \right| \\ &\leq \lim_{N \to \infty} \frac{1}{N} \frac{2}{|e^{2 \pi i k x} - 1|} \\ &= 0 \end{align*} $$ noting that since $x$ is irrational $e^{2 \pi i k x} \neq 1$ for any $k \neq 0$. On the other hand, for $k = 0$ we have $e^0 = 1$, so $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1 = 1. $$ It follows that $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N e^{2 \pi i k n x} = \int_0^1 e^{2 \pi i k t} \, \mathrm{d}t. $$
Now for any continuous function $f$ on $\mathbb{R}$ with period $1$, there is a sequence of complex numbers $\{c_k\}_{-\infty}^\infty$ such that $$ f(t) = \sum_{k = -\infty}^\infty c_k e^{2 \pi i k t}. $$ So with a little justification of the interchange between sum and integral, $$ \begin{align*} \int_0^1 f(t) \, \mathrm{d}t &= \int_0^1 \sum_{k = -\infty}^\infty c_k e^{2 \pi i k t} \, \mathrm{d}t \\ &= \sum_{k = -\infty}^\infty c_k \int_0^1 e^{2 \pi i k t} = c_0. \end{align*} $$ And correspondingly, for irrational $x \in [0,1]$, $$ \begin{align*} \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(n x) &= \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N \sum_{k = -\infty}^\infty c_k e^{2 \pi i k n x} \\ &= \sum_{k = -\infty}^\infty c_k \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N e^{2 \pi i k n x} \\ &= c_0. \end{align*} $$
It follows that $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(n x) = \int_0^1 f(t)\, \mathrm{d}t. $$
Here's a sketch of another way to try to do it, if for some odd reason you don't want to try the better way using Stone-Weierstrass.
Assume that you have the conclusion of the hint, as demonstrated in the previous answer.
Consider the family of functions, for $N = 1, 2, 3, \dots$ $$ h_N(t) = \frac{1}{N}\{ f(t+\alpha) + f(t+2\alpha) + \dots + f(t+N\alpha) \} - \int_0^1 f(t)\ dt $$
Using the hint, show that the sequence $$\widehat{h_N}(k) = \int_0^1 h_N(t)\ dt$$ converges to $0$ in $l^2(\mathbb{Z})$.
We know $l^2(\mathbb{Z})$ and $L^2([0, 1])$ are Hilbert space isometric, because the trigonometric polynomials are dense in $L^2([0, 1])$, and so it follows that $h_N$ converges to $0$ in $L^2([0, 1])$.
We have to show that in fact this yields pointwise convergence, which is general not true but in this case is because of the continuity of $f$.
Here's an easy fact to verify: Suppose you have a sequence and a value, such that every subsequence has a sub-subsequence which converges to that value. Then, the original sequence also converges to that value.
Pick $t_*$ in $[0, 1]$, and pick an arbitrary subsequence of $\{h_N(t_*)\}$.
The corresponding subsequence of $\{h_N\}$ converges in $L^2$ to $0$, since $\{h_N\}$ itself does.
We know from $L^2$-space theory that there is a sub-subsequence which converges pointwise almost everywhere to $0$.
There are a couple of different ways to proceed from here. One way is to notice that the sub-subsequence of $\{h_N\}$ is equicontinuous. Then you can apply Arzela's theorem to find a sub-sub-subsequence which converges uniformly to a continuous limit function $\phi(t)$ on $[0, 1]$.
By Dominated Convergece, this sub-sub-subsequence also converges in $L^2$ to $\phi$.
So, $\phi$ equals $0$ almost everywhere, and by continuity they are in fact equal everywhere.
The upshot is that $\phi(t) = 0$ for all $t$, so for our arbitrary $t_*$ and our arbitrary subsequence of $\{h_N(t_*)\}$, we found a sub-sub-subsequence which converges pointwise at $t_*$ to $0$.
So $\{h_N(t_*)\}$ converges to $0$, but $t_*$ was arbitrary so $\{h_N(t)\}$ converges to $0$ for every $t$.
Choosing $t = 0$ gives the result we are looking for.
