If $f$ is a continuous periodic function with irrational period and if $\sum_n\frac{|f(n)|}{n}<\infty$, then $f$ is identically zero. Please show that if $f:\mathbb{R}\to\mathbb{R}$ is a continuous periodic function with irrational period and if $\sum_n\frac{|f(n)|}{n}<\infty$, then $f$ is identically zero.
(For example, using this we know $\sum_{n\ge1}\frac{|\sin n|}{n}$ diverges.)
The book mentions the "so-called equidistribution criterion"; I'm not sure if it's referring to this: http://en.wikipedia.org/wiki/Equidistribution_theorem ) 
 A: Yes, the reference is correct.
Assume $f$ is nonzero and $p$ its period. Then there exist $\epsilon>0$ and  $0<a<b<p$ with $|f(x)|>\epsilon$ for all $x\in [a,b]$. For $m$ sufficiently big, the proportion of numbers $n\in \{m+1, m+2, \ldots, 2m\}$ such that $n\bmod p\in [a,b]$ is approximately $\frac{b-a}p$, hence 
$$ \sum_{n=m+1}^{2m}\frac{|f(n)|}{n}\gtrsim\frac{b-a}pm\cdot \frac\epsilon{2m}=\frac{(b-a)\epsilon}{2p}$$
and hence the full series diverges. To make this precise, one must be more specific, what "approximately" means in the above.
A: Suppose that $f$ has period $\xi$ and that $\sum_n \frac{|f(n)|}{n} < +\infty$. By Kronecker's lemma, it follows that $\lim_{n \to \infty} \frac{1}{n} \sum_n |f(n)| = 0$. By the ergodic theorem (in this case equivalent to Weyl's equidistribution theorem), we also have that $\lim_{n \to \infty} \frac{1}{n} \sum_n |f(n)| = \frac{1}{\xi}\int_0^{\xi} |f(x)|dx$. Thus we conclude that $\int_0^{\xi} |f(x)|dx=0$, and thus $f=0$ almost everywhere. By continuity $f=0$ everywhere.
A: If there is a point $x_0$ such that $|f(x_0)|$ > 0 then there exists $
\epsilon $ such that if $|x-x_0|<\epsilon$ then $f(x) > f(x_0)/2$ by continuity.  Now look at all translates of the interval $[x_0-\epsilon,x_0+\epsilon]$ by the period $\alpha$ of the function.  Now the equidistribution theorem guarantees these intervals cover an integer value at least once every $\alpha / \epsilon = C$ translates.  Taking these we get a sequence of integers $n_0, n_1,...$ such that $n_k < Ck$ and $|f(n_i)|> |f(x_0)|/2$. Now combining these $|f(n_k)|/n_k > |f(x_0)|/(2Ck) = C'k^{-1}$.  So comparing the sum with the $C'$ times harmonic series, we see it must diverge.
