Proving that a function is continuously differentiable using decay of fourier series Let $\mathbb{S}^1=\mathbb{R}\backslash\mathbb{Z}$.
Let $\alpha$ be an irrational number, and consider the equation
$$g(x+\alpha)-g(x)=p(x), x\in \mathbb{S}^1$$
for an unknown function $g$, with a given function $p\in C^\infty(\mathbb{S}^1)$, such that $$\int_{\mathbb{S}^1} p(x)dx=0$$
Give a condition on $\alpha$ that guarantees $g\in C^1(\mathbb{S}^1)$ for any such function $p$.
$\textbf{Thoughts}$
Using Fourier series I was able to deduce that $$\hat{g}(n)=\frac{\hat{p}(n)}{e^{in\alpha}-1}, n \ne 0$$
I was thinking to prove that $g$ is continuously differentiable it might be enough to prove that $\{n\hat{g}(n)\}$ is absolutely summable. We also have arbitrary decay for $\hat{p}(n)$ in that sense for any $k>0$ $\hat{p}(n)\leq \frac{C_k}{n^k}$
Although I am a bit concerned about the choice of $\alpha$ since for irrational $\alpha$, $\{n\alpha\}$ is equidistributed so we can have a subsequence converging to 1. Perhaps there's a way out. Any help is appreciated.

Yo need to use the fact that $|e^{2 \pi i \alpha n} - 1|$ is comparable to the infimum over all natural numbers $m$ of 
$$
  |m - \alpha n | = n \, \Big| \frac{m}{n} - \alpha \Big|
$$
That quantity is measuring how googd can your number by approximated by rational numbers. You want to have some lower bound of the form:
$$
  \Big| \frac{m}{n} - \alpha \Big| > \frac{C}{n^d}.
$$
That holds for algebraic numbers of order $d$ is I recall correctly.
See the wikipedia for Lioville theorem. 
