Solving $f(x)-f(x+\alpha)=g(x)$ 
Problem: Suppose $g:\mathbb{R} \to \mathbb{R}$ has period $1$ and is $\mathcal{C}^{\infty}$. Let $\alpha \in \mathbb{R}$. Consider the following equation:
\begin{equation}
f(x)-f(x+\alpha)=g(x)
\end{equation}

I am asked to:
$1$. Let $\alpha \in \mathbb{Q}$. Find necessary and sufficient condition on $g$ such that the equation has a solution $\mathcal{C}^{\infty}$ with period $1$.
$2$. Let $\alpha \in \mathbb{R} \smallsetminus \mathbb{Q}$ such that exist $\gamma >0$ and $\tau >0$ with $|\alpha - \frac{p}{q}|>\gamma q^{-2-\tau}$ for all $\frac{p}{q} \in \mathbb{Q}$, $q \geq 1$. Find necessary and sufficient condition on $g$ such that the equation has a solution $\mathcal{C}^{\infty}$ with period $1$
$3$. Prove that the set of $\alpha$ of the second point has to measure $0$ in $\mathbb{R}$.
Attempt:
If we define $$\tau_{\alpha}f(x)=f(x-\alpha) 
 \space \text{for}\space f:\mathbb{R} \to \mathbb{R}$$
then we would study $(\mathbb{I}-\tau_{-\alpha})f=g$ and a necessary condition for the first point is that $$\sum_{j=0}^kg(x+j\alpha)=0$$ where $k$ is such that $k\alpha \in \mathbb{Z}$.
 A: As @levap writes in the comment we use Fourier series. Define for $a>0$ the operator $\delta_a f(x):=f(ax)$. Thus we have that $\delta_{\frac{1}{2\pi}}f$ and $\delta_{\frac{1}{2\pi}}g$ are $2\pi$-periodic. Denoting $c_n(h)=\frac{1}{2\pi} \int_0^{2\pi}h(x)e^{-inx}dx$ the Fourier coefficient we have that from $(\mathbb{I}-\tau_{-\alpha})f=g$ then:
\begin{equation}
\delta_{\frac{1}{2\pi}} \circ (\mathbb{I}-\tau_{-\alpha})f= \delta_{\frac{1}{2\pi}}g
\end{equation}
and thus $f(\frac{x}{2\pi})-f(\frac{x}{2\pi}+\alpha)=g(\frac{x}{2\pi})$. Taking the Fourier coefficients:
$$c_n(\delta_{\frac{1}{2\pi}}f)-c_n(\delta_{\frac{1}{2\pi}} \tau_{-\alpha} f)=c_n(\delta_{\frac{1}{2\pi}}g)$$
where $c_n(\delta_{\frac{1}{2\pi}} \tau_{-\alpha} f)=e^{2\pi i n \alpha} c_n(\delta_{\frac{1}{2\pi}}f)$.
We then get:
$$(1-e^{2\pi i n \alpha})c_n(\delta_{\frac{1}{2\pi}}f)=c_n(\delta_{\frac{1}{2\pi}}g)$$
Let's prove the three points separately:

*

*$\alpha \in \mathbb{Q}$.
In this case for $n$ such that $n\alpha \in \mathbb{N}$ then $c_n(\delta_{\frac{1}{2\pi}}g)=0$ and we define $c_n(\delta_{\frac{1}{2\pi}}f)=0$. Otherwise we define:
$$c_n(\delta_{\frac{1}{2\pi}}f)=\frac{c_n(\delta_{\frac{1}{2\pi}}g)}{(1-e^{2\pi i n \alpha})}$$
We will discuss the convergence later.


*Thanks to the property of $\alpha$ we have that $|1-e^{2\pi i q \alpha}| \geq \sqrt{2-2\cos (\frac{2\pi \gamma}{q^{1+\tau}})}\geq \frac{\sqrt{2}\pi \gamma}{q^{1+\tau}}(1+o(1))$ for $q \to +\infty$ and then the following inequality holds:
$$|c_n(\delta_{\frac{1}{2\pi}}f)| \leq \frac{n^{1+\tau}}{\sqrt{2}\pi \gamma}(1+o(1)) |c_n(\delta_{\frac{1}{2\pi}}g)|$$
and this allows us to conclude that $f$ is well defined and $f \in \mathcal{C}^{\infty}$. The same estimate can be used to conclude the previous point. In fact if $q \nmid n$ then $n\alpha = \frac{k}{h}$ where $h \geq 2$.


*Let $A_n=\{ x \in \mathbb{R} |x| \leq n \}$. Then:
$$B_n:=\bigcap\limits_{\tau \in \mathbb{N},\gamma \in \{\frac{1}{m}:m \in \mathbb{N} \}} (\bigcup\limits_{p \in \mathbb{N},q \in \mathbb{N}} \{x \in A_n: |x-\frac{p}{q}| \leq \gamma q^{-2-\tau} \})$$
has measure $0$. In fact if $B_{n,\tau,\gamma}=\bigcup\limits_{p \in \mathbb{N},q \in \mathbb{N}} \{x \in A_n: |x-\frac{p}{q}| \leq \gamma q^{-2-\tau} \}$ we have $|B_{n,\tau,\gamma}| \leq 2\gamma n q^{-2} \sum\limits_{q \in \mathbb{N}} q^{-\tau}$ which proves $|B_n|=0$ for all $n \in \mathbb{N}$.
