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: $$$$f(x)-f(x+\alpha)=g(x)$$$$

$$1$$. 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 $$k\alpha \in \mathbb{Z}$$.

• Part (a) really calls for using Fourier series. In terms of Fourier coefficients, your equation is equivalent to $\left( 1 - e^{2\pi i n \alpha} \right) \hat{f}(n) = \hat{g}(n)$. So if $\alpha = \frac{p}{q}$ where $p,q$ have no common factors then a necessary condition on $g$ is that $\hat{g}(qk) = 0$ for all $k \in \mathbb{Z}$. This is also a sufficient condition as you define $g$ as $g(x) = \sum_{z \in \mathbb{Z}, z \notin q\mathbb{Z}} \frac{\hat{f}(n)}{1 - e^{2 \pi i n \alpha}} e^{2\pi i n x}$ and verify that this sum converges uniformly together with all derivatives if $f$ is smooth. Aug 16, 2021 at 20:11
• @levap You’re right. Maybe in the last equation you have exchanged $f$ with $g$. We want to find $f$ given $g$, not vice versa. Aug 17, 2021 at 7:29

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:

$$$$\delta_{\frac{1}{2\pi}} \circ (\mathbb{I}-\tau_{-\alpha})f= \delta_{\frac{1}{2\pi}}g$$$$

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:

1. $$\alpha \in \mathbb{Q}$$. 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.

2. 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 $$h \geq 2$$.

3. 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 $$n \in \mathbb{N}$$.