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

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    $\begingroup$ 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. $\endgroup$
    – levap
    Aug 16, 2021 at 20:11
  • $\begingroup$ @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. $\endgroup$ Aug 17, 2021 at 7:29

1 Answer 1

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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:

  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 $n \in \mathbb{N}$.

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