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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.

Sanchit
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  • There are absolutely summable Fourier series that give non $C^1$ functions. Think of $$ f(\theta) = \sum_{n = 1}^\infty \frac{\sin(3^n \theta)}{2^n},$$ so that strategy wouldn't work. Whatever solution to the problem you choose you will be forced to use the irrationality of $\alpha$ since the proposition is false otherwise. – Adrián González Pérez Sep 09 '18 at 12:58
  • Huh that's interesting. I was thinking that if I prove enough decay for the coefficients, that would be enough to prove that the sequence ${n\hat{g}(n)}$ and hence ${\hat{g}(n)}$ would be absolutely summable giving us a $C^1$ function a.e. I was thinking along the lines of https://math.stackexchange.com/questions/351866/decay-of-fourier-coefficients-and-smoothness?rq=1 – Sanchit Sep 09 '18 at 13:45
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    My bad. I read read the $n$ term. Yes, if you prove that $\hat{g}(n) n$ is absolutely summable, then $g$ has one continuous derivative. – Adrián González Pérez Sep 09 '18 at 13:49
  • Do you any hints on how to take care of the fact that the $e^{in\alpha}-1$ can be arbitrarily close to $0$ (since ${n\alpha}$ are equidistributed)? I was hoping that the arbitrary decay for $\hat{p}(n)$ can be useful somehow. Thanks though – Sanchit Sep 09 '18 at 13:59
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    Yes, your intuition is correct. The branch of matematics that studies how well/bad $\alpha n$ approximates an integer $m$, for irrational $n$ is called "Diophantine appraximation" – Adrián González Pérez Sep 09 '18 at 14:03
  • Great! Thanks for the answer. I'll try to understand it and accept soon. – Sanchit Sep 09 '18 at 14:05

1 Answers1

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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.

Adrián González Pérez
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  • I think it is all clear except for one thing. How do you see $|e^{2\pi i\alpha n}-1|$ is comparable to the infimum you mentioned. I also find it interesting that this question showed up in a qualifying exam :) – Sanchit Sep 09 '18 at 14:24
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    The quantity $|e^ {2 \pi i \alpha n} - 1|$ is the Euclidean distance in the complex plane between $e^ {2 \pi i \alpha n}$ and $1$. When it is small enough it can be compared to the "arc" in the unit circle between the angles $2 \pi i \alpha n$ and $0$. But the length in the circle is the same as the length in the line modulo the integers. – Adrián González Pérez Sep 09 '18 at 14:55