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Suppose that $F : \mathbb{R} \to \mathbb{R}$ is a continuous and $k$-times continuously differentiable $2\pi$-periodic function, with $k\ge0$. Suppose that $f: \mathbb{R}\to\mathbb{R}$ is a $2\pi$-periodic locally summable function such that \begin{equation*} F(x) = f(x+1)-f(x-1). \end{equation*} holds for a.e. $x \in \mathbb{R}$.

What is the maximum amount of (essential) regularity for $f$ that we can guarantee?

I tried the following. Take the Fourier transform of both sides. We get \begin{equation*} \forall n \in \mathbb{Z}, \qquad \hat{F}(n)= 2i\sin(n) \hat{f}(n) \;. \end{equation*} This means that, if for each $n\in \mathbb{Z}$ we denote by $e_n$ the $n$-character, i.e., $e_n:\mathbb{R} \to \mathbb{C}, x\mapsto e^{inx}$, we have that \begin{equation*} f = \hat{f}(0)+ \sum_{n \in \mathbb{Z}, n \neq 0} \frac{\hat{F}(n)}{2i\sin(n) }e_n \end{equation*} where this series has to be interpreted in distributional sense. Now, it seems that it isn't exactly known how fast the sequence $(\sin(n))_{n \in \mathbb{\mathbb{N}}}$ clusters around zero, with the known gap between the upper and the lower bound that is quite high (see, e.g., the comments to this question). As a consequence, it seems that this approach would guarantee some regularity for $f$ (e.g., that after changing the values of $f$ on a suitable set of measure zero we obtain a continuous function) only given that $F$ is very regular, so that $|\hat{F}(n)|\to0, n\to\infty$ quite fast.

Any other ideas?

Bob
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  • I realized that also the vice-versa should hold (in a sense that has to be quantified), implying that any guarantee we can obtain about the regularity of $f$ gives us information about how fast $(\sin(n))_{n \in \mathbb{N}}$ clusters around $0$. Were this true, the problem seems far beyond any elementary intuition without any other assumptions about $F$ or $f$ – Bob Feb 19 '22 at 13:36

1 Answers1

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Comment.
Note, this $f$ is not $2\pi$-periodic.
Let $\varphi$ be an arbitrary locally summable function on $[0,1)$. For $x \in \mathbb R$, write $\lfloor x \rfloor$ for the integer part of $x$ and $\{x\} = x-\lfloor x \rfloor$ for the fractional part of $x$. Then $$ f(x) = (-1)^{\lfloor x \rfloor}\;\varphi\big(\{x\}\big) $$ satisfies $f(x-1)-f(x+1) = 0$ for all $x$, and $f$ agrees with $\varphi$ on $[0,1)$.

GEdgar
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