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I know that the sequence $A_n = \displaystyle\sum_{k=1}^n \sin k$ is bounded. I suspect the following general proposition is true, and it could then immediately imply $A_n$ is bounded, as an alternative to the methods used to prove $A_n$ being bounded, as for example, here and here.

Proposition: Let $f:\mathbb{R}^+ \to\mathbb{R}\ $ be a continuous (and therefore bounded on $[0,1]$) function with period $1$ and $\displaystyle \int_{0}^{1} f(x) dx = 0.\ $ Let $\alpha$ be a positive irrational number. Then $A_n:=\displaystyle\sum_{k=1}^{n} f(\alpha k)\ $ is bounded.

Maybe this can be proved using Weyl's criterion, although I think we need a property of the sequence $\ \left(\alpha n\right)_{n\in\mathbb{N}}\ $ stronger than equidistribution, like low-discrepancy, to prove the theorem, but I am not sure.

Adam Rubinson
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  • I think you will be going around in circle here, because the Birkhoff means $\frac1n\sum_{j=1}^n f(j\alpha+x)$ converges with rate $O(1/n)$ (a.e.-$x$) is precisely the condition there exists $g\in L^\infty$ such that $f(x)=g(x+\alpha)-g(x)$. – user10354138 Oct 03 '23 at 11:28
  • What's "the Birkhoff"? – Adam Rubinson Oct 03 '23 at 11:34
  • It is not "the Birkhoff" but "the Birkhoff mean", i.e., an operator that take an ergodic transformation $T$ and some $L^1$ function $f$, and returns the $L^1$ function that is averaging the $f$ over the first $n$ iterates $f, f\circ T,\dots, f\circ T^{n-1}$. – user10354138 Oct 03 '23 at 11:42
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    @user10354138 But it is intriguing how completely differently the first comment (almost) parses if one assumes that "the Birkhoff" is a thing :) – Hagen von Eitzen Oct 04 '23 at 15:32
  • @HagenvonEitzen Funnily enough on my browser window the line wraps after the word Birkhoff :) Maybe I should have added a link to Birkhoff's ergodicity theorem. – user10354138 Oct 04 '23 at 16:28

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