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Let $\alpha,x\in\Bbb R\setminus\{0\}$ $^*$ with $\alpha>0$. Show that there infinitely many $n\in\Bbb N$ such that

$$\lvert\sin(\alpha^n\pi x)\rvert\ge\frac12$$

(note that the choice of $\frac12$ was arbitrary, any real number $0<b<1$ will work.)

This is the last step in a proof of the almost-nowhere differentiability of a class of everywhere continuous functions which are differentiable at a countable set of points, and I'm too frustrated to see whatever clever trick makes this obvious. Please help.

$^*$ I believe that the correct condition for $x$ is actually $\alpha^n x\notin\Bbb Z$. I don't know enough number theory to be sure. Whatever the case, "the set of points $x$ such that $\lim_{n\to\infty}\lvert\sin(\alpha^n\pi x)\rvert=0$" is always countable, and can be ignored for the sake of the proof - just assume $\alpha^nx\notin\Bbb Z$. However, I should probably know where $\lvert\sin(\alpha^n\pi x)\rvert$ converges either way, so I'll ask that as a separate question.

R. Burton
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  • Surely you want to add more conditions, because if you take $x$ an integer and $\alpha=2$ (or any even integer), your sinus term always has value $0$ – charmd Sep 25 '22 at 19:49
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    what is $\alpha$? is it a positive integer? – trytryandtry Sep 25 '22 at 20:05
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    @trytryandtry $\alpha$ is a positive real greater than $1$. – R. Burton Sep 25 '22 at 20:05
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    @charmd Quite right. I think the necessary condition is $(\exists n\in\Bbb N)(\sin(\alpha^n\pi x)\ne 0)$. This is a countable set, but I suppose it still counts. – R. Burton Sep 25 '22 at 20:12
  • Related to your question: Wikipedia mentions that the sequence $(\alpha^n)_{n \ge 1}$ is uniformly distributed mod $1$ (stronger that what you need to prove) for almost all $\alpha>1$. See this page: https://en.wikipedia.org/wiki/Equidistributed_sequence#Metric_theorems (a proof in German is linked) – charmd Sep 25 '22 at 20:45
  • This doesn't really answer the question you asked, but if your goal is to show that there are only countably many $\alpha$ such that $$\lim_{n\to\infty}\lvert\sin(\alpha^n\pi x)\rvert=0,$$ have you considered letting $z = e^{i \alpha^n\pi x}$ and showing that $z^4$ almost always doesn't converge to $1$? – Joseph Camacho Sep 25 '22 at 20:46
  • If your original problem is something else, you can also probably do some similar complex number shenanigans – Joseph Camacho Sep 25 '22 at 20:48
  • @JosephCamacho I did think of trying something like that, but I want to avoid using complex numbers if possible. I know it's silly, but some people seem to think that real analysis should exclusively make use of the real numbers. – R. Burton Sep 25 '22 at 20:52
  • The wikipedia page from above also mentions an explicit counterexample to your question, meaning that "for almost all $\alpha$" is probably the best you can achieve. Take $\alpha=\varphi$ the golden ratio and $x = \frac{2}{\sqrt{5}}$. Then $x \alpha^n = 2 F_n - \frac{2 \psi^n}{\sqrt{5}}$ with $\psi = \frac{1-\sqrt{5}}{2}$ and $F_n$ the n-th Fibonacci number. $\sin\big(\frac{2\varphi^n \pi }{\sqrt{5}}\big) = \sin\big(\frac{-2\psi^n \pi }{\sqrt{5}}\big)\to 0$ since $|\psi|<1$, yet the sinus value is never exactly zero. – charmd Sep 25 '22 at 20:59
  • You can still prove a "for almost all $\alpha$" result, with a few problematic cases for which you may or may not still have a "for almost all $x$" result (not sure about it; at least for $\alpha=\varphi$, the $\sin$ does not converge to zero for almost all $x$). Example of problematic $\alpha$ are Pisot numbers like $\varphi$ (https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number) – charmd Sep 25 '22 at 21:06
  • @R.Burton I will soon have more time to write an answer; could you indicate if it's imperative for you that the results hold for all $(\alpha, x)$ except countably many? Or whether "except for a negligible set" is enough? – charmd Sep 28 '22 at 09:23

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