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If $x$ is irrational number, what is $\limsup\limits_{n\to\infty} \cos (2\pi n x)$?

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  • $\begingroup$ I know the answer is 1. but I do not know how I can prove it. $\endgroup$ – Mary Oct 15 '15 at 1:22
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Hint: If $x$ is irrational, then the set of fractional parts of $nx$ is dense in the interval $[0,1]$.

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  • $\begingroup$ Can you clarify it more. How I am going to use these fractional parts? $\endgroup$ – Mary Oct 15 '15 at 1:49
  • $\begingroup$ That means that for any positive $\epsilon$, however small, there will be an $n$, and an integer $m$, such that $|2\pi nx =2\pi m|\lt \epsilon$. $\endgroup$ – André Nicolas Oct 15 '15 at 1:59
  • $\begingroup$ $|2πnx - 2πm|<ϵ$ $\endgroup$ – Mary Oct 15 '15 at 2:11
  • $\begingroup$ Can I apply this argument to rational numbers? $\endgroup$ – Mary Oct 15 '15 at 2:16
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    $\begingroup$ Yes, sorry, the = key is too close to the - key, and because of an odd eyesight difficulty I sometimes do not see the mistake while proofreading. $\endgroup$ – André Nicolas Oct 15 '15 at 2:18
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As you stated in a comment, the answer should be $1$. Let's think a little about it: We know that $\cos(y)$ has period $2\pi$, so when we consider $\cos(2\pi y)$ instead, we end up with a function with period $1$ (it is always good to normalize things).

We are then considering the sequence $x,2x,3x,\ldots,nx,\ldots$, and evaluating $\cos(2\pi y)$ at these points. Since the period is $1$, the only thing that matters is the fractional part of $nx$. Since $x$ is irrational, we should be able to make the fractional part of $nx$ arbitrarily close to whatever we want in $[0,1]$, for $n$ very large.

In order to formalize this, I'm going to use one of my favorite questions: Characterizing Dense Subgroups of the Reals.

Lemma: The set $B=\left\{nx+k:n,k\in\mathbb{Z}\right\}$ is dense in $\mathbb{R}$.

Proof: See https://math.stackexchange.com/a/889316/58818 and the comment.

Now for the solution of the exercise: The inequality $\limsup \cos(2\pi n x)\leq 1$ is trivial. For the converse, we already know that $B$ as above is dense in $\mathbb{R}$, so $2\pi B=\left\{2\pi y:y\in B\right\}$ is also dense in $\mathbb{R}$ and by continuity, $\cos(2\pi B)$ is dense in $\cos(\mathbb{R})=[0,1]$. But $$\cos(2\pi B)=\left\{\cos(2\pi nx+2\pi k):n,k\in\mathbb{Z}\right\}=\left\{\cos(2\pi nx):n\in\mathbb{Z}\right\}=\left\{\cos(2\pi n x):n\in\mathbb{N}\right\},$$ where the last equality follows from the fact that $\cos(y)=\cos(-y)$.

Therefore, we can find a sequence $(n_k)_k$ such that $\cos(2\pi x n_k)$ is strictly increasing and goes to $1$. Then the sequence $(n_k)$ is injective, and going to a subsequence if necessary, we may assume that $(n_k)$ is strictly increasing, and it follows that $1=\lim \cos(2\pi x n_k)\leq\limsup\cos(2\pi n x)$.

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