My Blog: Anh-Tourguide: An Other Limit Supremum Of Sin Function

In the topic we showed that for any irrational $\alpha$ the limit

$\mathop {\lim }\limits_{n \to \infty } \sin \left( {n\alpha \pi } \right)$

does not exist. In this topic, we consider the following limit

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right)$ .

To be precise, we prove that

$\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$

for almost every $x \in [0,2\pi]$ .

Solution. Let

$\displaystyle A = \left\{ {x \in \left( {0,2\pi } \right): \frac{x} {\pi } \notin \mathbb{Q}} \right\}$ .

Then $A$ is a measurable set of measure $2\pi$ . Moreover, for any $x \in A$ ,

$\displaystyle\mathop {\overline {\lim } }\limits_{n \to \infty } \sin \left( {nx} \right) = 1$ .

Indeed for any $x \in A$ , since

$\displaystyle\left\{ {k\frac{x} {\pi } - 2l: l \in \mathbb{Z}} \right\}$

is dense subgroup of $\mathbb R$ there are sequences $\{k_n\}$ and $\{l_n\}$ of $\mathbb Z$ such that

$\displaystyle \mathop {\lim }\limits_{n \to \infty } \left( {{k_n}\frac{x} {\pi } - {l_n}} \right) = \frac{1} {2}$ .

Since

$\displaystyle \frac{1} {2} \notin \left\{ {k\frac{x} {\pi } - 2l: k,l \in \mathbb{Z}} \right\}$

$\{k_n\}$ admits a subsequence $\{k'_n\}$ either increasing to $+\infty$ or decreasing to $-\infty$ . If $\mathop {\lim }\limits_{n \to \infty } {{k'}_n} = + \infty$ then

$\mathop {\lim }\limits_{n \to \infty } \sin \left( {{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \left( {...$

Otherwise $\mathop {\lim }\limits_{n \to \infty } \left( { - 3{{k'}_n}} \right) = + \infty$ and

$\mathop {\lim }\limits_{n \to \infty } \sin \left( { - 3{{k'}_n}x} \right) = \mathop {\lim }\limits_{n \to \infty } \sin \lef...$

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An Other Limit Supremum Of Sin Function

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