Ngô Quốc Anh

August 18, 2009

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