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}.


\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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