Proof of $\limsup\sin nx=1, n\rightarrow \infty \forall x\in \mathbb{R}.$

Question

I have a problem: need to prove $$\limsup_{n\to\infty}\sin nx=1$$ for all x without set which measure is equal zero.

We need to come up with a sequence with limit that equals to one, but I don't know how to do this.

Answer 1

This isn't true. Let $x = m\pi$ where $m$ is an integer. The sequence $\{\sin(nx)\}_{n=0}^\infty$ has zero as every term.

Answer 2

Let $f(x) = \sin 2\pi x$ and $a_{n} = \left< \frac{nx}{2\pi} \right>$, where $\langle x \rangle = x - \lfloor x \rfloor$ denotes the fractional part of $x$.

Then whenever $x/2\pi \in \Bbb{R}\setminus\Bbb{Q}$, it is not hard to prove that there exists a subsequence $(a_{n_k})$ which converges to $1/4$. The simplest way of proving this, as far as I know, is to exploit the pigeonhole principle. (c.f. Dirichlet's approximation theorem.) Thus

$$\sin n_{k} x = f(a_{n_{k}}) \xrightarrow[k\to\infty]{} f(\tfrac{1}{4}) = 1. $$

Since $\sin n x \leq 1$ for all $n$, it follows that $\varlimsup_{n\to\infty} \sin nx = 1$ whenever $x/2\pi$ is irrational, and in particular, for a.e. $x$.

P.s. Here is a key result that is relevant to this answer:

Lemma. If $\alpha \in \Bbb{R}$ is irrational, then the set of limit points of $a_{n} = \langle \alpha n \rangle$ is $[0, 1]$.

Proof. Let $A$ denote the set of limit points of $(a_{n})$. Since $a_{n} \in [0, 1]$, we have $A \subset [0, 1]$. Now fix any positive integer $N$ and for each $1 \leq k \leq N$ let

$$ P_{k} = \bigcup_{n\in\Bbb{Z}} \left[ n+\frac{k-1}{N}, n+\frac{k}{N} \right) = \left\{ x \in \Bbb{R} : \tfrac{k-1}{N} \leq \langle x \rangle < \tfrac{k}{N} \right\}. $$

Then $P_{1}, \cdots, P_{N}$ is a partition of $\Bbb{R}$. Now by the pigeonhole principle, there exist $1 \leq i < j \leq N+1$ such that both $ i\alpha$ and $j\alpha$ are contained in the same partition $P_{k}$. Thus for some $m \in \Bbb{Z}$,

$$ 0 < \left| (j-i)\alpha - m \right| < \frac{1}{N}. $$

(Indeed, $m = [j\alpha] - [i\alpha]$ works.) Thus by a simple argument, we can prove the following observation: For any $1 \leq k \leq N$, there exists infinitely many positive integer $n$ such that $n \alpha \in P_{k}$.

Now let $x \in [0, 1]$. Then for any $N \geq 1$, $x \in P_{k}$ for some $N$ and thus there exists infinitely many $n$ satisfying $|x - a_{n}| < \frac{1}{N}$. This proves that $x \in A$, proving the claim.