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Let $n_{k+1}>n_k \in \mathbb{N}$, let $\alpha \in (0,+\infty)\setminus 2\pi\mathbb{Q}$. Prove that $\sin(\alpha n_k)$ doesn't converge.

Context: showing there exists a sequence $f_n \in C[0,1]$ with no uniformly convergent subsequence.

Proof: assume $\exists \lim_k \sin(\alpha n_k)$. Then $\exists \lim_k \sin(\alpha n_k+b)=\lim_k\sin(\alpha n_k)=:l$.

Now, $\sin(\alpha n_k+b)=\sin(\alpha n_k)\cos(b) + \cos(\alpha n_k)\sin(b)$.

Therefore $(\sin(\alpha n_k+b)-\sin(\alpha n_k)\cos(b))^2= \cos(\alpha n_k)^2\sin(b)=(1-\sin(\alpha n_k)^2)\sin(b)^2$.

Taking the limit yields $l^2(1-\cos(b))^2=(1-l^2)\sin(b)^2$.

If $l \neq \pm 1$ then $\displaystyle \frac{l^2}{1-l^2}=\frac{\sin(b)^2}{(1-\cos(b))^2}$, which is not a constant. Otherwise take the reciprocal.

Question: why "$\exists \lim_k \sin(\alpha n_k)\implies\exists \lim_k \sin(\alpha n_k+b)=\lim_k\sin(\alpha n_k)$"?

The above proof is based on this post

Lilla
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    Not true, try $\alpha=2\pi,e$ and $n_k=k!$. Concerning your original problem: consider $f_n(x)=x^n$. No subsequence can converge uniformly, because the limit function is not continuous. –  Nov 13 '20 at 20:24
  • @Professor Vector. Why don't you consider $n_k=k$ instead of $n_k=k!$ ? – pabodu Nov 14 '20 at 06:30
  • @pabodu if Prof. Vector will said $n_k=k$ then the sequence ${\sin(2\pi e k)}_{k\in \mathbb{N}}$ will be dense in $[-1,1]$ so the sequence cannot converge to a number. However with $n_k=k!$ and the definition of $e$ its easy to check that the fractional part of $e k!$ converge to zero, so $\sin(2\pi e k!)\to 0$ – Masacroso Nov 14 '20 at 12:47

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Let $\beta :=\alpha /2\pi$, then from the conditions of the exercise we knows that $\beta $ is irrational (and positive). Now, by the Kronecker approximation theorem we knows that the set $\{\{\beta n\}: n\in \mathbb{N}\}$ is dense in $[0,1]$, where $\{\beta n\}:=\beta n-\lfloor \beta n \rfloor$. Therefore $\{2\pi\{\beta n\}:n\in \mathbb{N}\}$ is dense in the set $[0,2\pi]$, so for every $x\in[0,2\pi]$ there exists an strictly increasing sequence $n_1,n_2,\ldots $ of natural numbers such that $\lim_{k\to\infty}2\pi\{\beta n_k\}=x$.

By last note that $$ \sin (\alpha n_k)=\sin (2\pi\beta n_k)=\sin (2\pi(\lfloor \beta n_k \rfloor+\{\beta n_k\}))=\sin (2\pi\{\beta n_k\}) $$

So the statement that you want to prove is false in general.

Masacroso
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  • Thus, the problem can be reformulated as follows: Prove that there exists an increasing sequence ${n_k}$ of natural numbers such that $\sin(\alpha n_k)$ converges. – pabodu Nov 14 '20 at 06:36
  • @pabodu yes, the reformulation that you proposed will be a valid exercise – Masacroso Nov 14 '20 at 12:42