Is $\{ \sin(\sqrt {nx} \mid n\in \mathbb N\}$ is equicontinuous in $C[0,1]$?

Question

Is $\{\sin(\sqrt {nx}\mid n\in \mathbb N\}$ is equicontinuous in $C[0,1]$?

As near $0$, its derivative is unbounded, so function is not even uniformly continuous . So not equicontinuous .

Please tell me I am wrong or not

Any help will be appreciated.

Answer 1

Hint: Show that for any $\delta>0$, there exists an $n \in \Bbb N$ for which $f(x) = \sin(\sqrt{nx})$ satisfies $$ |f(\delta) - f(0)| \geq \frac 12. $$

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Regarding your comment on the question itself: we could use the sum-to-product identity to note that $$ \sin(\sqrt{nx}) - \sin(\sqrt{ny}) = 2\sin(\frac 12 (\sqrt{nx} - \sqrt{ny}))\cos(\frac 12 (\sqrt{nx} + \sqrt{ny})). $$ Alternatively, we can simply use the mean value theorem.