
> 
> *Is it true that a sequence of real numbers $\{a\_n\}$ converges to zero if and only if the sequences $\{\sin^2(nh)a\_n\}$ $(h \in \mathbb{R})$ all converge to zero?*
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In case the answer is affirmative (and please skip the obvious implication):


**1.** How far can we replace the condition "for every value of $h \in \mathbb{R}$" by a (much) weaker condition "for every value of $h \in A \subset \mathbb{R}$"? Will, for example, some countable or finite set $A$ suffice?


**2.** Is there anything special with the function $\sin^2(x)$ or can we replace it with (many) other functions? 


The topmost question arose from a strategy to prove the following little nut: 



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> *If $f : \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function such that the function $\Delta\_h f(x) := f(x + h) - f(x)\ $ is smooth for every value of $h$, then $f$ is itself smooth.* 
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This problem is in turn related to this previous [MO problem](https://mathoverflow.net/questions/167336/).


**Edit June 2, 2014.** A colleague just "reminded me" that the OP is a consequence of the 


**Cantor-Lebesgue Theorem.** *Let $A\_n(x) := c\_n e^{inx} + c\_{-n}e^{-inx}$. If 
$A\_n(x)$ tends to zero as $n$ tends to infinity for all $x$ in some set of positive measure, then $c\_n$ and $c\_{-n}$ tend to zero as $n$ tends to infinity.*


See this [paper](http://www.jstor.org/stable/2320583?__redirected) by R.L. Cooke for a nice presentation.



A question on null sequences
The following theorem about uniform distribution on the circle $\mathbb T^1$ is well known (and is due to Weyl): given an increasing sequence of positive integers $k\_n$, there is $z\in\mathbb T^1$ such that $z^{k\_n}$ is uniformly distributed in $\mathbb T^1$ (using the complex notation). In fact, this statement is true if you replace $\mathbb T^1$ by any compact connected second countable abelian group and almost every $z$ is good in the sense of Haar measure.


Now to your question. Assume that $a\_n$ does not converge to zero. Then it has a subsequence $a\_{k\_n}$ which is bounded away from zero. By the theorem mentioned above there is $h\in\mathbb R$ with $\sin^2(hk\_n)$ running densely through $[0,1]$. Consequently, $\sin^2(hk\_n)a\_{k\_n}$ has a subsequence bounded away from zero and so it does not converge to zero.


