# A question on null sequences

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?

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:

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.

This problem is in turn related to this previous MO problem.

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 by R.L. Cooke for a nice presentation.

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.
• By the way, what is special about $\sin$ is that it corresponds to the projection of $\mathbb T^1$ onto $[-1,1]$ as a part of the imaginary axis and that it takes dense subsets of $\mathbb T^1$ onto dense subsets of $[-1,1]$. This should give an idea on how to search for other functions. May 30 '14 at 13:19
• I checked this in the book Uniform distribution of sequences'' by Kuipers and Niederreiter. The circle case is due to Weyl (Theorem 4.1, page 32). The group case is due to Hartman and Ryll-Nardzevski (at the bottom of page 279, the same book). Can you help me out what the Ryll-Nardzevski fixed point theorem says? May 30 '14 at 13:51