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I'd like to prove\begin{align} \lim_{n\rightarrow\infty}a_n=0 \iff \lim_{n\rightarrow\infty}a_n\sin(n t)=0 \quad \forall t\in[0,1]\end{align}

Since $\sin$ is bounded one of the implications is trivial. For the other one letting $t=\frac{\pi}{4 l}, l\in\mathbb{N}$ implies $a_{n_k}\rightarrow 0$ for all subsequences satisfying $(n_k)_{k\in\mathbb{N}}\subset\mathbb{N}\setminus 4 l \mathbb{N}$. Does this lead anywhere?

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up vote 2 down vote accepted

Suppose $\limsup_{n \to \infty} \left| a_n\right| > \epsilon > 0$, i.e. there are infinitely many $n$ with $|a_n| > \epsilon$. Let $T_n = \{t \in [0,1]: |a_k \sin(kt)| > \epsilon \text{ for some } k > n\}$. For any $k \ge 4$ there is $t \in [0,1]$ with $\sin(kt)=1$, so all $T_n$ are nonempty. Now if $|a_k \sin(kt)| > \epsilon$, so is $|a_k \sin(k(t+j\pi/k)|$ for any integer $j$. Any nonempty interval $(a,b) \subset [0,1]$ contains some $t+j\pi/k$ if $k$ is sufficiently large, i.e. $T_n$ is dense in $[0,1]$. Moreover, $T_n$ is easily seen to be an open set. The Baire Category Theorem then shows that $\bigcap_n T_n$ is nonempty. But if $t \in \bigcap_n T_n$, it means $|a_k \sin(kt)| > \epsilon$ for infinitely many $k$, so $\lim_{n \to \infty} a_n \sin(nt) \ne 0$ for this $t$.

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I had something fairly similar but I demanded $\lvert a_k \sin kt\lvert >\epsilon$ for all $k>n$, which turned out to be empty sets... nice solution! – Julian Jul 7 '12 at 7:08

Proving by contrapositive: Suppose that $\operatorname{lim}a_n \neq 0.$ Then, letting $t=\pi/4$ we find that the sequence $(a_n\operatorname{sin}(nt))$ cannot converge to $0$, since the $\operatorname{sin}(nt)$ terms forever cycle through $\pm\frac{1}{\sqrt{2}},\pm1,0.$ Thus, $\exists t\in [0,1]$ such that $\operatorname{lim}(a_n\operatorname{sin}(nt))\neq0,$ which is the negation of the condition on the right hand side.

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Let $a_n=1$ if $4|n$, and let $a_n=1/n$ otherwise. – Benji Jul 6 '12 at 19:57
Nice example! I tacitly assumed that $(a_n)$ is convergent. – Andrew Jul 6 '12 at 20:03

Suppose $(a_n)$ is such that for all $t\in I=[0,1],~\lim~a_n\sin(nt)$ exists and $=0$. You are in effect saying that the sequence of functions $f_n:I\rightarrow \mathbb R, t\mapsto a_n\sin(nt)$ converges simply (direct translation from french, maybe "pointwise" in english) to the $0$ funtion. What you want to show is that this implies uniform convergence of the sequence $(f_n)$ to the zero function (since for $n\geq2>\pi/2,~\|f_n\|_{\infty}=|a_n|$.)

Suppose $(a_n)$ doesn't tend to $0$. Then there is a positive real number $\epsilon>0$ and a strictly increasing sequence of positive integers $\varphi(n)\uparrow +\infty$ such that $|a_{\varphi(n)}|>\epsilon$ for all $n$. Without loss of generality, we may assume that for all $n,~\varphi(n+1)\geq 100\varphi(n)$ and $\varphi(0)>100\pi$ ($100$ isn't special, it is just large enough for our purpose.)

For $x\in \mathbb R$ and $r>0$, we call $[x-r,x+r]$ the segment centered at $x$ and of length $2\times r$. We construct a sequence of nested closed intervals $(I_n)$ centered around maxima of $|f_{\varphi(n)}|$ and of length $2\times\frac{\pi}{3\varphi(n)}$.

The first one is defined to be $I_0\subset [0,1]$ centered around one of the maxima of $|f_{\varphi(0)}|$ of the form $\frac{\pi/2+k\pi}{\varphi(0)}$ and of length $2\times\frac{\pi}{3\varphi(0)}$. Since $\varphi(0)$ is quite large, $f_{\varphi(0)}$ has plenty of time to complete many of full oscillations, so there is room enough to fit $I_0$ into $I$.

Suppose the $n^{th}$ such nested interval $I_n$ has been constructed. Because $\varphi(n+1)\geq 100\varphi(n)$, the $\varphi(n+1)^{th}$ function has time to do several full oscillations inside $I_n$, and we just pick one point in $x_{n+1}\in I_n$ that realizes a maxima for $|f_{\varphi(n+1)}|$, is of the form $\frac{\pi/2+k\pi}{\varphi(n+1)}$, and is close enough to the center of $I_n$ so that the segment centered at this maxima and or length $2\times\frac{\pi}{3\varphi(n+1)}$ is completely contained in $I_n$.

Saying that the maxima of $|f_{\varphi(n)}|$ we choose are of this particular form is not necessary, since by hopothesis the $a_{\varphi(n)}$ are non zero, and so the maxima of $|f_{\varphi(n)}|$ are those of $|\sin(~\cdot\times\varphi(n))|$. I include this extra information only to facilitate the verification of a calculation further down which relies upon the fact that $|\sin(x)|\geq 1/2$ on $[\pi/2-\pi/3,\pi/2+\pi/3]$.

We thus get a nested sequence of closed intervals whose length tend quickly to $0$, and by compactness there exists $t\in I$ with $$\bigcap_{n\in\mathbb N}I_n=\lbrace t\rbrace.$$ Our choice of $I_n$ (the fact that it is centered around a maxima of $|f_{\varphi(n)}|$ and of length $2\times\frac{\pi}{3\varphi(n)}$ ensures that for all $s\in I_n,~|\sin(\varphi(n)s)|\geq 1/2$. Since for all $n$ we have $t\in I_n$, we have for all $n$ that $|f_{\varphi(n)}(t)|\geq |a_n|/2>\epsilon/2$ which contradicts the assumption that $|f_n(t)|$ must tend to $0$.

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@Julian do you understand my answer? – Olivier Bégassat Jul 6 '12 at 22:34
I'm still trying to figure it out. The key point in you argument is basically that there exists a increasing $\phi: \mathbb{N}\rightarrow \mathbb{N}$ and a $t\in [0,1]$ s.t. $\lvert\sin t\phi(n)\lvert\geq 1/2$ for all $n$ which is basically also the point off Roberts argument. – Julian Jul 7 '12 at 7:35

If the $a_n \sin(nt)$ converge pointwise to 0, so do $|a_n| |\sin(nt)|$. It's a fact from measure theory (Egorov's Theorem) that if a sequence of measurable functions converge pointwise on $[0,1]$ to any function $f(t)$ then they converge almost uniformly to $f(t)$, meaning that for any $\epsilon > 0$ one can remove a set $A_{\epsilon}$ of measure $< \epsilon$ and the functions will converge uniformly to $f(t)$ on $[0,1] - A_{\epsilon}$.

In this case $f(t)$ is just zero, so off of $A_{\epsilon}$ the functions $|a_n| |\sin(nt)|$ converge uniformly to zero. But these functions are bigger than ${1 \over 2}|a_n|$ on most of $[0,1]$. Show if you choose $\epsilon$ small enough, you get that $|a_n| \rightarrow 0$.

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Would the following be correct? $0\leftarrow\sup_{t\in [0,1]\setminus A_\epsilon} \lvert a_n \sin nt \lvert\geq \sup_{t\in [0,1]} \lvert a_n \sin nt\lvert+\lvert a_n\lvert \epsilon=\lvert a_n\lvert(1+\epsilon)$ where the inequality comes from the fact that $a_n \sin nt$ ist Lipschitz continous with constant $\epsilon \lvert a_n\lvert$ and the distance of points in $A_\epsilon$ to $[0,1]\setminus A_\epsilon$ is surely smaller than $\epsilon$. – Julian Jul 7 '12 at 7:03
I meant Lipschitz constant $\lvert a_n \lvert$ of course... – Julian Jul 7 '12 at 9:04
The Lipschitz constant of that function is actually $n|a_n|$. Instead I'd just use that $|\sin(nt)| \geq {1 \over 2}$ on $2/3$ of its period. So for example on the interval $[0,1], |a_n\sin(nt)| \geq {1 \over 2}|a_n|$ on a set of measure at least ${2 / 3} - {1 /2\pi n}$. The ${1 / 2\pi n}$ here is the length of a period of $|a_n \sin(nt)|$, which I'm removing because it might get "cutoff" on the right side of $[0,1]$. – Zarrax Jul 7 '12 at 13:26

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