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Let $a_n$ be a sequence of real numbers such that $\exp(it\cdot a_n)$ converges to 1 for all real t. Show that $a_n$ converges to 0.

One can show this by letting $X_n$ be degenerate random variables which take value $a_n$ with prob 1 and then using results about characteristic functions and convergence. But this seems like overkill and it doesn't provide any intuition for what's going on.

It seems like there should be a simpler proof based on basic properties of continuous functions and convergent sequences. Any ideas ?

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Here is my intuition. For any particular value of t this says that a_n gets closer and closer to some element of the Z-lattice spanned by 2pi/t. By taking sufficiently many Q-independent values of t we should be able to force a_n to get farther and farther away from the origin (or to converge to 0) in order to get closer to many different Z-lattices simultaneously But without any uniformity assumptions I am having trouble making this rigorous. – Qiaochu Yuan Nov 6 '10 at 10:45
No pre-set countable set of $t$'s will do - for a finite number of $t_j$'s and any $1/n$ you can find $a_n$ such that $exp(i t_j a_n) < 1/n$, and then use a diagonal argument. Note that $a_n=n!$ works for all rationals. You could try to use $t$'s dependent on $a$'s.. – Max Nov 6 '10 at 11:18
up vote 5 down vote accepted

Step 1 The sequence $(a_n)$ cannot be bounded, if it doesn't converge to 0. Suppose the contrary, then up to a subsequence $(a_{\sigma(n)})$ converges to some value $A$. Let $t = \pi/A$. Then $\exp i t a_\sigma(n) \to -1$ contradicting convergence assumption. (Another way to see this step is to note that if $(a_n)$ were bounded, then the sequence $(\exp i t a_n)$ is equicontinuous, and on any compact interval the convergence must be uniform.)

Step 2 The sequence cannot be unbounded. Assume the contrary, then up to a subsequence we can choose $a_{\sigma(n+1)} > 3 a_{\sigma(n)}$. Without loss of generality we can assume the subsequence is positive. Let $t_n = 2\pi a_{\sigma(n)}^{-1}$. Define a sequence of closed intervals recursively. Let $I_1 = [t_1/4, 3t_1/4]$. By construction $\exists m_2 \in \mathbb{Z}$ such that $m_2t_2, (m_2+1)t_2 \in I_1$. Let $I_2 = [ (m_2 + 1/4) t_2, (m_2 + 3/4) t_2]$.

At each step $I_n$ has length $t_n / 2$. Hence there exists $m_{n+1}$ such that $m_{n+1}t_{n+1}, (m_{n+1}+1)t_{n+1} \in I_n$. And so we construct $I_{n+1}$ analogously.

In particular, we have $I_n \supset I_{n+1} \supset I_{n+2} \ldots$. Furthermore, by construction, $\Re e^{it a_{\sigma(n)}} \leq 0$ for $t \in I_n$.

Then for $t\in \cap_1^\infty I_n$, we have that $\exp i t a_j$ is bounded away from 1 on a subsequence, and hence contradicts the assumption of convergence.

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+1 Nice elementary solution. – Nuno Nov 6 '10 at 19:14
Thanks, Willie. I came up with a fairly similar solution myself. – Cosmonut Nov 8 '10 at 3:18
The difficult part was proving that $a_n$ must be bounded. I noticed that was the tough part in your proof as well. – Cosmonut Nov 8 '10 at 3:19

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