TITLE: If $e^{itx_n}$ converges for every $t\in\mathbb R$, then does $x_n$ converge?
QUESTION [11 upvotes]: Question : Is the following true for any real number sequence $x_n$ ?
"If $e^{itx_n}$ converges for every $t\in\mathbb R$, then $x_n$ converges."
Note that $i^2=-1$.

Motivation : We know that the converse of the above proposition is true. However, I'm facing difficulty for the above proposition. Can anyone help?

REPLY [3 votes]: Let $G$ be a locally compact topological group. The dual $\hat G$ of $G$ is the group of continuous homomorphisms $G \to S^1$.
Given a real number $s$, we may define a continuous homomorphism $\varphi_s :\mathbf R \to S^1$ by $t \mapsto e^{ist}$. The map $s \mapsto \varphi_s$ is a map $\mathbf R \to \hat{\mathbf R}$. This map is an isomorphism of locally compact abelian groups.
Therefore, the sequence $\{x_n\}$ converges in $\mathbf R$ if and only if the sequence $\{\varphi_{x_n}\}$ converges in $\hat{\mathbf R}$ if and only if the sequence $e^{itx_n}$ converges for every $t \in \mathbf R$.