If $e^{itx_n}$ converges for every $t\in\mathbb R$, then does $x_n$ converge?

Solution 1:

Yes, $x_n$ converges.

Denote $$f(t):=\lim_{n\to\infty}e^{itx_n},\quad t\in\Bbb R.$$ Clearly $f$ is Lebesgue measurable on $\Bbb R$, so by dominated convergence theorem, $$\lim_{n\to\infty}\int_{t_1}^{t_2}e^{itx_n}dt=\int_{t_1}^{t_2}f(t)dt,\quad \forall~ t_1<t_2.$$ Note that $|f(t)|=1$, $\forall t\in\Bbb R$, so there exist $a<b$, such that $$\int_a^bf(t)dt\ne 0\Longrightarrow \int_a^be^{itx_n}dt\ne 0 \quad\text{for large }n.$$ Then by fundamental theorem of calculus,
$$x_n=\frac{i(e^{iax_n}-e^{ibx_n})}{\int_a^be^{itx_n}dt} \quad\text{for large }n.$$ Letting $n\to\infty$, we have $$\lim_{n\to\infty}x_n=\frac{i(f(a)-f(b))}{\int_a^bf(t)dt}.$$

Solution 2:

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 $t \in \mathbf R$.