If $e^{iux_n}$ converges for $u\in\mathbb{R}$, $x_n$ converge to a finite limit

Question

Let $x_n$ be a sequence of real numbers such that $e^{iux_n}$ converges as $n\to\infty$ for almost all $u\in\mathbb{R}$. Then $x_n$ converges to a finite limit.

I know a proof by probability theory (Protter's "Stochastic Integration and Differential Equations"), but I want to show this theorem by analysis way. I don't know what to do.

Thank you for your cooperation.

Answer 1

Note that $e^{iu(x_n-x_m)}=\frac {e^{iux_n}}{e^{iux_m}} \to 1$ as $n, m \to \infty$ for almost all $u$. By DCT we get $\frac 1 2 \int_{-1}^{1} e^{iu(x_n-x_m)} du \to 1$. Take the real part and evaluate the integral. You will get $\frac {\sin (x_n-x_m)} {x_n-x_m} \to 1$. I will leave it to you to use some properties of $\sin x$ to show that $x_n-x_m\to 0$ as $n,m \to \infty$. Thus $(x_n)$ is Cauchy, hence convergent.

Some details for the last part: Let $\epsilon >0$ and consider the function $1-\frac {\sin x } x$ on $[\epsilon, \infty)$. This function is positive (why?) and continuous. It tends to $1$ as $ x \to \infty$. These facts imply that it has a positive minimum. Call this minimum $\delta$. It follows that $ 1-\frac {\sin x} x \geq \delta$ whenever $x \geq \epsilon$. Hence $ 1-\frac {\sin x } x < \delta$ implies $x < \epsilon$. Take $x=|x_n-x_m|$ in this.

Answer 2

By Euler's formula, $$ e^{iux_n} = \cos ux_n + i\sin ux_n. $$ Since $e^{iux_n}$ converges, both its real and imaginary parts must converge. Can you proceed from there?