Prove that $\displaystyle \lim_{n \to \infty}\mathbb E(e^{iuX_n})=\lim_{n \to\infty}\mathbb E(e^{iuX})$ where $X$ is the mean square limit of $X_n$

Question

Suppose $\{X_n, n \in \mathbb{N}\}$ is a sequence of random variables with finite second moment, and $X_n$ convergences in mean square to $X$, i.e., $\displaystyle \lim_{n \rightarrow \infty} \mathbb E(|X_n - X|^2)=0$, then prove that $$\displaystyle \lim_{n \rightarrow \infty}\mathbb E\left(e^{iuX_n}\right)=\lim_{n \rightarrow \infty}\mathbb E\left(e^{iuX}\right)$$

Here is how my textbook goes:

\begin{align}\left|\mathbb E\left(e^{iuX_n}\right)-\mathbb E\left(e^{iuX}\right)\right|&=\left|\mathbb E\left(e^{iuX_n}-e^{iuX}\right)\right|=\left|\mathbb E\left(e^{iuX}\left(1-e^{iu(X_n-X)}\right)\right)\right|\\[0.1cm]& \le \mathbb E\left|1-e^{iu(X_n-X)}\right|\\[0.1cm]&\le \mathbb E\left|1-\cos{\left(u(X_n-X)\right)}-i\sin{\left(u(X_n-X)\right)}\right|\\[0.1cm]&\le \mathbb E\left|u(X_n-X)\right|\end{align}

But I couldn't understand how the last $\le$ holds. Could you please shed some light on it?

Answer 1

Just expand: $E|1-\cos(A)-i\sin(A)|=E\sqrt{(1-\cos(A))^2+\sin^2(A)}=E\sqrt{2(1-\cos(A))}=2E|\sin(A/2)|$

Here $A=u(X_n-X)$ and for all $x$, $|\sin(x)|\leq |x|$.

Hence $2E|\sin(\frac{u(X_n-X)}{2})|\leq E|u(X_n-X)|$