Prove that as $u \to 0$, $\frac{2-\varphi(u)-\varphi(-u)}{u^2} \to E(X^2)$ X, a real random variable, $\varphi(u)$ is defined as a characteristic function for X, $\varphi$: $R\to C$, $\varphi(u)=E(e^{iuX})$ for all u$\in R$.
Our professor gave us a hint that 1-cos $\theta$ =2$sin^2(\theta/2)$.
I really didn't get this problem or this hint. It seems like there is some gap in my knowledge, because I couldn't see the connection of this characteristic function containing complex number and the trigometric function.
 A: From the identity $\cos \theta = \frac{1}{2}(e^{i\theta} + e^{-i\theta})$, we have
\begin{align*}
\frac{2-\varphi(u)-\varphi(-u)}{u^2}
&= \Bbb{E}\bigg[ \frac{2}{u^2}\left(1 - \frac{e^{iuX} + e^{-iuX}}{2} \right) \bigg]\\
&= \Bbb{E}\bigg[ \frac{2(1 - \cos (uX))}{u^2} \bigg]\\
&= \Bbb{E}\bigg[ \frac{4\sin^2(uX/2)}{u^2} \bigg].
\end{align*}
In order to utilize this computation, we observe from the inequality $\left| \sin\theta \right| \leq \left|\theta\right|$ that
$$ 0 \leq \frac{4\sin^2(uX/2)}{u^2} \leq X^2
\quad \text{and} \quad
\lim_{u\to0} \frac{4\sin^2(uX/2)}{u^2} = X^2. $$
Thus if $\Bbb{E}[X^2] < \infty$, then by the dominated convergence theorem we have
$$ \lim_{u \to 0} \frac{2-\varphi(u)-\varphi(-u)}{u^2}
= \Bbb{E}\bigg[ \lim_{u \to 0} \frac{4\sin^2(uX/2)}{u^2} \bigg]
= \Bbb{E}[X^2] $$
as desired. When $\Bbb{E}[X^2] = \infty$, then by the Fatou's lemma
$$ \liminf_{u \to 0} \frac{2-\varphi(u)-\varphi(-u)}{u^2}
\geq \Bbb{E}\bigg[ \liminf_{u \to 0}  \frac{4\sin^2(uX/2)}{u^2} \bigg]
= \Bbb{E}[X^2]
= \infty. $$
Therefore the conclusion holds unconditionally.
