Show that $|1-\exp\{ix\}|^{2}=2(1-\cos x) \leq x^{2}$ for all $x \in \mathbb{R}$. Use this to show that $|1-\varphi_X(u)|\leq E\{|uX|\}$, where $\varphi_X(u) =E\{\exp(i\langle u,X\rangle)\}$ is the characteristic function.
We were given the hint that the power series for $\cos x$ is an alternating series. However, I'm not sure where that comes into play. For the first part, by using the definition of a square of a modulus, Euler's formula, and a trig identity, I got that $|1-\exp\{ix\}|^{2}=2-\sin^{2}x$. Now, $\sin^{2}x \leq x^{2}$, right? And so, $2-\sin^{2}x \leq 2-x^{2} \leq x^{2}$, right? So, I'm not sure hiw it's supposed to come into play in the first part, if at all.
The second part, showing that $|1-\varphi_X(u)|\leq E\{|uX|\}$ I am having the most trouble with. Any help you could give would be most appreciated!
$$ 1-\frac{x^2}{2} < \cos x < 1 - \frac{x^2}{2} + \frac{x^4}{24} \text{ if }x\text{ is near }0. $$ You couldn't draw this conclusion if the series didn't have terms alternating in sign and (if $x$ is near $0$) getting smaller as the degree grows.
Hence $$ x^2 - \frac{x^4}{12}\le2(1-\cos x) \le x^2. $$
You should carefully distinguish between $x$ and $X$. You're not doing that.
Before I say more, could you clarify whether you intend $X\sim N(0,1)$, as I'm guessing?