How can I prove $\left|\frac{e^{it_p x_j}-1}{t_p}\right| \leq 2|x|$?

Question

Background:

I am trying to understand why, if $X$ is a random variable with $\mathbb{E}\big[|X|\big]< +\infty$ and $\mu=\mathbb{P}^{X}$, then $$\frac{\partial}{\partial x_j}\overline{\mu}(u)=i\int e^{i \langle u,x \rangle} x_j \mu(dx).$$

My problem:

Suppose $t_p \to 0$ and $x \in \mathbb{R}^n$. How can I prove that $$\left|\frac{e^{it_p x_j}-1}{t_p}\right| \leq 2|x|,$$ where $x_j$ is the $j$-th coordinate of $x$?

Answer 1

In fact you can get $|x|$ as a bound. Use the fact that $|e^{i\theta} -1| \leq |\theta|$ for all real numbers $\theta$.

[$\int_0^{1} e^{it\theta} dt=\frac 1 {i \theta} (e^{i\theta}-1)$ and LHS is bounded in modulus by $\int_0^{1}dt=1$].