Well-definedness of the characteristic function of a compound Poisson variable

Question

I am reading about compound Poisson variables and cannot get through the following statement.

Let $\nu$ be a non-zero finite measure on $\mathbb{R}\setminus\{0\}$. Assume that

$$\int \min\{|u|,1\}\nu(du) < \infty. \tag{$\ast$}$$

We say that the distribution of a random variable $S$ belongs to a class $\mathcal{P}_{1,0}$ if its characteristic function writes as

$$\mathbb{E}e^{itS} = \exp\left\{ \int (e^{itu}-1)\nu(du) \right\}$$

for some measure $\nu$ satisfying $(\ast)$. Note that for any real $t$ the integral on the right-hand side is well defined since $|e^{itu}-1| \leq \min\{|t||u|, 2\}$.

Question 1: It is easy to see that $|e^{itu} -1|$ is bounded by 2. How to derive the other bound $|t||u|$ and what is its significance?

Question 2: How to make use of $(\ast)$ and actually show that the integral is well defined?

$$\int|e^{itu}-1|\nu(du) \leq \int \min\{|t||u|, 2\} \nu(du) \leq \dots$$

Thank you!

Regards, Ivan

Answer 1

The inequality

$$|e^{\imath \, tu}-1 \leq |t| \cdot |u| \tag{1}$$

follows easily from the fact that

$$e^{\imath \, tu}-1 = \frac{1}{\imath} \int_0^{tu} e^{\imath x} \, dx.$$

From $(1)$ we see that

$$\begin{align*} \int |e^{\imath \, t u}-1| \, \nu(du) \leq \int \min\{|t|\cdot |u|,2\} \nu(du) = |t| \int_{-\frac{2}{t}}^{\frac{2}{t}} |u| \, \nu(du) +2 \int_{|u|>\frac{2}{t}} \nu(du) \end{align*}$$

As $$\int \min\{|u|,1\} \nu(du)<\infty, \tag{2}$$ we conclude that the integral on the left-hand side is finite.

As @drhab already pointed out, the condition $(2)$ is superfluous if $\nu$ is a finite measure as the finiteness of the integral follows directly from $$ \int |e^{\imath \, tu}-1| \, \nu(du) \leq \int 2 \, \nu(du)<\infty.$$ If $\nu$ is not necessarily a finite measure but a ($\sigma$-finite) measure satisfying $(2)$, then the above calculation proves that the characteristic function is well-defined.

Answer 2

If $a\in\mathbb{R}$ then: $$\left|e^{ia}-1\right|=\left|\cos a-1+i\sin a\right|=\sqrt{\left(\cos a-1\right)^{2}+\sin^{2}a}=\sqrt{2-2\cos a}=\sqrt{2-2\left(1-2\sin^{2}\left(\frac{1}{2}a\right)\right)}=2\left|\sin\left(\frac{1}{2}a\right)\right|\leq\left|a\right|$$

That could help for your first question.

I don't see its functionality. $\left|e^{itu}-1\right|\leq2$ is enought to ensure that: $$\int\left|e^{itu}-1\right|\nu\left(du\right)\leq2\nu\left(\mathbb{R}\backslash\left\{ 0\right\} \right)<\infty$$