A question about the proof of the Levy-Khintchine representation Theorem

Question

I'm studying Infinitely Divisible random variables using this Lecture Notes. And I have a question that is driving me crazy.

In the proof of the "only if" part of the Levy-Khintchine representation (Theorem 5.3), the Theorem 5.13 contains an essential step. In this Theorem 5.13, the truncation function $h'$ must satisfy conditions (\ref{hhdjj}):
\begin{equation}\label{hhdjj}\tag{5.4} h'(x)=1+o(|x|) \quad as\quad |x|\to 0\quad \hbox{ and }\quad h'(x)=O(1/|x|), \quad as \quad x\to\infty. \end{equation}

However, in the proof of the "only if" part of the Theorem 5.3 (page 17), They approximate $\rho$ by a sequence of Compound Poisson random variables $\rho_n$ whose triple Levy-Khintchine representation and the use of Theorem 5.13 are described in the following print (Bottom of page 17):

So my big question is that they can't use Theorem 5.13 directly because $h\equiv 0$ doesn't satisfy the first condition in (\ref{hhdjj}).

Am I making a faux pas by reading the notes?

I appreciate your help.

Answer 1

$\newcommand\R{\mathbb R}\newcommand\ip[1]{\langle #1 \rangle}$In these notes, two related definitions of truncation functions are given. In Definition 5.6, a truncation function is defined as a bounded function $h\colon\R^d\to\R^d$ such that $h(x) = x$ in a neighborhood of $0$. In Definition 5.7, a truncation function $h'$ is defined as in your post. The relation between $h$ and $h'$ is given by the formula $h(x)=xh'(x)$ for all $x\in\R^d$.

In the proof of Theorem 5.13 in these notes, a truncation function $h$ as in Definition 5.6 is used. You are of course right that $h\equiv 0$ is not a valid truncation function according to either definition.

However, since in that proof $\rho$ is a probability measure, we can rewrite formula (5.25) in the notes as $$\hat\rho_n(u)= \exp\Big(ib_n+\int_{\R^d}\big(e^{i\ip{u,x}}-1-i\ip{u,h(x)}\big)\nu_n(dx)\Big),$$ where $h$ is any truncation function according to Definition 5.6 (e.g. $h(x)\equiv x\,1(|x|\le1)$), $\nu_n:=t_n^{-1}\rho^{t_n}$, and $b_n:=\int_{\R^d}\ip{u,h(x)}\,\nu_n(dx)\in\R$ (since the measure $\nu_n$ is finite). Then $\rho_n$ is infinitely divisible and has the Lévy--Khintchine representation with the triplet $(b_n,0,\nu_n)_h$, as desired.

(I would suggest using, instead of these notes, one of a number of known, more authoritative sources on infinitely divisible distributions.)