# Local subexponentiality and infinitely divisible distributions

Muneya Matsui  Department of Business Administration, Nanzan University, 18
Yamazato-cho, Showa-ku, Nagoya 466-8673, Japan<EMAIL_ADDRESS>and
Toshiro Watanabe  Center for Mathematical Sciences, The University of Aizu,
Ikkimachi Tsuruga, Aizu-Wakamatsu, Fukushima, 965-8580, Japan.
<EMAIL_ADDRESS>
###### Abstract.

We completely characterize $\Delta$\- and local subexponentialities of
positive-half compound Poisson distributions and extend the characterization
on two-sided distributions. Moreover, $\Delta$-subexponentiality of infinitely
divisible distributions is characterized with new conditions, and local
subexponentiality is newly characterized in the two-sided case. In the process
closedness properties of these subexponentialities are derived, particularly
for distributions on $\mathbb{R}$. Most results are obtained by exploiting
monotonic-type assumptions. We apply our results to distributions of supremum
of a random work and a randomly stopped iid sum.

###### Key words and phrases:

infinitely divisible, long-tailedness, $\Delta$-subexponentiality, local
subexponentiality, asymptotic to a non-increasing function, almost decreasing

###### 2010 Mathematics Subject Classification:

60E07, 60G70, 62F12

Muneya Matsui’s research is partly supported by the JSPS Grant-in-Aid for
Scientific Research C (19K11868).

##  1\. Introduction

Studies on the subexponentiality of infinitely divisible distributions (IDs
for abbreviation and ID in the singular form) have been initiated by Embrecht
et al. [5], where the subexponentiality of one-sided distributions was
completely characterized. Pakes [12] extended the result into distributions on
the real line. Here the “characterization of subexponentiality” means the
equivalence of three assertions: $(\mathrm{i})$ the truncated Lévy measure is
subexponential, $(\mathrm{ii})$ the corresponding ID is subexponential, and
$(\mathrm{iii})$ Lévy measure is long-tailed and its tail is asymptotically
equivalent to that of the corresponding ID. Since then, under various
assumptions and with several different versions, subexponential properties of
IDs have been investigated for forty years. However, not a few issues remain
unsolved (see the detailed explanation below).

Recently, two monotonic-type conditions: asymptotic to a non-increasing
function (a.n.i. for short) and almost decreasing (al.d. for short) are found
to be useful for characterizing subexponential densities of IDs, particularly
in the two-sided case, see Matsui [11]. The latter assumption is known to be
crucial for analyzing two-sided subexponential densities, see Foss et al. [9,
Section 4.3] and Finkelshtein and Tkachov [8].

In this paper, we will exploit the two conditions in different versions of
subexponentiality and extend the boundaries of the subexponential
characterization on IDs significantly. Firstly we adopt four different
definitions of subexponentiality and characterize their relations (see Section
2). In particular we show that these definitions are equivalent under the
a.n.i. condition.

Secondly, we focus on $\Delta$\- and local subexponentialities and derive
properties such as the closedness in convolution, asymptotic equivalence,
factorization and convolution root. The closedness property has been one of
central problems in the literature (see Section 3 together with references
their in). $\Delta$-subexponentiality has been introduced by Asmussen et al.
[2] and is now applied in a rather wide area, while local subexponentiality by
Watanabe and Yamamuro [21] is of theoretical importance, i.e. it has a close
inseparable relation with $\gamma$\- subexponentiality with $\gamma>0$.

Based on the obtained properties in Section 3, we characterize $\Delta$\- and
local subexponentialities of IDs on the whole real line.
$\Delta$-subexponentiality was only partially characterized, and local
subexponentiality was completed only on positive-half IDs and was not
investigated at all in the two-sided case (see the table and explanation
below). We completely characterize $\Delta$\- and local subexponentialities on
positive-half compound Poisson distributions, where we found that no
additional assumptions are needed. Moreover, with new and more simple
assumptions we characterize these subexponentialities on IDs on $\mathbb{R}$.
Specifically, we extend the boundaries of $\Delta$-subexponential
characterization from that of the self-decomposable distribution to that of
$s$-self-decomposable distribution (Jurek class [10]). These main results are
stated in Section 4.

As applications, we characterize $\Delta$-subexponentiality of the
distribution of supremum of a random walk and that of randomly stopped iid sum
in Section 5, which are closely related with classical ruin theory and queuing
theory ([9, Section 5]).

In the remainder of this section we introduce the four definitions of
subexponentiality and review past researches, associating with
characterization of these subexponentialities on subclasses of IDs. Then, we
explain more precisely what we obtained in our paper. This gives a good
perspective of the results in this paper, and one could see remaining open
problems also. Afterwards, we provide other definitions and notions necessary
in the main part.

Let $F,G,H$ be probability distribution functions on $\mathbb{R}$ and denote
by $F\ast G$ the convolution of $F$ and $G$:

$F\ast G(x)=\int_{-\infty}^{\infty}F(x-y)G(dy)$

and denote by $F^{\ast n}$ the $n$th convolution with itself. The tail
probability of $F$ is denoted by $\overline{F}(x)=1-F(x)$. Let $f,g,h$ be the
corresponding probability density functions on $\mathbb{R}$ and we use the
same notations for the convolution as those for distributions, e.g.

$f\ast g(x)=\int_{-\infty}^{\infty}f(x-y)g(y)dy\quad\text{or}\quad f^{\ast
n}(x)\quad\text{for the $n$th convolution}.$

Throughout the paper, let $\Delta:=(0,c]$ with $c>0$, and for any $x$ and for
any non-negative integer $n$. For convenience we sometimes write
$F(x+\Delta):=F(x,x+c]=F(x+c)-F(x)$. Moreover, for functions
$\alpha,\beta:\mathbb{R}\to\mathbb{R}_{+}$, $\alpha(x)\sim\beta(x)$ means that
$\lim_{x\to\infty}\alpha(x)/\beta(x)\to 1$.

######  Definition 1.1.

$(\mathrm{i})$ A non-negative measurable function $\alpha(x)$ belongs to the
class ${\mathcal{L}}$ if there exists $x_{0}>0$ such that $\alpha(x)>0,\,x\geq
x_{0}$ and for any fixed $y>0$ $\alpha(x+y)\sim\alpha(x)$.
$(\mathrm{ii})$ The density $f$ of $F$ is subexponential on $\mathbb{R}$,
denoted by ${\mathcal{S}}$, if $f\in{\mathcal{L}}$ and $f^{\ast 2}(x)\sim
2f(x)$.
$(\mathrm{iii})$ $F$ belongs to the class ${\mathcal{L}}_{\Delta}$ if
$F(x+\Delta)\in{\mathcal{L}}$.
$(\mathrm{iv})$ $F$ belongs to the class ${\mathcal{S}}_{\Delta}$ if
$F\in{\mathcal{L}}_{\Delta}$ and $F^{\ast 2}(x+\Delta)\sim 2F(x+\Delta)$.
$(\mathrm{v})$ $F$ belongs to the class ${\mathcal{L}}_{loc}$ if
$F\in{\mathcal{L}}_{\Delta}$ for all $\Delta:=(0,c]$ with $c>0$.
$(\mathrm{vi})$ $F$ belongs to the class ${\mathcal{S}}_{loc}$ if
$F\in{\mathcal{S}}_{\Delta}$ for all $\Delta:=(0,c]$ with $c>0$.
$(\mathrm{vii})$ $F(dx)=f(x)dx$ belongs to the class ${\mathcal{L}}_{ac}$ if
$f(x)\in{\mathcal{L}}$.
$(\mathrm{viii})$ $F(dx)=f(x)dx$ belongs to the class ${\mathcal{S}}_{ac}$ if
$f(x)\in{\mathcal{S}}$.

Table 1. Past researches on subexponential characterization | one-sided CP | two-sided CP | one-sided ID | two-sided ID | SD
---|---|---|---|---|---
${\mathcal{S}}$ | [5] | [12] | [5] | [12] | $--$
${\mathcal{S}}_{\Delta}$ | [2, 15, 20] | [20] | [2, 15] | [20] | [21]
${\mathcal{S}}_{loc}$ | [21] |  | [21] |  | [21]
${\mathcal{S}}_{ac}$ | [14] | [11] | [18] | [11] | [21]

In the literature, the subexponentiality for the compound Poisson distribution
(CP for short) and the ID are characterized with three equivalent relations
$(\mathrm{i})$-$(\mathrm{iii})$ above. We will review past researches in terms
of this equivalence, taking
${\mathcal{S}},\,{\mathcal{S}}_{\Delta},\,{\mathcal{S}}_{loc},\,{\mathcal{S}}_{ac}$
in order. In Table 1, we summarize these past researches, where hatching cells
implies that the subexponentiality was completely characterized in these cases
and SD is the abbreviation for the self-decomposable distribution.

${\mathcal{S}}$: the characterization for the distribution has been completed
by [5] and [12].

${\mathcal{S}}_{\Delta}$: Asmussen et al. [2] characterized two equivalences
with some conditions for both the CP and the ID in one-sided case. This was
extended to the three equivalences by Wang et al. [15] with different
conditions. In the general two-sided case Watanabe and Yamamuro [20] showed
three equivalences with less conditions than before.

${\mathcal{S}}_{loc}$: completely characterized in the one-sided ID case
(Watanabe and Yamamuro [21]).

${\mathcal{S}}_{ac}$: in the positive-half case, Shimura and Watanabe [14]
treated the CP, while Watanabe [18] studied the subexponential density on the
ID. The difference between the CP and the ID is that the treatment of problems
arising from the behavior of Lévy measures around the origin which explodes in
the ID case. Recently the two-sided cases in both the CP and the ID were
investigated by [11], where a kind of monotonic condition is shown to be
necessary.

In addition for the class of self-decomposable distributions it is known that
the three equivalent relations hold without any conditions in all definitions
of subexponentiality (see Theorem 1.3 and Lemma 4.2 in [21]).

After the review of past researches, now we revisit our achievement in detail.
Recall that our main focus in on ${\mathcal{S}}_{\Delta}$ and
${\mathcal{S}}_{loc}$. We complete the characterization of both
${\mathcal{S}}_{\Delta}$ and ${\mathcal{S}}_{loc}$ in the positive-half CP
case without additional assumptions. In the two-sided case, we characterize
three equivalences for ${\mathcal{S}}_{\Delta}$ with less conditions than
before, particularly we remove the long tailed assumption on Lévy measure,
which was assumed in the preceding results [2, 20]. As a by-product, we
recovered the classical result of Embrechts and Hawkes [6] in the one-sided
case and extended it to that for the two-sided case. We also study
${\mathcal{S}}_{loc}$ in the two-side case, which has not been investigated at
all.

For IDs we characterize ${\mathcal{S}}_{\Delta}$ with different conditions,
i.e. we replace the moment condition of [20] with the al.d. condition on
increments of both the distribution and the Lévy measure, with which we only
need to check the positive-half sides. Moreover, based on the two monotonic-
type conditions, we derive new characterization of ${\mathcal{S}}_{\Delta}$
without using the long-tailed assumption on the Lévy measure. With this result
we newly characterize $\Delta$-subexponentiality of the $s$-self-decomposable
distribution. Again our study of ${\mathcal{S}}_{loc}$ in the two-sided case
has not been investigated before. In summary, we achieve a major progress in
the characterization problems of subexponentialities on IDs.

We close this section with definitions need in the latter sections. We are
starting with the monotonic-type conditions tailored for each definition of
subexponentiality.

######  Definition 1.2.

We say that a function $\alpha:\mathbb{R}\to\mathbb{R}_{+}$ is asymptotic to a
non-increasing function $($a.n.i.$)$ denoted by $\alpha\in{\mathcal{A}}$ if
$\alpha$ is locally bounded and positive on $[x_{0},\infty)$ for some
$x_{0}>0$, and

(1.1) $\displaystyle\sup_{t\geq
x}\alpha(t)\sim\alpha(x)\quad\text{and}\quad\inf_{x_{0}\leq t\leq
x}\alpha(t)\sim\alpha(x).$

$(\mathrm{i})$ $F$ belongs to the class ${\mathcal{A}}_{\Delta}$ if
$F(x+\Delta)\in{\mathcal{A}}$.
$(\mathrm{ii})$ $F$ belongs to the class ${\mathcal{A}}_{loc}$ if
$F(x+\Delta)\in{\mathcal{A}}$ for all $\Delta$.
$(\mathrm{iii})$ $F$ belongs to the class ${\mathcal{A}}_{ac}$ if $F$ is
absolutely continuous and its density $f\in{\mathcal{A}}$.

######  Definition 1.3.

We say that a function $\alpha:\mathbb{R}\to\mathbb{R}_{+}$ is almost
decreasing $($al.d.$)$ denoted by $\alpha\in{\mathcal{D}}$ if there exists
$x_{0}>0$ and $K>0$ such that

(1.2) $\displaystyle\alpha(x+y)\leq K\alpha(x)\quad\text{for all}\
x>x_{0},\,y>0.$

$(\mathrm{i})$ $F$ belongs to the class ${\mathcal{D}}_{\Delta}$ if
$F(x+\Delta)\in{\mathcal{D}}$.
$(\mathrm{ii})$ $F$ belongs to the class ${\mathcal{D}}_{loc}$ if
$F(x+\Delta)\in{\mathcal{D}}$ for all $\Delta$.
$(\mathrm{iii})$ $F$ belongs to the class ${\mathcal{D}}_{ac}$ if $F$ is
absolutely continuous and its density $f\in{\mathcal{D}}$.

Obviously
${\mathcal{A}}_{ac}\subset{\mathcal{A}}_{loc}\subset{\mathcal{A}}_{\Delta}$
and
${\mathcal{D}}_{ac}\subset{\mathcal{D}}_{loc}\subset{\mathcal{D}}_{\Delta}$
and we have following lemma, whose proof is elementary and omitted.

######  Lemma 1.4.

If $\alpha$ is al.d. and $\alpha(x)\sim\beta(x)$, then $\beta$ is al.d.

The precise definition of the ID $\mu$ is given with its characteristic
function (ch.f.) denoted by $\widehat{\mu}$,

(1.3)
$\displaystyle\widehat{\mu}(z)=\exp\Big{\\{}\int_{-\infty}^{\infty}(e^{izy}-1-izy{\bf
1}_{\\{|y|\leq 1\\}})\nu(dy)+iaz-\frac{1}{2}b^{2}z^{2}\Big{\\}},$

where $a\in\mathbb{R},\,b\geq 0$ and $\nu$ is the Lévy measure satisfying
$\nu(\\{0\\})=0$ and $\int_{-\infty}^{\infty}(1\wedge x^{2})\nu(dx)<\infty$.
We denote by ${\rm ID}(\mathbb{R})$ and ${\rm ID}(\mathbb{R}_{+})$ the class
of all IDs on $\mathbb{R}$ and that of all IDs on $\mathbb{R}_{+}$,
respectively.

Other miscellaneous definitions are as follows. Let $f_{+}$ be the density of
the conditional distribution $F_{+}$ of $F$ on $\mathbb{R}_{+}$:
$f_{+}(x)={\bf 1}_{\mathbb{R}_{+}}(x)f(x)/\overline{F}(0),\,x\in\mathbb{R}$.
$F^{+}$ is the distribution given by $F^{+}(x)=F(x)$ for $x\geq 0$ and
$F^{+}(x)=0$ for $x<0$. For non-negative functions $\alpha(x)$ and $\beta(x)$,
a notation $\alpha(x)\asymp\beta(x)$ implies that

$0<\liminf_{x\to\infty}\big{(}\alpha(x)/\beta(x)\big{)}\leq\limsup_{x\to\infty}\big{(}\alpha(x)/\beta(x)\big{)}<\infty.$

The set of integers and the set of non-negative integers are respectively
denoted by ${\mathbb{Z}}$ and ${\mathbb{Z}}_{+}$.

##  2\. Relations between Subexponential definitions under ${\mathcal{A}}$ and
${\mathcal{D}}$

In the following two lemmas we provide equivalent conditions of a.n.i.
assumptions, and characterize a relation between ${\mathcal{A}}_{ac}$ and
${\mathcal{L}}_{ac}$, respectively. Then we investigate the relations between
subexponential definitions ${\mathcal{S}}_{\Delta},{\mathcal{S}}_{loc}$ and
${\mathcal{S}}_{ac}$.

######  Lemma 2.1.

Let $\alpha(x)\in{\mathcal{A}}$, and then

(2.4) $\displaystyle\text{there exists a positive non-increasing function}\
\beta\ \text{such that}\ \alpha(x)\sim\beta(x).$

Conversely, if $\alpha(x)\to 0$ as $x\to\infty$, then (2.4) implies
$\alpha\in{\mathcal{A}}$. In particular, $f\in{\mathcal{L}}$ with the
condition (2.4) implies that $f\in{\mathcal{A}}$, and
$F\in{\mathcal{A}}_{\Delta}$ if and only if $F(x+\Delta)$ satisfies (2.4), so
that $F\in{\mathcal{A}}_{loc}$ if and only if $F(x+\Delta)$ satisfies (2.4)
for all $\Delta$.

######  Lemma 2.2.

$F\in{\mathcal{L}}_{\Delta}\cap{\mathcal{A}}_{ac}$ implies
$F\in{\mathcal{L}}_{ac}$.

Generally, neither $F_{+}\in{\mathcal{S}}_{\Delta}$ nor
$F^{+}\in{\mathcal{S}}_{\Delta}$ may imply $F\in{\mathcal{S}}_{\Delta}$, which
is the same as that $F_{+}\in{\mathcal{S}}_{ac}$ or
$F^{+}\in{\mathcal{S}}_{ac}$ may not imply $F\in{\mathcal{S}}_{ac}$. Under the
al.d. assumption $F^{+}\in{\mathcal{S}}_{\Delta}$ is known to imply
$F\in{\mathcal{S}}_{\Delta}$ [9, Lemma 27]. The same can hold with
$F_{+}\in{\mathcal{S}}_{\Delta}$.

######  Proposition 2.3.

Let $F\in{\mathcal{D}}_{\Delta}$, then
$F^{+}\in{\mathcal{S}}_{\Delta}\Leftrightarrow
F_{+}\in{\mathcal{S}}_{\Delta}\Leftrightarrow F\in{\mathcal{S}}_{\Delta}$, so
that under $F\in{\mathcal{D}}_{loc}$,
$F^{+}\in{\mathcal{S}}_{loc}\Leftrightarrow
F_{+}\in{\mathcal{S}}_{loc}\Leftrightarrow F\in{\mathcal{S}}_{loc}$.

Since ${\mathcal{A}}_{\Delta}\subset{\mathcal{D}}_{\Delta}$ and
${\mathcal{A}}_{loc}\subset{\mathcal{D}}_{loc}$, the results in Proposition
2.3 follow under ${\mathcal{A}}_{\Delta}$ and ${\mathcal{A}}_{loc}$,
respectively.

Now we characterize the relations between subexponential definitions.

######  Proposition 2.4.

Assume that $F$ has a density. Under $F\in{\mathcal{A}}_{ac}$, the following
are equivalent.

$F_{+}\in{\mathcal{S}}_{\Delta},\,F^{+}\in{\mathcal{S}}_{\Delta},\,F\in{\mathcal{S}}_{\Delta},\,F_{+}\in{\mathcal{S}}_{loc},\,F^{+}\in{\mathcal{S}}_{loc},\,F\in{\mathcal{S}}_{loc},\,F_{+}\in{\mathcal{S}}_{ac},\,F^{+}\in{\mathcal{S}}_{ac},\,F\in{\mathcal{S}}_{ac}.$

If $F$ is a distribution on $\mathbb{R}_{+}$, then under
$F\in{\mathcal{L}}_{ac}$ the following are equivalent.

$F\in{\mathcal{S}}_{\Delta},\,F\in{\mathcal{S}}_{loc},\,F\in{\mathcal{S}}_{ac}.$

##  3\. Properties of ${\mathcal{S}}_{\Delta}$ and ${\mathcal{S}}_{loc}$

In this section we focus on what we call “usual properties” of
${\mathcal{L}}_{\Delta}$ and ${\mathcal{S}}_{\Delta}$ which are common in the
literature of the subexponentiality and which are useful to derive the main
theorems. One of the central problems there is the closedness property of
${\mathcal{L}}_{\Delta}$ and ${\mathcal{S}}_{\Delta}$ under various
operations. In the positive-half case some of these properties were
investigated by [9, Sections 4.5-4.8]. Since ${\mathcal{L}}_{\Delta}$ is
unchanged in both one- and two-sided cases, we concentrate on properties for
${\mathcal{S}}_{\Delta}$. We derive several new results (versions of
convolution root and factorization) and moreover, extend all results of [9,
Subsection 4.6] into the two-sided distribution. In the derivation, both
a.n.i. and al.d. assumptions are exploited.

We start with an extension of [9, Theorem 4.21].

######  Lemma 3.1.

Suppose that $F\in{\mathcal{L}}_{\Delta}$ for some $\Delta$. Let $\xi_{1}$ and
$\xi_{2}$ be two independent r.v.’s with common distribution $F$. Then the
following assertions are equivalent:
$(\mathrm{i})$ $F\in{\mathcal{S}}_{\Delta}$.
$(\mathrm{ii})$ For every function $\alpha$ such that $\alpha(x)\to\infty$ as
$x\to\infty$ and $\alpha(x)<x/2$,

(3.5) $\displaystyle{\mathbb{P}}\big{(}\xi_{1}+\xi_{2}\in
x+\Delta,\,\xi_{1}\leq-\alpha(x)\big{)}=\int_{(-\infty,-\alpha(x)]}F(x+\Delta-y)F(dy)$
$\displaystyle=o(F(x+\Delta)),$ (3.6)
$\displaystyle\text{and}\qquad{\mathbb{P}}\big{(}\xi_{1}+\xi_{2}\in
x+\Delta,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x)\big{)}$
$\displaystyle=o(F(x+\Delta)).$

$(\mathrm{iii})$ There exists a function $\alpha$ such that
$\alpha(x)<x/2,\,\alpha(x)\to\infty$ as $x\to\infty$, $F(x+\Delta)$ is
$\alpha$-insensitive and both (3.5) and (3.6) hold.

If additionally $F$ is a distribution on $\mathbb{R}_{+}$, then equivalence of
the three holds without the condition (3.5) in $(\mathrm{ii})$ and
$(\mathrm{iii})$. Moreover, we could replace the condition (3.6) of
$(\mathrm{iii})$ with

(3.7)
$\displaystyle\int_{(\alpha(x),x-\alpha(x)]}F(x+\Delta-y)F(dy)=o(F(x+\Delta)).$

The second one is the closedness of ${\mathcal{S}}_{\Delta}$ under
factorization.

######  Proposition 3.2 (factrization).

Let $H=F\ast G\in{\mathcal{S}}_{\Delta}$ and $G(x+\Delta)=o(H(x+\Delta))$. If
$F\in{\mathcal{A}}_{\Delta}$ and $[$$H\in{\mathcal{D}}_{\Delta}$ or
$G(x+\Delta)=o(F(x+\Delta))$$]$, then $F\in{\mathcal{S}}_{\Delta}$.

Next we show several versions of [the closedness under convolution root] with
a.n.i. assumptions.

######  Proposition 3.3 (convolution root: positive-half).

Let $F$ be a distribution on ${\mathbb{R}}_{+}$. If
$F\in{\mathcal{A}}_{\Delta}$ and $F^{*N}\in{\mathcal{S}}_{\Delta}$, then
$F\in{\mathcal{S}}_{\Delta}$. Thus, if $F\in{\mathcal{A}}_{loc}$ and
$F^{*N}\in{\mathcal{S}}_{loc}$, then $F\in{\mathcal{S}}_{loc}$.

######  Proposition 3.4 (convolution root).

If $F\in{\mathcal{A}}_{\Delta}$ and
$F^{*N}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$, then
$F\in{\mathcal{S}}_{\Delta}$. Consequently, if $F\in{\mathcal{A}}_{loc}$ and
$F^{*N}\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$, then
$F\in{\mathcal{S}}_{loc}$.

######  Corollary 3.5 (convolution root for density).

Let $f\in{\mathcal{A}}$. If $f^{*N}\in{\mathcal{S}}\cap{\mathcal{D}}$, then
$f\in{\mathcal{S}}$.

######  Remark 3.6.

$(\mathrm{i})$ There are many studies about closedness of subexponentiality
with the convolution root operation. The closedness of ${\mathcal{S}}$ for
positive-half distributions was shown by [5] and that for two-sided ones was
proved by [16]. However, the classes
${\mathcal{S}}_{ac},{\mathcal{S}}_{\Delta}$ and ${\mathcal{S}}_{loc}$ are
known not to be closed under convolution roots, see [22, Corollary 1.1] and of
[17, Corollary 1.1]. Then the next natural question has been that under what
conditions the closedness holds. Propositions 3.3, 3.4 and Corollary 3.5 give
some solutions to the problem. Notice that the closedness of under convolution
roots often holds within the class of compound Poisson distributions or IDs.
We discuss this topic with unsolved future problems in Section 4.2, see Remark
4.11.
$(\mathrm{ii})$ Corollary 3.5 is an improvement of [11, Theorem 3.13], where
either $f^{\ast k}\in{\mathcal{A}}$ for all $1\leq k\leq N$, or
$f\in{\mathcal{L}}$ is required.

The following results are two-sided extension of one-sided properties:
Theorems 4.22, 4.23, Corollary 4.24 and Theorem 4.25 of [9]. In particular
Kesten bound (Proposition 3.10) is useful to derive results in the following
sections.

######  Proposition 3.7 (asymptotic equivalence).

Suppose that $F\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$G\in{\mathcal{L}}_{\Delta}$ and $G(x+\Delta)\asymp F(x+\Delta)$ as
$x\to\infty$. Then $G\in{\mathcal{S}}_{\Delta}$. In particular, if
$G(x+\Delta)\sim cF(x+\Delta)$ as $x\to\infty$ for some $c>0$, then
$G\in{\mathcal{S}}_{\Delta}$. Moreover, if additionally $G$ is a distribution
on $\mathbb{R}_{+}$, then the assertion holds without assumption
${\mathcal{D}}_{\Delta}$ on $F$.

######  Proposition 3.8 (convolution).

Suppose that $F\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. Let
$G_{1},G_{2}$ be two distributions such that $G_{1}(x+\Delta)/F(x+\Delta)\to
c_{1}$ and $G_{2}(x+\Delta)/F(x+\Delta)\to c_{2}$ as $x\to\infty$ for some
constants $c_{1},c_{2}\geq 0$. Then

(3.8) $\frac{G_{1}*G_{2}(x+\Delta)}{F(x+\Delta)}\to
c_{1}+c_{2}\quad\text{as}\quad x\to\infty.$

Further, if $c_{1}+c_{2}>0$ then $G_{1}*G_{2}\in{\mathcal{S}}_{\Delta}$.

Following corollary is immediately derived by Proposition 3.8.

######  Proposition 3.9.

Suppose that $F\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. Let $G$
be a distribution such that $G(x+\Delta)/F(x+\Delta)\to c\geq 0$ as
$x\to\infty$. Then, for any $n\geq 2$, $G^{*n}(x+\Delta)/F(x+\Delta)\to nc\geq
0$ as $x\to\infty$. If $c>0$, then $G^{*n}\in{\mathcal{S}}_{\Delta}$.

######  Proposition 3.10 (Kesten bound).

Suppose that $F\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. Then, for
any $\varepsilon>0$, there exists $x_{0}=x_{0}(\varepsilon)>0$ and
$c(\varepsilon)>0$ such that, for any $x>x_{0}$ and for any $n\geq 1$,

$F^{*n}(x+\Delta)\leq c(\varepsilon)(1+\varepsilon)^{n}F(x+\Delta).$

The last two results are about the closedness under factorization when a
negligible distribution has a point mass at the origin. The negligible
distribution includes a compound Poisson distribution.

######  Proposition 3.11 (factrization).

Let $H=G*F\in{\mathcal{S}}_{\Delta}$ such that
$G(dx)=p_{G}\delta_{0}(dx)+(1-p_{G})G_{1}(dx)$ with $p_{G}\in(2^{-1},1)$ where
$G_{1}$ is a distribution. Suppose that $H\in{\mathcal{D}}_{\Delta}$ or
$[\,F,G,H$ are distributions on $\mathbb{R}_{+}\,]$. Then,
$G(x+\Delta)=o(H(x+\Delta))$ implies $F\in{\mathcal{S}}_{\Delta}$, and indeed
$H(x+\Delta)\sim F(x+\Delta).$

######  Proposition 3.12 (factrization: compound Poisson).

Let $H=G*F\in{\mathcal{S}}_{\Delta}$ such that $G$ is a compound Poisson
distribution. Suppose that $H\in{\mathcal{D}}_{\Delta}$ or $[\,F,G,H$ are
distributions on $\mathbb{R}_{+}\,]$. Then, $G(x+\Delta)=o(H(x+\Delta))$
implies $F\in{\mathcal{S}}_{\Delta}$, and indeed $H(x+\Delta)\sim
F(x+\Delta).$

##  4\. Main results

We investigate ${\mathcal{S}}_{\Delta}$ and ${\mathcal{S}}_{loc}$ separately
in the compound Poisson case (Section 4.1) and in the infinitely divisible
case (Section 4.2).

###  4.1. Compound Poisson case

Let $G$ be a distribution on $\mathbb{R}$. For $\lambda>0$ define the compound
Poisson probability measure by

(4.9)
$\displaystyle\mu(dx)=e^{-\lambda}\sum_{n=0}^{\infty}(\lambda^{n}/n!)G^{\ast
n}(dx),$

where $G^{\ast 0}$ is the Dirac measure.

######  Theorem 4.1.

Let $\mu$ be a compound Poisson given by (4.9). The following are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$ $G\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$\mu(x+\Delta)/G(x+\Delta)\sim\lambda$.
If additionally $G$ is a distribution on $\mathbb{R}_{+}$, then the following
assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}$, $(\mathrm{ii})$
$G\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}$
and $\mu(x+\Delta)/G(x+\Delta)\sim\lambda$.

An immediate corollary follows from the fact
${\mathcal{A}}_{\Delta}\subset{\mathcal{D}}_{\Delta}$.

######  Corollary 4.2.

Let $\mu$ be a compound Poisson given by (4.9). The following are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{A}}_{\Delta}$,
$(\mathrm{ii})$ $G\in{\mathcal{S}}_{\Delta}\cap{\mathcal{A}}_{\Delta}$,
$(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}\cap{\mathcal{A}}_{\Delta}$ and
$\mu(x+\Delta)/G(x+\Delta)\sim\lambda$.

In the following corollary we recovered the classical one-sided result of
Embrechts and Hawkes [6], and moreover extended it to the characterization of
the two-sided distribution.

######  Corollary 4.3.

Let $\Delta=(0,1]$ and let $\mu$ be an ID on $\mathbb{Z}$ with Lévy measure
$\lambda G$ where $\lambda>0$ and $G$ is a distribution. The following
assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$ $G\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$\mu(x+\Delta)/G(x+\Delta)\sim\lambda$.
If additionally $G$ is a distribution on ${\mathbb{Z}}_{+}$, then the
following assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}$, $(\mathrm{ii})$
$G\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}$
and $\mu(x+\Delta)/G(x+\Delta)\sim\lambda$.

The results in Theorem 4.1 are immediately extended to those for the local
subexponentiality. Notice that the two-sided case below has not been studied
before.

######  Corollary 4.4.

Let $\mu$ be a compound Poisson with Lévy measure $\lambda G$ where
$\lambda>0$ and $G$ is a distribution. The following assertions are
equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$,
$(\mathrm{ii})$ $G\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$,
$(\mathrm{iii})$ $G\in{\mathcal{L}}_{loc}\cap{\mathcal{D}}_{loc}$ and
$\mu(x+\Delta)/G(x+\Delta)\sim\lambda$ for any $\Delta$.
If additionally $G$ is a distribution on $\mathbb{R}_{+}$, then the following
assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{loc}$, $(\mathrm{ii})$
$G\in{\mathcal{S}}_{loc}$, $(\mathrm{iii})$ $G\in{\mathcal{L}}_{loc}$ and
$\mu(x+\Delta)/G(x+\Delta)\sim\lambda$ for nay $\Delta$.

######  Remark 4.5.

The positive distribution part of Corollary 4.4 has two different proofs. One
is an immediate consequence of the positive part of Theorem 4.1, and the other
pass is given by a more general result [21, Theorem 1.1]. We briefly explain
this. For $(\mathrm{ii})\Leftrightarrow(\mathrm{iii})$ observe that by
Proposition 3.7, $G\in{\mathcal{S}}_{loc}\Leftrightarrow
G_{1}\in{\mathcal{S}}_{loc}$ where $G_{1}$, given in the proof of Theorem 4.1,
corresponds to the conditional Lévy measure $\nu_{(1)}$ of [21, Theorem 1.1].
Thus $(2)\Leftrightarrow(3)$ in [21, Theorem 1.1] yields
$(\mathrm{ii})\Leftrightarrow(\mathrm{iii})$ where the second condition of
$(3)$ is $\mu(x+\Delta)\sim\lambda G(x+\Delta)$ for all $\Delta$. The part
$(\mathrm{i})\Leftrightarrow(\mathrm{ii})$ directly follows from
$(1)\Leftrightarrow(2)$ of [21, Theorem 1.1] since
$G\in{\mathcal{S}}_{loc}\Leftrightarrow G_{1}\in{\mathcal{S}}_{loc}$.

###  4.2. Infinitely divisible case

Define the normalized Lévy measure $\nu_{(1)}$ as $\nu_{(1)}(dx)={\bf
1}_{(1,\infty)}(x)\nu(dx)/\nu((1,\infty))$.

######  Theorem 4.6.

Let $\mu\in{\rm ID}(\mathbb{R})$. Under the condition
$\nu_{(1)}\in{\mathcal{L}}_{\Delta}$, the following are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$
$\nu_{(1)}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{iii})$
$\nu_{(1)}\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$\mu(x+\Delta)\sim\nu(x+\Delta)$.

######  Remark 4.7.

In Theorem 4.6, the moment condition $[\,\int_{\mathbb{R}}e^{-\varepsilon
y}\nu(dy)<\infty$ for some $\varepsilon>0\,]$ in [20, Theorem 1.2] is removed,
but one has to pay the price that the condition ${\mathcal{D}}_{\Delta}$ is
required in both $\mu$ and $\nu_{(1)}$. The merit is that we only need to
check the positive-half side of $\mu$ or $\nu$.

Notice that the assumption $\nu_{(1)}\in{\mathcal{L}}_{\Delta}$ of Theorem 4.6
is too strong, and thus we derive the three equivalent relations in different
conditions.

######  Theorem 4.8.

Let $\mu\in{\rm ID}(\mathbb{R})$ and for a constant $c_{1}>1$ let
$\mu_{c_{1}}$ be a compound Poisson with the Lévy measure
$\nu_{c_{1}}(dx)={\bf 1}_{\\{c_{1}<x\\}}\nu(dx)$ Suppose that there exists
$c_{1}>1$ such that $\mu_{c_{1}}\in{\mathcal{A}}_{\Delta}$. Then following
assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$ $\nu_{(1)}\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$
$\nu_{(1)}\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$\mu(x+\Delta)\sim\nu(x+\Delta)$.

Now we could treat the $s$-self-decomposable distribution which is a more
general class in ${\rm ID}(\mathbb{R})$ than the self-decomposable
distribution ([10]), namely its Lévy density $g$ of $\nu(dx)=g(x)dx$ satisfies
that $g(x)$ is non-increasing on $(0,\infty)$ and is non-decreasing on
$(-\infty,0)$ ([10, Theorem 2.2 (b)]).

######  Theorem 4.9.

Let $\mu\in{\rm ID}(\mathbb{R})$ be an $s$-self-decomposable distribution with
the Lévy measure $\nu(dx)$. Then following assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$ $\nu_{(1)}\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$
$\nu_{(1)}\in{\mathcal{L}}_{\Delta}$ and $\mu(x+\Delta)\sim\nu(x+\Delta)$.

We state the result for ${\mathcal{S}}_{loc}$ corresponding to Theorem 4.6.

######  Theorem 4.10.

Let $\mu\in{\rm ID}(\mathbb{R})$. The following assertions are equivalent.
$(\mathrm{i})$ $\mu\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$,
$(\mathrm{ii})$ $\nu_{(1)}\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$,
$(\mathrm{iii})$ $\nu_{(1)}\in{\mathcal{L}}_{loc}\cap{\mathcal{D}}_{loc}$ and
$\mu(x+\Delta)\sim\nu(x+\Delta)$ for any $\Delta$.

######  Remark 4.11.

Recall that the classes ${\mathcal{S}}_{ac},{\mathcal{S}}_{\Delta}$ and
${\mathcal{S}}_{loc}$ are not closed under convolution roots, see Remark 3.6.
However, we find from Theorem 4.1 and Corollary 4.4 that the classes
${\mathcal{S}}_{\Delta}$ in the one-sided case and
${\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ are closed within compound
Poisson distributions. Moreover, due to Theorem 4.10 and [21, Theorem 1.1] ,
the classes ${\mathcal{S}}_{loc}$ in one-sided case and
${\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$ are closed within the classes of
${\rm ID}(\mathbb{R}_{+})$ and ${\rm ID}(\mathbb{R})$ respectively. Thus we
raise a natural open question:

Problem: Are the classes ${\mathcal{S}}_{ac}$ and ${\mathcal{S}}_{\Delta}$ in
the one-sided case and the classes ${\mathcal{S}}_{ac}\cap{\mathcal{D}}_{ac}$
and ${\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ closed under
convolution roots within the classes of ${\rm ID}(\mathbb{R}_{+})$ and ${\rm
ID}(\mathbb{R})$ respectively $?$

##  5\. Application

Needless to say, a random walk and a randomly stopped sum have applications in
a variety of fields. We will observe that $\Delta$-subexponentiality is the
characterized in the two classical themes. Thus, the same results hold for
local subexponentiality.

###  5.1. Supremum of a random walk

We are starting with $\Delta$-subexponentiality of compound negative binomial
distributions, which include compound geometric distributions as a subclass.

######  Theorem 5.1.

Let $0<\lambda<1$, $a>0$ and $G$ be a distribution. Define a compound negative
binomial distribution $F_{a}$ as

$F_{a}:=\sum_{n=0}^{\infty}{a+n-1\choose
a-1}(1-\lambda)^{a}\lambda^{n}G^{*n}.$

Then the following assertions are equivalent.
$(\mathrm{i})$ $F_{a}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{ii})$ $G\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$F_{a}(x+\Delta)\sim a\lambda/(1-\lambda)G(x+\Delta)$.
If additionally $G$ is a distribution on $\mathbb{R}_{+}$, then the following
assertions are equivalent,
$(\mathrm{i})$ $F_{a}\in{\mathcal{S}}_{\Delta}$, $(\mathrm{ii})$
$G\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$ $G\in{\mathcal{L}}_{\Delta}$
and $F_{a}(x+\Delta)\sim a\lambda/(1-\lambda)G(x+\Delta)$.

Next, we define the distribution $\pi$ of the supremum of a random walk. Let
$\\{X_{n}\\}_{n=1}^{\infty}$ be iid random variables with the common
distribution $G$ on $\mathbb{R}$. Let $\\{S_{n}\\}_{n=0}^{\infty}$ be a random
walk on $\mathbb{R}$ defined by $S_{0}:=0$ and $S_{n}:=\sum_{k=1}^{n}X_{k}$
for $n\geq 1$. Let $\pi$ be the distribution of the supremum $M$ of
$\\{S_{n}\\}$, that is, $M:=\sup_{n\geq 0}S_{n}.$ Denote by ${\bf
1}_{(0,\infty)}(x)$ the indicator function of the set $(0,\infty).$ Define the
measure $\nu$ on $(0,\infty)$ and a quantity $B$ as

$\nu(dx):={\bf 1}_{(0,\infty)}(x)\sum_{n=1}^{\infty}n^{-1}G^{*n}(dx)$

and

$B:=\sum_{n=1}^{\infty}n^{-1}P(S_{n}>0)=\nu((0,\infty)).$

It is well known that $M<\infty$ a.s. if and only if $B<\infty$ and that if
$B<\infty$, then $\pi$ is a compound Poisson distribution on $\mathbb{R}_{+}$
with Lévy measure $\nu$. A sufficient condition for $M<\infty$ a.s. is that
$-\infty<{\mathbb{E}}[X_{1}]<0$, (for more details, see [7, Section XII.7]).
Define $\lambda$ as $\lambda:=1-e^{-B}$ when $B<\infty$. We denote by $\lambda
G_{0}$ the defective distribution of the first ascending ladder height in the
random walk $\\{S_{n}\\}$. Here, $G_{0}$ is a distribution on $\mathbb{R}_{+}$
with $G_{0}(\\{0\\})=0$ given by

$G_{0}=-\lambda^{-1}\sum_{n=1}^{\infty}\big{(}(-1)^{n}/n!\big{)}\nu^{\ast
n}\quad\Leftrightarrow\quad\nu=\sum_{n=1}^{\infty}(\lambda^{n}/n)G_{0}^{\ast
n}.$

It is also well known that

$\pi=\sum_{n=0}^{\infty}(1-\lambda)\lambda^{n}G_{0}^{*n},$

which is a compound geometric distribution, $(a=1)$ in $F_{a}$ (see, e.g. [9,
Section 5.3] or [1, Chapter X, 3.]). The distribution $\pi$ is important in
classical ruin theory and queuing theory (see Asmussen and Albrecher [1]). We
say that $F$ is non-lattice if the support of $F$ is not on any lattice.

######  Theorem 5.2.

Suppose that $-\infty<{\mathbb{E}}[X_{1}]<0$ and that $G$ is non-lattice. Then
the following assertions are equivalent.
$(\mathrm{i})$ $\pi\in{\mathcal{S}}_{\Delta}$, $(\mathrm{ii})$
$B^{-1}\nu\in{\mathcal{S}}_{\Delta}$, $(\mathrm{iii})$
$B^{-1}\nu\in{\mathcal{L}}_{\Delta}$ and $\pi(x+\Delta)\sim\nu(x+\Delta)$,
$(\mathrm{iv})$ $G_{0}\in{\mathcal{S}}_{\Delta}$, $(\mathrm{v})$
$G_{0}\in{\mathcal{L}}_{\Delta}$ and
$\pi(x+\Delta)\sim\lambda/(1-\lambda)G_{0}(x+\Delta)$.

###  5.2. Randomly stopped sums

Let $\\{X_{n}\\}_{n=1}^{\infty}$ and $S_{n},\,n\geq 0$ be respectively the
random sequence and its partial sum as defined above. Independently we take a
counting r.v. $\tau$ on ${\mathbb{Z}}_{+}$. Then the distribution $S_{\tau}$
is given by

(5.10) $\displaystyle
F^{\ast\tau}:=\sum_{n=0}^{\infty}{\mathbb{P}}(\tau=n)F^{\ast n},$

where $F^{\ast 0}=\delta_{0}$. The following is a two-sided extension of [9,
Theorem 3.1].

######  Theorem 5.3.

Suppose $F\in{\mathcal{L}}_{\Delta}$ and ${\mathbb{E}}[\tau]<\infty$.
$(\mathrm{i})$ If $F\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
${\mathbb{E}}[(1+\delta)^{\tau}]<\infty$ for some $\delta>0$, then

(5.11) $\displaystyle{\mathbb{P}}(S_{\tau}\in
x+\Delta)/F(x+\Delta)\to{\mathbb{E}}[\tau]\quad\text{as}\quad x\to\infty.$

$(\mathrm{ii})$ If ${\mathbb{P}}(\tau>1)>0$ and further the relation (5.11)
holds, then $F\in{\mathcal{S}}_{\Delta}$.

The proof is quite similar to that of [9, Theorem 3.1] if one uses the two-
sided Kesten bound (Proposition 3.10), and thus it is omitted.

##  6\. Proofs

Let $c$ be a generic positive constant whose value is not of interest. Notice
that we abuse this $c$ also for the definition of $\Delta$ and we make
cautions where the double use may possibly cause confusion.

###### Proof of Lemma 2.1.

Obviously the a.n.i. properties imply (2.4), and we show the converse. We
observe that for sufficiently large $x>0$,

$1\leq\frac{\sup_{t\geq x}\alpha(t)}{\alpha(x)}=\sup_{t\geq
x}\frac{\alpha(t)}{\beta(x)}\frac{\beta(x)}{\alpha(x)}\leq\sup_{t\geq
x}\frac{\alpha(t)}{\beta(t)}\frac{\beta(x)}{\alpha(x)}\to 1$

as $x\to\infty$. Moreover, since $\alpha(x)$ is positive on $[x_{0},\infty)$
for some $x_{0}>0$, and $\alpha(x)\to 0$ as $x\to\infty$, there exists
$y_{x}\leq x$ such that

$1\geq\frac{\inf_{x_{0}\leq t\leq x}\alpha(t)}{\alpha(x)}=\inf_{y_{x}\leq
t\leq
x}\frac{\alpha(t)}{\beta(x)}\frac{\beta(x)}{\alpha(x)}\geq\inf_{y_{0}\leq
t\leq x}\frac{\alpha(t)}{\beta(t)}\frac{\beta(x)}{\alpha(x)},$

where $y_{0}\in(x_{0},y_{x}]$ is arbitrary. Letting $x\to\infty$, so that one
may let $y_{x}\to\infty$ and then $y_{0}\to\infty$, the right converges to $1$
and we obtain the result. It is known that $f\in{\mathcal{L}}$ implies that
$f(x)\to 0$ as $x\to\infty$ (see [9, p.76]). Next we see that $F(x+\Delta)\to
0$ as $x\to\infty$. But this follows from

$\sum_{n=-\infty}^{\infty}F(x+nc+\Delta)=1,\quad x>0$

together with the Cauchy criterion for summability. The final result is
trivial if we take arbitrary $\Delta$ in the previous argument. ∎

###### Proof of Lemma 2.2.

Take a sufficiently large $x_{0}>0$ such that $\inf_{t\in[x_{0},x]}f(t)\sim
f(x)$. For any $y\in\mathbb{R}$

$\displaystyle\frac{f(x+y)}{f(x)}$ $\displaystyle=\frac{\sup_{t\geq
x+y}f(t)}{\inf_{s\in[x_{0},x]}f(s)}\frac{f(x+y)}{\sup_{t\geq
x+y}f(t)}\frac{\inf_{s\in[x_{0},x]}f(s)}{f(x)}$
$\displaystyle\geq\frac{F(x+y+\Delta)}{F(x-c+\Delta)}\frac{f(x+y)}{\sup_{t\geq
x+y}f(t)}\frac{\inf_{s\in[x_{0},x]}f(s)}{f(x)}.$

Moreover,

$\displaystyle\frac{f(x+y)}{f(x)}$
$\displaystyle=\frac{\inf_{t\in[x_{0},x+y]}f(t)}{\sup_{s\geq
x}f(s)}\frac{f(x+y)}{\inf_{t\in[x_{0},x+y]}f(t)}\frac{\sup_{s\geq
x}f(s)}{f(x)}$
$\displaystyle\leq\frac{F(x+y-c+\Delta)}{F(x+\Delta)}\frac{f(x+y)}{\inf_{t\in[x_{0},x+y]}f(t)}\frac{\sup_{s\geq
x}f(s)}{f(x)}.$

Now since $F\in{\mathcal{L}}_{\Delta}$ and $f\in{\mathcal{A}}$, by taking
$\lim_{x\to\infty}$ in both inequalities, we have

$1=\lim_{x\to\infty}\frac{F(x+y+\Delta)}{F(x-c+\Delta)}\leq\lim_{x\to\infty}\frac{f(x+y)}{f(x)}\leq\lim_{x\to\infty}\frac{F(x+y-c+\Delta)}{F(x+\Delta)}=1.$

Thus we have $f\in{\mathcal{L}}$. ∎

###### Proof of Proposition 2.3.

Since the assertion of ${\mathcal{S}}_{loc}$ follows from that of
${\mathcal{S}}_{\Delta}$, we only prove the result for
${\mathcal{S}}_{\Delta}$. It suffices to see that
$F_{+}\in{\mathcal{S}}_{\Delta}\Rightarrow F\in{\mathcal{S}}_{\Delta}$.
Although we can use $F_{+}(x+\Delta)\asymp F^{+}(x+\Delta)$ together with [9,
Lemma 4.20] for the proof, we give a direct proof for the latter use. Notice
that the proof of [9, Lemma 4.20] was omitted by similarity.

We observe that

$F_{+}(x)=(F(x)-F(0-))/\overline{F}(0-),$

where $F(x-)=\lim_{y\uparrow x}F(y)$, so that $\overline{F}(0-)=1-F(0-)$ and

(6.12) $\displaystyle F_{+}^{\ast 2}(x+\Delta)$
$\displaystyle=\frac{1}{\overline{F}(0-)^{2}}\Big{(}\int_{[0,x]}F(x+\Delta-y)F(dy)$
$\displaystyle\hskip 71.13188pt+\int_{(x,x+c]}(F(x+c-y)-F(0-))F(dy)\Big{)}.$

The condition $F_{+}\in{\mathcal{S}}_{\Delta}$ implies that

(6.13) $\displaystyle\frac{F_{+}^{\ast
2}(x+\Delta)}{F_{+}(x+\Delta)}=\frac{F_{+}^{\ast
2}(x+\Delta)\overline{F}(0-)}{F(x+\Delta)}\to 2$

as $x\to\infty$. Noticing (6.12) we write

$\displaystyle\frac{F^{\ast 2}(x+\Delta)}{F(x+\Delta)}$
$\displaystyle=\int_{(-\infty,\infty)}\frac{F(x+\Delta-y)}{F(x+\Delta)}F(dy)$
$\displaystyle=\frac{F_{+}^{\ast
2}(x+\Delta)\overline{F}(0-)^{2}}{F(x+\Delta)}+2\int_{(-\infty,0)}\frac{F(x+\Delta-y)}{F(x+\Delta)}F(dy),$

where we exploit the identity by Fubini:

(6.14)
$\displaystyle\begin{split}\int_{(x,\infty)}F(x-y)F(dy)&=\int_{\mathbb{R}^{2}}{\bf
1}_{\\{x<y\leq x-s\\}}(F\times F)(d(s,y))\\\
&=\int_{(-\infty,0)}F(x-s)F(ds)-F(x)F(0-).\end{split}$

Since we have (6.13), it suffices to show that the second integral term
converges to $2F(0-)$. Since $F\in{\mathcal{L}}_{\Delta}$, there exists a
function $\alpha$ such that $\alpha(x)<x/2$, $\alpha(x)\to\infty$ as
$x\to\infty$ and $F(x+\Delta)$ is $\alpha$-insensitive. Then

$2\int_{(-\alpha(x),0)}\frac{F(x+\Delta-y)}{F(x+\Delta)}F(dy)\to
2F(0-),\quad\text{as}\quad x\to\infty,$

while by $F\in{\mathcal{D}}_{\Delta}$,

$2\int_{(-\infty,-\alpha(x)]}\frac{F(x+\Delta-y)}{F(x+\Delta)}F(dy)\leq
2K\int_{(-\infty,-\alpha(x)]}F(dy)\to 0\quad\text{as}\quad x\to\infty.$

Now we obtain the desired result. ∎

###### Proof of Proposition 2.4.

In view of Proposition 2.3 and
${\mathcal{S}}_{\Delta}\supset{\mathcal{S}}_{loc}\supset{\mathcal{S}}_{ac}$ it
suffices to show that $F_{+}\in{\mathcal{S}}_{\Delta}\Rightarrow
F\in{\mathcal{S}}_{ac}$. We observe by Lemma 2.2 that $f_{+}\in{\mathcal{L}}$,
so that $f_{+}^{\ast 2}\in{\mathcal{L}}$. Then the uniform convergence
property of ${\mathcal{L}}$ yields

$F_{+}(x+\Delta)\sim cf_{+}(x)\ \text{and}\ F_{+}^{\ast 2}(x+\Delta)\sim
cf_{+}^{\ast 2}(x).$

Hence, $f_{+}\in{\mathcal{S}}$ if and only if
$F_{+}\in{\mathcal{S}}_{\Delta}$. Finally, by the a.n.i. property
$f_{+}\in{\mathcal{S}}$ implies $f\in{\mathcal{S}}$. The remaining result is
trivial if we put $F=F_{+}$. ∎

###### Proof of Lemma 3.1.

$(\mathrm{i})\Rightarrow(\mathrm{ii})$. Define the event
$B=\\{\xi_{1}+\xi_{2}\in x+\Delta\\}$. We have

(6.15) $\displaystyle{\mathbb{P}}(B)=$ $\displaystyle
2{\mathbb{P}}\big{(}B,\,-\alpha(x)<\xi_{1}\leq\alpha(x)\big{)}+2{\mathbb{P}}\big{(}B,\,\xi_{1}\leq-\alpha(x)\big{)}$
$\displaystyle\quad+{\mathbb{P}}\big{(}B,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x)\big{)}.$

By Fatou’s lemma

(6.16)
$\displaystyle\liminf_{x\to\infty}\frac{{\mathbb{P}}\big{(}B,\,-\alpha(x)<\xi_{1}\leq\alpha(x)\big{)}}{F(x+\Delta)}=\liminf_{x\to\infty}\int_{(-\alpha(x),\alpha(x)]}\frac{F(x+\Delta-y)}{F(x+\Delta)}F(dy)\geq
1.$

Hence from (6.15) and (6.16) and the definition of ${\mathcal{S}}_{\Delta}$,
we obtain (3.6).
$(\mathrm{ii})\Rightarrow(\mathrm{iii})$. This part is trivial since the
condition $F\in{\mathcal{L}}_{\Delta}$ implies the existence of a function
$\alpha$ w.r.t. which $F(x+\Delta)$ is $\alpha$-insensitive.
$(\mathrm{iii})\Rightarrow(\mathrm{i})$. We again use the decomposition (6.15)
for $B$ as defined above. Then

$\displaystyle{\mathbb{P}}\big{(}B,-\alpha(x)<\xi_{1}\leq\alpha(x)\big{)}$
$\displaystyle=\int_{(-\alpha(x),\alpha(x)]}F(x-y+\Delta)F(dy)$
$\displaystyle\sim F(x+\Delta)\int_{(-\alpha(x),\alpha(x)]}F(dy)$
$\displaystyle\sim F(x+\Delta)$

and so (3.5) and (3.6) together with (6.15) imply ${\mathcal{S}}_{\Delta}$ of
$F$.

If $F$ is a distribution on $\mathbb{R}_{+}$, then
${\mathbb{P}}\big{(}B,\,\xi_{1}\leq-\alpha(x)\big{)}=0$ and (3.5) is
automatically satisfied. For the last assertion, we observe that

$\displaystyle{\mathbb{P}}(\xi_{1}+\xi_{2}\in
x+\Delta,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x))$
$\displaystyle={\mathbb{E}}\big{[}{\mathbb{E}}[{\bf
1}_{\\{\alpha(x)\vee(x-\xi_{2})<\xi_{1}\leq x+c-\xi_{2}\\}}\mid\xi_{2}]{\bf
1}_{\\{\xi_{2}>\alpha(x)\\}}\big{]}$
$\displaystyle=\int_{(\alpha(x),x-\alpha(x)]}F(x+\Delta-y)F(dy)+\int_{(x-\alpha(x),x+c-\alpha(x)]}\big{(}F(x+c-y)-F(\alpha(x))\big{)}F(dy).$

The second integral has the bound

$F(\alpha(x)+\Delta)F(x+\Delta-\alpha(x))=o(F(x+\Delta)),$

where we notice that $F(x+\Delta)$ is $\alpha$-insensitive and
$F(\alpha(x)+\Delta)\to 0$ by the integrability of $F$. Thus (3.7) implies
(3.6). ∎

###### Proof of Proposition 3.2.

Let $z\in\mathbb{R}$. For sufficiently large $d>0$

$\displaystyle H(x+\Delta+z)$
$\displaystyle\geq\int_{[z,x+z-d]}F(x+\Delta+z-y)G(dy)$
$\displaystyle\geq\inf_{y\in[d,x]}F(y+\Delta)G([z,x+z-d]).$

Since $F\in{\mathcal{A}}_{\Delta}$,

$\displaystyle\liminf_{x\to\infty}\frac{H(x+\Delta)}{F(x+\Delta)}$
$\displaystyle=\lim_{z\to-\infty}\liminf_{x\to\infty}\frac{H(x+\Delta)}{H(x+\Delta+z)}\frac{H(x+\Delta+z)}{\inf_{y\in[d,x]}F(y+\Delta)}\frac{\inf_{y\in[d,x]}F(y+\Delta)}{F(x+\Delta)}$
$\displaystyle\geq\lim_{z\to-\infty}G([z,\infty))=1,$

which implies

(6.17) $\displaystyle\limsup_{x\to\infty}\frac{F(x+\Delta)}{H(x+\Delta)}\leq
1.$

We will prove

(6.18) $\displaystyle\liminf_{x\to\infty}\frac{F(x+\Delta)}{H(x+\Delta)}\geq
1.$

Let $\xi_{1}$ and $\xi_{2}$ be r.v.’s whose distribution functions are
respectively given by $G$ and $F$, and define the event
$B=\\{\xi_{1}+\xi_{2}\in x+\Delta\\}$. We take a function $\alpha(x)$ such
that $\alpha(x)\to\infty$, $0<\alpha(x)<x/2$ and $H$ is $\alpha$-insensitive.
We study

$\displaystyle{\mathbb{P}}(B)$
$\displaystyle={\mathbb{P}}\big{(}B,\,\xi_{1}\leq\alpha(x)\big{)}+{\mathbb{P}}\big{(}B,\,\xi_{2}\leq\alpha(x)\big{)}+{\mathbb{P}}\big{(}B,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x)\big{)}.$

Write

$\displaystyle{\mathbb{P}}(B,\,\xi_{1}\leq\alpha(x))$
$\displaystyle=\Big{(}\int_{(-\infty,-\alpha(x))}+\int_{[-\alpha(x),\alpha(x)]}\Big{)}F(x+\Delta-y)G(dy)$
$\displaystyle=:I_{1}(x)+I_{2}(x).$

Since $F\in{\mathcal{A}}_{\Delta}$ and (6.17), we have

(6.19) $\displaystyle\frac{I_{1}(x)}{H(x+\Delta)}\leq\frac{\sup_{y\leq
0}F(x+\Delta-y)}{F(x+\Delta)}\frac{F(x+\Delta)}{H(x+\Delta)}G(-\alpha(x))\to
0$

as $x\to\infty$. For the second integral, write

$\displaystyle\frac{I_{2}(x)}{H(x+\Delta)}\leq\frac{H(x+\Delta-\alpha(x))}{H(x+\Delta)}\frac{F(x+\Delta-\alpha(x))}{H(x+\Delta-\alpha(x))}\frac{\sup_{y\leq\alpha(x)}F(x+\Delta-y)}{F(x+\Delta-\alpha(x))}G([-\alpha(x),\alpha(x)]).$

The terms other than $F(x+\Delta-\alpha(x))/H(x+\Delta-\alpha(x))$ converge to
$1$, and we obtain

(6.20)
$\displaystyle\liminf_{x\to\infty}\frac{I_{2}(x)}{H(x+\Delta)}\leq\liminf_{x\to\infty}\frac{F(x+\Delta)}{H(x+\Delta)}.$

When $H\in{\mathcal{D}}_{\Delta}$, we observe that

$\displaystyle\frac{{\mathbb{P}}(B,\xi_{2}\leq\alpha(x))}{H(x+\Delta)}$
$\displaystyle=\int_{(-\infty,\alpha(x)]}\frac{G(x+\Delta-y)}{H(x+\Delta-y)}\frac{H(x+\Delta-y)}{H(x+\Delta)}F(dy)$
$\displaystyle\leq
K\sup_{y\leq\alpha(x)}\frac{G(x+\Delta-y)}{H(x+\Delta-y)}\frac{H(x+\Delta-\alpha(x))}{H(x+\Delta)}F(\alpha(x)),$

where $K>0$ is a constant of ${\mathcal{D}}_{\Delta}$. Since
$G(x+\Delta)=o(H(x+\Delta))$ and $H\in{\mathcal{L}}_{\Delta}$, we obtain

(6.21)
$\displaystyle\limsup_{x\to\infty}{\mathbb{P}}(B,\xi_{2}\leq\alpha(x))/H(x+\Delta)=0.$

When $G(x+\Delta)=o(F(x+\Delta))$, we notice that for sufficiently large $x>0$

$\displaystyle\frac{{\mathbb{P}}(B,\xi_{2}\leq\alpha(x))}{H(x+\Delta)}$
$\displaystyle=\int_{(-\infty,\alpha(x)]}\frac{G(x+\Delta-y)}{F(x+\Delta-y)}\frac{F(x+\Delta-y)}{F(x+\Delta-\alpha(x))}\frac{F(x+\Delta-\alpha(x))}{H(x+\Delta)}F(dy)$
$\displaystyle\leq
c\sup_{y\leq\alpha(x)}\frac{G(x+\Delta-y)}{F(x+\Delta-y)}F(\alpha(x)),$

where we notice that $F\in{\mathcal{A}}_{\Delta}$,
$H\in{\mathcal{L}}_{\Delta}$ and (6.17). Thus (6.21) again holds in this case.

Finally, write

$\displaystyle{\mathbb{P}}(B,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x))$
$\displaystyle={\mathbb{E}}\big{[}{\mathbb{E}}[{\bf
1}_{\\{\alpha(x)<\xi_{1}\leq x+c-\xi_{2}\\}}-{\bf 1}_{\\{\alpha(x)<\xi_{1}\leq
x-\xi_{2}\\}}\mid\xi_{2}]{\bf 1}_{\\{\xi_{2}>\alpha(x)\\}}\big{]}$
$\displaystyle=\int_{(\alpha(x),x-\alpha(x)]}G(x+\Delta-y)F(dy)+\int_{(x-\alpha(x),x+c-\alpha(x)]}\big{(}G(x+c-y)-G(\alpha(x))\big{)}F(dy)$
$\displaystyle\leq\int_{(\alpha(x),x-\alpha(x)]}G(x+\Delta-y)F(dy)+G(\alpha(x)+\Delta)F(x+\Delta-\alpha(x)).$

Since $H\in{\mathcal{L}}_{\Delta}$, (6.17) and $G$ is integrable, the last
term in the last line is $o(H(x+\Delta))$. For the integral in the last line,
let $\xi_{3}$ be the r.v. with distribution $H$ independent of
$(\xi_{1},\xi_{2})$, and we use the inequalities,

$\displaystyle\int_{(\alpha(x),x-\alpha(x)]}H(x+\Delta-y)F(dy)$
$\displaystyle\leq{\mathbb{P}}(\xi_{2}+\xi_{3}\in
x+\Delta,\,\xi_{2}>\alpha(x),\xi_{3}>\alpha(x))$
$\displaystyle\leq\int_{(\alpha(x),x-\alpha(x)]}F(x+\Delta-y)H(dy)+F(\alpha(x)+\Delta)H(x+\Delta-\alpha(x)).$

Then the integral is bounded as

$\displaystyle\int_{(\alpha(x),x-\alpha(x)]}\frac{G(x+\Delta-y)}{H(x+\Delta-y)}H(x+\Delta-y)F(dy)$
$\displaystyle\leq\sup_{\alpha(x)<y\leq
x-\alpha(x)}\frac{G(x+\Delta-y)}{H(x+\Delta-y)}\Big{(}\int_{(\alpha(x),x-\alpha(x)]}H(x+\Delta-y)H(dy)$
$\displaystyle\hskip
156.49014pt+F(\alpha(x)+\Delta)H(x+\Delta-\alpha(x))\Big{)},$

where in the final step we use the inequalities above and (6.17). The second
quantity in the last line is $o(H(x+\Delta))$ by the integrability of $F$. Now
since Lemma 3.1 $\mathrm{(}\mathrm{ii})$ together with
$H\in{\mathcal{S}}_{\Delta}$ implies (3.7), we obtain
${\mathbb{P}}(B,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x))/H(x+\Delta)\to 0$ as
$x\to\infty$. Now this together with (6.19), (6.20) and (6.21) yields the
desired result. ∎

###### Proof of Proposition 3.3.

Let $F\in{\mathcal{A}}_{\Delta}$. Then, $F$ satisfies Assumptions A, B, and C
of [19, Theorem 4.1], and the result follows. ∎

###### Proof of Proposition 3.4.

Recall that $F_{+}$ is the conditional distribution on $\mathbb{R}_{+}$ and
that $F^{+}(dx)=\overline{F}(0-)F_{+}(dx)+F(0-)\delta_{0}(dx)$. We further let
$F_{-}$ be the conditional distribution on $(-\infty,0)$, so that
$F(dx)=\overline{F}(0-)F_{+}(dx)+F(0-)F_{-}(dx)$. Our strategy is to show
$(F^{+})^{\ast N}(x+\Delta)\sim F^{\ast N}(x+\Delta)$ and apply Propositions
3.3 and 2.3. The key is Lemma 6.1 below, which we will use several times.

We start by the following inequality. For an integer $k:1\leq k\leq N$, there
exist $c_{k}>0$ and $a_{k}\in\mathbb{R}$ such that

(6.22) $\displaystyle F_{+}^{\ast k}(x+\Delta)\leq
c_{k}\overline{F}(0-)^{-N}(F^{\ast N}(x+a_{k}+\Delta)+F^{\ast
N}(x+a_{k}+c+\Delta)).$

If $F_{+}=0$, this is trivial and otherwise we put
$F_{+}^{\ast(N-k)}(a_{k}+\Delta)=c_{k}^{-1}>0$ for some $a_{k}\in\mathbb{R}$.
Then, (6.22) follows from

$F_{+}^{\ast(N-k)}(a_{k}+\Delta)F_{+}^{\ast k}(x+\Delta)\leq F_{+}^{\ast
N}(x+a_{k}+2\Delta)\leq\overline{F}(0-)^{-N}F^{\ast N}(x+a_{k}+2\Delta).$

######  Lemma 6.1.

Suppose that $F^{\ast N}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$
and $F\in{\mathcal{A}}_{\Delta}$. Then, for any integer $k:1\leq k\leq N$ we
can take a constant $A>0$ and choose functions $\varepsilon_{k}(A,x)\geq 0$
independent of $y$ such that for $x>kA$ and $y>0$

$F_{+}^{\ast k}(x+\Delta)-F_{+}^{\ast k}(x+y+\Delta)\geq-\varepsilon_{k}(A,x)$

and moreover,

$\lim_{A\to\infty}\limsup_{x\to\infty}\frac{\varepsilon_{k}(A,x)}{F^{\ast
n}(x+\Delta)}=0.$

###### Proof.

The proof is by induction. For $k=1$ take sufficiently large $A>0$ such that
for $x>A$ and $y>0$

$F_{+}(x+\Delta)-F_{+}(x+y+\Delta)\geq-\varepsilon(A)F_{+}(x+\Delta),$

where $\varepsilon(A)\geq 0$ and $\lim_{A\to\infty}\varepsilon(A)=0$, which is
possible by $F_{+}\in{\mathcal{A}}_{\Delta}$. By (6.22) and $F^{\ast
N}\in{\mathcal{L}}_{\Delta}$

$\limsup_{x\to\infty}\frac{F_{+}(x+\Delta)}{F^{\ast N}(x+\Delta)}<\infty.$

Thus, the lemma holds with $k=1$. Assume that the assertions hold with $1\leq
k\leq n\leq N-1$ and we show that also with $k=n+1$. Let $X_{1},X_{2},X_{3}$
and $X_{4}$ be independent r.v.’s with the distribution $F_{+},F_{+}^{\ast
n},F_{+}^{\ast(N-1)}$ and $F_{+}^{\ast(N-n)}$, respectively. Let
${\mathbb{P}}(X_{3}+X_{4}\in a+\Delta)=C>0$ for some $a\in\mathbb{R}_{+}$. We
evaluate
$F_{+}^{\ast(n+1)}(x+\Delta)-F_{+}^{\ast(n+1)}(x+y+\Delta),\,x>(n+1)A,\ y>0,$
and obtain a convenient representation. We use the decomposition

$\displaystyle F_{+}^{\ast(n+1)}(x+c)$
$\displaystyle=\big{(}\int_{[0,A]}+\int_{(A,x+c-A]}\big{)}F_{+}^{\ast
n}(x+c-u)F_{+}(du)$
$\displaystyle\quad+\int_{[0,A]}(F_{+}(x+c-u)-F_{+}(x+c-A))F_{+}^{\ast
n}(dy),$

where we use Fubini as in (6.14), and derive

(6.23) $\displaystyle F_{+}^{\ast(n+1)}(x+\Delta)$
$\displaystyle=\int_{[0,A]}F_{+}^{\ast
n}(x+\Delta-u)F_{+}(du)+\int_{[0,A]}F_{+}(x+\Delta-u)F_{+}^{\ast n}(du)$
$\displaystyle\quad+\int_{(A,x-A]}F_{+}^{\ast n}(x+\Delta-u)F_{+}(du)$
$\displaystyle\quad+\int_{(x-A,x+c-A]}(F_{+}^{\ast n}(x+c-u)-F_{+}^{\ast
n}(A))F_{+}(du),$

where the last two terms are equal to

(6.24) $\displaystyle{\mathbb{P}}(X_{1}+X_{2}\in x+\Delta,\,X_{1}>A,X_{2}>A)$

by the same calculation as for
${\mathbb{P}}(B,\,\xi_{1}>\alpha(x),\,\xi_{2}>\alpha(x))$ in the proof of
Proposition 3.2. At there $(\alpha(x),G,F)$ are replaced with $(A,F_{+}^{\ast
n},F_{+})$.

Now applying (6.23) for $x>(n+1)A$ we have

$\displaystyle F_{+}^{\ast(n+1)}(x+\Delta)-F_{+}^{\ast(n+1)}(x+y+\Delta)$
$\displaystyle=\int_{[0,A]}(F_{+}^{\ast n}(x-u+\Delta)-F_{+}^{\ast
n}(x-u+y+\Delta))F_{+}(du)$
$\displaystyle\quad+\int_{[0,A]}(F_{+}(x-u+\Delta)-F_{+}(x-u+y+\Delta))F_{+}^{\ast
n}(du)$ $\displaystyle\quad+{\mathbb{P}}(X_{1}+X_{2}\in
x+\Delta,X_{1}>A,X_{2}>A)-{\mathbb{P}}(X_{1}+X_{2}\in
x+y+\Delta,X_{1}>A,X_{2}>A).$

Here, since $X_{1}+X_{2}$ and $X_{3}+X_{4}$ are independent, the last quantity
is bounded as

$\displaystyle{\mathbb{P}}(X_{1}+X_{2}\in x+y+\Delta,X_{1}>A,X_{2}>A)$
$\displaystyle\leq C^{-1}{\mathbb{P}}(X_{1}+X_{2}\in
x+y+\Delta,X_{1}>A,X_{2}>A,X_{3}+X_{4}\in a+\Delta)$ $\displaystyle\leq
C^{-1}{\mathbb{P}}(X_{1}+X_{3}+X_{2}+X_{4}\in
x+a^{\prime}+2\Delta,X_{1}+X_{3}>A,X_{2}+X_{4}>A)(=:I_{3}),$

where $a^{\prime}=a+y$ and thus

$\displaystyle F_{+}^{\ast(n+1)}(x+\Delta)-F_{+}^{\ast(n+1)}(x+y+\Delta)$
$\displaystyle\geq-\int_{[0,A]}\varepsilon_{n}(A,x-u)F_{+}(du)-\int_{[0,A]}\varepsilon_{1}(A,x-u)F_{+}(du)-I_{3}$
$\displaystyle=:-I_{1}-I_{2}-I_{3}.$

Notice that $F^{\ast N}(x-u+\Delta)/F^{\ast N}(x+\Delta)$ converges to $1$
uniformly in $u\in[0,A]$ as $x\to\infty$, and thus by the induction hypothesis

$\lim_{A\to\infty}\limsup_{x\to\infty}\frac{I_{1}}{F^{\ast
N}(x+\Delta)}\leq\lim_{A\to\infty}\limsup_{y\to\infty}\frac{\varepsilon_{n}(A,y)}{F^{\ast
N}(y+\Delta)}=0$

and

$\lim_{A\to\infty}\limsup_{x\to\infty}\frac{I_{2}}{F^{\ast
N}(x+\Delta)}\leq\lim_{A\to\infty}\limsup_{y\to\infty}\frac{\varepsilon_{1}(A,y)}{F^{\ast
N}(y+\Delta)}=0.$

Let $Y_{1}$ and $Y_{2}$ be independent and identical r.v.’s with the
distribution $F^{\ast N}$. By the similar identity as for (6.24) and
$F_{+}^{\ast N}(dx)\leq\overline{F}(0-)^{-N}F^{\ast N}(dx)$ , we obtain

$\displaystyle\frac{I_{3}}{F^{\ast N}(x+a^{\prime}+\Delta)}$
$\displaystyle\leq\overline{F}(0-)^{-2N}\int_{[A,x+a^{\prime}-A]}\frac{F^{\ast
N}(x+a^{\prime}+\Delta-u)}{F^{\ast N}(x+a^{\prime}+\Delta)}F^{\ast N}(du)$
$\displaystyle\quad+\overline{F}(0-)^{-2N}\int_{[A,x+a^{\prime}-A]}\frac{F^{\ast
N}(x+a^{\prime}+c+\Delta-u)}{F^{\ast N}(x+a^{\prime}+\Delta)}F^{\ast N}(du)$
$\displaystyle\quad+\overline{F}(0-)^{-2N}\frac{F^{\ast N}(A+2\Delta)F^{\ast
N}(x+a^{\prime}+2\Delta-A)}{F^{\ast N}(x+a^{\prime}+\Delta)}.$

Since $F^{\ast N}\in{\mathcal{S}}_{\Delta}$, by (3.6) of Lemma 3.1 the two
integrals converge to $0$ as $x\to\infty$ and then $A\to\infty$ uniformly in
$a^{\prime}>a$. By the integrability condition, the last quantity also
converges to $0$ as $x\to\infty$ and then $A\to\infty$ uniformly in
$a^{\prime}>a$. Thus by $F^{\ast N}\in{\mathcal{D}}_{\Delta}$ $I_{3}$ is
bounded by $I_{3}^{\prime}$ independent of $y>0$ and

$\lim_{A\to\infty}\limsup_{x\to\infty}\frac{I_{3}^{\prime}}{F^{\ast
N}(x+\Delta)}=0.$

Now by letting $\varepsilon_{n+1}(A,x)=I_{1}+I_{2}+I_{3}^{\prime}$, we obtain
the assertion for $k=n+1$. ∎

Now we proceed to the proof and for convenience we write $q=\overline{F}(0-)$.
For $x>NA$ we observe that

$\displaystyle(F^{+})^{\ast N}(x+\Delta)-F^{\ast N}(x+\Delta)$
$\displaystyle=\sum_{k=1}^{N-1}\binom{N}{k}q^{k}(1-q)^{N-k}\int_{(-\infty,0)}\big{(}F_{+}^{\ast
k}(x+\Delta)-F_{+}^{\ast k}(x-y+\Delta)\big{)}F_{-}^{\ast(N-k)}(dy)$
$\displaystyle\geq-\sum_{k=1}^{N-1}\binom{N}{k}q^{k}(1-q)^{N-k}\varepsilon_{k}(A,x).$

Thus, we see that

(6.25) $\displaystyle\liminf_{x\to\infty}\frac{(F^{+})^{\ast
N}(x+\Delta)}{F^{\ast N}(x+\Delta)}\geq
1-\sum_{k=1}^{N-1}\binom{N}{k}q^{k}(1-q)^{N-k}\lim_{A\to\infty}\limsup_{x\to\infty}\frac{\varepsilon_{k}(A,x)}{F^{\ast
N}(x+\Delta)}=1.$

Moreover, for $x>(N+1)A$

$\displaystyle(F^{+})^{\ast N}(x+\Delta)-F^{\ast N}(x-A+\Delta)$
$\displaystyle=\sum_{k=1}^{N}\binom{N}{k}q^{k}(1-q)^{N-k}\int_{(-\infty,0)}\big{(}F_{+}^{\ast
k}(x+\Delta)-F_{+}^{\ast k}(x-A-y+\Delta)\big{)}F_{-}^{\ast(N-k)}(dy)$
$\displaystyle\leq J_{1}+J_{2},$

where

$J_{1}=\sum_{k=1}^{N-1}\binom{N}{k}q^{k}(1-q)^{N-k}\int_{(-A,0)}\varepsilon_{k}(A,x-A-y)F_{-}^{\ast(N-k)}(dy)+q^{N}\varepsilon_{N}(A,x-A)$

and

$J_{2}=\sum_{k=1}^{N-1}\binom{N}{k}q^{k}(1-q)^{N-k}\int_{(-\infty,-A]}F_{+}^{\ast
k}(x+\Delta)F_{-}^{\ast(N-k)}(dy).$

Note that $F^{\ast N}\in{\mathcal{S}}_{\Delta}$ and that $F^{\ast
N}(x-A+\Delta)/F^{\ast N}(x+\Delta)$ converges to $1$ as $x\to\infty$. We see
from Lemma 6.1 that

(6.26) $\displaystyle\lim_{A\to\infty}\limsup_{x\to\infty}\frac{J_{1}}{F^{\ast
N}(x+\Delta)}=0.$

We obtain from (6.22) and $F^{\ast N}\in{\mathcal{L}}_{\Delta}$ that

(6.27) $\displaystyle\lim_{A\to\infty}\limsup_{x\to\infty}\frac{J_{2}}{F^{\ast
N}(x+\Delta)}=0.$

Now in view of (6.25), (6.26) and (6.27) we have

$\limsup_{x\to\infty}\frac{(F^{+})^{\ast N}(x+\Delta)}{F^{\ast
N}(x+\Delta)}\leq\lim_{x\to\infty}\frac{F^{\ast N}(x-A+\Delta)}{F^{\ast
N}(x+\Delta)}+\lim_{A\to\infty}\limsup_{x\to\infty}\frac{J_{1}+J_{2}}{F^{\ast
N}(x+\Delta)}=1$

and thus $(F^{+})^{\ast N}(x+\Delta)\sim F^{\ast N}(x+\Delta)$. By Proposition
3.7 this implies $(F^{+})^{\ast N}\in{\mathcal{S}}_{\Delta}$, so that
$F^{+}\in{\mathcal{S}}_{\Delta}$ follows from Propositions 3.3. Since
$F\in{\mathcal{A}}_{\Delta}$ we see from Proposition 2.3 that
$F\in{\mathcal{S}}_{\Delta}$. ∎

###### Proof of Corollary 3.5.

Let $F(dx):=f(x)dx$ and $f\in{\mathcal{A}}$. Then we have
$F\in{\mathcal{A}}_{\Delta}$. Since $f^{*N}\in{\mathcal{S}}\cap{\mathcal{D}}$
implies $F^{*N}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$, we see
from Proposition 3.4 that $F\in{\mathcal{S}}_{\Delta}$. Hence we find from
Proposition 2.4 that $f\in{\mathcal{S}}$. ∎

###### Proof of Proposition 3.7.

Choose a function $\alpha(x)$ such that $\alpha(x)<x/2$, $\alpha(x)\to\infty$
as $x\to\infty$ and the function $G(x+\Delta)$ is $\alpha$-insensitive. Let
$\zeta_{1},\zeta_{2}$ be independent identical r.v.’s with the distribution
$G$. By Lemma 3.1, it suffices to prove (3.5) and (3.6) with $\xi_{1},\xi_{2}$
and $F$ there replaced with $\zeta_{1},\zeta_{2}$ and $G$ here. The proof of
(3.6) is exactly the same as that of [9, Proposition 4.22] and is omitted. We
prove (3.6). By the assumption, we have

(6.28) $\displaystyle\begin{split}{\mathbb{P}}\big{(}\zeta_{1}+\zeta_{2}\in
x+\Delta,\zeta_{1}\leq-\alpha(x)\big{)}&=\int_{(-\infty,-\alpha(x)]}G(x-y+\Delta)G(dy)\\\
&\leq c\int_{(-\infty,-\alpha(x)]}F(x-y+\Delta)G(dy)\\\
&=o(F(x+\Delta))=o(G(x+\Delta)).\end{split}$

Thus we have proved the assertion. The proof of the final part is done by
noticing that $F\in{\mathcal{S}}_{\Delta}$ implies
$F_{+}\in{\mathcal{S}}_{\Delta}$ and applying [9, Theorem 4.22]. ∎

###### Proof of Proposition 3.8.

Let $\zeta_{1}$ and $\zeta_{2}$ be independent r.v.’s with distributions
$G_{1}$ and $G_{2}$, respectively. Let $\alpha$ be a function as in Lemma 3.1
and $F(x+\Delta)$ is $\alpha$-insensitive. Define the event
$B=\\{\zeta_{1}+\zeta_{2}\in x+\Delta\\}$. Then

$\displaystyle{\mathbb{P}}(B)=$
$\displaystyle{\mathbb{P}}\big{(}B,\,-\alpha(x)\leq\zeta_{1}\leq\alpha(x)\big{)}+{\mathbb{P}}\big{(}B,\,-\alpha(x)\leq\zeta_{2}\leq\alpha(x)\big{)}$
$\displaystyle+{\mathbb{P}}\big{(}B,\,\zeta_{1}<-\alpha(x)\big{)}+{\mathbb{P}}\big{(}B,\,\zeta_{2}<-\alpha(x)\big{)}+{\mathbb{P}}\big{(}B,\,\zeta_{1}>\alpha(x),\,\zeta_{2}>\alpha(x)\big{)}.$

By the dominated convergence theorem, we have

$\displaystyle\lim_{x\to\infty}\frac{{\mathbb{P}}\big{(}B,\,-\alpha(x)\leq\zeta_{1}\leq\alpha(x)\big{)}}{F(x+\Delta)}$
$\displaystyle=\int_{[-\alpha(x),\alpha(x)]}\lim_{x\to\infty}\frac{G_{2}(x-y+\Delta)}{F(x-y+\Delta)}\frac{F(x-y+\Delta)}{F(x+\Delta)}G_{1}(dy)=c_{2}$

and similarly

$\displaystyle\lim_{x\to\infty}\frac{{\mathbb{P}}\big{(}B,\,-\alpha(x)\leq\zeta_{2}\leq\alpha(x)\big{)}}{F(x+\Delta)}=c_{1}.$

By the same argument as in (6.28), we obtain that

${\mathbb{P}}\big{(}B,\,\zeta_{i}<-\alpha(x)\big{)}=o(F(x+\Delta))\quad
i=1,2.$

By the same argument as that in the proof of [9, Proposition 4.22], we see
that

${\mathbb{P}}\big{(}B,\,\zeta_{1}>\alpha(x),\,\zeta_{2}>\alpha(x)\big{)}=o(F(x+\Delta)).$

Thus, (3.8) holds. The final assertion follows from Proposition 3.8. ∎

###### Proof of Proposition 3.10.

For a sequence of iid r.v.’s $\\{\xi_{n}\\}$ with distribution $F$, we put
$S_{n}=\sum_{k=1}^{n}\xi_{k}$. For $x_{0}\geq 0$ and $k\geq 1$ we write

$A_{k}:=A_{k}(x_{0})=\sup_{x>x_{0}}\frac{F^{*k}(x+\Delta)}{F(x+\Delta)}.$

Define the event $B=\\{\xi_{1}+\xi_{2}\in x+\Delta\\}$. Let $\varepsilon>0$ be
any fixed constant.

By Lemma 3.1, there exist $x_{0}>0$ and $C>0$ such that $x_{0}>2C$ and, for
any $x>x_{0}$,

${\mathbb{P}}(B,\xi_{2}\leq
C)+{\mathbb{P}}(B,\xi_{1}>C,\xi_{2}>C)\leq(1+\varepsilon/2)F(x+\Delta).$

Since we have

${\mathbb{P}}(B,\xi_{1}\leq C)\leq{\mathbb{P}}(B,\xi_{2}>x-C),$

we observe that for any $x>x_{0}$,

$\displaystyle{\mathbb{P}}(B,\xi_{2}\leq x-x_{0})$
$\displaystyle\leq{\mathbb{P}}(B,\xi_{2}\leq x-C)$
$\displaystyle={\mathbb{P}}(B)-{\mathbb{P}}(B,\xi_{2}>x-C)$
$\displaystyle\leq{\mathbb{P}}(B)-{\mathbb{P}}(B,\xi_{1}\leq C)$
$\displaystyle={\mathbb{P}}(B,\xi_{2}\leq
C)+{\mathbb{P}}(B,\xi_{1}>C,\xi_{2}>C),$

so that we can choose $x_{0}>0$ such that for any $x>x_{0}$,

(6.29) ${\mathbb{P}}(B,\xi_{2}\leq x-x_{0})\leq(1+\varepsilon/2)F(x+\Delta).$

For any $n>1$ and $x>x_{0}$, we write

$\displaystyle{\mathbb{P}}(S_{n}\in x+\Delta)$
$\displaystyle={\mathbb{P}}(S_{n}\in x+\Delta,\xi_{n}\leq
x-x_{0})+{\mathbb{P}}(S_{n}\in x+\Delta,\xi_{n}>x-x_{0})$
$\displaystyle=:P_{1}(x)+P_{2}(x),$

where, by the choice (6.29) of $x_{0}$ and by the definition of $A_{n-1}$,

$\displaystyle P_{1}(x)$
$\displaystyle=\int_{(-\infty,x-x_{0}]}{\mathbb{P}}(S_{n-1}\in
x-y+\Delta)F(dy)$ $\displaystyle\leq
A_{n-1}\int_{(-\infty,x-x_{0}]}F(x-y+\Delta)F(dy)$
$\displaystyle=A_{n-1}{\mathbb{P}}(B,\xi_{2}\leq x-x_{0})$ (6.30)
$\displaystyle\leq A_{n-1}(1+\varepsilon/2)F(x+\Delta).$

Moreover,

$\displaystyle P_{2}(x)$
$\displaystyle=\int_{(-\infty,x_{0}+c]}{\mathbb{P}}(\xi_{n}\in
x-y+\Delta,\xi_{n}>x-x_{0}){\mathbb{P}}(S_{n-1}\in dy)$
$\displaystyle\leq\sup_{-\infty<t\leq x_{0}+c}F(x-t+\Delta).$

Thus, if $x>2x_{0}$, then

$P_{2}(x)\leq L_{1}F(x+\Delta),$

where

$L_{1}:=\sup_{-\infty<t\leq
x_{0}+c,\,y>2x_{0}}\frac{F(y-t+\Delta)}{F(y+\Delta)}.$

If $x_{0}<x\leq 2x_{0}$, then $P_{2}(x)\leq 1$ implies

$\frac{P_{2}(x)}{F(x+\Delta)}\leq\frac{1}{\inf_{x_{0}<x\leq
2x_{0}}F(x+\Delta)}=:L_{2}.$

Since $F\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$, $L_{1}$ and
$L_{2}$ are finite for $x_{0}$ sufficiently large. Put $L=\max(L_{1},L_{2})$
and then, for any $x>x_{0}$

(6.31) $P_{2}(x)\leq LF(x+\Delta).$

It follows from (6.30) and (6.31) that for $n>1$

$A_{n}\leq A_{n-1}(1+\varepsilon/2)+L.$

Now by induction, it follows that

$A_{n}\leq
A_{1}(1+\varepsilon/2)^{n-1}+L\sum_{k=0}^{n-2}(1+\varepsilon/2)^{k}\leq
Ln(1+\varepsilon/2)^{n-1},$

which implies the result. ∎

###### Proof of Proposition 3.11.

With the form $G$, we may write

(6.32) $\displaystyle 1=p_{G}F(x+\Delta)/H(x+\Delta)+(1-p_{G})F\ast
G_{1}(x+\Delta)/H(x+\Delta),$

and put

$\overline{C}:=\limsup_{x\to\infty}F(x+\Delta)/H(x+\Delta)\quad\text{and}\quad\underline{C}:=\liminf_{x\to\infty}F(x+\Delta)/H(x+\Delta),$

which are well-defined since $H(x+\Delta)\geq p_{G}F(x+\Delta)$.

Assume the first condition ($H\in{\mathcal{D}}_{\Delta}$). Take an insensitive
function $\alpha$ for $H(x+\Delta)$ and consider

$\displaystyle\frac{F\ast G_{1}(x+\Delta)}{H(x+\Delta)}$
$\displaystyle=\Big{(}\int_{(-\infty,-\alpha(x))}+\int_{[-\alpha(x),\alpha(x)]}+\int_{(\alpha(x),\infty)}\Big{)}\frac{F(x+\Delta-y)}{H(x+\Delta)}G_{1}(dy)$
$\displaystyle=:I_{1}(x)+I_{2}(x)+I_{3}(x).$

By Fatou’s lemma and $H(x+\Delta)\in{\mathcal{L}}$,

(6.33) $\displaystyle\liminf_{x\to\infty}I_{2}(x)$
$\displaystyle=\liminf_{x\to\infty}\int_{[-\alpha(x),\alpha(x)]}\frac{F(x+\Delta-y)}{H(x+\Delta-y)}\frac{H(x+\Delta-y)}{H(x+\Delta)}G_{1}(dy)$
$\displaystyle\geq\int_{\mathbb{R}}\liminf_{x\to\infty}\frac{F(x+\Delta-y)}{H(x+\Delta-y)}{\bf
1}_{\\{y\in[-\alpha(x),\alpha(x)]\\}}G_{1}(dy)\geq\underline{C},$

where we may take a continuous $\alpha(x)$ if needed. Again by Fatou’s lemma

(6.34)
$\displaystyle\limsup_{x\to\infty}I_{2}(x)\leq\int_{\mathbb{R}}\limsup_{x\to\infty}\frac{F(x+\Delta-y)}{H(x+\Delta-y)}{\bf
1}_{\\{y\in[-\alpha(x),\alpha(x)]\\}}G_{1}(dy)\leq\overline{C}.$

Let $\zeta_{1}$ and $\zeta_{2}$ be independent r.v.’s with the distributions
$F$ and $G_{1}$, respectively. Then we observe

$\displaystyle I_{3}(x)$
$\displaystyle=\frac{{\mathbb{P}}(\zeta_{1}+\zeta_{2}\in
x+\Delta,\zeta_{2}\geq\alpha(x))}{H(x+\Delta)}\leq\frac{{\mathbb{P}}(\zeta_{1}+\zeta_{2}\in
x+\Delta,\zeta_{1}\leq x+c-\alpha(x))}{H(x+\Delta)}.$

Since $H(dx)\geq p_{G}F(dx)$ and $G_{1}(x+\Delta)=o(H(x+\Delta))$ we have

(6.35) $\displaystyle\limsup_{x\to\infty}I_{3}(x)$
$\displaystyle\leq\limsup_{x\to\infty}\int_{(-\infty,x-\alpha(x)+c]}\frac{G_{1}(x+\Delta-y)}{H(x+\Delta-y)}\frac{H(x+\Delta-y)}{H(x+\Delta)}p_{G}^{-1}H(dy)$
$\displaystyle\leq\limsup_{x\to\infty}\frac{G_{1}(x+\Delta)}{H(x+\Delta)}c\limsup_{x\to\infty}\frac{H^{\ast
2}(x+\Delta)}{H(x+\Delta)}=0.$

By the al.d. property of $H(x+\Delta)$, it follows that

(6.36) $\displaystyle\limsup_{x\to\infty}I_{1}(x)$
$\displaystyle=\limsup_{x\to\infty}\int_{(-\infty,-\alpha(x))}\frac{F(x+\Delta-y)}{H(x+\Delta-y)}\frac{H(x+\Delta-y)}{H(x+\Delta)}G_{1}(dy)$
$\displaystyle\leq
c\limsup_{x\to\infty}\int_{(-\infty,-\alpha(x))}G_{1}(dy)=0.$

Collecting (6.33)-(6.36), we obtain

(6.37) $\displaystyle\liminf_{x\to\infty}\frac{F\ast
G_{1}(x+\Delta)}{H(x+\Delta)}\geq\underline{C}\quad\text{and}\quad\limsup_{x\to\infty}\frac{F\ast
G_{1}(x+\Delta)}{H(x+\Delta)}\leq\overline{C}.$

Now taking $\liminf_{x\to\infty}$ and $\limsup_{x\to\infty}$ on both sides of
(6.32), we have

(6.38) $\displaystyle p_{G}\overline{C}+(1-p_{G})\underline{C}\ \leq 1\ \leq\
p_{G}\underline{C}+(1-p_{G})\overline{C}$

and thus

$0\leq(1-2p_{G})(\overline{C}-\underline{C}).$

The assumption $p_{G}\in(2^{-1},1)$ implies $\overline{C}=\underline{C}$.
Moreover from (6.38), $\overline{C}=\underline{C}=1.$ Then noticing
$H(x+\Delta)\geq p_{G}F(x+\Delta)$ for all $x\in\mathbb{R}$, we have
$F\in{\mathcal{S}}_{\Delta}$ from Proposition 3.7.

For the proof under the second condition, we only indicate different points.
The quantities $\overline{C}$ and $\underline{C}$ are the same. We work on
another form

$\displaystyle\frac{F\ast G_{1}(x+\Delta)}{H(x+\Delta)}$
$\displaystyle=\int_{[0,\alpha(x)]}\frac{F(x+\Delta-y)}{H(x+\Delta)}G_{1}(dy)$
$\displaystyle\quad+\Big{(}\int_{(\alpha(x),x+c]}\frac{F(x+c-y)}{H(x+\Delta)}G_{1}(dy)-\int_{(\alpha(x),x]}\frac{F(x-y)}{H(x+\Delta)}G_{1}(dy)\Big{)}$
$\displaystyle=:J_{1}(x)+J_{2}(x).$

Similarly as before we obtain

$\liminf_{x\to\infty}J_{1}(x)\geq\underline{C}\quad\text{and}\quad\limsup_{x\to\infty}J_{1}(x)\leq\overline{C}.$

For the $J_{2}(x)$, we use the following identity

$\displaystyle\int_{(\alpha(x),x]}F(x-y)G_{1}(dy)$
$\displaystyle=\int_{\mathbb{R}_{+}\times(\alpha(x),x]}{\bf 1}_{\\{z\leq
x-y\\}}(F\times G_{1})(d(z,y))$
$\displaystyle=\int_{\mathbb{R}_{+}\times\mathbb{R}_{+}}{\bf
1}_{\\{\alpha(x)<y\leq x-z\\}}{\bf 1}_{\\{z<x-\alpha(x)\\}}(F\times
G_{1})(d(z,y))$
$\displaystyle=\int_{[0,x-\alpha(x)]}(G_{1}(x-z)-G_{1}(\alpha(x)))F(dz),$

so that

$\displaystyle J_{2}(x)$
$\displaystyle=\int_{[0,x-\alpha(x)]}\frac{G_{1}(x+\Delta-z)}{H(x+\Delta)}F(dz)+\int_{(x-\alpha(x),x+c-\alpha(x)]}\frac{G_{1}(x+c-z)-G_{1}(\alpha(x))}{H(x+\Delta)}F(dz)$
$\displaystyle\leq\int_{[0,x-\alpha(x)]}\frac{G_{1}(x+\Delta-z)}{H(x+\Delta-z)}\frac{H(x+\Delta-z)}{H(x+\Delta)}F(dz)+\frac{F(x+\Delta-\alpha(x))}{H(x+\Delta)}G_{1}(\alpha(x)+\Delta).$

Since $H(dx)\geq p_{G}F(dx)$ and $G(\alpha(x)+\Delta)\to 0$ as $x\to\infty$,

$J_{2}(x)\leq
o(1)\int_{[0,x-\alpha(x)]}\frac{H(x+\Delta-y)}{H(x+\Delta)}H(dy)+o(1)=o(1).$

The remaining proof is the same as before. ∎

###### Proof of Proposition 3.12.

By similarity, we only give the proof under that $H\in{\mathcal{D}}_{\Delta}$.
We assume non-degeneracy for $G$, since otherwise the proof is obvious. Let
$\lambda$ be the Poisson parameter of $G$. If $\lambda<\log 2$, then since
$e^{-\lambda}>2^{-1}$ the result is immediate from Proposition 2.14. If
$\lambda\geq\log 2$, we take an integer $n$ such that $\lambda/n<\log 2$, and
define a compound Poisson distribution

$G_{1/n}(dx)=e^{-\lambda/n}\delta_{0}(dx)+(1-e^{-\lambda/n})G_{1/n,1}(dx)$

such that $G=G_{1/n}^{\ast n}$. Notice that

$G(x+\Delta)=\big{(}e^{-\lambda/n}\delta_{0}+(1-e^{-\lambda/n})G_{1/n,1}\big{)}^{\ast
n}(x+\Delta)\geq ne^{-\lambda(n-1)/n}(1-e^{-\lambda/n})G_{1/n,1}(x+\Delta)$

and thus $G(x+\Delta)=o(H(x+\Delta))$ implies
$G_{1/n}(x+\Delta)=o(H(x+\Delta))$. Now we apply Proposition 2.14 to
$G_{1/n}\ast(G_{1/n}^{\ast(n-1)}\ast F)$ with $G_{1/n}$ be the negligible
part, and obtain that $G_{1/n}^{\ast(n-1)}\ast F\in{\mathcal{S}}_{\Delta}$ and
$H(x+\Delta)\sim G_{1/n}^{\ast(n-1)}\ast F(x+\Delta)$. Since
$H\in{\mathcal{D}}_{\Delta}$, so is $G_{1/n}^{\ast(n-1)}\ast F$ by Lemma 1.4.
We iterate this step until we reach $F\in{\mathcal{S}}_{\Delta}$ and
$H(x+\Delta)\sim F(x+\Delta)$. ∎

##  7\. Proofs for Section 4

Note that we abuse the notation $c$, which has two meanings. One is for $c$ of
$\Delta$ and the other is arbitrary positive constant whose values are not of
interest. This makes no confusion.

###### Proof of Theorem 4.1.

We give the proof for the general two-sided case first, and then we indicate
the difference for the positive-half case since the proof is quite similar.

$[\,(\mathrm{i})$ implies $(\mathrm{ii})$ and $(\mathrm{iii})\,]$ We choose
$c_{1}>0$ such that $\overline{G}(c_{1})=\Lambda_{1}<\log 2/\lambda$ and
define another compound Poisson by

(7.39)
$\displaystyle\widehat{\mu}_{1}(z)=\exp\Big{(}\lambda\Lambda_{1}\int_{(c_{1},\infty)}(e^{izx}-1)G_{1}(dy)\Big{)},\quad
G_{1}(x)=G(x)/\Lambda_{1},$

so that
$\mu=\mu_{1}\ast\mu_{2},\,\text{i.e.}\,\widehat{\mu}(z)=\widehat{\mu}_{1}(z)\widehat{\mu}_{2}(z),$
where

$\widehat{\mu}_{2}(z)=\exp\Big{(}\lambda\Lambda_{2}\int_{(-\infty,c_{1}]}(e^{izx}-1)G(dy)/\Lambda_{2}\Big{)},\quad\Lambda_{2}=G(c_{1}).$

Since $\mu_{2}$ has exponential moments of any orders ([13, Theorem 25.3]), we
have for any $\gamma>0$

(7.40) $\displaystyle e^{\gamma x}\mu_{2}(x+\Delta)\leq\int_{x}^{x+c}e^{\gamma
y}\mu_{2}(dy)\to 0\quad\text{as}\ x\to\infty,$

so that $\mu(x+\Delta)=o(e^{-\gamma x})$. Now by Proposition 3.12 with
$(H,G,F)$ there replaced with $(\mu,\mu_{2},\mu_{1})$ here, we obtain

(7.41)
$\displaystyle\mu_{1}\in{\mathcal{S}}_{\Delta}\quad\text{and}\quad\mu(x+\Delta)\sim\mu_{1}(x+\Delta).$

This implies $\mu_{1}\in{\mathcal{D}}_{\Delta}$ by Lemma 1.4.

We will see that $\mu_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$
implies $G_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. Write

(7.42)
$\displaystyle\mu_{1}(dx)=e^{-\lambda\Lambda_{1}}\delta_{0}(dx)+(1-e^{-\lambda\Lambda_{1}})\mu_{10}(dx),$

where

$\mu_{10}(dx)=(e^{\lambda\Lambda_{1}}-1)^{-1}\sum_{n=1}^{\infty}\big{(}(\lambda\Lambda_{1})^{n}/n!\big{)}G_{1}^{\ast
n}(dx).$

Consider the Laplace transform of $\mu_{10}$,

$L_{\mu_{10}}(z)=(e^{\lambda\Lambda_{1}L_{G_{1}}(z)}-1)/(e^{\lambda\Lambda_{1}}-1),$

which yields,

$\lambda\Lambda_{1}L_{G_{1}}(z)=-\sum_{n=1}^{\infty}n^{-1}(1-e^{\lambda\Lambda_{1}})^{n}L_{\mu_{10}}^{n}(z).$

Thus, by the inversion

(7.43)
$\displaystyle\lambda\Lambda_{1}G_{1}(x+\Delta)=-\sum_{n=1}^{\infty}n^{-1}(1-e^{\lambda\Lambda_{1}})^{n}\mu_{10}^{\ast
n}(x+\Delta).$

Take a sufficiently small $\varepsilon>0$ such that
$(e^{\lambda\Lambda_{1}}-1)(1+\varepsilon)<1$. Since
$\mu_{10}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$, by Proposition
3.10 there exists $C(\varepsilon)$ such that $\mu_{10}^{\ast n}(x+\Delta)\leq
C(\varepsilon)(1+\varepsilon)^{n}\mu_{01}(x+\Delta)$ for $x$ sufficiently
large. Applying the dominated convergence in (7.43), we obtain

(7.44)
$\displaystyle\lim_{x\to\infty}G_{1}(x+\Delta)/\mu_{10}(x+\Delta)=(1-e^{-\lambda\Lambda_{1}})/(\lambda\Lambda_{1}).$

Now correcting the results (7.39),(7.41),(7.42) and (7.44), we see

$\lim_{x\to\infty}\frac{\mu(x+\Delta)}{G(x+\Delta)}=\lim_{x\to\infty}\frac{\mu(x+\Delta)}{\mu_{10}(x+\Delta)}\frac{\mu_{10}(x+\Delta)}{G_{1}(x+\Delta)}\frac{G_{1}(x+\Delta)}{G(x+\Delta)}=\lambda.$

By Proposition 3.7 and Lemma 1.4, we obtain $G\in{\mathcal{S}}_{\Delta}$ and
$G\in{\mathcal{D}}_{\Delta}$ respectively. Thus $(\mathrm{ii})$ and
$(\mathrm{iii})$ are proved.

$[\,(\mathrm{iii})$ implies $(\mathrm{ii})\,]$ In view of (4.9), we write

$G^{\ast
2}(x+\Delta)=(2/\lambda^{2})\big{\\{}e^{\lambda}\mu(x+\Delta)-\sum_{n\neq
2}^{\infty}(\lambda^{n}/n!)G^{\ast n}(x+\Delta)\big{\\}}.$

Dividing this by $G(x+\Delta)$ and taking $\limsup_{x\to\infty}$ on both
sides, we have by Fatou’s lemma

$\displaystyle\limsup_{x\to\infty}\frac{G^{\ast 2}(x+\Delta)}{G(x+\Delta)}$
$\displaystyle\leq(2/\lambda^{2})\Big{\\{}\lambda e^{\lambda}-\sum_{n\neq
2}^{\infty}(\lambda^{n}/n!)\liminf_{x\to\infty}\frac{G^{\ast
n}(x+\Delta)}{G(x+\Delta)}\Big{\\}}$ $\displaystyle=(2/\lambda^{2})\\{\lambda
e^{\lambda}-\lambda\sum_{n\neq 1}^{\infty}\lambda^{n}/n!\\}=2,$

where we use [9, Corollary 4.19] together with $G\in{\mathcal{L}}_{\Delta}$.
Again by [9, Corollary 4.19] with $n=2$, we obtain $G^{\ast 2}(x+\Delta)\sim
2G(x+\Delta)$, which is $(\mathrm{ii})$.

$[\,(\mathrm{ii})$ implies $(\mathrm{iii})$ and $(\mathrm{i})\,]$ Notice that
Proposition 3.10 holds with $G$. Thus applying $G^{\ast
n}(x+\Delta)/G(x+\Delta)\to n$ as $x\to\infty$ and the dominated convergence
to

$\frac{\mu(x+\Delta)}{G(x+\Delta)}=e^{-\lambda}\sum_{n=1}^{\infty}(\lambda^{n}/n!)\frac{G^{\ast
n}(x+\Delta)}{G(x+\Delta)},$

we prove $(\mathrm{iii})$. Then a combination of $(\mathrm{ii})$ and
$(\mathrm{iii})$ yields $\mu\in{\mathcal{S}}_{\Delta}$ by Proposition 3.7.
Here $\mu\in{\mathcal{D}}_{\Delta}$ follows from that of $G$.

Next we proceed to the proof for the positive half case. As stated before we
only indicate the difference. For the part [$(\mathrm{i})$ implies
$(\mathrm{ii})$ and $(\mathrm{iii})$], we change $\widehat{\mu}_{2}(z)$ to

$\widehat{\mu}_{2}(z)=\exp\Big{(}\lambda\Lambda_{2}\int_{(0,c_{1}]}(e^{izy}-1)G(dy)/\Lambda_{2}\Big{)}\quad\text{with}\quad\Lambda_{2}=G(c_{1}).$

Then using Proposition 3.12 for positive-half distributions, we obtain
$\mu_{1}\in{\mathcal{S}}_{\Delta}$. On the way for proving that
$\mu_{1}\in{\mathcal{S}}_{\Delta}$ implies $G_{1}\in{\mathcal{S}}_{\Delta}$,
we replace Proposition 3.10 with [9, Theorem 4.25]. Now if Proposition 3.7 is
replaced by Theorem [9, Theorem 4.22] in the final part, the proof is
completed. Nothing should be changed for the part [$(\mathrm{iii})$ implies
$(\mathrm{ii})$]. For the part [$(\mathrm{ii})$ implies $(\mathrm{iii})$ and
$(\mathrm{i})$], it suffices to replace Propositions 3.10 and 3.7 respectively
with Theorems 4.25 and 4.22 in [9]. ∎

###### Proof of Theorem 4.6.

We decompose $\mu\in{\rm ID}(\mathbb{R})$ as
$\mu=\mu_{1}\ast\mu_{2},\,\text{i.e.}\,\widehat{\mu}(z)=\widehat{\mu}_{1}(z)\,\widehat{\mu}_{2}(z),$
where

$\widehat{\mu}_{2}(z)=\exp\Big{(}\int_{(-\infty,-1)}(e^{izx}-1)\nu(dy)\Big{)}$

is the ch.f. of a compound Poisson distribution. Observe that
$\mu_{2}(x+\Delta)=o(e^{-\gamma x})$ and

$\int_{\mathbb{R}}e^{-\gamma x}\mu_{1}(dx)<\infty$

for any $\gamma>0$. Suppose that $(\mathrm{i})$ holds. We obtain from
Proposition 3.12 that
$\mu_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and
$\mu(x+\Delta)\sim\mu_{1}(x+\Delta)$. Define the condition
$(\mathrm{iii})^{\prime}$ as

$(\mathrm{iii})^{\prime}\qquad\nu_{(1)}\in{\mathcal{L}}_{\Delta}\cap{\mathcal{D}}_{\Delta}\quad\text{and}\quad\mu_{1}(x+\Delta)\sim\nu(x+\Delta).$

We see from Theorem 1.2 of [20] that the condition
$\mu_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ is equivalent to
$(\mathrm{ii})$ and $(\mathrm{iii})^{\prime}$. Since
$\mu(x+\Delta)\sim\mu_{1}(x+\Delta)$ under
$\mu_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$,
$(\mathrm{iii})^{\prime}$ is equivalent to $(\mathrm{iii})$. Conversely,
suppose the condition
$\mu_{1}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ and then
$(\mathrm{i})$ follows from Proposition 3.8. ∎

###### Proof of Theorem 4.8.

$[\,(\mathrm{i})$ implies $(\mathrm{ii})$ and $(\mathrm{iii})\,]$ Define
$\mu_{2}\in{\rm ID}(\mathbb{R})$ such that $\mu_{2}$ satisfies
$\mu=\mu_{c_{1}}\ast\mu_{2}$. By the same reasoning given as (7.40),
$\mu_{2}(x+\Delta)=O(e^{-\gamma x})$. Then by Proposition 3.2 with
$(H,F,G)=(\mu,\mu_{c_{1}},\mu_{2})$ we obtain
$\mu_{c_{1}}\in{\mathcal{S}}_{\Delta}$. Let $\mu_{1}$ be a compound Poisson
with the Lévy measure $\nu_{(1)}\nu((1,\infty))$ and write
$\mu_{1}=\mu_{3}\ast\mu_{c_{1}}$ where $\mu_{3}$ is a compound Poisson with
the Lévy measure $\nu_{3}(dx)={\bf 1}_{\\{1<x\leq c_{1}\\}}\nu(dx)$. Similarly
as for $\mu_{2}$ we have $\mu_{3}(x+\Delta)=O(e^{-\gamma x})$ for any
$\gamma>0$, so that $\mu_{3}(x+\Delta)=o(\mu_{c_{1}}(x+\Delta))$. Then by
Proposition 3.8 with $(F,G_{1},G_{2})=(\mu_{c_{1}},\mu_{c_{1}},\mu_{3})$ and
$(c_{1},c_{2})=(1,0)$, we obtain $\mu_{1}\in{\mathcal{S}}_{\Delta}$ and
$\mu_{1}(x+\Delta)\sim\mu_{c_{1}}(x+\Delta)$. Now since $\mu_{1}$ is a
compound Poisson on $\mathbb{R}_{+}$, due to Theorem 4.1,
$\nu_{(1)}\in{\mathcal{S}}_{\Delta}$ and
$\mu_{1}(x+\Delta)\sim\nu_{(1)}(x+\Delta)\nu((1,\infty))$. Thus
$(\mathrm{ii})$ follows. In view of $\mu(x+\Delta)\sim\mu_{1}(x+\Delta)$ and
$\nu(x+\Delta)\sim\nu_{(1)}(x+\Delta)\nu((1,\infty))$, we have already proved
$(\mathrm{iii})$. Here $\nu_{(1)}\in{\mathcal{D}}_{\Delta}$ follows from
$\mu_{c_{1}}\in{\mathcal{D}}_{\Delta}$ together with the asymptotic
equivalence $\mu(x+\Delta)\sim\mu_{1}(x+\Delta)\sim\mu_{c_{1}}(x+\Delta)$ (cf.
Lemma 1.4).

$[\,(\mathrm{iii})$ implies $(\mathrm{ii})\,]$ The results immediately follow
from Theorem 4.6.

$[\,(\mathrm{ii})$ implies $(\mathrm{i})\,]$ Let
$\nu_{(c_{1})}(dx)=\nu_{c_{1}}(dx)/\nu((c_{1},\infty))$. By Proposition 3.7
and $\nu_{(1)}(x+\Delta)\nu((1,\infty))\sim\nu_{c_{1}}(x+\Delta)$,
$\nu_{(c_{1})}\in{\mathcal{S}}_{\Delta}$ follows. Then applying
$\nu_{(c_{1})}^{\ast n}(x+\Delta)/\nu_{(c_{1})}(x+\Delta)\to n$ as
$x\to\infty$ together with Kesten bound ([9, Theorem 4.25]) we obtain by the
dominated convergence that

$\displaystyle\lim_{x\to\infty}\frac{\mu_{c_{1}}(x+\Delta)}{\nu_{(c_{1})}(x+\Delta)}=e^{-\nu((c_{1},\infty))}\sum_{n=0}^{\infty}\frac{(\nu((c_{1},\infty)))^{n}}{n!}\lim_{x\to\infty}\frac{\nu_{(c_{1})}^{\ast
n}(x+\Delta)}{\nu_{(c_{1})}(x+\Delta)}=\nu((c_{1},\infty)).$

Thus $\mu_{c_{1}}\in{\mathcal{S}}_{\Delta}$. Recall that
$\mu_{c_{1}}\in{\mathcal{A}}_{\Delta}$ by assumption and
$\mu_{2}(x+\Delta)=O(e^{-\gamma x})$ for any $\gamma>0$. Now by Proposition
3.8 with $(F,G_{1},G_{2})=(\mu_{c_{1}},\mu_{c_{1}},\mu_{2})$ and
$(c_{1}=1,c_{2}=0)$, we obtain $\mu\in{\mathcal{S}}_{\Delta}$. Finally
$\mu\in{\mathcal{D}}_{\Delta}$ follows from
$\mu(x+\Delta)\sim\mu_{c_{1}}(x+\Delta)$ (cf. Lemma 1.4). ∎

###### Proof of Theorem 4.9.

$[\,(\mathrm{ii})$ implies $(\mathrm{i})$ and $(\mathrm{iii})\,]$ Fix
$c_{1}>1$ and we use $\mu_{c_{1}}$, $\nu_{c_{1}}$ and $\nu_{(c_{1})}$ of
Theorem 4.8. Since
$\nu_{(1)}(x+\Delta)\nu((1,\infty))\sim\nu_{(c_{1})}(x+\Delta)\nu((c_{1},\infty))$,
$\nu_{(c_{1})}\in{\mathcal{S}}_{\Delta}$ follows form [9, Theorem 4.23]. By
definition of $s$-self-decomposability we have
$\nu_{c_{1}}\in{\mathcal{A}}_{\Delta}$. Recall that $\mu_{c_{1}}$ is a
compound Poisson with the Lévy measure $\nu_{c_{1}}$ and thus by Theorem 4.1
$(\mathrm{iii})$ $\mu_{c_{1}}(x+\Delta)\sim\nu_{c_{1}}(x+\Delta)$, so that
$\mu_{c_{1}}\in{\mathcal{A}}_{\Delta}$. Thus Theorem 4.8 implies
$(\mathrm{i})$ and $(\mathrm{iii})$.

$[\,(\mathrm{iii})$ implies $(\mathrm{ii})$ and $(\mathrm{i})\,]$ Obviously
$\nu_{(1)}\in{\mathcal{A}}_{\Delta}$ and the results follows from Theorem 4.6.

$[\,(\mathrm{i})$ implies $(\mathrm{ii})\,]$ We prepare three IDs. Recall that
we may write $\nu(dx)=g(x)dx$ since $s$-self-decomposability implies the
existence of a Lévy density. Let $\mu_{\overline{c}_{1}}\in{\rm
ID}(\mathbb{R})$ with $a=b=0$ and the Lévy measure

$\nu_{\overline{c}_{1}}(dx)=({\bf 1}_{\\{x\in[-c_{1},c_{1}]\\}}g(c_{1})+{\bf
1}_{\\{|x|>c_{1}\\}}g(|x|))dx$

and let $\mu_{0}\in{\rm ID}(\mathbb{R})$ with the same $a$ and $b$ as those of
$\mu$ and the Lévy measure

$\nu_{0}(dx)=\big{(}{\bf 1}_{\\{x\in[-c_{1},c_{1}],\,g(x)-g(c_{1})\geq
0\\}(g(x)-g(c_{1}))}+{\bf 1}_{\\{x<-c_{1}\\}}g(x)\big{)}dx$

and let $\mu_{-}\in{\rm ID}(\mathbb{R})$ with $a=b=0$ and the Lévy measure

$\nu_{-}(dx)=\big{(}{\bf 1}_{\\{x<-c_{1}\\}}g(-x)+{\bf
1}_{\\{x\in[-c_{1},0],\,g(c_{1})-g(x)\geq 0\\}}(g(c_{1})-g(x))\big{)}dx.$

Here observe that $\mu_{-}(x+\Delta)=O(e^{-\gamma x})$ and
$\mu_{0}(x+\Delta)=O(e^{-\gamma x})$ for any $\gamma>0$. Then
$\mu\ast\mu_{-}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ follows
from Proposition 3.8 with $(F,G_{1},G_{2})=(\mu,\mu,\mu_{-})$ and
$(c_{1}=1,c_{2}=0)$ (cf. Lemma 1.4). Notice that $\nu_{\overline{c}_{1}}$ is
symmetric and unimodal, so is the compound Poisson $\mu_{\overline{c}_{1}}$
and $\mu_{\overline{c}_{1}}\in{\mathcal{A}}_{\Delta}$. Since
$\mu_{\overline{c}_{1}}\ast\mu_{0}=\mu\ast\mu_{-}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$
by Proposition 3.2,
$\mu_{\overline{c}_{1}}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{A}}_{\Delta}$.
Recall that $\mu_{\overline{c}_{1}}$ is a compound Poisson with Lévy measure
$\nu_{\overline{c}_{1}}$. Write
$\nu_{(\overline{c}_{1})}(dx)=\nu_{\overline{c}_{1}}(dx)/(2(c_{1}g(c_{1})+G(c_{1},\infty)))$
and then due to Corollary 4.2,
$\nu_{(\overline{c}_{1})}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{A}}_{\Delta}$
follows. Now since
$\nu_{\overline{c}_{1}}(x+\Delta)\sim\nu_{(1)}(x+\Delta)\nu((1,\infty))$, we
have $\nu_{(1)}\in{\mathcal{S}}_{\Delta}$. ∎

###### Proof of Theorem 4.10.

We decompose $\mu$ as in the proof of Theorem 4.6. Suppose that $(\mathrm{i})$
holds. We obtain from Proposition 3.12 that
$\mu_{1}\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$ and
$\mu(x+\Delta)\sim\mu_{1}(x+\Delta)$ for any $\Delta$. Introduce the condition

$(\mathrm{iii})^{\prime}\qquad\nu_{(1)}\in{\mathcal{L}}_{loc}\cap{\mathcal{D}}_{loc}\quad\text{and}\quad\mu_{1}(x+\Delta)\sim\nu(x+\Delta)\quad\text{for
any}\ \Delta.$

We see from Corollary 2.1 of [21] that the condition
$\mu_{1}\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$ is equivalent to
$(\mathrm{ii})$ and $(\mathrm{iii})^{\prime}$. Since
$\mu(x+\Delta)\sim\mu_{1}(x+\Delta)$ for any $\Delta$,
$(\mathrm{iii})^{\prime}$ is equivalent to $(\mathrm{iii})$. Conversely,
suppose the condition $\mu_{1}\in{\mathcal{S}}_{loc}\cap{\mathcal{D}}_{loc}$.
Then $(\mathrm{i})$ follows from Proposition 3.8. ∎

###### Proof of Theorem 5.1.

We give the proof only for the general two-sided case, since that for the
positive-half case is similar. As in the proof of Theorem 4.1 we can prove
that $(\mathrm{iii})$ implies $(\mathrm{ii})$ and that $(\mathrm{ii})$ implies
$(\mathrm{iii})$ and $(\mathrm{i})$. Thus it is enough to prove that
$(\mathrm{i})$ implies $(\mathrm{ii})$. Define $\delta:=-\log(1-\lambda)$ and
the distribution $\nu_{0}$ as

$\nu_{0}:=\delta^{-1}\sum_{n=1}^{\infty}(\lambda^{n}/n)\,G^{*n}.$

Then, as shown in the proof of [16, Theorem 1.8] (cf. the proof of [5,
Corollary 3]) that

$F_{a}=e^{-\delta a}\sum_{n=0}^{\infty}((\delta
a)^{n}/n!)\,\nu_{0}^{*n}\quad\text{and}\quad
G=-\lambda^{-1}\sum_{n=1}^{\infty}\big{(}(-\delta)^{n}/n!\big{)}\,\nu_{0}^{*n}.$

Suppose that $F_{a}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$ for
some $a>0$, or equivalently for all $a>0$. Then we obtain from Theorem 4.1
that $\nu_{0}\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. Now due to
Propositions 3.9 and 3.10, the dominated convergence theorem yields

$\lim_{x\to\infty}\frac{G(x+\Delta)}{\nu_{0}(x+\Delta)}=\delta
e^{-\delta}/\lambda>0$

and hence by Proposition 3.7 that
$G\in{\mathcal{S}}_{\Delta}\cap{\mathcal{D}}_{\Delta}$. ∎

###### Proof of Theorem 5.2.

Let $F_{a}=\pi,$ $G=G_{0}$, and $a=1$ in Theorem 5.1. Then the proof is clear
from Theorems 4.1 and 5.1. ∎

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