Two Hypotheses on the Exponential Class in the Class Of O-subexponential Infinitely Divisible Distributions

Abstract

Two hypotheses on the class ${\mathcal {L}}(\gamma )$ in the class $\mathcal {OS}\cap \mathcal {ID}$ are discussed. Two weak hypotheses on the class ${\mathcal {L}}(\gamma )$ in the class $\mathcal {OS}\cap \mathcal {ID}$ are proved. A necessary and sufficient condition in order that, for every $t>0$, the t-th convolution power of a distribution in the class $\mathcal {OS}\cap \mathcal {ID}$ belongs to the class ${\mathcal {L}}(\gamma )$ is given. Sufficient conditions are given for the validity of two hypotheses on the class ${\mathcal {L}}(\gamma )$.

1 Introduction and Results

In what follows, we denote by ${\mathbb {R}}$ the real line and by ${\mathbb {R}}_{+}$ the half line $[0,\infty )$. Denote by ${\mathbb {N}}$ the totality of positive integers and by $a{\mathbb {N}}$ the set $\{a,2a,3a,\dots \}$. The symbol $\delta _a(\hbox {d}x)$ stands for the delta measure at $a \in {\mathbb {R}}$. Let $\eta$ and $\rho$ be probability distributions on ${\mathbb {R}}$. We denote by $\eta *\rho$ the convolution of $\eta$ and $\rho$ and by $\rho ^{n*}$ n-th convolution power of $\rho$ with the understanding that $\rho ^{0*}(\hbox {d}x)=\delta _0(\hbox {d}x)$. Denote by ${{\bar{\xi }}}(x)$ the tail $\xi ((x,\infty ))$ of a measure $\xi$ on ${\mathbb {R}}$ for $x \in {\mathbb {R}}$. Let $\gamma \ge 0$. We define the $\gamma$-exponential moment ${{\widehat{\xi }}}(\gamma )$ as

$$\begin{aligned} {{\widehat{\xi }}}(\gamma ):=\int _{-\infty }^{\infty }e^{\gamma x}\xi (\hbox {d}x). \end{aligned}$$

If ${{\widehat{\xi }}}(\gamma ) < \infty ,$ we define the Fourier–Laplace transform ${{\widehat{\xi }}}(\gamma +iz)$ for $z\in {\mathbb {R}}$ as

$$\begin{aligned} {{\widehat{\xi }}}(\gamma +iz):=\int _{-\infty }^{\infty }e^{(\gamma + iz)x}\xi (\hbox {d}x). \end{aligned}$$

An integral $\int _a^bg(x)\rho (\hbox {d}x)$ means $\int _{a+}^{b+}g(x)\rho (\hbox {d}x)$. For positive functions $f_1(x)$ and $g_1(x)$ on $[A,\infty )$ for some $A \in {\mathbb {R}}$, we define the relation $f_1(x)\sim g_1(x)$ by $\lim _{x \rightarrow \infty }f_1(x)/g_1(x)=1$ and the relation $f_1(x)\asymp g_1(x)$ by

$$\begin{aligned} 0<\liminf _{x \rightarrow \infty }f_1(x)/g_1(x)\le \limsup _{x \rightarrow \infty }f_1(x)/g_1(x)< \infty . \end{aligned}$$

Let $\gamma \ge 0$. A distribution $\rho$ on ${\mathbb {R}}$ belongs to the class ${\mathcal {L}}(\gamma )$ if ${\overline{\rho }}(x)>0$ for all $x >0$ and, for every $a \in {\mathbb {R}}$,

$$\begin{aligned} {\overline{\rho }}(x+a)\sim e^{-\gamma a}{\overline{\rho }}(x). \end{aligned}$$

A distribution $\rho$ on ${\mathbb {R}}$ belongs to the class ${\mathcal {S}}(\gamma )$ if $\rho \in {\mathcal {L}}(\gamma )$, ${{\widehat{\rho }}}(\gamma ) < \infty$, and

$$\begin{aligned} \overline{\rho ^{2*}}(x) \sim 2{{\widehat{\rho }}}(\gamma ){\overline{\rho }}(x). \end{aligned}$$

A distribution $\rho$ on ${\mathbb {R}}$ belongs to the class $\mathcal {OL}$ if ${\overline{\rho }}(x)>0$ for $x >0$ and, for all $a \ge 0$,

$$\begin{aligned} {\overline{\rho }}(x-a) \asymp {\overline{\rho }}(x). \end{aligned}$$

A distribution $\rho$ on ${\mathbb {R}}$ belongs to the class $\mathcal {OS}$ if ${\overline{\rho }}(x)>0$ for all $x >0$ and

$$\begin{aligned} \overline{\rho ^{2*}}(x) \asymp {\overline{\rho }}(x). \end{aligned}$$

Note that the class $\mathcal {OS}$ is included in the class $\mathcal {OL}$. A distribution $\rho$ on ${\mathbb {R}}$ belongs to the class ${\mathcal {S}}_{\sharp }$ if $\rho \in \mathcal {OS}$ and

$$\begin{aligned} \limsup _{A \rightarrow \infty }\limsup _{x \rightarrow \infty }\frac{{\overline{\rho }}(x-A){{\bar{\rho }}}(A)+\int _A^{x-A}{{\bar{\rho }}}(x-u)\rho (\hbox {d}u)}{{{\bar{\rho }}}(x)}=0. \end{aligned}$$

The class ${\mathcal {S}}_{\sharp }$ includes $\cup _{\gamma \ge 0}{\mathcal {S}}(\gamma )$, and it is closed under convolution powers. A finite measure $\xi$ satisfies the Wiener condition if ${{\widehat{\xi }}}(iz)\ne 0$ for every $z \in {\mathbb {R}}$. Denote by ${\mathcal {W}}$ the totality of finite measures on ${\mathbb {R}}$ satisfying the Wiener condition. We denote by $\mathcal {ID}$ the class of all infinitely divisible distributions on ${\mathbb {R}}$. For $\mu \in \mathcal {ID}$, denote by $\nu$ its Lévy measure. Under the assumption that ${{\bar{\nu }}}(c)>0$ for every $c >0,$ define $\nu _1(\hbox {d}x):=1_{(1,\infty )}(x)\nu (\hbox {d}x)/{{\bar{\nu }}}(1).$ Let $\mu \in \mathcal {ID}$. We define a compound Poisson distribution $\mu _1$ with $c={{\bar{\nu }}}(1)$ as

$$\begin{aligned} \mu _1(\hbox {d}x):=e^{-c}\sum _{k=0}^{\infty }\frac{c^k}{k!}\nu _1^{k*}(\hbox {d}x). \end{aligned}$$

Denote by $\mu ^{t*}$ the t-th convolution power of $\mu \in \mathcal {ID}$ for $t >0$. Note that $\mu ^{t*}$ is the distribution of $X_t$ for a certain Lévy process $\{X_t\}$ on ${\mathbb {R}}$. Let $\gamma \ge 0$. Define $T(\mu ,\gamma )$ as

$$\begin{aligned} T(\mu ,\gamma ):=\{t>0: \mu ^{t*} \in {\mathcal {L}}(\gamma )\}. \end{aligned}$$

Since the class ${\mathcal {L}}(\gamma )$ is closed under convolutions by Theorem 3 of Embrechts and Goldie [2], $T(\mu ,\gamma )$ is empty or an additive semigroup in $(0,\infty )$. We see from Lemma 2.2 that for $\mu \in \mathcal {OS}\cap \mathcal {ID}$, there are positive integers n such that $\nu _1^{n*} \in \mathcal {OS}$. Let $n_0$ be the positive integer defined by (2.1). Note that we do not yet know an example of $\mu \in \mathcal {OS}\cap \mathcal {ID}$ such that $n_0 \ge 3.$

A class ${\mathcal {C}}$ of distributions is called closed under convolution roots if $\rho ^{n*} \in {\mathcal {C}}$ for some $n \in {\mathbb {N}}$ implies $\rho \in {\mathcal {C}}$. We see from Shimura and Watanabe [11] that the class $\mathcal {OS}$ is not closed under convolution roots, but from Watanabe and Yamamuro [15] that the class $\mathcal {OS}\cap \mathcal {ID}$ is closed under convolution roots. Embrechts et al. [4] in the one-sided case and Watanabe [13] in the two-sided case proved that the class ${\mathcal {S}}(0)$ is closed under convolution roots, and Embrechts and Goldie [2] conjectured that the class ${\mathcal {L}}(\gamma )$ with $\gamma \ge 0$ is closed under convolution roots, but Shimura and Watanabe [12] showed that the class ${\mathcal {L}}(\gamma )$ with $\gamma \ge 0$ is not closed under convolution roots. Moreover, Watanabe and Yamamuro [16] proved that the class ${\mathcal {S}}_{ac}$ of all absolutely continuous distributions on ${\mathbb {R}}$ with subexponential densities is not closed under convolution roots. Embrechts and Goldie [3] conjectured that the class ${\mathcal {S}}(\gamma )$ with $\gamma >0$ is closed under convolution roots. Watanabe [13] proved that ${\mathcal {S}}(\gamma )\cap \mathcal {ID}$ with $\gamma \ge 0$ is closed under convolution roots, but Watanabe [14] showed that the class ${\mathcal {S}}(\gamma )$ with $\gamma >0$ is not closed under convolution roots. We add the following. Klüppelberg [5] showed that the class $\mathcal {OS}$ is closed under convolutions. The class ${\mathcal {S}}(\gamma )$ is closed under convolution powers for $\gamma \ge 0$, but Leslie [7], for $\gamma = 0$, and Klüppelberg and Villasenor [6], for $\gamma > 0$, proved that the class ${\mathcal {S}}(\gamma )$ is not closed under convolutions.

We consider the following two hypotheses on the class ${\mathcal {L}}(\gamma )$ in the class $\mathcal {OS}\cap \mathcal {ID}$:

Hypothesis I Let $\gamma \ge 0$. For every $\mu \in \mathcal {OS}\cap \mathcal {ID}$, if $\mu ^{n*} \in {\mathcal {L}}(\gamma )$ for some $n \in {\mathbb {N}}$, then $\mu ^{(n+1)*} \in {\mathcal {L}}(\gamma )$.

Hypothesis II Let $\gamma \ge 0$. For every $\mu \in \mathcal {OS}\cap \mathcal {ID}$, if $\mu ^{n*} \in {\mathcal {L}}(\gamma )$ for some $n \in {\mathbb {N}}$, then $\mu \in {\mathcal {L}}(\gamma )$.

We also consider the weak version of the above hypotheses:

Hypothesis I$^\prime$ Let $\gamma \ge 0$. For every $\mu \in \mathcal {OS}\cap \mathcal {ID}$, if $\mu ^{n*}, \mu ^{(n+1)*} \in {\mathcal {L}}(\gamma )$ for some $n \in {\mathbb {N}}$, then $\mu ^{(n+2)*} \in {\mathcal {L}}(\gamma )$.

Hypothesis II$^\prime$ Let $\gamma \ge 0$. For every $\mu \in \mathcal {OS}\cap \mathcal {ID}$, if $\mu ^{n*}, \mu ^{(n+1)*} \in {\mathcal {L}}(\gamma )$ for some $n \in {\mathbb {N}}$, then $\mu \in {\mathcal {L}}(\gamma )$.

Let $\gamma \ge 0$. Define

$$\begin{aligned} {\mathcal {A}}(\gamma ):= & {} \{\mu \in \mathcal {OS}\cap \mathcal {ID}:T(\mu ,\gamma )=(0,\infty )\}; \\ {\mathcal {B}}(\gamma ):= & {} \{\mu \in \mathcal {OS}\cap \mathcal {ID}:T(\mu ,\gamma )=\emptyset \}; \end{aligned}$$

and

$$\begin{aligned} {\mathcal {C}}(\gamma ):=\{\mu \in \mathcal {OS}\cap \mathcal {ID}:T(\mu ,\gamma )=a_0{\mathbb {N}}\,\, \text{ with } \text{ some } \,a_0 >0\}. \end{aligned}$$

Theorem 1.1

Let $\gamma \ge 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. We have the following:

• (i) $\mathcal {OS}\cap \mathcal {ID}={\mathcal {A}}(\gamma )\cup {\mathcal {B}}(\gamma )\cup {\mathcal {C}}(\gamma ).$ Thus, Hypotheses I$^\prime$ and II$^\prime$ are true.

• (ii) The relation $\mu \in {\mathcal {A}}(\gamma )$ holds if and only if, for all $a \ge 0$,

  $$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\nu _1}(x-a)-\overline{\nu _1}(x)}{\overline{\nu _1^{n_0*}}(x)} = 0. \end{aligned}$$

  (1.1)

  If $\mu \in {\mathcal {A}}(\gamma )$, then $\nu _1^{n*} \notin {\mathcal {L}}(\gamma )\cap \mathcal {OS}$ for $1 \le n \le n_0-1$ and $\nu _1^{n*} \in {\mathcal {L}}(\gamma )\cap \mathcal {OS}$ for $n \ge n_0$.

Corollary 1.1

Let $\gamma \ge 0$. Then the following are equivalent:

• (1) Hypothesis I is true.

• (2) Hypothesis II is true.

• (3) ${\mathcal {C}}(\gamma )$ is empty.

• (4) For every $\mu \in \mathcal {OS}\cap \mathcal {ID}$ it holds that, for every $2t \in T(\mu ,\gamma )$ and for every $a \ge 0$,

  $$\begin{aligned} \limsup _{x \rightarrow \infty }\limsup _{\lambda \rightarrow \infty }\frac{|\int _x^{\lambda -x}(e^{-\gamma a}\overline{\mu _1^{t*}}(\lambda -a-u)-\overline{\mu _1^{t*}}(\lambda -u))\mu _1^{t*}(\hbox {d}u)|}{\overline{\mu _1^{t*}}(\lambda )}=0.\qquad \quad \end{aligned}$$

  (1.2)

Remark 1.1

Let $\gamma = 0$. Then, ${\mathcal {C}}(0)$ is empty and Hypotheses I and II are true. The relation $\mu \in {\mathcal {A}}(0)$ holds if and only if

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{\nu _1((x,x+1])}{\overline{\nu _1^{n_0*}}(x)} = 0. \end{aligned}$$

If $\mu \in {\mathcal {A}}(0)$, then $\nu _1^{n*} \notin {\mathcal {L}}(0)\cap \mathcal {OS}$ for $1 \le n \le n_0-1$ and $\nu _1^{n*} \in {\mathcal {L}}(0)\cap \mathcal {OS}$ for $n \ge n_0$. Xu et al. showed in Theorem 2.2 of [18] an example of $\mu \in {\mathcal {A}}(0)$ with $n_0=2$.

For $\gamma > 0$, we cannot yet answer the question whether Hypotheses I and II are true. However, under some additional assumptions in terms of Lévy measure, we establish that ${\mathcal {C}}(\gamma )$ is empty.

Proposition 1.1

Let $\gamma > 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. Suppose that, for every $a \ge 0$,

$$\begin{aligned} \liminf _{x\rightarrow \infty }e^{-\gamma a}{{\bar{\nu }}}_1(x-a)/{{\bar{\nu }}}_1(x) \ge 1. \end{aligned}$$

(1.3)

Then, we have either $T(\mu ,\gamma )= (0,\infty )$ or $\emptyset$.

Remark 1.2

Cui et al. [1] proved a result analogous to the above proposition under a stronger assumption. Xu et al. showed in Theorem 1.1 of [19] an example of the case where $T(\mu ,\gamma )\ne \emptyset$ in the above proposition.

Proposition 1.2

Let $\gamma > 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. Suppose that $\nu _1^{2*} \in {\mathcal {L}}(\gamma )$ and the real part of ${{\widehat{\nu }}}_1(\gamma +iz)$ is not zero for every $z \in {\mathbb {R}}$. Then, either $T(\mu ,\gamma )= (0,\infty )$ or $\emptyset$.

Proposition 1.3

Let $\gamma > 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. Suppose that there exists $n_1 \in {\mathbb {N}}$ such that $\nu _1^{n_1*} \in {\mathcal {S}}_{\sharp }$. Then, either $T(\mu ,\gamma )= (0,\infty )$ or $\emptyset$. The equality $T(\mu ,\gamma )=(0,\infty )$ holds if and only if $\nu _1 \in {\mathcal {S}}(\gamma )$.

Remark 1.3

Watanabe made in Theorem 1.1 of [14] a distribution $\eta \in {\mathcal {S}}_{\sharp }$ such that $\eta ^{n*} \in {\mathcal {S}}(\gamma )$ for every $n \ge 2$ but $\eta \notin {\mathcal {S}}(\gamma )$. Thus, taking this $\eta$ as $\nu _1$, then Proposition 1.3 holds with $T(\mu ,\gamma )= \emptyset$.

2 Preliminaries

In this section, we give several basic results as preliminaries. Pakes [8] proved the following.

Lemma 2.1

(Lemmas 2.1 and 2.5 of [8]) Let $\mu \in \mathcal {ID}$. Then we have $\mu \in {\mathcal {L}}(\gamma )$ if and only if $\mu _1 \in {\mathcal {L}}(\gamma )$.

Watanabe and Yamamuro [15] proved the following.

Lemma 2.2

(Proposition 3.1 of [15]) Suppose that $\mu \in \mathcal {ID}$. Then, we have $\mu \in \mathcal {OS}$ if and only if there is $n \in {\mathbb {N}}$ such that $\nu _1^{n*} \in \mathcal {OS}$ and $\overline{\mu _1^{t*}}(x) \asymp \overline{\nu _1^{n*}}(x)$ for any $t >0$.

For $\mu \in \mathcal {OS}\cap \mathcal {ID}$, define $n_0 \in \mathbb N$ as

$$\begin{aligned} n_0:=\min \{n \in {\mathbb {N}}: \nu _1^{n*} \in \mathcal {OS}\}. \end{aligned}$$

(2.1)

Lemma 2.3

Let $\mu \in \mathcal {OS}\cap \mathcal {ID}$.

• (i) There exists $C(a) >0$ such that, for all $a \ge 0$ and all $x >0$,

  $$\begin{aligned} \overline{\nu _1^{n_0*} }(x-a)\le C(a) \overline{\nu _1^{n_0*} }(x). \end{aligned}$$

• (ii) There exists $K >1$ such that, for all $n \in {\mathbb {N}}$ and all $x >0$,

  $$\begin{aligned} \overline{\nu _1^{n*} }(x)\le K^n \overline{\nu _1^{n_0*} }(x). \end{aligned}$$

Proof

Assertion (i) is clear since $\nu _1^{n_0*} \in \mathcal {OS}\subset \mathcal {OL}.$ We see from Proposition 2.4 of Shimura and Watanabe [11] that there exists $K_1 >1$ such that, for all $k \in {\mathbb {N}}$ and all $x >0$,

$$\begin{aligned} \overline{\nu _1^{(kn_0)*} }(x)\le K_1^k \overline{\nu _1^{n_0*} }(x). \end{aligned}$$

Note that, for $m \le n$,

$$\begin{aligned} \overline{\nu _1^{m*} }(x)\le \overline{\nu _1^{n*} }(x). \end{aligned}$$

Hence, we have, for $0 \le j \le n_0-1$ and for all $k \in \mathbb N$, with $K=K_1^{2/n_0}>1$

$$\begin{aligned} \overline{\nu _1^{(kn_0+j)*} }(x)\le K^{(kn_0+j)} \overline{\nu _1^{n_0*} }(x). \end{aligned}$$

This inequality holds for $k=0$ too. Thus, assertion (ii) is true. $\square$

Under the assumption that $\zeta \in \mathcal {OS}\subset \mathcal {OL}$, we define the following. Let

$$\begin{aligned} d^*:=\limsup _{x \rightarrow \infty } \frac{\overline{\zeta ^{2*}}(x)}{{\overline{\zeta }}(x)}<\infty . \end{aligned}$$

Let $\Lambda$ be the totality of increasing sequences $\{\lambda _n\}_{n=1}^{\infty }$ with $\lim _{n \rightarrow \infty } \lambda _n = \infty$ such that, for every $x \in {\mathbb {R}}$, the following limit exists and is finite:

$$\begin{aligned} m(x;\{\lambda _n\}):=\lim _{n \rightarrow \infty }\frac{{{\bar{\zeta }}}(\lambda _n-x)}{{{\bar{\zeta }}}(\lambda _n)}. \end{aligned}$$

(2.2)

Define, for each sequence $\{x_n\}_{n=1}^{\infty }$ with $\lim _{n \rightarrow \infty } x_n = \infty$, $T_n(y)$ as

$$\begin{aligned} T_n(y):=\frac{{{\bar{\zeta }}}(x_n-y)}{{{\bar{\zeta }}}(x_n)}. \end{aligned}$$

Since $\{T_n(y)\}_{n=1}^{\infty }$ is a sequence of increasing functions, uniformly bounded on every finite interval, by Helly’s selection principle, there exists an increasing subsequence $\{\lambda _n\}$ of $\{x_n\}$ with $\lim _{n \rightarrow \infty } \lambda _n = \infty$ such that everywhere on ${\mathbb {R}}$ (2.2) holds. The limit function $m(x;\{\lambda _n\})$ is increasing and is finite. That is, $\{\lambda _n\} \in \Lambda$. It follows that, under the assumption that $\zeta \in \mathcal {OS}$, there exists an increasing subsequence $\{\lambda _n\} \in \Lambda$ of $\{x_n\}$ for each sequence $\{x_n\}_{n=1}^{\infty }$ with $\lim _{n \rightarrow \infty } x_n = \infty$.

Lemma 2.4

Suppose that $\zeta \in \mathcal {OS}$. Then, we have the following.

• (i) If $\{\lambda _n\} \in \Lambda$, then $\{\lambda _n-a\} \in \Lambda$ for every $a \in {\mathbb {R}}$.

• (ii) For $\{\lambda _n\} \in \Lambda$,

  $$\begin{aligned} \int _{-\infty }^{\infty }m(x;\{\lambda _n\})\zeta (\hbox {d}x) < \infty \end{aligned}$$

  and

  $$\begin{aligned} \lim _{a \rightarrow \infty }m(a;\{\lambda _n\}){{\bar{\zeta }}}(a)=0. \end{aligned}$$

  In particular, if $\zeta \in \mathcal {OS}\cap {\mathcal {L}}(\gamma )$, then $m(x;\{\lambda _n\})=e^{\gamma x}$ and ${{\widehat{\zeta }}}(\gamma ) < \infty .$

Proof

We prove (i). Suppose that $\{\lambda _n\} \in \Lambda .$ We have, for $x,a \in {\mathbb {R}}$,

$$\begin{aligned} \lim _{n \rightarrow \infty }\frac{{{\bar{\zeta }}}(\lambda _n-a-x)}{{{\bar{\zeta }}}(\lambda _n-a)}=\frac{m(x+a;\{\lambda _n\})}{m(a;\{\lambda _n\})}. \end{aligned}$$

Thus, $\{\lambda _n-a\} \in \Lambda .$ Next, we prove (ii). Let $\rho$ be a distribution on ${\mathbb {R}}$. Note that, for $x >2A,$

$$\begin{aligned} \overline{\rho ^{2*}}(x)=2\int _{-\infty }^{A+}{{\bar{\rho }}}(x-u)\rho (\hbox {d}u)+{\overline{\rho }}(x-A){{\bar{\rho }}}(A)+\int _A^{x-A}{{\bar{\rho }}}(x-u)\rho (\hbox {d}u).\nonumber \\ \end{aligned}$$

(2.3)

We see from (2.3) that, for $\{\lambda _n\} \in \Lambda$ and $s >0$,

$$\begin{aligned} \begin{aligned} d^*&\ge \limsup _{n \rightarrow \infty } \frac{\overline{\zeta ^{2*}}(\lambda _n)}{{\overline{\zeta }}(\lambda _n)}\\&\ge 2\limsup _{n \rightarrow \infty }\int _{-\infty }^{s+}\frac{{{\bar{\zeta }}}(\lambda _n-x)}{{{\bar{\zeta }}}(\lambda _n)}\zeta (\hbox {d}x)\\&\ge 2\int _{-\infty }^{s+}m(x;\{\lambda _n\})\zeta (\hbox {d}x). \end{aligned} \end{aligned}$$

As $s \rightarrow \infty$, we have

$$\begin{aligned} \int _{-\infty }^{\infty }m(x;\{\lambda _n\})\zeta (\hbox {d}x) < \infty . \end{aligned}$$

Since $m(x;\{\lambda _n\})$ is increasing in x, we have

$$\begin{aligned} \begin{aligned}&\lim _{a \rightarrow \infty }m(a;\{\lambda _n\}){{\bar{\zeta }}}(a)\\&\quad \le \lim _{a \rightarrow \infty }\int _{a+}^{\infty }m(x;\{\lambda _n\})\zeta (\hbox {d}x)=0. \end{aligned} \end{aligned}$$

Hence, if $\zeta \in \mathcal {OS}\cap {\mathcal {L}}(\gamma )$, then $m(x;\{\lambda _n\})=e^{\gamma x}$ and ${{\widehat{\zeta }}}(\gamma ) < \infty .$ Thus, we have proved the lemma. $\square$

Pakes [8, 9] asserted and Watanabe [13] finally proved the following.

Lemma 2.5

(Theorem 1.1 of [13]) Let $\gamma \ge 0$. Then $\mu \in \mathcal {ID}\cap {\mathcal {S}}(\gamma )$ if and only if $\nu _1 \in {\mathcal {S}}(\gamma )$.

Lemma 2.6

Let $\gamma \ge 0$. Suppose that $\rho \in {\mathcal {S}}_{\sharp }$.

• (i) If ${{\bar{\eta }}}(x) \asymp {{\bar{\rho }}}(x)$, then $\eta \in {\mathcal {S}}_{\sharp }$.

• (ii) $\rho \in {\mathcal {S}}(\gamma )$ if and only if $\rho \in {\mathcal {L}}(\gamma )$.

Proof

Suppose that $\rho \in {\mathcal {S}}_{\sharp }$. We prove (i). If ${{\bar{\eta }}}(x) \asymp {{\bar{\rho }}}(x)$, then there is $C >0$ such that ${{\bar{\eta }}}(x)\le C {{\bar{\rho }}}(x)$ for $x \in {\mathbb {R}}$. By using integration by parts in the second inequality, we obtain that

$$\begin{aligned} \begin{aligned}&{{\bar{\eta }}}(x-A){{\bar{\eta }}}(A)+\int _A^{x-A}{{\bar{\eta }}}(x-u)\eta (\hbox {d}u)\\&\quad \le C^2{{\bar{\rho }}}(x-A){{\bar{\rho }}}(A)+C \int _A^{x-A}{{\bar{\rho }}}(x-u)\eta (\hbox {d}u)\\&\quad \le 2C^2{{\bar{\rho }}}(x-A){{\bar{\rho }}}(A)+C^2\int _A^{x-A}{{\bar{\rho }}}(x-u)\rho (\hbox {d}u). \end{aligned} \end{aligned}$$

Thus, we see that

$$\begin{aligned} \limsup _{A \rightarrow \infty }\limsup _{x \rightarrow \infty }\frac{({\overline{\eta }}(x-A){{\bar{\eta }}}(A)+\int _A^{x-A}{{\bar{\eta }}}(x-u)\eta (\hbox {d}u))}{{{\bar{\eta }}}(x)}=0. \end{aligned}$$

That is, $\eta \in {\mathcal {S}}_{\sharp }$. Next we prove (ii). If $\rho \in {\mathcal {S}}(\gamma )$, then clearly $\rho \in {\mathcal {L}}(\gamma )$. Note that, for $x >2A,$ (2.3) holds. If $\rho \in {\mathcal {S}}_{\sharp }\cap {\mathcal {L}}(\gamma )$, then we have

$$\begin{aligned} \begin{aligned}&\lim _{x \rightarrow \infty }\frac{\overline{\rho ^{2*}}(x)}{{{\bar{\rho }}}(x)}\\&\quad =\lim _{A \rightarrow \infty }2\int _{-\infty }^{A+}\lim _{x \rightarrow \infty }\frac{{{\bar{\rho }}}(x-u)}{{{\bar{\rho }}}(x)}\rho (\hbox {d}u)\\&\quad =2{{\widehat{\rho }}}(\gamma ) < \infty . \end{aligned} \end{aligned}$$

Thus, we see that $\rho \in {\mathcal {S}}(\gamma )$. $\square$

Watanabe [14] extended Wiener’s approximation theorem in [17] as follows.

Lemma 2.7

(Lemma 2.6 of Watanabe [14]) Let $\xi$ be a finite measure on ${\mathbb {R}}$. The following are equivalent:

• (1) $\xi \in {\mathcal {W}}.$

• (2) If, for a bounded measurable function g(x) on ${\mathbb {R}}$,

  $$\begin{aligned} \int _{-\infty }^{\infty }g(x-t)\xi (dt)=0\quad \hbox {for a.e.}\,\, \,x\in {\mathbb {R}}, \end{aligned}$$

  then $g(x)=0$  for a.e. $x\in {\mathbb {R}}.$

3 Convolution Lemmas

In this section, we give important lemmas on convolutions.

Lemma 3.1

Let $\gamma \ge 0$. Suppose that $\zeta \in \mathcal {OS}$. For $j=1,2$, let $\rho _j$ be distributions on ${\mathbb {R}}_+$ satisfying

$$\begin{aligned} {{\bar{\rho }}}_j(x) \le C_j{{\bar{\zeta }}}(x)\,\, \text{ with } \text{ some } \,\, C_j>0 \text{ for } \text{ all } x >0. \end{aligned}$$

(3.1)

Let $\{\lambda _n\} \in \Lambda .$

• (i) Let $\lambda _n > a+x$ and $x >0$. We have, for every $a \ge 0$,

  $$\begin{aligned} e^{-\gamma a}\overline{\rho _1*\rho _2}(\lambda _n-a)-\overline{\rho _1*\rho _2}(\lambda _n)=:\sum _{j=1}^{4}I_j, \end{aligned}$$

  (3.2)

  where

  $$\begin{aligned} I_1:= & {} -\int _{\lambda _n-a-x}^{\lambda _n-x}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y), \\ I_2:= & {} \overline{\rho _1}(x)(e^{-\gamma a}\overline{\rho _2}(\lambda _n-a-x)-\overline{\rho _2}(\lambda _n-x)), \\ I_3:= & {} \int _{0-}^{(\lambda _n-a-x)+}(e^{-\gamma a}\overline{\rho _1}(\lambda _n-a-y)-\overline{\rho _1}(\lambda _n-y))\rho _2(\hbox {d}y), \end{aligned}$$

  and

  $$\begin{aligned} I_4:= & {} \int _{0-}^{x+}(e^{-\gamma a}\overline{\rho _2}(\lambda _n-a-y)-\overline{\rho _2}(\lambda _n-y))\rho _1(\hbox {d}y). \end{aligned}$$

• (ii) We have for $j=1,2$

  $$\begin{aligned} \limsup _{x\rightarrow \infty }\limsup _{n \rightarrow \infty }\frac{|I_j|}{{{\bar{\zeta }}}(\lambda _n)}=0. \end{aligned}$$

  (3.3)

Proof

By using integration by parts, we have

$$\begin{aligned} \begin{aligned}&\overline{\rho _1*\rho _2}(\lambda _n-a)\\&\quad =\int _{0-}^{(\lambda _n-a-x)+}\overline{\rho _1}(\lambda _n-a-y)\rho _2(\hbox {d}y)\\&\qquad +\int _{\lambda _n-a-x}^{\lambda _n-a}\overline{\rho _1}(\lambda _n-a-y)\rho _2(\hbox {d}y)+\overline{\rho _2}(\lambda _n-a)\\&\quad =\int _{0-}^{(\lambda _n-a-x)+}\overline{\rho _1}(\lambda _n-a-y)\rho _2(\hbox {d}y)+\int _{0-}^{x+}\overline{\rho _2}(\lambda _n-a-y)\rho _1(\hbox {d}y)\\&\qquad +\overline{\rho _1}(x)\overline{\rho _2}(\lambda _n-a-x), \end{aligned} \end{aligned}$$

and

$$\begin{aligned}&\overline{\rho _1*\rho _2}(\lambda _n)\\&\quad =\int _{0-}^{(\lambda _n-a-x)+}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y) +\int _{\lambda _n-a-x}^{\lambda _n-x}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y)\\&\quad \quad +\int _{\lambda _n-x}^{\lambda _n}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y) +\overline{\rho _2}(\lambda _n)\\&\quad =\int _{0-}^{(\lambda _n-a-x)+}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y)+\int _{\lambda _n-a-x}^{\lambda _n-x}\overline{\rho _1}(\lambda _n-y)\rho _2(\hbox {d}y)\\&\quad \quad +\int _{0-}^{x+}\overline{\rho _2}(\lambda _n-y)\rho _1(\hbox {d}y)+\overline{\rho _1}(x)\overline{\rho _2}(\lambda _n-x). \end{aligned}$$

Thus, assertion (i) is valid. We have by Lemma 2.4 for $j=1,2$

$$\begin{aligned} \begin{aligned}&\limsup _{x\rightarrow \infty }\limsup _{n \rightarrow \infty }\frac{|I_j|}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad \le \limsup _{x \rightarrow \infty }\overline{\rho _1}(x)\limsup _{n \rightarrow \infty }\frac{\overline{\rho _2}(\lambda _n-a-x)}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad \le C_1C_2\limsup _{x \rightarrow \infty }{{\bar{\zeta }}}(x)m(x;\{\lambda _n-a\})m(a;\{\lambda _n\})=0. \end{aligned} \end{aligned}$$

Lemma 3.2

Let $\gamma \ge 0$. Suppose that $\zeta \in \mathcal {OS}$. For $j=1,2$, let $\rho _j$ be distributions on ${\mathbb {R}}_+$ satisfying (3.1). Suppose further that for $j=1,2$ and every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_j(x-a)-{{\bar{\rho }}}_j(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

Then, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _1*\rho _2}(x-a)- \overline{\rho _1*\rho _2}(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

(3.4)

Proof. Let $\{\lambda _n\} \in \Lambda .$ By the assumption for $j=1$, there is $\epsilon (x)>0$ such that $\epsilon (x) \rightarrow 0$ as $x \rightarrow \infty$ and

$$\begin{aligned} |e^{-\gamma a}\overline{\rho _1}(\lambda _n-a-y)-\overline{\rho _1}(\lambda _n-y)|\le \epsilon (x)\zeta (\lambda _n-y) \end{aligned}$$

for $0 \le y \le \lambda _n-a-x.$ Thus, we have

$$\begin{aligned} \begin{aligned}&\limsup _{x \rightarrow \infty }\limsup _{n \rightarrow \infty }\frac{|I_3|}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad \le \limsup _{x \rightarrow \infty }\epsilon (x) \limsup _{n \rightarrow \infty }\frac{\int _{0-}^{(\lambda _n-a-x)+}{{\bar{\zeta }}}(\lambda _n-y)\rho _2(\hbox {d}y)}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad \le \limsup _{x \rightarrow \infty }\epsilon (x)\limsup _{n \rightarrow \infty }\frac{ {{\bar{\zeta }}}(\lambda _n)+\int _{a+x}^{\lambda _n}\overline{\rho _2}(\lambda _n-y)\zeta (\hbox {d}y)}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad \le \limsup _{x \rightarrow \infty }\epsilon (x) \limsup _{n \rightarrow \infty }\frac{{{\bar{\zeta }}}(\lambda _n)+C_2\overline{\zeta ^{2*}}(\lambda _n)}{{{\bar{\zeta }}}(\lambda _n)}=0. \end{aligned} \end{aligned}$$

(3.5)

As in the above argument, we have

$$\begin{aligned} \limsup _{x \rightarrow \infty }\limsup _{n \rightarrow \infty }\frac{|I_4|}{{{\bar{\zeta }}}(\lambda _n)}=0. \end{aligned}$$

Thus, by (3.2) and (3.3) of Lemma 3.1, we have proved (3.4). $\square$

Lemma 3.3

Let $\gamma \ge 0$. Suppose that $\zeta \in \mathcal {OS}$. For $j=1,2$, let $\rho _j$ be distributions on ${\mathbb {R}}_+$ satisfying (3.1). Suppose further that, for $j=1,2,$ and for every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_j(x-a)-{{\bar{\rho }}}_j(x)}{{{\bar{\zeta }}}(x)} \ge 0. \end{aligned}$$

Then, we have, for every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _1*\rho _2}(x-a)-\overline{\rho _1*\rho _2}(x)}{{{\bar{\zeta }}}(x)} \ge 0. \end{aligned}$$

(3.6)

Proof

Let $\{\lambda _n\} \in \Lambda .$ Let $\epsilon >0$ and $a \ge 0$ be arbitrary, and let $n \in {\mathbb {N}}$ and $x \in (0,\lambda _n-a)$ be sufficiently large such that

$$\begin{aligned} e^{-\gamma a}\overline{\rho _1}(\lambda _n-a-y)-\overline{\rho _1}(\lambda _n-y) \ge -\epsilon {{\bar{\zeta }}}(\lambda _n-y) \end{aligned}$$

for $0 \le y \le \lambda _n-a-x$ and

$$\begin{aligned} e^{-\gamma a}\overline{\rho _2}(\lambda _n-a-y)-\overline{\rho _2}(\lambda _n-y)\ge -\epsilon {{\bar{\zeta }}}(\lambda _n-y) \end{aligned}$$

for $0 \le y \le x.$ By (3.2) and (3.3) of Lemma 3.1, we have only to prove that

$$\begin{aligned} \sum _{j=3}^{4}\liminf _{x \rightarrow \infty }\liminf _{n \rightarrow \infty }\frac{I_j}{{{\bar{\zeta }}}(\lambda _n)} \ge 0. \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} I_3&\ge -\epsilon \int _{0-}^{(\lambda _n-a-x)+}{{\bar{\zeta }}}(\lambda _n-y)\rho _2(\hbox {d}y)\\&\ge -\epsilon \left( {{\bar{\zeta }}}(\lambda _n)+\int _{a+x}^{\lambda _n}\overline{\rho _2}(\lambda _n-y)\zeta (\hbox {d}y)\right) \\&\ge -\epsilon \left( {{\bar{\zeta }}}(\lambda _n)+C_2\overline{\zeta ^{2*}}(\lambda _n)\right) , \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} I_4&\ge -\epsilon \int _{0-}^{x+}{{\bar{\zeta }}}(\lambda _n-y)\rho _1(\hbox {d}y)\\&\ge -\epsilon \left( {{\bar{\zeta }}}(\lambda _n)+\int _{\lambda _n-x}^{\lambda _n}\overline{\rho _1}(\lambda _n-y)\zeta (\hbox {d}y)\right) \\&\ge -\epsilon \left( {{\bar{\zeta }}}(\lambda _n)+C_1\overline{\zeta ^{2*}}(\lambda _n)\right) . \end{aligned} \end{aligned}$$

Thus, we see that

$$\begin{aligned} \liminf _{n \rightarrow \infty }\frac{I_3}{{{\bar{\zeta }}}(\lambda _n)}\ge -\epsilon (1+C_2d^*), \end{aligned}$$

and

$$\begin{aligned} \liminf _{n \rightarrow \infty }\frac{I_4}{{{\bar{\zeta }}}(\lambda _n)}\ge -\epsilon (1+C_1d^*). \end{aligned}$$

Since $\epsilon >0$ is arbitrary, we established, for $j =3,4$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\liminf _{n \rightarrow \infty }\frac{I_j}{{{\bar{\zeta }}}(\lambda _n)}\ge 0. \end{aligned}$$

Thus, we have proved (3.6). $\square$

Lemma 3.4

Let $\gamma \ge 0$. Suppose that $\zeta \in \mathcal {OS}\cap {\mathcal {L}}(\gamma )$. For $j=1,2$, let $\rho _j$ be distributions on ${\mathbb {R}}_+$ satisfying (3.1). Suppose further that, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_1(x-a)-{{\bar{\rho }}}_1(x)}{{{\bar{\zeta }}}(x)} = 0 \end{aligned}$$

and, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _1*\rho _2}(x-a)- \overline{\rho _1*\rho _2}(x)}{{{\bar{\zeta }}}(x)} = 0 \end{aligned}$$

(3.7)

and that $e^{\gamma x}\rho _1(\hbox {d}x) \in {\mathcal {W}}$. Then, we have, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_2(x-a)-{{\bar{\rho }}}_2(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

(3.8)

Proof

Let $\Lambda _2$ be the totality of increasing sequences $\{\lambda _n\}_{n=1}^{\infty }$ with $\lim _{n \rightarrow \infty } \lambda _n = \infty$ such that, for every $x \in {\mathbb {R}}$, the following limit exists and is finite:

$$\begin{aligned} m_2(x;\{\lambda _n\}):=\lim _{n \rightarrow \infty }\frac{{{\bar{\rho }}}_2(\lambda _n-x)}{{{\bar{\zeta }}}(\lambda _n)}. \end{aligned}$$

We have $\Lambda _2\subset \Lambda$. As for $\Lambda$, it follows that, under the assumption that $\zeta \in \mathcal {OS}$ and $\overline{\rho _2}(x) \le C_2{{\bar{\zeta }}}(x)$, there exists an increasing subsequence $\{\lambda _n\} \in \Lambda _2$ of $\{x_n\}$ for each sequence $\{x_n\}_{n=1}^{\infty }$ with $\lim _{n \rightarrow \infty } x_n = \infty$. Let $\{\lambda _n\} \in \Lambda _2.$ Recall from Lemma 2.4 that $m(x;\{\lambda _n\}) =e^{\gamma x}$ and ${{\widehat{\zeta }}}(\gamma ) < \infty .$ As in the proof of Lemma 3.2, we have (3.5). We find that, for every $a\in {\mathbb {R}}$,

$$\begin{aligned} \begin{aligned} l(x)&:=\lim _{n \rightarrow \infty }\frac{I_4}{{{\bar{\zeta }}}(\lambda _n)}\\&=\int _{0-}^{x+}(e^{-\gamma a}m_2(a+y;\{\lambda _n\})-m_2(y;\{\lambda _n\}))\rho _1(\hbox {d}y). \end{aligned} \end{aligned}$$

Define $M_2(y;\{\lambda _n\}):=e^{-\gamma y}m_2(y;\{\lambda _n\})$. Then $M_2(y;\{\lambda _n\})\le C_2$ on ${\mathbb {R}}$. Note that

$$\begin{aligned} l(x)=\int _{0-}^{x+}(M_2(a+y;\{\lambda _n\})-M_2(y;\{\lambda _n\}))e^{\gamma y}\rho _1(\hbox {d}y). \end{aligned}$$

We see from (3.2), (3.3) of Lemma 3.1, (3.5), and (3.7) that, for every $a\in {\mathbb {R}}$,

$$\begin{aligned} \lim _{x \rightarrow \infty }l(x)=\int _{0-}^{\infty }(M_2(a+y;\{\lambda _n\})-M_2(y;\{\lambda _n\}))e^{\gamma y}\rho _1(\hbox {d}y)=0. \end{aligned}$$

Thus, we obtain that, for every $a, b \in {\mathbb {R}}$,

$$\begin{aligned} \int _{0-}^{\infty }(M_2(a+b+y;\{\lambda _n\})-M_2(b+y;\{\lambda _n\}))e^{\gamma y}\rho _1(\hbox {d}y)=0. \end{aligned}$$

Since $e^{\gamma y}\rho _1(\hbox {d}y) \in {\mathcal {W}}$, we find from Lemma 2.7 that, for every $a\in {\mathbb {R}}$,

$$\begin{aligned} M_2(a+b;\{\lambda _n\})=M_2(b;\{\lambda _n\})\,\, \text{ for } \text{ a.e. } \,b \in {\mathbb {R}}. \end{aligned}$$

Since the function $m_2(x;\{\lambda _n\})$ is increasing, the functions $M_2(x+;\{\lambda _n\})$ and $M_2(x-;\{\lambda _n\})$ exist for all $x \in {\mathbb {R}}$. Taking $b_n=b_n(a)\downarrow 0$ and $b_n=b_n(a)\uparrow 0$, we have

$$\begin{aligned} M_2(a+;\{\lambda _n\})=M_2(0+;\{\lambda _n\})\, \text{ and } \, M_2(a-;\{\lambda _n\})=M_2(0-;\{\lambda _n\}). \end{aligned}$$

As $a\uparrow 0$ in the first equality, we see that

$$\begin{aligned} M_2(0-;\{\lambda _n\})=M_2(0+;\{\lambda _n\}) \end{aligned}$$

and hence, for every $a\in {\mathbb {R}}$,

$$\begin{aligned} M_2(a;\{\lambda _n\})=M_2(0;\{\lambda _n\}). \end{aligned}$$

Since $\{\lambda _n\} \in \Lambda _2$ is arbitrary, we have (3.8). $\square$

Lemma 3.5

Let $\gamma \ge 0$. Suppose that $\zeta \in \mathcal {OS}$. For $j=1,2$, let $\rho _j$ be distributions on ${\mathbb {R}}_+$ satisfying (3.1). Suppose further that, for $j=1,2,$ and for every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_j(x-a)-{{\bar{\rho }}}_j(x)}{{{\bar{\zeta }}}(x)} \ge 0. \end{aligned}$$

If we have, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _1*\rho _2}(x-a)-\overline{\rho _1*\rho _2}(x)}{{{\bar{\zeta }}}(x)} = 0, \end{aligned}$$

then, for $j=1,2,$ and for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_j(x-a)-{{\bar{\rho }}}_j(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

Proof

Suppose that, for some $a >0$,

$$\begin{aligned} \limsup _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _2}(x-a)-\overline{\rho _2}(x)}{{{\bar{\zeta }}}(x)} >0. \end{aligned}$$

Then there is $\{\lambda _n\} \in \Lambda$ such that, for some $a >0$,

$$\begin{aligned} \lim _{n \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho _2}(\lambda _n-a)-\overline{\rho _2}(\lambda _n)}{{{\bar{\zeta }}}(\lambda _n)} =: \delta _0 >0. \end{aligned}$$

So there is $\delta _1 >0$ such that, for some $a > 0$,

$$\begin{aligned} \liminf _{n \rightarrow \infty }\frac{e^{-\gamma (a+\delta _1)}\overline{\rho _2}(\lambda _n-a)-\overline{\rho _2}(\lambda _n)}{{{\bar{\zeta }}}(\lambda _n)} =: \delta _2>0. \end{aligned}$$

Take $y_0$ such that $x>y_0 >\delta _1$ and $\rho _1((y_0-\delta _1,y_0]) >0$. Let $\lambda _n^{\prime }:=\lambda _n+y_0$ and $a^{\prime }:=a+\delta _1$. Then we have

$$\begin{aligned} \begin{aligned}&\int _{y_0-\delta _1}^{y_0}(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _2}(\lambda _n^{\prime }-y))\rho _1(\hbox {d}y)\\&\quad \ge \rho _1((y_0-\delta _1,y_0])(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n-a)-\overline{\rho _2}(\lambda _n)). \end{aligned} \end{aligned}$$

(3.9)

Let $\lambda _n^{\prime } > a^{\prime }+x$ and $x >0$. Define J as

$$\begin{aligned} J:=e^{-\gamma a^{\prime }}\overline{\rho _1*\rho _2}(\lambda _n^{\prime }-a^{\prime })-\overline{\rho _1*\rho _2}(\lambda _n^{\prime }). \end{aligned}$$

Then we have as in assertion (i) of Lemma 3.1

$$\begin{aligned} J=\sum _{j=1}^{4} I_j^{\prime }, \end{aligned}$$

where

$$\begin{aligned} I_1^{\prime }:= & {} -\int _{\lambda _n^{\prime }-a^{\prime }-x}^{\lambda _n^{\prime }-x}\overline{\rho _1}(\lambda _n^{\prime }-y)\rho _2(\hbox {d}y),\\ I_2^{\prime }:= & {} \overline{\rho _1}(x)(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-x)-\overline{\rho _2}(\lambda _n^{\prime }-x)),\\ I_3^{\prime }:= & {} \int _{0-}^{(\lambda _n^{\prime }-a^{\prime }-x)+}(e^{-\gamma a^{\prime }}\overline{\rho _1}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _1}(\lambda _n^{\prime }-y))\rho _2(\hbox {d}y), \end{aligned}$$

and

$$\begin{aligned} I_4^{\prime }:=\int _{0-}^{x+}(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _2}(\lambda _n^{\prime }-y))\rho _1(\hbox {d}y). \end{aligned}$$

For $1\le j\le 3,$ let

$$\begin{aligned} J_j:=I_j^{\prime }, \end{aligned}$$

and let

$$\begin{aligned} I_4^{\prime }=\sum _{j=4}^{6}J_j, \end{aligned}$$

where

$$\begin{aligned} J_4:= & {} \int _{0-}^{(y_0-\delta _1)+}(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _2}(\lambda _n^{\prime }-y))\rho _1(\hbox {d}y), \\ J_5:= & {} \int _{y_0}^x(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _2}(\lambda _n^{\prime }-y))\rho _1(\hbox {d}y), \end{aligned}$$

and

$$\begin{aligned} J_6:=\int _{y_0-\delta _1}^{y_0}(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n^{\prime }-a^{\prime }-y)-\overline{\rho _2}(\lambda _n^{\prime }-y))\rho _1(\hbox {d}y). \end{aligned}$$

Then we have

$$\begin{aligned} J=\sum _{j=1}^{6} J_j. \end{aligned}$$

As in the proof of Lemma 3.3, we see from the assumption and (3.9) that

$$\begin{aligned} \begin{aligned} 0&=\lim _{n\rightarrow \infty }\frac{J}{{{\bar{\zeta }}}(\lambda _n^{\prime })}\\&\ge \sum _{j=1}^{6}\liminf _{x\rightarrow \infty }\liminf _{n\rightarrow \infty }\frac{J_j}{{{\bar{\zeta }}}(\lambda _n^{\prime })}\\&\ge \liminf _{n\rightarrow \infty }\frac{J_6}{{{\bar{\zeta }}}(\lambda _n^{\prime })}\\&\ge \liminf _{n\rightarrow \infty }\rho _1((y_0-\delta _1,y_0])\frac{(e^{-\gamma a^{\prime }}\overline{\rho _2}(\lambda _n-a)-\overline{\rho _2}(\lambda _n))}{{{{\bar{\zeta }}}(\lambda _n^{\prime })}}\\&=\rho _1((y_0-\delta _1,y_0])\frac{\delta _2}{m(-y_0;\{\lambda _n\})} >0. \end{aligned} \end{aligned}$$

This is a contradiction. Thus, we have, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_2(x-a)-{{\bar{\rho }}}_2(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

By the analogous argument, we have for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}_1(x-a)-{{\bar{\rho }}}_1(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

Thus, we have proved the lemma. $\square$

Lemma 3.6

Let $\gamma \ge 0$. Let $\rho$ be a distribution on ${\mathbb {R}}_+$. Suppose that $\rho \in \mathcal {OS}$ and, for every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }e^{-\gamma a}{{\bar{\rho }}}(x-a)/{{\bar{\rho }}}(x) \ge 1. \end{aligned}$$

(3.10)

Then, for some positive integer $n\ge 2$, $\rho ^{n*} \in {\mathcal {L}}(\gamma )$ implies that $\rho \in {\mathcal {L}}(\gamma )$.

Proof

Let $\zeta :=\rho$. Then we see from Lemma 3.3 that, for every $k \in {\mathbb {N}}$ and every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\rho ^{k*}}(x-a)-\overline{\rho ^{k*}}(x)}{{{\bar{\rho }}}(x)} \ge 0. \end{aligned}$$

Thus, we find that $\rho _1:=\rho$ and $\rho _2:=\rho ^{(n-1)*}$ satisfy the assumptions of Lemma 3.5. Hence, we have by Lemma 3.5, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\rho }}}(x-a)-{{\bar{\rho }}}(x)}{{{\bar{\rho }}}(x)} = 0. \end{aligned}$$

That is, $\rho \in {\mathcal {L}}(\gamma )$. $\square$

Remark 3.1

For $\gamma = 0$, the assumption (3.10) necessarily holds, but for $\gamma > 0$, without the assumption (3.10) the lemma does not hold. For $\gamma > 0$, Watanabe [14] made a distribution $\eta \in \mathcal {OS}$ such that $\eta ^{n*} \in {\mathcal {L}}(\gamma )$ for every $n \ge 2$ but $\eta \notin {\mathcal {L}}(\gamma )$.

4 Proof of Results

In this section, we prove the results stated in Sect. 1.

Lemma 4.1

Let $\gamma \ge 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. If, for every $a \ge 0$, (1.1) holds, then, for all $n\in {\mathbb {N}}$ and every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\nu _1^{n*}} (x-a)-\overline{\nu _1^{n*}}(x)}{ \overline{\nu _1^{n_0*}}(x)}=0, \end{aligned}$$

(4.1)

and we have $T(\mu ,\gamma )=(0,\infty )$.

Proof

By induction, we see from Lemma 3.2 that if (1.1) holds for every $a \ge 0$, then, for all $n\in {\mathbb {N}}$ and every $a \ge 0$, we have (4.1). We have with $c:={{\bar{\nu }}}(1)$, for $t>0$,

$$\begin{aligned} \mu _1^{t*}:=e^{-ct}\sum _{k=0}^{\infty }\frac{(ct)^k}{k!}\nu _1^{k*}. \end{aligned}$$

Suppose that, for all $n\in {\mathbb {N}}$ and every $a \ge 0$, (4.1) holds. Let $\epsilon > 0$ be arbitrary. By Lemma 2.3, we can choose sufficiently large $N \in {\mathbb {N}}$ such that, for $\epsilon > 0$,

$$\begin{aligned} e^{-ct}\sum _{k=N+1}^{\infty }\frac{(ct)^k}{k!}\frac{|e^{-\gamma a}\overline{\nu _1^{k*}} (x-a)-\overline{\nu _1^{k*}}(x)|}{ \overline{\nu _1^{n_0*}}(x)} < \epsilon . \end{aligned}$$

We find from (4.1) that, for every $a \ge 0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }e^{-ct}\sum _{k=1}^{N}\frac{(ct)^k}{k!}\frac{e^{-\gamma a}\overline{\nu _1^{k*}} (x-a)-\overline{\nu _1^{k*}}(x)}{ \overline{\nu _1^{n_0*}}(x)} =0. \end{aligned}$$

Thus, we see that, for every $a \ge 0$ and for every $t>0$,

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\mu _1^{t*}}(x-a)-\overline{\mu _1^{t*}}(x)}{ \overline{\nu _1^{n_0*}}(x)} =0. \end{aligned}$$

Since $\overline{\mu _1^{t*}}(x)\asymp \overline{\nu _1^{n_0*}}(x)$ for every $t>0$, we have $T(\mu ,\gamma )=(0,\infty )$. $\square$

Lemma 4.2

Let $\gamma \ge 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. If 0 is a limit point of $T(\mu ,\gamma )$, then, for every $a \ge 0$, (1.1) holds.

Proof. Suppose that 0 is a limit point of $T(\mu ,\gamma )$. Then, there exists a strictly decreasing sequence $\{t_n\}_{n=1}^{\infty }$ in $T(\mu ,\gamma )$ converging to 0 as $n \rightarrow \infty$. We have with $c:={{\bar{\nu }}}(1)$

$$\begin{aligned} \mu _1^{t_n*}:=e^{-ct_n}\sum _{k=0}^{\infty }\frac{(ct_n)^k}{k!}\nu _1^{k*}. \end{aligned}$$

Since $\{t_n\}_{n=1}^{\infty }$ in $T(\mu ,\gamma )$ and $\overline{\mu _1^{t_n*}}(x) \asymp \overline{\nu _1^{n_0*}}(x)$ from Lemma 2.2, we see that, for every $a \ge 0$,

$$\begin{aligned} \begin{aligned}&\lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\mu _1^{t_n*}}(x-a)-\overline{\mu _1^{t_n*}}(x)}{ \overline{\nu _1^{n_0*}}(x)}\\&\quad =\lim _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\mu _1^{t_n*}}(x-a)-\overline{\mu _1^{t_n*}}(x)}{ \overline{\mu _1^{t_n*}}(x)}\frac{ \overline{\mu _1^{t_n*}}(x)}{\overline{\nu _1^{n_0*}}(x)}=0. \end{aligned} \end{aligned}$$

Thus, we obtain from Lemma 2.3 that, for every $a \ge 0$,

$$\begin{aligned} \begin{aligned}&\limsup _{x \rightarrow \infty }|\frac{e^{-\gamma a}\overline{\nu _1}(x-a)-\overline{\nu _1}(x)}{ \overline{\nu _1^{n_0*}}(x)}| \\&\quad =\limsup _{n \rightarrow \infty }\limsup _{x \rightarrow \infty }|\frac{e^{ct_n}}{ct_n}\frac{e^{-\gamma a}\overline{\mu _1^{t_n*}}(x-a)-\overline{\mu _1^{t_n*}}(x)}{ \overline{\nu _1^{n_0*}}(x)} -\frac{e^{-\gamma a}\overline{\nu _1}(x-a)-\overline{\nu _1}(x)}{ \overline{\nu _1^{n_0*}}(x)}|\\&\quad \le \limsup _{n \rightarrow \infty }\limsup _{x \rightarrow \infty }\sum _{k=2}^{\infty }\frac{(ct_n)^{(k-1)}}{k!}\frac{e^{-\gamma a}\overline{\nu _1^{k*}}(x-a)+\overline{\nu _1^{k*}}(x)}{ \overline{\nu _1^{n_0*}}(x)} =0. \end{aligned} \end{aligned}$$

Thus, we have (1.1) for every $a \ge 0$. $\Box$

Lemma 4.3

Let $\gamma \ge 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. If $t_0, t_1 \in T(\mu ,\gamma )$ with $t_1 >t_0$, then $t_1-t_0 \in T(\mu ,\gamma )$. If $T(\mu ,\gamma )$ has a limit point, then $T(\mu ,\gamma ) =(0,\infty )$. If $T(\mu ,\gamma )$ has the minimum $a_0 >0$, then $T(\mu ,\gamma ) =a_0{\mathbb {N}}$.

Proof

Suppose that $t_0, t_1 \in T(\mu ,\gamma )$ with $t_1 >t_0$. Let $\zeta :=\rho _1:=\mu ^{t_0*}$ and $\rho _2:=\mu ^{(t_1-t_0)*}$. The distribution $e^{\gamma x}\rho _1(\hbox {d}x)/{{\widehat{\rho }}}_1(\gamma )$ is an exponentially tilted infinitely divisible distribution and hence itself is infinitely divisible, thus having a non-vanishing characteristic function. That is, $e^{\gamma x}\rho _1(\hbox {d}x) \in {\mathcal {W}}$. See (iii) of Theorem 25.17 of Sato [10]. Thus, we see from Lemma 3.4 that $\mu ^{(t_1-t_0)*}\in {\mathcal {L}}(\gamma )$. Thus, if $T(\mu ,\gamma )$ has a limit point, then 0 is a limit point of $T(\mu ,\gamma )$, and hence, by Lemmas 4.1 and 4.2, $T(\mu ,\gamma ) =(0,\infty )$. If $T(\mu ,\gamma )$ has the minimum $a_0 >0$, then clearly $a_0{\mathbb {N}}\subset T(\mu ,\gamma )$ and $T(\mu ,\gamma ){\setminus } a_0{\mathbb {N}} =\emptyset$. $\square$

Proof of Theorem 1.1

Assertion (i) is clear from Lemmas 4.1, 4.2, and 4.3. The first part of assertion (ii) is due to Lemmas 4.1 and 4.2. Suppose that $\mu \in {\mathcal {A}}(\gamma )$. If $n < n_0,$ then $\nu _1^{n*} \not \in \mathcal {OS}$ simply because of the definition of $n_0$. If $n \ge n_0$ and x is large, then $\overline{\nu _1^{n*}}(x) \ge \overline{\nu _1^{n_0*}}(x)$, and hence, (4.1) implies that $\nu _1^{n*} \in {\mathcal {L}}(\gamma )$. $\square$

Proof of Corollary 1.1

Suppose that ${\mathcal {C}}(\gamma )$ is not empty. Then there is the minimum $a_0 >0$ in $T(\mu ,\gamma )$ for $\mu \in {\mathcal {C}}(\gamma )$. Since $a_0>0$ is a period of $T(\mu ,\gamma )$, for $n=2$, $\mu ^{a_0*}=(\mu ^{(a_0/n)*})^{n*} \in {\mathcal {L}}(\gamma )$ but $(\mu ^{(a_0/n)*})^{(n+1)*} \notin {\mathcal {L}}(\gamma )$ and $\mu ^{(a_0/n)*} \notin {\mathcal {L}}(\gamma )$. Thus, Hypotheses I and II are not true. Suppose that ${\mathcal {C}}(\gamma )$ is empty. Then, obviously, Hypotheses I and II are true. Thus, (1), (2), and (3) are equivalent. We prove the equivalence of (3) and (4). Suppose that ${\mathcal {C}}(\gamma )$ is empty. Then for every $\mu \in \mathcal {OS}\cap \mathcal {ID}$ it holds that, for every $2t \in T(\mu ,\gamma )$, $\mu _1^{t*}\in {\mathcal {L}}(\gamma )$, and hence, for all $a \ge 0$, (1.2) holds. Conversely, suppose that ${\mathcal {C}}(\gamma )$ is not empty and, for $a_0=2t \in T(\mu ,\gamma )$ with $\mu \in {\mathcal {C}}(\gamma )$ and for all $a \ge 0$, (1.2) holds. Letting $\rho _1:=\rho _2:=\mu _1^{t*}$, $\zeta :=\mu _1^{2t*}$, define $\Lambda _2$ as in Lemma 3.4 and let $\{\lambda _n\} \in \Lambda _2\subset \Lambda$. We have (3.3) by Lemma 3.1 for $j=1,2$. We have $I_3+I_4=2I_4+I_5$, where

$$\begin{aligned} I_5:=\int _x^{\lambda _n-a-x}(e^{-\gamma a}\overline{\rho _1}(\lambda _n-a-y)-\overline{\rho _1}(\lambda _n-y))\rho _2(\hbox {d}y), \end{aligned}$$

We have by the assumption (1.2) for every $a \ge 0$

$$\begin{aligned} \limsup _{x\rightarrow \infty }\limsup _{n \rightarrow \infty }\frac{|I_5|}{{{\bar{\zeta }}}(\lambda _n)}=0. \end{aligned}$$

Define $M_2(y;\{\lambda _n\}):=e^{-\gamma y}m_2(y;\{\lambda _n\})$. Thus, we find from (3.2), (3.3), and $2t \in T(\mu ,\gamma )$ that, for every $a \ge 0$,

$$\begin{aligned} \begin{aligned}&\lim _{x\rightarrow \infty }\lim _{n \rightarrow \infty }\frac{I_4}{{{\bar{\zeta }}}(\lambda _n)}\\&\quad =\int _{0-}^{\infty }(e^{-\gamma a}m_2(a+y;\{\lambda _n\})-m_2(y;\{\lambda _n\}))\rho _1(\hbox {d}y)\\&\quad =\int _{0-}^{\infty }(M_2(a+y;\{\lambda _n\})-M_2(y;\{\lambda _n\}))e^{\gamma y}\rho _1(\hbox {d}y)=0. \end{aligned} \end{aligned}$$

The distribution $e^{\gamma x}\rho _1(\hbox {d}x)/{{\widehat{\rho }}}_1(\gamma )$ is an exponentially tilted infinitely divisible distribution and hence itself is infinitely divisible, thus having a non-vanishing characteristic function. That is,

$$\begin{aligned} e^{\gamma y}\rho _1(\hbox {d}y)=e^{\gamma y}\mu _1^{t*}(\hbox {d}y)\in {\mathcal {W}}. \end{aligned}$$

As in the proof of Lemma 3.4, we have $\rho _2=\mu _1^{t*}\in {\mathcal {L}}(\gamma )$. This is a contradiction. Thus, (3) and (4) are equivalent. $\square$

Proof of Remark 1.1

Let $\gamma =0$. Then we see from Lemma 3.6 that Hypothesis II is true. Thus, ${\mathcal {C}}(0)$ is empty, and hence, Remark 1.1 follows from Theorem 1.1. $\square$

Proof of Proposition 1.1

Let $\gamma > 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. Suppose that (1.3) holds for every $a \ge 0$. Let $\zeta :=\nu _1^{n_0*}$. Then, by induction, we see from (1.3) and Lemma 3.3 that, for every $n \in {\mathbb {N}}$ and every $a \ge 0$,

$$\begin{aligned} \liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\nu _1^{n*}}(x-a)-\overline{\nu _1^{n*}}(x)}{\overline{\nu _1^{n_0*}}(x)} \ge 0. \end{aligned}$$

Let $\epsilon >0$ be arbitrary. Thus, letting $N \in {\mathbb {N}}$ sufficiently large, we have, for every $t >0$ and for every $a \ge 0$,

$$\begin{aligned} \begin{aligned}&\liminf _{x \rightarrow \infty }\frac{e^{-\gamma a}\overline{\mu _1^{t*}}(x-a)-\overline{\mu _1^{t*}}(x)}{ \overline{\nu _1^{n_0*}}(x)}\\&\quad =\liminf _{x \rightarrow \infty }e^{-ct}\sum _{k=1}^N\frac{(ct)^k}{k!}\frac{e^{-\gamma a}\overline{\nu _1^{k*}} (x-a)-\overline{\nu _1^{k*}}(x)}{ \overline{\nu _1^{n_0*}}(x)} \\&\quad \quad - \limsup _{x \rightarrow \infty }e^{-ct}\sum _{k=N+1}^{\infty }\frac{(ct)^k}{k!}\frac{e^{-\gamma a}\overline{\nu _1^{k*}} (x-a)+\overline{\nu _1^{k*}}(x)}{ \overline{\nu _1^{n_0*}}(x)}\ge -\epsilon . \end{aligned} \end{aligned}$$

Since $\epsilon >0$ is arbitrary and $\overline{\nu _1^{n_0*}}(x)\asymp \overline{\mu _1^{(t/n)*}}(x)$ for every $n \in {\mathbb {N}}$, we obtain that $\rho :=\mu _1^{(t/n)*}$ satisfies $\rho \in \mathcal {OS}$ and (3.10) holds. Hence, we find from Lemma 3.6 that if $t \in T(\mu ,\gamma )$, then $t/n \in T(\mu ,\gamma )$ for every $n \in {\mathbb {N}}$. Thus, by Lemmas 4.1 and 4.2, either $T(\mu ,\gamma )= (0,\infty )$ or $\emptyset$. $\square$

Proof of Proposition 1.2

Suppose that $\nu _1^{2*} \in {\mathcal {L}}(\gamma )$ and the real part of ${{\widehat{\nu }}}_1(\gamma +iz)$ is not 0 for every $z \in {\mathbb {R}}$. If $t \in T(\mu ,\gamma )$, then

$$\begin{aligned} \mu _1^{t*}=e^{-ct}\sum _{k=0}^{\infty }\frac{(ct)^k}{k!}\nu _1^{k*}\in {\mathcal {L}}(\gamma )\cap \mathcal {OS}. \end{aligned}$$

(4.2)

Define distributions $\eta _1$ and $\eta _2$ on ${\mathbb {R}}_+$ as

$$\begin{aligned} \eta _1:=(\cosh (ct))^{-1}\sum _{k=0}^{\infty }\frac{(ct)^{2k}}{(2k)!}\nu _1^{(2k)*} \end{aligned}$$

and

$$\begin{aligned} \eta _2:=(\sinh (ct))^{-1}\sum _{k=0}^{\infty }\frac{(ct)^{2k+1}}{(2k+1)!}\nu _1^{(2k+1)*}. \end{aligned}$$

We see from Proposition 3.1 of Shimura and Watanabe [11] that $\eta _j\in \mathcal {OS}$ and $\overline{\eta _j}(x)\asymp \overline{\nu _1^{n_0*}}(x)$ for $j=1,2$. Let $\epsilon >0$ be arbitrary. We obtain from Lemma 2.3 that there is a positive integer $N=N(a,\epsilon ,t)$ such that

$$\begin{aligned} \limsup _{x \rightarrow \infty }(\cosh (ct))^{-1}\sum _{k=N+1}^{\infty }\frac{(ct)^{2k}}{(2k)!}\frac{e^{-\gamma a}\overline{\nu _1^{(2k)*}}(x-a)+\overline{\nu _1^{(2k)*}}(x)}{\overline{\nu _1^{n_0*}}(x)}<\epsilon . \end{aligned}$$

Since $\nu _1^{(2k)*}\in {\mathcal {L}}(\gamma )$ for every $k \ge 0$, we have, for every $a \ge 0$ and every $t >0$,

$$\begin{aligned} \limsup _{x \rightarrow \infty }(\cosh (ct))^{-1}\sum _{k=0}^{N}\frac{(ct)^{2k}}{(2k)!}\frac{|e^{-\gamma a}\overline{\nu _1^{(2k)*}}(x-a)-\overline{\nu _1^{(2k)*}}(x)|}{\overline{\nu _1^{n_0*}}(x)}=0. \end{aligned}$$

Thus, with some $C=C(t) >0$ we have, for every $a \ge 0$ and every $t >0$,

$$\begin{aligned}&\limsup _{x \rightarrow \infty }\frac{|e^{-\gamma a}\overline{\eta _1}(x-a)-\overline{\eta _1}(x)|}{\overline{\eta _1}(x)} \\&\quad \le \limsup _{x \rightarrow \infty }(\cosh (ct))^{-1}\sum _{k=0}^{N}\frac{(ct)^{2k}}{(2k)!}\frac{|e^{-\gamma a}\overline{\nu _1^{(2k)*}}(x-a)-\overline{\nu _1^{(2k)*}}(x)|}{C\overline{\nu _1^{n_0*}}(x)}\\&\quad \quad +\,\limsup _{x \rightarrow \infty }(\cosh (ct))^{-1}\sum _{k=N+1}^{\infty }\frac{(ct)^{2k}}{(2k)!}\frac{e^{-\gamma a}\overline{\nu _1^{(2k)*}}(x-a)+\overline{\nu _1^{(2k)*}}(x)}{C\overline{\nu _1^{n_0*}}(x)}\le \epsilon /C. \end{aligned}$$

Since $\epsilon >0$ is arbitrary, we have

$$\begin{aligned} \eta _1\in {\mathcal {L}}(\gamma )\cap \mathcal {OS}. \end{aligned}$$

(4.3)

Since

$$\begin{aligned} \sinh (ct)\eta _2=e^{ct}\mu _1^{t*} -\cosh (ct)\eta _1, \end{aligned}$$

we have by (4.2) and (4.3)

$$\begin{aligned} \eta _2\in {\mathcal {L}}(\gamma )\cap \mathcal {OS}. \end{aligned}$$

Let $\zeta :=\rho _1:=\eta _2$ and $\rho _2:=\nu _1.$ Then, by argument similar to the proof of (4.3),

$$\begin{aligned} \rho _1*\rho _2=(\sinh (ct))^{-1}\sum _{k=0}^{\infty }\frac{(ct)^{2k+1}}{(2k+1)!}\nu _1^{(2k+2)*}\in {\mathcal {L}}(\gamma )\cap \mathcal {OS}. \end{aligned}$$

Since the real part of ${{\widehat{\nu }}}_1(\gamma +iz)$ is not 0 for every $z \in {\mathbb {R}}$,

$$\begin{aligned} 2\sinh (ct){{\widehat{\rho }}}_1(\gamma +iz)=\exp (ct{{\widehat{\nu }}}_1(\gamma +iz))-\exp (-ct{{\widehat{\nu }}}_1(\gamma +iz))\ne 0 \end{aligned}$$

for every $z \in {\mathbb {R}}$, that is, $e^{\gamma x}\rho _1(\hbox {d}x) \in {\mathcal {W}}$. Thus, we see from Lemma 3.4 that

$$\begin{aligned} \lim _{x \rightarrow \infty }\frac{e^{-\gamma a}{{\bar{\nu }}}_1(x-a)-{{\bar{\nu }}}_1(x)}{{{\bar{\zeta }}}(x)} = 0. \end{aligned}$$

Since ${{\bar{\zeta }}}(x)\asymp \overline{\nu _1^{n_0*}}(x)$, we see from Theorem 1.1 that $T(\mu ,\gamma )=(0,\infty )$. Thus, we have proved the proposition. $\square$

Proof of Proposition 1.3

Let $\gamma > 0$ and $\mu \in \mathcal {OS}\cap \mathcal {ID}$. Suppose that $\nu _1^{n_1*} \in {\mathcal {S}}_{\sharp }$. Since $\overline{\mu ^{t*}}(x)\asymp \overline{\nu _1^{n_1*} }(x)$, we have $\mu ^{t*}\in {\mathcal {S}}_{\sharp }$ for every $t >0$. Thus, we see from Lemmas 2.5 and 2.6 that if $T(\mu ,\gamma )\ne \emptyset$, then $\nu _1 \in {\mathcal {S}}(\gamma )$ and hence $T(\mu ,\gamma )=(0,\infty )$. That is, either $T(\mu ,\gamma )= (0,\infty )$ or $\emptyset$. Moreover, $T(\mu ,\gamma )=(0,\infty )$ if and only if $\nu _1 \in {\mathcal {S}}(\gamma )$. $\square$