#  Bivariate Bernstein-gamma functions, potential measures, and asymptotics of
exponential functionals of Lévy processes

M. Minchev  Email<EMAIL_ADDRESS>Faculty of Mathematics and
Informatics, Sofia University ”St. Kliment Ohridski”, 5, James Bourchier
blvd., 1164 Sofia, Bulgaria  M. Savov  Email<EMAIL_ADDRESS><EMAIL_ADDRESS>Faculty of Mathematics and Informatics, Sofia
University ”St. Kliment Ohridski”, 5, James Bourchier blvd., 1164 Sofia,
Bulgaria  Institute of Mathematics and Informatics, Bulgarian Academy of
Sciences, Akad. Georgi Bonchev str., Block 8, Sofia 1113, Bulgaria

###### Abstract

Let $\xi$ be a Lévy process and
$I_{\xi}(t):=\int_{0}^{t}e^{-\xi_{s}}\mathrm{d}s$, $t\geq 0,$ be the
exponential functional of Lévy processes on deterministic horizon. Given that
$\lim_{t\to\infty}\xi_{t}=-\infty$ we evaluate for general functions $F$ an
upper bound on the rate of decay of $\mathbb{E}\left[F(I_{\xi}(t))\right]$
based on an explicit integral criterion. When
$\mathbb{E}\left[\xi_{1}\right]\in\left(-\infty,0\right)$ and
$\mathbb{P}\left(\xi_{1}>t\right)$ is regularly varying of index $\alpha>1$ at
infinity, we show that the law of $I_{\xi}(t)$, suitably normed and rescaled,
converges weakly to a probability measure stemming from a new generalisation
of the product factorisation of classical exponential functionals. These
results substantially improve upon the existing literature and are obtained
via a novel combination between Mellin inversion of the Laplace transform of
$\mathbb{E}\left[I^{-a}_{\xi}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]$, $a\in(0,1)$, $x\in(0,\infty],$ and Tauberian theory
augmented for integer-valued $\alpha$ by a suitable application of the one-
large jump principle in the context of the de Haan theory. The methodology
rests upon the representation of the aforementioned Mellin transform in terms
of the recently introduced bivariate Bernstein-gamma functions for which we
develop the following new results of independent interest (for general $\xi$):
we link these functions to the $q$-potentials of $\xi$; we show that their
derivatives at zero are finite upon the finiteness of the aforementioned
integral criterion; we offer neat estimates of those derivatives along complex
lines. These results are useful in various applications of the exponential
functionals themselves and in different contexts where properties of bivariate
Bernstein-gamma functions are needed. $\xi$ need not be non-lattice.

Keywords: Exponential functionals of Lévy processes, Special functions,
Bernstein and bivariate Bernstein-gamma functions, Mellin and Laplace
transforms, Tauberian theorems, de Haan theory, Regular variation, Asymptotic
behaviour

MSC2020 Classification: 60G51, 60E07, 40E05, 33E03, 44A10, 44A99

###### Contents

  1. 1 Introduction and motivation
  2. 2 Preliminaries, methodology, background, literature review, and an open problem
    1. 2.1 Notation
    2. 2.2 Bivariate Bernstein-gamma functions and Mellin transform of exponential functionals
    3. 2.3 Methodology
    4. 2.4 Background and previous results
    5. 2.5 Open problems
  3. 3 Main results on exponential functionals
  4. 4 Derivatives of bivariate Bernstein–gamma functions
  5. 5 Proofs on Bernstein-gamma functions
  6. 6 Proofs on exponential functionals
  7. 7 Proofs of auxiliary results
  8. A Some facts from regular variation theory
    1. A.1 Classical Karamata theory
    2. A.2 de Haan theory
  9. Acknowledgments

##  1 Introduction and motivation

Let $\xi=\left(\xi_{s}\right)_{s\geq 0}$ be a one-dimensional Lévy process,
and for any $t\in\left(0,\infty\right)$, set

$I_{\xi}(t):=\int_{0}^{t}e^{-\xi_{s}}\mathrm{d}s$ (1.1)

for the exponential functional of Lévy processes on deterministic horizon.
When $t=\infty$ or $t=\mathbf{e}_{q}$ for some $q>0$ and $\mathbf{e}_{q}\sim
Exp(q)$ independent of $\xi$, we have the classical exponential functional of
Lévy processes. Due to their role in different areas of probability theory the
latter objects have been extensively studied in the last thirty years, see [3,
7, 15, 19, 20, 23, 24, 25, 26, 28, 34, 40] for a list of references covering
the main results in the area and the different methods by which they have been
obtained. Compared to the breadth and depth of the results concerning
$I_{\xi}(\infty)$ and $I_{\xi}(\mathbf{e}_{q})$, the knowledge about the
distributional properties of $I_{\xi}(t)$ for fixed $t$ is quite meagre
despite their widespread appearance in probability theory as well. Its law has
a closed but unwieldy form when $\xi$ is a Brownian motion with drift, see [2,
39] and the survey [18]. The moments of $I_{\xi}(t)$, including complex ones,
are only available as a series for a class of subordinators, i.e. $\xi$ being
a non-decreasing Lévy process, see [5, 22, 32]. In fact, [32] considers
stochastic processes with independent increments only, but the explicit
formulae are for the positive integer moments of $I_{\xi}(t)$ when $\xi$ is a
subordinator. Paper [22] also offers some new convolutional identities for the
moments of $I_{\xi}(t)$ when $\xi$ is a general Lévy process which yield that
$\mathbb{E}\left[I^{-1/2}_{\xi}(t)\right]$ is independent of $\xi$ when the
latter is a symmetric Lévy process, which is reminiscent of the invariance of
the law of the first ladder time of symmetric random walks. The most advanced
knowledge for exponential functionals of Lévy processes on deterministic
horizon is available for the asymptotic behaviour of distributional quantities
of $I_{\xi}(t)$, as $t\to\infty$, provided $I_{\xi}(\infty)=\infty$ almost
surely. These results are predominantly driven by applications, especially in
the area of branching processes in Lévy random environments, see e.g. [4, 21,
36]. Whilst postponing the detailed discussion of these results to Section
2.4, we highlight that our main contributions on exponential functionals of
Lévy processes on deterministic horizon are in the case when $\xi$ drifts to
minus infinity and therefore $I_{\xi}(\infty)=\infty$. Under the additional
assumptions that $\mathbb{E}\left[\xi_{1}\right]\in\left(-\infty,0\right)$ and
$\mathbb{P}\left(\xi_{1}>t\right)$ is regularly varying at infinity with index
$\alpha>1$, we show that the measures

$\frac{y^{-a}\mathbb{P}\left(I_{\xi}(t)\in\mathrm{d}y\right)}{\mathbb{P}\left(\xi_{1}>t\right)},\quad
a\in\left(0,1\right),$ (1.2)

converge weakly, as $t\to\infty$, to a finite positive measure supported by
$\left(0,\infty\right)$. Via Mellin inversion, we give a semi-explicit
analytic description of the cumulative distribution function of the limit
measure, which also yields the asymptotic behaviour of
$\mathbb{E}\left[I^{-a}_{\xi}(t)\right]$, see Theorem 3.1 and the ensuing
corollaries. The latter allows us to derive a neat probabilistic
representation of the normed limit measure which turns out to be the law of a
curious generalization of the main product factorisation for classical
exponential functionals, see Theorem 3.4. As a result, we improve
substantially upon the recent work of W. Xu [36], which obtains results in the
very same context, but under several additional assumptions, which do not seem
to allow for the deduction of the aforementioned weak convergence, see Section
2.4 for a detailed discussion. Moreover, via the probabilistic factorisation,
we obtain novel analytic properties for the limit measure such as existence
and smoothness of its density. Also, dropping the requirement for regular
variation of the upper tail and finiteness of the expectation, we obtain
general upper bounds for the decay of quantities of the type
$\mathbb{E}\left[F(I_{\xi}(t))\right]$, as $t\to\infty$, which frequently
suffice for applications such as some of those in [4, 21, 36]. The rate of
decay depends on the finiteness of an explicit integral criterion, see Theorem
3.6. We also offer very different methodology to the one in [36]. It is based
on a mixture of analytic and probabilistic techniques. We discuss it in more
detail in Section 2.3 and touch upon it further below in this part.

One of the main ingredients in the proof of the aforementioned results is the
new information, which is of independent interest, that we obtain on the
functional-analytic properties of the recently introduced in [5] class of
special functions called bivariate Bernstein-gamma functions. It is well-known
that the bivariate Bernstein-gamma functions generated by the two Wiener-Hopf
factors (Bernstein functions) of the underlying Lévy process $\xi$ represent
the Mellin transform of the classical exponential functional, and therefore in
turn the Laplace transform in $t$ of the exponential functional on
deterministic horizon, see [5, 22, 28]. For this reason, bivariate Bernstein-
gamma functions with their functional-analytical properties have proved a key
tool for the study of both classical exponential functionals and exponential
functionals on deterministic horizon, see e.g. [5, 22, 28]. In fact, being a
natural extension to the classical Gamma function in the context of right-
continuous self-similar Markov processes, these functions, predominantly
through their role in the understanding of the exponential functionals of Lévy
processes, are very useful in various other areas of probability, see e.g. for
applications in the spectral theory of non-self-adjoint Markov semigroups [27,
29, 30] and [17] for the study of extinction times in non-Markovian setting.
In this work we show that bivariate Bernstein-gamma functions related to the
Wiener-Hopf factors of transient Lévy processes are $n$ times differentiable
at zero in the first variable if and only if an explicit integral criterion is
finite, see Theorem 4.3. Moreover, we link the derivatives of these functions
to the convolutions and the derivatives of the $q$-potential measures of the
underlying Lévy process, see Lemma 4.1, thereby deriving some new results for
as basic objects as these potentials. This connection is not unexpected as
classical exponential functionals, whose Mellin transform is represented in
terms of Bernstein-gamma functions, are related to the potential measure in
the recent work [3], which offers an integral equation for the density of the
exponential functional in which the potential measure appears. When the
potential measure has a bounded density, we obtain upper bounds for the
densities of the convolutions of the potential measure, see Theorem 4.6. Using
these results we derive universal and seemingly quite handy estimates for the
derivatives of the Mellin transforms of the classical exponential functionals
and their decay along complex lines, see Corollary 4.4.

The results on the derivatives of bivariate Bernstein-gamma functions play a
crucial role in a convoluted Tauberian approach we adopt in order to derive
the weak convergence of the measures defined in (1.2). The method requires a
particularly intricate analysis when the regular variation index $\alpha$ of
$\mathbb{P}\left(\xi_{1}>t\right)$ is an integer. Then we have to appeal to
the de Haan theory and in order to utilize it successfully we have to resort
to probabilistic considerations which complement the analytical approach. The
latter in turn reveals that one may have a more direct way to study the
moments of the measure (1.2) but at the cost of a loss of explicitness and
precision. Similar probabilistic approach has been adopted in the predecessor
of our work, see [36]. We discuss further our methodology in Section 2.3.

The paper is organized as follows: Section 2 introduces the necessary
notation, main objects of interest and previous results, outlines our
methodology, and offers a discussion on an open problem. The results on
finiteness of the derivatives of bivariate Bernstein-gamma functions are
presented in Section 4 and proved in Section 5. In Section 6 we prove our main
result, Theorem 3.1, the factorisation Theorem 3.4, and the general estimates
of Theorem 3.6. Section 7 provides the proofs of auxiliary results. Appendix A
contains classic facts about slowly varying functions and Tauberian theorems,
as well as it presents the de Haan class of slowly varying functions.

##  2 Preliminaries, methodology, background, literature review, and an open
problem

###  2.1 Notation

We use $\mathbb{C}$ to denote the complex plane with $z$ standing for a
complex variable with a real part $\mathtt{Re}\left(z\right)$ and an imaginary
part $\mathtt{Im}\left(z\right)$, and $\mathbb{R}_{+}$ for the set of non-
negative real numbers, $[0,\infty)$. For a set $A\subset\mathbb{R}$, we employ
$\mathbb{C}_{A}$ for the subset of the complex plane
$\mathbb{C}_{A}:=\left\\{z\in\mathbb{C}:\mathtt{Re}\left(z\right)\in
A\right\\}$. Further, unless stated otherwise, $n,k,l$ will be positive
integers. For any non-negative random variable $X$ we denote by
$\mathcal{M}_{X}(z):=\mathbb{E}\left[X^{z-1}\right]$ its Mellin transform,
which is defined at least on the complex line
$\mathbb{C}_{\left\\{1\right\\}}=\left\\{z\in\mathbb{C}:\mathtt{Re}\left(z\right)=1\right\\}$.
If $\mu$ is a measure, we define for better readability
$\mu(a,b):=\mu((a,b))$. Next, for a function $f$, we formally define its
Laplace transform, for $q\geq 0$, as
$\widehat{f}(q):=\int_{0}^{\infty}e^{-qt}f(t)\mathrm{d}t$. For asymptotic
behaviour, we use the classic Landau notation and also
$f\stackrel{{\scriptstyle a}}{{\sim}}g$ to denote $\lim_{x\to a}f(x)/g(x)=1$.
We recall that a function $\phi$ is a Bernstein function if and only if it can
be represented as

$\phi(z)=\phi(0)+dz+\int_{0}^{\infty}\left(1-e^{-zy}\right)\mu(\mathrm{d}y),\quad
z\in\mathbb{C}_{\left[0,\infty\right)},$ (2.1)

where $\phi(0)\geq 0$, $d\geq 0$, and $\mu$ is a sigma-finite positive measure
such that $\int_{0}^{\infty}\min\left\\{1,y\right\\}\mu(\mathrm{d}y)<\infty$.
Note that each Bernstein function is the Laplace exponent of a potentially
killed subordinator with killing rate $\phi(0)$, linear drift $d$ and jump
measure $\mu$. In the same fashion, a bivariate Bernstein function $\kappa$,
given by

$\kappa(\zeta,z)=\kappa(0,0)+d_{1}\zeta+d_{2}z+\int_{0}^{\infty}\int_{0}^{\infty}\left(1-e^{-\zeta
y_{1}-zy_{2}}\right)\mu(\mathrm{d}y_{1},\mathrm{d}y_{2}),\quad(\zeta,z)\in\mathbb{C}_{\left[0,\infty\right)}\times\mathbb{C}_{\left[0,\infty\right)},$
(2.2)

is the Laplace exponent of a potentially killed bivariate subordinator, where
$\kappa(0,0)$, $d_{1}$ and $d_{2}$ are non-negative and have the same meaning
as above, and $\mu$ is the bivariate jump measure, which satisfies
$\int_{0}^{\infty}\int_{0}^{\infty}\min\\{1,\sqrt{y_{1}^{2}+y_{2}^{2}}\\}\mu(\mathrm{d}y_{1},\mathrm{d}y_{2})<\infty$.

###  2.2 Bivariate Bernstein-gamma functions and Mellin transform of
exponential functionals

From [5, Definition 2.5, Theorem 2.8], we know that for any given bivariate
Laplace exponent $\kappa\not\equiv 0$, the equation

$W_{\kappa}(\zeta,z+1)=\kappa(\zeta,z)W_{\kappa}(\zeta,z),\quad\quad
W_{\kappa}(\zeta,1)=1,\zeta\in\mathbb{C}_{\left[0,\infty\right)},z\in\mathbb{C}_{\left(0,\infty\right)},$
(2.3)

has a unique bivariate holomorphic solution in
$\mathbb{C}_{\left(0,\infty\right)}\times\mathbb{C}_{\left(0,\infty\right)}$.
Moreover, for any $q\geq 0$, the function $W_{\kappa}(q,\cdot)$ is analytic on
$\mathbb{C}_{\left(0,\infty\right)}$, and it is a Mellin transform of a
positive random variable. Note that due to (2.3) the bivariate Bernstein-gamma
functions are a natural extension to both the classical Gamma function with
multiplier a general bivariate Bernstein function and the already well-known
(univariate) Bernstein-gamma functions. As mentioned in the introduction,
these functions have arisen from the studies of non-continuous self-similar
Markov processes.

From now on we work with the case when the bivariate Bernstein-gamma functions
are based on the Wiener-Hopf factors of Lévy processes. Recall that a Lévy
process $\xi$ with a Lévy -Khintchine exponent $\Psi$ is killed at rate
$q\geq 0$ if, for some independent of $\xi$ exponential random variable
$\mathbf{e}_{q}\sim Exp(q)$, we set $\xi_{t}=\infty$ for $t\geq\mathbf{e}_{q}$
with $\mathbf{e}_{0}=\infty$ a.s. The Lévy -Khintchine exponent of the
killed Lévy process is

$\Psi_{q}(z)=\log_{0}\mathbb{E}\left[e^{z\xi_{1}}\right]=\Psi(z)-q=\gamma
z+\frac{\sigma^{2}}{2}z^{2}+\int_{-\infty}^{\infty}\left(e^{zy}-1-zy\mathbbm{1}_{\left\\{|y|\leq
1\right\\}}\right)\Pi(\mathrm{d}y)-q,$ (2.4)

where we use $\log_{0}$ for the principal branch of the complex logarithm,
$\gamma\in\mathbb{R}$ and $\sigma^{2}\geq 0$ are respectively the linear term
and the Brownian component of $\xi$, $\Pi$ is the Lévy measure of the process
and satisfies
$\int_{-\infty}^{\infty}\min\\{1,y^{2}\\}\Pi(\mathrm{d}y)<\infty$, and $q$ is
the killing rate. For any potentially killed Lévy process, we have the Wiener-
Hopf factorisation

$\begin{split}&\Psi_{q}(z)=\Psi(z)-q=-\phi_{+}(q,-z)\phi_{-}(q,z)=-h(q)\kappa_{+}\left(q,-z\right)\kappa_{-}\left(q,z\right),\quad
z\in i\mathbb{R},\end{split}$ (2.5)

where, for $q\geq 0$,

$\begin{split}&h(q):=\exp\left(-\int_{0}^{\infty}\left(\frac{e^{-t}-e^{-qt}}{t}\right)\mathbb{P}\left(\xi_{t}=0\right)\mathrm{d}t\right),\end{split}$
(2.6)

which in the case of a transient Lévy process $\xi$ is non-zero by [6, Theorem
12], and, for $q\geq 0$ and $\mathtt{Re}(z)\geq 0$,

$\begin{split}&\kappa_{\pm}(q,z)=c_{\pm}\exp\left(\int_{0}^{\infty}\int_{\left[0,\infty\right)}\left(\frac{e^{-t}-e^{-qt-
zx}}{t}\right)\mathbb{P}\left(\pm\xi_{t}\in\mathrm{d}x\right)\mathrm{d}t\right)\end{split}$
(2.7)

are the bivariate Laplace exponents of the ascending/descending ladder times
and ladder heights processes which form bivariate subordinators, see [6,
Corollary VI.10] and [11, p.27] for more information. We note that the
constants $c_{\pm}>0$ depend on the choice of the local time which we can
choose appropriately so that $c_{+}c_{-}=1$. Furthermore, we can choose
$\phi_{+}(q,z)=\kappa_{+}\left(q,z\right)/c_{+}$ and
$\phi_{-}(q,z)=h(q)\kappa_{-}\left(q,z\right)/c_{-}$, i.e., for $q\geq 0$ and
$\mathtt{Re}(z)\geq 0$,

$\begin{split}&\phi_{+}(q,z)=\exp\left(\int_{0}^{\infty}\int_{[0,\infty)}\left(\frac{e^{-t}-e^{-qt-
zx}}{t}\right)\mathbb{P}\left(\xi_{t}\in\mathrm{d}x\right)\mathrm{d}t\right),\\\
&\phi_{-}(q,z)=\exp\left(\int_{0}^{\infty}\int_{\left(0,\infty\right)}\left(\frac{e^{-t}-e^{-qt-
zx}}{t}\right)\mathbb{P}\left(\xi_{t}\in-\mathrm{d}x\right)\mathrm{d}t\right).\end{split}$
(2.8)

We note that $h(q)\equiv 1$ provided that $\xi$ is not a compound Poisson
process (CPP), and that in this case $h(q)\equiv 1$ and
$\phi_{\pm}\equiv\kappa_{\pm}$ are bivariate Laplace exponents as well. When
$h(q)\not\equiv 1$, for any fixed $q\geq 0$,
$\phi_{q,\pm}(z):=\phi_{\pm}(q,z)$ are Laplace exponents of univariate
subordinators.

Let us denote, for $q\geq 0$,
$I_{\Psi_{q}}:=I_{\xi}(\mathbf{e}_{q})=\int_{0}^{e_{q}}e^{-\xi_{s}}\mathrm{d}s$.
From [28, Theorem 2.4] we know that

$\mathcal{M}_{I_{\Psi_{q}}}(z)=\phi_{-}(q,0)\frac{\Gamma(z)}{W_{\phi_{q,+}}(z)}W_{\phi_{q,-}}(1-z)=:\phi_{-}(q,0)M_{\Psi}\left({q},{z}\right)\quad\text{at
least for $z\in\mathbb{C}_{(0,1)}$,}$ (2.9)

where for the fixed $q$, the functions $W_{\phi_{q,\pm}}$ solve (2.3), and
therefore they are Bernstein-gamma functions. To provide a link between the
Bernstein-gamma functions associated to the Wiener-Hopf factors $\phi_{\pm}$
and the probabilistic $\kappa_{\pm}$, we use that by [28, Theorem 4.1 (5)],
for any $c>0$, $z\in\mathbb{C}_{(0,\infty)}$, and a Bernstein function $\phi$,
$W_{c\phi}(z)=c^{z-1}W_{\phi}(z)$, so we obtain from the choice of
$\phi_{q,\pm}$ made above that

$W_{\phi_{q,+}}(z)=c_{+}^{1-z}W_{\kappa_{+}}(q,z),\quad\text{and}\quad
W_{\phi_{q,-}}(z)=c_{-}^{1-z}h^{z-1}(q)W_{\kappa_{-}}(q,z).$ (2.10)

Substituting in (2.9), we get

$\mathcal{M}_{I_{\Psi_{q}}}(z)=h^{z}(q)\kappa_{-}(q,0)\frac{\Gamma(z)}{W_{\kappa_{+}}(q,z)}W_{\kappa_{-}}(q,1-z).$
(2.11)

###  2.3 Methodology

Recall that $I_{\xi}(t):=\int_{0}^{t}e^{-\xi_{s}}\mathrm{d}s$. From (2.9), for
$a\in\left(0,1\right)$, we have that the Laplace transform for the moments in
$t$ is given by

$
\int_{0}^{\infty}e^{-qt}\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\mathrm{d}t=\frac{\mathcal{M}_{I_{\Psi_{q}}}(1-a)}{q}=\frac{M_{\Psi}\left({q},{1-a}\right)}{\phi_{+}(q,0)},$
(2.12)

where we have utilized that from (2.5) it holds that
$q=\phi_{+}(q,0)\phi_{-}(q,0)$, and we have employed (2.9). Relation (2.12) is
the starting point of our proof as it opens the way to apply Tauberian
theorems to determine the asymptotic behaviour of
$\mathbb{E}\left[I_{\xi}^{-a}(t)\right]$. However, in the setting
$\mathbb{E}\left[\xi_{1}\right]<0$, it is valid that

$\int_{0}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\mathrm{d}t<\infty,$

see Theorem 3.6, and a direct application of Tauberian theorems is impossible.
For this purpose, depending on the index of regular variation $\alpha>1$ of
$\mathbb{P}\left(\xi_{1}>t\right)$, we resort to investigating several
repeated tails, e.g. when $\alpha\in\left(n,n+1\right]$, $n\geq 1$, we
consider

$V(t)=\int_{t}^{\infty}\int_{s_{1}}^{\infty}\dots\int_{s_{n-1}}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(s_{n})\right]\mathrm{d}s_{n}\dots\mathrm{d}s_{1}$

and via Tauberian methods we deduct the asymptotic behaviour of $V(t),$ as
$t\to\infty$. When $\alpha\neq n+1$ by the monotone density theorem we infer
the asymptotics of $\mathbb{E}\left[I_{\xi}^{-a}(t)\right]$. Dealing with
Laplace transforms of repeated tails requires the consideration of the
derivatives in $q$ of $M_{\Psi}\left({q},{1-a}\right)$, see (2.12), including
at $q=0$, which through (2.9) depend on the derivatives in $q$ of the involved
bivariate Bernstein-gamma functions. Via an integral representation of the
bivariate Bernstein-gamma functions which links these functions to the
convolutions and the derivatives of the $q$-potential measures of $\xi$, we
derive results of independent interest for both the aforementioned derivatives
and the potential measures.

The outstanding case is when $\alpha=n+1$ is an integer. Then the results
concerning the repeated tails can be deduced as above, but the monotone
density theorem fails. Indeed, $\int_{0}^{t}V(s)\mathrm{d}s$ is slowly varying
at infinity, see beneath (6.17), and Theorem A.10 does not hold. In this case,
we are forced to resort to the more complicated de Haan theory, which requires
a preliminary estimate of the type
$\mathbb{E}\left[I_{\xi}^{-a}(t)\right]=\mathrm{O}\left(\mathbb{P}\left(\xi_{1}>t\right)\right)$,
and which we derive by purely probabilistic means such as for example the one-
large jump principle. This allows us to use a generalised monotone density
theorem, see Theorem A.14. The rest is similar to the non-integer $\alpha$
case.

Having established the asymptotics of
$\mathbb{E}\left[I_{\xi}^{-a}(t)\right]$, we proceed to study the behaviour of

$\frac{\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]}{\mathbb{P}\left(\xi_{1}>t\right)}\quad\text{for any fixed
}x\in\left(0,\infty\right).$

We do so via a Mellin inversion representation of the Laplace transform

$\begin{split}x\mapsto\frac{1}{q}\mathbb{E}\left[I_{\xi}^{-a}(\mathbf{e}_{q})\mathbbm{1}_{\left\\{I_{\xi}(\mathbf{e}_{q})\leq
x\right\\}}\right]&=\int_{0}^{\infty}e^{-qt}\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]\mathrm{d}t\\\ &=-\frac{1}{2\pi
i\phi_{+}({q,0})}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}M_{\Psi}\left({q},{z+1-a}\right)\mathrm{d}z,\end{split}$

and a subsequent investigation of $n$th repeated tails of
$\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]$ as in the case above, which corresponds to the limit case
$x=\infty$. Note that here the differentiation of the Laplace transform
involves differentiation under the sign of the integral. This requires the
additional understanding of the growth rate of the derivatives of bivariate
Bernstein-gamma functions along complex contours which we develop in Theorem
4.3 and Corollary 4.4. The case when $\xi$ is a compound Poisson process with
positive drift requires further refinement of the aforementioned estimates,
see item (iii) of Corollary 4.4. Then we are able to apply the same Tauberian,
including the de Haan theory, arguments as in the case of the moments above.
We emphasize that, for $\alpha\in\left(n,n+1\right]$, the main carrier of the
asymptotics is the behaviour of $\left(\phi_{+}({q,0})\right)^{(n)}$ at zero.
As a result, we prove that

$\frac{\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]}{\mathbb{P}\left(\xi_{1}>t\right)}$

converges to the cumulative distribution function of a finite measure
$\nu_{a}$ and compute its Mellin transform, see Theorems 3.1 and 3.4. The
latter allows us to deduce analytic properties for the density of the limiting
measure and to obtain a product factorisation of the random variable behind
the normed $\nu_{a}$, that is $\tilde{\nu}_{a}$, in terms of well-known
independent random variables, namely the classical exponential functional of
the subordinator pertaining to the Wiener-Hopf factor $\phi_{0,+}$ and the
remainder term pertaining to $\phi_{0,-}$, see Theorem 3.4.

The bounds on the rate of decay of
$\mathbb{E}\left[\left|F(I_{\xi}(t))\right|\right]$ in Theorem 3.6 are simple
to derive from (2.12) once we have established the finiteness of the
derivatives at $q=0$ of the right-hand side of (2.12). The latter yields
finiteness of the derivatives of the integral to the left-hand side.

###  2.4 Background and previous results

We discuss the available results in the domain of the asymptotic behaviour of
exponential functionals of Lévy processes on deterministic horizon.

When $\mathbb{E}\left[\xi_{1}\right]=0$, the Lévy process oscillates and
$I_{\xi}\left(\infty\right)=\infty$. In this case, under the assumptions that
$\mathbb{E}\left[\xi^{2}_{1}\right]<\infty$ and the existence of some positive
exponential moments for $\xi$, the authors of [21] show that
$\lim_{t\to\infty}\sqrt{t}\mathbb{E}\left[I^{-a}_{\xi}(t)\right]=C\in\left(0,\infty\right)$.
Furthermore, under the additional assumption for the existence of some
negative exponential moments, they have obtained a result of the type
$\lim_{t\to\infty}\sqrt{t}\mathbb{E}\left[F\left(I_{\xi}(t)\right)\right]=C_{F}$
for a limited class of functions $F$. Similar results in the same setting are
obtained in [16]. The requirements for the existence of exponential moments,
for the finiteness of the variance, and for the limited form of $F$ have been
completely removed in [28, Theorem 2.20]. Therein even oscillating Lévy
processes $\xi$ with no first moment and with Spitzer’s condition are studied
in full generality.

When $\xi$ is symmetric with infinite activity, it is shown in [22, Corollary
2.3.] that $\mathbb{E}\left[I^{-1/2}_{\xi}(t)\right]=1/\sqrt{t}$. This is one
of the few cases where a precise evaluation in a closed form of a fractional
moment of exponential functional of Lévy processes on deterministic horizon is
available. It fits the asymptotics mentioned in the previous paragraph.

The case when $\mathbb{E}\left[\xi_{1}\right]<0$ and $\xi$ drifts to minus
infinity is considered in [21]. Under an assumption for the existence of
negative exponential moments of some order, they show that the quantities
considered in the paragraph above have explicit rate of decay. The case
$\mathbb{E}\left[\xi_{1}\right]<0$ and $\mathbb{P}\left(\xi_{1}>t\right)$
being regularly varying at infinity of index $\alpha>1$, which is precisely
our setting, is considered in [36]. However, there the author assumes
additionally that $\mathbb{P}\left(\xi_{1}\in\left[t,t+\delta\right)\right)$
is regularly varying at infinity for any $\delta>0$, that is
$\mathbb{P}\left(\xi_{1}>t\right)$ is additionally locally regularly varying.
Then upon further assuming that $F$ is bounded, positive, non-increasing,
locally-Lipschitz, and $\sup_{x>0}x^{a}F(x)<\infty$ for some
$a\in\left(0,1\right)$, see [36, Conditions 4.1 and 4.2], then (3.4) holds
true. We note that in our Corollary 3.1 we only assume
$\sup_{x>0}x^{a}F(x)<\infty$ for some $a\in\left(0,1\right)$ and $F$ being
continuous. Since the result in [36] is proved in a probabilistic way, the
constant has a probabilistic representation, see [36, (4.3)]. We offer a
characterization of the limiting measure and the related constants in terms of
special functions which in many cases are easier to evaluate. Moreover, the
Mellin inversion representation of the limiting measure yields analytical
properties of this measure that cannot be captured by probabilistic means and
offers in turn even a neat new probabilistic representation of the normed
limiting measure as the law of the product of two independent well-known in
the literature and easier to study random variables. We do not discard lattice
Lévy processes either which would be necessitated by local regular variation.
We wish to highlight that [37] obtains similar results to the results in [36]
for exponential functionals of random walks. The respective conditions in [37]
match those in [36].

In all the aforementioned works that deal with asymptotic behaviour of
$\mathbb{E}\left[F(I_{\xi}(t))\right]$ there are results that explicitly or
implicitly yield some bounds on the rate of decay of
$\mathbb{E}\left[F(I_{\xi}(t))\right]$. However, they require as a minimum the
existence of exponential moments or fine structure of the tail
$\mathbb{P}\left(\xi_{1}>t\right)$ as in [36]. Here, we offer such estimates
on the rate of decay based on the finiteness of an explicit general integral
criterion.

We are not aware of any results linking the derivatives of bivariate
Bernstein-gamma functions and convolutions and derivatives of the
$q-$potential measures of the underlying Lévy process. We expect that this
connection will prove useful in other contexts.

###  2.5 Open problems

A close inspection of our proof, see Section 6, shows that
$\mathbb{P}\left(\xi_{1}>t\right)$ being regularly varying at infinity with
index $\alpha>1$ implies that $\mathbb{P}\left(\xi_{t}>0\right)$ is regularly
varying with index $\alpha-1$, see (6.5), which in turn yields that the
derivative of the Lévy -Khintchine exponent of the ascending ladder time is
regularly varying at zero, that is $\phi^{(n)}_{+}(q,0)$ is regularly varying
of order $\left\\{\alpha\right\\}-1$, see (6.8). A natural question is whether
we can start only with assumption of the behaviour of the latter derivative.
In fact, the only point where we need the regular variation of
$\mathbb{P}\left(\xi_{1}>t\right)$ is when we apply the de Haan theory via the
probabilistic estimate, see the proof of Lemma 6.3, which relies on the one
large jump principle. Should the latter be avoided, then one might prove
results like ours only upon requirements on some derivative of
$\phi_{+}(q,0)$. Due to the link to bivariate Bernstein-gamma functions, we
believe this is a worthy line for future research which could further expand
our knowledge not only on exponential functionals but on bivariate Bernstein-
gamma functions themselves too.

The curious generalisation of the classical product factorisation of
exponential functionals, see Theorem 3.4, leads us to the question whether
such a factorisation is more general in the limit as now it depends on the
assumption of regular tails. However, classical factorisations are universal.
Unfortunately, our approach does not offer a route to answer this.

##  3 Main results on exponential functionals

We state the main result on exponential functionals.

######  Theorem 3.1.

Let $\xi$ be a Lévy process with a finite negative mean and

$\mathbb{P}\left(\xi_{1}>t\right)\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{\ell(t)}{t^{\alpha}}$
(3.1)

for $\alpha>1$ and $\ell$ a slowly varying at infinity function. Then, for any
$a\in(0,1)$,

$\frac{\mathbb{P}(I_{\xi}(t)\in\mathrm{d}y)t^{\alpha}}{y^{a}\ell(t)}\xrightarrow[t\to\infty]{w}\nu_{a}(\mathrm{d}y),$
(3.2)

where $\nu_{a}$ is a finite measure, supported on $(0,\infty)$, with a
distribution function given by

$\nu_{a}\left((0,x]\right)=\frac{1}{C}\int_{\mathtt{Re}z=b}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z\quad\text{with}\quad
C=-2\pi i\phi_{+}(0,0)\left(-\mathbb{E}\left[\xi_{1}\right]\right)^{\alpha}$
(3.3)

and for all $b\in\mathbb{C}_{(a-1,0)}$.

Relations of the statement of Theorem 3.1 to previous results can be found in
subsection 2.4. A direct consequence of the last theorem are the following
results.

######  Corollary 3.2.

Under the conditions of Theorem 3.1, for every function
$F:(0,\infty)\to\mathbb{R}$ such that, for some $a\in(0,1)$, $x\mapsto
x^{a}F(x)$ is bounded and continuous,

$\frac{t^{\alpha}}{\ell(t)}\mathbb{E}\left[F\left(I_{\xi}(t)\right)\right]\xrightarrow[t\to\infty]{}\int_{(0,\infty)}y^{a}F(y)\nu_{a}(\mathrm{d}y)<\infty.$
(3.4)

######  Corollary 3.3.

Under the conditions of Theorem 3.1, for any $a\in(0,1)$,

$\frac{t^{\alpha}}{\ell(t)}\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\xrightarrow[t\to\infty]{}\frac{M_{\Psi}\left({0},{1-a}\right)}{\phi_{+}(0,0)\left(-\mathbb{E}\left[\xi_{1}\right]\right)^{\alpha}}<\infty.$
(3.5)

Relations of the statements of Theorem 3.1 and Corollaries 3.2-3.3 to previous
results can be found in subsection 2.4.

Theorem 3.1 and Corollary 3.3 allow the following neat probabilistic
reformulation which contains an extension of the classical factorisation of
exponential functionals, see Remark 3.5 for more details. Following [28,
Theorem 2.22], for a positive random variable $X$ and $a\in\mathbb{R}$ such
that $\mathbb{E}\left[X^{a}\right]<\infty$, we introduce the size-biased
transform via

$\mathbb{E}\left[f(\mathcal{B}_{a}X)\right]=\frac{\mathbb{E}\left[X^{a}f(X)\right]}{\mathbb{E}\left[X^{a}\right]}$
(3.6)

for any bounded, continuous function $f$ and provided
$\mathbb{E}\left[X^{a}\right]<\infty$.

######  Theorem 3.4.

Under the conditions of Theorem 3.1, for any $a\in\left(0,1\right)$,

$\mathcal{B}_{-a}I_{\xi}(t)\stackrel{{\scriptstyle
d}}{{\xrightarrow[t\to\infty]{}}}\mathcal{B}_{-a}I_{\phi_{+}}\times\mathcal{B}_{1-a}Y^{-1}_{\phi_{-}},$
(3.7)

where $\times$ stands for product of independent random variables,
$I_{\phi_{+}}$ is the classical exponential functional pertaining to the
killed subordinator with Bernstein function $\phi_{+}:=\phi_{0,+}$, and
$Y_{\phi_{-}}$ is the positive random variable whose Mellin transform is
$W_{\phi_{-}}$ with Bernstein function $\phi_{-}:=\phi_{0,-}$, see (2.10).
Finally, the law of
$\mathcal{B}_{-a}I_{\phi_{+}}\times\mathcal{B}_{1-a}Y^{-1}_{\phi_{-}}$ and
therefore $\nu_{a}$ has infinitely differentiable bounded densities as long as
$\xi$ is not a compound Poisson process with positive drift.

######  Remark 3.5.

$Y_{\phi_{-}}$ is known as the remainder random variable and
$\mathcal{B}_{1}Y^{-1}_{\phi_{-}}$ is a factor in the general factorisation of
classical exponential functionals, see [28, Theorem 2.22]. This corresponds to
$a=0$, which is not applicable here as the process $\xi$ drifts to $-\infty$.
Nonetheless, if $\xi$ drifts to $\infty$, then
$I_{\Psi}=I_{\xi}(\infty)=I_{\phi_{+}}\times\mathcal{B}_{1}Y^{-1}_{\phi_{-}}<\infty$,
see [28, Theorem 2.22], and clearly then (3.7) holds for all
$a\in\left[0,1\right)$. In this sense, the result above is an extension of the
classical factorisation for exponential functionals, which in this case is
valid for the limit and only for $a\in(0,1)$. This raises the question whether
a result as in (3.7) may hold beyond the assumptions of this theorem. We note
that the Mellin transform of $\nu_{a}$ in Theorem 3.1 is proportional to
$M_{\Psi}(z-a)$ and from there and the representation (2.9) combined with the
results in [28] one can proceed to establish a number of functional-analytic
properties of $\nu_{a}$. Due to the already lengthy manuscript, we leave this
aside.

The next result has no general assumptions on $\xi$ apart from it being
transient and despite the fact that it only offers
$\mathrm{o}\left(\cdot\right)$-type of estimate, it frequently suffices, see
e.g. the applications in the area of branching processes in Lévy random
environments in [4, 21, 36]. Following [13], for $x>0$ and the linear term
$\gamma$, defined in (2.4), we introduce the function

$A(x)=\gamma+\Pi\left(1,\infty\right)-\Pi\left(-\infty,-1\right)+\int_{1}^{x}\left(\Pi\left(y,\infty\right)-\Pi\left(-\infty,-y\right)\right)\mathrm{d}y.$
(3.8)

From [13, Lemma 13 $(ii)$], in the case where
$\lim_{t\to\infty}\xi_{t}=-\infty$, if
$\mathbb{E}\left[|\xi_{1}|\right]<\infty$, then $\lim_{x\to\infty}A(x)$
$=\mathbb{E}\left[\xi_{1}\right]<0$; and if $\mathbb{E}\left[\xi_{1}\right]$
is not finite, $\lim_{x\to\infty}A(x)=-\infty$. Since
$\lim_{x\to\infty}\left|A(x)\right|=\left|A(\infty)\right|>0$, there exists
$x_{0}$ such that $A(x)$ is non-zero for $x\geq x_{0}$. Since the integral
criterion in (4.6) below is in fact borrowed from [13, (1.14)], which requires
integrability only at infinity, the possibility that it may be the case that
$A(x)=0$ for some $x\in(1,x_{0}]$ can without loss of generality be avoided by
redefining $A(x)$ to be constant over $(1,x_{0}]$. From now on we work as if
$x_{0}=1$. Now we are ready to state our final result for this section.

######  Theorem 3.6.

Let $\xi$ be a Lévy process such that $\lim_{t\to\infty}\xi_{t}=-\infty$. If

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty$
(3.9)

for some $n\geq 1$, then, for every measurable function
$F:(0,\infty)\to\mathbb{R}$ such that for some $a\in(0,1)$, $x\mapsto
x^{a}F(x)$ is bounded, we have that at infinity

$\mathbb{E}\left[\left|F(I_{\xi}(t))\right|\right]=\mathrm{o}\left(t^{-n}\right),$
(3.10)

and
$\int_{0}^{\infty}t^{n}\mathbb{E}\left[\left|F(I_{\xi}(t))\right|\right]\mathrm{d}t<\infty$.
Finally, only upon the assumptions that $\lim_{t\to\infty}\xi_{t}=-\infty$ and
$x\mapsto x^{a}F(x)$ is bounded, it holds that
$\int_{0}^{\infty}\mathbb{E}\left[\left|F(I_{\xi}(t))\right|\right]\mathrm{d}t<\infty$.

##  4 Derivatives of bivariate Bernstein–gamma functions

In this section, we study the finiteness of the derivatives of
$W_{\kappa_{\pm}}$ and related quantities, which play a key role in the proofs
in Section 6. However, since they are of independent interest, we state them
in this separate section. We start with an integral representation of
$W_{\kappa_{\pm}}$ recalling that, for $q\geq 0$,

$U_{q}(\mathrm{d}x)=\int_{0}^{\infty}e^{-qt}\mathbb{P}\left(\xi_{t}\in\mathrm{d}x\right)\mathrm{d}t$
(4.1)

are the $q$-potential measures of $\xi$ and $U^{*n}_{q}$ are their respective
formal convolutions. $U:=U_{0}$ is called the potential measure which is a
well-defined Radon measure when $\xi$ is transient.

######  Lemma 4.1.

Let $\xi$ be a Lévy process with Laplace exponent $\Psi$, and $\kappa_{\pm}$
be the bivariate Laplace exponents associated with $\Psi$, introduced in
(2.5). Then, for any $z\in\mathbb{C}_{(-1,\infty)}$ and $q>0$,

$\begin{split}&\log\left(W_{\kappa_{\pm}}(q,z+1)\right)=z\ln\kappa_{\pm}(q,1)+\int_{[0,\infty)}\left(e^{-zy}-1-z(e^{-y}-1)\right)\frac{\mathcal{W}_{\pm}(q,\mathrm{d}y)}{e^{y}-1},\end{split}$
(4.2)

where $\log$ is a complex logarithm, and the measures
$\mathcal{W}_{\pm}(q,\mathrm{d}y)$ do not charge $0$ and are defined on
$\mathbb{R}\setminus\\{0\\}$ as

$\mathcal{W}_{\pm}(q,\mathrm{d}y)=\int_{0}^{\infty}\frac{e^{-qt}}{t}\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)\mathrm{d}t.$
(4.3)

Also, for any $z\in\mathbb{C}_{(-1,\infty)}$ and $q>0$, we have the
representation

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\kappa_{\pm}}(q,1+z)}}{W_{\kappa_{\pm}}(q,1+z)}=\int_{[0,\infty)}\frac{1-e^{-zy}}{e^{y}-1}U_{q}(\pm\mathrm{d}y).\end{split}$
(4.4)

Finally, for any $q_{0}>0$, all derivatives of
$\mathcal{W}_{\pm}(q,\mathrm{d}y)$ in $q$ exist at $q_{0}$ and are equal to

$\begin{split}\mathcal{W}^{(n)}_{\pm}(q_{0},\mathrm{d}y)&=(-1)^{n}\int_{0}^{\infty}t^{n-1}e^{-q_{0}t}\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)\mathrm{d}t\\\
&=(-1)^{n}(n-1)!U^{*n}_{q_{0}}(\pm\mathrm{d}y)\\\
&=-U^{(n-1)}_{q_{0}}(\pm\mathrm{d}y),\end{split}$ (4.5)

where $\left(\partial^{n-1}/\partial
q^{n-1}\right)U_{q_{0}}=U^{(n-1)}_{q_{0}}$ are the measure derivatives of
$U_{q}$ at $q=q_{0}$.

######  Remark 4.2.

Note that given the integral representation [28, Theorem 4.7 (4.15)], the new
information contained in (4.2) is the representation (4.3) in terms of the
harmonic measure of the underlying Lévy process $\xi$ and the link in (4.4)
with the $q$-potential measures. However, this minor improvement is quite
useful when dealing with $W_{\kappa_{\pm}}$, since the input/analytic data we
have are the Lévy process $\xi$ and its Lévy -Khintchine exponent. The
differentiation of the measures is in the sense of weak limits of the
respective difference ratios, i.e.
$\left(\mathcal{W}_{\pm}\left(q_{0}+h,\mathrm{d}y\right)-\mathcal{W}_{\pm}\left(q_{0},\mathrm{d}y\right)\right)/h$
as $h\to 0$. We note that the measures in (4.3) are known as the $q$-harmonic
measures of $\xi$. A nice link between the harmonic measures and the law of
exponential functionals of subordinators has been established in [1]. Here, it
is clear that this link extends to general exponential functionals of Lévy
processes.

Lemma 4.1 allows us to study the derivatives of $W_{\kappa_{\pm}}$. We have
the following result.

######  Theorem 4.3.

Let $\xi$ be a Lévy process with Lévy -Khintchine exponent $\Psi$ such that
$\lim_{t\to\infty}\xi_{t}=-\infty$. Then, for any $n\geq 0$,
$z\in\mathbb{C}_{(-1,\infty)}$, and $q\geq 0$,

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty\iff\frac{\partial^{n+1}}{\partial{q}^{n+1}}{W_{\kappa_{\pm}}}(q,z+1){\text{
and }}\frac{\partial^{n+1}}{\partial{q}^{n+1}}{W_{\phi_{q,\pm}}}(z+1)\text{
are finite},$ (4.6)

where for $q=0$ the derivatives are understood as right derivatives. Upon the
finiteness of the integral condition in (4.6), the derivatives are right-
continuous at $q=0$.

######  Corollary 4.4.

Let $\xi$ be a Lévy process with Lévy -Khintchine exponent $\Psi$ such that
$\lim_{t\to\infty}\xi_{t}=-\infty$, and assume that, for some $n\geq 0$,

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty.$

Then, for any $0\leq k\leq n+1$, $z\in\mathbb{C}_{(0,1)}$, and $q\geq 0$:

  1. (i)

the derivatives
$\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}$ are
finite and are right-continuous at zero;

  2. (ii)

there exist polynomials $P_{\mathtt{Re}\left(z\right),k}$ of degree $k$ such
that

$\left|\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}\right|\leq
P_{\mathtt{Re}\left(z\right),k}\left(\left|z\right|\right)\left|M_{\Psi}\left({q},{z}\right)\right|;$

  3. (iii)

if the potential measure of $\xi$ has a bounded density with respect to the
Lebesgue measure, there exist polynomials $P_{\mathtt{Re}\left(z\right),k}$ of
degree $k$ such that

$\left|\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}\right|\leq
P_{\mathtt{Re}\left(z\right),k}\left(\left|\ln|z|\right|\right)\left|M_{\Psi}\left({q},{z}\right)\right|.$

If $\mathbb{E}\left[\xi_{1}\right]\in\left(-\infty,0\right)$, $n\geq 1$, and
for any $0<\beta<n$,

$\int_{(1,\infty)}x^{\beta+1}\Pi(\mathrm{d}x)<\infty=\int_{(1,\infty)}x^{n+1}\Pi(\mathrm{d}x),$
(4.7)

then item (i) holds for any $0\leq k\leq n+1$ and for all $q>0$, and items
(ii)-(iii) are valid for any $0\leq k\leq n$ and for all $q\geq 0$. For
$k=n+1$, $q>0$, and $z\in\mathbb{C}_{(0,1)}$, items (ii)-(iii) are modified to

  1. (ii’)

for any $\delta>0$, there exist polynomials
$P_{\mathtt{Re}\left(z\right),n,\delta}$ of degree $n+1$ such that, for all
$q>0$,

$\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{M_{\Psi}\left({q},{z}\right)}\right|\leq\left(U^{*n}_{q}(\mathbb{R}_{+})+q^{-\delta}\right)P_{\mathtt{Re}\left(z\right),n,\delta}\left(\left|z\right|\right)\left|M_{\Psi}\left({q},{z}\right)\right|;$

  2. (iii’)

if the potential measure of $\xi$ has a bounded density with respect to the
Lebesgue measure, for any $\delta>0$, there exist polynomials
$P_{\mathtt{Re}\left(z\right),n,\delta}$ of degree $n+1$ such that, for all
$q>0$,

$\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{M_{\Psi}\left({q},{z}\right)}\right|\leq\left(U^{*n}_{q}(\mathbb{R}_{+})+q^{-\delta}\right)P_{\mathtt{Re}\left(z\right),n,\delta}\left(\left|\ln|z|\right|\right)\left|M_{\Psi}\left({q},{z}\right)\right|.$

In both cases, $\lim_{q\to 0}U^{*n}_{q}(\mathbb{R}_{+})=\infty$.

######  Remark 4.5.

We note that according to [13, Theorem 1],

$\int_{1}^{\infty}t^{n-1}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t<\infty\iff\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty.$
(4.8)

Furthermore, from (4.5), (5.16), and (6.6), we see that

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty\iff
U^{*n}\left(\mathbb{R}_{+}\right)<\infty\iff\left|U^{(n-1)}\left(\mathbb{R}_{+}\right)\right|\iff\left|\phi^{(n)}_{+}(0)\right|<\infty.$
(4.9)

In the case where $-\infty<\mathbb{E}\left[\xi_{1}\right]<0$, as noted, we
have that $\lim_{x\to\infty}A(x)=\mathbb{E}\left[\xi_{1}\right]<0$, and the
integral criterion from $\eqref{eq:derW}$ simplifies further to

$\begin{split}\int_{1}^{\infty}t^{n-1}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t<\infty&\iff\int_{(1,\infty)}x^{n+1}\Pi(\mathrm{d}x)<\infty\iff\left|\phi^{(n)}_{+}(0)\right|<\infty\\\
&\iff\mathbb{E}\left[\xi^{n+1}_{1}\mathbbm{1}_{\left\\{\xi_{1}>0\right\\}}\right]<\infty\iff
U^{*n}\left(\mathbb{R}_{+}\right)<\infty,\end{split}$ (4.10)

where the equivalence before the last one is due to [33, p.159].

The last result shows that, under the condition of Corollary 4.4 and in the
setting of item $(\ref{it:cor_der_Mq_iii})$, we have a general estimate for
the densities of the convolutions of the $q$-potential measures.

######  Theorem 4.6.

Let $\xi$ be a Lévy process with Lévy -Khintchine exponent $\Psi$ such that
$\lim_{t\to\infty}\xi_{t}=-\infty$. Also, assume that $U$ has a bounded
density $u$ on $\mathbb{R}$ with $C:=\sup_{x\in\mathbb{R}}u(x)$. Then, for any
$k\geq 1$ and $q>0$, the potential densities of $U_{q}^{*k}$ exist and are
locally bounded, since they satisfy, for any $x\in\mathbb{R}$ and $q>0$,

$u_{q}^{*k}(x)\leq kCU^{*(k-1)}_{q}\left(\min\\{0,x\\},\infty\right)$ (4.11)

with the convention $U^{*0}\equiv 1$. Next, if for some $n\geq 1$,

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty,$

then, for any $1\leq k\leq n+1$, we have that the potential densities of
$U^{*k}$ exist and are locally bounded since they satisfy (4.11) for any
$x\in\mathbb{R}$, and also $U^{*l}(0,\infty)<\infty$ for any $1\leq l\leq n$.

######  Remark 4.7.

To be precise, we note that in the last theorem we work with the canonical
$q$-excessive versions of the respective densities, see [6, Proposition I.12].
This allows us to state universal and not almost everywhere results. The proof
of (4.11) can be easily modified to improve the bound to

$u_{q}^{*k}(x)\leq
C\sum_{j=0}^{k-1}U^{*j}_{q}\left(x,\infty\right)U^{*(k-1-j)}_{q}(0,\infty),$

but we use the former as it is more compact. Further, we note that Theorems
4.3 and 4.6 can be extended to include the case
$\lim_{t\to\infty}\xi_{t}=\infty$: by considering the process
$\widehat{\xi}=-\xi$, which drifts to $-\infty$, and has
$\widehat{\kappa}_{\pm}=\kappa_{\mp}$, $\widehat{A}=-A$, we see that the
corresponding integral criterion would be

$\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(-\mathrm{d}x)<\infty,$

and in (4.11), $U_{q}^{*j}\left(\min\\{0,x\\},\infty\right)$ should be changed
to $U_{q}^{*j}\left(-\infty,\max\\{0,x\\}\right)$. Similarly, Corollary 4.4
can also be adjusted accordingly for the case where $\xi$ drifts to infinity.

##  5 Proofs on Bernstein-gamma functions

We proceed with the proofs for Section 4 by stating a preliminary proposition,
which links the derivatives of the $q$-potentials, see (4.1), with their
convolutions. We have not been able to detect this result in the literature,
but it is natural in view of the resolvent equation and even clearer in the
case of Lévy processes.

######  Proposition 5.1.

Let $U_{q}$ be the potential measures of a Lévy process. Then, for every
$q\geq 0$ and $n\geq 1$,

$U_{q}^{*n}(\mathrm{d}x)=\frac{1}{(n-1)!}(-1)^{n-1}U^{(n-1)}_{q}(\mathrm{d}x)=\frac{1}{(n-1)!}\int_{0}^{\infty}e^{-qt}t^{n-1}\mathbb{P}(\xi_{t}\in\mathrm{d}x)\mathrm{d}t,$
(5.1)

with the measures necessarily finite for $q>0$ and possibly infinite for some
or all $n$ when $q=0$.

We prove Proposition 5.1, as well as the next technical lemma, in Section 7.

######  Lemma 5.2.

For $z\in\mathbb{C}_{(-1,\infty)}$, the functions

$u_{z}(y):=\frac{e^{-zy}-1-z(e^{-y}-1)}{e^{y}-1}\quad\text{and}\quad
v_{z}(y):=ze^{-y}-u_{z}(y)=\frac{1-e^{-zy}}{e^{y}-1}$ (5.2)

are bounded and continuous on $\left[0,\infty\right)$. Furthermore, there
exist
$C_{\mathtt{Re}\left(z\right)},C_{1,\mathtt{Re}\left(z\right)},C_{2,\mathtt{Re}\left(z\right)},\epsilon_{1,\mathtt{Re}\left(z\right)}$,
$\epsilon_{2,\mathtt{Re}\left(z\right)}>0$ such that, for all $y\geq 0$ and
$x\geq C_{\mathtt{Re}\left(z\right)}$,

$|v_{z}(y)|\leq
C_{1,\mathtt{Re}\left(z\right)}|z|e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y}\quad\text{and}\quad|v_{z}(x)|\leq
C_{2,\mathtt{Re}\left(z\right)}e^{-\epsilon_{2,\mathtt{Re}\left(z\right)}x}.$
(5.3)

In the next theorem we also outline some basic facts about potential measures,
which can be found in [31]. Its first part is [31, p. 101, Corrolary 3.3.6],
and the second one is a combination of [31, p. 169, Theorem 5.3.1], [31, p.
171, Theorem 5.3.4], and [31, p. 173, Theorem 5.3.8].

######  Theorem 5.3.

Let $\xi$ be a Lévy process which drifts to $-\infty$.

  1. (i)

The measures $U$ are Radon measures, i.e., for every compact
$K\subset\mathbb{R}$, $U(K)<\infty$.

  2. (ii)

(renewal theorem) If $\xi$ is non-lattice, then it holds that vaguely
$\lim_{x\to\infty}U(x-\mathrm{d}y)=0\mathrm{d}y$, and
$\lim_{x\to-\infty}U(x-\mathrm{d}y)=-\mathrm{d}y/\mathbb{E}\left[\xi_{1}\right]$
with the convention $1/\infty=0$. In the case of a lattice process,
$\mathrm{d}y$ should be replaced by the counting measure on the respective
lattice.

We start with the proof of Lemma 4.1.

###### Proof of Lemma 4.1.

Recall that $\phi_{q,\pm}(z):=\phi_{\pm}(q,z)$. From [28, Theorem 4.7 (4.15)],
we know that, for $z\in\mathbb{C}_{(-1,\infty)}$,

$\log_{0}\left(W_{\phi_{q,\pm}}(1+z)\right)=z\ln\phi_{q,\pm}(1)+\int_{[0,\infty)}\left(e^{-zy}-1-z(e^{-y}-1)\right)\frac{k_{\pm}(q,\mathrm{d}y)}{y(e^{y}-1)},$
(5.4)

where the measures $k_{\pm}(q,\mathrm{d}y)$ are defined in [28, Theorem 4.7
(4.15)] for any $q>0$, and, from [28, (5.37)], have Laplace transforms, for
$\lambda>0$,

$\begin{split}\int_{[0,\infty)}e^{-\lambda
y}k_{\pm}(q,\mathrm{d}y)=\frac{\frac{\partial}{\partial{\lambda}}{\phi_{\pm}(q,\lambda)}}{\phi_{\pm}(q,\lambda)}.\end{split}$

On the other hand, for $q>0$ and $\lambda>0$,
$\phi_{+}(q,\lambda)=\kappa_{+}\left(q,\lambda\right)/c_{+}$;
$\phi_{-}(q,\lambda)=h(q)\kappa_{-}\left(q,\lambda\right)/c_{-}$, and
differentiating with respect to (w.r.t.) $z$ the logarithm of (2.7), we get

$\displaystyle\frac{\frac{\partial}{\partial{\lambda}}{\phi_{\pm}(q,\lambda)}}{\phi_{\pm}(q,\lambda)}=\frac{\frac{\partial}{\partial{\lambda}}{\kappa_{\pm}(q,\lambda)}}{\kappa_{\pm}(q,\lambda)}$
$\displaystyle=\int_{0}^{\infty}\int_{[0,\infty)}e^{-qt}ye^{-\lambda
y}\frac{\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)}{t}\mathrm{d}t$
$\displaystyle=\int_{[0,\infty)}e^{-\lambda
y}y\int_{0}^{\infty}e^{-qt}\frac{\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)}{t}\mathrm{d}t.$

Identifying the two Laplace transforms above, we arrive at

$k_{\pm}(q,\mathrm{d}y)=y\int_{0}^{\infty}e^{-qt}\frac{\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)}{t}\mathrm{d}t=:y\mathcal{W}_{\pm}\left(q,\mathrm{d}y\right),$
(5.5)

as defined in (4.3) of the lemma itself. Plugging this into (5.4) and using
the relations (2.10), we demonstrate the validity of (4.2). Next, we prove
(4.5) noting that we understand the derivatives as measures, which are the
weak limit of the respective difference ratios, i.e.
$(\mathcal{W}_{\pm}\left(q_{0}+h,\mathrm{d}y\right)$ $-$
$\mathcal{W}_{\pm}\left(q_{0},\mathrm{d}y\right))/h$. From (5.5), it is
immediately seen that, for $q>0$ and any $n\geq 0$, by monotone convergence,

$\mathcal{W}^{(n)}_{\pm}\left(q,\mathrm{d}y\right)=(-1)^{n}\int_{0}^{\infty}t^{n-1}e^{-qt}\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)\mathrm{d}t=-\frac{\partial^{n-1}}{\partial{q}^{n-1}}{U^{(n-1)}_{q}}(\pm\mathrm{d}y),$
(5.6)

where the last expression holds by the very definition of the $q$-potential
measure, see (4.1). Using (5.1) of Proposition 5.1, we derive all identities
in (4.5). Finally, we prove (4.4): first we note that from [28, Theorem 4.1],
for each $q>0$, $W_{\phi_{q,\pm}}$ are analytic and zero-free on
$\mathbb{C}_{(0,\infty)}$, and according to [5, Theorem 2.8],
$W_{\kappa_{\pm}}(q,z)$ are analytic on
$\mathbb{C}_{\left(0,\infty\right)}\times\mathbb{C}_{\left(0,\infty\right)}$.
Using the defined in Lemma 5.2 bounded and continuous function $u_{z}$, see
(5.2), by differentiating (5.4) and using (5.5), we get that, for $q>0$ and
$z\in\mathbb{C}_{(-1,\infty)}$,

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\phi_{q,\pm}}(1+z)}}{W_{\phi_{q,\pm}}(1+z)}=z\frac{\frac{\partial}{\partial{q}}{\phi_{q,\pm}}(1)}{\phi_{q,\pm}(1)}+\int_{[0,\infty)}u_{z}(y)\mathcal{W}^{\prime}_{\pm}(q,\mathrm{d}y).\end{split}$
(5.7)

Substituting the relations (2.10), we get that

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\kappa_{\pm}}(q,1+z)}}{W_{\kappa_{\pm}}(q,1+z)}=z\frac{\frac{\partial}{\partial{q}}{\kappa_{\pm}}(q,1)}{\kappa_{\pm}(q,1)}+\int_{[0,\infty)}u_{z}(y)\mathcal{W}^{\prime}_{\pm}(q,\mathrm{d}y).\end{split}$
(5.8)

For the first term in the sum above, employing (2.7), we arrive at

$\frac{\frac{\partial}{\partial{q}}{\kappa_{\pm}}(q,1)}{\kappa_{\pm}(q,1)}=\int_{0}^{\infty}\int_{[0,\infty)}e^{-qt}e^{-y}\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)\mathrm{d}t=\int_{[0,\infty)}e^{-y}U_{q}(\pm\mathrm{d}y).$
(5.9)

The finiteness of the last quantity for $U_{0}=U$ is guaranteed by the vague
convergence stated in Theorem 5.3, and also for $q>0$, because in the sense of
measures $U_{q}(\pm\mathrm{d}y)\leq U(\pm\mathrm{d}y)$, see (4.1). Also, from
(5.6), we see that
$\mathcal{W}^{\prime}_{\pm}(q,\mathrm{d}y)=-U_{q}(\pm\mathrm{d}y)$, so once
again by the renewal Theorem 5.3 and the exponential decay of $u_{z}(y)$ as
$y\to\infty$, see Lemma 5.2, we conclude that

$\int_{[0,\infty)}u_{z}(y)\mathcal{W}^{\prime}_{\pm}(q,\mathrm{d}y)=-\int_{[0,\infty)}u_{z}(y)U_{q}(\pm\mathrm{d}y)$

is well-defined for $q\geq 0$. Plugging the last expression and (5.9) into
(5.8), we arrive at

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\kappa_{\pm}}(q,1+z)}}{W_{\kappa_{\pm}}(q,1+z)}=\int_{[0,\infty)}\left(ze^{-y}-u_{z}(y)\right)U_{q}(\pm\mathrm{d}y)=\int_{[0,\infty)}\frac{1-e^{-zy}}{e^{y}-1}U_{q}(\pm\mathrm{d}y),\end{split}$
(5.10)

which is (4.4). This concludes the proof of the lemma. ∎

Now we proceed with the proof of the main result of Section 4, Theorem 4.3.

###### Proof of Theorem 4.3.

We first prove the forward direction for the derivatives of
$W_{\kappa_{\pm}}$.

Let us recall again that from [5, Theorem 2.8], $W_{\kappa_{\pm}}(q,z)$ are
analytic on
$\mathbb{C}_{\left(0,\infty\right)}\times\mathbb{C}_{\left(0,\infty\right)}$,
and thus all derivatives of $W_{\kappa_{\pm}}(q,1+z)$ exist at any $q>0$ and
any $z\in\mathbb{C}_{(-1,\infty)}$. For each such pair $\left(q,z\right)$,
using the defined in Lemma 5.2 bounded and continuous function $v_{z}$, see
(5.2), (4.4) of Lemma 4.1 gives us

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\kappa_{\pm}}(q,1+z)}}{W_{\kappa_{\pm}}(q,1+z)}=\int_{[0,\infty)}\frac{1-e^{-zy}}{e^{y}-1}U_{q}(\pm\mathrm{d}y)=:\int_{[0,\infty)}v_{z}(y)U_{q}(\pm\mathrm{d}y).\end{split}$
(5.11)

From (5.3), we have that for any fixed $z$ such that $\mathtt{Re}(z)>-1$,
there exist $\epsilon:=\epsilon_{\mathtt{Re}\left(z\right)}>0$ and
$C:=C_{z}>0$ such that

$\left|v_{z}(y)\right|\leq Ce^{-\epsilon y}.$ (5.12)

From [28, Theorem 4.1 (1)], it is known that $W_{\kappa_{\pm}}(q,z)$ are zero
free on $q\geq 0$, $\mathtt{Re}\left(z\right)>0$. Therefore, for $q>0$, we can
use the dominated convergence theorem to get from (5.11)

$\begin{split}\frac{\partial^{n}}{\partial{q}^{n}}{}\frac{\frac{\partial}{\partial{q}}{W_{\kappa_{\pm}}(q,1+z)}}{W_{\kappa_{\pm}}(q,1+z)}&=\frac{\partial^{n}}{\partial
q^{n}}\int_{[0,\infty)}v_{z}(y)\int_{0}^{\infty}e^{-qt}\mathbb{P}\left(\xi_{t}\in\pm\mathrm{d}y\right)\mathrm{d}t\\\
&=\int_{[0,\infty)}v_{z}(y)U^{(n)}_{q}(\pm\mathrm{d}y).\end{split}$ (5.13)

Taking limit as $q\to 0$, from (5.12) and using
$|U_{q}^{(n)}(\pm\mathrm{d}y)|\leq|U^{(n)}(\pm\mathrm{d}y)|$ in the sense of
measures, we observe the two implications

$\begin{split}\int_{[0,\infty)}e^{-\epsilon
y}\left|U^{(n)}(\pm\mathrm{d}y)\right|<\infty&\implies\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{W_{\kappa_{\pm}}(0,1+z)}\right|<\infty\\\
&\implies\lim_{q\to
0}\frac{\partial^{n+1}}{\partial{q}^{n+1}}{W_{\kappa_{\pm}}(q,1+z)}=\frac{\partial^{n+1}}{\partial{q}^{n+1}}{W_{\kappa_{\pm}}(0,1+z)}.\end{split}$
(5.14)

However, from (4.5), we know that $n$th derivative of $U$ is a constant
multiple of the $(n+1)$-st convolution of $U$, so to obtain the finiteness of
the considered derivatives of $W_{\kappa_{\pm}}$ at $0$, it would be enough to
prove that

$\int_{[0,\infty)}e^{-\epsilon y}U^{*k}(\pm\mathrm{d}y)<\infty,\quad 1\leq
k\leq n+1.$ (5.15)

By (4.5),

$U^{*k}(\pm\mathrm{d}y)=\int_{0}^{\infty}t^{k-1}\mathbb{P}(\xi_{t}\in\pm\mathrm{d}y)\mathrm{d}t,$

so if we check (5.15) for $k=n+1$, this would imply the result for all $k\leq
n$. If $n=0$, (5.15) follows by the same application of the renewal theorem as
for (5.9), since Theorem 5.3 holds. Assume now that $n\geq 1$. From the
assumption (4.6), its equivalent statement (4.8), and (4.5), we see that

$\begin{split}\int_{(1,\infty)}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)<\infty\iff
U^{*n}\left(\mathbb{R}_{+}\right)=\frac{1}{(n-1)!}\int_{0}^{\infty}t^{n-1}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t<\infty.\end{split}$ (5.16)

Plugging the expression

$U^{*(n+1)}(\pm\mathrm{d}y)=\int_{-\infty}^{\infty}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)$
(5.17)

into (5.15) for $k=n+1$, we derive

$\begin{split}\int_{[0,\infty)}e^{-\epsilon
y}&\int_{-\infty}^{\infty}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)=\int_{-\infty}^{\infty}\int_{[0,\infty)}e^{-\epsilon
y}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)\\\
&=\int_{[0,\infty)}\int_{[0,\infty)}e^{-\epsilon
y}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)+\int_{(-\infty,0)}\int_{[0,\infty)}e^{-\epsilon
y}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)\\\
&=:I^{\pm}_{1}+I^{\pm}_{2}.\end{split}$ (5.18)

To establish (5.15), we will check that $I^{\pm}_{1}$ and $I^{\pm}_{2}$ are
finite. For $I^{\pm}_{1}$, we have that

$\begin{split}I^{\pm}_{1}&=\int_{[0,\infty)}\sum_{m\geq
0}\int_{[m,m+1)}e^{-\epsilon y}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)\\\
&\leq\sum_{m\geq 0}e^{-\epsilon
m}\int_{[0,\infty)}\int_{[m,m+1)}U(\pm\mathrm{d}y-x)U^{*n}(\mathrm{d}x)\\\
&\leq\sum_{m\geq 0}e^{-\epsilon m}\int_{[0,\infty)}U{\left([\pm
m-x,\pm(m+1)-x)\right)}U^{*n}(\mathrm{d}x)\leq\sup_{v\in\mathbb{R}}U\left([v,v+1)\right)U^{*n}\left(\mathbb{R}^{+}\right)\frac{1}{1-e^{-\epsilon}}.\end{split}$

Therefore, from (5.16) and the finiteness of the last supremum by Theorem 5.3,
we conclude that $I^{\pm}_{1}<\infty$. We proceed to study $I^{\pm}_{2}$ for
which we need to introduce some more notation: let, for $x>0$,

$T_{-x}:=\inf\left\\{t\geq 0:\xi_{t}<-x\right\\},$

which is a.s. finite because $\lim_{t\to\infty}\xi_{t}=-\infty$. We thus
obtain that

$\begin{split}U^{*n}\left([-x,0)\right)&=\int_{0}^{\infty}t^{n-1}\mathbb{P}\left(\xi_{t}\in\left[-x,0\right)\right)\mathrm{d}t\leq\mathbb{E}\left[T^{n}_{-x}\right]+\int_{0}^{\infty}\mathbb{E}\left[(t+T_{-x})^{n-1}\right]\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t.\end{split}$ (5.19)

However, from the finiteness of the integrals in (5.16) and [13, Theorem 1
(1.15)], which triggers $\mathbb{E}\left[T^{n}_{-x}\right]<\infty$, we arrive
via Hölder’s inequality at

$\begin{split}U^{*n}\left([-x,0)\right)&\leq\mathbb{E}\left[T^{n}_{-x}\right]+\sum_{j=0}^{n-1}\binom{n-1}{j}\mathbb{E}\left[T^{n-1-j}_{-x}\right]\int_{0}^{\infty}t^{j}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t\leq
C_{n}\max\left\\{1,\mathbb{E}\left[T^{n}_{-x}\right]\right\\},\end{split}$
(5.20)

where $C_{n}>0$ is some finite constant, so we see also that $U^{*n}$ is
$\sigma$-finite. We note that as $T_{-x}$ goes to infinity as $x$ increases,
since by assumption $\lim_{t\to\infty}\xi_{t}=-\infty$, the last inequality
implies that, for some $C^{\prime}>1$, if $x\geq C^{\prime}$, then
$U^{*n}\left([-x,0)\right)\leq C_{n}\mathbb{E}\left[T^{n}_{-x}\right].$
Therefore, by [13, (1.25)], for $x\geq C^{\prime}$ and some $D_{n}>0$,

$U^{*n}\left([-x,0)\right)\leq C_{n}\mathbb{E}\left[T^{n}_{-x}\right]\leq
D_{n}\left(\frac{x}{|A(x)|}\right)^{n},$ (5.21)

where we recall that $A$ may have been suitably redefined to be positive for
$x\geq 1$, see below (3.8). Next, since $\xi$ drifts to $-\infty$, we can
apply [38, Lemma 2.2], to get

$U([m+x,m+x+1))\leq c_{1}\mathbb{P}\left(\rho_{m+x,m+x+1}<\infty\right)$

for some $c_{1}>0$ and, for $a<b$, $\rho_{a,b}:=\inf\left\\{t\geq
0:\xi_{t}\in\left[a,b\right]\right\\}$. Therefore, we immediately deduct that,
for $m\geq 0$,

$U([m+x,m+x+1))\leq c_{1}\mathbb{P}\left(\mathcal{H}_{\tau}\geq m+x\right)\leq
c_{1}\mathbb{P}\left(\mathcal{H}_{\tau}\geq x\right),$ (5.22)

where $\left(\mathcal{H}_{t}\right)_{t\geq 0}$ is the conservative ascending
ladder height process, which, due to the fact that
$\lim_{t\to\infty}\xi_{t}=-\infty$, is killed at an independent exponential
time $\tau\sim Exp(\lambda)$ with $\lambda>0$ so that
$\mathcal{H}_{\tau-}=\sup_{t\geq 0}\xi_{t}$. We note that by quasi-left-
continuity, $\mathcal{H}_{\tau-}=\mathcal{H}_{\tau}$ a.s., see [6, Proposition
I.7], and we will use the latter for smoother presentation. Hence, employing
(5.21) and (5.22) in (5.18), we next get

$\begin{split}&I^{+}_{2}=\int_{(-\infty,0)}\int_{[0,\infty)}e^{-\epsilon
y}U(\mathrm{d}y-x)U^{*n}(\mathrm{d}x)=\int_{(0,\infty)}\int_{[0,\infty)}e^{-\epsilon
y}U(\mathrm{d}y+x)U^{*n}(-\mathrm{d}x)\\\ &\leq\sum_{m\geq 0}e^{-\epsilon
m}\int_{(0,\infty)}U([m+x,m+x+1))U^{*n}(-\mathrm{d}x)\leq c_{1}\sum_{m\geq
0}e^{-\epsilon m}\int_{(0,\infty)}\mathbb{P}(\mathcal{H}_{\tau}\geq
x)U^{*n}(-\mathrm{d}x)\\\
&=\frac{c_{1}}{1-e^{-\epsilon}}\int_{(0,\infty)}\int_{[x,\infty)}\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}y)U^{*n}(-\mathrm{d}x)=\frac{c_{1}}{1-e^{-\epsilon}}\int_{(0,\infty)}U^{*n}\left([-x,0)\right)\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)\\\
&\leq\frac{c_{1}D_{n}}{1-e^{-\epsilon}}\left(c_{2}+\int_{(1,\infty)}\left(\frac{x}{|A(x)|}\right)^{n}\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)\right)\end{split}$
(5.23)

for $c_{2}:=U^{*n}\left([-C^{\prime},0)\right)<\infty$ with $C^{\prime}$
defined prior to (5.21). If $\mathbb{E}\left[\xi_{1}\right]$ is finite, then
$\left|A(\infty)\right|=-\mathbb{E}\left[\xi_{1}\right]\in\left(0,\infty\right)$,
so it is enough to prove that

$\int_{(0,\infty)}x^{n}\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)=\mathbb{E}\left[H_{\tau}^{n}\right]=\mathbb{E}\left[\left(\sup_{t\geq
0}\xi_{t}\right)^{n}\right]<\infty,$

which is true by [13, Theorem 5, (1.33)] and assumption (4.6). Next, let us
consider the case where $\mathbb{E}\left[\xi_{1}\right]=-\infty$, so $|A(x)|$
is unbounded as well. Let us define $\Pi_{\mathcal{H}}$ to be the Lévy measure
of $\mathcal{H}$ and
$A_{\mathcal{H}}(x):=\int_{0}^{x}\overline{\Pi}_{\mathcal{H}}(y)\mathrm{d}y$,
for which we know from [13, (2.15)] and [14, p.34] that

$|A(x)|\asymp\int_{1}^{x}\overline{\Pi}_{-}(y)\mathrm{d}y\asymp
A_{\mathcal{H}}(x)$ (5.24)

with ”$\asymp$” denoting that the ratio of the two sides is bounded away from
zero and infinity. Hence, our assumption (4.6) entails that

$\int_{(1,\infty)}\left(\frac{x}{A_{\mathcal{H}}(x)}\right)^{n+1}\Pi(\mathrm{d}x)=:\int_{(1,\infty)}\left(g(x)\right)^{n+1}\Pi(\mathrm{d}x)<\infty,$
(5.25)

where we have used once again the convention that if $x/A_{\mathcal{H}}(x)$ is
not well-defined over $(1,x_{0}]$ for some $x_{0}>1$, we define it there as
some constant. Moreover, as
$g(x)=x/A_{\mathcal{H}}(x)=1/\int_{0}^{1}\overline{\Pi}_{\mathcal{H}}(xy)\mathrm{d}y$,
the function $g$ is increasing, tends to infinity at infinity, and has a
finite limit at 0. Also, by (5.24), $A_{\mathcal{H}}(\infty)$ is a positive
constant or infinity, so

$g^{n}(x)=\mathrm{O}\left(x^{n}\right),$ (5.26)

and, also from [14, (7.21)], we have the following connection between
$\overline{\Pi}_{\mathcal{H}}$ and $\Pi$: for large $x$, there exist a
constant $c_{3}>0$ such that

$\overline{\Pi}_{\mathcal{H}}(x)\leq
c_{3}\int_{(x,\infty)}g(y)\Pi(\mathrm{d}y).$ (5.27)

We are ready to prove that $I_{2}^{+}$ is finite in the considered case: by
(5.23), (5.24), and the last discussion, it would be enough to prove that
$\int_{(1,\infty)}g^{n}(x)\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)$ is
finite. To do this, first note that because
$g^{n}(x)\leq\int_{(1,x)}\mathrm{d}\left(g^{n}(y)\right)+g^{n}(2)$, we have to
check that

$\begin{split}\int_{(1,\infty)}\left(\int_{(1,x)}\mathrm{d}\left(g^{n}(y)\right)+g^{n}(2)\right)\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)&\leq
g^{n}(2)+\int_{(1,\infty)}\mathbb{P}(\mathcal{H}_{\tau}>y)\mathrm{d}\left(g^{n}(y)\right)\\\
&=g^{n}(2)+\int_{(1,\infty)}\int_{0}^{\infty}\lambda e^{-\lambda
t}\mathbb{P}(\mathcal{H}_{t}>y)\mathrm{d}t\mathrm{d}\left(g^{n}(y)\right)\end{split}$
(5.28)

is finite. To bound the integral, let us split the process into two
components, according to the size of the jumps:
$\mathcal{H}=\mathcal{H}^{s}+\mathcal{H}^{l}$ for $\mathcal{H}^{l}$ the
process of the jumps larger than some $c>0$ and $\mathcal{H}^{s}$, the
component with jumps at most $c$, and then use the bound
$\mathbb{P}(\mathcal{H}_{t}>y)\leq\mathbb{P}(\mathcal{H}^{s}_{t}>y/2)+\mathbb{P}(\mathcal{H}^{l}_{t}>y/2)$
in (5.28). For $\mathcal{H}^{s}$, observe that by Markov’s inequality, for
$\beta>0$,

$\mathbb{P}(\mathcal{H}^{s}_{t}>y/2)\leq e^{-\beta
y/2}\left(\mathbb{E}\left[e^{\beta\mathcal{H}^{s}_{1}}\right]\right)^{t},$

where the expectation is finite, because the Lévy measure has bounded support,
see [33, Theorem 25.3]. Choosing $\beta=\beta(c,\lambda)$ small enough, we
have $\mathbb{E}\left[e^{\beta\mathcal{H}^{s}_{1}}\right]\leq e^{\lambda/2}$,
so by (5.26),

$\int_{0}^{\infty}\int_{(1,\infty)}\lambda e^{-\lambda
t}\mathbb{P}(\mathcal{H}^{s}_{t}>y/2)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\leq\int_{0}^{\infty}\lambda
e^{-\lambda t/2}\int_{(1,\infty)}e^{-\beta
y/2}\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t<\infty.$

For the process of large jumps, first observe that

$\begin{split}\int_{(1,\infty)}\int_{c_{4}\ln y}^{\infty}\lambda e^{-\lambda
t}\mathbb{P}(\mathcal{H}^{l}_{t}>y/2)\mathrm{d}t\mathrm{d}\left(g^{n}(y)\right)\leq\int_{(1,\infty)}y^{-\lambda
c_{4}}\mathrm{d}\left(g^{n}(y)\right),\end{split}$

where we used $\mathbb{P}(\mathcal{H}^{l}_{t}>y/2)\leq 1$, so again by (5.26),
choosing sufficiently large $c_{4}$, the last integral is finite. For the
remaining integral, with $N_{t}:=\\#\text{jumps of $\mathcal{H}^{l}$ in
}[0,t]\sim Poi(t\overline{\Pi}_{\mathcal{H}}(c))$, we can choose $c>0$ as
large as we wish so that $e^{\overline{\Pi}_{\mathcal{H}}(c)}-1\leq\lambda/2$.
Fix $D>0$ such that for $x\geq 1/(8D)>c$ relation (5.27) holds. Then, we have

$\begin{split}\int_{(1,\infty)}&\int_{0}^{c_{4}\ln y}\lambda e^{-\lambda
t}\mathbb{P}(\mathcal{H}^{l}_{t}>y/2)\mathrm{d}t\mathrm{d}\left(g^{n}(y)\right)=\int_{0}^{\infty}\lambda
e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}\mathbb{P}\left(\mathcal{H}^{l}_{t}>y/2\right)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&=\int_{0}^{\infty}\lambda e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}\mathbb{P}\left(\mathcal{H}^{l}_{t}>y/2;N_{t}\leq
2Dy\right)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\hphantom{{}=\int_{0}^{\infty}}+\int_{0}^{\infty}\lambda e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}\mathbb{P}\left(\mathcal{H}^{l}_{t}>y/2;N_{t}>2Dy\right)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t=:J_{1}+J_{2}.\end{split}$
(5.29)

First, we have by the classical Markov inequality and the choice of $c$

$\begin{split}J_{2}&\leq\int_{0}^{\infty}\lambda e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}\mathbb{P}\left(N_{t}>2Dy\right)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\leq\int_{0}^{\infty}\lambda e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}e^{-2Dy}e^{t\left(e^{\overline{\Pi}_{\mathcal{H}}(c)}-1\right)}\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\leq\int_{0}^{\infty}\lambda e^{-\lambda
t/2}\mathrm{d}t\int_{({1},\infty)}e^{-2Dy}\mathrm{d}\left(g^{n}(y)\right)<\infty,\end{split}$
(5.30)

because $e^{-2Dy}\leq y^{-\lambda c_{4}}$ for large $y$ and the finiteness
discussed above. Also, with the quantity $k(y)=\max\left\\{l\geq 1:l\leq
2Dy\right\\}$,

$\begin{split}J_{1}&\leq\int_{0}^{\infty}\lambda e^{-\lambda
t}\int_{\left(e^{t/c_{4}},\infty\right)}\sum_{k=1}^{k(y)}\mathbb{P}\left(\mathcal{H}^{l}_{t}>y/2\middle|N_{t}=k\right)\mathbb{P}(N_{t}=k)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\leq\int_{0}^{\infty}\lambda e^{-\lambda
t}\sum_{k=1}^{\infty}\int_{\left(e^{t/c_{4}},\infty\right)}\mathbbm{1}_{\left\\{k\leq
2Dy\right\\}}\mathbb{P}(N_{t}=k)k\frac{\overline{\Pi}_{\mathcal{H}}\left(y/(2k)\right)}{\overline{\Pi}_{\mathcal{H}}\left(c\right)}\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\leq\frac{c_{2}}{\overline{\Pi}_{\mathcal{H}}\left(c\right)}\int_{0}^{\infty}\lambda
e^{-\lambda
t}\sum_{k=1}^{\infty}k\mathbb{P}(N_{t}=k)\int_{\left(e^{t/c_{4}},\infty\right)}\mathbbm{1}_{\left\\{k<4Dy\right\\}}\int_{(y/(2k),\infty)}g(x)\Pi(\mathrm{d}x)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t\\\
&\leq\frac{c_{2}}{\overline{\Pi}_{\mathcal{H}}\left(c\right)}\int_{0}^{\infty}\lambda
e^{-\lambda
t}\sum_{k=1}^{\infty}k\mathbb{P}(N_{t}=k)\int_{\left(\max\left\\{e^{t/c_{4}},k/(4D)\right\\},\infty\right)}\int_{(y/(2k),\infty)}g(x)\Pi(\mathrm{d}x)\mathrm{d}\left(g^{n}(y)\right)\mathrm{d}t,\end{split}$
(5.31)

where, since $y/2k\geq 1/(4D)>1/(8D)>c$, by the choice of $D$ we apply (5.27)
for the third inequality, and for the second inequality we utilize the fact
that if $\mathcal{H}^{l}_{t}>y/2$ and there are exactly $k$ jumps up to time
$t$, then at least one of them has to be larger than $y/(2k)$ with the
probability of such jump given by
$\overline{\Pi}_{\mathcal{H}}\left(y/(2k)\right)/\overline{\Pi}_{\mathcal{H}}\left(c\right)$.
For clarity, let us separate the following term from (5.31):

$\begin{split}&\int_{\left(\max\left\\{e^{t/c_{4}},k/(4D)\right\\},\infty\right)}\int_{(y/(2k),\infty)}g(x)\Pi(\mathrm{d}x)\mathrm{d}\left(g^{n}(y)\right)\\\
&\quad\quad=\int_{\left(\max\left\\{e^{t/c_{4}}/(2k),1/(8D)\right\\},\infty\right)}g(x)\int_{\left(\max\left\\{e^{t/c_{4}},k/(4D)\right\\},2kx\right)}\mathrm{d}\left(g^{n}(y)\right)\Pi(\mathrm{d}x)\\\
&\quad\quad\leq\int_{\left(1/(8D),\infty\right)}g(x)\left(\frac{2kx}{A_{\mathcal{H}}(2kx)}\right)^{n}\Pi(\mathrm{d}x)\leq(2k)^{n}\int_{(1/(8D),\infty)}g^{n+1}(x)\Pi(\mathrm{d}x)=\mathrm{O}\left(k^{n}\right),\end{split}$

where we have used that $A_{\mathcal{H}}$ is decreasing, the definition of
$g$, and the finiteness in (5.25). Substituting the last in (5.31), because,
e.g. by Hölder’s inequality,
$\mathbb{E}\left[N_{t}^{n+1}\right]=\mathrm{O}\left(t^{n+1}\right)$, we can
conclude that $J_{1}<\infty$. From this and (5.30), we can deduce that the
integral in (5.29) is finite which yields that $I_{2}^{+}$ is finite as well.

Thus, it remains to demonstrate that $I^{-}_{2}<\infty$. Note that upon change
of variables $x\to-x$ and $-y+x\to-z$ we get that

$\begin{split}I^{-}_{2}&=\int_{(-\infty,0)}\int_{[0,\infty)}e^{-\epsilon
y}U(-\mathrm{d}y-x)U^{*n}(\mathrm{d}x)=\int_{(0,\infty)}e^{-\epsilon
x}\int_{[-x,\infty)}e^{-\epsilon z}U(-\mathrm{d}z)U^{*n}(-\mathrm{d}x)\\\
&=\int_{(0,\infty)}e^{-\epsilon x}\int_{[0,\infty)}e^{-\epsilon
z}U(-\mathrm{d}z)U^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{[-x,0)}e^{-\epsilon z}U(-\mathrm{d}z)U^{*n}(-\mathrm{d}x)\\\
&=:J_{1}+J_{2}.\end{split}$ (5.32)

Clearly, from (5.21),

$\begin{split}J_{1}&=\int_{[0,\infty)}e^{-\epsilon
z}U(-\mathrm{d}z)\int_{(0,\infty)}e^{-\epsilon
x}U^{*n}(-\mathrm{d}x)=\epsilon\int_{[0,\infty)}e^{-\epsilon
z}U(-\mathrm{d}z)\int_{0}^{\infty}U^{*n}\left(-y,0\right)e^{-\epsilon
y}\mathrm{d}y\\\ &\leq\epsilon\int_{(0,\infty)}e^{-\epsilon
z}U(-\mathrm{d}z)\int_{0}^{\infty}\left(U^{*n}\left(-C^{\prime},0\right)+D_{n}\left(\frac{y}{|A(y)|}\right)^{n}\mathbbm{1}_{\left\\{y>1\right\\}}\right)e^{-\epsilon
y}\mathrm{d}y<\infty,\end{split}$ (5.33)

since the integral with respect to $U(-\mathrm{d}z)$ is finite thanks to
Theorem 5.3, and the other one, from (5.24) and the properties of $g$, see
(5.26). For $J_{2}$, by splitting the inner integration at $x/2$, we get

$\begin{split}J_{2}&=\int_{(0,\infty)}e^{-\epsilon x}\int_{(0,x]}e^{\epsilon
z}U(\mathrm{d}z)U^{*n}(-\mathrm{d}x)\\\ &\leq\int_{(0,\infty)}e^{-\epsilon
x/2}U\left(\left(0,x/2\right]\right)U^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{(x/2,x)}e^{\epsilon z}U(\mathrm{d}z)U^{*n}(-\mathrm{d}x)\\\ &\leq
c_{5}\int_{(0,\infty)}\max\left\\{x,1\right\\}e^{-\epsilon
x/2}U^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{(x/2,x)}e^{\epsilon z}U(\mathrm{d}z)U^{*n}(-\mathrm{d}x),\end{split}$

where $U\left(\left(0,x/2\right]\right)\leq c_{5}\max\left\\{x,1\right\\}$ for
$x>0$ and some $c_{5}>0$ follows easily from Theorem 5.3. The first integral
in the last expression is finite since $xe^{-\epsilon x/2}$ decays faster to
$0$ than $e^{-\epsilon x/3}$, and we can then proceed as in (5.33). Next, we
estimate the second integral, using $\lfloor\cdot\rfloor$ for the floor
function, and employing (5.22) in the second inequality, as follows

$\displaystyle\int_{(0,\infty)}e^{-\epsilon x}\int_{(x/2,x)}e^{\epsilon
z}U(\mathrm{d}z)U^{*n}(-\mathrm{d}x)$
$\displaystyle\leq\int_{(0,\infty)}e^{-\epsilon x}\sum_{m=\lfloor
x/2\rfloor}^{\lfloor x\rfloor}e^{\epsilon(m+1)}U([m,m+1))U^{*n}(-\mathrm{d}x)$
$\displaystyle\leq c_{1}e^{\epsilon}\int_{(0,\infty)}e^{-\epsilon
x}\sum_{m=\lfloor x/2\rfloor}^{\lfloor x\rfloor}e^{\epsilon
m}\mathbb{P}\left(\mathcal{H}_{\tau}\geq m\right)U^{*n}(-\mathrm{d}x)$
$\displaystyle\leq c_{1}e^{\epsilon}\int_{(0,\infty)}e^{-\epsilon
x}\sum_{m=0}^{\lfloor x\rfloor}e^{\epsilon
m}\mathbb{P}\left(\mathcal{H}_{\tau}\geq\lfloor
x/2\rfloor\right)U^{*n}(-\mathrm{d}x)$ $\displaystyle\leq
c_{6}\int_{(0,\infty)}\mathbb{P}\left(\mathcal{H}_{\tau}\geq\lfloor
x/2\rfloor\right)U^{*n}(-\mathrm{d}x)$ $\displaystyle\leq
c_{6}\left(\int_{(0,6]}U^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}\mathbb{P}\left(\mathcal{H}_{\tau}>x/3\right)U^{*n}(-\mathrm{d}x)\right)$
$\displaystyle=c_{6}\left(U^{*n}\left([-6,0)\right)+\int_{(0,\infty)}U^{*n}(-3x,0)\mathbb{P}\left(\mathcal{H}_{\tau}\in\mathrm{d}x\right)\right),$

where we have defined $c_{6}:=c_{1}e^{2\epsilon}/(e^{\epsilon}-1)$. The last
is expression is finite because $U^{*n}\left([-6,0)\right)$ is finite by
(5.20), and for the remaining, we can proceed exactly as in (5.23): if
$A(\infty)<\infty$, we have a direct connection with
$\mathbb{E}\left[H_{\tau}^{n}\right]<\infty$, and in the other case, we can
work with $A_{\mathcal{H}}$, which is monotone and therefore, by (5.21), for
large $x$,

$U^{*n}(-3x,0)\leq D_{n}\left(\frac{3x}{|A_{\mathcal{H}}(3x)|}\right)^{n}\leq
D_{n}3^{n}\left(\frac{x}{|A_{\mathcal{H}}(x)|}\right)^{n},$

and we reduce to (5.23).

Therefore, $J_{2}<\infty$, and from (5.33) and (5.32), we get that
$I^{-}_{2}<\infty$. From the previous arguments, all $I^{\pm}_{1},$
$I^{\pm}_{2}$ are finite, and from (5.18), we obtain that (5.15) holds true.
Thus far we have proved that the considered derivatives of $W_{\kappa_{\pm}}$
are finite. To conclude that this leads to the same conclusion for the
derivatives of the same order of $W_{\phi_{q,\pm}}$, we recall (2.10):

$W_{\phi_{q,+}}(z)=c_{+}^{1-z}W_{\kappa_{+}}(q,z),\quad\text{and}\quad
W_{\phi_{q,-}}(z)=c_{-}^{1-z}h^{z-1}(q)W_{\kappa_{-}}(q,z),$

with $h$ defined in (2.6) as

$\begin{split}&h(q):=\exp\left(-\int_{0}^{\infty}\left(\frac{e^{-t}-e^{-qt}}{t}\right)\mathbb{P}\left(\xi_{t}=0\right)\mathrm{d}t\right).\end{split}$

If $\xi$ is not a CPP, $h\equiv 1$ and our claim is obvious. Let us assume now
that $\mathbb{P}(\xi_{t}=0)\neq 0$ on some Lebesgue non-negligible set. Since
$\xi$ is transient, the integral above is finite for every $q\geq 0$ by [6,
Theorem VI.12], and therefore $h$ is strictly positive for each $q\geq 0$. For
higher derivatives of $h$, note that by the definition of $h$ and (5.1),

$(\ln
h(q))^{(k)}=(-1)^{k+1}\int_{[0,\infty)}e^{-qt}t^{k-1}\mathbb{P}(\xi_{t}=0)\mathrm{d}t=U^{(k-1)}_{q}\left(\left\\{0\right\\}\right)=(-1)^{k-1}(k-1)!U_{q}^{*k}(\\{0\\}).$
(5.34)

The last is valid for $1\leq k\leq n+1$, as we have already proven (5.15)
which implies for any $\epsilon>0$ that

$U_{q}^{*k}(\\{0\\})\leq\int_{[0,\infty)}e^{-\epsilon
y}U^{*k}(\pm\mathrm{d}y)<\infty,$

and therefore, for these $k$, $(\ln h(q))^{(k)}$ are finite and non-zero for
each $q\geq 0$. Therefore $h^{(k)}(q)$ are finite as well, which concludes the
proof of the forward part of the theorem.

For the backward part of the theorem, see (4.6), by (5.13) and (2.10), it
would suffice to show that if the integral in (4.6) is infinite, then
$U^{*(n+1)}$, and, respectively, $U^{(n)}$, do not define a $\sigma$-finite
measure. Under the assumption that

$\int_{1}^{\infty}\left(\frac{x}{\left|A(x)\right|}\right)^{n+1}\Pi(\mathrm{d}x)=\infty,$

[13, Theorem 4] yields that, with $T_{a}:=\inf\left\\{t\geq
0:\xi_{t}>a\right\\}$ for $a>0$, it holds that

$\mathbb{E}\left[T^{n}_{a}\mathbbm{1}_{\left\\{T_{a}<\infty\right\\}}\right]=\infty.$
(5.35)

Thus, picking $a<b$, with $\xi^{\prime}_{t}=\xi_{t+T_{a}}-\xi_{T_{a}}$ on
$T_{b}<\infty$,

$\begin{split}U^{*(n+1)}\left(a,b\right)&=\int_{0}^{\infty}t^{n}\mathbb{P}\left(\xi_{t}\in\left(a,b\right)\right)\mathrm{d}t=\mathbb{E}\left[\int_{T_{a}}^{\infty}t^{n}\mathbbm{1}_{\left\\{\xi_{t}\in\left(a,b\right)\right\\}}\mathrm{d}t\right]\\\
&=\mathbb{E}\left[\mathbbm{1}_{\left\\{T_{a}<\infty\right\\}}\int_{0}^{\infty}\left(t+T_{a}\right)^{n}\mathbbm{1}_{\left\\{\xi^{\prime}_{t}\in\left(a-\xi_{T_{a}},b-\xi_{T_{a}}\right)\right\\}}\mathrm{d}t\right]\\\
&\geq\mathbb{E}\left[T^{n}_{a}\mathbbm{1}_{\left\\{T_{a}<\infty\right\\}}\int_{0}^{\infty}\mathbbm{1}_{\left\\{\xi^{\prime}_{t}\in\left(a-\xi_{T_{a}},b-\xi_{T_{a}}\right)\right\\}}\mathrm{d}t\right].\end{split}$
(5.36)

Assume, temporarily, that $\xi$ does not live on a lattice. Then clearly there
exists $K>0$ such that $\mathbb{P}\left(a\leq\xi_{T_{a}}\leq K+a\right)\geq
1/2$, and therefore from (5.36), we get that

$\begin{split}U^{*(n+1)}\left(a,b\right)&\geq\frac{1}{2}\inf\left\\{a\leq
y\leq
K+a:U\left(a-y,b-y\right)\right\\}\mathbb{E}\left[T^{n}_{a}\mathbbm{1}_{\left\\{T_{a}<\infty\right\\}}\right].\end{split}$
(5.37)

Since $\inf\left\\{a\leq y\leq K+a:U\left(a-y,b-y\right)\right\\}>0$, we get
from (5.35) that $U^{*(n+1)}\left(a,b\right)=\infty$. If $\xi$ lives on a
lattice, then the result is proved in the case where $\left(a,b\right)$
contains a point of this lattice. This suffices. Therefore, our theorem is
proved. ∎

###### Proof of Corollary 4.4.

(i) First, let us recall that (2.9) defines

$M_{\Psi}\left({q},{z}\right):=\frac{\Gamma(z)}{W_{\phi_{q,+}}(z)}W_{\phi_{q,-}}(1-z).$

By Theorem 4.3, $\frac{\partial^{k}}{\partial{q}^{k}}{W_{\phi_{q,\pm}}(z)}$
are right-continuous and finite for $q\geq 0$, $\mathtt{Re}\left(z\right)>0$,
and $0\leq k\leq n+1$. Moreover, from [28, Theorem 4.1 (1)],
$W_{\phi_{q,+}}(z)$ is non-zero for $q\geq 0$ and
$\mathtt{Re}\left(z\right)>0$, so differentiating the equality above, we
obtain the desired result.

(ii) The claim is trivial for $k=0$. Fix $k\in\\{1,\dots,n+1\\}$, and write,
for $z\in\mathbb{C}_{(0,1)}$, $M_{\Psi}\left({q},{z}\right)=e^{I(q,z)}$ with

$\begin{split}I(q,z)&:=\log_{0}(\Gamma(z))+\log_{0}\left(W_{\phi_{q,-}}(1-z)\right)-\log_{0}\left(W_{\phi_{q,+}}(z)\right),\end{split}$
(5.38)

which is in fact equal to $\log_{0}(M_{\Psi}\left({q},{z}\right))$ or
$\log_{0}(M_{\Psi}\left({q},{z}\right))+2\pi i$, which gives the same value
after applying the exponential function. Next, substituting
$\phi_{q,+}(1)=\kappa_{+}(q,1)/c_{+}$ and (5.9) into (5.7), we obtain similar
to (5.10) that

$\frac{\frac{\partial}{\partial{q}}{W_{\phi_{q,+}}(1+z)}}{W_{\phi_{q,+}}(1+z)}=z\frac{\frac{\partial}{\partial{q}}{\kappa_{+}(q,1)}}{\kappa_{+}(q,1)}+\int_{[0,\infty)}u_{z}(y)\mathcal{W}^{\prime}_{+}(q,\mathrm{d}y)=\int_{[0,\infty)}v_{z}(y)U_{q}(\mathrm{d}y)$

with $v_{z}$ is defined in (5.2) and is bounded and continuous by Lemma 5.2.
Similarly, because $\phi_{q,-}(1)=h(q)\kappa_{-}(q,1)/c_{-}$, the relations
(5.7), (5.9), and (5.34) gives us

$\begin{split}\frac{\frac{\partial}{\partial{q}}{W_{\phi_{q,-}}(1+z)}}{W_{\phi_{q,-}}(1+z)}&=z\left(\frac{h^{\prime}(q)}{h(q)}+\frac{\frac{\partial}{\partial{q}}{\kappa_{-}(q,1)}}{\kappa_{-}(q,1)}\right)+\int_{[0,\infty)}u_{z}(y)\mathcal{W}^{\prime}_{-}(q,\mathrm{d}y)\\\
&=zU_{q}\left(\left\\{0\right\\}\right)+\int_{[0,\infty)}v_{z}(y)U_{q}(-\mathrm{d}y).\end{split}$

Using the last two equations in (5.38), we get that

$\frac{\partial^{k}}{\partial{q}^{k}}{I}(q,z)=\int_{[0,\infty)}v_{-z}(y)U^{(k-1)}_{q}\left(-\mathrm{d}y\right)-\int_{[0,\infty)}v_{z-1}(y)U^{(k-1)}_{q}\left(\mathrm{d}y\right)-zU_{q}^{(k-1)}(\\{0\\}),$
(5.39)

where $v_{z}$ is defined in (5.2). We will bound the quantities in the last
equality: we first note that from Proposition 5.1,

$\left|U^{(k-1)}_{q}(\mathrm{d}v)\right|\leq\left|U^{(k-1)}(\mathrm{d}v)\right|,$

and as the conditions of Theorem 4.3 are fulfilled by assumption, we know from
its proof, see (5.15), and (5.1), that, for each $\epsilon>0$ and any $1\leq
k\leq n+1$,

$\int_{[0,\infty)}e^{-\epsilon
y}\left|U^{(k-1)}(\pm\mathrm{d}y)\right|=(k-1)!\int_{[0,\infty)}e^{-\epsilon
y}\left|U^{*k}(\pm\mathrm{d}y)\right|<\infty.$ (5.40)

Therefore, by (5.40) and with some $\epsilon>0$, we have that

$\left|zU^{(k-1)}\left(\\{0\\}\right)\right|\leq|z|\int_{[0,\infty)}e^{-\epsilon
y}\left|U^{(k-1)}(\pm\mathrm{d}y)\right|\leq c_{k}|z|,$ (5.41)

with $c_{k}\geq 0$. Next, by Lemma 5.2, there exist constants
$c_{1,\mathtt{Re}\left(z\right)},c_{2,\mathtt{Re}\left(z\right)},\epsilon_{1,\mathtt{Re}\left(z\right)},\epsilon_{2,\mathtt{Re}\left(z\right)}>0$
such that for $z\in\mathbb{C}_{(0,1)}$,

$\left|v_{-z}(y)\right|\leq
c_{1,\mathtt{Re}\left(z\right)}\left|z\right|e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y},\quad\text{and}\quad\left|v_{z-1}(y)\right|\leq
c_{2,\mathtt{Re}\left(z\right)}\left|1-z\right|e^{-\epsilon_{2,\mathtt{Re}\left(z\right)}y}.$
(5.42)

Substituting $\eqref{eq:hh}$ and $\eqref{eq: vz bounds in proof}$ in (5.39),
we get that

$\begin{split}&\left|\frac{\partial^{k}}{\partial{q}^{k}}{I}(q,z)\right|\leq\int_{[0,\infty)}\left|v_{-z}(y)\right|\left|U^{(k-1)}_{q}\left(-\mathrm{d}y\right)\right|+\int_{[0,\infty)}\left|v_{z-1}(y)\right|\left|U^{(k-1)}_{q}\left(\mathrm{d}y\right)\right|+c_{k}|z|\\\
&\,\leq
c_{1,\mathtt{Re}\left(z\right)}\left|z\right|\int_{[0,\infty)}e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y}\left|U^{(k-1)}\left(-\mathrm{d}y\right)\right|+c_{2,\mathtt{Re}\left(z\right)}\left(1+\left|z\right|\right)\int_{[0,\infty)}e^{-\epsilon_{2,\mathtt{Re}\left(z\right)}y}\left|U^{(k-1)}\left(\mathrm{d}y\right)\right|+c_{k}|z|\\\
&\,=:Q_{k,\mathtt{Re}\left(z\right)}\left(\left|z\right|\right),\end{split}$
(5.43)

where $Q_{k,\mathtt{Re}\left(z\right)}$ is a polynomial of first degree. To
conclude, let us note that by Faà di Bruno’s formula,

$\left(e^{f}\right)^{(k)}=e^{f}\sum_{m_{1}+2m_{2}+\cdots+km_{k}=k}\frac{k!}{m_{1}!\,m_{2}!\,\,\cdots\,m_{k}!}\prod_{j=1}^{k}\left(\frac{f^{(j)}}{j!}\right)^{m_{j}},$
(5.44)

and since we have $M_{\Psi}\left({q},{z}\right)=e^{I(q,z)}$, by using the
bound (5.43), we obtain that

$\left|\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}\right|=\left|\frac{\partial^{k}}{\partial{q}^{k}}{e^{I(q,z)}}\right|\leq\left|e^{I(q,z)}\right|P_{k,\mathtt{Re}\left(z\right)}\left(\left|z\right|\right)=\left|M_{\Psi}\left({q},{z}\right)\right|P_{k,\mathtt{Re}\left(z\right)}\left(\left|z\right|\right)$

with $P_{k,\mathtt{Re}\left(z\right)}$ a polynomial of degree $k$, which is
exactly the statement of the lemma.

(iii) Under the assumption that there exists a bounded $u$ such that
$U(\mathrm{d}x)=u(x)\mathrm{d}x$, we will obtain a stronger bound in (5.43).
Since the integral criterion from Theorem 4.6 holds, we know that the
densities of $U^{*k}(\mathrm{d}x)$, that is $u^{*k}$, exist and are locally
bounded for $1\leq k\leq n+1$. Next, by (5.1) of Proposition 5.1, we conclude
that for $1\leq k\leq n+1$, $U^{(k-1)}(\mathrm{d}x)$ has also a density
$u_{k-1}(x)=k!(-1)^{k-1}u^{*k}(x)$, which is therefore locally bounded. Note
that
$|U^{(k-1)}\left(\left\\{0\right\\}\right)|=\int_{0}^{\infty}t^{k-1}\mathbb{P}\left(\xi_{t}=0\right)\mathrm{d}t=0$.
Next, since $\mathtt{Re}(-z)=-\mathtt{Re}\left(z\right)\in(-1,0)$, by Lemma
5.2 there exist constants
$C_{\mathtt{Re}}\left(z\right)>0,c_{3,\mathtt{Re}\left(z\right)},\epsilon_{3,\mathtt{Re}\left(z\right)}$
such that for $y>C_{\mathtt{Re}\left(z\right)}$, $|v_{-z}(y)|\leq
c_{3,\mathtt{Re}\left(z\right)}e^{-y\epsilon_{3,\mathtt{Re}\left(z\right)}}$,
so if $1/|z|<C_{\mathtt{Re}\left(z\right)}$,

$\begin{split}\int_{0}^{\infty}\left|v_{-z}(y)\right|&\left|U^{(k-1)}_{q}\left(-\mathrm{d}y\right)\right|=\int_{[0,1/|z|)\cup[1/|z|,C_{\mathtt{Re}\left(z\right)})\cup[C_{\mathtt{Re}\left(z\right)},\infty)}\left|v_{-z}(y)\right|\left|U^{(k-1)}_{q}\left(-\mathrm{d}y\right)\right|\\\
&\leq\frac{1}{|z|}c_{1,\mathtt{Re}\left(z\right)}\left|z\right|\int_{0}^{\infty}e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y}\left|U^{(k-1)}\left(-\mathrm{d}y\right)\right|+\int_{1/|z|}^{C_{\mathtt{Re}\left(z\right)}}\left|\frac{1-e^{zy}}{e^{y}-1}u_{k-1}(-y)\mathrm{d}y\right|\\\
&\hphantom{{}=\leq\frac{1}{|z|}c_{1,\mathtt{Re}\left(z\right)}\left|z\right|\int_{0}^{\infty}e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y}\left|U^{(k-1)}\left(-\mathrm{d}y\right)\right|}+c_{3,\mathtt{Re}\left(z\right)}\int_{0}^{\infty}e^{-\epsilon_{3,\mathtt{Re}\left(z\right)}y}\left|U^{(k-1)}\left(-\mathrm{d}y\right)\right|\\\
&\leq
c_{4,\mathtt{Re}\left(z\right)}+\int_{1/|z|}^{C_{\mathtt{Re}\left(z\right)}}\left|\frac{1-e^{zy}}{e^{y}-1}u_{k-1}(-y)\right|\mathrm{d}y,\end{split}$
(5.45)

for some $c_{4,\mathtt{Re}\left(z\right)}>0$ by (5.40), and in the first bound
on the region $0\leq y<1/|z|$, we have used the first bound of (5.42). We note
that if $1/|z|\geq C_{\mathtt{Re}\left(z\right)}$, we can bound the quantity
of interest as above, but without having the integral over
$[1/|z|,C_{\mathtt{Re}\left(z\right)})$, resulting in a constant bound by
$c_{4,\mathtt{Re}\left(z\right)}$. For the last integral in (5.45), because
$|e^{zy}-1|\leq e^{y\mathtt{Re}\left(z\right)}+1\leq
2e^{y\mathtt{Re}\left(z\right)}$ for the considered $z\in\mathbb{C}_{(0,1)}$,
and $u_{k-1}$ is locally bounded, we get that

$\begin{split}\int_{1/|z|}^{C_{\mathtt{Re}}\left(z\right)}\left|\frac{1-e^{zy}}{e^{y}-1}u_{k-1}(-y)\right|\mathrm{d}y&\leq\sup_{x\in\left[-C_{\mathtt{Re}\left(z\right)},0\right]}\left\\{|u_{k-1}(x)|\right\\}\int_{1/|z|}^{C_{\mathtt{Re}\left(z\right)}}\frac{2e^{y\mathtt{Re}\left(z\right)}}{y}\mathrm{d}y\leq
c_{1}\int_{1/|z|}^{C_{\mathtt{Re}\left(z\right)}}\frac{1}{y}\mathrm{d}y\\\
&=c_{1}\left(\ln\left(C_{\mathtt{Re}\left(z\right)}\right)+\ln|z|\right),\end{split}$
(5.46)

for $c_{1}:=c_{1}(\mathtt{Re}\left(z\right))>0$, so (5.45) gives a linear in
$\ln|z|$ bound for the first summand on the first line of (5.43), with
coefficients that depend only on $\mathtt{Re}\left(z\right)$. For the second
one, as $\mathtt{Re}(z-1)\in(-1,0)$, with the same approach we can obtain a
linear in $\ln(|1-z|)$ bound and because
$|1-z|\leq|z|+1\leq|z|(1+1/\mathtt{Re}\left(z\right))$, we would obtain a
linear in $\ln|z|$ bound, with coefficients depending only on $\mathtt{Re}z$,
and not $|z|$. Repeating the arguments which prove $(\ref{it:cor_der_Mq_ii})$
from (5.43), we get the desired polynomial in $\ln|z|$ bound.

We proceed with the last two items, (ii’) and (iii’). We shall be economical
due to the similarities to the proof above. Setting

$R(\beta):=\int_{(1,\infty)}x^{\beta+1}\Pi(\mathrm{d}x),$ (5.47)

we have by assumption, see (4.7), that $R(\beta)<R(n)=\infty$ for
$\beta\in\left(0,n\right)$. Also,

$\lim_{q\to 0}U^{*n}(\mathbb{R}_{+})=\lim_{q\to
0}\int_{0}^{\infty}t^{n-1}e^{-qt}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t=\infty$

follows from (4.9). Next, $R(\beta)<\infty$ implies
$\int_{0}^{\infty}t^{\beta-1}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t<\infty$, see [13, Theorem 1] applied to $-\xi$, and from
[13, (1.25)] applied to $-\xi$, and because
$|A(\infty)|\in\left(0,\infty\right)$ since
$\mathbb{E}\left[\xi_{1}\right]\in\left(-\infty,0\right)$, we get that

$\mathbb{E}\left[T^{\beta}_{-x}\right]\leq
C_{\beta,1}\max\left\\{x^{\beta},1\right\\}$ (5.48)

for some finite constant $C_{\beta,1}\geq 0$ and with $T_{-x}$ the first
passage time below $-x$. Then, a slight modification of (5.19) yields, for
$x>0$,

$\begin{split}U_{q}^{*n}\left([-x,0)\right)&=\int_{0}^{\infty}t^{n-1}e^{-qt}\mathbb{P}\left(\xi_{t}\in\left[-x,0\right)\right)\mathrm{d}t\\\
&\leq\mathbb{E}\left[\int_{0}^{T_{-x}}t^{n-1}e^{-qt}\mathrm{d}t\right]+\mathbb{E}\left[e^{-qT_{-x}}\int_{0}^{\infty}e^{-qt}(t+T_{-x})^{n-1}\mathbbm{1}_{\left\\{\tilde{\xi}_{t}\geq
0\right\\}}\mathrm{d}t\right],\end{split}$ (5.49)

where $T_{-x}<\infty$ a.s., and by the strong Markov property
$\tilde{\xi}:=\left(\xi_{t+T_{-x}}-\xi_{T_{-x}}\right)_{t\geq 0}$ is
independent of $T_{-x}$ and has the law of $\xi$. Next, using the obvious
inequality that for any $\delta>0$ small enough. there is
$D_{\delta,1}<\infty$ such that for any $x>0$,

$\int_{0}^{x}t^{n-1}e^{-t}\mathrm{d}t\leq D_{\delta,1}x^{n-\delta},$

we conclude that

$\mathbb{E}\left[\int_{0}^{T_{-x}}t^{n-1}e^{-qt}\mathrm{d}t\right]=q^{-n}\mathbb{E}\left[\int_{0}^{qT_{-x}}t^{n-1}e^{-t}\mathrm{d}t\right]\leq
D_{\delta,1}q^{-\delta}\mathbb{E}\left[T^{n-\delta}_{-x}\right].$

Applying the latter and (5.48) in (5.49), we arrive with some
$D_{\delta,2}<\infty$ at

$U_{q}^{*n}\left([-x,0)\right)\leq
D_{\delta,2}q^{-\delta}\max\left\\{x^{n-\delta},1\right\\}+\mathbb{E}\left[\int_{0}^{\infty}e^{-qt}(t+T_{-x})^{n-1}\mathbbm{1}_{\left\\{\tilde{\xi}_{t}\geq
0\right\\}}\mathrm{d}t\right].$

Expanding the last brackets, and treating terms like in (5.20), we deduct

$\begin{split}U_{q}^{*n}\left([-x,0)\right)&\leq
D_{\delta,2}q^{-\delta}\max\left\\{x^{n-\delta},1\right\\}+\sum_{j=0}^{n-1}\binom{n-1}{j}\mathbb{E}\left[T^{n-1-j}_{-x}\right]\int_{0}^{\infty}e^{-qt}t^{j}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t.\end{split}$

From (4.1) and (4.9), and like in (5.20), applying Hölder’s inequality, we get

$\begin{split}U_{q}^{*n}\left([-x,0)\right)&\leq
D_{\delta,2}q^{-\delta}\max\left\\{x^{n-\delta},1\right\\}+\int_{0}^{\infty}e^{-qt}t^{n-1}\mathbb{P}\left(\xi_{t}\geq
0\right)\mathrm{d}t+C_{n}\max\left\\{1,\mathbb{E}\left[T^{n-\delta}_{-x}\right]\right\\}\\\
&\leq
C_{\delta,n}q^{-\delta}\max\left\\{x^{n-\delta},1\right\\}+U^{*n}_{q}(\mathbb{R}_{+}),\end{split}$
(5.50)

where in the first inequality we have taken out from the sum the summand with
index $n-1$, we have estimated the rest with $q=0$, and we have applied that
by (4.9), for $j<n$, $U^{*j}(\mathbb{R}_{+})<\infty$ since $R(j)<\infty$. For
the second inequality, we have employed once again (5.48) for the expectation
term, and from (5.1) we have related the free integral to
$U_{q}^{*n}(\mathbb{R}_{+})$. Given that $R(k)<\infty$ for $k<n,$ all
estimates for $\frac{\partial^{k}}{\partial{q}^{k}}{I}(q,z)$ for $k\leq n$ can
be reused from the previous part. Therefore, we need to estimate only the
$(n+1)$-st derivative, which from (5.39) is equal to

$\frac{\partial^{n+1}}{\partial{q}^{n+1}}{I}(q,z)=\int_{[0,\infty)}v_{-z}(y)U^{(n)}_{q}\left(-\mathrm{d}y\right)-\int_{[0,\infty)}v_{z-1}(y)U^{(n)}_{q}\left(\mathrm{d}y\right)-zU_{q}^{(n)}(\\{0\\}).$

Substituting (5.1) to relate $U^{(n)}_{q}$ to $U^{*(n+1)}_{q}$, we get

$\begin{split}\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{I}(q,z)\right|\leq
n!\left(\int_{[0,\infty)}\left|v_{-z}(y)\right|U^{*(n+1)}_{q}\left(-\mathrm{d}y\right)+\int_{[0,\infty)}\left|v_{z-1}(y)\right|U^{*(n+1)}_{q}\left(\mathrm{d}y\right)+|z|U^{*(n+1)}_{q}(\\{0\\})\right).\end{split}$

From Lemma 5.2, there exist constants
$c_{1,\mathtt{Re}\left(z\right)},c_{2,\mathtt{Re}\left(z\right)},\epsilon_{1,\mathtt{Re}\left(z\right)},\epsilon_{2,\mathtt{Re}\left(z\right)}>0$
such that, for $z\in\mathbb{C}_{(0,1)}$,

$\left|v_{-z}(y)\right|\leq
c_{1,\mathtt{Re}\left(z\right)}\left|z\right|e^{-\epsilon_{1,\mathtt{Re}\left(z\right)}y},\quad\text{and}\quad\left|v_{z-1}(y)\right|\leq
c_{2,\mathtt{Re}\left(z\right)}\left|1-z\right|e^{-\epsilon_{2,\mathtt{Re}\left(z\right)}y}.$

Therefore, for some constants $c_{1}:=c_{1}(\mathtt{Re}\left(z\right))>0$ and
$\epsilon:=\min\left\\{\epsilon_{1,\mathtt{Re}\left(z\right)},\epsilon_{2,\mathtt{Re}\left(z\right)}\right\\}>0$,
we arrive at

$\begin{split}\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{I}(q,z)\right|&\leq
c_{1}\left(2|z|\int_{[0,\infty)}e^{-\epsilon
y}U^{*(n+1)}_{q}\left(-\mathrm{d}y\right)+|1-z|\int_{[0,\infty)}e^{-\epsilon
y}U^{*(n+1)}_{q}\left(\mathrm{d}y\right)|z|\right)\\\ &\leq
c_{1}\left(2|z|\left(I^{-}_{1}(q)+I_{2}^{-}(q)\right)+|1-z|\left(I^{+}_{1}(q)+I^{+}_{2}(q)\right)\right),\end{split}$
(5.51)

where we have borrowed the notation from (5.18), and the multiplier term $2$
comes from bounding the atom at zero by an integral over $[0,\infty)$.
$I^{\pm}_{1}(q)$ can be estimated precisely as below (5.18) to get

$I^{\pm}_{1}(q)\leq\sup_{v\in\mathbb{R}}U_{q}\left([v,v+1)\right)U_{q}^{*n}\left(\mathbb{R}_{+}\right)\frac{1}{1-e^{-\epsilon}}\leq\sup_{v\in\mathbb{R}}U\left([v,v+1)\right)U_{q}^{*n}\left(\mathbb{R}_{+}\right)\frac{1}{1-e^{-\epsilon}}\leq
c_{2}U_{q}^{*n}\left(\mathbb{R}_{+}\right),$ (5.52)

for
$c_{2}:=\sup_{v\in\mathbb{R}}U\left([v,v+1)\right)/(1-e^{-\epsilon})<\infty$
since the supremum is finite by the renewal Theorem 5.3. Next, we get as in
(5.23)

$\begin{split}I^{+}_{2}(q)&=\int_{(-\infty,0)}\int_{[0,\infty)}e^{-\epsilon
y}U_{q}(\mathrm{d}y-x)U_{q}^{*n}(\mathrm{d}x)=\int_{(0,\infty)}\int_{[0,\infty)}e^{-\epsilon
y}U_{q}(\mathrm{d}y+x)U_{q}^{*n}(-\mathrm{d}x)\\\ &\leq\sum_{m\geq
0}e^{-\epsilon m}\int_{(0,\infty)}U([m+x,m+x+1))U^{*n}_{q}(-\mathrm{d}x)\\\
&\leq\frac{c_{3}}{1-e^{-\epsilon}}\int_{(0,\infty)}U^{*n}\left([-x,0)\right)\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)\end{split}$

for some $c_{3}>0$, where in the first inequality we have used that in terms
of measures $U_{q}\leq U$. Substituting the bound from (5.50), we find that

$\begin{split}I^{+}_{2}(q)&\leq\frac{c_{3}}{1-e^{-\epsilon}}\int_{(0,\infty)}\left(C_{\delta,n}q^{-\delta}\max\left\\{x^{n-\delta},1\right\\}+U^{*n}_{q}(\mathbb{R}_{+})\right)\mathbb{P}(\mathcal{H}_{\tau}\in\mathrm{d}x)\\\
&\leq c_{4}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right),\end{split}$
(5.53)

for $c_{4}:=c_{4}(\mathtt{Re}\left(z\right),n,\delta)>0$, since
$R(n-\delta)<\infty$ implies via [13, 1.33] applied to $-\xi$ that
$\mathbb{E}\left[\left(\sup_{t\geq 0}\xi_{t}\right)^{n-\delta}\right]<\infty$,
and, as already used, $\mathcal{H}_{\tau}=\sup_{t\geq 0}\xi_{t}$ a.s. Finally,
as in (5.32),

$\begin{split}I^{-}_{2}(q)&=\int_{(0,\infty)}e^{-\epsilon
x}\int_{[0,\infty)}e^{-\epsilon
z}U_{q}(-\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{[-x,0)}e^{-\epsilon z}U_{q}(-\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)\\\
&=:J_{1}(q)+J_{2}(q).\end{split}$ (5.54)

For $J_{1}$, following (5.33), we get that

$\begin{split}J_{1}(q)&=\int_{[0,\infty)}e^{-\epsilon
z}U_{q}(-\mathrm{d}z)\int_{(0,\infty)}e^{-\epsilon
x}U_{q}^{*n}(-\mathrm{d}x)=\epsilon\int_{[0,\infty)}e^{-\epsilon
z}U_{q}(-\mathrm{d}z)\int_{0}^{\infty}U_{q}^{*n}\left(-y,0\right)e^{-\epsilon
y}\mathrm{d}y\\\ &\leq\epsilon\int_{(0,\infty)}e^{-\epsilon
z}U(-\mathrm{d}z)\int_{0}^{\infty}\left(C_{\delta,n}q^{-\delta}\max\left\\{y^{n-\delta},1\right\\}+U^{*n}_{q}(\mathbb{R}_{+})\right)e^{-\epsilon
y}\mathrm{d}y\\\ &\leq
c_{5}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right),\end{split}$ (5.55)

where in the first inequality we have used the bound in (5.50) and $U_{q}\leq
U$. The finite constant $c_{5}:=c_{5}(\mathtt{Re}\left(z\right),n,\delta)$
incorporates the finiteness of the first integral by the renewal Theorem 5.3
as well as the others involved explicit constants. Next, for $J_{2}$, as below
(5.33) we have

$\begin{split}J_{2}(q)&=\int_{(0,\infty)}e^{-\epsilon
x}\int_{(0,x]}e^{\epsilon z}U_{q}(\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)\\\
&\leq\int_{(0,\infty)}e^{-\epsilon
x/2}U((0,x/2])U_{q}^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{(x/2,x)}e^{\epsilon z}U_{q}(\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)\\\
&\leq c_{6}\int_{(0,\infty)}e^{-\epsilon
x/3}U_{q}^{*n}(-\mathrm{d}x)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{(x/2,x)}e^{\epsilon z}U_{q}(\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)\\\
&\leq
c_{7}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right)+\int_{(0,\infty)}e^{-\epsilon
x}\int_{(x/2,x)}e^{\epsilon
z}U_{q}(\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x),\end{split}$

for $c_{6}>0,c_{7}:=c_{7}(\mathtt{Re}\left(z\right),n,\delta)<\infty$, where
using the arguments below (5.33), we have bounded
$U_{q}\left(\left(0,x/2\right]\right)\leq U((0,x/2])\leq
c_{8}\max\\{x,1\\}\leq c_{9}e^{\epsilon x/6}$ for some finite $c_{8},c_{9}$
and we have treated $\int_{(0,\infty)}e^{-\epsilon
x/3}U_{q}^{*n}(-\mathrm{d}x)$ precisely as in (5.55), possibly at the cost of
a different constant. Third, again precisely as for $J_{2}$ below (5.33), we
can obtain

$\begin{split}\int_{(0,\infty)}e^{-\epsilon x}\int_{(x/2,x)}e^{\epsilon
z}U_{q}(\mathrm{d}z)U_{q}^{*n}(-\mathrm{d}x)\leq
c_{10}\left(U_{q}^{*n}\left([-6,0)\right)+\int_{(0,\infty)}U_{q}^{*n}(-3x,0)\mathbb{P}\left(\mathcal{H}_{\tau}\in\mathrm{d}x\right)\right)\end{split}$

for $c_{10}:=c_{10}(\mathtt{Re}\left(z\right))\geq 0$, so using the bound
(5.50), and repeating the estimate as in (5.53), we get by substitution above
in $J_{2}(q)$ that

$J_{2}(q)\leq c_{11}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right),$

for $c_{11}:=c_{11}(\mathtt{Re}\left(z\right),n,\delta)\geq 0$. Feeding the
latter together with the bound for $J_{1}(q)$, see (5.55), into
$I_{2}^{-}(q)$, (5.54), we get with some
$c_{12}:=c_{12}(\mathtt{Re}\left(z\right),n,\delta)<\infty$,

$I^{-}_{2}(q)\leq c_{12}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right).$

Combining the latter with (5.53) and (5.52), and adding them in (5.51), we
obtain

$\begin{split}\left|\frac{\partial^{n+1}}{\partial{q}^{n+1}}{I}(q,z)\right|&\leq
c_{13}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right)\left(|1-z|+|z|\right)\end{split}$

for $c_{13}:=c_{13}(\mathtt{Re}\left(z\right),n,\delta)\geq 0$. Combining the
latter with the bounds for the lower order derivatives of $I_{q}$, (5.43), and
Faa di Bruno’s formula, (5.44), we obtain (ii’).

The proof of (iii’) has similarities: again, we only need to estimate the
$(n+1)$-st $q$-derivative of $I$, as for lower orders we can use the estimates
from the proof of (iii), see (5.45) and below. Adapting suitably (5.45) and
(5.46), we get that, for some finite $\epsilon,c_{14},c_{15}>0$, which depend
on $\mathtt{Re}\left(z\right)$,

$\begin{split}\int_{0}^{\infty}\left|v_{-z}(y)\right|\left|U^{(n)}_{q}\left(-\mathrm{d}y\right)\right|\leq
c_{14}\int_{0}^{\infty}e^{-\epsilon
y}&\left|U_{q}^{*(n+1)}\left(-\mathrm{d}y\right)\right|\\\
&+\sup_{x\in\left[-c_{15},0\right]}\left\\{|u_{q}^{*(n+1)}(x)|\right\\}\left(\ln\left(c_{15}\right)+\ln|z|\right).\end{split}$
(5.56)

We have already seen in the proof of (ii’), see (5.51) and below, that

$\int_{0}^{\infty}e^{-\epsilon y}U^{*(n+1)}_{q}\left(-\mathrm{d}y\right)\leq
c_{16}\left(q^{-\delta}+U^{*n}_{q}(\mathbb{R}_{+})\right)$ (5.57)

with $c_{16}:=c_{16}(\mathtt{Re}\left(z\right),n,\delta)$, and for the
supremum, by (4.11) from Theorem 4.6, we have that, for
$C:=\sup_{x\in\mathbb{R}}|u(x)|<\infty$,

$\begin{split}\sup_{x\in\left([-c_{15},0)\right)}\left\\{|u_{q}^{*(n+1)}(x)|\right\\}&\leq(n+1)CU_{q}^{*n}\left(-c_{15},\infty\right).\end{split}$

Substituting (5.50) in the former, we get that

$\begin{split}\sup_{x\in\left([-c_{15},0)\right)}\left\\{|u_{q}^{*(n+1)}(x)|\right\\}\leq(n+1)C\left(C_{\delta,n}q^{-\delta}\max\left\\{c_{15}^{\delta},1\right\\}+2U_{q}^{*n}\left(\mathbb{R}_{+}\right)\right),\end{split}$

so, for some finite
$c_{17}:=c_{17}\left(\mathtt{Re}\left(z\right),n,\delta\right)$,

$\sup_{x\in\left([-c_{15},0)\right)}\left\\{|u_{q}^{*(n+1)}(x)|\right\\}\leq
c_{17}\left(q^{-\delta}+U_{q}(\mathbb{R}_{+})\right).$

Substituting the last inequality and (5.57) into (5.56), the desired result,
(iii’), follows as in the proof of (iii), see below (5.46). ∎

###### Proof of Theorem 4.6.

We prove by induction that for $k=1,2,\dots,n+1$,

$u_{q}^{*k}(x)\leq kCU^{*(k-1)}_{q}(\min\\{0,x\\},\infty).$ (5.58)

The result in the base case $k=1$ holds because $U_{q}(\mathrm{d}y)\leq
U(\mathrm{d}y)$ implies that
$\sup_{x\in\mathbb{R}}u_{q}(x)\leq\sup_{x\in\mathbb{R}}u(x)=:C$ a.e. which
leads to inequality everywhere when working with the canonical $u_{q}$, see
[6, Proposition I.11.(ii)]. Let (5.58) hold for some $l\geq 1$. Then,

$\begin{split}u^{*(l+1)}_{q}(x)&=\int_{-\infty}^{\infty}u_{q}(y)u^{*l}_{q}(x-y)\mathrm{d}y\\\
&\leq
C\int_{-\infty}^{x}u_{q}^{*l}(x-y)\mathrm{d}y+lC\int_{x}^{\infty}u_{q}(y)U^{*(l-1)}_{q}\left(\min\left\\{0,x-y\right\\},\infty\right)\mathrm{d}y\\\
&=CU_{q}^{*l}\left(0,\infty\right)+lC\int_{x}^{\infty}U_{q}^{*(l-1)}\left(x-y,\infty\right)U_{q}\left(\mathrm{d}y\right)\\\
&\leq
CU_{q}^{*l}\left(0,\infty\right)+lC\int_{-\infty}^{\infty}U_{q}^{*(l-1)}\left(x-y,\infty\right)U_{q}\left(\mathrm{d}y\right)\\\
&=CU_{q}^{*l}\left(0,\infty\right)+lCU_{q}^{*l}\left(x,\infty\right)\leq(l+1)CU^{*l}_{q}\left(\min\left\\{0,x\right\\},\infty\right),\end{split}$

as needed, which settles (4.11). Next, under the integral condition in the
theorem’s statement, by taking a limit $q\to 0$ in (5.58) and using the
monotone convergence theorem, we conclude the proof for the case $q=0$ since
from (5.16) and (5.21) we have that $U^{*k}(x,\infty)<\infty$ for any
$x\in\mathbb{R}$ and any $1\leq k\leq n$. ∎

##  6 Proofs on exponential functionals

We state three auxiliary results before we commence the proof of Theorem 3.1.
Their proofs can be found in Section 7.

######  Proposition 6.1.

Let $f:[0,\infty)\to[0,\infty)$, and define $f_{0}:=f$. For $x\geq 0$ and
$n\geq 1$, define

$f_{n}(x):=\int_{x}^{\infty}\int_{s_{1}}^{\infty}\dots\int_{s_{n-1}}^{\infty}f(s_{n})\mathrm{d}s_{n}\dots\mathrm{d}s_{1}.$

Then, for $q\geq 0$,

$\widehat{f_{n}}(q)=\frac{(-1)^{n}}{(n-1)!}\int_{0}^{1}\widehat{f}^{(n)}(qv)(1-v)^{k-1}\mathrm{d}v$
(6.1)

in the extended with $\infty=\infty$ sense.

######  Proposition 6.2.

Let $f$ be a regularly varying function at zero with index
$\beta\in\left(-1,0\right]$ and $n\geq 1$ fixed. Then

$\int_{0}^{1}f(qv)(1-v)^{n-1}\mathrm{d}v\stackrel{{\scriptstyle
0}}{{\sim}}{f(q)}\int_{0}^{1}v^{\beta}(1-v)^{n-1}\mathrm{d}v=f(q)\frac{\Gamma(n)\Gamma(\beta+1)}{\Gamma(n+\beta+1)}.$
(6.2)

######  Lemma 6.3.

Let $\xi$ be a Lévy process with a finite negative mean and
$\mathbb{P}(\xi_{1}>t)\stackrel{{\scriptstyle\infty}}{{\sim}}\ell(t)/t^{\alpha}$
for some $\alpha>1$. Then

$\mathbb{E}\left[I_{\xi}^{-a}(t)\right]=\mathrm{O}\left(\mathbb{P}(\xi_{1}>t)\right)=\mathrm{O}\left(\frac{\ell(t)}{t^{\alpha}}\right).$
(6.3)

###### Proof of Theorem 3.1.

We start with the observation that by [35, Theorem 1.1], (3.1) is equivalent
to

$\overline{\Pi}_{+}(t)\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{\ell(t)}{t^{\alpha}}$
(6.4)

with $\overline{\Pi}_{+}(t):=\Pi(t,\infty)$, and that by [12, Theorem 4],
applied to the process $\xi_{t}-t\mathbb{E}\left[\xi_{1}\right]$ with
$c=-\mathbb{E}\left[\xi_{1}\right]>0$ therein, this assumption also leads to
the second relation in

$\mathbb{P}(\xi_{t}\geq
0)\stackrel{{\scriptstyle\infty}}{{\sim}}\mathbb{P}(\xi_{t}>0)\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\frac{\ell(t)}{t^{\alpha-1}},$
(6.5)

with $c_{1}:=\left(-\mathbb{E}{\xi_{1}}\right)^{-\alpha}$. The first is
obvious when the Lévy process has infinite activity as then
$\mathbb{P}\left(\xi_{t}=0\right)=0$. Otherwise, we can apply [12, Theorem 4]
with $c_{\delta}:=-\mathbb{E}\left[\xi_{1}\right]-\delta$,
$-\mathbb{E}\left[\xi_{1}\right]>\delta>0,$ in which case we arrive at

$\mathbb{P}\left(\xi_{t}>0\right)\leq\mathbb{P}\left(\xi_{t}\geq
0\right)\leq\mathbb{P}\left(\xi_{t}>-\delta
t\right)\stackrel{{\scriptstyle\infty}}{{\sim}}c^{-\alpha}_{\delta}\frac{\ell(t)}{t^{\alpha-1}}.$

Setting $\delta\to 0$, we verify the first asymptotic relation in (6.5).

Let us define for $x\geq 0$ the function
$\\{x\\}:=\sup\left\\{y\in(0,1]:x-y\in\mathbb{Z}\right\\}$. This is a slight
variation of the usual fractional part function with the difference that it is
1 in the integers and not 0.

Next, let $\phi_{+}({q}):=\phi_{+}(q,0)$ for $q\geq 0$. By choosing $z=0$ in
(2.8) and differentiating $n:=\alpha-\\{\alpha\\}\geq 1$ times w.r.t. $q$, we
get that, for $q>0$,

$\left(\ln\phi_{+}\right)^{(n)}(q)=(-1)^{n+1}\int_{0}^{\infty}e^{-qt}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t.$ (6.6)

We aim to obtain the asymptotic behaviour of the integral on the right as
$q\to 0$ in order to extract information about $\phi_{+}$ via a Tauberian
theorem. For this purpose, let us observe that by (6.5) and the choice of $n$,
we get that, for $x\to\infty$,

$\int_{0}^{x}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\int_{1}^{x}t^{n-1}t^{-\alpha+1}\ell(t)\mathrm{d}t=c_{1}\int_{1}^{x}t^{-\\{\alpha\\}}\ell(t)\mathrm{d}t.$
(6.7)

Now we can differentiate naturally two cases: if $\alpha$ is non-integer, we
can apply Karamata’s theorem, Theorem A.5, in (6.7), and continue with a
standard Tauberian theorem, Theorem A.7. In the integer case, we will use a
more subtle method, based on the fact that

$x\mapsto\int_{1}^{x}t^{-1}\ell(t)\mathrm{d}t$

is in the de Haan class $\Pi_{\ell}$, see the Appendix A and [8, Chapter 3]
for more information on this class.

Case 1: $\alpha$ is non-integer, i.e.
$\alpha\in(1,\infty)\setminus\mathbb{N}$. Take $\alpha\in\left(n,n+1\right)$
for some $n\geq 1$. Let us now apply the aforementioned Karamata’s theorem,
A.5, in (6.7), which implies that

$\int_{0}^{x}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\frac{x^{1-\\{\alpha\\}}}{1-\\{\alpha\\}}\ell(x),$

and so, by a standard Tauberian theorem A.7, we get from (6.6) that

$\left(\ln\phi_{+}\right)^{(n)}(q)\stackrel{{\scriptstyle
0}}{{\sim}}(-1)^{n+1}c_{1}\Gamma\left(1-\\{\alpha\\}\right)q^{\\{\alpha\\}-1}\ell(1/q).$
(6.8)

The latter is infinite at $0$ and we can also see that
$\left(\ln\phi_{+}\right)^{(k)}(0+)$ are finite for $k<n$ which follows from
the integral representation (6.6) and the finiteness of the non-vanishing
integral in (4.9). Thus, as by Faà di Bruno’s formula, with non-negative
integers $m_{i}$,

$\begin{split}&\left(\ln\phi_{+}\right)^{(n)}=\\\
&\hphantom{{}=(l}\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\\!\\!\\!\\!\frac{n!}{m_{1}!\,m_{2}!\,\,\cdots\,m_{n}!}\frac{(-1)^{m_{1}+\cdots+m_{n}-1}(m_{1}+\cdots+m_{n}-1)!}{\phi_{+}^{m_{1}+\cdots+m_{n}}}\prod_{j=1}^{n}\left(\frac{\phi_{+}^{(j)}}{j!}\right)^{m_{j}},\end{split}$
(6.9)

and the fact that, since the process drifts to $-\infty$, $\phi_{+}({0})$ is
finite, we can see that

$\left|\phi_{+}^{{(k)}}\left({0+}\right)\right|<\infty\text{ for }0\leq k\leq
n-1,\quad\text{ and
}\quad\left|\phi_{+}^{{(n)}}\left({0+}\right)\right|=\infty.$ (6.10)

From the last three results, as the only term of
$\left(\ln\phi_{+}\right)^{(n)}$ containing $\phi_{+}^{(n)}$ is
$\phi_{+}^{(n)}/\phi_{+}$, i.e. when $m_{n}=1$ and $m_{j}=0$ for $j\leq n-1$
in the sum, we get that

$\phi_{+}^{{(n)}}\left({q}\right)\stackrel{{\scriptstyle
0}}{{\sim}}(-1)^{n+1}c_{1}\Gamma\left(1-\\{\alpha\\}\right)\phi_{+}({0})q^{\\{\alpha\\}-1}\ell\left(1/q\right).$
(6.11)

Define for brevity $g_{a}(t):=\mathbb{E}\left[I^{-a}_{\xi}(t)\right]$, which
we note is monotone decreasing, and we recall that from (2.12), for
$a\in(0,1)$,

$\begin{split}\widehat{g_{a}}^{(n)}(q)&=\frac{\partial^{n}}{\partial{q}^{n}}{\int_{0}^{\infty}}e^{-qt}g_{a}(t)\mathrm{d}t=(-1)^{n}\int_{0}^{\infty}t^{n}e^{-qt}g_{a}(t)\mathrm{d}t\\\
&=\left(\frac{M_{\Psi}\left({q},{1-a}\right)}{\phi_{+}({q})}\right)^{(n)}=\sum_{k=0}^{n}\binom{n}{k}\left(\frac{1}{\phi_{+}}\right)^{(k)}(q)\frac{\partial^{n-k}}{\partial{q}^{n-k}}{M_{\Psi}\left({q},{1-a}\right)}.\end{split}$
(6.12)

Since $\left|\mathbb{E}\left[\xi_{1}\right]\right|<\infty$ and
$\alpha\in\left(n,n+1\right)$, we know from (4.10) that
$\int_{(1,\infty)}x^{n-1+1}\Pi(\mathrm{d}x)<\infty$. Hence, by (i) of
Corollary 4.4, all derivatives of $M_{\Psi}\left({q},{1-a}\right)$ in (6.12)
are finite even for $q=0$. Furthermore, from [28, (7.64)], we know that for
$z\in\mathbb{C}_{(0,1)}$, $\lim_{q\to
0}M_{\Psi}\left({q},{z}\right)=M_{\Psi}\left({0},{z}\right)$, so combining
this with (6.10), we obtain that

$\widehat{g_{a}}^{(n)}(q)\stackrel{{\scriptstyle
0}}{{\sim}}\left(\frac{1}{\phi_{+}}\right)^{(n)}(q)M_{\Psi}\left({0},{1-a}\right),$
(6.13)

and using again Faà di Bruno’s formula, for $x>0$,

$\begin{split}&\left(\frac{1}{\phi_{+}}\right)^{(n)}(q)\\\
&\hphantom{{}=(l}=\sum_{m_{1}+2m_{2}+\cdots+nm_{n}=n}\frac{n!}{m_{1}!\,m_{2}!\,\,\cdots\,m_{n}!}\frac{(-1)^{m_{1}+\cdots+m_{n}}(m_{1}+\cdots+m_{n})!}{(\phi_{+}(q))^{m_{1}+\cdots+m_{n}+1}}\prod_{j=1}^{n}\left(\frac{\phi_{+}^{(j)}(q)}{j!}\right)^{m_{j}},\end{split}$
(6.14)

we can see that if we expand the derivative in (6.13), there is only one
infinite at zero term, which is exactly the one that contains
$\phi_{+}^{{(n)}}\left({q}\right)$. It is obtained for $m_{n}=1$ in
(LABEL:eq:FdB_1/f), so substituting (6.11), we get that

$\begin{split}\widehat{g_{a}}^{(n)}(q)\stackrel{{\scriptstyle
0}}{{\sim}}-\frac{\phi_{+}^{{(n)}}\left({q}\right)}{\phi_{+}^{2}(q)}M_{\Psi}\left({0},{1-a}\right)\stackrel{{\scriptstyle
0}}{{\sim}}q^{\\{\alpha\\}-1}\ell_{1}(q)\end{split}$ (6.15)

with

$\ell_{1}(q):=(-1)^{n}c_{1}\Gamma\left(1-\\{\alpha\\}\right)\phi^{-1}_{+}(0)M_{\Psi}\left({0},{1-a}\right)\ell(1/q)\in
SV_{0},$ (6.16)

where $SV_{0}$ denotes the class of slowly varying at zero functions, see
Appendix A. Now, let us define $g_{a,n}$ in the spirit of Proposition 6.1:

$g_{a,n}(x):=\int_{x}^{\infty}\int_{s_{1}}^{\infty}\dots\int_{s_{n-1}}^{\infty}g_{a}(s_{n})\mathrm{d}s_{n}\dots\mathrm{d}s_{2}\mathrm{d}s_{1},$
(6.17)

which is well-defined since by Lemma 6.3 and Potter’s bounds, see Proposition
A.3, we have
$g_{a}(t)=\mathrm{O}\left(\ell(t)/t^{\alpha}\right)=\mathrm{O}\left(1/t^{n+\left\\{\alpha\right\\}/2}\right)$.
Therefore, from Proposition 6.1 itself,

$\widehat{g_{a,n}}(q)=\frac{(-1)^{n}}{(n-1)!}\int_{0}^{1}\widehat{g_{a}}^{(n)}(qv)(1-v)^{n-1}\mathrm{d}v.$

Next from (6.15) we can invoke Proposition 6.2 with $\beta=\\{\alpha\\}-1$ to
obtain the asymptotic behaviour of $\widehat{g_{a,n}}$:

$\widehat{g_{a,n}}(q)\stackrel{{\scriptstyle
0}}{{\sim}}\frac{(-1)^{n}}{(n-1)!}\left(\int_{0}^{1}v^{\\{\alpha\\}-1}(1-v)^{n-1}\mathrm{d}v\right)q^{\\{\alpha\\}-1}\ell_{1}(q)\stackrel{{\scriptstyle
0}}{{\sim}}\frac{(-1)^{n}\Gamma(\\{\alpha\\})}{\Gamma(\alpha)}q^{\\{\alpha\\}-1}\ell_{1}(q).$

Therefore, from the standard Tauberian theorem A.7,

$\int_{0}^{t}g_{a,n}(y)\mathrm{d}y\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{(-1)^{n}\Gamma(\\{\alpha\\})}{\Gamma(\alpha)\Gamma(2-\\{\alpha\\})}t^{1-\\{\alpha\\}}\ell_{1}\left(\frac{1}{t}\right).$

Next, we can apply the monotone density theorem A.10 to get

$g_{a,n}(t)\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{(-1)^{n}\Gamma(\\{\alpha\\})}{\Gamma(\alpha)\Gamma(1-\\{\alpha\\})}t^{-\\{\alpha\\}}\ell_{1}\left(\frac{1}{t}\right)=\frac{c_{1}M_{\Psi}\left({0},{1-a}\right)\Gamma(\\{\alpha\\})}{\phi_{+}({0})\Gamma(\alpha)}t^{-\\{\alpha\\}}\ell(t),$

where we have substituted the definition of $\ell_{1}$, see (6.16). Finally,
taking into account the monotonicity of the inner integrals defining $g_{a,n}$
in (6.17), after $n$ consecutive applications of the second part of the
monotone density theorem, A.10, in the last asymptotic relation, we determine,
recalling $\alpha=n+\\{\alpha\\}$, that

$g_{a}(t)=\mathbb{E}\left[I^{-a}_{\xi}(t)\right]\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{c_{1}M_{\Psi}\left({0},{1-a}\right)}{\phi_{+}({0})}t^{-\alpha}\ell(t).$
(6.18)

To prove the weak convergence of the normalised quantities
$\mathbb{E}\left[I^{-a}_{\xi}(t)\right]$ as $t\to\infty$, let us now consider
the distribution functions, for $x>0$,

$g_{a}^{x}(t):=\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right].$ (6.19)

Define the Laplace transform of this function via
$f^{q}_{a}(x):=\widehat{g_{a}^{x}}(q)$. From [28, (7.46)], it can be seen that
the Mellin transform of $f_{a}^{q}$, for $z\in\mathbb{C}_{(a-1,0)}$, is

$\begin{split}\mathcal{M}_{\widehat{f_{a}^{q}}}(z)&=\int_{0}^{\infty}x^{z-1}\int_{0}^{\infty}e^{-qt}\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]\mathrm{d}t\mathrm{d}x\\\
&=\frac{1}{q}\int_{0}^{\infty}x^{z-1}\mathbb{E}\left[I_{\Psi_{q}}^{-a}\mathbbm{1}_{\left\\{I_{\Psi_{q}}\leq
x\right\\}}\right]\mathrm{d}x=-\frac{1}{\phi_{+}({q})z}M_{\Psi}\left({q},{z+1-a}\right),\end{split}$
(6.20)

where we recall that
$I_{\Psi_{q}}:=I_{\xi}(\mathbf{e}_{q})=\int_{0}^{e_{q}}e^{-\xi_{s}}\mathrm{d}s$.
We can try to get a similar result to (6.18) by Mellin inversion and Tauberian
theorems. We first argue that the Mellin transform is well-defined: from [28,
Theorem 2.3], the decay of $M_{\Psi}$ is at least polynomial along complex
lines in $\mathbb{C}_{(0,1)}$. To be more precise, writing $z=b+iy$ with
$b\in(a-1,0)$, the aforementioned theorem implies that if $\xi$ is not a CPP
with a positive drift, for any $\beta\geq 0$ and $q\geq 0$,

$\lim_{|y|\to\infty}|y|^{\beta}|M_{\Psi}\left({q},{b+1-a+iy}\right)|=0,$
(6.21)

and otherwise, there exists $\epsilon>0$ such that

$\lim_{|y|\to\infty}|y|^{\epsilon}|M_{\Psi}\left({q},{b+1-a+iy}\right)|=0.$
(6.22)

Therefore, we can invert the Mellin transform from (6.20), for $b\in(a-1,0)$,

$\begin{split}f_{a}^{q}(x)=\widehat{g_{a}^{x}}(q)&=-\frac{1}{2\pi
i}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{\phi_{+}({q})z}M_{\Psi}\left({q},{z+1-a}\right)\mathrm{d}z\\\
&=-\frac{x^{-b}}{2\pi\phi_{+}({q})}\int_{-\infty}^{\infty}\frac{x^{-iy}}{b+iy}M_{\Psi}\left({q},{b+1-a+iy}\right)\mathrm{d}y\\\
&=:-\frac{x^{-b}}{2\pi\phi_{+}({q})}\int_{-\infty}^{\infty}h(q,y)\mathrm{d}y,\end{split}$
(6.23)

as $\eqref{eq:decay_M}$ and $\eqref{eq:decay_M_CPP}$ ensure that for some
$\epsilon>0$, $\lim_{|y|\to\infty}|y|^{1+\epsilon}|h(q,y)|=0$, and thus the
last integral is finite. Next, if we wish to apply again Proposition 6.1 for
the $n$th derivative of $\widehat{g_{a}^{x}}(q)$, it would be enough to check
whether the integral of the respective derivatives (right-derivatives for
$q=0)$ w.r.t. $q$ of $h(q,y)$ is absolutely integrable.

Case 1.1: $\xi$ is not a CPP with positive drift. Let us write for simplicity
$z=b+1-a+iy$. Note that because $\mathtt{Re}\left(z\right)\in(0,1)$ and the
considerations below (6.12) are valid, then by (ii) of Corollary 4.4, for each
$0\leq k\leq n+1$, there exist polynomials $P_{\mathtt{Re}\left(z\right),k}$
of degree $k$ such that

$\left|\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}\right|\leq
P_{\mathtt{Re}\left(z\right),k}(|z|)|M_{\Psi}\left({q},{z}\right))|,$

and as we noted earlier, see (6.21), for $q\geq 0$, the decay of $M_{\Psi}$ is
super-polynomial along complex lines, so the last result proves that
$\int_{-\infty}^{\infty}|\frac{\partial^{k}}{\partial{q}^{k}}{h}(q,y)|\mathrm{d}y$
is finite.

Case 1.2: $\xi$ is a CPP with positive drift. The difference in this case is
that we have only a polynomial decrease of $M_{\Psi}$, see (6.22). However,
now every point is possible for $\xi$, so [6, Proposition II.11, Theorem
II.16] entails that the potential of the process has a bounded density w.r.t.
the Lebesgue measure, and therefore, by item (iii) of Corollary 4.4, for each
$0\leq k\leq n+1$, there exist polynomials $P_{\mathtt{Re}\left(z\right),k}$
of degree $k$ such that

$\left|\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}\right|\leq
P_{\mathtt{Re}\left(z\right),k}(\ln|z|)|M_{\Psi}\left({q},{z}\right))|,$

and similarly to the first case, we obtain that
$\int_{-\infty}^{\infty}|\frac{\partial^{k}}{\partial{q}^{k}}{h}(q,y)|\mathrm{d}y$
is finite for $q\geq 0$.

Summing up, in both subcases, 1.1 and 1.2, we can differentiate $n$ times the
equation (6.23), to obtain, for $q>0$,

$\widehat{g_{a}^{x}}^{(n)}(q)=\left(\frac{h_{a}^{x}(q)}{\phi_{+}({q})}\right)^{(n)}\text{\quad
with \quad}h_{a}^{x}(q):=-\frac{1}{2\pi
i}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}M_{\Psi}\left({q},{z+1-a}\right)\mathrm{d}z,$
(6.24)

which is in the form of (6.12), but with $h_{a}^{x}(q)$ in place of
$M_{\Psi}(q,1-a)$. Next, using the logic below [28, (7.48)], we get from [28,
(7.66)] that, for any $\epsilon>0$ in subcase 1.1. and for some $\epsilon>0$
in subcase 1.2,

$\limsup_{|b|\to\infty}|b|^{\epsilon}\sup_{0\leq q\leq
1}|M_{\Psi}\left({q},{z+1-a}\right)-M_{\Psi}\left({0},{z+1-a}\right)|=0,$
(6.25)

and from the other claims of [28, Lemma 7.3] that

$\lim_{q\to 0+}h_{a}^{x}(q)=h_{a}^{x}(0).$

Moreover, using again [28, (7.66)], for any $\epsilon>0$ in subcase 1.1. and
for some $\epsilon>0$ for subcase 1.2, we additionally deduce from the bounds
in items (ii)-(iii) of Corollary 4.4 that

$\limsup_{q\to 0+}\left|\left(h^{x}_{a}(q)\right)^{(k)}\right|<\infty$ (6.26)

for any $k\leq n+1$. Therefore, as the only infinite term at zero in (6.24) is
$\phi_{+}^{{(n)}}\left({0}\right)$, we can repeat the arguments from (6.12) to
(6.15) to get

$\widehat{g_{a}^{x}}^{(n)}(q)\stackrel{{\scriptstyle
0}}{{\sim}}-\frac{\phi_{+}^{{(n)}}\left({q}\right)}{\phi^{2}_{+}(q)}h_{a}^{x}(0)\stackrel{{\scriptstyle
0}}{{\sim}}\frac{h_{a}^{x}(0)}{M_{\Psi}\left({0},{1-a}\right)}q^{\\{\alpha\\}-1}\ell_{1}(q).$

Using the same chain of arguments as between (6.15) and (6.18), we obtain also
an analogue of (6.18) for $g_{a}^{x}$ simply by replacing $M_{\Psi}$ with
$h_{a}^{x}$:

$g_{a}^{x}(t)=\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{c_{1}h_{a}^{x}(0)}{\phi_{+}({0})}t^{-\alpha}\ell(t).$
(6.27)

Therefore, combining with the similar (6.18), we get that, for any $x>0$,

$\lim_{t\to\infty}\frac{\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{I_{\xi}(t)\leq
x\right\\}}\right]}{t^{\alpha}\ell(t)}=\frac{c_{1}h_{a}^{x}(0)}{\phi_{+}({0})},\quad\text{and}\quad\lim_{t\to\infty}\frac{\mathbb{E}\left[I_{\xi}^{-a}(t)\right]}{t^{\alpha}\ell(t)}=\frac{c_{1}M_{\Psi}\left({0},{1-a}\right)}{\phi_{+}({0})}.$
(6.28)

Let us define on $(0,\infty)$ the measures
$\nu_{a,t}(\mathrm{d}y):=y^{-a}\mathbb{P}(I_{\xi}^{-a}(t)\in\mathrm{d}y)t^{\alpha}/\ell(t)$.
As we see from (6.28), these measures have finite masses, so as we seek to
obtain weak convergence, we will normalise them and use the classic
Portmanteau theorem, see [9, Corollary 8.2.10]. Let
$\widetilde{\nu}_{a,t}(\mathrm{d}y):=\nu_{a,t}(\mathrm{d}y)/\nu_{a,t}(0,\infty)$.
From (6.28),

$\lim_{t\to\infty}\widetilde{\nu}_{a,t}(\mathrm{d}y)\left((0,x]\right)=\frac{h_{a}^{x}(0)}{M_{\Psi}\left({0},{1-a}\right)},$

hence we only need to verify that the last defines a distribution function of
a probability measure, for which it would be enough to check, substituting the
definition of $h_{a}^{x}$, (6.24), that

$\lim_{x\to\infty}h_{a}^{x}(0)=\lim_{x\to\infty}\frac{-1}{2\pi
i}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z=M_{\Psi}\left({0},{1-a}\right).$
(6.29)

We do this by the residue theorem: by [28, (2.10)], the function $z\mapsto
x^{-z}M_{\Psi}\left({0},{z+1-a}\right)/z$ is meromorphic on
$\mathbb{C}_{(a-1,a)}$, with a single simple pole at 0\. Its residue there is
exactly $M_{\Psi}\left({0},{1-a}\right)$, so integrating by the negatively
oriented contour of the rectangle
$\gamma_{R}:=\\{z\in\mathbb{C}|-b_{1}\leq\mathtt{Re}\left(z\right)\leq
b_{1},-R\leq\mathtt{Im}z\leq R\\}$ with the choice
$-b=b_{1}\in(0,\min\\{a,1-a\\})$ and $R>0$, gives us

$-2\pi
iM_{\Psi}\left({0},{1-a}\right)=\lim_{R\to\infty}\oint_{\gamma_{R}}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z.$
(6.30)

Along the horizontal strips of the contour, i.e. $\mathtt{Im}z=\pm R$, we use
the bound

$|M_{\Psi}\left({q},{z}\right)|\leq
M_{\Psi}\left({q},{\mathtt{Re}\left(z\right)}\right),$

which holds when the quantities are well-defined, and is based on the results
[29, Theorem 6.1 (3), Proposition 6.8], which state, respectively, that
$W_{\phi_{q,-}}(z)$ and $\Gamma(z)/W_{\phi_{q,+}}(z)$ are Mellin transforms of
positive random variables. Indeed, if we call these random variables
$V_{\phi_{q,-}}$ and $I_{\phi_{q,+}}$, we can see that

$\begin{split}\left|M_{\Psi}\left({q},{z}\right)\right|&=\left|\frac{\Gamma(z)}{W_{\phi_{q,+}}(z)}W_{\phi_{q,-}}(1-z)\right|=\left|\mathbb{E}\left[I_{\phi_{q,+}}^{z-1}\right]\mathbb{E}\left[V_{\phi_{q,-}}^{-z}\right]\right|\leq\mathbb{E}\left[\left|I_{\phi_{q,+}}^{z-1}\right|\right]\mathbb{E}\left[\left|V_{\phi_{q,-}}^{-z}\right|\right]\\\
&=\mathbb{E}\left[I_{\phi_{q,+}}^{\mathtt{Re}\left(z\right)-1}\right]\mathbb{E}\left[V_{\phi_{q,-}}^{-\mathtt{Re}\left(z\right)}\right]=M_{\Psi}\left({q},{\mathtt{Re}\left(z\right)}\right).\end{split}$
(6.31)

This gives us

$\begin{split}&\left|\int_{-b_{1}}^{b_{1}}\frac{x^{-y-iR}}{y+iR}M_{\Psi}\left({0},{y+1-a+iR}\right)-\frac{x^{-y+iR}}{y-iR}M_{\Psi}\left({0},{y+1-a-iR}\right)\mathrm{d}y\right|\\\
&\hphantom{{}=\int_{-b_{1}}^{b_{1}}\frac{x^{y+iR}}{y+iR}M_{\Psi}\left({0},{y+1-a+iR}\right)+\frac{x^{y-iR}}{y-iR}}\leq\frac{2}{R}\int_{-b_{1}}^{b_{1}}x^{-y}M_{\Psi}\left({0},{y+1-a}\right)\mathrm{d}y=\mathrm{O}\left(\frac{1}{R}\right),\end{split}$

because on $[-b_{1},b_{1}]$, $y\mapsto x^{-y}$ is obviously continuous, and
also $y\mapsto$ $M_{\Psi}\left({0},{y+1-a}\right)$ is continuous there by [28,
(2.10)]. Therefore, by dominated convergence along the horizontal strips of
$\gamma_{R}$, by (6.30),

$\begin{split}-2\pi
iM_{\Psi}\left({0},{1-a}\right)&=\lim_{R\to\infty}\oint_{\gamma_{R}}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z\\\
&=\int_{\mathtt{Re}\left(z\right)=-b_{1}}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z-\int_{\mathtt{Re}\left(z\right)=b_{1}}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z.\end{split}$

It remains to prove that the integral along $\mathtt{Re}\left(z\right)=b_{1}$
goes to zero as $x\to\infty$. Using once again the polynomial decay of
$M_{\Psi}\left({0},{\cdot}\right)$ along complex lines, see (6.21) and (6.22),
and the fact that $b_{1}>0$, we have that

$\begin{split}\left|\int_{\mathtt{Re}\left(z\right)=b_{1}}\frac{x^{-z}}{z}M_{\Psi}\left({0},{z+1-a}\right)\mathrm{d}z\right|&=\left|\int_{-\infty}^{\infty}\frac{x^{-b_{1}-iy}}{-b_{1}+iy}M_{\Psi}\left({0},{-b_{1}+1-a+iy}\right)\mathrm{d}y\right|\\\
&\leq
x^{-b_{1}}{\int_{-\infty}^{\infty}\frac{1}{\max\left\\{b_{1},|y|\right\\}}\left|M_{\Psi}\left({0},{-b_{1}+1-a+iy}\right)\right|\mathrm{d}y}<\infty,\end{split}$

so by dominated convergence, the last tends to $0$ when $x\to\infty$ as
desired, i.e. (6.29) holds. Therefore, by the Portmanteau theorem, we have the
weak convergence

$\widetilde{\nu}_{a,t}(\mathrm{d}y)=\frac{\nu_{a,t}(\mathrm{d}y)}{\nu_{a,t}\left((0,\infty]\right)}\xrightarrow[t\to\infty]{w}\widetilde{\nu}_{a}(\mathrm{d}y)$

with $\widetilde{\nu}_{a}(\mathrm{d}y)$ a probability measure with
distribution function $h_{a}^{x}(0)/M_{\Psi}\left({0},{1-a}\right)$.
Therefore, as, by (6.28),
$\lim_{t\to\infty}\nu_{a,t}\left((0,\infty]\right)=c_{1}M_{\Psi}\left({0},{1-a}\right)/\phi_{+}(0)$,
we obtain the statement of the theorem, (3.2),

$\nu_{a,t}(\mathrm{d}y)=\frac{\mathbb{P}(I_{\xi}^{-a}(t)\in\mathrm{d}y)t^{\alpha}}{y^{a}\ell(t)}\xrightarrow[t\to\infty]{w}\nu_{a}(\mathrm{d}y)$
(6.32)

with $\nu_{a}$ defined by $\nu_{a}((0,x])=c_{1}h_{a}^{x}(0)/\phi_{+}(0)$.

Case 2: $\alpha$ is integer, i.e. $\alpha\in\\{2,3,\dots\\}$.

We note that now $n=\alpha-1\geq 1$ and $\mathbb{P}(\xi_{t}\geq
0)\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\ell(t)/t^{n}$. The analogue of
(6.10) in this case is that

$\left|\phi_{+}^{{(k)}}\left({0+}\right)\right|<\infty,\quad k=0,1,\dots,n-1,$
(6.33)

and the difference is that we do not have for sure that the $n$th derivative
of $\phi_{+}$ is infinite at zero. However, by (6.6), since
$t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\stackrel{{\scriptstyle\infty}}{{\sim}}\ell(t)/t$, by Proposition A.11,

$\left(\ln\phi_{+}\right)^{(n)}(q)=\int_{0}^{\infty}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t\in SV_{0}$ (6.34)

and, as before, using Faà di Bruno’s formula, (6.9), we get that
$\phi_{+}^{{(n)}}\left({q}\right)\in SV_{0}$, so for small $q$, by Potter’s
bounds, see Proposition A.3,
$\left|\phi_{+}^{{(n)}}\left({q}\right)\right|=\mathrm{O}\left(q^{-1/2}\right)$.
Now, from Faà di Bruno’s formula for $1/\phi_{+}$, see (LABEL:eq:FdB_1/f), it
follows that

$\left|\left(\frac{1}{\phi+}\right)^{(n)}(q)\right|=\mathrm{O}\left(q^{-1/2}\right).$
(6.35)

Next, let us define, for $q\geq 0$, $x\in(0,1)$, and $\lambda\geq 1$,

$\widetilde{M}_{n}(q,x):=\frac{\partial^{n}}{\partial{q}^{n}}{M_{\Psi}\left({q},{x}\right)}-\frac{\partial^{n}}{\partial{q}^{n}}{M_{\Psi}\left({0},{x}\right)},\quad
K_{\lambda,n}(q):=\left(1/\phi_{+}\right)^{(n)}\left(q/\lambda\right)-\left(1/\phi_{+}\right)^{(n)}(q).$
(6.36)

By (i) from Corollary 4.4,
$\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({0},{x}\right)}$ are
finite for $0\leq k\leq n$, so $\widetilde{M}_{k}(q,x)$ are well-defined for
these $k$. Moreover, by the mean value theorem, for $0\leq k\leq n-1$, there
exist $\widetilde{\eta}_{k}\in(0,q)$ such that

$\left|\widetilde{M}_{k}(q,x)\right|=\left|q\frac{\partial^{k+1}}{\partial{q}^{k+1}}{M_{\Psi}\left({\widetilde{\eta}_{k}},{x}\right)}\right|=\mathrm{O}\left(q\right)=\mathrm{O}\left(q^{1/2}\right).$

In a similar spirit, by (6.33), $K_{\lambda,k}(q)$ are well defined for $0\leq
k\leq n-1$, and by the mean value theorem, for these $k$ and $q>0$, there
exist $\eta_{k}\in(q/\lambda,q)$ such that

$\left|K_{\lambda,k}(q)\right|=\left|q\left(\frac{1}{\lambda}-1\right)\left(\frac{1}{\phi_{+}}\right)^{(k+1)}(\eta_{k})\right|=\mathrm{O}\left(q^{1/2}\right)$

by (6.33) and (6.35). To summarise:

$\left|\widetilde{M}_{k}(q,x)\right|=\mathrm{O}\left(q^{1/2}\right),\left|K_{\lambda,k}(q)\right|=\mathrm{O}\left(q^{1/2}\right)\quad\text{for}\quad
0\leq k\leq
n-1;\quad\text{and}\quad\widetilde{M}_{n}(q,x)=\mathrm{o}\left(1\right).$
(6.37)

Let

$G_{\lambda,n}(q):=\widehat{g_{a}}^{(n)}(q/\lambda)-\widehat{g_{a}}^{(n)}(q).$
(6.38)

We can use once again (6.12) and then (6.33) and the terms of order
$\mathrm{O}\left(q^{1/2}\right)$ from (6.37) to obtain that, for $q\to 0$,

$\begin{split}&G_{\lambda,n}(q)=\sum_{k=0}^{n}\binom{n}{k}\bigg{(}K_{\lambda,k}(q)\frac{\partial^{n-k}}{\partial{q}^{n-k}}{M_{\Psi}\left({0},{1-a}\right)}\\\