&\hphantom{{}=\sum_{k=0}^{n}\binom{n}{k}}+\left(1/\phi_{+}\right)^{(k)}\left({q/\lambda}\right)\widetilde{M}_{n-k}(q/\lambda,1-a)-\left(1/\phi_{+}\right)^{(k)}(q)\widetilde{M}_{n-k}(q,1-a)\bigg{)}\\\
&=K_{\lambda,n}(q)M_{\Psi}\left({0},{1-a}\right)+\mathrm{O}\left(q^{1/2}\right)+\left(1/\phi_{+}\right)^{(n)}(q/\lambda)\widetilde{M}_{0}(q/\lambda,1-a)-\left(1/\phi_{+}\right)^{(n)}(q)\widetilde{M}_{0}(q,1-a)\\\
&\hphantom{{}=\sum_{k=0}^{n}\binom{n}{k}}+(1/\phi_{+})(q/\lambda)\widetilde{M}_{n}(q/\lambda,1-a)-(1/\phi_{+})(q)\widetilde{M}_{n}(q,1-a).\end{split}$
(6.39)

For the last 4 terms above, we calculate

$\begin{split}&\left(1/\phi_{+}\right)^{(n)}(q/\lambda)\widetilde{M}_{0}(q/\lambda,1-a)-\left(1/\phi_{+}\right)^{(n)}(q)\widetilde{M}_{0}(q,1-a)\\\
&\hphantom{{}=\left(1/\phi_{+}\right)^{(n)}}+(1/\phi_{+})(q/\lambda)\widetilde{M}_{n}(q/\lambda,1-a)-(1/\phi_{+})(q)\widetilde{M}_{n}(q,1-a)\\\
&\quad=K_{\lambda,n}(q)\widetilde{M}_{0}(q/\lambda,1-a)+\left(1/\phi_{+}\right)^{(n)}(q)\left(\widetilde{M}_{0}(q/\lambda,1-a)-\widetilde{M}_{0}(q,1-a)\right)\\\
&\quad\hphantom{{}=\left(1/\phi_{+}\right)^{(n)}}+K_{\lambda,0}(q)\widetilde{M}_{n}(q/\lambda,1-a)+(1/\phi_{+})(q)\left(\widetilde{M}_{n}(q/\lambda,1-a)-\widetilde{M}_{n}(q,1-a)\right)\\\
&\quad=K_{\lambda,n}(q)\mathrm{O}\left(q^{1/2}\right)+\mathrm{O}\left(q^{-1/2}\right)\mathrm{O}\left(q\right)+\mathrm{O}\left(q^{1/2}\right)\mathrm{o}\left(1\right)+\mathrm{O}\left(1\right)\left(\widetilde{M}_{n}(q/\lambda,1-a)-\widetilde{M}_{n}(q,1-a)\right)\\\
&\quad=\mathrm{o}\left(K_{\lambda,n}(q)\right)+\mathrm{O}\left(q^{1/2}\right)+\mathrm{O}\left(1\right)\left(\widetilde{M}_{n}(q/\lambda,1-a)-\widetilde{M}_{n}(q,1-a)\right),\end{split}$
(6.40)

where we have substituted (6.35) and (6.37), and have used the mean value
theorem once again for $\widetilde{M}_{0}(\cdot,1-a)$. For the last
difference, substituting the definition of $\widetilde{M_{n}}$, (6.36), by the
mean value theorem, there exists an $\eta_{q}\in(q/\lambda,q)$ such that

$\begin{split}\widetilde{M}_{n}\left(\frac{q}{\lambda},1-a\right)-\widetilde{M}_{n}(q,1-a)&=\frac{\partial^{n}}{\partial{q}^{n}}{M_{\Psi}\left({\frac{q}{\lambda}},{1-a}\right)}-\frac{\partial^{n}}{\partial{q}^{n}}{M_{\Psi}\left({q},{1-a}\right)}\\\
&=q\left(\frac{1}{\lambda}-1\right)\frac{\partial^{n+1}}{\partial{q}^{n+1}}{M_{\Psi}\left({\eta_{q}},{x}\right)},\end{split}$

which from item (ii’) of Corollary 4.4 with $\delta=1/2$ is
$\mathrm{O}\left(q(U^{*n}_{q}(\mathbb{R}_{+})+q^{-1/2})\right)$. Moreover,
from Proposition 5.1, we know that
$U^{*n}_{q}(\mathbb{R}_{+})=\int_{0}^{\infty}e^{-qt}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t$, which in our case, see (6.34), is a slowly varying function as
$q\to 0$, and is therefore $\mathrm{O}\left(q^{-\delta}\right)$ for every
$\delta>0$ by Proposition A.3. This leads to

$\widetilde{M}_{n}(q/\lambda,1-a)-\widetilde{M}_{n}(q,1-a)=\mathrm{O}\left(q\left(U^{*n}_{q}(\mathbb{R}_{+})+q^{-1/2}\right)\right)=\mathrm{O}\left(q\left(q^{-1/2}\right)\right)=\mathrm{O}\left(q^{1/2}\right).$
(6.41)

Substituting the above and (6.40) into (6.39), we get that

$G_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)M_{\Psi}\left({0},{1-a}\right)+\mathrm{o}\left(K_{\lambda,n}(q)\right)+\mathrm{O}\left(q^{1/2}\right).$
(6.42)

Now, we try to obtain the asymptotics of $K_{\lambda,n}$ via a link with the
representation (6.6): first, let us note that from it, for $q>0$,

$\begin{split}\left(\ln\phi_{+}\right)^{(n)}(q/\lambda)-\left(\ln\phi_{+}\right)^{(n)}(q)&=(-1)^{n+1}\int_{0}^{\infty}\left(e^{-qt/\lambda}-e^{-qt}\right)t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t\\\ &=(-1)^{n+1}q\int_{0}^{\infty}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\int_{t/\lambda}^{t}e^{-qs}\mathrm{d}s\mathrm{d}t\\\
&=(-1)^{n+1}q\int_{0}^{\infty}e^{-qs}\int_{s}^{\lambda
s}t^{n-1}\mathbb{P}(\xi_{t}\geq 0)\mathrm{d}t\mathrm{d}s.\end{split}$ (6.43)

Moreover, by (6.5) and Proposition A.12, $\int_{s}^{\lambda
s}t^{n-1}\mathbb{P}(\xi_{t}\geq
0)\mathrm{d}t\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\ln(\lambda)\ell(s)$,
so, similar to (6.8), by Karamata’s theorem A.5 and a standard Tauberian
theorem A.7,

$\left(\ln\phi_{+}\right)^{(n)}(q/\lambda)-\left(\ln\phi_{+}\right)^{(n)}(q)\stackrel{{\scriptstyle
0}}{{\sim}}(-1)^{n+1}c_{1}\ln(\lambda)\ell(1/q).$ (6.44)

Next, note that we can represent

$\begin{split}\left(\ln
f\right)^{(n)}=\left(-f(1/f)^{\prime}\right)^{(n-1)}=-\sum_{k=0}^{n-1}\binom{n-1}{k}f^{(k)}(1/f)^{(n-k)},\end{split}$
(6.45)

so we have that

$\begin{split}\left(\ln\phi_{+}\right)^{(n)}(q/\lambda)&-\left(\ln\phi_{+}\right)^{(n)}(q)\\\
&=-\sum_{k=0}^{n-1}\binom{n-1}{k}\left(K_{\lambda,n-k}(q)\phi^{(k)}_{+}(q/\lambda)+\left(\phi_{+}^{{(k)}}\left({q/\lambda}\right)-\phi_{+}^{{(k)}}\left({q}\right)\right)\left(1/\phi_{+}({q})\right)^{(n-k)}\right)\\\
&=-K_{\lambda,n}(q)\phi_{+}({q/\lambda})+\mathrm{O}\left(q^{1/2}\right),\end{split}$
(6.46)

where we have used (6.33), (6.37), and the mean value theorem, combined with
(6.33) and (6.35). Substituting in (6.44), we get that

$K_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}(-1)^{n}c_{1}\ln(\lambda)\phi^{-1}_{+}(0)\ell(1/q).$ (6.47)

Thus, from (6.42), because slowly varying functions grow slower than
polynomials, see again Proposition A.2, we get that

$G_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)M_{\Psi}\left({0},{1-a}\right)\stackrel{{\scriptstyle
0}}{{\sim}}(-1)^{n}M_{\Psi}\left({0},{1-a}\right)c_{1}\ln(\lambda)\phi^{-1}_{+}(0)\ell(1/q).$
(6.48)

Let us now define, for $\lambda\geq 1$,

$R(t):=\int_{0}^{t}g_{a,n}(s)\mathrm{d}s,\quad and\quad
R_{\lambda}(t):=\int_{t}^{\lambda t}g_{a,n}(s)\mathrm{d}s,$ (6.49)

which are well-defined by Lemma 6.3. We have that, for $q>0$,

$\begin{split}q\widehat{R_{\lambda}}(q)=q\int_{0}^{\infty}e^{-qt}\int_{t}^{\lambda
t}g_{a,n}(s)\mathrm{d}s\mathrm{d}t=q\int_{0}^{\infty}g_{a,n}(s)\int_{s/\lambda}^{s}e^{-qt}\mathrm{d}t\mathrm{d}s=\widehat{g_{a,n}}\left(q/\lambda\right)-\widehat{g_{a,n}}(q),\end{split}$
(6.50)

which by Proposition 6.1 and the definition of $G_{\lambda,n}$, see (6.38), is
equal to

$\frac{(-1)^{n}}{(n-1)!}\int_{0}^{1}G_{\lambda,n}(qv)(1-v)^{n-1}\mathrm{d}v,$

and thus, by (6.48) and Proposition 6.2,

$q\widehat{R_{\lambda}}(q)\stackrel{{\scriptstyle
0}}{{\sim}}\frac{c_{1}M_{\Psi}\left({0},{1-a}\right)\ln(\lambda)}{n!\phi_{+}({0})}\ell\left(1/q\right).$
(6.51)

Next, in the last result we will use a generalised Tauberian theorem for
$\widehat{R_{\lambda}}$, Theorem A.8, for which we will need to check the
regularity condition (A.5):

$\lim_{\mu\to 1+}\liminf_{x\to\infty}\inf_{1\leq
t\leq\mu}\frac{R_{\lambda}(tx)-R_{\lambda}(x)}{\ell(x)}\geq 0.$

Since $R$ is increasing and $g_{a,n}$ is decreasing, we have that, for
$\mu\geq t\geq 1$,

$\begin{split}R_{\lambda}(tx)-R_{\lambda}(x)&=R(\lambda tx)-R(\lambda
x)-(R(tx)-R(x))\geq-(R(tx)-R(x))\geq-(R(\mu x)-R(x))\\\ &=-\int_{x}^{\mu
x}g_{a,n}(s)\mathrm{d}s\geq(1-\mu)xg_{a,n}(x),\end{split}$

so it would be enough to check that
$xg_{a,n}(x)=\mathrm{O}\left(\ell(x)\right)$ as $x\to\infty$. This is indeed
true because from Lemma 6.3, $\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\leq
c_{2}\ell(t)/t^{n+1}$ for some $c_{2}>0$, and using this in the definition of
$g_{a,n}$, (6.17), we have that, for large $x$,

$xg_{a,n}(x)=x\int_{x}^{\infty}\int_{s_{1}}^{\infty}\dots\int_{s_{n-1}}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(s_{n})\right]\mathrm{d}s_{n}\dots\mathrm{d}s_{2}\mathrm{d}s_{1}\leq
c_{3}\ell(x),$ (6.52)

where $c_{3}>0$, and the calculation is valid by Theorem A.6. Therefore, from
the extended Tauberian theorem A.8,

$R_{\lambda}(t)=R(\lambda
t)-R(t)\stackrel{{\scriptstyle\infty}}{{\sim}}c_{1}\frac{M_{\Psi}\left({0},{1-a}\right)\ln(\lambda)}{n!\phi_{+}({0})}\ell\left(t\right),$

which means by Definition A.13 that $R\in\Pi_{\ell}$, and hence, by (6.49) and
Theorem A.14,

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

After $n$ applications of the monotone density theorem A.10 in the last
equivalence, we get that

$g_{a}(t)=\mathbb{E}\left[I^{-a}_{\xi}(t)\right]\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{c_{1}M_{\Psi}\left({0},{1-a}\right)}{\phi_{+}({0})}t^{-(n+1)}\ell(t),$
(6.53)

which, as $\alpha=n+1$, is the same result as in the first case, see (6.18).
The remaining arguments are also similar to past ones: first, note that (6.24)
holds: for $q>0$ and $b\in(a-1,0)$,

$\widehat{g_{a}^{x}}^{(n)}(q)=\left(\frac{h_{a}^{x}(q)}{\phi_{+}({q})}\right)^{(n)}\text{\quad
with \quad}h_{a}^{x}(q):=-\frac{1}{2\pi
i}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}M_{\Psi}\left({q},{z+1-a}\right)\mathrm{d}z,$
(6.54)

and let us define, similar to $G_{\lambda,n}$, see (6.38),

$G^{x}_{\lambda,n}(q):=\widehat{g_{a}^{x}}^{(n)}(q/\lambda)-\widehat{g_{a}^{x}}^{(n)}(q)$

with $g_{a}^{x}$ defined in (6.19). We will use similar arguments to the ones
that starting with (6.12), lead from (6.38) to (6.48), i.e. from

$\widehat{g_{a}}^{(n)}(q)=\left(\frac{M_{\Psi}\left({q},{1-a}\right)}{\phi_{+}({q})}\right)^{(n)}\quad\text{to}\quad
G_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)M_{\Psi}\left({0},{1-a}\right),$

to obtain that, using $h_{a}^{x}(q)$ instead of
$M_{\Psi}\left({q},{1-a}\right)$, that (6.54) leads to

$G^{x}_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)h_{a}^{x}(0).$

We will follow in detail the mentioned chain of arguments: define first an
analogue of the defined in (6.36) $\widetilde{M_{k}}$: for $k\leq n$,

$\widetilde{H}_{n}(q,x):=\left(h_{a}^{x}\right)^{(n)}(q)-\left(h_{a}^{x}\right)^{(n)}(0),$

which is well defined by (6.26). Therefore, as in (6.37), by the mean value
theorem

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

so (6.39) and (6.40) hold with $\widetilde{H}$ instead of $\widetilde{M}$, and
we are left again with the last term from (6.40): by the mean value theorem,
for some $\eta_{q}\in(q/\lambda,q)$,

$\begin{split}\widetilde{H}_{n}\left(\frac{q}{\lambda},1-a\right)-\widetilde{H}_{n}(q,1-a)&=\left(h_{a}^{x}\right)^{(n)}\left(\frac{q}{\lambda}\right)-\left(h_{a}^{x}\right)^{(n)}(q)=q\left(\frac{1}{\lambda}-1\right)\left(h_{a}^{x}\right)^{(n+1)}\left(\eta_{q}\right)\\\
&=\frac{q(\lambda-1)}{2\lambda\pi
i}\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}\frac{\partial^{n+1}}{\partial{q}^{n+1}}{M_{\Psi}\left({\eta_{q}},{z+1-a}\right)}\mathrm{d}z.\end{split}$
(6.55)

We will show that the last is $\mathrm{O}\left(q^{1/2}\right)$ as $q$ goes to
$0$. Consider the case where $\xi$ is not a CPP with a positive drift. From
item (ii’) of Corollary 4.4, as in (6.41), writing $z=b+iy$, we have that, for
small $q$,

$\begin{split}\left|\int_{\mathtt{Re}\left(z\right)=b}\frac{x^{-z}}{z}\frac{\partial^{n+1}}{\partial{q}^{n+1}}{M_{\Psi}\left({\eta_{q}},{z+1-a}\right)}\mathrm{d}z\right|\leq\eta_{q}^{-1/2}\int_{-\infty}^{\infty}\frac{x^{-b}}{|z|}P_{\mathtt{Re}\left(z\right),n}(|z|)\left|M_{\Psi}\left({\eta_{q}},{b+1-a+iy}\right)\right|\mathrm{d}y.\end{split}$

We have that the integral is finite by the uniform convergence (6.25) and the
super-polynomial decay of $\left|M_{\Psi}\right|$, (6.21). Moreover,
$\eta_{q}^{-1/2}=\mathrm{O}\left(q^{-1/2}\right)$ because
$\eta_{q}\in(q/\lambda,q)$, so substituting in (6.55) gives us the desired
result. The case where $\xi$ is a CPP with positive drift is similar, using
the at most logarithmic growth of the polynomial of (iii’) of Corollary 4.4
and the polynomial decay of $\left|M_{\Psi}\right|$, (6.22). Therefore, an
analogue of (6.41) holds, and we get the modified (6.42):

$G_{\lambda,n}^{x}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)h_{a}^{x}(0)+\mathrm{o}\left(K_{\lambda,n}(q)\right)+\mathrm{O}\left(q^{1/2}\right),$

which, by (6.47), gives

$G^{x}_{\lambda,n}(q)\stackrel{{\scriptstyle
0}}{{\sim}}K_{\lambda,n}(q)h_{a}^{x}(0).$

Continuing the analogy, following (6.49), let us define

$R^{x}(t):=\int_{0}^{t}g^{x}_{a,n}(s)\mathrm{d}s,\quad\text{and}\quad
R^{x}_{\lambda}(t):=\int_{t}^{\lambda t}g^{x}_{a,n}(s)\mathrm{d}s,$

which are well-defined because $g^{x}_{a,n}\leq g_{a,n}$. The same inequality
also gives that $yg^{x}_{a,n}(y)\leq
yg_{a,n}(y)=\mathrm{O}\left(\ell(y)\right)$ as $y\to\infty$, see (6.52), so
repeating the arguments from (6.50) to (6.53), we obtain the similar to (6.53)
result,

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

Then, in the same way as from (6.27) to (6.32), we obtain the same result:

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

∎

###### Proof of Theorem 3.4.

Recall that for a positive random variable $X$, its Mellin transform is given
formally by $\mathcal{M}_{X}(z)=\mathbb{E}\left[X^{z-1}\right]$. Set in the
notation of Theorem 3.1 and the proof above
$\tilde{\nu}_{a}((0,x]):=\nu_{a}((0,x])/\nu_{a}(0,\infty)$, and let $W$ be the
positive random variable with law $\tilde{\nu}_{a}$. Then, from (3.3) of
Theorem 3.1 and (3.5) of Corollary 3.5, we see that, for any
$x\in\left(0,\infty\right)$,

$\tilde{\nu}_{a}((0,x])=-\frac{1}{2\pi
i}\int_{\mathtt{Re}z=b}\frac{x^{-z}}{z}\frac{M_{\Psi}\left({0},{z+1-a}\right)}{M_{\Psi}\left({0},{1-a}\right)}\mathrm{d}z$

for $b\in\mathbb{C}_{(a-1,0)}$. Since this formula is precisely the formula
for Mellin inversion, we conclude that

$\mathcal{M}_{F}(z)=-\frac{1}{z}\frac{M_{\Psi}\left({0},{z+1-a}\right)}{M_{\Psi}\left({0},{1-a}\right)}=-\frac{1}{z}\frac{\Gamma(z+1-a)W_{\phi_{0,+}}(1-a)}{W_{\phi_{0,+}}(z+1-a)\Gamma(1-a)}\frac{W_{\phi_{0,-}}(a-z)}{W_{\phi_{0,-}}(a)},$

where simply $F(x):=\tilde{\nu}_{a}((0,x])$ is the cumulative distribution
function of $W$ and we have substituted the definition of $M_{\Psi}$, (2.9).
Now, setting $\phi_{\pm}:=\phi_{0,\pm}$, we further derive from above

$\mathcal{M}_{F}(z)=-\frac{1}{z}\frac{\mathbb{E}\left[I^{z}_{\phi_{+}}I^{-a}_{\phi_{+}}\right]}{\mathbb{E}\left[I^{-a}_{\phi_{+}}\right]}\frac{\mathbb{E}\left[(Y^{-1}_{\phi_{-}})^{z}(Y^{-1}_{\phi_{-}})^{1-a}\right]}{\mathbb{E}\left[(Y^{-1}_{\phi_{-}})^{1-a}\right]},$
(6.56)

where $Y_{\phi_{-}}$ is the random variable for $q=0$ in the definition of the
bivariate Bernstein-gamma function, see (2.3), or just below [28, (4.2)], and
$I_{\phi_{+}}$ is the exponential functional of the killed subordinator
(because $\phi_{+}(0)>0$) pertaining to $\phi_{+}$, see [28, Theorem 2.22].
The latter arguments are common knowledge in the area of exponential
functionals. Clearly, then (6.56) can be further rewritten as

$\mathcal{M}_{F}(z)=-\frac{1}{z}\mathcal{M}_{\mathcal{B}_{-a}I_{\phi_{+}}}(z+1)\mathcal{M}_{\mathcal{B}_{1-a}Y^{-1}_{\phi_{-}}}(z+1)=-\frac{1}{z}\mathcal{M}_{\mathcal{B}_{-a}I_{\phi_{+}}\times\mathcal{B}_{1-a}Y^{-1}_{\phi_{-}}}(z+1),$
(6.57)

with the operator $\mathcal{B}_{\cdot}$ standing for the size-biased versions
of the respective random variables, see (3.6). Recalling that the relation
between the Mellin transform of the cumulative distribution function and the
law of random variable, see [28, (7.12)], matches the expression at the right-
hand side in (6.57), we conclude that $W\stackrel{{\scriptstyle
d}}{{=}}\mathcal{B}_{-a}I_{\phi_{+}}\times\mathcal{B}_{1-a}Y^{-1}_{\phi_{-}}$
and (3.7) is established. Given that (6.57) holds, then the infinite
differentiability of $F$ follows immediately from [28, item (3) from Theorem
2.4] since it applies with $N_{\Psi}=\infty$ therein, see [28, (2.18) of
Theorem 2.3]. This concludes the proof of the theorem. ∎

###### Proof of Theorem 3.6.

Obviously, it suffices to prove the theorem for $F(x)=x^{-a}$,
$a\in\left(0,1\right)$. Then, (2.12) reads off

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

If (3.9) holds for some $n\geq 1$, then, from item (i) of Corollary 4.4, we
have that, for $0\leq k\leq n+1$,
$\frac{\partial^{k}}{\partial{q}^{k}}{M_{\Psi}\left({q},{z}\right)}$ are
finite and right-continuous at $q=0$, and from (4.9) we know that
$\left|\phi^{(n)}_{+}(0)\right|<\infty$. From the dominated convergence
theorem, we immediately deduce that, for $q>0$,

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

which augmented by the monotone convergence theorem and the finiteness and the
right-continuity of the derivatives yields

$(-1)^{n}\int_{0}^{\infty}t^{n}\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\mathrm{d}t=\left.\frac{\partial^{n}}{\partial{q}^{n}}{\frac{M_{\Psi}\left({q},{1-a}\right)}{\phi_{+}(q,0)}}\right|_{q=0}.$

The finiteness of the right-hand side and the fact that
$\mathbb{E}\left[I_{\xi}^{-a}(t)\right]$ is monotone decreasing leads to

$\sum_{m\geq
1}\mathbb{E}\left[I_{\xi}^{-a}(m+1)\right]m^{n}\leq\int_{0}^{\infty}t^{n}\mathbb{E}\left[I_{\xi}^{-a}(t)\right]\mathrm{d}t<\infty,$

which settles the claim in (3.10). The final claim follows from the expression
for the derivatives above with $n=0$, the finiteness of $\phi_{+}(0,0)$, and
the monotone convergence theorem. ∎

##  7 Proofs of auxiliary results

###### Proof of Proposition 5.1.

We proceed by induction for the case $q>0$. The base case $n=1$ is valid from
the definition of potential measures. For $n\in\mathbb{N}$, we calculate

$\begin{split}U_{q}^{*(n+1)}(\mathrm{d}x)&=\int_{-\infty}^{\infty}U_{q}^{*n}(\mathrm{d}y)U_{q}(\mathrm{d}x-y)\\\
&=\frac{1}{(n-1)!}\int_{-\infty}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}e^{-q(t_{1}+t_{2})}t_{1}^{n-1}\mathbb{P}(\xi_{t_{1}}\in\mathrm{d}y)\mathbb{P}(\xi_{t_{2}}\in\mathrm{d}x-y)\mathrm{d}t_{1}\mathrm{d}t_{2}\\\
&=\frac{1}{(n-1)!}\int_{0}^{\infty}\int_{0}^{t}e^{-qt}t_{1}^{n-1}\mathbb{P}(\xi_{t}\in\mathrm{d}x)\mathrm{d}t_{1}\mathrm{d}t\\\
&=\frac{1}{n!}\int_{0}^{\infty}e^{-qt}t^{n}\mathbb{P}(\xi_{t}\in\mathrm{d}x)\mathrm{d}t,\end{split}$

where we have used the induction hypotheses for $U_{q}^{*n}$, and afterwards,
we have employed that

$\int_{-\infty}^{\infty}\mathbb{P}(\xi_{t_{2}}\in\mathrm{d}x-y)\mathbb{P}\left(\xi_{t_{1}}\in\mathrm{d}y\right)=\mathbb{P}\left(\xi_{t_{1}+t_{2}}\in\mathrm{d}x\right),$

and changed variables as follows $t_{1}+t_{2}=t,t_{1}=t_{1}$, and finally we
have exchanged the order of integration. The case $q=0$ is valid by taking a
limit and using the monotone convergence theorem. ∎

###### Proof of Lemma 5.2.

For $\mathtt{Re}(z)>-1$, we can see that $u$ is bounded and continuous on
$[0,\infty)$ by the Taylor estimates

$\begin{split}u_{z}(y)\stackrel{{\scriptstyle
0}}{{\sim}}\frac{z^{2}-z}{2}y\quad\text{and}\quad
u_{z}(y)\stackrel{{\scriptstyle\infty}}{{\sim}}\frac{(z-1)\mathbbm{1}_{\left\\{\mathtt{Re}(z)\geq
0\right\\}}+\mathbbm{1}_{\left\\{\mathtt{Re}(z)<0\right\\}}}{e^{y+\mathtt{Re}(z)y\mathbbm{1}_{\left\\{\mathtt{Re}(z)<0\right\\}}}}.\end{split}$
(7.1)

As $v_{z}(y):=ze^{-y}-u_{z}(y)$, the same holds for $v_{z}$. For the precise
bound, we start by using the estimation lemma for complex integrals:

$\begin{split}|v_{z}(y)|=\left|\frac{1-e^{-zy}}{e^{y}-1}\right|=\left|\frac{\int_{0}^{zy}e^{-x}\mathrm{d}x}{e^{y}-1}\right|\leq\frac{|zy|\max\left\\{1,e^{-y\mathtt{Re}\left(z\right)}\right\\}}{e^{y}-1}.\end{split}$
(7.2)

If $\mathtt{Re}\left(z\right)\geq 0$, as $y/(e^{y}-1)\leq e^{-y/2}$, we
immediately obtain the first part of (5.3) with
$C_{1,\mathtt{Re}\left(z\right)}=1$, and
$\epsilon_{1,\mathtt{Re}\left(z\right)}=1/2$. For the second part, bounding,
for $y>1=:C_{\mathtt{Re}}\left(z\right)$,

$|v_{z}(y)|=\left|\frac{1-e^{-zy}}{e^{y}-1}\right|\leq\frac{1+e^{-\mathtt{Re}\left(z\right)y}}{e^{y}-1}\leq\frac{2}{e^{y}-1}\leq\frac{2}{e^{y/2}},$

checking directly the last inequality, we obtain the result with
$C_{2,\mathtt{Re}\left(z\right)}=2$, and
$\epsilon_{2,\mathtt{Re}\left(z\right)}=1/2$.

Next, if $\mathtt{Re}\left(z\right)\in(-1,0)$, (7.2) gives

$|v_{z}(y)|\leq|z|\frac{ye^{-y\mathtt{Re}\left(z\right)}}{e^{y}-1}.$

We see that there exists $C^{\prime}_{\mathtt{Re}\left(z\right)}>0$, such that
for $y>C^{\prime}_{\mathtt{Re}\left(z\right)}$, $e^{y}-1\geq
ye^{(-\mathtt{Re}\left(z\right)+(1+\mathtt{Re}\left(z\right))/2)y}$ and for
$0\leq y\leq C^{\prime}_{\mathtt{Re}\left(z\right)}$, we can just bound from
above $y\mapsto ye^{-y\mathtt{Re}\left(z\right)}/(e^{y}-1)$ with its supremum,
which is finite since the limit at 0 is finite and the function is continuous.
We obtain that for some $C^{\prime\prime}_{\mathtt{Re}\left(z\right)}>0$, and
$\epsilon_{\mathtt{Re}\left(z\right)}=(1+\mathtt{Re}\left(z\right))/2>0$, we
have that

$\frac{ye^{-y\mathtt{Re}\left(z\right)}}{e^{y}-1}\leq\begin{cases}C^{\prime\prime}_{\mathtt{Re}\left(z\right)},&\text{if
$y<C^{\prime}_{\mathtt{Re}\left(z\right)}$};\\\
e^{-\epsilon_{\mathtt{Re}\left(z\right)}y},&\text{if $y\geq
C^{\prime}_{\mathtt{Re}\left(z\right)}$}.\end{cases}$

This leads to the desired result with
$C_{1,\mathtt{Re}\left(z\right)}=\max\left\\{1,C^{\prime\prime}_{\mathtt{Re}\left(z\right)}e^{\epsilon_{\mathtt{Re}\left(z\right)}C^{\prime}_{\mathtt{Re}\left(z\right)}}\right\\}$
and
$\epsilon_{1,\mathtt{Re}\left(z\right)}=\epsilon_{\mathtt{Re}}\left(z\right)$.
For the second part of (5.3),

$|v_{z}(y)|=\left|\frac{1-e^{-zy}}{e^{y}-1}\right|\leq\frac{1+e^{-\mathtt{Re}\left(z\right)y}}{e^{y}-1}\leq\frac{2e^{-\mathtt{Re}\left(z\right)y}}{e^{y}-1}\leq\frac{2e^{-\mathtt{Re}\left(z\right)y}}{2e^{(-\mathtt{Re}\left(z\right)+\epsilon_{\mathtt{Re}}\left(z\right))y}},$

for sufficiently large $y>C_{\mathtt{Re}}\left(z\right)$ for some
$C_{\mathtt{Re}}\left(z\right)$, as
$-\mathtt{Re}\left(z\right)+\epsilon_{\mathtt{Re}}\left(z\right)=(1-\mathtt{Re}\left(z\right))/2<1$,
so the second part of (5.3) is true with $C_{2,\mathtt{Re}\left(z\right)}=1$
and
$\epsilon_{2,\mathtt{Re}\left(z\right)}=\epsilon_{\mathtt{Re}}\left(z\right)$.
∎

###### Proof of Proposition 6.1.

We proceed by induction. For $k=1$, by Tonelli’s theorem, we have that

$\begin{split}\widehat{f_{1}}(q)&=\int_{0}^{\infty}e^{-qs}\int_{s}^{\infty}f(y)\mathrm{d}y\mathrm{d}s=\int_{0}^{\infty}f(y)\int_{0}^{y}e^{-qs}\mathrm{d}s\mathrm{d}y=\frac{1}{q}\int_{0}^{\infty}f(y)\left(1-e^{-qy}\right)\mathrm{d}y\\\
&=\frac{1}{q}\left(\widehat{f}(0)-\widehat{f}(q)\right)=-\frac{1}{q}\int_{0}^{q}\widehat{f}^{\prime}(u)\mathrm{d}u=-\int_{0}^{1}\widehat{f}^{\prime}(qv)\mathrm{d}v.\end{split}$

For the induction step, we use similar logic: assume that the result is true
for $n=1,2,\dots,k$. Then, using it for $n=1$ and the function $f_{k}$, we get

$\begin{split}\widehat{f_{k+1}}(q)&=\int_{0}^{\infty}e^{-qs}\int_{s}^{\infty}f_{k}(y)\mathrm{d}y\mathrm{d}s=-\int_{0}^{1}\widehat{f_{k}}^{\prime}(qv)\mathrm{d}v.\end{split}$

Now differentiating (6.1) for $n=k$ in $q$, we substitute in the last equality

$\begin{split}-\int_{0}^{1}\widehat{f_{k}}^{\prime}(qv)\mathrm{d}v&=\frac{(-1)^{k+1}}{(k-1)!}\int_{0}^{1}\int_{0}^{1}y\widehat{f}^{(k+1)}(qvy)(1-y)^{k-1}\mathrm{d}v\\\
&=\frac{(-1)^{k+1}}{(k-1)!}\int_{0}^{1}\int_{0}^{1}w\widehat{f}^{(k+1)}(qvw)(1-w)^{k-1}\mathrm{d}w\mathrm{d}v\\\
&=\frac{(-1)^{k+1}}{(k-1)!}\int_{0}^{1}\widehat{f}^{(k+1)}(qy)\int_{y}^{1}(1-w)^{k-1}\mathrm{d}w\mathrm{d}y\\\
&=\frac{(-1)^{k+1}}{k!}\int_{0}^{1}\widehat{f}^{(k+1)}(qy)(1-y)^{k}\mathrm{d}y,\end{split}$

where we used the substitution $y=vw$. This concludes the induction step and,
respectively, establishes the claim of the proposition. ∎

###### Proof of Proposition 6.2.

Fix $\varepsilon\in\left(0,1\right)$ and from the assumptions let
$|f(q)|\stackrel{{\scriptstyle 0}}{{\sim}}q^{\beta}\ell(q)$ with the slowly
varying function $\ell$ being positive in a neighbourhood of zero. Then, from
Theorem A.2 applied to $\ell$ for the compact interval
$\left[\varepsilon,1\right]$, we have that

$\lim_{q\to
0}\frac{\int_{\varepsilon}^{1}f(qv)(1-v)^{n-1}\mathrm{d}v}{f(q)}=\int_{\varepsilon}^{1}v^{\beta}(1-v)^{n-1}\mathrm{d}v.$

With
$g(w):=f(w^{-1})\stackrel{{\scriptstyle\infty}}{{\sim}}w^{-\beta}\ell(w^{-1})$
and $\ell^{\prime}(w):=\ell(w^{-1})$, we compute that

$\begin{split}\limsup_{q\to
0}&\left|\frac{\int_{0}^{\varepsilon}f(qv)(1-v)^{n-1}\mathrm{d}v}{f(q)}\right|\leq\limsup_{q\to
0}\frac{\int_{0}^{\varepsilon
q}\left|f(w)\right|\mathrm{d}w}{q\left|f(q)\right|}=\limsup_{q\to
0}\frac{\int_{\varepsilon^{-1}q^{-1}}^{\infty}\left|f(w^{-1})\right|w^{-2}\mathrm{d}w}{q\left|f(q)\right|}\\\
&=\limsup_{q\to
0}\frac{\int_{\varepsilon^{-1}q^{-1}}^{\infty}\left|g(w)\right|w^{-2}\mathrm{d}w}{q\left|g(q^{-1})\right|}=\limsup_{y\to\infty}\frac{y\int_{\varepsilon^{-1}y}^{\infty}\left|g(w)\right|w^{-2}\mathrm{d}w}{\left|g(\varepsilon^{-1}y)\right|}\frac{\left|g(\varepsilon^{-1}y)\right|}{\left|g(y)\right|}\\\
&=\limsup_{y\to\infty}\frac{\int_{\varepsilon^{-1}y}^{\infty}\left|\ell^{\prime}(w)\right|w^{-2-\beta}\mathrm{d}w}{\left|\varepsilon^{\beta}y^{-\beta-1}\ell^{\prime}(\varepsilon^{-1}y)\right|}\frac{\left|g(\varepsilon^{-1}y)\right|}{\left|g(y)\right|}=\varepsilon\limsup_{y\to\infty}\frac{\int_{\varepsilon^{-1}y}^{\infty}\left|\ell^{\prime}(w)\right|w^{-2-\beta}\mathrm{d}w}{\left|\varepsilon^{\beta+1}y^{-\beta-1}\ell^{\prime}(\varepsilon^{-1}y)\right|}\frac{\left|g(\varepsilon^{-1}y)\right|}{\left|g(y)\right|}\\\
&=\frac{\varepsilon^{1+\beta}}{1+\beta},\end{split}$

where in the last relation we have employed Theorem A.6 with
$\alpha=-\beta-2<-1$, since $\beta\in\left(-1,0\right]$, and the fact that $g$
is regularly varying with index $-\beta$. From the two relations, we get that,
for any $\varepsilon\in\left(0,1\right)$,

$\limsup_{q\to
0}\left|\frac{\int_{0}^{1}f(qv)(1-v)^{n-1}\mathrm{d}v}{f(q)}-\int_{0}^{1}v^{\beta}(1-v)^{n-1}\mathrm{d}v\right|\leq\frac{\varepsilon^{1+\beta}}{1+\beta}+\int_{0}^{\varepsilon}v^{\beta}(1-v)^{n-1}\mathrm{d}v.$

Since $\varepsilon$ can be set to tend to $0$ and $\beta\in\left(-1,0\right]$,
we conclude the proof. ∎

###### Proof of Lemma 6.3.

Let us note with $\Delta\xi_{t}$ the jump in time $t$ of $\xi$ and define
$J_{1}^{(\epsilon t)}$ $:=$ $\inf\\{s\geq 0:\Delta\xi_{s}\geq\epsilon t\\}$
for $\epsilon>0$, i.e. the time of the first jump of size at least $\epsilon
t$. We split the considered expectation according to the time of this jump:

$\mathbb{E}\left[I^{-a}_{\xi}(t)\right]=\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon
t)}\leq
t\right\\}}\right]+\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon
t)}>t\right\\}}\right].$ (7.3)

For the first term, we have that

$\begin{split}\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J^{(\epsilon
t)}_{1}\leq
t\right\\}}\right]&=\int_{0}^{t}\mathbb{E}\left[I_{\xi}^{-a}(t)\middle|J_{1}^{(\epsilon
t)}=s\right]\overline{\Pi}_{+}(\epsilon t)e^{-\overline{\Pi}_{+}(\epsilon
t)s}\mathrm{d}s\\\ &\leq\overline{\Pi}_{+}(\epsilon
t)\int_{0}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(t)\middle|J_{1}^{(\epsilon
t)}=s\right]\mathrm{d}s.\end{split}$

In the following, we will use the notation $\xi^{(A)}$ for the process $\xi$
from which we have removed jumps larger than $A>0$. It is immediate that
$\xi^{(A)}\leq\xi$, and thus, for each $t>0$,

$I_{\xi^{(A)}}(t)\geq I_{\xi}(t)\quad\text{and}\quad I^{-a}_{\xi^{(A)}}(t)\leq
I_{\xi}^{-a}(t).$ (7.4)

Therefore, for $0<s\leq t$, we can see that

$\mathbb{E}\left[I_{\xi}^{-a}(t)\middle|J_{1}^{(\epsilon
t)}=s\right]\leq\mathbb{E}\left[I_{\xi}^{-a}(s)\middle|J_{1}^{(\epsilon
t)}=s\right]=\mathbb{E}\left[I_{\xi^{(\epsilon
t)}}^{-a}(s)\right]\leq\mathbb{E}\left[I_{\xi}^{-a}(s)\right].$

Moreover, as in (6.4), by [35, Theorem 1.1] and the assumption of regularly
varying tail, for large $t$, $\overline{\Pi}(\epsilon t)\leq
c_{1}\mathbb{P}(\xi_{1}>t)$ for some $c_{1}>0$, so we obtain that for large
$t$,

$\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon t)}\leq
t\right\\}}\right]\leq
c_{1}\mathbb{P}(\xi_{1}>t)\int_{0}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(s)\right]\mathrm{d}s=\mathrm{O}\left(\mathbb{P}(\xi_{1}>t)\right),$
(7.5)

as the integral is finite from (2.12).

We continue with bounding
$\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon
t)}>t\right\\}}\right]$. Once again, writing

$\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon
t)}>t\right\\}}\right]=\mathbb{P}\left(J_{1}^{(\epsilon
t)}>t\right)\mathbb{E}\left[I_{\xi}^{-a}(t)\middle|J_{1}^{(\epsilon
t)}>t\right]\leq\mathbb{E}\left[I_{\xi}^{-a}(t)\middle|J_{1}^{(\epsilon
t)}>t\right]=\mathbb{E}\left[I_{\xi^{(\epsilon t)}}^{-a}(t)\right],$

we are left with estimating $I_{\xi^{(\epsilon t)}}$. Let
$J_{1}^{(1)}:=\inf\\{s\geq 0:\Delta\xi_{s}\geq 1\\}\sim
Exp(\overline{\Pi}_{+}(1))$, and split, for some $c_{2}>0$,

$\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon
t)}}(t)\right]=\mathbb{P}\left(J_{1}^{(1)}\geq c_{2}\ln
t\right)\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon t)}}(t)\middle|J_{1}^{(1)}\geq
c_{2}\ln t\right]+\mathbb{P}\left(J_{1}^{(1)}<c_{2}\ln
t\right)\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon
t)}}(t)\middle|J_{1}^{(1)}<c_{2}\ln t\right].$ (7.6)

As $J_{1}^{(1)}\sim Exp(\overline{\Pi}_{+}(1))$, let us choose
$c_{2}=(\alpha+1)/\overline{\Pi}_{+}(1)$, so we have
$\mathbb{P}\left(J_{1}^{(1)}\geq c_{2}\ln t\right)=1/t^{\alpha+1}$. Therefore,
by (7.4), we have for the first term in (7.6) that, for large $t$,

$\begin{split}\mathbb{P}\left(J_{1}^{(1)}\geq c_{2}\ln
t\right)\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon t)}}(t)\middle|J_{1}^{(1)}\geq
c_{2}\ln
t\right]&\leq\frac{1}{t^{\alpha+1}}\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon
t)}}(c_{2}\ln t)\middle|J_{1}^{(1)}\geq c_{2}\ln t\right]\\\
&=\frac{1}{t^{\alpha+1}}\mathbb{E}\left[I^{-a}_{\xi^{(1)}}(c_{2}\ln
t)\right]\\\
&\leq\frac{1}{t^{\alpha+1}}\mathbb{E}\left[I_{\xi}^{-a}(1)\right]=\mathrm{O}\left(\frac{\ell(t)}{t^{\alpha}}\right)=\mathrm{O}\left(\mathbb{P}(\xi_{1}\geq
t)\right),\end{split}$

where we used that $1/t=\mathrm{O}\left(\ell(t)\right)$ by Proposition A.3, so
we have obtained a bound of the needed order.

To work with the second term of (7.6), we disintegrate $J_{1}^{(1)}\sim
Exp(\overline{\Pi}_{+}(1))$ and use the Markov property at $J_{1}^{(1)}$: for
$\eta$ an independent copy of $\xi$,

$\begin{split}\mathbb{P}\left(J_{1}^{(1)}<c_{2}\ln
t\right)&\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon
t)}}(t)\middle|J_{1}^{(1)}<c_{2}\ln t\right]\\\
&=\mathbb{P}\left(J_{1}^{(1)}<c_{2}\ln t\right)\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[I^{-a}_{\xi^{(\epsilon
t)}}(t)\middle|J_{1}^{(1)}=s\right]\overline{\Pi}_{+}(1)e^{-\overline{\Pi}_{+}(1)s}\mathrm{d}s\\\
&\leq\overline{\Pi}_{+}(1)\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[\left(I_{\xi^{(1)}}(s)+e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}I_{\eta^{(\epsilon t)}}(t-s)\right)^{-a}\right]\mathrm{d}s,\end{split}$
(7.7)

where we have used that, conditionally on $\\{J_{1}^{(1)}=s\\}$,

$I^{-a}_{\xi^{(\epsilon
t)}}(t)=I_{\xi^{(1)}}(s)+e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}\int_{0}^{t-s}e^{-\left(\xi^{(\epsilon t)}_{s+v}-\xi^{(\epsilon
t)}_{s}\right)}\mathrm{d}v$

and $\left(\xi^{(\epsilon t)}_{s+v}-\xi^{(\epsilon t)}_{s}\right)_{v\geq 0}$
is an independent copy of $\xi^{(\epsilon t)}$. Now, let us consider two
cases: $\eta^{(\epsilon t)}_{t/4}\leq t\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/8$ and the opposite. Note that $\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]\leq\mathbb{E}\left[\eta_{1}\right]=\mathbb{E}\left[\xi_{1}\right]<0$,
so $t\mathbb{E}\left[\eta_{1}^{(\epsilon t)}\right]/8\to-\infty$ as
$t\to\infty$. We will show that in the first case, the value of
$\eta^{(\epsilon t)}_{t/4}$ is small enough to make $I_{\eta^{(\epsilon t)}}$
sufficiently large, and afterwards that the probability of the second case is
sufficiently small.

Case 1: Under $A_{t}:=\left\\{\eta^{(\epsilon t)}_{t/4}\leq
t\mathbb{E}\left[\eta_{1}^{(\epsilon t)}\right]/8\right\\}$.

In this case, and for $s\leq c_{2}\ln t$ and large $t$, we have by the Markov
property at time $t/4$ and (7.4) that

$\begin{split}\mathbbm{1}_{A_{t}}I_{\eta^{(\epsilon
t)}}(t-s)&=\mathbbm{1}_{A_{t}}\int_{0}^{t-s}e^{-\eta^{(\epsilon
t)}_{v}}\mathrm{d}v\geq\mathbbm{1}_{A_{t}}\int_{t/4}^{t/2}e^{-\eta^{(\epsilon
t)}_{v}}\mathrm{d}v=\mathbbm{1}_{A_{t}}e^{-\eta^{(\epsilon
t)}_{t/4}}\int_{0}^{t/4}e^{-{\widetilde{\eta}}^{(\epsilon
t)}_{v}}\mathrm{d}v\\\ &\geq e^{-t\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/8}I_{\widetilde{\eta}^{(\epsilon t)}}(t/4)\geq
e^{-t\mathbb{E}\left[\eta_{1}\right]/8}I_{\widetilde{\eta}}(t/4)\geq
e^{-t\mathbb{E}\left[\eta_{1}\right]/8}I_{\widetilde{\eta}}(1),\end{split}$

so

$\mathbb{E}\left[\mathbbm{1}_{A_{t}}I^{-a}_{\eta^{(\epsilon
t)}}(t-s)\right]\leq
e^{at\mathbb{E}\left[\eta_{1}\right]/8}\mathbb{E}\left[I^{-a}_{\widetilde{\eta}}(1)\right]=:e^{-c_{3}at}c_{4},$

where $\widetilde{\eta}$ is an independent copy of $\eta$ and $c_{3},c_{4}>0$.
We note that the $c_{4}$ is well-defined, i.e. the last expectation is finite,
by [28, (2.36)]. Therefore,

$\begin{split}\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[\left(I_{\xi^{(1)}}(s)+e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}I_{\eta^{(\epsilon
t)}}(t-s)\right)^{-a}\mathbbm{1}_{A_{t}}\right]\mathrm{d}s&\leq\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[\left(e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}I_{\eta^{(\epsilon
t)}}(t-s)\right)^{-a}\mathbbm{1}_{A_{t}}\right]\mathrm{d}s\\\ &\leq
e^{-c_{3}at}c_{4}\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[\left(e^{-\xi^{(1)}_{s}-\epsilon
t}\right)^{-a}\right]\mathrm{d}s\\\ &:=c_{4}e^{-c_{5}t}\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[e^{a\xi^{(1)}_{s}}\right]\mathrm{d}s\end{split}$ (7.8)

for $c_{5}:=a(c_{3}-\epsilon)$, which is positive for sufficiently small
$\epsilon$ and where we have estimated $\Delta^{(\epsilon t)}_{s}\leq\epsilon
t$. As for the last integral above, because the Lévy measure of $\xi^{(1)}$ is
supported on $(-\infty,1]$, by [33, Theorem 25.3], one can see that the
characteristic function of $\xi^{(1)}$ in any $a\geq 0$, $\Psi^{(1)}(a)$, is
finite, so for large $t$,

$\int_{0}^{c_{2}\ln
t}\mathbb{E}\left[e^{a\xi^{(1)}_{s}}\right]\mathrm{d}s=\int_{0}^{c_{2}\ln
t}e^{s\Psi^{(1)}(a)}\mathrm{d}s\leq c_{2}\ln te^{c_{2}\ln
t\left|\Psi^{(1)}(a)\right|},$

so the behaviour of the last expression in (7.8) is dominated by
$e^{-c_{5}t}$, and thus is $\mathrm{O}\left(\ell(t)/t^{\alpha}\right)$.

Case 2: Under $B_{t}:=\left\\{\eta^{(\epsilon
t)}_{t/4}>t\mathbb{E}\left[\eta_{1}^{(\epsilon t)}\right]/8\right\\}$.

First, let us bound the expectation from (7.7), conditioned on this case:
using once again (7.4), we calculate that

$\begin{split}\int_{0}^{c_{2}\ln
t}\mathbbm{1}_{B_{t}}&\mathbb{E}\left[\left(I_{\xi^{(1)}}(s)+e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}I_{\eta^{(\epsilon t)}}(t-s)\right)^{-a}\right]\mathrm{d}s\\\
&\hphantom{{}=\sum_{k=0}^{n}\binom{n}{k}}=\int_{0}^{c_{2}\ln
t}\mathbb{P}(B_{t})\mathbb{E}\left[\left(I_{\xi^{(1)}}(s)+e^{-\xi_{s}^{(1)}-\Delta_{s}^{(\epsilon
t)}}I_{\eta^{(\epsilon t)}}(t-s)\right)^{-a}\middle|B_{t}\right]\mathrm{d}s\\\
&\hphantom{{}=\sum_{k=0}^{n}\binom{n}{k}}\leq\mathbb{P}(B_{t})\int_{0}^{\infty}\mathbb{E}\left[I^{-a}_{\xi^{(1)}}(s)\right]\mathrm{d}s\leq\mathbb{P}(B_{t})\int_{0}^{\infty}\mathbb{E}\left[I_{\xi}^{-a}(s)\right]\mathrm{d}s.\end{split}$
(7.9)

Therefore, because, as already noted, the last integral is finite from (2.12),
a bound of $\mathbb{P}(B_{t})$ of order $\ell(t)/t^{\alpha}$ would be enough
to conclude. We will obtain such a bound via a Chernoff type argument. For
$\lambda>0$, by Markov’s inequality for $x\mapsto e^{\lambda x}$ and using the
Lévy -Khintchine formula, we have

$\mathbb{P}(B_{t})=\mathbb{P}\left(\eta^{(\epsilon
t)}_{t/4}>t\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/8\right)\leq\exp\left(t\Psi^{(\epsilon t)}(\lambda)/4-\lambda
t\mathbb{E}\left[\eta_{1}^{(\epsilon t)}\right]/8\right),$ (7.10)

where we recall, see (2.4), that

$\begin{split}\Psi^{(\epsilon
t)}(\lambda)=\lambda\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]+\frac{1}{2}\lambda^{2}\sigma^{2}+\int_{(-\infty,\epsilon
t]}\left(e^{\lambda x}-1-\lambda x\right)\Pi(\mathrm{d}x).\end{split}$ (7.11)

We would choose $\lambda$ to depend on $t$ in $\eqref{eq:Chernoff}$ as

$\lambda_{t}:=\frac{\ln t^{(\alpha-1-\epsilon)/4}}{\epsilon t},$

and we will work with $0<\epsilon<\alpha-1$. For better readability, we will
not use the explicit expression of $\lambda_{t}$ if not needed. What would be
of vital importance is that $\lambda_{t}\to 0$ when $t\to\infty$ and that the
highest order terms in (7.11) would be of order $\lambda_{t}$.

Let us start with bounding the integral in (7.11): since
$\mathbb{E}\left[\xi_{1}\right]=\mathbb{E}\left[\eta_{1}\right]$ is finite, we
know that $\int_{-\infty}^{-1}x\Pi(x)\mathrm{d}x$ is finite, see e.g. [33,
Corollary 25.8], so we can choose $K_{1}<-1$ such that

$-\int_{(-\infty,K_{1}]}x\Pi(\mathrm{d}x)\leq-\mathbb{E}\left[\eta_{1}\right]/4\leq-\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/4.$

Next, split the integral from (7.11) into three regions:

$\int_{(-\infty,\epsilon
t]}\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x)=\int_{(-\infty,K_{1}]\cup(K_{1},1]\cup(1,\epsilon
t]}\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x).$

Over the first region, by the choice of $K_{1}<-1$ and since
$e^{\lambda_{t}x}-1\leq 0$ for $x<0$ and large $t$,

$\int_{(-\infty,K_{1}]}\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x)\leq-\lambda_{t}\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/{4}.$ (7.12)

For bounding over the second region, note that because
$e^{x}-1-x\stackrel{{\scriptstyle 0}}{{\sim}}x^{2}/2$, it is true that for
sufficiently small $x$, $e^{x}-1-x\leq x^{2}$. Therefore, as $\lambda_{t}\to
0$ and $K_{1}$ is fixed, for large $t$,

$\begin{split}\int_{(K_{1},1]}\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x)&\leq\lambda_{t}^{2}\int_{(K_{1},-1]\cup(-1,1]}x^{2}\Pi(\mathrm{d}x)\\\
&\leq\lambda_{t}^{2}K_{1}^{2}\int_{(K_{1},1]}\min\left\\{1,x^{2}\right\\}\Pi(\mathrm{d}x)=\mathrm{O}\left(\lambda_{t}^{2}\right)=\mathrm{o}\left(1/t\right)\end{split}$
(7.13)

as by construction $\lambda_{t}=\mathrm{O}\left(\ln t/t\right)$. Over the last
region, after integration by parts,

$\begin{split}\int_{(1,\epsilon
t]}&\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x)=\lambda_{t}\int_{(1,\epsilon
t]}\int_{0}^{x}\left(e^{\lambda_{t}y}-1\right)\mathrm{d}y\Pi(\mathrm{d}x)\\\
&=\lambda_{t}\int_{0}^{1}\int_{(1,\epsilon
t]}\left(e^{\lambda_{t}y}-1\right)\Pi(\mathrm{d}x)\mathrm{d}y+\lambda_{t}\int_{1}^{\epsilon
t}\int_{(y,\epsilon
t]}\left(e^{\lambda_{t}y}-1\right)\Pi(\mathrm{d}x)\mathrm{d}y\\\
&=\lambda_{t}\int_{0}^{1}\left(e^{\lambda_{t}y}-1\right)\left(\overline{\Pi}_{+}(1)-\overline{\Pi}_{+}(\epsilon
t)\right)\mathrm{d}y+\lambda_{t}\int_{1}^{\epsilon
t}\left(e^{\lambda_{t}y}-1\right)\left(\overline{\Pi}_{+}(y)-\overline{\Pi}_{+}(\epsilon
t)\right)\mathrm{d}y.\end{split}$ (7.14)

Similarly to a previous argument, as $e^{y}-1\stackrel{{\scriptstyle
0}}{{\sim}}y$, for large $t$ we have

$\lambda_{t}\int_{0}^{1}\left(e^{\lambda_{t}y}-1\right)\left(\overline{\Pi}_{+}(1)-\overline{\Pi}_{+}(\epsilon
t)\right)\mathrm{d}y\leq\lambda_{t}\overline{\Pi}_{+}(1)\int_{0}^{1}2\lambda_{t}y\mathrm{d}y=\lambda_{t}^{2}\overline{\Pi}_{+}(1)=\mathrm{O}\left(\lambda_{t}^{2}\right)=\mathrm{o}\left(1/t\right)$
(7.15)

so our final step would be to bound the last integral from (7.14). As
previously noted, from (6.4), there exists $C^{\ast}>0$ such that, for large
$t$, $\overline{\Pi}_{+}(t)\leq C^{\ast}\mathbb{P}(\xi_{1}>t)\leq
2C^{\ast}\ell(t)/t^{\alpha}\leq 1/t^{\alpha-\epsilon}$ and we recall that
$\alpha-\epsilon>1$. Let us fix $K_{2}>1$ such that for $t>K_{2}$, the last
inequality holds, i.e. $\overline{\Pi}_{+}(t)\leq 1/t^{\alpha-\epsilon}$.
Therefore, for large $t$, because
$\sqrt{t}\lambda_{t}=\mathrm{o}\left(1\right)$,

$\begin{split}\lambda_{t}\int_{1}^{\epsilon
t}\left(e^{\lambda_{t}y}-1\right)&\left(\overline{\Pi}_{+}(y)-\overline{\Pi}_{+}(\epsilon
t)\right)\mathrm{d}y=\lambda_{t}\int_{(1,K_{2}]\cup(K_{2},\sqrt{t}]\cup(\sqrt{t},\epsilon
t]}\left(e^{\lambda_{t}y}-1\right)\left(\overline{\Pi}_{+}(y)-\overline{\Pi}_{+}(\epsilon
t)\right)\mathrm{d}y\\\
&\leq\lambda_{t}\left(\overline{\Pi}_{+}(1)\int_{1}^{K_{2}}2\lambda_{t}y\mathrm{d}y+\int_{K_{2}}^{\sqrt{t}}2\lambda_{t}y\frac{1}{y^{\alpha-\epsilon}}\mathrm{d}y+e^{\lambda_{t}\epsilon
t}\int_{\sqrt{t}}^{\infty}\frac{1}{y^{\alpha-\epsilon}}\mathrm{d}y\right)\\\
&=\lambda_{t}\left(c_{6}\lambda_{t}+c_{7}\lambda_{t}t^{(\epsilon-\alpha+2)/2}+c_{8}e^{\lambda_{t}\epsilon
t}t^{(\epsilon-\alpha+1)/2}\right)\\\
&=\lambda_{t}\left(c_{6}\lambda_{t}+c_{9}t^{(\epsilon-\alpha)/2}\ln
t+c_{8}t^{(\epsilon-\alpha+1)/4}\right)=\mathrm{o}\left(1/t\right)\end{split}$
(7.16)

where we have substituted that $\lambda_{t}=\ln
t^{(\alpha-1-\epsilon)/4}/(\epsilon t)$ and implicitly have defined
$c_{6},c_{7},c_{8},c_{9}\geq 0$. Using (7.15) and (7.16) in (7.14), we get

$\int_{(1,\epsilon
t]}\left(e^{\lambda_{t}x}-1-\lambda_{t}x\right)\Pi(\mathrm{d}x)=\mathrm{o}\left(1/t\right)$
(7.17)

Substituting the bounds (7.12), (7.13) and (7.17) into (7.11), we obtain

$\Psi^{(et)}(\lambda_{t})\leq 3\lambda_{t}\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]/{4}+\mathrm{o}\left(1/t\right)$

and plugging this into (7.10), along with $\lambda_{t}=\ln
t^{(\alpha-1-\epsilon)/4}/(\epsilon t)$, we can see that for large $t$

$\mathbb{P}(B_{t})\leq\exp\left(\frac{1}{16}\lambda_{t}t\mathbb{E}\left[\eta_{1}^{(\epsilon
t)}\right]+\mathrm{o}\left(1\right)\right)\leq\exp\left(\frac{1}{32}\lambda_{t}t\mathbb{E}\left[\eta_{1}\right]\right)=\exp\left(\frac{(\alpha-1-\epsilon)\mathbb{E}\left[\eta_{1}\right]\ln
t}{32\epsilon}\right).$

Choosing $\epsilon$ sufficiently small, explicitly,
$\epsilon<(1-\alpha)\mathbb{E}\left[\eta_{1}\right]/(32(\alpha+1)-\mathbb{E}\left[\eta_{1}\right]$),
we get that, for large $t$,

$\mathbb{P}(B_{t})\leq\exp\left(-(\alpha+1)\ln(t)\right)=1/t^{\alpha+1}=\mathrm{O}\left(\ell(t)/t^{\alpha}\right)=\mathrm{O}\left(\mathbb{P}\left(\xi_{1}>t\right)\right),$

which concludes the proof that
$\mathbb{E}\left[I_{\xi}^{-a}(t)\mathbbm{1}_{\left\\{J_{1}^{(\epsilon
t)}>t\right\\}}\right]=\mathbb{E}\left[I_{\xi^{(\epsilon t)}}^{-a}(t)\right]$
is
$\mathrm{O}\left(\ell(t)/t^{\alpha}\right)=\mathrm{O}\left(\mathbb{P}\left(\xi_{1}>t\right)\right)$.
Using this in (7.3) along with (7.5), we obtain the statement of the lemma,
see (6.3). ∎

##  Appendix A Some facts from regular variation theory

In this appendix, we provide basic facts about the classical Karamata theory
of slowly varying functions and an extension of it over a class of functions,
named after de Haan. The presentation is based on, respectively, Chapter I and
Chapter III of the extensive work [8].

###  A.1 Classical Karamata theory

######  Definition A.1.

Let $\ell$ be a positive measurable function, defined on some neighbourhood
$[X,\infty)$ of infinity, and satisfying for all $\lambda>0$,

$\lim_{x\to\infty}\frac{\ell(\lambda x)}{\ell(x)}=1,$

then $\ell$ is said to be slowly varying at infinity. We note the class of
these functions as $SV_{\infty}$. In a similar manner, if $\ell$ is defined in
a positive/negative neighbourhood of $0$ and satisfying

$\lim_{x\to 0+/0-}\frac{\ell(\lambda x)}{\ell(x)}=1,$

$\ell$ is said to be slowly varying at $0+$/$0-$. We note the class of these
functions as $SV_{0}$.

Basic properties, such as comparison with polynomials and linear properties,
can be found in [8, Proposition 1.3.6].

######  Theorem A.2 (Uniform Convergence Theorem, [8, Theorem 1.2.1]).

If $\ell\in SV_{\infty}$, then for all $\lambda>0$,

$\lim_{x\to\infty}\frac{\ell(\lambda x)}{\ell(x)}=1,\text{ uniformly on each
compact set in $(0,\infty)$}.$ (A.1)

As we will often use the fact that slowly varying functions grow slower than
polynomials, we state the following easy consequence of Potter’s theorem, see
[8, Theorem 1.5.6].

######  Proposition A.3.

If $\ell\in SV_{\infty}$, then for any chosen constants $A>1$, $\delta>0$,
there exist $X=X(A,\delta)$ such that for all $x\geq X$,

$\ell(x)\leq Ax^{\delta}\quad\text{and}\quad\ell(x)\geq Ax^{-\delta}.$ (A.2)

######  Remark A.4.

If $\ell\in SV_{0}$, applying Theorem A.2 for $\ell^{\prime}(x)=\ell(1/x)\in
SV_{\infty}$, we can see that (A.1) holds after the modification $x\to 0$.
Similarly, (A.2) leads to $\ell(x)=\mathrm{O}\left(x^{-\delta}\right)$ for any
$\delta>0$ as $x\to 0$.

######  Theorem A.5 (Karamata’s Theorem, [8, Proposition 1.5.8]).

If $\ell\in SV_{\infty}$, $X$ is so large that $\ell(x)$ is locally bounded in
$[X,\infty)$, and $\alpha>-1$, then

$\int_{X}^{x}t^{\alpha}\ell(t)\mathrm{d}t\stackrel{{\scriptstyle\infty}}{{\sim}}x^{\alpha+1}\ell(x)/(\alpha+1).$

######  Theorem A.6 (Karamata’s Theorem, [8, Proposition 1.5.10]).

If $l$ is a slowly varying function and $\alpha<-1$ then
$\int^{\infty}t^{\alpha}l(t)dt$ converges and

$\lim_{x\to\infty}\frac{x^{\alpha+1}l(x)}{\int_{x}^{\infty}t^{\alpha}l(t)dt}=-\alpha-1.$

######  Theorem A.7 (Karamata Tauberian Theorem, [8, Theorem 1.7.1]).

Let $U$ be a non-decreasing right-continuous function on $\mathbb{R}$ with
$U(x)=0$ for all $x<0$. If $\ell\in SV_{\infty}$ and $c\geq 0,\rho\geq 0$, the
following are equivalent:

$\displaystyle U(x)$
$\displaystyle\stackrel{{\scriptstyle\infty}}{{\sim}}cx^{\rho}\ell(x)/\Gamma(1+\rho),$
(A.3) $\displaystyle\int_{[0,\infty)}e^{-sx}\mathrm{d}U(x)=:\widehat{U}(s)$
$\displaystyle\stackrel{{\scriptstyle 0}}{{\sim}}cs^{-\rho}\ell(1/s).$ (A.4)

When $c=0$, (A.3) is to be interpreted as
$U(x)=\mathrm{o}\left(x^{\rho}\ell(x)\right)$; similarly for (A.4).

######  Theorem A.8 (extended Karamata Tauberian Theorem, [8, Theorem 1.7.6]).

Assume $U(\cdot)\geq 0,c\geq
0,\rho>-1,s\widehat{U}(s):=s\int_{0}^{\infty}e^{-sx}U(x)\mathrm{d}x$
convergent for $s>0$, and $\ell\in SV_{\infty}$. Then, under the condition

$\lim_{\lambda\to 1+}\liminf_{x\to\infty}\inf_{1\leq
t\leq\lambda}\frac{U(tx)-U(x)}{x^{\rho}\ell(x)}\geq 0,$ (A.5)

it is true that

$s\widehat{U}(s)\stackrel{{\scriptstyle
0}}{{\sim}}cs^{-\rho}\ell(1/s)\quad\text{and}\quad$

implies

$U(x)\stackrel{{\scriptstyle\infty}}{{\sim}}cx^{\rho}\ell(x)/\Gamma(1+\rho).$

######  Remark A.9.

The reverse implication in the last theorem is true without the assumption
(A.5).

######  Theorem A.10 (Monotone Denstity theorem, [8, Theorems 1.7.2 - 2’]).

Let $U(x)=\int_{0}^{x}u(y)\mathrm{d}y$. If
$U(x)\stackrel{{\scriptstyle\infty}}{{\sim}}cx^{\rho}\ell(x)$, where
$c\in\mathbb{R}$, $\rho\in\mathbb{R}$, $\ell\in SV_{\infty}$, and if $u$ is
ultimately monotone, then

$u(x)\stackrel{{\scriptstyle\infty}}{{\sim}}c\rho x^{\rho-1}\ell(x).$

Let $V(x)=\int_{x}^{\infty}v(x)dx$ with $v$ integrable at infinity. Let
$V(x)\stackrel{{\scriptstyle\infty}}{{\sim}}cx^{\rho}\ell(x)$ for $\rho<0$ and
let $v$ be ultimately non-increasing. Then

$v(x)\stackrel{{\scriptstyle\infty}}{{\sim}}-c\rho x^{\rho-1}\ell(x).$

###  A.2 de Haan theory

We begin with a motivating example: let $\ell\in SV_{\infty}$ and choose $X$
so that $\ell\in L^{1}_{\text{loc}}[X,\infty)$. In Karamata’s theorem A.5, the
situation where $\alpha=-1$, i.e. when considering the asymptotic behaviour of

$m(x):=\int_{X}^{x}\ell(t)\mathrm{d}t/t,$

represents a special boundary case. We have the following result.

######  Proposition A.11 ([8, Proposition 1.5.9a]).

The function $m\in SV_{\infty}$ and

$\lim_{x\to\infty}\frac{m(x)}{\ell(x)}=\infty.$

However, it turns out that there exist more precise links between $m$ and
$\ell$.

######  Proposition A.12 ([8, p. 127]).

For any $\lambda>0$,

$\frac{1}{\ell(x)}\int_{x}^{\lambda
x}\ell(t)\mathrm{d}t/t=\lim_{x\to\infty}\frac{m(\lambda
x)-m(x)}{\ell(x)}=\ln\lambda.$

The class of all functions which exhibit similar behaviour to $m$, constitutes
the de Haan class $\Pi_{\ell}$.

######  Definition A.13.

We say that a measaruble function $f$ is in the de Haan class $\Pi_{\ell}$ if
for any $\lambda>0$,

$\lim_{x\to\infty}\frac{f(\lambda x)-f(x)}{\ell(x)}=c\ln\lambda,$

with some non-zero constant $c$, called the $\ell$-index of $f$.

The following theorem from [8] refers the work of de Haan, [10], as its
backbone. It characterises the class $\Pi_{\ell}$ via a monotone density
representation.

######  Theorem A.14 ([8, Theorem 3.6.8]).

Let $U(x)=U(X)+\int_{X}^{x}u(t)\mathrm{d}t$, where $u$ is monotone on some
neighbourhood of infinity, and let $\ell\in SV_{\infty}$. Then
$U\in\Pi_{\ell}$ with $\ell$-index $c$ if and only if

$u(x)\stackrel{{\scriptstyle\infty}}{{\sim}}cx^{-1}\ell(x).$

## Acknowledgments

This study is financed by the European Union’s NextGenerationEU, through the
National Recovery and Resilience Plan of the Republic of Bulgaria, project No
BG-RRP-2.004-0008. The authors are thankful to Pierre Patie for his usual
unusual resourcefulness in spotting probabilistic relations as in Theorem 3.4.

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