2.1. Laplace transforms and convolutions

Throughout, ${\mathcal{L}}$ denotes the (one-sided) Laplace transform

(2.1) \begin{align} {\mathcal{L}}\mu(\beta) \,:\!=\, {\mathcal{L}}[\mu](\beta) \,:\!=\, \int_{[0,\infty)} {\mathrm{e}}^{-\beta x} \,\mu(\mathrm{d} x),\end{align}

for a measure $\mu$ on $[0,\infty)$ and every $\beta\geq 0$ for which the integral converges. As usual, if $\mu$ is absolutely continuous with respect to Lebesgue measure with a locally integrable density f, i.e. $\mu(\mathrm{d} x) = f(x)\mathrm{d} x$ , we write $\mu\ll \mathrm{d} x$ , identify $\mu$ and f, and write ${\mathcal{L}} f={\mathcal{L}}\mu$ .

Let $\mu$ and $\nu$ be two measures with support in ${\unicode{x211D}}_+$ such that the Laplace transforms ${\mathcal{L}}\mu$ and ${\mathcal{L}}\nu$ converge for some $\beta_0>0$ . Then they also converge for all $\beta\in[\beta_0,\infty)$ , and we have

(2.2) \begin{align} {\mathcal{L}}[\mu*\nu](\beta) = {\mathcal{L}}\mu(\beta) {\mathcal{L}}\nu(\beta) \quad\text{on}\,\,[\beta_0,\infty),\end{align}

for the convolution of the measures

(2.3) \begin{align} (\mu*\nu)(B)\,:\!=\,\int_{{\unicode{x211D}}}\int_{{\unicode{x211D}}} {\unicode{x1D7D9}}_B(x+y) \,\mu(\mathrm{d} x)\,\nu(\mathrm{d} y),\quad\text{for all Borel sets}\,\, B\subset{\unicode{x211D}}.\end{align}

If $\mu$ and $\nu$ are absolutely continuous with respect to Lebesgue measure with locally integrable densities f and g, then (2.2) becomes

\begin{align*} {\mathcal{L}}[f*g](\beta) = {\mathcal{L}} f(\beta) {\mathcal{L}} g(\beta),\end{align*}

where the convolution is given by

\begin{align*} f*g(x) = \int_{\unicode{x211D}} f(x-y)g(y)\mathrm{d} y = \int_{[0,x]} f(x-y)g(y)\mathrm{d} y, \quad x\geq 0.\end{align*}

The second equality is due to the fact that the supports of f and g are contained in ${\unicode{x211D}}_+$ . If $f,g\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ , then for all $x\geq 0$

(2.4) \begin{align} \|f*g\|_{L^1([0,x])}\leq \|f\|_{L^1([0,x])}\|g\|_{L^1([0,x])},\end{align}

where $\|f\|_{L^1([0,x])}\,:\!=\,\int_{[0,x]} |f(y)|\mathrm{d} y$ . Finally, if $f\in AC[0,\infty)$ or $f\in AC(0,\infty)$ is absolutely continuous, integration by parts yields

(2.5) \begin{align} {\mathcal{L}}[\partial f](\beta)=\beta {\mathcal{L}} f(\beta)-f(0{+}).\end{align}

2.2. Spectrally negative Lévy processes

Throughout this article, $(L_t)_{t\geq 0}$ denotes a spectrally negative Lévy process, $(\mathbb{P}_x)_{x\in{\unicode{x211D}}}$ the law of $(L_t)_{t\geq0}$ , and $({\mathcal{F}}_t)_{t\geq0}$ the natural filtration satisfying the usual assumptions. A spectrally negative Lévy process is a one-dimensional Lévy process with no positive jumps and with Laplace exponent given by

\begin{align*} \psi(\beta) \,:\!=\, \frac{1}{t} \log \mathbb{E}\big[{\mathrm{e}}^{\beta L_t}\big] = c\beta + \sigma^2\beta^2 + \int_{({-}\infty,0)}\left({\mathrm{e}}^{\beta x}-1-\beta x {\unicode{x1D7D9}}_{(0, 1)}(|x|) \right) \widetilde\nu(\mathrm{d} x), \quad \beta\geq 0,\end{align*}

where $\big(c, 2 \sigma^2, \widetilde{\nu}\big)$ is the characteristic triplet of $(L_t)_{t\geq 0}$ . To simplify notation, we denote by $\nu$ the reflection of $\widetilde\nu$ , i.e. $\nu(B)=\widetilde\nu({-}B)$ for all measurable sets $B\subset{\unicode{x211D}}$ , which yields

(2.6) \begin{align} \psi(\beta) = c\beta + \sigma^2\beta^2 + \int_{(0,\infty)} \left({\mathrm{e}}^{-\beta x}-1+\beta x{\unicode{x1D7D9}}_{(0, 1)}(x) \right) \nu(\mathrm{d} x), \quad \beta\geq 0.\end{align}

Depending on the characteristics $\big(c, 2 \sigma^2, \widetilde{\nu}\big)$ we may simplify the representation of the Laplace exponent $\psi$ ; see Lemma 2.1. Although such representations appear frequently in the literature, we provide short arguments for them, which also serves to fix notation and ideas.

Proof. To see (i), set $c^{\prime}=c+\int_{(0, 1)} y\,\nu(\mathrm{d} y)$ and compute

\begin{align*} \psi(\beta) &=c^{\prime}\beta+\sigma^2\beta^2 +\int_{(0,\infty)} ({\mathrm{e}}^{-\beta x}-1)\nu(\mathrm{d} x) \\ &= c^{\prime}\beta+\sigma^2\beta^2 -\beta\int_{ (0,\infty)} \int_{ [0,x]} {\mathrm{e}}^{-\beta y}\,\mathrm{d} y \, \nu(\mathrm{d} x)\\ &=c^{\prime}\beta+\sigma^2\beta^2 -\beta\int_{ [0,\infty)}{\mathrm{e}}^{-\beta y}\int_{[y,\infty)}\nu(\mathrm{d} x) \mathrm{d} y \\ &=c^{\prime}\beta+\sigma^2\beta^2 -\beta{\mathcal{L}}\bar\nu(\beta).\end{align*}

Further, (ii) follows in a similar way if we set $c^{\prime\prime}=c-\int_{[1,\infty)} y\,\nu(\mathrm{d} y)$ and use

\begin{align*} \psi(\beta) &=c^{\prime\prime} \beta + \sigma^2 \beta^2 +\int_{ (0,\infty)}\left( {\mathrm{e}}^{-\beta x} -1+\beta x\right) \nu(\mathrm{d} x)\\ &=c^{\prime\prime}\beta+\sigma^2\beta^2 -\beta\int_{ (0,\infty)} \int_{[0,x]}\big( {\mathrm{e}}^{-\beta y}-1\big)\mathrm{d} y \, \nu(\mathrm{d} x)\\ &=c^{\prime\prime}\beta+\sigma^2\beta^2 +\beta^2\int_{ (0,\infty)} \int_{[0,x]}\int_{[0,y]} {\mathrm{e}}^{-\beta z} \,\mathrm{d} z \, \mathrm{d} y \, \nu(\mathrm{d} x)\\ &=c^{\prime\prime}\beta+\sigma^2\beta^2 -\beta^2\int_{ [0,\infty)} {\mathrm{e}}^{-\beta z} \int_{ [z,\infty)}\int_{[y,\infty)} \nu(\mathrm{d} x) \mathrm{d} y \, \mathrm{d} z\\ &=c^{\prime\prime}\beta+\sigma^2\beta^2+\beta^2{\mathcal{L}}\bar{\bar\nu}(\beta).\end{align*}

This completes the proof.

From now on we will always use the constant $c\in{\unicode{x211D}}$ to denote the location parameter if $\psi$ is in the form (2.6), and $c^{\prime},c^{\prime\prime}\in{\unicode{x211D}}$ if we write $\psi$ as in Lemma 2.1(i)–(ii).

Note that whenever $\int_{[1,\infty)}y\nu(\mathrm{d} y)<\infty$ , the function $\bar{\bar\nu}$ is well-defined, and integration by parts implies that in this case

\begin{align*} \int_{(0, 1)}y\nu(\mathrm{d} y)=\infty \quad \Longleftrightarrow \quad \bar{\bar\nu}(0{+})=\infty.\end{align*}

Further background information on (spectrally negative) Lévy processes can be found in [Reference Bertoin5, Reference Sato27].

2.3. Scale functions

The q-scale function of a spectrally negative Lévy process $(L_t)_{t\geq 0}$ is uniquely determined via its Laplace transform given in (1.1). If $\widetilde\nu \equiv 0$ , i.e. if $(L_t)_{t\geq 0}$ is a Brownian motion with drift, the Laplace transform in (1.1) can be inverted explicitly (cf. [Reference Kuznetsov, Kyprianou and Rivero18]), leading to the well-known formula

(2.7) \begin{gather} W^{(q)}(x) = \frac{1}{\sqrt{c^2 +2q\sigma^2}} \left({\mathrm{e}}^{\left(\sqrt{c^2 +2q\sigma^2} - c\right){x}/{\sigma^2}} - {\mathrm{e}}^{-\left(\sqrt{c^2 +2q\sigma^2} + c\right){x}/{\sigma^2}} \right), \quad x\geq 0.\end{gather}

We will therefore restrict ourselves to processes with $\widetilde\nu \not\equiv 0$ . Moreover, we will exclude subordinators, i.e. monotone Lévy processes. In particular, if $\sigma^2=0$ and $\int_{(0, 1)} x\,\nu(\mathrm{d} x)<\infty$ , we assume $c\geq \int_{(0, 1)} x\nu(\mathrm{d} x)$ .

For a spectrally negative Lévy process $L=(L_t)_{t\geq0}$ with Laplace exponent $\psi$ and characteristic triplet $\big(c,2\sigma^2,\widetilde\nu\big)$ it is well known (cf. [Reference Kuznetsov, Kyprianou and Rivero18, Section 3.3], [Reference Chan, Kyprianou and Savov8], and [Reference Kyprianou19, Chapter 8.2]) that under the exponential change of measure

\begin{align*}\frac{\mathrm{d} {\mathbb{P}}_x^{\Phi(q)}}{\mathrm{d} {\mathbb{P}}_x}\Bigg|_{{\mathcal{F}}_t}={\mathrm{e}}^{\Phi(q)(L_t-x)-qt}, \quad q\geq0,\end{align*}

the process $\big(L,{\mathbb{P}}^{\Phi(q)}\big)$ is again a spectrally negative Lévy process whose Laplace exponent is given by $\psi_{\Phi(q)}(\beta)=\psi(\beta+\Phi(q))-q$ and whose Lévy measure is $\widetilde\nu_{\Phi(q)}(\mathrm{d} x)={\mathrm{e}}^{\Phi(q)x}\widetilde\nu(\mathrm{d} x)$ . The shape of $\psi_{\Phi(q)}$ , together with (1.1), yields $W^{(q)}(x)={\mathrm{e}}^{\Phi(q)x}W_{\Phi(q)}(x)$ , where $W_{\Phi(q)}$ denotes the 0-scale function of $\big(L,{\mathbb{P}}^{\Phi(q)}\big)$ . This relation often allows us to assume without loss of generality that $q=0$ , and we will use this observation below in Section 4.1 for the proof of Theorem 4.1.

The surveys [Reference Avram, Grahovac and Vardar-Acar2] and [Reference Kuznetsov, Kyprianou and Rivero18] contain a thorough introduction to scale functions and some of their applications.

3.1. Lévy processes with a Gaussian component

We start by presenting a series expansion for scale functions of spectrally negative Lévy processes that have a Gaussian component. Results of this kind have a long history. For example, if we combine in [Reference Chan, Kyprianou and Savov8] the proof of Theorem 1 with Corollary 9, we arrive at a series representation for $\partial W^{(0)}$ , and this leads to an (implicit) expansion for $\partial W^{(q)}$ using the relations between $W^{(0)}$ and $W^{(q)}$ . Also note that, setting $\nu\equiv 0$ in Theorem 3.1 below, we recover (2.7). At this point we note that the moment condition $\int_{[1,\infty)} y\,\nu(\mathrm{d} y)<\infty$ appearing in Theorem 3.1 is for notational convenience only, and can be discarded at the expense of heavier notation; see also Remark 3.1 below.

Theorem 3.1. Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process with triplet $\big(c, 2 \sigma^2,\widetilde\nu\big)$ such that $\sigma^2>0$ and $\int_{[1,\infty)} y\,\nu(\mathrm{d} y)<\infty$ . For $q\geq 0$ set

\begin{align*}f(x) \,:\!=\, -c^{\prime\prime}+ qx - \bar{\bar\nu}(x) = \textstyle\int_{[1,\infty)} y\,\nu(\mathrm{d} y) - c + qx - \bar{\bar\nu}(x), \quad x\geq0; \end{align*}

then the q-scale function of $(L_t)_{t\geq 0}$ is given by

(3.1) \begin{align} W^{(q)} = {\operatorname{id}} *\sum_{n=0}^\infty \frac{f^{*n}}{(\sigma^2)^{n+1}}, \end{align}

and the series on the right-hand side converges uniformly on compact subsets of ${\unicode{x211D}}_+$ to a limit in $L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ .

Proof. By Lemma 2.1(ii), the Laplace exponent of L is given by $\psi(\beta)=c^{\prime\prime}\beta+\sigma^2\beta^2+\beta^2{\mathcal{L}}\bar{\bar\nu}(\beta)$ with $c^{\prime\prime}=c-\int_{[1,\infty)} y\,\nu(\mathrm{d} y)$ . From the definition of the scale function we thus get for all $q\geq 0$

\begin{equation}\begin{aligned} {\mathcal{L}}\big[W^{(q)}\big](\beta) &=\frac1{\beta^2\!\left(c^{\prime\prime}\beta^{-1} + \sigma^2 + {\mathcal{L}}\bar{\bar\nu}(\beta)-q\beta^{-2}\right)}. \nonumber \end{aligned}\end{equation}

For $\beta>1$ large enough we have $\sigma^2 > \big|c^{\prime\prime}\beta^{-1} + {\mathcal{L}}\bar{\bar\nu}(\beta) -q\beta^{-2}\big|$ , and so we may expand the fraction appearing on the right-hand side into a geometric series:

(3.2) \begin{align} {\mathcal{L}}\big[W^{(q)}\big](\beta) &=\frac1{\beta^2} \sum_{n=0}^\infty \frac{\left({-}c^{\prime\prime}\beta^{-1} - {\mathcal{L}}\bar{\bar\nu}(\beta) + q\beta^{-2}\right)^n}{(\sigma^2)^{n+1}} =\frac1{\beta^2}\sum_{n=0}^\infty \frac{\left({\mathcal{L}} f(\beta)\right)^n}{(\sigma^2)^{n+1}} \end{align}

with $f(x) = -c^{\prime\prime} - \bar{\bar\nu}(x)+ qx $ as in Theorem 3.1, and for sufficiently large $\beta > 1$ .

Note that it is enough to know the (existence of the) Laplace transform for large values of $\beta\in[\Phi(q),\infty)$ to characterize $W^{(q)}$ ; see e.g. [Reference Widder30, Theorem 6.3]. Thus, to prove the representation for $W^{(q)}$ , it remains to show that the summation and the Laplace transform in (3.2) can be interchanged. To do so, let us write for a moment $g(x) \,:\!=\, \sigma^{-2}f(x)$ . Then for every $m\in{\unicode{x2115}}$

\begin{align*} \left| {\mathcal{L}}\!\left(\sum_{n=0}^\infty g^{*n}\right)-\sum_{n=0}^m {\mathcal{L}}\!\left(g^{*n}\right)\right| &= \left| {\mathcal{L}}\!\left(\sum_{n=0}^\infty g^{*n}\right)-{\mathcal{L}}\!\left(\sum_{n=0}^m g^{*n}\right) \right|\\ &\leq {\mathcal{L}}\!\left(\sum_{n=m+1}^\infty |g|^{*n}\right) = \sum_{n=m+1}^\infty {\mathcal{L}}[|g|]^n. \end{align*}

For sufficiently large arguments $\beta > 1$ we know that ${\mathcal{L}}[|g|](\beta)<1$ , and so the series on the right-hand side converges, which allows us to interchange the summation and the Laplace transform. Moreover, the above calculation shows that $\sum_{n=1}^\infty g^{*n}$ converges in the weighted space $L^1 \big({\mathrm{e}}^{-\beta y}\,\mathrm{d} y\big)$ , and so it is well-defined in, say, $L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ . Further, from Hölder’s inequality it follows that

\begin{align*} \left|\left ({\operatorname{id}} * \sum_{n=1}^\infty g^{*n}\right)\!(x)\right| \leq x\left\|\sum_{n=1}^\infty g^{*n}\right\|_{L^1[0,x]}, \quad x\geq0. \end{align*}

By Lemma 5.3, the series converges uniformly on compact subsets of ${\unicode{x211D}}_+$ . Finally, since ${\mathcal{L}}[{\operatorname{id}}](\beta) = \beta^{-2}$ and ${\operatorname{id}} \ast g^{\ast 0} = {\operatorname{id}} \ast \delta_0={\operatorname{id}}$ , we get

\begin{align*} W^{(q)} = {\operatorname{id}} * \sum_{n=0}^\infty \frac{f^{*n}}{(\sigma^2)^{n+1}}, \end{align*}

as stated.

Remark 3.1. The condition $\int_{[1,\infty)} y\,\nu(\mathrm{d} y)<\infty$ ensures that the integrated tail $\bar{\bar\nu}$ is finite. We can obtain similar results without this condition, if we split the Lévy measure into $\nu^{(0)}=\nu\vert_{[0,z)}$ and $\nu^{(\infty)}=\nu\vert_{[z,\infty)}$ for some fixed $z\geq 1$ , and replace $\bar{\bar\nu}$ by $\bar{\bar\nu}^{(0)}(x)=\int_{[x,z)} \bar\nu(y)\mathrm{d} y$ . The Laplace exponent $\psi$ is then given by

\begin{align*} \psi(\beta) = c^{\prime\prime}_{z}\beta + \sigma^2\beta^2 + \beta^2{\mathcal{L}}\big[\bar{\bar\nu}^{(0)}\big](\beta) + {\mathcal{L}}\big[\nu^{(\infty)}\big](\beta) - \big\|\nu^{(\infty)}\big\|, \end{align*}

where $c^{\prime\prime}_{z} \,:\!=\, c - \int_{[1,\infty)} x\nu^{(0)}(\mathrm{d} x)$ and $\big\|\nu^{(\infty)}\big\| = \nu^{(\infty)}\left([0,\infty)\right)$ is the total mass. Thus, with the same argument as in the proof of Theorem 3.1,

(3.3) \begin{align} W^{(q)} = {\operatorname{id}} *\sum_{n=0}^\infty \frac{f^{*n}}{(\sigma^2)^{n+1}}, \end{align}

with f such that

\begin{align*} {\mathcal{L}} f(\beta)= -c_z^{\prime\prime} \beta^{-1} - {\mathcal{L}}\big[\bar{\bar\nu}^{(0)}\big](\beta) - \beta^{-2} {\mathcal{L}}\big[\nu^{(\infty)}\big](\beta) + \beta^{-2}\big(q+\big\|\nu^{(\infty)}\big\|\big), \end{align*}

i.e. with f given by

\begin{align*} f(x)= -c_z^{\prime\prime} - \bar{\bar\nu}^{(0)}(x) - \int_{[z,x]} ( x-y )\nu^{(\infty)}(\mathrm{d} y) + \big(q+\big\|\nu^{(\infty)}\big\|\big)x, \quad x\geq0. \end{align*}

Note that the integral term in f vanishes on [0, z); therefore, it can be omitted when evaluating $W^{(q)}$ on [0, z). The above procedure thus yields a series expansion of the same form as (3.1) for bounded subsets of ${\unicode{x211D}}_+$ , if we pick $z\geq 0$ sufficiently large and adjust c ^′′ and $\bar{\bar\nu}^{(0)}$ accordingly.

Remark 3.2. A different construction of the series expansions yields further useful formulae for the q-scale function. Consider the setting of Theorem 3.1. Instead of (3.2) we can use

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta) = \frac1{c^{\prime\prime}\beta + \sigma^2\beta^2 + \beta^2{\mathcal{L}}\bar{\bar\nu}(\beta)-q} = \sum_{n=0}^\infty \frac{({-}1)^n\big(\beta^2{\mathcal{L}}\bar{\bar\nu}(\beta)\big)^n}{\big(c^{\prime\prime}\beta + \sigma^2\beta^2 - q\big)^{n+1}}. \end{align*}

Denote by

\begin{equation*}a_{1,2}=-\frac{c^{\prime\prime}}{2\sigma^2}\pm\sqrt{\Big(\frac{c^{\prime\prime}}{2\sigma^2}\Big)^2+\frac{q}{\sigma^2}}\end{equation*}

the roots of the polynomial in the denominator. This allows us to invert the Laplace transform, leading to

\begin{align*} W^{(q)} =\sum_{n=0}^\infty \frac{({-}1)^n}{(\sigma^2)^{n+1}} {\mathcal{L}}^{-1}\left(\frac{\beta^n}{(\beta+a_1)^{n+1}}\right) * {\mathcal{L}}^{-1}\left(\frac{\beta^n}{(\beta+a_2)^{n+1}}\right) * \bar{\bar\nu}^{*n}, \end{align*}

with the inverse Laplace transforms of the rational functions given by

\begin{align*} {\mathcal{L}}^{-1}\!\left(\frac{\beta^n}{(\beta+a)^{n+1}}\right)=\frac1{n!}\partial^n\!\left(x^n{\mathrm{e}}^{-ax}\right) \quad\text{for all}\,\, n\in{\unicode{x2115}},\; a\in{\unicode{x211D}}; \end{align*}

cf. [Reference Widder30]. If, moreover, the Lévy measure $\nu$ has finite total mass $\|\nu\|<\infty$ , we get

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta) = \frac1{c^{\prime}\beta+\sigma^2\beta^2+{\mathcal{L}}\nu(\beta)-(q+\|\nu\|)} = \sum_{n=0}^\infty \frac{({-}1)^n({\mathcal{L}}\nu(\beta))^n}{\big(c^{\prime}\beta+\sigma^2\beta^2-(q+\|\nu\|)\big)^{n+1}} \end{align*}

with $c^{\prime}=c+\int_{(0, 1)} x\nu(\mathrm{d} x)$ . Thus, we have

(3.4) \begin{gather} W^{(q)} = \sum_{n=0}^\infty \frac{({-}1)^n}{(\sigma^2)^{n+1}}\left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_3 x}\right) * \left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_4 x}\right)*\nu^{*n} \end{gather}

with

\begin{equation*}a_{3,4}=-\frac{c^{\prime}}{2\sigma^2}\pm\sqrt{\Big(\frac{c^{\prime}}{2\sigma^2}\Big)^2+\frac{q+\|\nu\|}{\sigma^2}}.\end{equation*}

The convolution

\begin{equation*} \left(\frac{1}{n!}x^{n}{\mathrm{e}}^{-a_3 x}\right)*\left(\frac{1}{n!}x^{n}{\mathrm{e}}^{-a_4 x}\right) \end{equation*}

can be explicitly worked out. Rather than stating the lengthy expression, we refer to [Reference Landriault and Willmot21], where the case of finite Lévy measures is thoroughly studied for both $\sigma^2>0$ and $\sigma^2=0$ .

We close this section with two explicit examples in the spirit of the previous remark. Suppose that the Lévy measure $\nu$ is bounded away from zero, i.e. there exists some $\varepsilon>0$ such that ${\operatorname{supp}}(\nu) \cap [0,\varepsilon) = \emptyset$ ; in particular, $\nu$ is a finite measure. In this case, the series expansion in (3.4) becomes a closed-form expression, since ${\operatorname{supp}}(\nu^{*n})\cap[0,n\varepsilon) = \emptyset$ for any $n\in{\unicode{x2115}}$ . As usual, we use $\lfloor x \rfloor \,:\!=\, \max \{y\in{\unicode{x2124}}\::\: y\leq x \}$ to denote the largest integer below $x\in{\unicode{x211D}}$ .

Example 3.1. (Geometrically distributed jumps.) Let $\nu= \sum_{k=1}^\infty (1-p)^{k-1}p\, \delta_{k}$ be a geometric distribution with success probability $p\in (0, 1)$ . It is well known that the n-fold convolution $\nu^{*n}$ is a negative binomial distribution, i.e.

\begin{align*} \nu^{\ast n}=\sum_{k=n}^\infty \binom{k-1}{k-n} (1-p)^{k-n}p^k\, \delta_k, \quad n\in {\unicode{x2115}}. \end{align*}

Hence for $q\geq 0$ the q-scale function of the spectrally negative Lévy process with Laplace exponent

\begin{align*} \psi(\beta)=c^{\prime}\beta+\sigma^2\beta^2+\int_{ (0,\infty)} \left({\mathrm{e}}^{-\beta x}-1\right) \nu(\mathrm{d} x) \end{align*}

and $\sigma^2>0$ is given by

\begin{align*} W^{(q)}(x) = \sum_{n=0}^{\lfloor x\rfloor}\sum_{k=n}^{\lfloor x\rfloor} ({-}1)^n \frac{ \binom{k-1}{k-n} (1-p)^{k-n}p^k}{(\sigma^2)^{n+1}}\left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_3 x}\right)*\left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_4 x}\right)(x-k),\qquad x\geq 0, \end{align*}

with $a_3,a_4$ as in Remark 3.2.

Example 3.2. (Zero-truncated Poisson distributed jumps.) Let $\nu$ denote the distribution of a Poisson random variable $\xi$ with parameter $\mu>0$ . In order to determine the q-scale functions using Remark 3.2, we need to know the probability mass functions $f_n$ of $\nu^{*n}$ , $n\in{\unicode{x2115}}$ . For $k\neq 0$ they are given by

(3.5) \begin{align} f_n(k) = \frac{{\mathbb{P}}\!\left(\sum_{i=1}^n \xi_i = k \:\&\: \forall i \leq n \,:\, \xi_i>0 \right)}{{\mathbb{P}}\!\left(\forall i \leq n \,:\, \xi_i >0\right)} \,=\!:\, \frac{z(k,n)}{(1-{\mathrm{e}}^{-\mu})^n} , \quad k\in{\unicode{x2115}}\setminus\{0\}, \end{align}

where $\xi_i$ , $i\in{\unicode{x2115}}\setminus\{0\}$ are independent and identically distributed copies of $\xi$ . For $n=0$ we have, by definition, $\nu^{*0}=\delta_0$ . The numerator z(k, n) in (3.5) can be computed recursively: obviously, we have $z(k,n)=0$ if $k<n\in{\unicode{x2115}}\setminus\{0\}$ , and $z(k,1)=\frac{\mu^s}{s!}{\mathrm{e}}^{-\mu}$ if $k\in{\unicode{x2115}}\setminus\{0\}$ . Otherwise, we have

\begin{align*} z(k,n) &={\mathbb{P}}\!\left(\sum_{i=1}^n \xi_i = k\right) - \sum_{\ell=1}^{n-k-1} \binom{n}{\ell} {\mathbb{P}}\!\left(\sum_{i=1}^{\ell} \xi_i =0\right)z(k,n-\ell)\\ &= \frac{(n\mu)^k}{k!}{\mathrm{e}}^{-n\mu} - \sum_{\ell=1}^{n-k-1} \binom{n}{\ell} {\mathrm{e}}^{-\ell\mu}z(k,n-\ell). \end{align*}

Setting $z(0,0)=1$ and $z(k,0)=0$ , $k\in{\unicode{x2115}}\setminus\{0\}$ , the q-scale function of the spectrally negative Lévy process with Laplace exponent

\begin{align*} \psi(\beta)=c^{\prime}\beta+\sigma^2\beta^2+\int_{(0,\infty)} \left({\mathrm{e}}^{-\beta x}-1\right) \nu(\mathrm{d} x) \end{align*}

and $\sigma^2>0$ is given by

\begin{align*} W^{(q)}(x) = \frac{1}{c} \sum_{n=0}^{\lfloor x \rfloor} \sum_{k=n}^{\lfloor x \rfloor}\frac{({-}1)^n z(n,k)}{(\sigma^2)^{n+1}\left(1-{\mathrm{e}}^{-\mu}\right)^k}\left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_3 x}\right)*\left(\frac{x^{n}}{n!}{\mathrm{e}}^{-a_4 x}\right)(x-k),\qquad x\geq 0, \end{align*}

with $a_3,a_4$ as in Remark 3.2.

3.2. Lévy processes with paths of bounded variation

If $(L_t)_{t\geq 0}$ has paths of bounded variation, the following series expansions for the q-scale functions hold true.

Theorem 3.2. Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process with triplet $\big(c,0,\widetilde\nu \big)$ such that $\int_{(0, 1)} y\,\nu(\mathrm{d} y)<\infty$ . Recall $c^{\prime}=c+\int_{(0, 1)} y\,\nu(\mathrm{d} y)$ . For $q\geq 0$ set

\begin{align*} f(x) \,:\!=\, q + \bar\nu(x), \quad x\geq0; \end{align*}

then the q-scale function of $(L_t)_{t\geq 0}$ is given by

(3.6) \begin{align} W^{(q)}={\unicode{x1D7D9}}*\sum_{n=0}^\infty\frac{ f^{*n}}{\left(c^{\prime}\right)^{n+1}}, \end{align}

and the series on the right-hand side converges uniformly on compact subsets of ${\unicode{x211D}}_+$ .

Proof. By Lemma 2.1(i), the Laplace exponent of $L_t$ is given by $\psi(\beta)=c^{\prime}\beta-\beta{\mathcal{L}}\bar\nu(\beta)$ with $c^{\prime}=c + \int_{(0, 1)} y\,\nu(\mathrm{d} y)$ , and we obtain for $q\geq 0$

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta) = \frac1{c^{\prime}\beta -\beta{\mathcal{L}}\bar\nu(\beta)-q} = \frac1{\beta\!\left(c^{\prime}-{\mathcal{L}}\bar\nu(\beta)-q\beta^{-1}\right)}. \end{align*}

As in the proof of Theorem 3.1, we have $c^{\prime}>\left|{\mathcal{L}}\bar\nu(\beta)-q\beta^{-1}\right|$ for sufficiently large $\beta> 1$ , and thus

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta) &= \frac1\beta\sum_{n=0}^\infty \frac{\left({\mathcal{L}}\bar\nu(\beta)+q\beta^{-1}\right)^n}{(c^{\prime})^{n+1}} = \frac1\beta\sum_{n=0}^\infty \frac{\left({\mathcal{L}} f(\beta)\right)^n}{(c^{\prime})^{n+1}} \end{align*}

with $f(x) = q + \bar\nu(x)$ . The rest of the proof coincides with the proof of Theorem 3.1, with the slight difference that we now use

\begin{align*} \left|\left ({\unicode{x1D7D9}} * \sum_{n=1}^\infty \frac{f^{*n}}{(c^{\prime})^{n+1}}\right)(x)\right| \leq \left\|\sum_{n=1}^\infty \frac{f^{*n}}{(c^{\prime})^{n+1}}\right\|_{L^1[0,x]}, \quad x\geq0, \end{align*}

to obtain uniform convergence on compact subsets of ${\unicode{x211D}}_+$ .

Remark 3.3. The formula (3.6) may be regarded as a generalization of the Pollaczek–Khinchin representation of ruin probabilities in the Cramér–Lundberg compound Poisson model. Let $(L_t)_{t\geq 0}$ be a spectrally negative compound Poisson process with positive drift; i.e. its Laplace exponent is of the form

\begin{align*} \psi(\beta) = c\beta + \lambda\int_{(0,\infty)} \left({\mathrm{e}}^{-\beta x} -1\right)\! \Pi(\mathrm{d} x), \end{align*}

where $c,\lambda>0$ and $\Pi((0,\infty))=1$ . Under the net-profit condition

\begin{align*} \psi^{\prime}(0{+}) = c- \lambda \int_{(0,\infty)} x\Pi(\mathrm{d} x) >0, \end{align*}

the ruin probability $r(x) \,:\!=\,{\mathbb{P}}(\exists t\geq 0\::\: x+L_t < 0)$ is given by the Pollaczek–Khinchin formula

(3.7) \begin{align} r(x) = 1-\psi^{\prime}(0{+})\int_{(0,x)} \sum_{n=0}^\infty\frac{\lambda^n\bar\Pi^{*n}(y)}{c^{n+1}}\,\mathrm{d} y, \quad x\geq 0, \end{align}

with $\bar{\Pi}(y)=\int_{[y,\infty)} \Pi( \mathrm{d} x)$ ; see e.g. [Reference Asmussen and Albrecher1, Eq. IV(2.2)] or [Reference Kyprianou19, Eq. (1.15)]. Moreover, we have (cf. [Reference Kyprianou19, Theorem 8.1])

\begin{align*} r(x) = 1-\psi^{\prime}(0{+}) W^{(0)}(x), \end{align*}

which yields the following representation of the 0-scale function:

\begin{align*} W^{(0)}(x) = \int_{(0,x)} \sum_{n=0}^\infty\frac{\lambda^n\bar\Pi^{*n}(y)}{c^{n+1}}\,\mathrm{d} y = {\unicode{x1D7D9}} * \sum_{n=0}^\infty\frac{\lambda^n\bar\Pi^{*n}(x)}{c^{n+1}}. \end{align*}

Theorem 3.6 works, however, both for all $q\geq 0$ and for a larger class of jump measures, and it does not require any assumption on $\psi^{\prime}(0{+})$ .

In the actuarial context, and in renewal theory, one often uses the following definition of the convolution:

$$\begin{align*} f\circledast g(x) &\,:\!=\, \int_{[0,x]} f(x-y)\mathrm{d} g(y), \, \cr & f^{\circledast n}(x) :\!=\, \int_{[0,x]} f^{\circledast (n-1)}(x-y)\mathrm{d} f(y), \quad\text{and}\quad f^{\circledast 0} \,:\!=\, {\unicode{x1D7D9}}, \end{align*}$$

for any $n\in{\unicode{x2115}}$ and whenever these integrals are defined. In particular, $f\circledast g = f*g^{\prime}$ if g is absolutely continuous with derivative g ^′. With this notation, (3.7) can be rewritten in the more familiar form

\begin{align*} r(x)=1-\psi^{\prime}(0{+})\sum_{n=0}^\infty\frac{(\lambda\mu)^n \Pi_I^{\circledast n}(x)}{c^{n+1}} = 1-\left(1-\frac{\lambda \mu}{c} \right)\sum_{n=0}^\infty\left(\frac{\lambda\mu}{c}\right)^n \Pi_I^{\circledast n}(x) \end{align*}

with $\Pi_I(x)\,:\!=\,\mu^{-1}\int_{(0,x)} \bar\Pi(y)\mathrm{d} y = \mu^{-1}{\unicode{x1D7D9}}{*}\bar\Pi(x)$ and the first moment $\mu =\int_{ (0,\infty)} x\Pi(\mathrm{d} x)$ of $\Pi$ .

3.3. Lévy processes with paths of unbounded variation without a Gaussian component

It remains to consider spectrally negative Lévy processes with paths of unbounded variation and $\sigma^2=0$ . The following theorem provides a series expansion of the q-scale involving an auxiliary function h such that $h*\bar{\bar\nu}(0{+})=1$ . The existence of h as well as its properties and some examples will be discussed below. Since we are only interested in the behaviour of h near $x=0$ h need not be unique. Indeed, any $h_a\,:\!=\, h \cdot {\unicode{x1D7D9}}_{[0,a]}$ , with $a>0$ , will be as good as h.

Theorem 3.3. Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process with triplet $\big(c,0,\widetilde\nu\big)$ such that $\int_{(0,1]} y\,\nu(\mathrm{d} y) = \infty$ and $\int_{[1,\infty)} y\,\nu(\mathrm{d} y) < \infty$ . Recall that $c^{\prime\prime}=c-\int_{[1,\infty)} y\,\nu(\mathrm{d} y)$ . If there exists some $h\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ such that $\partial \big(h\ast \bar{\bar\nu}\big) \in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ and $h*\bar{\bar\nu}(0{+})=1$ , then the q-scale function of $(L_t)_{t\geq 0}$ is given by

(3.8) \begin{align} W^{(q)}=H*\sum_{n=0}^\infty f^{*n},\quad q\geq 0. \end{align}

Here, $H(x) \,:\!=\, \int_{(0,x)} h(y)\mathrm{d} y = {\unicode{x1D7D9}}\ast h(x)$ is the primitive of h, and

\begin{align*} f(x) \,:\!=\, qH(x)- c^{\prime\prime}h(x)-\partial\big(h*\bar{\bar\nu}\big)(x),\quad x\geq0. \end{align*}

The series on the right-hand side of (3.8) converges uniformly on compact subsets of ${\unicode{x211D}}_+$ .

Proof. Let h be the function from the statement of the theorem, and assume in addition that $h\in L^1({\unicode{x211D}}_+)$ , which ensures that the Laplace transform exists. The identity (2.5) for the Laplace transforms shows that

\begin{align*} {\mathcal{L}}\big[\partial\big(h*\bar{\bar\nu}\big)\big](\beta) = \beta{\mathcal{L}} h(\beta) {\mathcal{L}}\bar{\bar\nu}(\beta)- h*\bar{\bar\nu}(0{+}) = \beta{\mathcal{L}} h(\beta) {\mathcal{L}}\bar{\bar\nu}(\beta)-1. \end{align*}

By assumption, $\nu_h\,:\!=\,\partial\big(h*\bar{\bar\nu}\big)$ exists. Thus, the Laplace exponent $\psi$ can be written as

\begin{align*} \psi(\beta) = c^{\prime\prime}\beta + \frac\beta{{\mathcal{L}} h(\beta)}\left( {\mathcal{L}}\nu_h(\beta) +1\right), \end{align*}

and from the definition of the scale function, we get for all $q\geq 0$

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta) = \frac{{\mathcal{L}} h(\beta)}\beta\frac1{\left( c^{\prime\prime}{\mathcal{L}} h(\beta) + {\mathcal{L}}\nu_h(\beta) + 1 - q\beta^{-1}{\mathcal{L}} h(\beta) \right)}. \end{align*}

Since $\nu_h, h\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ we obtain

\begin{align*} {\mathcal{L}}\big[W^{(q)}\big](\beta)=\frac{{\mathcal{L}} h(\beta)}{\beta}\sum_{n=0}^\infty \left({\mathcal{L}} f(\beta)\right)^n, \end{align*}

for large enough $\beta> 1$ and with $f(x) = q\int_{ (0,x)} h(y)\mathrm{d} y - c^{\prime\prime} h(x) - \partial\big(h*\bar{\bar\nu}\big)(x)$ as in Theorem 3.3. Using the same arguments as in the proof of Theorem 3.1, we can invert the Laplace transform and obtain the formula (3.8).

If $h\in L^1_{\mathrm{loc}}({\unicode{x211D}}{+})\setminus L^1({\unicode{x211D}}_+)$ , we set $ h_a =h\cdot{\unicode{x1D7D9}}_{[0,a]}$ for some $a>0$ and observe that $h_a$ meets the requirements stated in the theorem and that ${\mathcal{L}}( h_a )$ is well-defined. Thus, $W^{(q)}$ is given by the series expansion (3.8) with respect to $h_a$ . Replacing h by $h_a$ does not affect f(x); hence $W^{(q)}(x)$ if $x\in [0,a]$ . Since $a>0$ is arbitrary, the proof is complete.

Resolvents, Bernstein functions, and renewal equations

In order to find a suitable function h in the setting of Theorem 3.3, i.e. $h\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ such that $\partial \big(h\ast \bar{\bar\nu}\big) \in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ and $h*\bar{\bar\nu}(0{+})=1$ , let us consider the auxiliary problem

(3.9) \begin{gather} \rho * \bar{\bar\nu} \equiv 1,\end{gather}

where $\rho = \rho(\mathrm{d} x)$ is a measure on $[0,\infty)$ and $\bar{\bar\nu}$ is the integrated tail function. Equations of this type are also used in the theory of Volterra integral equations, where the measure $\rho$ appearing in (3.9) is called the resolvent of the function $\bar{\bar\nu}$ ; cf. [Reference Gripenberg, London and Staffans16]. Taking Laplace transforms on both sides of (3.9) yields

(3.10) \begin{gather} {\mathcal{L}}[\rho](\beta) = \frac{1}{\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)},\quad \beta > 0.\end{gather}

The calculations in the proof of Lemma 2.1 in our current setting show that $\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)$ $=\int_{(0,\infty)} \left(1-{\mathrm{e}}^{-x\beta}\right)\bar\nu(x)\mathrm{d} x$ . Hence $\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)$ is a Bernstein function ( ${\operatorname{BF}}$ ) with Lévy triplet $(0,0,\bar\nu(x)\mathrm{d} x)$ ; cf. [Reference Schilling, Song and Vondraček28, Theorem 3.2]. Thus, its reciprocal ${1}/\big(\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)\big)$ is a potential (cf. [Reference Schilling, Song and Vondraček28, Definition 5.24]), and in particular it is a completely monotone function. From Bernstein’s theorem (cf. [Reference Schilling, Song and Vondraček28, Theorem 1.4]), we thus see that (3.10), and hence (3.9), always has a measure-valued solution $\rho$ ; see also [Reference Gripenberg15]. These observations also allow for an important probabilistic interpretation of (3.9): $\rho \ast \bar{\bar{\nu}}=1$ is the probability that a subordinator $(S_t)_{t\geq 0}$ with BF $\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)$ and Lévy triplet $(0,0,\bar\nu(x)\mathrm{d} x)$ crosses any level $x>0$ ; see Bertoin [Reference Bertoin5, Chapter III.2]. The measure $\rho$ is then the potential or zero-resolvent of $(S_t)_{t\geq 0}$ .

For our purpose, however, only $L^1_{\mathrm{loc}}({\unicode{x211D}}_+)$ -solutions of (3.9) are of interest. The following lemma constructs such solutions in the setting of Theorem 3.3.

From the probabilistic interpretation of $\rho*\bar{\bar\nu}$ , the fact that $\rho$ is absolutely continuous is not surprising. One can argue as follows: since the Lévy measure $\bar\nu(x)\mathrm{d} x$ of the subordinator $(S_t)_{t\geq 0}$ is absolutely continuous, this property is inherited by the transition densities ${\mathbb{P}}(S_t\in dx)$ (cf. Sato [Reference Sato27, Theorem 27.7]) and all q-resolvent measures [Reference Sato27, p. 242], in particular by the zero-resolvent measure $\rho$ . The following proof, based on the renewal equation, gives a short and direct approach to this, complementing also Neveu’s renewal theorem; cf. [Reference Bertoin6, Proposition 1.7].

Proof of Lemma 3.1. Set $\phi(\beta) \,:\!=\, \int_{(0,\infty)} \left(1-e^{-x\beta}\right)\bar\nu(x)\mathrm{d} x$ and $\bar N(x) = x\bar\nu(x)$ . Under the assumptions of Theorem 3.3 we know that $\bar{\bar\nu}(0{+})=\infty$ . We have already seen that there exists a measure $\rho$ on $[0,\infty)$ solving (3.9). Since $\bar{\bar\nu}(0{+})=\infty$ , the measure $\rho$ cannot have atoms: indeed, if $\rho \geq c\delta_{y}$ for some $c>0$ and $y\in [0,\infty)$ , this would lead to the contradiction

\begin{align*} 1 = \rho*\bar{\bar\nu}(x) \geq c\delta_y*\bar{\bar\nu}(x) = c\bar{\bar\nu}(x-y) \xrightarrow{x\downarrow y} c\bar{\bar\nu}(0{+}) = \infty. \end{align*}

Moreover, $\rho$ satisfies ${\mathcal{L}}[\rho](\beta) = \big(\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)\big)^{-1} = 1/\phi(\beta)$ . Therefore,

\begin{align*} \partial{\mathcal{L}}[\rho] = \partial\frac 1{\phi} = -\frac 1{\phi^2}\partial\phi = -{\mathcal{L}}[\rho]{\mathcal{L}}[\rho]{\mathcal{L}}[y\bar\nu(y)] = -{\mathcal{L}}[\rho*\rho*\bar N]. \end{align*}

On the other hand, $\partial{\mathcal{L}}[\rho] = -{\mathcal{L}}[y\rho]$ , and the uniqueness of the Laplace transform shows that

(3.11) \begin{gather} x\rho(\mathrm{d} x) = \rho*\rho*\bar N(x)\mathrm{d} x. \end{gather}

This implies $\rho(\mathrm{d} x) = \rho\{0\}\delta_0(\mathrm{d} x) + r(x)\mathrm{d} x$ with the density function $r(x) = x^{-1} (\rho *\rho*\bar N)(x)$ on $(0,\infty)$ . Because of $\bar{\bar\nu}(0{+})=\infty$ we know that $\rho\{0\}=0$ .

Further, recall that ${\operatorname{supp}}(\rho*\sigma)=\overline{{\operatorname{supp}} \, \rho +{\operatorname{supp}} \, \sigma}$ holds for any two measures $\rho, \sigma$ . Thus we see from $\rho*\bar{\bar\nu}=1$ that $0\in{\operatorname{supp}} \, \bar{\bar\nu}\cap{\operatorname{supp}} \, \rho$ ; in particular, ${\operatorname{supp}} \, \rho$ and ${\operatorname{supp}} \, \bar{\bar\nu}$ contain some neighbourhood of 0. Using again (3.11), ${\operatorname{supp}} \, \rho = \overline{{\operatorname{supp}} \, \rho+{\operatorname{supp}} \, \rho+{\operatorname{supp}} \, \bar N}$ , and $0\in{\operatorname{supp}} \, \bar N$ (this follows from $0\in{\operatorname{supp}} \, \bar{\bar\nu}$ ), we conclude that ${\operatorname{supp}}\rho$ is unbounded and even that ${\operatorname{supp}}\rho=[0,\infty)$ , since it contains a neighbourhood of 0.

Finally, by the very definition of the convolution,

\begin{align*} r(x) = \frac 1x \int_{(0,x)}\int_{(0,x-y)} \bar N(x-y-z)\,\rho(\mathrm{d} z)\,\rho(\mathrm{d} y), \end{align*}

proving that $r(x)>0$ since ${\operatorname{supp}} \, \rho=[0,\infty)$ and $0\leq \bar N\not\equiv 0$ . Thus $\rho$ is positive as well.

The proof of further regularity properties of solutions to (3.9) seems to be quite difficult, with the exception of complete monotonicity ( ${\operatorname{CM}}$ ), which is treated in the following corollary. Although this result is known (cf. [Reference Rao, Song and Vondraček22] or [Reference Song and Vondraček29, Rem 2.2]), we provide a short alternative proof for the reader’s convenience.

Remark 3.5. With the exception of Corollary 3.1, higher-order regularity properties of the solution to $\rho*\bar{\bar\nu}\equiv 1$ seem to be quite difficult to prove; see also [Reference Gripenberg14] for a related study. The following observation points in an interesting direction. Recall that a Bernstein function $f\in {\operatorname{BF}}$ is a special Bernstein function (see [Reference Schilling, Song and Vondraček28, Chapter 11]) if the conjugate function $\beta/f(\beta)$ is also a Bernstein function; iterating this argument, it becomes ‘self-improving’ and shows that $\beta/f(\beta)$ is even a special Bernstein function. A subordinator whose Laplace exponent is a special Bernstein function is a special subordinator. Note that complete Bernstein functions are special, and we have used this (implicitly) in Corollary 3.1.

Assume, for a moment, that we know $\beta{\mathcal{L}}[\rho](\beta)\in {\operatorname{BF}}$ . Then ${\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)=1/(\beta{\mathcal{L}}[\rho](\beta))$ is the potential of a special subordinator; hence $\beta{\mathcal{L}}[\rho](\beta)$ is a special Bernstein function. Thus $\beta/\beta{\mathcal{L}}[\rho](\beta) = 1/{\mathcal{L}}[\rho](\beta) = \beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)$ is also a special Bernstein function. This, in turn, means that ${\mathcal{L}}[\rho](\beta) = 1/\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)$ is the potential of a special subordinator, and according to [Reference Schilling, Song and Vondraček28, Theorem 11.3], $\rho$ is of the form $\rho(\mathrm{d} x) = c\delta_0(\mathrm{d} x) + r(x)\mathrm{d} x$ for some constant $c\geq 0$ and a decreasing function r(x). If we assume, as before, that $\nu(0,\infty)=\infty$ , $\rho$ has no atoms, so $c=0$ and $\rho(\mathrm{d} x) = r(x) \,\mathrm{d} x$ .

From $\int_{(0,\infty)} \big(1-e^{-\beta x}\big) \mu(\mathrm{d} x) = \beta {\mathcal{L}}[\bar{\mu}](\beta)$ we see that $\beta{\mathcal{L}}[\rho](\beta)$ is a Bernstein function only if $\rho$ has a decreasing density. This means the above argument becomes a vicious circle, unless we can come up with an independent criterion: let us consider the case that $x\mapsto\bar\nu(x)$ is logarithmically convex, i.e.

\begin{align*} \bar\nu(\lambda x + (1-\lambda)y)\leq \bar\nu(x)^\lambda\bar\nu(y)^{1-\lambda} \quad\text{for all $x,y>0$ and $\lambda \in (0, 1)$}. \end{align*}

Since $\bar{\bar\nu}(x)$ has a negative derivative, $\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)=\int_{(0,\infty)} \left(1-e^{-\beta x}\right)\bar\nu(x)\mathrm{d} x$ is a Bernstein function and, because of log-convexity, even a special Bernstein function (see [Reference Schilling, Song and Vondraček28, Theorem 11.11]); hence the conjugate function $1/{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta) = \beta{\mathcal{L}}[\rho](\beta)$ is a Bernstein function. Together with the above consideration, this proves the following:

If $\nu(0,\infty)=\infty$ and $x\mapsto\bar\nu(x)$ is logarithmically convex, the solution to $\rho*\bar{\bar\nu}\equiv 1$ is of the form $\rho(\mathrm{d} x) = r(x)\mathrm{d} x$ with a decreasing density $r(x)>0$ on $(0,\infty)$ .

See also [Reference Kyprianou, Rivero and Song20, Lemma 2.2] for a similar result stated in terms of potential measures of subordinators.

The following lemma provides a particular representation of the resolvent of $\bar{\bar\nu}$ as a continuous function; in fact, it is the Radon–Nikodym derivative of the scale function of some Lévy process.

Proof. From Lemma 2.1(ii) we know that the Laplace exponent of $(\widetilde{L}_t)_{t\geq 0}$ is given by

\begin{align*} \widetilde{\psi}(\beta) = \psi(\beta) - {\mathbb{E}}[L_1] \beta = c^{\prime\prime}\beta + \beta^2 {\mathcal{L}}\big[\bar{\bar\nu}\big](\beta) - c^{\prime\prime} \beta = \beta^2 {\mathcal{L}}\big[\bar{\bar\nu}\big](\beta). \end{align*}

By the definition of the scale function, we see ${\mathcal{L}} \widetilde{W} = \big(\beta^2{\mathcal{L}}\bar{\bar\nu}\big)^{-1}$ , and with (2.5) it follows that

\begin{align*} {\mathcal{L}} [\widetilde{W}^{\prime}](\beta) = \frac 1{\beta{\mathcal{L}}\big[\bar{\bar\nu}\big](\beta)} \iff {\mathcal{L}} [ \widetilde{W}^{\prime}](\beta)\cdot {\mathcal{L}}\big[\bar{\bar\nu}\big](\beta) = \frac 1{\beta}, \end{align*}

since $\widetilde{W}(0{+})=0$ . Finally, by [Reference Kuznetsov, Kyprianou and Rivero18, Lemma 2.4] we have $\widetilde W\in {\mathcal{C}}^1(0,\infty)$ , and thus $\rho\in{\mathcal{C}}(0,\infty)$ .

The above considerations prove that in the setting of Theorem 3.3 there always exists a resolvent of the integrated tail function $\bar{\bar\nu}$ . This immediately leads to the following simple corollary of Theorem 3.3.

Corollary 3.2. Let $(L_t)_{t\geq 0}$ be as in Theorem 3.3 and let $\rho\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ be a resolvent of $\bar{\bar\nu}$ . Then the q-scale function of $(L_t)_{t\geq 0}$ is given by

(3.12) \begin{align} W^{(q)} = R*\sum_{n=0}^\infty (qR-c^{\prime\prime}\rho)^{*n},\quad q\geq 0, \end{align}

where $R(x) \,:\!=\, \int_0^x \rho(y)\mathrm{d} y = {\unicode{x1D7D9}}\ast\rho (x)$ for all $x\in{\unicode{x211D}}_+$ . The series on the right-hand side of (3.12) converges uniformly on compact subsets of ${\unicode{x211D}}_+$ .

The deep relationship between scale functions and renewal equations has been explained in detail in [Reference Chan, Kyprianou and Savov8]. From the definition of scale functions (1.1) one immediately obtains that for a spectrally negative Lévy process $(L_t)_{t\geq 0}$ with triplet $\big(c,2\sigma^2,\widetilde\nu\big)$ and $\int_{[1,\infty)}y\nu(\mathrm{d} y)<\infty$ ,

(3.13) \begin{align}c^{\prime\prime}W^{(0)} + \sigma^2 \partial W^{(0)} + \bar{\bar\nu}*\partial W^{(0)} = 1.\end{align}

In the case $\sigma^2>0$ and with the notation of Remark 3.3, this equation is a renewal equation on ${\unicode{x211D}}_+$ of the type

(3.14) \begin{align} f=1+f\circledast g \end{align}

with $f(x)=\sigma^2\partial W^{(0)}(x)$ and $g(x)= -2\sigma^{-2} \int_0^x\bar{\bar\nu}(y)\mathrm{d} y - 2c^{\prime\prime}\sigma^{-2}x$ . This has been discussed in [Reference Chan, Kyprianou and Savov8], where (3.14) is solved for $\partial ^{(0)}$ to infer smoothness properties of $W^{(0)}$ . However, if $\sigma^2=0$ and $\int_{(0, 1)}y\nu(\mathrm{d} y)=\infty$ , then (3.13) yields

\begin{align*} c^{\prime\prime}W^{(0)} = 1 - \frac{\bar{\bar\nu}}{c^{\prime\prime}}*\partial \big(c^{\prime\prime}W^{(0)}\big) = 1 - \frac{\bar{\bar\nu}}{c^{\prime\prime}} \circledast \big(c^{\prime\prime}W^{(0)}\big), \end{align*}

which is not a renewal equation in the sense of [Reference Chan, Kyprianou and Savov8], since $\bar{\bar\nu}(0{+})\neq0$ . Nevertheless, our approach to obtaining series expansions for $W^{(0)}$ in such cases still provides a technique for solving this kind of renewal equation on ${\unicode{x211D}}_+$ , namely

(3.15) \begin{align} f=1+g*\partial f =1+ g\circledast f,\end{align}

where we do not necessarily assume that $g(0{+})=0$ . This is illustrated in the next theorem, whose proof follows the lines of the proof of Theorem 3.3.

Proof. It is enough to prove (ii). First note that $f(0{+})=0$ , since $f(x)=\int_{[0,x]}u(y)\mathrm{d} y$ with locally integrable integrand

\begin{equation*}u(y)\,:\!=\,h*\sum_{n=1}^\infty ({-}1)^n(h+ g_h)^{*(n-1)}.\end{equation*}

Convolving both sides of (3.15) with h and applying Lemma 5.1(i) yields

\begin{align*} h*f &={\unicode{x1D7D9}}*h+h*g*\partial f = {\unicode{x1D7D9}}*h+\partial(h*g* f)= {\unicode{x1D7D9}}*h + \partial(h*g)*f+f.\end{align*}

Applying the Laplace transform and solving for ${\mathcal{L}}[f](\beta)$ we get

\begin{align*} {\mathcal{L}}[f](\beta) = \frac{{\mathcal{L}}[{\unicode{x1D7D9}}*h](\beta)}{{\mathcal{L}}[h](\beta) - {\mathcal{L}}[g_h](\beta) - 1} &= -{\mathcal{L}}[{\unicode{x1D7D9}}*h](\beta) \sum_{n=0}^\infty \left({\mathcal{L}}[h-g_h](\beta)\right)^n\\ &= -{\mathcal{L}}[{\unicode{x1D7D9}}*h](\beta) \sum_{n=0}^\infty {\mathcal{L}}\big[(h-g_h)^{*n}\big](\beta),\end{align*}

from which the assertion follows by the uniqueness of the Laplace transform.

Examples

Although the existence of a resolvent $\rho$ in Corollary 3.2 is clear, there seems to be no general method of constructing $\rho$ explicitly. Fortunately, there are quite a few cases where a suitable function h (as in Theorem 3.3) can be constructed. In particular, this is true if $\bar{\bar\nu}$ is regularly varying at 0. Recall that a measurable function $f\,:\,(0,\infty)\to{\unicode{x211D}}$ is regularly varying at 0 with index $\gamma\in {\unicode{x211D}}$ if

\begin{align*} \lim_{x\to 0+}\frac{f(\lambda x)}{f(x)}=\lambda^{-\gamma} \quad \text{exists for all }\lambda >0.\end{align*}

We write $\mathcal R_\gamma$ for the space of regularly varying functions with index $\gamma$ . Our standard reference is the monograph [Reference Bingham, Goldie and Teugels7].

Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process as in Theorem 3.3, and assume that $\bar{\bar\nu} \, \text{is in} \, \mathcal{R}_\gamma$ for some $\gamma\in(0, 1)$ . From [Reference Bingham, Goldie and Teugels7, Theorem 1.8.3] and its proof we know that there exists a completely monotone function k such that $\lim_{x\to 0} \bar{\bar\nu}(x)/k(x)= 1.$ With the arguments of the previous subsection, this implies that there exists a unique completely monotone function h for which $h*k=1$ . Set $k_\varepsilon\,:\!=\,\bar{\bar\nu}-k$ . Since $\lim_{x\to0+}{k_\varepsilon(x)}/{k(x)}=0$ , we obtain

(3.16) \begin{equation}\begin{aligned} |h*k_\varepsilon(x)| &= \left| \int_{ (0,x)} h(x-y) k_\varepsilon (y) \,\mathrm{d} y \right|\\ &\leq \int_{ (0,x)} h(x-y)\left|\frac{k_\varepsilon(y)}{k(y)}\right|k(y)\mathrm{d} y \leq \sup_{y\leq x} \left|\frac{k_\varepsilon(y)}{k(y)}\right| \xrightarrow[x\to 0+]{}0;\end{aligned}\end{equation}

hence,

(3.17) \begin{equation} h\ast \bar{\bar\nu}(0{+})=1.\end{equation}

We may thus use this function h in Theorem 3.3 to obtain the representation of the q-scale function given in the following corollary.

Corollary 3.3. Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process as in Theorem 3.3, and assume that the Lévy measure $\nu$ is such that $\bar{\bar\nu}\in\mathcal{R}_\gamma \cap {\mathcal{C}}^1(0,\infty)$ for some $\gamma\in(0, 1)$ . Let $k_\varepsilon$ and h be the functions in (3.16). Then

\begin{align*} W^{(q)}=H*\sum_{n=0}^\infty \big(qH-c^{\prime\prime}h -\partial(h*k_\varepsilon)\big)^{*n} \end{align*}

where $H(x) \,:\!=\, \int_{(0,x)} h(y)\mathrm{d} y = {\unicode{x1D7D9}}\ast h (x)$ for all $x\in{\unicode{x211D}}_+$ .

Proof. We check the conditions stated in Theorem 3.3. First of all, $h\in L_{\mathrm{loc}}^1({\unicode{x211D}}_+)$ as it is the resolvent of $k\in L^1_{\mathrm{loc}}({\unicode{x211D}}_+)$ . Furthermore, we note that

\begin{align*} h\ast \bar{\bar\nu} = h\ast (k + k_\varepsilon) = 1 + h\ast k_\varepsilon, \end{align*}

which implies $\partial\big(h\ast \bar{\bar\nu}\big) = \partial (h\ast k_\varepsilon)$ . By assumption, $k_\varepsilon\in {\mathcal{C}}^1(0,\infty)$ . Thus, $h*k_\varepsilon\in{\mathcal{C}}^1(0,\infty)$ by Lemma 5.1(ii), and as $h*k_\varepsilon(0{+})=0$ we obtain $\partial(h*k_\varepsilon)\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ . Finally, in view of (3.16), $h\ast \bar{\bar\nu}(0{+}) = 1+ h\ast k_\varepsilon(0{+}) = 1$ , finishing the proof.

In the following, we restrict ourselves to jump measures $\nu$ satisfying $\bar{\bar\nu}\in\mathcal{R}_\gamma$ , $\gamma\in(0, 1)$ , and for which, in addition, there exists a constant $C>0$ such that

(3.18) \begin{equation} \lim_{x\to 0+}\frac{ Cx^{-\gamma}}{\bar{\bar\nu}(x)}=1.\end{equation}

Clearly, these are special cases of the jump measures treated in Corollary 3.3, but we can now use techniques from fractional calculus which will simplify our calculations. Recall that the Riemann–Liouville fractional integral $I_{0+}^\alpha\, f$ of order $\alpha\in(0, 1)$ of a function $f\,:\,{\unicode{x211D}}_+\to{\unicode{x211D}}_+$ is defined by

\begin{align*} \big(I_{0+}^\alpha\, f\big)(x) \,:\!=\,\frac1{\Gamma(\alpha)} \int_{ (0,x)} \frac{f(y)}{(x-y)^{1-\alpha}} \,\mathrm{d} y = \frac1{\Gamma(\alpha)}((\cdot)^{\alpha-1}*f)(x),\quad x\geq 0,\end{align*}

and the Riemann–Liouville fractional derivative $D_{0+}^\alpha$ of order $\alpha\in(0, 1)$ of $f\,:\,{\unicode{x211D}}_+\to{\unicode{x211D}}_+$ is defined by

\begin{align*} \big(D_{0+}^\alpha f\big)(x)\,:\!=\,\partial \big(I_{0+}^{1-\alpha} f\big)(x).\end{align*}

A thorough introduction to fractional calculus is found in the monograph [Reference Samko, Kilbas and Marichev26].

Corollary 3.4. Let $(L_t)_{t\geq 0}$ be a spectrally negative Lévy process as in Theorem 3.3, assume that the Lévy measure $\nu$ satisfies $\bar{\bar\nu}\in{\mathcal{C}}^1(0,\infty)$ and (3.18) for some $C>0$ , and let $\gamma\in(0, 1)$ . Set $h(x)\,:\!=\, x^{\gamma-1}$ and $H(x)\,:\!=\,{\unicode{x1D7D9}}*h(x)= \gamma^{-1}x^{\gamma}$ for $x>0$ . The q-scale function of $(L_t)_{t\geq 0}$ is given by

(3.19) \begin{align} W^{(q)} = H*\sum_{n=0}^\infty \left(\frac{\sin\!(\gamma\pi)}{C\pi}\right)^{n+1} f^{*n}, \quad q\geq 0, \end{align}

with $f(x)\,:\!=\,qH(x)- c^{\prime\prime}h(x)- \Gamma(\gamma)D^{1-\gamma}_{0+} \bar{\bar\nu}(x)$ for $x>0$ .

Proof. Set $k(x)=C x^{-\gamma}$ such that $\lim_{x\to 0} \bar{\bar\nu}(x)/k(x)= 1$ , and define $k_\varepsilon\,:\!=\,\bar{\bar\nu}-k$ . Clearly, $k_\varepsilon = o(x^{-\gamma})$ . Set $h(x)= x^{\gamma-1}$ . The same computation as in (3.16) yields

\begin{align*} |h\ast k_\varepsilon(x)|\to 0 \quad\text{as }x\to0. \end{align*}

Moreover,

\begin{align*} h\ast k(x) &= \int_{(0,x)} h(x-y) k(y) \,\mathrm{d} y = \int_{(0, 1)}h(x(1-z)) k(xz) x \,\mathrm{d} z\\ &= C\cdot \int_{(0, 1)} (1-z)^{\gamma-1} z^{-\gamma} \,\mathrm{d} z = C\cdot B(1-\gamma,\gamma), \end{align*}

where B(x, y), $x,y> 0$ , denotes the beta function. Since $B(1-\gamma,\gamma) = {\pi}/{\sin\!(\gamma \pi)}$ , we have

\begin{align*} h\ast \bar{\bar\nu} (0{+}) = h\ast k_\varepsilon(0{+}) + h\ast k(0{+}) = C\frac{\pi}{\sin\!(\gamma \pi)}. \end{align*}

We may thus apply Theorem 3.3 with $\widetilde{h}(x)\,:\!=\,\frac{\sin\!(\gamma \pi)}{C\pi} h(x)$ . Clearly, $\widetilde{h}\in L_{\mathrm{loc}}^1({\unicode{x211D}}_+)$ , and the calculations above show $\widetilde{h}\ast \bar{\bar\nu} (0{+})=1$ . Furthermore, $\partial\big(\widetilde{h}\ast \bar{\bar\nu}\big) \in L_{\mathrm{loc}}^1({\unicode{x211D}}_+)$ , since

\begin{align*} \partial\big(h\ast \bar{\bar\nu}\big) (x) = \partial \Big(\Gamma(\gamma) I_{0+}^{\gamma} \bar{\bar\nu}(x) \Big) = \Gamma(\gamma)D^{1-\gamma}_{0+} \bar{\bar\nu}(x), \end{align*}

with $D^{1-\gamma}_{0+} \bar{\bar\nu}\in {\mathcal{C}}^\gamma$ as $\bar{\bar\nu} \in {\mathcal{C}}^1$ by assumption; cf. [Reference Samko, Kilbas and Marichev26, Lemma 2.2]. This yields the representation of $W^{(q)}$ .

Example 3.3. (Spectrally negative stable processes.) Consider a spectrally negative stable Lévy process with stability index $\alpha\in(1,2)$ and Laplace exponent $\psi(\beta)=\beta^\alpha$ . Then as shown in [Reference Bertoin3] (see also [Reference Hubalek and Kyprianou17]), for all $q\geq 0$ ,

(3.20) \begin{align} W^{(q)}=\alpha x^{\alpha-1} E^{\prime}_\alpha\big(qx^\alpha\big),\quad x\geq 0, \end{align}

where $E_\alpha(x)=\sum_{k=0}^\infty {\Gamma(1+\alpha k)}^{-1} x^k$ , $x\geq 0$ , is the Mittag-Leffler function.

A lengthy but otherwise straightforward calculation shows that this formula agrees with (3.19): use that $\bar{\bar\nu}$ fulfils the limit relation (3.18) with $C= 1/{\Gamma(2-\alpha)}$ and $\gamma=\alpha-1$ , and moreover $\int^x_{0+} (x-y)^{\alpha-1}y^{-\alpha}\,\mathrm{d} y = \pi/\sin\!(\alpha\pi)$ for any $x>0$ and $\alpha\in(0, 1)$ .

The following observation allows us to construct further explicit examples of scale functions. Consider a spectrally negative Lévy process $(L_t)_{t\geq0}$ with Laplace exponent $\psi_L$ and suppose that we know its scale functions $W^{(q)}_L$ . Further, let $(C_t)_{t\geq0}$ be an independent compound Poisson process with positive jumps and without drift; we assume that the jumps have intensity $\lambda>0$ and are distributed according to the probability measure $\Pi$ . Then $(L_t-C_t)_{t\geq0}$ is a spectrally negative Lévy process whose Laplace exponent is given by

\begin{align*} \psi_{L-C}(\beta)=\psi_L(\beta)+ \lambda {\mathcal{L}}[\Pi](\beta)-\lambda.\end{align*}

As for any $\lambda$ we have $|\lambda {\mathcal{L}}[\Pi](\beta){\mathcal{L}}\big[W_L^{(q+\lambda)}(\beta)|<1$ for $\beta$ large enough, the q-scale function of $(L_t-C_t)_{t\geq0}$ is thus defined by

\begin{align*} {\mathcal{L}}\Big[W_{L-C}^{(q)}\Big] = \frac1{\psi_{L-C}-q} = \frac{{\mathcal{L}}\Big[W_L^{(q+\lambda)}\Big]}{1+ \lambda {\mathcal{L}}[\Pi]{\mathcal{L}}\Big[W_L^{(q+\lambda)}\Big]} = {\mathcal{L}}\Big[W_L^{(q+\lambda)}\Big]\sum_{k=0}^\infty\left({-} \lambda {\mathcal{L}}[\Pi]{\mathcal{L}}\Big[W_L^{(q+\lambda)}\Big]\right)^k.\end{align*}

This immediately implies

(3.21) \begin{align} W^{(q)}_{L-C}=W_L^{(q+\lambda)}*\sum_{k=0}^\infty \lambda^k \left({-}\Pi*W_L^{(q+\lambda)}\right)^{*k}.\end{align}

Example 3.4. (Stable process plus Poisson process.) Let $(L_t)_{t\geq0}$ be a spectrally negative $\alpha$ -stable process with $\alpha\in(1,2)$ and let $\Pi=\delta_1$ . For each $n\in{\unicode{x2115}}$ and $q\geq 0$ we then have from (3.21) that

\begin{align*} W^{(q)}_{L-C}(x) =W_L^{(q+\lambda)}*\sum_{k=0}^\infty ({-}\lambda)^k \left(W_L^{(q+\lambda)}({\cdot}-1)\right)^{*k}, \end{align*}

where $W_L^{(q+\lambda)}$ is given by (3.20). In fact, the sum on the right-hand side is finite, as $W_L^{(q+\lambda)}(x-1)=0$ for $x\in[0,1]$ , and thus,

\begin{align*} \left(W_L^{(q+\lambda)}({\cdot}-1)\right)^{*k}(x)=0\quad\text{for all}\,\, k>x. \end{align*}

4.1. Proof of Theorem 4.1

We will use the series expansion of Theorem 3.3 in order to prove Theorem 4.1. At first sight, this seems rather circumstantial in comparison with the representation in Corollary 3.2. The trouble is that for the convolution equation $\rho * f = 1$ rather little is known about how the smoothness of $\rho$ and f are related to each other; see Section 3.3 above.

In order to apply Theorem 3.3, we need to construct a suitable function h. We are going to show that, in general, a suitable approximation of the resolvent of $N_1$ is good enough for our purposes. By assumption, the resolvent $\rho_1$ of $N_1$ is continuously differentiable. The following lemma will help us to approximate $\rho^{\prime}_1$ and then construct a suitable function h in the subsequent Lemma 4.2.

Proof. Let $\bigcup_{\lambda}U_\lambda = (0,\infty)$ be any cover of $(0,\infty)$ with open intervals. There is a locally finite, smooth subordinate partition of unity, i.e. a sequence $(\chi_n)_{n\geq 1}\subset{\mathcal{C}}_c^\infty(0,\infty)$ such that ${\operatorname{supp}}\chi_n\subset U_{\lambda(n)}$ for some $\lambda(n)$ , $\chi_n\geq 0$ , and $\sum_{n=1}^\infty\chi_n(x) = {\unicode{x1D7D9}}_{(0,\infty)}(x)$ , such that for every compact set K and $x\in K$ the sum only consists of finitely many summands; see e.g. [Reference Rudin25, Theorem 6.20].

Choose $U_n \,:\!=\, \big(2^{n-1},2^{n+1}\big)$ , $n\in{\unicode{x2124}}$ . Since g is continuous and strictly positive, $\varepsilon_n \,:\!=\, \inf_{\overline U_n}g > 0$ . By Weierstrass’s theorem, we can approximate any continuous function f on $\overline U_n$ uniformly by a polynomial $p_n$ such that $\sup_{\overline U_n}|f-p_n|\leq\varepsilon_n$ . Therefore, we get

\begin{align*} \Bigg|f - \sum_{n\in{\unicode{x2124}}}p_n\chi_n\Bigg| = \Bigg|\sum_{n\in{\unicode{x2124}}}(f-p_n)\chi_n\Bigg| \leq \sum_{n\in{\unicode{x2124}}}|f-p_n|\chi_n \leq \sum_{n\in{\unicode{x2124}}}\varepsilon_n \chi_n \leq \sum_{n\in{\unicode{x2124}}} g \chi_n = g. \end{align*}

Since the sum $\sum_{n\in{\unicode{x2124}}} \chi_n = {\unicode{x1D7D9}}_{(0,\infty)}$ is locally finite, we see that $\sum_{n\in{\unicode{x2124}}} p_n\chi_n\in {\mathcal{C}}_c^\infty(0,\infty)$ .

Lemma 4.1 now leads to a smooth approximation h of $\rho_1$ as follows.

Proof. By assumption, $\rho^{\prime}_1$ exists and is continuous. Using Lemma 4.1 we can find a smooth function $u\in{\mathcal{C}}^\infty(0,\infty)$ such that $|u(x)-\rho^{\prime}_1(x)|\leq x$ for all $x\in (0, 1)$ .

(i) By construction, $u-\rho^{\prime}_1\in L^1_{\mathrm{loc}}({\unicode{x211D}}_+)$ , and so we may define

\begin{align*} h(x)\,:\!=\, - \int_{(x,1)} u(y)\mathrm{d} y + \int_{(0, 1)} (u(y)-\rho^{\prime}_1(y))\mathrm{d} y + \rho_1(1). \end{align*}

This choice results in

\begin{align*} |h(x)-\rho_1(x)| = \left|\int_{(0,x)} \left(u(y)-\rho^{\prime}_1(y)\right) \mathrm{d} y\right| \leq \int_{(0,x)} y\,\mathrm{d} y = \frac 12 x^2,\quad x\in (0, 1), \end{align*}

which implies that h and $\rho_1$ are asymptotically equivalent at zero, since $\rho_{1}(0{+})\neq 0$ .

(ii) Note that

(4.2) \begin{equation} \bar{\bar\nu}*h = N_1*h+N_2*h = N_1*(h-\rho_1)+ N_1*\rho_1 +N_2*h = N_1*(h-\rho_1)+ {\unicode{x1D7D9}}+N_2*h.\end{equation}

Thus the claim follows from Lemma 5.2(ii) together with the assumption $N_2(0{+})=0$ .

(iii) Using dominated convergence we have for the right derivative

Since $N_1\in L^1_{\mathrm{loc}}({\unicode{x211D}}_+)$ , the right-hand side tends to zero as $x\to 0$ , and thus

\begin{align*} \lim_{x\to 0}\partial_+(N_1*h)(x)=0; \end{align*}

a similar calculation for the left derivative shows that $\partial_-(N_1*h)=0$ . Further, combining Lemma 4.2(i) with the assumptions on $\rho_1$ shows that $|h(x)|<Cx^{\alpha-1}$ for $x\in(0, 1)$ . Thus the assumptions on $N_2$ and Lemma 5.2(ii) ensure that

(4.3) \begin{align}\nonumber |\partial(N_2*h)(x)|& = |(\partial N_2)*h(x)|\\ & \leq C_1C_2\int_{(0,x)}(x-y)^{-\alpha}y^{\alpha-1}\,\mathrm{d} y \leq C_1C_2 B(1-\alpha,\alpha) \end{align}

for $x\in(0, 1)$ and with $C_1,C_2,\alpha>0$ as in Theorem 4.1.

(iv) This follows from (iii) and the fact that $h\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ . Indeed,

\begin{align*} \lim_{x\to0+}|h*\partial\big(h*\bar{\bar\nu}\big)(x)| \leq \lim_{x\to0+} \left(\sup_{z\in(0, 1)}|\partial\big(h*\bar{\bar\nu}\big)(z)|\right) \int_{(0,x)}|h(y)|\mathrm{d} y = 0. \end{align*}

This completes the proof.

We are now ready to prove the main result of this section.

Proof of Theorem 4.1. It suffices to consider the case $q=0$ , as otherwise an exponential change of measure allows us to represent $W^{(q)}$ in terms of $W^{(0)}$ as stated in Section 2.3.

The function h from Lemma 4.2 satisfies the requirements of Theorem 3.3. Therefore, we have the representation

\begin{align*} W\,:\!=\, W^{(0)} ={\unicode{x1D7D9}}*h*\sum_{n=0}^\infty f^{*n},\quad q\geq 0, \end{align*}

with $f(x) \,:\!=\, c^{\prime\prime}h(x)-\partial\big(h*\bar{\bar\nu}\big)(x)$ for $x\in (0,\infty)$ ; moreover,

(4.4) \begin{align} W^{\prime}=h*\sum_{n=0}^\infty f^{*n}. \end{align}

By assumption, $\bar\nu\in{\mathcal{C}}^{k_0}(0,\infty)$ . Hence, $\bar{\bar\nu}\in{\mathcal{C}}^{k_0+1}(0,\infty)$ , and, since $h\in L^1_{{\mathrm{loc}}}({\unicode{x211D}}_+)$ , Lemma 5 shows that $h*\bar{\bar\nu}\in{\mathcal{C}}^{k_0+1}(0,\infty)$ . Again with Lemma 5.1 we see that $f^{*n}\in{\mathcal{C}}^{k_0}(0,\infty)$ for all $n\in{\unicode{x2115}}\setminus\{0\}$ . Thus, $W\in{\mathcal{C}}^{k_0+1}(0,\infty)$ follows if we can show that the series (4.4) converges in the space ${\mathcal{C}}^{k_0}(0,\infty)$ , i.e. locally uniformly with all derivatives up to order $k_0$ .

Lemma 4.2(i), together with the assumptions on $\rho_1$ , shows that there exist constants $C,\alpha>0$ such that $|h(x)|<Cx^{\alpha-1}$ for $x\in(0, 1)$ . As $\partial(\bar{\bar\nu}*h)$ is bounded on (0, 1) (see Lemma 4.2(iii)), there exist constants $\bar{C},\bar{\alpha}>0$ such that $|f(x)|<\bar{C}x^{\bar\alpha-1}$ for $x\in(0, 1)$ .

Hence, the assumptions of Corollary 5.1 are satisfied, and setting $m\,:\!=\,\lfloor 2/\bar\alpha\rfloor +1$ we have for all $k\leq k_0$ and $n\geq k_0m$

\begin{align*} \partial^k f^{*n}=\big(\partial f^{*m}\big)^{*k}*f^{*(n-km)}.\end{align*}

By Lemma 5.2(ii) we have $\partial^kf^{*(n+m)}(0)=0$ for all $k\leq k_0$ . Thus, we can use Lemma 5.1(i) to obtain

\begin{align*} \partial^k \big(h*f^{*(n+m)}\big) = h*\big(\partial f^{*m}\big)^{*k}*f^{*(n-(k-1)m)}.\end{align*}

Returning to the series (4.4), this means that for $N> (k_0+1)m$

(4.5) \begin{align} \partial^k \bigg(h*\sum_{n=0}^N f^{*n}\bigg) = \partial^k\bigg(h*\sum_{n=0}^{(k_0+1)m} f^{*n}\bigg) + h*\left(\partial f^{*m}\right)^{*k} *\!\!\!\!\!\!\!\! \sum_{{n=(k_0+1)m+1}}^{N} f^{*(n-km)}.\end{align}

The first term is independent of N. If we pull out $f^{*m}$ from the sum appearing in the second term, we see that

\begin{align*} h*\big(\partial f^{*m}\big)^{*k}*\!\!\!\!\!\!\!\! \sum_{{n=(k_0+1)m+1}}^{N} f^{*(n-km)} = h*\big(\partial f^{*m}\big)^{*k}*f^{*m}*\!\!\!\!\!\!\!\! \sum_{{n=k_0m+1}}^{N-m} f^{*(n-km)}.\end{align*}

The first three convolution factors can be bounded by

\begin{align*} S\,:\!=\,\sup_{x\leq z} \big|h*\big(\partial f^{*m}\big)^{*k}*f^{*m}(x)\big| < \infty \quad\text{for all $z>0$};\end{align*}

here we use Lemma 5.2 and the continuity of h, $\big(\partial f^{*m}\big)^{*k}$ , and $f^{*m}$ . Thus, with Hölder’s inequality,

\begin{align*} \left|h*\big(\partial f^{*m}\big)^{*k}*f^{*m}*\!\!\!\!\!\!\!\! \sum_{{n=k_0m+1}}^{N-m} f^{*(n-km)}(x)\right| \leq S \sum_{n=k_0m+1}^{N-m} \left\|f^{*(n-km)}\right\|_{L^1[0, z]}\end{align*}

for all $x\in[0,z]$ . Lemma 5.3 implies that the sum on the right-hand side converges uniformly for all $x\in [0,z]$ , and so the series

\begin{align*} \partial^k \bigg(h*\sum_{n=0}^\infty f^{*n}\bigg)\end{align*}

converges locally uniformly in $(0,\infty)$ . This proves $W\in{\mathcal{C}}^{k_0+1}(0,\infty)$ .

Assume now that $\kappa(\rho_1)=2$ . Lemma 4.2(i) implies that $\kappa(h)=2$ , as well. Thus, Lemma 5.2 gives $h^{*2}(0{+})=0$ . Plugging the definition of f into the series (4.4) shows that for $x>0$

\begin{align*} W^{\prime}(x)&=\lim_{N\to\infty} h*\sum_{n=0}^N \big(c^{\prime\prime}h-\partial\big(h*\bar{\bar\nu}\big)\big)^{*n}(x)\\ &=\lim_{N\to\infty} \sum_{n=0}^N \sum_{\ell=0}^n ({-}1)^\ell \binom n\ell (c^{\prime\prime}h)^{*(n-\ell)}*h*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*\ell}.\end{align*}

Each summand contains a factor of the form $h*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*\ell}$ for some $\ell\in{\unicode{x2115}}$ , and we will now see how smooth it is. The case $\ell=0$ is trivial, as $h\in{\mathcal{C}}^\infty(0,\infty)$ . For $\ell=1$ we have, by Lemma 5.1,

(4.6) \begin{align} h*\partial\big(h*\bar{\bar\nu}\big)=\partial\big(h^{*2}*\bar{\bar\nu}\big)-h=\partial\big(h^{*2}\big)*\bar{\bar\nu}-h \in{\mathcal{C}}^{k_0+1}(0,\infty).\end{align}

Finally, if $\ell\geq2$ , we use

\begin{align*} h*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*\ell}=h*\partial\big(h*\bar{\bar\nu}\big)*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*(\ell-1)}.\end{align*}

With (4.6), Lemma 4.2(iv), and Lemma 5.1 we obtain

\begin{align*} \partial\!\left(h*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*\ell}\right) = \partial\!\left(h*\partial\big(h*\bar{\bar\nu}\big)\right)*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*(\ell-1)} \in{\mathcal{C}}^{k_0}(0,\infty).\end{align*}

This shows that each summand of

\begin{align*} \sum_{n=0}^N \sum_{\ell=0}^n({-}1)^\ell \binom n\ell (c^{\prime\prime}h)^{*(n-\ell)}*h*\big(\partial\big(h*\bar{\bar\nu}\big)\big)^{*\ell}\end{align*}

is $(k_0+1)$ times continuously differentiable on $(0,\infty)$ . As in the first case we see that this smoothness in inherited by the series (4.4). Consequently, $W\in{\mathcal{C}}^{k_0+2}(0,\infty)$ , and the proof is complete.

4.2. Discussion

We begin the section with an example that illustrates Theorem 4.1.

Example 4.1. Denote by $\big(L^{(i)}_t\big)_{t\geq0}$ , $i=1,2$ , two independent spectrally negative Lévy processes with characteristic triplets $\big(c^{(i)},0,\widetilde\nu^{(i)}\big)$ , $i=1,2$ . The characteristic triplet of the sum $L_t\,:\!=\,L^{(1)}_t+L^{(2)}_t$ is then $\big(c^{(1)}+c^{(2)},\widetilde\nu^{(1)}+\widetilde\nu^{(2)}\big)$ . For $i=1,2$ , denote by $N_i$ the integrated tail function of the reflection $\nu^{(i)}$ of the Lévy measure of $L^{(i)}_t$ .

Let $N_1$ be completely monotone, and assume that there exist constants $C>0$ , $\alpha\in(0, 1) $ such that $N_1(x)\geq Cx^{-\alpha}$ for $x\in(0, 1)$ . By Corollary 3.1, $N_1$ admits a completely monotone resolvent $\rho_1\in{\mathcal{C}}^\infty(0,\infty)$ for which

\begin{align*} \rho_1(x) \frac{C}{1-\alpha}= x^{ \alpha-1 }\rho_1(x) \int_{(0,x)} Cy^{-\alpha}\,\mathrm{d} y \leq \int_{(0,x)} \rho_1(x-y) N_1(y)\mathrm{d} y =1 \end{align*}

for $x\in(0, 1)$ . Thus, $\rho_1(x)<\frac{1-\alpha}Cx^{\alpha-1}$ on (0, 1), and the condition (i) in Theorem 4.1 is met.

This shows that the smoothness of the scale functions $W^{(q)}$ of the sum $L_t$ depends on the smoothness of $N_2$ only, i.e. on the Lévy measure of $L^{(2)}_t$ . For example, $L^{(2)}_t$ could be such that $-L^{(2)}_t$ is a subordinator, and with Lévy measure $\nu^{(2)}$ such that $\bar\nu^{(2)}<C_2x^{-\alpha}$ on (0, 1) for some constant $C_2>0$ . The condition (ii) of Theorem 4.1 is then easily verified.

We conclude that in this case, if $\bar\nu^{(2)}\in {\mathcal{C}}^{k_0}(0,\infty) $ for $k_0\in{\unicode{x2115}}$ , then $W^{(q)}\in{\mathcal{C}}^{k_0+1}(0,\infty)$ , and even $W^{(q)}\in{\mathcal{C}}^{k_0+2}(0,\infty)$ , if additionally $\alpha\geq \frac12$ .

The following simple corollary of Theorem 4.1 shows that the implication

\begin{align*} \bar\nu\in{\mathcal{C}}^n(0,\infty) \implies W^{(q)}\in{\mathcal{C}}^{n+\kappa(\rho_1)}(0,\infty)\end{align*}

holds, if the 0-scale function of the compensated process is at least twice continuously differentiable and of bounded growth.

Both Theorem 4.1, and the two previously mentioned results from [Reference Chan, Kyprianou and Savov8] on the smoothness of scale functions of spectrally negative Lévy processes with either a Gaussian component or paths of bounded variation, require a growth condition of the type

\begin{align*} |f(x)|<Cx^{\alpha-1},\quad x\in(0, 1),\end{align*}

for some constants $C,\alpha>0$ on one of the occurring terms. Getting rid of this requirement seems to be the biggest challenge in a possible proof of Doney’s conjecture.

In this regard, the next corollary gives a hint as to how the growth condition on $\rho_1$ in Theorem 4.1 can be avoided by imposing stronger conditions on the component $N_2$ .

Proof. Without loss of generality we assume that $q=0$ and consider $W\,:\!=\,W^{(0)}$ . By Remark 4.1 we can assume $N_2(0{+})=0$ , which then yields that the function f in the series expansion (4.4) of W ^′ has to be replaced by

(4.7) \begin{equation} f(x)= -\partial\big(h*\big(\bar{\bar\nu}+c^{\prime\prime}\big)\big)(x). \end{equation}

Thus, the term c ^′′ h(x) vanishes owing to the assumptions on $N_2$ .

We can now follow the lines of the proof of Theorem 4.1 and observe that the growth condition on $\rho_1$ appears at exactly two places.

Firstly, in the proof of Lemma 4.2(iii) it appears in the argument (4.3), which can now be replaced by

\begin{align*} |\partial(N_2*h)(x)| = |\partial N_2*h(x)|\leq C \end{align*}

by Lemma 5.2. Thus, Lemma 4.2(iii) still holds true, and therefore, $\partial\big(h*\big(\bar{\bar\nu}+c^{\prime\prime}\big)\big)(x)$ is bounded on (0, 1). Note that $\bar{\bar\nu}+c^{\prime\prime}$ plays the role of $\bar{\bar\nu}$ ; see Remark 4.1.

Secondly, it is needed to apply Corollary 5.1 to the function f in the series expansion of W. But the requirements of Corollary 5.1 are already met, as f stays bounded in (0, 1) by the argument above.

Finally, we mention here that the close connection between scale functions and Volterra resolvents of the first kind as discussed in Section 3.3 allows for a smoothness result on the latter as well.

Proof. The assumptions on f ensure that $\partial^2 f(x)\mathrm{d} x$ is the reflection of the Lévy measure of a spectrally negative Lévy process with paths of unbounded variation, while the requirements for $\rho$ in (i) imply that the spectrally negative Lévy process with Laplace exponent

\begin{align*} \psi(\beta)=\beta^2{\mathcal{L}}[f](\beta) \end{align*}

fulfils the assumptions of Theorem 4.1. The assertion (i) now follows directly as $\rho=\partial W^{(0)}$ , while (ii) follows directly from Doney’s conjecture without further assumptions on $\rho$ .