and L 1∗ ¯ Π+qs = L 1sL ¯
Let us first motivate the approach. Taking into account the series representation we divide u
q
into two parts:
u
q
x =
N −1
X
n=0
−1
n
δ
n+1
1 ∗ ¯ Π + q
∗n
x + X
n≥N
−1
n
δ
n+1
1 ∗ ¯ Π + q
∗n
x =:
N −1
X
n=0
−1
n
δ
n+1
1 ∗ ¯ Π + q
∗n
x + φ
N ,q
x. The goal of the following Fourier analysis is to derive a convenient integral representation for
φ
N ,q
. For this sake first note that it follows as in the proof of Proposition 1 that
φ
N ,q
is the unique locally bounded solution of the following renewal type equation:
φ
N ,q
x = −1
N
δ
N +1
1 ∗ ¯ Π + q
∗N
x − 1
δ φ
N ,q
x ∗ ¯ Π + qx.
4.3 If we were allowed to turn convolution into multiplication in Laplace domain for Res
0, Equation 4.3 transforms into
L φ
N ,q
s = −1
N
δ
N +1
L 1sL ¯ Π + qs
N
− 1
δ L φ
N ,q
sL ¯ Π + qs.
4.4 Solving 4.4 for L
φ
N ,q
, leads to L φ
N ,q
s = −1
N
L 1sL ¯ Π + qs
N
δ
N +1
1 +
1 δ
L ¯ Π + qs
=: g
N ,q
s 4.5
whenever the quotient is well-defined. If furthermore we were able to verify integrability conditions needed for Laplace inversion we obtain an integral representation for
φ
N ,q
. We are now going to check what is needed to turn this formal approach into rigorous statements.
For the rest of this section we set δ = 1 in order to simplify the notation.
For the first step it is shown that indeed Equation 4.3 turns into Equation 4.4. A priori this is not clear due to the possible singularity of ¯
Π at zero.
Lemma 2. There is
λ ≥ 0 such that L φ
N ,q
solves 4.4 on { λ + iθ : λ ≥ λ
, θ ∈ R}.
Proof. To show that the first transformation can be carried out we show for s = λ + iθ , λ bounded
from below by some λ
, that L ¯ Π+qs and L φ
N ,q
s are well defined to deduce L ¯ Π+q
∗n
s = L ¯
Π+qs
n
, L ¯ Π+q∗φ
N ,q
s = L ¯ Π+qsL φ
N ,q
s, and L 1∗ ¯ Π+qs = L 1sL ¯
Π+qs. To validate Laplace transformation of ¯
Π + q
n
for λ large enough, note that we may choose λ
such that
Z
∞
e
−λ x
¯ Π + qx d x 1
4.6
484
which is possible as R
1
¯ Πx d x ∞ and lim
x→∞
¯ Πx = 0. It now follows directly from Fubini’s
theorem that iterated convolutions of ¯ Π + q turn into multiplication under Laplace transforms. We
now show that φ
N ,q
can be Laplace transformed for which we use the fact that
∞
X
n=N
L ¯ Π + q
∗n
s =
∞
X
n=N
L ¯ Π + qs
n
∞ to justify the change of summation and integration in the following:
L φ
N ,q
s = Z
∞
e
−s x ∞
X
n=N
−1
n
¯ Π + q
∗n
x d x =
∞
X
n=N
−1
n
Z
∞
e
−s x
¯ Π + q
∗n
x ds ∞. 4.7 Now it is only left to show L
φ
N ,q
∗ ¯ Π + qs = L φ
N ,q
sL ¯ Π + qs and further L 1∗ ¯
Π + qs =
L 1sL ¯ Π + qs. As ¯
Π + q and φ
N ,q
can be Laplace transformed, we obtain from 4.6 and 4.7 the bound
Z
∞
Z
∞
e
−st
φ
N ,q
t − x ¯ Π + qx
d t d x =
Z
∞
Z
∞
e
−λt−x
φ
N ,q
t − x e
−λx
¯ Π + qx
d t d x ∞ enabling us to apply Fubini’s theorem once more to obtain L
φ
N ,q
∗ ¯ Π + qs = L φ
N ,q
sL ¯ Π +
qs. The final identity L 1 ∗ ¯
Π + qs = L 1sL ¯ Π + qs follows from similar arguments noting
that |L 1s| ≤
1 λ
. The second step in our analysis consists of showing that the convolution Equation 4.4 can indeed
be solved in Laplace domain leading to Equation 4.5. The following lemma is stronger then the previous as we can show that g
N ,q
s is well defined for any s = λ + iθ with λ 0 even though a priori we do not know that g
N ,q
is the Laplace transform of φ
N ,q
.
Lemma 3. Suppose that Π is non-trivial, then g
N ,q
λ + iθ is well-defined for λ 0. Proof. To show that g
N ,q
s is well-defined at s ∈ C with positive real-part, it suffices to show that L ¯
Π + qs 6= −1. Let us assume that L ¯
Π+qs = −1 for some s
= λ + iθ
with λ
0. Without loss of generality way may assume
θ 6= 0 as otherwise the contradiction follows trivally. The assumption necessarily
implies that I mL ¯
Π + qs =
Z
∞
e
−λ x
sin θ
x ¯ Π + qxd x = 0.
4.8 As ¯
Π is decreasing, we see that Z
∞
e
−λ x
¯ Π + qx d x ∞.
4.9
485
Dividing the integral of the absolutely integrable function e
−λ x
¯ Π + qx sinθ
x into pieces of the length of one period of the sine function and applying Fubini’s theorem, we obtain
0 = Z
∞
e
−λ x
¯ Π + qx sinθ
xd x =
Z
∞ 2
πθ ∞
X
k=0
1
[2kπθ ,2k+1
πθ
xe
−λ x
¯ Π + qx sinθ
xd x =
∞
X
k=0
Z
2k+1 π
θ0 2k
π θ0
e
−λ x
¯ Π + qx sinθ
xd x. As e
−λ x
¯ Πx is strictly decreasing, unless ¯
Π + qx = 0 and ¯ Π + qx is non-increasing each
summand Z
2k+1 π
θ0 2k
π θ0
e
−λ x
¯ Π + qx sinθ
x d x must be non-negative and hence vanish as the total sum is zero. In particular, this implies that
¯ Πx + q = 0 for all x 0 so that for q 0 a direct contradiction occurs. For q = 0 the contradiction
occurs as Π was assumed to be non-trivial. Thus g s is well-defined.
Remark 5. If furthermore E[X
1
] ∞, then g
N ,0
is well-defined also on the imaginary axis. This follows from the same proof noting that in this case 4.9 holds as well for
λ = 0. Indeed as each term
R
2k+1 π
θ0 2k
π θ0
¯ Πx sinθ
x d x has to vanish we conclude that Π has to be concentrated on {2k πθ
}
k≥1
. On the other hand, in this case, as
ReL ¯ Πs
= Z
∞
e
−λ x
cos θ
x ¯ Πxd x
= X
k≥0
¯ Π
2k π
θ Z
2k+1 π
θ0 2k
π θ0
cos θ
xd x = 0, we see that ReL ¯
Πs 6= −1 and thus g
N ,q
s is well-defined. The third step of our derivation of an integral representation for u
q
is an inversion approach for g
N ,q
. We now briefly discuss the connection to Fourier transforms which is crucial for the inversion: for integrable functions f define for x ∈ R
F f x = Z
∞ −∞
e
−i x t
f t d t. Apparently, the Fourier transform F appears when evaluating the Laplace transform on the imagi-
nary line only. Defining the auxiliary function r
λ
x = e
−λx
¯ Π + qx
486
the simple connection is F r
λ
θ = L ¯ Π + qλ + iθ .
Taking into account this close connection of Laplace and Fourier transforms, classical Fourier inver- sion for
λ 0 gives the inversion formula also known as Bromwich integral φ
N ,q
x = 1
2 π
Z
∞ −∞
L φ
N ,q
λ + iθ e
λ+iθ x
d θ
= 1
2 π
e
λx
Z
∞ −∞
g
N ,q
λ + iθ e
i θ x
d θ
if g
N ,q
λ + iθ is absolutely integrable with respect to θ . To prove the needed integrability we start with a simple estimate.
Lemma 4. For any a 0 and y ≤ a the estimate ¯
Π y ≤ Ca, ǫ y
−βΠ+ε
holds for all ǫ 0 with
βΠ + ǫ 1. Proof. First note that by the definition of the Blumenthal-Getoor index
R
1
y
βΠ+ǫ
Πd y ∞ for any
ǫ 0. The claim follows from the simple observation that for any α 0 there is δ 0 such that for any
τ δ α ≥
Z
δ τ
y
βΠ+ǫ
Πd y ≥ τ
βΠ+ǫ
¯ Πτ − ¯
Πδ as ¯
Π is decreasing. Letting τ go to zero, we deduce lim sup
τ→0
τ
βΠ+ǫ
¯ Πτ ≤ α.
The need for Assumption A comes from the following lemma and its consequences.
Lemma 5. For any
λ 0 and ε 0 the following estimate holds: |L ¯
Πλ + iθ | ≤ C
1 λ
: | θ | ≤ 1,
C| θ |
βΠ+ε−1
: | θ | 1,
where C = C ǫ 0.
Proof. We estimate the imaginary and real part of L ¯ Π separately. For the imaginary part we first
487
estimate for θ 0:
I mL ¯ Πλ + iθ
= θ
−1
Z
∞
sin yr
λ
yθ d y = θ
−1 ∞
X
k=0
Z
2k+1π 2k
π
r
λ
yθ − r
λ
y + πθ sin yd y ≤ θ
−1
π
∞
X
k=1
r
λ
2kθ − r
λ
2k + 2θ + θ
−1
Z
π
r
λ
yθ − r
λ
y + πθ d y ≤ θ
−1
Z
π
r
λ
yθ d y ≤ Cθ
βΠ+ǫ−1
Z
π
y
−βΠ+ǫ
d y = C θ
βΠ+ǫ−1
, where we have used Lemma 4 and that r
λ
is decreasing in the last inequality. Unfortunately, this uni- form in
λ upper bound is not suitable for all θ as the constant of Lemma 4 explodes as θ approaches zero. To circumvent this problem we derive a different upper bound that works everywhere equally
well but is not uniform in λ:
I mL ¯ Πλ + iθ
≤ θ
−1
Z
∞
r
λ
yθ d y = θ
−1
Z
θ
r
λ
yθ d y + θ
−1
Z
∞ θ
r
λ
yθ d y ≤ θ
−1
Z
θ
¯ Π yθ d y + θ
−1
¯ Π1
Z
∞ θ
e
− yλθ
d y ≤ θ
−1
C Z
θ
yθ
−βΠ+ǫ
d y + θ
−1
¯ Π1
θ λ
= C + C 1
λ ,
where we again used Lemma 4 but now y θ does not explode for small θ as we only integrate up
to θ . Having an estimate for positive θ we note that I mL ¯
Πλ + iθ as a function of θ is odd to deduce that
|I mL ¯ Πλ + iθ | ≤
C
1 λ
: | θ | ≤ 1,
C| θ |
βΠ+ǫ−1
: | θ | 1.
4.10
488
Similarly, we estimate the real part ReL ¯
Πλ + iθ =
Z
∞
cos θ yr
λ
yd y
= Z
π 2
cos yr
λ
yθ
−1
d y + θ
−1 ∞
X
k=1
Z
4k+3
π 2
4k+1
π 2
r
λ
yθ
−1
− r
λ
y + πθ
−1
cos yd y
≤ Z
π 2
r
λ
yθ
−1
d y ≤ C|θ |
βΠ+ǫ−1
for large | θ | and precisely as above for small |θ |. This finishes the proof of the lemma.
The upper bound can now be used to derive the necessary integrability of g
N ,q
.
Lemma 6. For arbitrary integer N larger than 0 and any λ 0 we have
Z
∞ −∞
g
N ,q
λ + iθ d
θ ∞ 4.11
for g
N ,q
defined in 4.5. Proof. As we have already found a good upper bound for L ¯
Πs in the previous lemma, it suffices to show that the denominator
p λ + iθ = 1 + L ¯
Π + qλ + iθ is bounded away from zero. In Lemma 3 we have shown that p
λ + iθ has no zeros for λ ≥ 0 and, hence, by continuity of p it suffices to show that as |
θ | tends to infinity pλ + iθ stays bounded away from zero. To this end it suffices to note that from Lemma 5
lim
|θ |→∞
L ¯ Π + qλ + iθ = lim
|θ |→∞
L ¯ Πλ + iθ +
q λ + iθ
= 0. Using the fact |L
1 λ + iθ | = |1λ + iθ | ≤ min
1 λ
,
1 |θ |
for λ 0, we employ Lemma 5 to obtain the upper bound
g
N ,q
λ + iθ ≤ C|L 1λ + iθ ||L ¯
Π + qλ + iθ |
N
= C 1
λ + iθ L ¯
Πλ + iθ + q
λ + iθ
N
≤ C
′ 1
λ
N +1
: | θ | ≤ 1,
C
′
|θ |
N βΠ+ǫ−1−1
: | θ | 1.
4.12 The right hand side is integrable in
θ as by assumption βΠ 1 and ǫ can be chosen sufficiently small so that N
βΠ + ǫ − 1 0. We are now in a position to derive the Laplace inversion representations for u
q
. 489
Proof of Proposition 2: For λ ≥ λ
, λ
satisfying R
∞
e
−λ x
¯ Π + qx d x 1 we can directly follow
the strategy explained before Lemma 2. The proof then follows directly from the definition of φ
N ,q
and Laplace inversion justified by Lemmas 2, 3, and 6. The proof of the proposition is complete if we can show that for arbitrary 0
λ λ Z
Γλ
e
s x
g
N ,q
s ds = Z
Γλ
e
s x
g
N ,q
s ds, where Γ
λ = {λ + iθ : θ ∈ R}. Laplace transforms are analytic see for instance Theorem 75.2 of [K88] and g
N ,q
has no singularity for λ 0 by Lemma 3, hence, Cauchy’s theorem applied to the
closed contour formed by the pieces Γλ
∩ {|θ | ≤ R}, Γλ ∩ {|θ | ≤ R},
ΦR = s : s = r + iR, r ∈ [λ, λ
] ,
˜ ΦR =
s : s = r − iR, r ∈ [λ, λ ]
, taken with the right orientation implies the claim. Note that the integrals over the horizontal pieces
vanish as R tends to infinity:
lim
R→∞
Z
ΦR
e
s x
g
N ,q
sds ≤ C e
λ x
λ − λ lim
R→∞
|R|
N βΠ+ǫ−1−1
, where we have used 4.12 for |
θ | 1 to estimate |g
N ,q
s| for R big enough. Choosing ǫ small enough so that
βΠ + ǫ − 1 0, the right hand side tends to zero. The same argument shows that the integral over ˜
ΦR vanishes. Proof of Corollary 2: As remarked after the corollary, the arguments of [CKS10] will not be repeated.
Instead, under Assumption
A we prove the Laplace inversion representation and deduce from this
the series representation of [CKS10]. We first show that the right and left derivatives of u
q
exist and are given by the representation of the theorem. First, right and left derivatives of the finite sum in 3.3 exist by termwise differentiating
the finite sum and using that
1 ∗ ¯ Π + q
∗n
x = R
x
¯ Π + q
∗n
y d y is differentiable from the left and the right with derivative
¯ Π + q
∗n
x− resp. ¯
Π + q
∗n
x+. As iterated convolutions are continuous, only the first summand is not everywhere differentiable.
To see that the integral is differentiable at x and to deduce the integral representation of u
q ′
note that
d d x
e
λx
1 2
π Z
∞ −∞
e
i θ x
g
N ,q
λ + iθ dθ = e
λx
1 2
π Z
∞ −∞
λ + iθ e
i θ x
g
N ,q
λ + iθ dθ = e
λx
1 2
π Z
∞ −∞
e
i θ x
h
N ,q
λ + iθ dθ . The differentiation under the integral is justified by dominated convergence and the upper bound
d d x
e
i θ x
g
N ,q
λ + iθ =
θ g
N ,q
λ + iθ ≤
C| θ |
1 λ
N +1
: | θ | ≤ 1,
C| θ |
N βΠ+ǫ−1
: | θ | 1,
490
derived in 4.12 which is integrable in θ for sufficiently small ǫ by our choice of N .
As a second step we now derive the pointwise series representation from the Laplace transform representation of u
q ′
. As for 4.12 we obtain the upper bound h
N ,q
λ + iθ ≤ C|L ¯
Πλ + iθ |
N
≤ C
1 λ
N
: | θ | ≤ 1,
C| θ |
N βΠ+ǫ−1
: | θ | 1,
4.13 for arbitrary
ǫ 0. To prove the corollary it suffices to show that for N tending to infinity, the Laplace inversion integral
e
λx
1 2
π Z
∞ −∞
e
i θ x
h
N ,q
λ + iθ dθ vanishes for fixed x
0 and λ 0. With our choice of ǫ, i.e. βΠ + ǫ − 1 0, 4.13 implies pointwise convergence e
i θ x
h
N ,q
λ + iθ → 0. We are done if we can justify the change of limit and integration. This comes from the uniform for N ≥ 2[
βΠ + ǫ − 1
−1
] + 2 in θ integrable upper bound
e
i θ x
h
N ,q
λ + iθ ≤
C : |
θ | ≤ 1, C|
θ |
−2
: | θ | 1,
and the dominated convergence theorem.