Master Dissertation

where
and
k ∗ (x) =
f ∗ (x) = k ∗ x
(x) +
a∗ h ∗ (s)ds,
x+1
f
x
∗ (x) − f ∗ a∗ +1
(s)ds +
a∗ h ∗ (x) = f ∗ (x + 1) − f ∗ (x).
Note that h∗ is continuous, since f ∗ is continuous.
Now (6) implies
f(x) = g(x) + f ∗ (x) = g(x) + g ∗ x
(x) +
a∗ Finally, let g ∗ (x) := g(x) + k ∗ (x).
f ∗ (s)ds
h ∗ (s)ds
Now we are finally ready to prove the representation theorem.
Proof of Theorem 6.2. Let f(x) := ln r(e−x ). By Lemma 6.7 f has the
following representation
That is
ln(r(e −x )) = g ∗ x
(x) +
a∗ ln r(t) = g ∗ (− ln t) +
− ln t
a ∗
h ∗ (s)ds.
h ∗ (s)ds.
Letting ν(t) := g ∗ (− ln(t)), substituting s := − ln t ′ and h(t ′ ) := h ∗ (− ln t ′ ),
Thus,
where t1 = e −a∗
t
ln r(t) = η(t) −
e−a∗ h(t ′ )
dt′
t ′
t1 h(t
r(t) = exp(η(t) +
t
′ )
t ′ dt′ ),
as claimed.
Note that the representation of 6.2 only holds for t < t1 but since we are
only concerned with the limit t → 0 we will usually not have any problem
using the representation theorem. The following theorem will be important
later when we shall split distributions.
Corollary 6.8. Let ρ be a regularly varying function and ɛ > 0, then
there exist some S0(ɛ) and non-negative constants C and C ′ such that
C ′ t ω+ɛ ≤ ρ(t) ≤ Ct ω−ɛ , for 0 ≤ t ≤ s0(ɛ). (6.11)
47