The class of slowly varying functions contains members with rather diverse limiting behavior for x→ ∞. There are functions converging to some positive constant, diverge (slow enough) for instance f(x) = log(x), as well as functions converging to zero (slow enough) likef(x) = 1/log(x). In order to impose some structure on this class we quantify the convergence rate ofL(tx)/L(t) to 1 as t→ ∞. More precisely, considering someL∈RV0, we seek for some positive function ˜A with limt→∞A(t) = 0 such that˜
exists for anyx >0 and is not constant. It turns out that ˜Ais necessarily regularly varying with index ρ≤0 and that only specific limits ψ are possible. This leads to the following definition.
Definition 2.3 (Second-order slow variation).
A slowly varying function L is said to be second-order slowly varying if there exists an auxiliary function A with |A| ∈RVρ (ρ≤0), such that
where for ρ= 0 the right-hand side of (2.26) reads as log(x).
We show that|A| ∈RVρ as well asψ(x) = (xρ−1)/ρfollows from (2.25) following the arguments in the proof of Theorem 1.9 in Geluk and de Haan (1987).
Since ψ is not constant there exists some x0 > 0 such that ψ(x0) 6= 0. Hence we can choose ˜A(t) := (L(tx0)/L(t)−1)/ψ(x0). In particular, we may assume without loss of generality ˜A to be measurable. Moreover, we have for anyx, y >0
t→∞lim
L(txy)−L(t)
L(t) ˜A(t) − L(ty)−L(t)
L(t) ˜A(t) = lim
t→∞
L(txy)−L(ty)
L(t) ˜A(t) · L(ty) ˜A(ty) L(ty) ˜A(ty)
= lim
t→∞
L(txy)−L(ty)
L(ty) ˜A(ty) · L(ty)
L(t) · A(ty)˜ A(t)˜ , which yields
ψ(xy)−ψ(y) =ψ(x) lim
t→∞
A(ty)˜
A(t)˜ . (2.27)
This implies the existence of φ(y) := limt→∞A(ty)/˜ A(t) for any˜ y > 0. In particular, note that φ is measurable since ˜A is measurable. Hence, we conclude that eitherφ(y) =yρ for some ρ ∈R or φ(y) = 0 for y > 0. However,φ(y) = 0 implies a constant limit ψ contradicting the assumption above. Moreover, for A˜∈ RVρ and ρ >0 we have limt→∞A(t) =˜ ∞. Therefore ˜A ∈RVρ with ρ ≤ 0 follows. From (2.27) we obtain the functional equation
ψ(xy)−ψ(y) =ψ(x)yρ for all x, y >0 (2.28) where ρ ≤ 0. If ρ = 0, this again yields Cauchy’s functional equation in its additive form. It follows ψ(y) =c·log(y) for somec6= 0 andy >0.
Ifρ <0, interchanging xand y in (2.28) and considering the difference of (2.28) and the resulting relation yields
ψ(x)(1−yρ) = ψ(y)1−xρ for all x, y >0.
This implies that ψ(x)/(1−xρ) is constant for any x > 0 and x 6= 1. Hence we can conclude that ψ(x) := c(xρ −1)/ρ for x > 0 with c 6= 0 and ρ < 0.
Note, that dividing by ρ allows a smooth transition from ρ < 0 to ρ = 0, since limρ→0(xρ −1)/ρ = log(x). Moreover, it should be mentioned that we have subsumed the constant c 6= 0 and the function ˜A ∈ RVρ, measuring the convergence rate ofL(tx)/L(t) to 1 ast→ ∞, into the functionAin the definition above. Since the sign ofcis unknown, we can only conclude that |A| ∈RVρ. If A possesses eventually a positive signL(tx)/L(t) approaches 1 from above, otherwise from below.
Remark. Goldie and Smith (1987) introduced the class of slowly varying functions of second-order terming it slow variation with remainder (see also Bingham et al.
1987 and Aljanˇci´c et al. 1974). Our definition corresponds to the one stated in Goldie and Smith (1987), denoted by (SR2). It should be mentioned that in contrast to ERV the auxiliary function A˜ of a slowly varying function of second-order has a clear meaning as a rate function tending to zero and possessing a ultimately constant sign indicating the ’direction’ from which L(tx)/L(t) approaches 1.
We can easily generalize the definition of slow variation of second-order to regular variation of second-order taking into account that anyf ∈RVα can be represented asf(x) = xαL(x), where L∈RV0. This gives rise to the following definition.
Definition 2.4 (Second-order regular variation).
Suppose f ∈RVα and there exists some auxiliary function A with limt→∞A(t) = 0 and ultimately constant sign, such that
t→∞lim
f(tx) f(t) −xα
A(t) =xαxρ−1
ρ (2.29)
holds for all x > 0 with ρ ≤ 0. Then, we call f second-order regularly varying with second-order parameter ρ. Notation: f ∈ 2RVα,ρ. In particular, we have
|A| ∈RVρ. For ρ= 0 the right hand side of (2.29) reads as xαlog(x).
Remark. Note that2RV0,ρ is the class of slowly varying functions of second-order.
In particular, it turns that 2RV0,0 is closely related to the class Π introduced by de Haan (1970). A quite extensive treatment of Π can be found in de Haan and Ferreira (2006).
Example 2.2. A simple example for f ∈2RV0,0 is given byf(t) := log(t), where a possible auxiliary function is given by A(t) = 1/log(t). Note that we also have f ∈Π with possible auxiliary function a(t) = 1, illustrating thatlimt→∞a(t)→0 does not necessarily hold in Π (or ERV0).
Example 2.3. As an example for f ∈ 2RV0,ρ (ρ 6= 0) we consider functions of the form f(x) := D1 +D2xρ with D1 > 0 and D2 6= 0. Note that since limx→∞f(x) = D1 >0 we have f ∈RV0. Setting A(t) := ρD1−1D2tρ, we obtain for any x >0
t→∞lim
f(tx) f(t) −1
A(t) = lim
t→∞
D1+D2(tx)ρ−D1−D2tρ
(D1+D2tρ)ρD−11 D2tρ = xρ−1 ρ .
Hence f ∈ 2RV0,ρ with possible auxiliary function A. Note that |A| ∈RVρ and limt→∞A(t) = 0.
Next, we detail some relations betweenRV,ERV and 2RV.
Theorem 2.9. Let f :R+ →R be some measurable, eventually positive function.
Then, we have the following relations:
(i) For α >0 we have f ∈ERVα ⇐⇒ f ∈RVα.
(ii) Iff ∈ERVα, then f ∈2RV0,α with an ultimately positive auxiliary function A. The converse is also true.
Proof. We start with (i). For f ∈ERVα with α >0 we obtain by Theorem 2.7 thatf ∈RVα. For the converse statement suppose f ∈RVα with α >0. Then, for any x >0,
t→∞lim f(tx)
f(t) =xα ⇐⇒ lim
t→∞
f(tx) f(t) −1
α = lim
t→∞
f(tx)−f(t)
αf(t) = xα−1 α . Since, by definition,f is eventually positive, there exists some positive function a with a(t)∼αf(t) such that
t→∞lim
f(tx)−f(t)
a(t) = xα−1 α . Hence, f ∈ERVα.
Next, we prove (ii) for α = 0. Therefore we require the following auxiliary result.
Corollary 2.1 (Corollary B.2.13, de Haan and Ferreira (2006)).
If f ∈ERV0, then limt→∞f(t) =f(∞)≤ ∞ exists. If the limit is infinite, then f ∈RV0. If the limit is finite, then f(∞)−f(t)∈RV0. Moreover,
a(t) =o(f(t)) as t → ∞.
and when f(∞)<∞,
a(t) =o(f(∞)−f(t)) as t → ∞.
According to Corollary 2.1 limt→∞f(t) :=f(∞)≤ ∞ exists if f ∈ERV0. Since f is ultimately positive we deduce 0≤ f(∞)≤ ∞. Additionally, if f(∞) <∞, f(∞)−f(t) ∈ RV0 follows. Hence, we can exclude f(∞) = 0, since otherwise
−f ∈RV0, contradicting the assumption that f is ultimately positive.
Thus, for f ∈ ERV0 and f ultimately positive, we have 0 < f(∞) ≤ ∞. For 0 < f(∞) < ∞ we immediately obtain f ∈ RV0. For f(∞) = ∞, we have f ∈RV0 by Corollary 2.1. Further, we have for any x >0
t→∞lim
f(tx)−f(t)
a(t) = log(x).
This is equivalent to
t→∞lim
f(tx) f(t) −1
f(t)−1a(t) = log(x).
Hence, f ∈2RV0,0 with a possible auxiliary function A(t) :=f(t)a(t). Note that A∈ RV0, since f, a∈ RV0. For the converse note that due to f ∈2RV0,0 with auxiliary functionA we have
t→∞lim
f(tx) f(t) −1
A(t) = log(x).
This yields
t→∞lim
f(tx)−f(t)
f(t)A(t) = log(x).
Sincef and A are both ultimately positive, there exists some positive function a with a(t)∼f(t)A(t) such that
t→∞lim
f(tx)−f(t)
a(t) = log(x).
Thusf ∈ERV0 follows.
Next, we prove (ii) for α <0. For f ∈ERVα with α <0 and positive auxiliary functiona∈RVα we have
t→∞lim
f(tx)−f(t)
a(t) = xα−1
α . (2.30)
This implies limt→∞f(tx)−f(t) = 0 or equivalently limt→∞f(tx)/f(t) = 1, i.e.
f ∈RV0. Moreover, we conclude from (2.30) that
t→∞lim
f(tx) f(t) −1
f(t)−1a(t) = xα−1
α for any x >0.
Observe that due to f ∈ RV0 and a ∈RVα with α < 0 we get limt→∞A(t) = 0 for A(t) := f(t)−1a(t). In particular,|A| ∈ RVα follows. Moreover A, possesses eventually a constant sign, sinceais a positive function andf possesses eventually a constant sign due to slow variation. Hence,f ∈2RV0,α with auxiliary function A.
The converse is similar. Any f ∈2RV0,α satisfies
t→∞lim
f(tx) f(t) −1
A(t) = xα−1
α for any x >0,
whereAis an auxiliary function with eventually constant sign satisfying limt→∞A(t) = 0 and|A| ∈RVα. Simple rearrangements yield
t→∞lim
f(tx)−f(t)
f(t)A(t) = xα−1
α for any x >0.
Moreover, since f ∈ RV0 and |A| ∈ RVα with α < 0, there exists a positive auxiliary functiona(t)∼f(t)A(t) such that a∈RVα. Hence f ∈ERVα follows.
Remark. Note that for f ∈ ERVα with α ≤ 0, the role of the parameter α is twofold. It shows that the underlying function is slowly varying, while additionally it characterizes the rate of convergence of f(tx)/f(t) towards one ast→ ∞. Thus, it is essentially a second-order parameter. In contrast, for f ∈ERVα with α >0, we can only deduce that f ∈RVα, i.e. that f(tx)/f(t) tends to xα as t→ ∞. In particular, observe that in this case no information about the convergence rate of f(tx)/f(t) towardsxα is available. Hence α is a first-order parameter in this case.
Thus, one can conclude that ERV contains rather asymmetric information about its members such that the parametrization of this class with only one parameter
α∈R might be confusing.
In contrast, the class 2RVα,ρ exhibits a clear separation between parameters of first- and second-order and therefore offers a more consistent parametrization.
As demonstrated by the last example ρ measures the rate of convergence of L(tx)/L(t) towards 1, whereL∈RV0results from the decompositionf(x) = xαL(x) for anyf ∈RVα,ρ. It may happen, thatL(tx)/L(t) converges to 1 faster than any negative power oft, i.e. for any ρ <0 we have
t→∞lim
L(tx) L(t) −1
t−ρ = 0.
In contrast, iff ∈2RV0,0, then f(tx)/f(t)→1 with a rate of convergence slower than any polynomial rate.
Next, we establish weighted uniform approximation forf ∈2RVα,ρ. This requires a slight modification of the auxiliary function.
Theorem 2.10 (Weighted uniform approximation for f ∈2RVα,ρ).
Let f ∈2RVα,ρ with α ∈R, ρ≤0 and some auxiliary function A. If ρ= 0, then
Proof. We follow the arguments of Cheng and Jiang (2001), who established a similar weighted uniform approximation for f ∈ERVα. Note that it is sufficient to prove the assertion for f ∈2RV0,ρ.
For ρ = 0, we have f ∈ ERV0. Therefore, according to Theorem B.2.12 in de Haan and Ferreira (2006), there exists someϕ∈RV0 such that
f(t) = ϕ(t) + Z t
t0
ϕ(s)ds
s . (2.31)
Further, setting A0(t) :=ϕ(t)(f(t))−1 yields
f(tx)
Moreover, note that forF(t) := 1t Rt
Thusϕ satisfies (2.31). In particular, we obtain A0(t) = 1− After some simple rearrangements, we get
|ρ|