Electronic Journal of Qualitative Theory of Differential Equations Proc. 7th Coll. QTDE, 2004, No. 1 1-32; http://www.math.u-szeged.hu/ejqtde/

ALMOST SURE SUBEXPONENTIAL DECAY RATES OF SCALAR ITÔ-VOLTERRA EQUATIONS

JOHN A. D. APPLEBY

Abstract. The paper studies the subexponential convergence of solutions of scalar Itô-Volterra equations. First, we consider linear equations with an instantaneous multiplicative noise term with intensity σ. If the kernel obeys

lim_{t→∞} k′(t)/k(t) = 0,

and another nonexponential decay criterion, and the solution X_{σ} tends to zero as t → ∞, then

lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} = 1 − Λ(|σ|), a.s.

where the random variable Λ(|σ|) → 0 as σ → ∞ a.s. We also prove a decay result for equations with a superlinear diffusion coefficient at zero. If the deterministic equation has solution which is uniformly asymptotically stable, and the kernel is subexponential, the decay rate of the stochastic problem is exactly the same as that of the underlying deterministic problem.

1. Introduction

In Appleby and Reynolds [2] the asymptotic stability of the scalar deterministic equation

(1)  x′(t) = −ax(t) + ∫_{0}^{t} k(t − s)x(s) ds, t ≥ 0

is studied when k is continuous, positive and integrable and k obeys the nonexponential decay criterion

(2)  lim_{t→∞} \frac{k′(t)}{k(t)} = 0.

If the zero solution of (1) is asymptotically stable, then k must be integrable and we have

lim inf_{t→∞} \frac{|x(t)|}{k(t)} ≥ \frac{|x(0)|}{a(a − ∫_{0}^{∞} k(s) ds)}.

Date: 27 September 2003. 1991 Mathematics Subject Classification. Primary: 34K20, 34K50, 60H10. Key words and phrases. almost sure exponential asymptotic stability, Itô-

Volterra equations.

This paper is in final form and no version of it will be submitted for publication elsewhere.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 1 In Appleby and Reynolds [5] a lower bound is found on the decay rate of a stochastic version of (1). For the linear equation

(3) dX(t) = (−aX(t) + ∫_{0}^{t} k(t − s)X(s) ds) dt + σX(t) dB(t),

the methods of [2] are extended to establish an almost sure lower bound on the decay rate of solutions of (3) under a weaker hypothesis on the kernel k than (2), namely

(4)  lim_{t→∞} sup_{0≤s≤T} |\frac{k(t − s)}{k(t)} − 1| = 0, for all T > 0.

It was shown in [5] if (3) has a solution which tends to zero on a set A of positive probability, then

lim sup_{t→∞} \frac{|X(t)|}{k(t)} = ∞, a.s. on A.

In this paper, we seek to use the approach of [5] to impose a sharper lower bound on the decay rate by reimposing the condition (2) on the kernel k. Then, if X(t) → 0 on a set A of positive probability, we prove for any fixed ε > 0 that

lim sup_{t→∞} \frac{|X(t)|}{k(t)t^{σ^{2}/(σ^{2}+2a)−ε}} = ∞, a.s. on A.

This result has an important corollary for solutions of (3). When a > ∫_{0}^{∞} k(s) ds, and k decays to zero polynomially according to

lim_{t→∞} \frac{log k(t)}{log t} = −α,

for some α > 1, then almost sure decay rate of the solution as the noise intensity becomes arbitrarily large is approximately tk(t), in the sense that

lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} = 1 − Λ(|σ|), a.s.

where Λ is a bounded nonnegative random variable with

lim_{|σ|→∞} Λ(|σ|) = 0 a.s.

In the second half of the paper, we concentrate on understanding the asymptotic behaviour of scalar Itô-Volterra equations where the statedependent diffusion term is a nonlinear function of the current state. Specifically, we consider the equation

dX(t) = (−aX(t) + ∫_{0}^{t} k(t − s)X(s) ds) dt + σ(X(t)) dB(t).

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 2 Intuitively, we might expect the linearisation of this equation to determine the asymptotic behaviour of solutions, and in terms of the conditions required to guarantee a.s. asymptotic stability, it suffices to study the stability of the deterministic linear equation. When the kernel is subexponential, the relationship between the size of the diffusion term close to zero and the speed at which the subexponential kernel decays seems to play a role in determining whether the solutions are a.s. subexponential. For instance, if k is regularly varying at infinity with index −α < −1, and σ(x) ∼ C|x|^{β} for some β > 1, it is sufficient to have α > 1 + (2(β − 1))^{−1} to ensure that the solution is a positive subexponential function, with

lim_{t→∞} \frac{X(t)}{k(t)} = \frac{∫_{0}^{∞} X(s) ds}{a − ∫_{0}^{∞} k(s) ds}.

However, if k is a subexponential function which decays more quickly, obeying

lim_{t→∞} \frac{log k(t)}{log t} = −∞

(and another technical condition), there does not seem to be such a restriction on the size of β. We explore and comment upon these questions at greater length in Sections 5-8.

2. Main Results for the Linear Problem

In this paper, (B(t))_{t≥0} is a standard one-dimensional Brownian motion on a complete filtered probability space (Ω, F, (F^{B}(t))_{≥0}, ℙ), where the filtration is the natural one, viz., F^{B}(t) = σ(B(s) : 0 ≤ s ≤ t). When almost sure events are referred in this paper, they are always ℙ-almost sure. Consider the scalar (stochastic) Itô-Volterra equation

(5) dX(t) = (−aX(t) + ∫_{0}^{t} k(t − s)X(s) ds) dt + σX(t) dB(t),

for t ≥ 0, where a and σ ≠ 0 are real constants. There is no loss of generality incurred by assuming that σ > 0. The kernel satisfies

(6)  k(t) ≥ 0, k ∈ L^{1}(0, ∞), k ∈ C[0, ∞).

As (5) is linear, we may assume X(0) = 1 without loss. The fact that (5) has a unique strong solution follows from, for example, Theorem 2E of Berger and Mizel [6].

To ensure that k is not exponentially integrable, we impose, as in [2], the following additional condition:

(7)  k ∈ C^{1}[0, ∞), k(t) > 0 for t ≥ 0, lim_{t→∞} \frac{k′(t)}{k(t)} = 0.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 3 Consequently, for every γ > 0,

(8)  lim_{t→∞} k(t)e^{γt} = ∞.

The last condition of (7) implies (4), as was pointed out in [2].

We now state the main result of the paper concerning the linear equation (5), and comment upon it.

Theorem 1. Let k satisfy (7), and σ ≠ 0. Suppose that the unique strong solution of (5) satisfies lim_{t→∞} X(t) = 0 on a set A of positive probability. Then, for every ε > 0

(9)  lim sup_{t→∞} \frac{|X(t)|}{k(t)t^{\frac{1}{1+2a/σ^{2}} −ε}} = ∞ a.s. on A.

Theorem 1 has an important corollary. In order to state it, we first recall the definition of a subexponential function, introduced in [3].

Definition 2. Let k ∈ C(ℝ^{+}; ℝ^{+})∩L^{1}(ℝ^{+}), with k(t) > 0 for all t ≥ 0. We say that k is a positive subexponential function if

lim_{t→∞} \frac{(k ∗ k)(t)}{k(t)} = 2 ∫_{0}^{∞} k(s) ds, (US1)

(US2)  lim_{t→∞} \frac{k(t − s)}{k(t)} = 1 for each fixed s > 0.

The class of positive subexponential functions is denoted by U.

A discussion of this class in contained in [3]. Note however that it contains for example, all positive and integrable functions which are regularly varying at infinity, as well as functions positive functions which obey k(t) ∼ Ce^{−t^{α}}, as t → ∞, for some C > 0 and α ∈ (0, 1). Also observe that a function which obeys (7) satisfies (US2) above.

In Appleby [1] it is shown that a > ∫_{0}^{∞} k(s) ds implies X(t) → 0 as t → ∞ a.s. In Appleby and Reynolds [4], it is shown that when k is a subexponential function, then

lim sup_{t→∞} \frac{X(t)}{k(t)t^{1+ε}} = 0, a.s.

Therefore, if k decays to zero polynomially, in the sense that

(10)  lim_{t→∞} \frac{log k(t)}{log t} = −α,

for α > 1, then almost sure decay rate of the solution as the noise intensity increases is approximately tk(t). We make this precise in the following theorem.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 4 Theorem 3. Suppose that k is a subexponential kernel which obeys (7) and (10) and suppose the zero solution of (1) is uniformly asymptotically stable. If the unique nontrivial strong solution of (5) is denoted by X_{σ}, then

(11)  lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} = 1 − Λ(|σ|), a.s.

where Λ(|σ|) is a bounded nonnegative random variable with

lim_{|σ|→∞} Λ(|σ|) = 0 a.s.

Therefore, the solution decays to zero t times more slowly than the deterministic solution as the noise intensity increases.

This mimics a result obtained in [4] for the subclass of subexponential kernels called superpolynomial kernels, which decay to zero more quickly than a polynomial in the sense that

lim_{t→∞} \frac{log k(t)}{log t} = −∞.

In [4] it is shown for this class that

lim sup_{t→∞} \frac{log X(t)}{log k(t)} = 1, a.s.

3. Proof of Theorem 1

In order to prove Theorem 1, we first need a technical result on the asymptotic behaviour of a scalar diffusion process. Introduce the process \overline{Y}_{ε} = \{Y_{ε}(t) : t ≥ 0\} which is the unique strong solution of the stochastic differential equation

(12)  d\overline{Y}_{ε}(t) = (1 − (a + ε)\overline{Y}_{ε}(t)) dt + σ\overline{Y}_{ε}(t) dB(t)

where \overline{Y}_{ε}(0) = 1.

Lemma 4. Let ε > 0 and a + σ^{2}/2 > 0. Then the process \overline{Y}_{ε} given by (12), with \overline{Y}_{ε}(0) = 1, obeys

(13)  lim sup_{t→∞} \frac{\overline{Y}_{ε}(t)}{t^{1/(1+2(a+ε)/σ^{2})}} = ∞, a.s.

Proof. A scale function p for the strictly positive process \overline{Y}_{ε} is

p(x) = e^{−2/σ^{2}} ∫_{1}^{x} e^{\frac{2}{σ^{2}} \frac{1}{ξ}} ξ^{\frac{2(a+ε)}{σ^{2}}} dξ, x > 0.

Since a + σ^{2}/2 > 0, ε > 0, we have 2(a + ε)/σ^{2} > −1. Therefore lim_{x→∞} p(x) = ∞. Using the substitution ζ = 1/ξ, we get

lim_{x→0^{+}} p(x) = −e^{−2/σ^{2}} ∫_{1}^{∞} e^{\frac{2}{σ^{2}} ζ}ζ^{−(2+2(a+ε)/σ^{2})} dζ = −∞.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 5 If m is a speed measure for \overline{Y}_{ε}, we have

m(0, ∞) = \frac{2}{σ^{2}} e^{2/σ^{2}} lim_{y→0^{+}} ∫_{y}^{∞} e^{− \frac{2}{σ^{2}} \frac{1}{ξ}} ξ^{−(2+2(a+ε)/σ^{2})} dξ.

Since 2(a + ε)/σ^{2} + 2 > 1, we have

∫_{1}^{∞} e^{− \frac{2}{σ^{2}} \frac{1}{ξ}} ξ^{−(2+2(a+ε)/σ^{2})} dξ < ∞.

Moreover, using the substitution ζ = 1/ξ once more gives

lim_{y→0^{+}} ∫_{y}^{1} e^{− \frac{2}{σ^{2}} \frac{1}{ξ}} ξ^{−(2+2(a+ε)/σ^{2})} dξ = ∫_{1}^{∞} e^{− \frac{2}{σ^{2}} ζ}ζ^{2(a+ε)/σ^{2}} dζ < ∞.

Thus m(0, ∞) < ∞.

Now by e.g., Proposition 5.5.22 in Karatzas and Shreve [8], as \overline{Y}_{ε} is the strong solution of a scalar stochastic differential equation with time independent coefficients which obey the usual nondegeneracy and local integrability conditions, and \overline{Y}_{ε} has a deterministic initial condition in (0, ∞), the conditions lim_{x→∞} p(x) = ∞, lim_{x→0^{+}} p(x) = −∞ imply that \overline{Y}_{ε} is recurrent on (0, ∞). Since it also has a finite speed measure m(0, ∞) < ∞, we may apply the result of Motoo [9] (see Itô and McKean [7], Chapter 4.12, equation 6) to the diffusion \overline{Y}_{ε}.

Motoo’s result tells us that if there is a positive and increasing function h such that h(t) → ∞ as t → ∞, and

(14)  ∫_{1}^{∞} \frac{1}{p(h(t))} dt = ∞,

then

lim sup_{t→∞} \frac{\overline{Y}_{ε}(t)}{h(t)} > 1, a.s.

To apply the result, note as 2(a + ε)/σ^{2} + 1 > 0, that L’Hôpital’s rule gives

lim_{x→∞} \frac{p(x)}{x^{2(a+ε)/σ^{2}+1}} = \frac{1}{e^{2/σ^{2}} (\frac{2(a+ε)}{σ^{2}} + 1)} > 0.

Let M > 0 and h(t) = Mt^{1/(1+2(a+ε)/σ^{2})}. Then

lim_{t→∞} \frac{p(h(t))^{−1}}{t^{−1}} = \frac{M^{1+2(a+ε)/σ^{2}}}{e^{2/σ^{2}} (2(a + ε)/σ^{2} + 1)} > 0,

so with this choice of h, (14) holds. Therefore

lim sup_{t→∞} \frac{\overline{Y}_{ε}(t)}{t^{1/(1+2(a+ε)/σ^{2})}} > M, a.s.

Letting M → ∞ through the integers yields (13). □ EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 6

We can now turn to the proof of Theorem 1.

Proof of Theorem 1. On account of the linearity of (3), we may choose X(0) = 1 without loss of generality. Referring to the proof of Theorem 1 in [5], we can show there is a pair of finite positive random variables L_{1}, T_{1} such that

(15)  \frac{X(t)}{k(t)} ≥ L_{1} \frac{ϕ(t)}{k(t)} ∫_{T_{1}}^{t} k(s)ϕ(s)^{−1} ds, t ≥ T_{1}

almost surely where (ϕ(t))_{t≥0} is the positive process given by

(16)  ϕ(t) = e^{−(a+σ^{2}/2)t+σB(t)}, t ≥ 0.

A careful reading of the proof of Theorem 1 in [5] shows that X(t) ≥ ϕ(t). Therefore, if X(t) → 0 as t → ∞ on a nontrivial set, it follows that ϕ(t) → 0 as t → ∞ on a nontrivial set. This implies that a + σ^{2}/2 > 0.

Next, define the process (Y (t))_{t≥0} by

(17)  Y (t) = ϕ(t)k(t)^{−1} ∫_{0}^{t} k(s)ϕ(s)^{−1} ds, t ≥ 0

and the random function Ỹ by

(18)  Ỹ (t) = ϕ(t)k(t)^{−1} ∫_{T_{1}}^{t} k(s)ϕ(s)^{−1} ds, t ≥ T_{1}.

As a + σ^{2}/2 > 0, (8) implies k(t)ϕ(t)^{−1} → ∞ as t → ∞, a.s., we therefore get

(19)  lim_{t→∞} \frac{Ỹ (t)}{Y (t)} = lim_{t→∞} \frac{∫_{T_{1}}^{t} k(s)ϕ(s)^{−1} ds}{∫_{0}^{t} k(s)ϕ(s)^{−1} ds} = 1, a.s.

Thus (15) and (17)–(19) now show that it suffices to prove

(20)  lim sup_{t→∞} \frac{Y (t)}{t^{\frac{1}{1+2a/σ^{2}} −ε}} = ∞, a.s.

for every ε > 0 in order to assure the result. The proof of (20) is the subject of Lemma 5 below. □

Lemma 5. Suppose k obeys (7) and a + σ^{2}/2 > 0. Then the process (Y (t))_{t≥0} defined by (17), obeys (20) for every ε > 0.

Proof. By (16) and (17), and the fact that k is positive and in C^{1}(0, ∞), it follows that Y obeys the stochastic differential equation

dY (t) = (1 − [a + \frac{k′(t)}{k(t)}] Y (t)) dt + σY (t) dB(t)

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 7 with Y (0) = 0. By (7), there exists 0 ≤ c < ∞ such that

c = sup_{t≥0} |\frac{k′(t)}{k(t)}| .

Therefore, the process does not explode in finite time, a.s. Therefore, for any fixed deterministic time T > 0, it follows that 0 < Y (T) < ∞, a.s., as Y is a strictly positive process on (0, ∞).

Next, let ε > 0. By (7), there exists T = T (ε) > 0 such that

|\frac{k′(t)}{k(t)}| ≤ \frac{ε}{2}, t ≥ T (ε).

Next, define the process Y_{ε} = {Y_{ε}(t) : t ≥ T (ε)} so that

(21) Y_{ε}(t) = Y (T (ε)) + ∫_{T (ε)}^{t} 1 − (a + ε)Y_{ε}(s) ds + ∫_{T (ε)}^{t} σY_{ε}(s) dB(s),

for t ≥ T (ε). We prove momentarily that

(22)  Y_{ε}(t) ≤ Y (t), t ≥ T (ε), a.s.

Also recall that Lemma 4 implies (13). Now we show that

(23)  lim_{t→∞} \frac{Y_{ε}(t)}{\overline{Y}_{ε}(t)} = 1, a.s.

where \overline{Y}_{ε} is defined by (12) with \overline{Y}_{ε}(0) = 1.

To see this, introduce the process (ϕ̃(t))_{t≥T (ε)} which obeys

ϕ̃(t) = e^{−(a+σ^{2}/2)(t−T (ε))+σ(B(t)−B(T (ε)))}, t ≥ T (ε).

Then ϕ̃(t) = ϕ(t)ϕ(T (ε))^{−1}, t ≥ T (ε), and the processes Y_{ε}, \overline{Y}_{ε} defined by (21), (22) are explicitly given by

\overline{Y}_{ε}(t) = ϕ(t)e^{−εt} + ∫_{0}^{t} e^{−ε(t−s)}ϕ(t)ϕ(s)^{−1} ds,
Y_{ε}(t) = ϕ̃(t)Y (T (ε))e^{−ε(t−T (ε))} + ∫_{T (ε)}^{t} e^{−ε(t−s)}ϕ̃(t)ϕ̃(s)^{−1} ds,

where \overline{Y}_{ε}(t) is defined for all t ≥ 0, and Y_{ε}(t) for all t ≥ T (ε), respectively. Thus, for t ≥ T (ε)

(24)  \frac{Y_{ε}(t)}{\overline{Y}_{ε}(t)} = \frac{Y (T (ε))e^{εT (ε)} + ∫_{T (ε)}^{t} e^{εs}ϕ̃(s)^{−1} ds}{ϕ(T (ε)) + ∫_{0}^{t} e^{εs}ϕ̃(s)^{−1} ds}.

Since a + σ^{2}/2 > 0, e^{εt}ϕ̃(t)^{−1} → ∞ as t → ∞ for any ε > 0. Therefore, applying L’Hôpital’s rule to the righthand side of (24) enables us to conclude (23).

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 8 Now, by (13), (22), and (23), as \overline{Y}_{ε}(t) > 0 for all t ≥ 0, a.s., we have

lim sup_{t→∞} \frac{Y (t)}{t^{1/(1+2(a+ε)/σ^{2})}} ≥ lim sup_{t→∞} \frac{Y_{ε}(t)}{t^{1/(1+2(a+ε)/σ^{2})}}

= lim sup_{t→∞} \frac{\overline{Y}_{ε}(t)}{t^{1/(1+2(a+ε)/σ^{2})}} \frac{Y_{ε}(t)}{\overline{Y}_{ε}(t)} = ∞, a.s.

Clearly, if

lim sup_{t→∞} \frac{Y (t)}{t^{1/(1+2(a+ε′)/σ^{2})}} = ∞, a.s.

for all ε′ > 0, this implies that (20) holds for every ε > 0.

We return finally to the proof of (22). Introduce the strictly positive process (ψ(t))_{t≥T (ε)} which is the unique strong solution of

ψ(t) = 1 + ∫_{T (ε)}^{t} σψ(s) dB(s), t ≥ T (ε),

and also define the processes (Z(t))_{t≥T (ε)} and (Z_{ε}(t))_{t≥T (ε)} by Z(t) = Y (t)ψ(t)^{−1}, Z_{ε}(t) = Y_{ε}(t)ψ(t)^{−1}. Then Z(T (ε)) = Z_{ε}(T (ε)), and Z, Z_{ε} are positive. Using integration by parts, we get

Z′(t) = ψ(t)^{−1} − (a + \frac{k′(t)}{k(t)}) Z(t), t > T (ε),
Z′_{ε}(t) = ψ(t)^{−1} − (a + ε) Z(t), t > T (ε).

Consider D(t) = Z(t) − Z_{ε}(t) for t ≥ T (ε). Clearly, D(T (ε)) = 0 and

D′(t) = −(a + ε)D(t) + (ε − \frac{k′(t)}{k(t)}) Z(t), t > T (ε).

Since (ε − k′(t)/k(t))Z(t) > 0 for all t ≥ T (ε), it follows that D(t) > 0 for all t > T (ε) a.s. Therefore Z(t) ≥ Z_{ε}(t) for t ≥ T (ε), so by construction, (22) holds. □

The result of Theorem 1 does not rely directly on the hypothesis that X(t) → 0 on a set of positive probability. In fact, by studying the proof of Theorem 1, it is apparent that the hypothesis a + σ^{2}/2 > 0 may be used in place of the asymptotic stability of the solution. Therefore, we have the following corollary of Theorem 1.

Corollary 6. Let k satisfy (7) and σ ≠ 0. If X is a nontrivial strong solution of (3), and a + σ^{2}/2 > 0, then

lim sup_{t→∞} \frac{|X(t)|}{k(t)t^{\frac{1}{1+2a/σ^{2}} −ε}} = ∞, a.s.

for every ε > 0.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 9

An interesting consequence of this result is the following: if k obeys (7) and there exists ε > 0 such that

(25)  lim_{t→∞} k(t)t^{\frac{1}{1+2a/σ^{2}} −ε} = ∞,

then every nontrivial solution of (3) obeys

lim sup_{t→∞} |X(t)| = ∞, a.s.

For a > 0, this result cannot arise for integrable kernels, as (25) is not consistent with k being integrable. However, if a < 0 (when the deterministic problem (1) is unstable) but a + σ^{2}/2 > 0 (so the stochastic problem without memory is almost surely asymptotically stable), the solution of (3) is unstable if the kernel decays too slowly.

This result also rules out a natural conjecture for the almost sure asymptotic stability of solutions of (3). All solutions of the deterministic equation x′(t) = −ax(t) tend to zero if and only if a > 0, while all solutions of the stochastic equation

dX(t) = −aX(t) dt + σX(t) dB(t)

tend to zero almost surely if and only if a + σ^{2}/2 > 0. A necessary and sufficient condition for all solutions of (1) to be uniformly asymptotically stable is a > ∫_{0}^{∞} k(s) ds. On the basis of these three stability results, one might therefore conjecture that all solutions of (3) would be asymptotically stable (on a set of positive probability) whenever a + σ^{2}/2 > ∫_{0}^{∞} k(s) ds. However, if k is a positive, continuously differentiable and integrable function which obeys

lim_{t→∞} \frac{k′(t)}{k(t)} = 0, lim_{t→∞} k(t)t^{\frac{σ^{2}}{σ^{2}+2a} −ε} = ∞ for some ε > 0

while a < 0 and a + σ^{2}/2 > ∫_{0}^{∞} k(s) ds > 0, we have

lim sup_{t→∞} |X(t)| = ∞ a.s.

Therefore the condition

a + σ^{2}/2 > ∫_{0}^{∞} k(s) ds

is not sufficient to ensure the asymptotic stability of solutions of (3), even on a set of positive probability.

4. Proof of Theorem 3

We now prove Theorem 3, which uses the result of Theorem 1. EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 10 Proof of Theorem 3. If the deterministic solution of (1) is uniformly asymptotically stable, then a > ∫_{0}^{∞} k(s) ds > 0. This condition ensures that lim_{t→∞} X_{σ}(t) = 0, a.s., where X_{σ} denotes the solution of (3). Since k is a subexponential function, a result in [4] tells us that

lim sup_{t→∞} \frac{|X_{σ}(t)|}{k(t)t^{1+ε}} = 0, a.s.

Since k obeys (10), this implies

lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log t} ≤ −α + 1 + ε, a.s.

Letting ε ↓ 0 through the rational numbers, and using (10) once again gives

(26)  lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} ≤ 1, a.s.

Since X_{σ}(t) → 0 as t → ∞ a.s., and k obeys (7), Theorem 3 implies

lim sup_{t→∞} \frac{|X_{σ}(t)|}{k(t)t^{\frac{1}{1+2a/σ^{2}} −ε}} = ∞, a.s.

for every ε > 0. Thus, as k satisfies (10), the last equation implies

lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log t} ≥ −α + \frac{1}{1 + 2a/σ^{2}} − ε, a.s.

Letting ε ↓ 0 through the rational numbers, and using (10) once again gives

(27)  lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} ≥ \frac{1}{1 + 2a/σ^{2}} , a.s.

Taking (26), (27) together, we see there exists a bounded non-negative random variable Λ(σ) such that 0 ≤ Λ(σ) ≤ \frac{2a}{σ^{2}+2a} a.s. and

lim sup_{t→∞} \frac{log |X_{σ}(t)|}{log(tk(t))} = 1 − Λ(σ), a.s.

while Λ(σ) = O(σ^{−2}) as |σ| → ∞, so Λ(σ) → 0 as σ → ∞ almost surely. □

5. Subexponential solutions of scalar nonlinear equations

In [5], it was shown that the decay rate of (3) differs from that of (1) when k is a subexponential function; in particular the a.s. rate of decay of the stochastic equation is slower. Moreover, although we have presented sharp upper bounds for the rate of decay for the equation (3) for superpolynomial functions, and also for regularly varying functions

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 11 in this paper, we have not exhibited exact rates of decay for stochastic equations with subexponential kernels. In this section, we ask whether it is possible for a class of scalar stochastic equations to exhibit a.s. subexponential asymptotic behaviour.

Intuitively, it would appear that the decay rate of (3) is slower than that of (1) when k is subexponential on account of the strength of the state-dependent stochastic perturbation as the solution approaches zero. Therefore, we might conjecture that the solution of a stochastic perturbation of (1) would have the same asymptotic behaviour as (1) if the state-dependent diffusion term is sufficiently small.

5.1. Problem to be studied; main results. In this section, we consider the scalar Itô-Volterra equation

(28) dX(t) = (−aX(t) + ∫_{0}^{t} k(t − s)X(s) ds) dt + σ(X(t)) dB(t)

with X(0) > 0. Here we assume σ(0) = 0, σ is locally Lipschitz continuous and has a global linear bound. We also assume that σ does not have a linear leading order term at the origin by imposing

(29)  lim_{x→0} \frac{σ(x)}{x} = 0.

By adapting results in [1], it is possible to show that the linearisation of (28), namely (1), has the same asymptotic behaviour as (28). This fact is made precise in the following theorem.

Theorem 7. Suppose that k is a positive, continuous and integrable function. Let X be the unique nontrivial strong solution of (28) with X(0) ≠ 0. Suppose the zero solution of (1) is uniformly asymptotically stable. If σ(0) = 0, σ is locally Lipschitz continuous and obeys a global linear bound, then

(30)  X ∈ L^{1}(ℝ^{+}), lim_{t→∞} X(t) = 0, a.s.

The proof is very similar to that of results presented in [1], so it is not given.

It is also possible to establish estimates on the decay rate of solutions of (28) when k is subexponential without making a stronger assumption on the nature of the nonlinearity of the function σ at zero.

Theorem 8. Suppose that k is a positive, continuous and integrable function. Let X be the unique nontrivial strong solution of (28) with X(0) ≠ 0. Suppose the zero solution of (1) is uniformly asymptotically stable. If σ(0) = 0, σ is locally Lipschitz continuous and obeys a global

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 12 linear bound, then

(31)  lim sup_{t→∞} \frac{X(t)}{k(t)} ≥ \frac{∫_{0}^{∞} X(s) ds}{a}, a.s.

Moreover, if k is subexponential, then for every ε > 0

(32)  lim sup_{t→∞} \frac{X(t)}{k(t)t^{1+ε}} = 0, a.s.

Again, since very similar results are proven in [4], we do not present a proof here.

A simple corollary of Theorem 8 is the following: if k is a superpolynomial function (31) and (32) can be combined to give the following sharp estimate on the asymptotic rate of decay of solutions of (28).

Corollary 9. Let k is a positive subexponential function which obeys

(33)  lim_{t→∞} \frac{log k(t)}{log t} = −∞,

and X be the unique nontrivial strong solution of (28) with X(0) ≠ 0. Suppose the zero solution of (1) is uniformly asymptotically stable. If σ(0) = 0, σ is locally Lipschitz continuous and obeys a global linear bound, then

(34)  lim sup_{t→∞} \frac{log X(t)}{log k(t)} = 1, a.s.

These results will be of great use in determining more precise decay rates of solutions of (28). In this section, the main emphasis is placed on determining conditions under which solutions of (28) have exactly the same asymptotic behaviour as (1), when k is a positive subexponential function. In other words, we determine sufficient conditions which ensure

(35)  lim_{t→∞} \frac{X(t)}{k(t)} = \frac{∫_{0}^{∞} X(s)}{a − ∫_{0}^{∞} k(s) ds}, a.s.

and also that a.a. paths of X are positive subexponential functions. Since solutions of (28) are nowhere differentiable, it is impossible to show that paths of X are smooth subexponential functions, as are the solutions of (1). However, it is possible to show that

(36)  lim_{t→∞} \frac{X(t)}{∫_{t}^{∞} X(s) ds} = 0, a.s.

To prove these results, we assume that σ has a polynomial leading order term at zero, in the sense that there is C ∈ [0, ∞) and β > 1 such that

(37)  lim sup_{x→∞} \frac{|σ(x)|}{|x|^{β}} = C.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 13 We prove results in the cases where (a) k is subexponential with polynomial asymptotic behaviour, obeying

(38)  lim_{t→∞} \frac{log k(t)}{log t} = −α,

for some α > 1, and (b) when k is subexponential with superpolynomial asymptotic behaviour, and obeys (33).

Theorem 10. Suppose k is a positive subexponential function which obeys (38) for some α > 1. Let σ(0) = 0, σ be locally Lipschitz continuous, obey a global linear bound, and obey (37) for some β > 1. If a > ∫_{0}^{∞} k(s) ds,

(39)  α > 1 + \frac{1}{2(β − 1)},

and X is the unique nontrivial strong solution of (28) with X(0) > 0, then it satisfies each of the following:

(i) X obeys (35), a.s., (ii) X ∈ U, a.s., (iii) X obeys (36), a.s.

In the superpolynomial case, we have the following result.

Theorem 11. Suppose k is a positive subexponential function which obeys (33) and also satisfies

(40)  lim sup_{t→∞} sup_{γ≥1} \frac{log k(t)}{log k(γt)} ≤ 1.

Let σ be locally Lipschitz continuous and globally linearly bounded function, with σ(0) = 0 which obeys (37) for some β > 1. If a > ∫_{0}^{∞} k(s) ds, and X is the unique nontrivial strong solution of (28) with X(0) > 0, it satisfies each of the following:

(i) X obeys (35), a.s., (ii) X ∈ U, a.s., (iii) X obeys (36), a.s.

The condition (40) indicates that − log k(t) behaves similarly to a regularly varying function at infinity of order zero. (40) is obeyed by many important superpolynomial and subexponential functions, such as k(t) ∼ e^{−t^{α}} for α ∈ (0, 1). We note also that (40) is satisfied if k is ultimately nonincreasing i.e., there exists T > 0 such that k is nonincreasing on [T, ∞).

An interesting open question is to ask to what extent the ancillary hypotheses in Theorems 10 and 11 are essential. For instance, is the constraint (39) purely technical, or does it reflect a requirement that

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 14 exact asymptotic estimates are possible only when the noise perturbation is sufficiently small and the rate of decay of the kernel sufficiently fast. In particular, we do not know whether necessary and sufficient conditions of the form (39) could be developed in Theorem 10 so that the limit

lim_{t→∞} \frac{X(t)}{k(t)}

exists.

5.2. Preliminary analysis. We start with some general observations which will be necessary in proving Theorems 10, 11.

By Theorem 7, the process X which is a solution of (28) is in L^{1}(ℝ^{+}) and X(t) → 0 as t → ∞, a.s. Therefore, as |σ(x)| ≤ L|x| for some L ≥ 0 and all x ∈ ℝ, it follows that |σ(X(t))| ≤ L|X(t)| for all t ≥ 0. Since X ∈ L^{1}(ℝ^{+}) and X(t) → 0 as t → ∞, it follows that X ∈ L^{2}(ℝ^{+}) a.s. Therefore

∫_{0}^{∞} σ(X(s))^{2} ds < ∞, a.s.

and consequently, by the martingale time change theorem

lim_{t→∞} ∫_{0}^{t} σ(X(s)) dB(s) = ∫_{0}^{∞} σ(X(s)) dB(s), a.s.

where the limit on the righthand side is a.s. finite. Define

Ω_{0} = {ω ∈ Ω : X(·, ω) ∈ L^{1}(ℝ^{+}), lim_{t→∞} X(t, ω) = 0},

and

Ω_{1} = {ω ∈ Ω : lim_{t→∞} (∫_{0}^{t} σ(X(s)) dB(s)) (ω) exists}.

For ω ∈ Ω_{1}, we may define the random function T (ω) : ℝ^{+} → ℝ by

T (t, ω) = − (∫_{0}^{∞} σ(X(s)) dB(s) − ∫_{0}^{t} σ(X(s)) dB(s)) (ω)

=: − (∫_{t}^{∞} σ(X(s)) dB(s)) (ω),

so that lim_{t→∞} T (t, ω) = 0. Next, introduce the process Y such that

Y (t) = X(0) + ∫_{0}^{t} σ(X(s)) dB(s).

Then for ω ∈ Ω_{0} ∩ Ω_{1}

lim_{t→∞} Y (t, ω) = X(0)(ω) + (∫_{0}^{∞} σ(X(s)) dB(s)) (ω) =: Y^{∗}(ω)

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 15 where Y^{∗}(ω) is finite. For each ω ∈ Ω_{0} ∩ Ω_{1}, we introduce

U(t, ω) = X(t, ω) − Y (t, ω) + Y^{∗}(ω) = X(t, ω) + T (t, ω),

so that lim_{t→∞} U(t, ω) = 0 for ω ∈ Ω_{0} ∩ Ω_{1}. Notice also that T (t, ω) = Y (t, ω) − Y^{∗}(ω). The process Z defined by Z(t) = X(t) − Y (t) obeys

Z(t) = Z(0) + ∫_{0}^{t} −aX(s) + ∫_{0}^{s} k(s − u)X(u) du ds.

Therefore, as k and X(ω) are continuous functions, the function Z(ω) is in C^{1}(ℝ^{+}). Therefore, for each ω ∈ Ω_{0}∩Ω_{1}, since U(t, ω) = Z(t, ω)+ Y^{∗}(ω), we have

U′(t, ω) = Z′(t, ω)
= −aX(t, ω) + ∫_{0}^{t} k(t − s)X(s, ω) ds

= −a(U(t, ω) + T (t, ω)) + ∫_{0}^{t} k(t − s)(U(s, ω) + T (s, ω)) ds

= −aU(t, ω) + ∫_{0}^{t} k(t − s)U(s, ω) ds + f(t, ω)

where

f(t, ω) = −aT (t, ω) + ∫_{0}^{t} k(t − s)T (s, ω) ds.

Since Ω_{0} ∩ Ω_{1} is almost sure, we have proven the following result.

Lemma 12. Suppose that k is a positive and integrable function and that a > ∫_{0}^{∞} k(s) ds. Then for each ω in an almost sure set Ω^{∗}, the realisation X(ω) can be represented as

(41)  X(t, ω) = U(t, ω) + T (t, ω)

where T (ω) is the function defined by

(42)  T (t, ω) = − (∫_{t}^{∞} σ(X(s)) dB(s)) (ω),

U(ω) solves the equation

(43)  U′(t, ω) = −aU(t, ω) + ∫_{0}^{t} k(t − s)U(s, ω) ds + f(t, ω),

and f(ω) is the function

(44)  f(t, ω) = −aT (t, ω) + ∫_{0}^{t} k(t − s)T (s, ω) ds.

We use Lemma 12 to obtain the asymptotic behaviour of (28) as follows: Theorem 8 yields an a priori upper estimate on the almost sure decay rate of solutions of (28), so there is an estimate on the decay rate of

t ↦ ∫_{t}^{∞} σ(X(s))^{2} ds

as σ(x)^{2} ≤ C′|x|^{2β} for x sufficiently small. This decay rate can then be linked to that of T by proving the following Lemma, whose proof is relegated to the Appendix.

Lemma 13. Suppose that B is a standard Brownian motion with natural filtration (F^{B}(t))_{t≥0}. Suppose that A = {A(t) : 0 ≤ t < ∞; F^{B}(t)} has continuous sample paths, and satisfies A(t) > 0 for all t ≥ 0, a.s., and A ∈ L^{2}(ℝ^{+}), a.s. Then

(45) lim sup_{t→∞} \frac{|∫_{t}^{∞} A(s) dB(s)|}{\sqrt{2 ∫_{t}^{∞} A(s)^{2} ds log log (∫_{t}^{∞} A(s)^{2} ds)^{−1}}} = 1, a.s.

Once an upper estimate on the decay rate of T has been so obtained, an upper estimate on the decay rate of f given by (44) is known, and hence, by a variation of parameters argument, the decay rate of U obeying (43) is determined. Since upper bounds on the decay rate of U and T are now known, by (41), we have a new upper estimate on the a.s. decay rate of X. If the new estimate on the decay rate is faster, the argument can be iterated as often as necessary to obtain ever sharper estimates on the rate of decay of the process. If at any stage in this iteration it can be shown that the decay rate of T to zero is faster than that of k, it is then possible to prove that X enjoys the same decay rate as k.

The proof of Theorem 11 requires one iteration of this argument, while that of Theorem 10 may require several iterations.

6. Subexponential solutions

We start by proving the last claim above; namely, if T decays quickly enough, then X(t)/k(t) tends to a well-defined finite limit as t → ∞. This then enables us to conclude that X is an almost surely positive subexponential function, which obeys (36).

Lemma 14. Let Ω^{∗} be an almost sure set, and T defined by (42) obey

(46)  lim_{t→∞} \frac{T (t)}{k(t)} = 0,

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 17 on Ω^{∗}. If k is a positive subexponential function and a > ∫_{0}^{∞} k(s) ds, then

lim_{t→∞} \frac{X(t)}{k(t)} = \frac{∫_{0}^{∞} X(s) ds}{a − ∫_{0}^{∞} k(s) ds}

on Ω^{∗}.

Proof. If (46) holds then T (ω) ∈ L^{1}(ℝ^{+}) because k ∈ L^{1}(ℝ^{+}). Thus by (44) and Theorem 4.1 in [3] (47)

lim_{t→∞} \frac{f(t, ω)}{k(t)} = −a lim_{t→∞} \frac{T (t, ω)}{k(t)} + lim_{t→∞} \frac{(k ∗ T (ω))(t)}{k(t)} = ∫_{0}^{∞} T (s, ω) ds.

Now, by (44), as T (ω) and k are integrable, we have

(48)  ∫_{0}^{∞} f(s, ω) ds = −(a − ∫_{0}^{∞} k(s) ds) ∫_{0}^{∞} T (s, ω) ds.

Therefore, by Corollary 6.3 in [3] the solution of (43) obeys

lim_{t→∞} \frac{U(t, ω)}{k(t)} = \frac{1}{(a − ∫_{0}^{∞} k(s) ds)^{2}} (U(0, ω) + ∫_{0}^{∞} f(s, ω) ds)

+ \frac{L_{k}f(ω)}{a − ∫_{0}^{∞} k(s) ds},

where L_{k}f(ω) = lim_{t→∞} f(t, ω)/k(t). Using (47), (48) this simplifies to give

lim_{t→∞} \frac{U(t, ω)}{k(t)} = \frac{U(0, ω)}{(a − ∫_{0}^{∞} k(s) ds)^{2}} ,

which, on account of (46), yields

(49)  lim_{t→∞} \frac{X(t)}{k(t)} = \frac{U(0, ω)}{(a − ∫_{0}^{∞} k(s) ds)^{2}} .

Now (50)

U(0, ω) = X(0, ω) − T (0, ω) = X(0, ω) + (∫_{0}^{∞} σ(X(s)) dB(s)) (ω).

But, as X ∈ L^{1}(ℝ^{+}), writing (28) in integral form, noting that

lim_{t→∞} ∫_{0}^{t} σ(X(s)) dB(s)

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 18 exists a.s., and that X(t) → 0 as t → 0 a.s., we may let t → ∞ to obtain

− X(0) = −a ∫_{0}^{∞} X(s) ds + ∫_{0}^{∞} k(s) ds ∫_{0}^{∞} X(s) ds

+ ∫_{0}^{∞} σ(X(s)) dB(s), a.s.

Hence

(51) (a − ∫_{0}^{∞} k(s) ds) ∫_{0}^{∞} X(s) ds = X(0) + ∫_{0}^{∞} σ(X(s)) dB(s),

so by (49), (50), and (51), we have the required result. □

Once this Lemma is proved, we can show that X is a positive subexponential function which obeys (36).

Lemma 15. Let Ω^{∗} be an almost sure set, and T defined by (42) obey (46) on Ω^{∗}. If k is a positive subexponential function, a > ∫_{0}^{∞} k(s) ds, and X(0) > 0, then all the conclusions of Theorem 10 and 11 hold.

Proof. By considering the line of proof of Theorem 1 in [5], it is possible to show that whenever X(0) > 0, then X(t) > 0 for all t ≥ 0, a.s. Therefore, as X ∈ L^{1}(ℝ^{+}) (by Theorem 7) it follows that

∫_{0}^{∞} X(s) ds > 0, a.s.

and so

lim_{t→∞} \frac{X(t)}{k(t)} = \frac{∫_{0}^{∞} X(s) ds}{a − ∫_{0}^{∞} k(s) ds} > 0, a.s.

Since k is a positive subexponential function, and

L_{k}X = lim_{t→∞} X(t)/k(t)

exists a.s., by Lemma 4.3 in [3], X satisfies (US1) and (US2). Since X is positive and

lim_{t→0^{+}} \frac{X(t)}{k(t)} = \frac{X(0)}{k(0)} > 0

Lemma 4.3 in [3] further enables us to conclude that X is positive subexponential, so (ii) of Theorem 10 and Theorem 11 follow. By Lemma 14, we know that part (i) of these Theorems also hold. It therefore is necessary to prove merely that part (iii) of Theorem 10 and 11 hold.

Since (35) holds, we have

lim_{t→∞} \frac{∫_{t}^{∞} X(s) ds}{∫_{t}^{∞} k(s) ds} = \frac{∫_{0}^{∞} X(s) ds}{a − ∫_{0}^{∞} k(s) ds}, a.s.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 19 Therefore, (36) holds if

(52)  lim_{t→∞} \frac{k(t)}{∫_{t}^{∞} k(s) ds} = 0.

Let N ∈ ℕ, so we have

\frac{1}{k(t + N)} ∫_{t}^{∞} k(s) ds ≥ ∫_{t}^{t+N} \frac{k(s)}{k(t + N)} ds = ∫_{0}^{N} \frac{k(t + N − s)}{k(t + N)} ds.

By property (US1) of positive subexponential functions, for each N ∈ ℕ, there is T (N) > 0 such that t > T (N) implies

\frac{k(t + N − s)}{k(t + N)} > \frac{1}{2}, 0 ≤ s ≤ N.

Therefore, for t > T (N)

\frac{1}{k(t + N)} ∫_{t}^{∞} k(s) ds ≥ \frac{N}{2} .

Thus, for fixed N ∈ ℕ, the fact that k obeys (US1) implies

lim_{t→∞} \frac{k(t + N)}{k(t)} = 1.

and so

lim inf_{t→∞} \frac{∫_{t}^{∞} k(s) ds}{k(t)} ≥ lim inf_{t→∞} \frac{k(t + N)}{k(t)} \frac{1}{k(t + N)} ∫_{t}^{∞} k(s) ds ≥ \frac{N}{2} .

Letting N → ∞ proves (52), and hence the result. □

7. Proof of Theorem 11

We are now in a position to establish (46) under the hypotheses of Theorem 11, which, by the analysis in the previous section, ensures that Theorem 11 holds.

Proof of Theorem 11. By Lemma 13, we have ∫_{0}^{∞} σ(X(s))^{2} ds < ∞ a.s., so by the definition of T, we have

lim sup_{t→∞} \frac{|T (t)|}{\sqrt{2 ∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1}}} = 1 a.s.

Suppose this holds on the almost sure set Ω_{2}, which also contains the sets Ω_{0} and Ω_{1} defined in Section 5. Therefore (46) is true once

(53) lim_{t→∞} \frac{∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1}}{k(t)^{2}} = 0, a.s.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 20 By (37) and the continuity of σ there is x^{∗} > 0 such that |x| < x^{∗} implies

|σ(x)| ≤ (1 + C)|x|^{β}.

Since X(t) → 0 as t → ∞ on Ω_{2}, there is T_{1}(ω) > 0 such that |X(t, ω)| < x^{∗} for all t > T_{1}(ω) and ω ∈ Ω_{2}. Hence, for t > T_{1}(ω),

σ(X(t, ω))^{2} ≤ (1 + C)^{2}|X(t, ω)|^{2β}.

By (32) in Theorem 8, for every ε > 0 there is an almost sure set Ω^{ε} such that for all ω ∈ Ω^{ε} there is a T_{2}(ω, ε) such that t > T_{2}(ω, ε) implies

|X(t, ω)| ≤ k(t)t^{1+ε}.

Now consider the almost sure set Ω̃^{ε} = Ω^{ε} ∩ Ω_{2}. Then for ω ∈ Ω̃^{ε}, if we define T_{3}(ω, ε) = T_{1}(ω) ∨ T_{2}(ω, ε), for all t > T_{3}(ω, ε), we have

(54)  σ(X(t, ω))^{2} ≤ (1 + C)^{2}k(t)^{2β}t^{2β(1+ε)}.

Next, for all ε ∈ (0, 1) there is x^{∗∗}(ε) > 0 such that x ∈ (0, x^{∗∗}(ε)) implies

x log log \frac{1}{x} ≤ x^{1−ε}.

Now, as ∫_{t}^{∞} σ(X(s))^{2} ds → 0 as t → ∞, for each ω ∈ Ω_{2} there is a T_{4}(ω, ε) > 0 such that t > T_{4}(ω, ε) implies ∫_{t}^{∞} σ(X(s, ω))^{2} ds < x^{∗∗}. Thus for t > T_{4}(ω, ε)

(55) (∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1}) (ω)

≤ (∫_{t}^{∞} σ(X(s))^{2} ds)^{1−ε} (ω).

Thus, for ω ∈ Ω̃^{ε} we may define T_{5}(ω, ε) = T_{3}(ω, ε) ∨ T_{4}(ω, ε), so that, for t > T_{5}(ω, ε) we have

(∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1}) (ω)

≤ (1 + C)^{2(1−ε)} (∫_{t}^{∞} k(s)^{2β}s^{2β(1+ε)} ds)^{1−ε} .

Therefore (53)— and consequently (46)— follows once

(56)  lim_{t→∞} \frac{(∫_{t}^{∞} k(s)^{2β}s^{2β(1+ε)} ds)^{1−ε}}{k(t)^{2}} = 0

for some ε ∈ (0, 1). We finally prove that the conditions (33), (40) imply (56). This is the subject of Lemma 16 which follows this proof. Hence the proof of (46) is complete, and so, by Lemma 14, part (i)

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 21 of Theorem 11 follows. The proofs of parts (ii), (iii) are given in Lemma 15, so Theorem 11 is established. □

We return now to the proof of (56).

Lemma 16. Suppose that k is a positive and integrable function which obeys (33) and (40). If β > 1, then there exists ε ∈ (0, 1) such that (56) holds.

Proof. Since k obeys (33), for every ε ∈ (0, 1) there is T_{1}(ε) > 0 such that t^{2β(1+ε)}k(t)^{2βε} < 1 for t > T_{1}(ε). Thus, for t > T_{1}(ε) we have t^{2β(1+ε)}k(t)^{2β} ≤ k(t)^{2β(1−ε)}. To prove (56) it is therefore enough to prove

(57)  lim_{t→∞} \frac{∫_{t}^{∞} k(s)^{2β(1−ε)} ds}{k(t)^{\frac{2}{1−ε}}} = 0

for some ε ∈ (0, 1). Next define λ(t) = − log k(t) so λ(t) → ∞ as t → ∞. Therefore (40) implies that for every ε ∈ (0, 1) there is a T_{2}(ε) > 0 such that for all s ≥ t > T_{2}(ε) we have

\frac{λ(t)}{λ(s)} < 1 + ε,

Now define μ(ε) = β(1 − ε)^{2} − (1 + ε) for ε ∈ (0, 1). Hence there exists ε^{∗} ∈ (0, 1) so that μ(ε^{∗}) = \frac{1}{2}(β − 1) > 0. Now, for s ≥ t > T_{2}(ε^{∗}), we have

β(1 − ε^{∗})^{2} − \frac{λ(t)}{λ(s)} > μ(ε^{∗}).

Therefore, because

\frac{1}{k(t)^{\frac{2}{1−ε}}} ∫_{t}^{∞} k(s)^{2β(1−ε)} ds

= ∫_{t}^{∞} exp (−\frac{2λ(s)}{1 − ε} \{β(1 − ε)^{2} − \frac{λ(t)}{λ(s)}\}) ds,

for all t > T_{2}(ε^{∗}) we have

\frac{1}{k(t)^{\frac{2}{1−ε^{∗}}}} ∫_{t}^{∞} k(s)^{2β(1−ε^{∗})} ds ≤ ∫_{t}^{∞} e^{− \frac{2λ(s)}{1−ε} μ(ε^{∗})} ds.

Next fix M = 2(1 − ε^{∗})/(β − 1). For all T > T_{3}(M) = T_{3}(ε^{∗}) we have λ(t) > M log t by (33). Now let T_{4}(ε^{∗}) = T_{2}(ε^{∗}) ∨ T_{3}(ε^{∗}). Therefore,

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 22 for t > T_{4}(ε^{∗})

\frac{1}{k(t)^{\frac{2}{1−ε^{∗}}}} ∫_{t}^{∞} k(s)^{2β(1−ε^{∗})} ds ≤ ∫_{t}^{∞} e^{− \frac{λ(s)(β−1)}{1−ε}} ds

≤ ∫_{t}^{∞} e^{− \frac{β−1}{1−ε} M log s} ds = \frac{1}{t}.

Therefore, by choosing ε^{∗} ∈ (0, 1) such that β(1 − ε^{∗})^{2} − (1 + ε^{∗}) = (β − 1)/2, we have proven (57) for ε = ε^{∗}. □

8. Proof of Theorem 10

In this section, we turn to the proof of Theorem 10. We start with the proof of a technical lemma.

Lemma 17. Suppose that g is a nonnegative and continuous function such that

(58)  lim sup_{t→∞} \frac{log g(t)}{log t} ≤ −λ_{1}

where λ_{1} > 0, and there is a continuous nonnegative and integrable function f such that

(59)  lim_{t→∞} \frac{log f(t)}{log t} = −λ_{2}

where λ_{2} > 1. Then

lim sup_{t→∞} \frac{log(f ∗ g)(t)}{log t} ≤ −(λ_{1} ∧ λ_{2}).

Proof. Let λ = λ_{1} ∧ λ_{2}, and 0 < ε < λ. Introduce h(t) = (1 + t)^{−λ+ε}. Notice that if we can prove

(60)  lim_{t→∞} \frac{(f ∗ g)(t)}{h(t)} = 0,

the result follows immediately by taking logarithms and letting ε ↓ 0.

We make some preliminary observations. First, the definition of h implies that

(61)  sup_{γ∈[1/2,1]} \frac{h(γt)}{h(t)} ≤ 2^{λ−ε}.

For any t ≥ 0, notice that we can write (62)

\frac{(f ∗ g)(t)}{h(t)} = ∫_{0}^{t/2} \frac{f(t − s)}{h(t − s)} \frac{h(t − s)}{h(t)}g(s) ds + ∫_{t/2}^{t} f(t − s)\frac{h(s)}{h(t)} \frac{g(s)}{h(s)} ds.

Due to (59), for every ε > 0 there is a T_{2}(ε) > 0 such that

(63)  f(t) ≤ (1 + t)^{−λ_{2}+ε/2}, t > T_{2}(ε).

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 23 We now consider the cases λ_{1} ∈ (0, 1] and λ_{1} > 1 separately.

First, let λ_{1} ∈ (0, 1]. On account of (58), for every ε there exists T_{1}(ε) > 0 and C(ε) > 0 such that for all t > T_{1}(ε)

(64) g(t) ≤ (1 + t)^{−λ_{1}+ε/2}, ∫_{0}^{t} g(s) ds ≤ C(ε)(1 + t)^{−λ_{1}+ε/2+1}.

Considering the first term on the righthand side of (62), and using (61),

(63) and (64), for t > 2T_{1}(ε) ∨ 2T_{2}(ε) we get

∫_{0}^{t/2} \frac{f(t − s)}{h(t − s)} \frac{h(t − s)}{h(t)}g(s) ds

≤ sup_{t/2≤s≤t} \frac{f(s)}{h(s)} sup_{t/2≤s≤t} \frac{h(s)}{h(t)} ∫_{0}^{t/2} g(s) ds

≤ (1 + t/2)^{−λ_{2}+ \frac{ε}{2} +λ−ε} · 2^{λ−ε} · C(ε)(1 + t/2)^{−λ_{1}+ \frac{ε}{2} +1}

= C(ε)2^{λ−ε}(1 + t/2)^{−(λ_{2}−1)},

since λ = λ_{1}. Therefore the first term on the righthand side of (62) tends to zero as t → ∞.

As to the second term on the righthand side of (62), as f is integrable, by employing (61), and (64), for t > 2T_{1}(ε), we have

∫_{t/2}^{t} f(t − s)\frac{h(s)}{h(t)} \frac{g(s)}{h(s)} ds. ≤ sup_{t/2≤s≤t} \frac{g(s)}{h(s)} sup_{t/2≤s≤t} \frac{h(s)}{h(t)} ∫_{0}^{t/2} f(s) ds

≤ (1 + t/2)^{−λ_{1}+ \frac{ε}{2} +λ−ε} · 2^{λ−ε} · ∫_{0}^{∞} f(s) ds

= 2^{λ−ε} ∫_{0}^{∞} f(s) ds (1 + t/2)^{−ε/2},

as λ = λ_{1}. Therefore the second term on the righthand side of (62) tends to zero as t → ∞, and so (60) holds and the result follows.

Next, let λ_{1} > 1. Then both f and g are integrable, and, since λ = λ_{2}, for t > 2T_{1}(ε) ∨ 2T_{2}(ε), we proceed to obtain the estimates

∫_{0}^{t/2} \frac{f(t − s)}{h(t − s)} \frac{h(t − s)}{h(t)}g(s) ds ≤ sup_{t/2≤s≤t} \frac{f(s)}{h(s)} sup_{t/2≤s≤t} \frac{h(s)}{h(t)} ∫_{0}^{∞} g(s) ds
≤ 2^{λ−ε} ∫_{0}^{∞} g(s) ds (1 + t/2)^{−ε/2},

and

∫_{t/2}^{t} f(t − s)\frac{h(s)}{h(t)} \frac{g(s)}{h(s)} ds ≤ 2^{λ−ε} ∫_{0}^{∞} f(s) ds (1 + t/2)^{−ε/2}.

Hence each term on righthand side of (62) vanishes as t → ∞, so (60) holds and the result follows. □

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 24

This result can now be used to obtain the proof of the following important lemma.

Lemma 18. Suppose k is a positive subexponential function which satisfies (38) for some α > 1, and T defined by (42) obeys

(65)  lim sup_{t→∞} \frac{log |T (t)|}{log t} ≤ −γ, a.s.

for some α > γ > 0. Let a > ∫_{0}^{∞} k(s) ds, and suppose σ is locally Lipschitz, globally linearly bounded, obeys (37) for some β > 1, and has σ(0) = 0. If X is the unique nontrivial strong solution of (28) with X(0) > 0, then it obeys

(66)  lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −γ, a.s.

Proof. By applying Lemma 17 pathwise, (38), (65), and the fact that α > γ > 0, together imply that

lim sup_{t→∞} \frac{log(k ∗ |T |)(t)}{log t} ≤ −γ, a.s.

By (44), the last inequality, and (65), it follows that

(67)  lim sup_{t→∞} \frac{log |f(t)|}{log t} ≤ −γ, a.s.

Next consider the resolvent z with z(0) = 1 which obeys

(68)  z′(t) = −az(t) + ∫_{0}^{t} k(t − s)z(s) ds, t ≥ 0.

Hence, as k is subexponential, and a > ∫_{0}^{∞} k(s) ds, it follows from [3] that

lim_{t→∞} \frac{z(t)}{k(t)} = \frac{1}{(a − ∫_{0}^{∞} k(s) ds)^{2}} .

Therefore, by (38), we have

(69)  lim_{t→∞} \frac{log z(t)}{log t} = −α.

Next, the solution of (43) can be represented in terms of that of (68) according to

U(t) = U(0)z(t) + ∫_{0}^{t} f(t − s)z(s) ds, t ≥ 0.

Applying Lemma 17 pathwise to the convolution term, and using (67), (69), and the fact that 0 < γ < α yields

lim sup_{t→∞} \frac{log(|f| ∗ z)(t)}{log t} ≤ −γ, a.s.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 25 Therefore, by (69), and the fact that 0 < γ < α, we have

(70)  lim sup_{t→∞} \frac{log |U(t)|}{log t} ≤ −γ, a.s.

On account of (65), (70) and (41), we get (66), as required. □

With this preparatory result established, we now state and prove the result that will enable us to successively improve estimates of the decay rate of solutions of (28).

Lemma 19. Suppose that k is a subexponential function which obeys (38) for some α > 1. Suppose a > ∫_{0}^{∞} k(s) ds, and that σ is a globally linearly bounded and locally Lipschitz continuous function which obeys (37) for some β > 1 and σ(0) = 0. Then the nontrivial strong solution of (28) obeys the following:

(a)

(71)  lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −(α − 1), a.s.

(b) Suppose that α > 1+(2β)^{−1} and there is c_{0} > (2β)^{−1} such that

(72)  lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{0}, a.s.

Then one of the following holds:

(i) If c_{0}β > 1/2 + α, then T defined by (42) obeys (46). (ii) If c_{0}β ≤ 1/2 + α, then

(73)  lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −(c_{0}β − 1/2), a.s.

Proof. The proof of part (a) follows from (32) in Theorem 8 and follows closely the argument of Theorem 3 thereafter, using the hypothesis (38).

To prove part (b), we will show that if (72) holds for some c_{0} > 1/(2β) then

(74)  lim sup_{t→∞} \frac{log R(t)}{log t} ≤ −2c_{0}β + 1, a.s.

where

R(t) = ∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1} .

We prove part (b) given (74) and return to its proof presently. If (74) holds, Lemma 13 implies

(75)  lim sup_{t→∞} \frac{|T (t)|}{\sqrt{2R(t)}} = 1 a.s.

Therefore, by (74), (75), we get

(76)  lim sup_{t→∞} \frac{log |T (t)|}{log t} ≤ −(c_{0}β − 1/2), a.s.

To conclude the proof of part (b), suppose first that c_{0}β > 1/2 + α. Then

lim sup_{t→∞} \frac{log |T (t)|}{log t} ≤ −c_{0}β + 1/2 < −α,

almost surely, so, by (38), (46) holds.

In the other case, when c_{0}β ≤ 1/2 + α, because (76) holds and k obeys (38), Lemma 18 implies (73).

We now return to the proof of (74). If

lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{0}, a.s.,

then, for every ε ∈ (0, c_{0}) there is T_{1}(ω, ε) such that for t > T_{1}(ω, ε) we have

X(t, ω) ≤ t^{−c_{0}+ε}.

on an almost sure set. Revisiting the proof of Theorem 11, we see that there is T_{2}(ω) such that t > T_{2}(ω) implies |σ(X(t, ω))| ≤ (1 + C)|X(t, ω)|^{β}. Hence for every t > T_{3}(ω, ε) := T_{1}(ω, ε) ∨ T_{2}(ω) in some almost sure set, we have

σ(X(t))^{2} ≤ (1 + C)^{2}t^{−2βc_{0}+2βε}.

Since ∫_{t}^{∞} σ(X(s))^{2} ds → 0 as t → ∞, a.s., by again appealing to the proof of Theorem 11 for all t > T_{4}(ω, ε), (55) is true. Now, as c_{0} > 1/(2β), once ε > 0 can be chosen so small that −2βc_{0}+2βε < −1, if we take t > T_{5}(ω, ε) = T_{3}(ω, ε) ∨ T_{4}(ω, ε), it follows that for all ω in an almost sure set that we have

R(t, ω) = (∫_{t}^{∞} σ(X(s))^{2} ds log log (∫_{t}^{∞} σ(X(s))^{2} ds)^{−1}) (ω)

≤ (∫_{t}^{∞} σ(X(s))^{2} ds)^{1−ε} (ω)

≤ (∫_{t}^{∞}(1 + C)^{2}s^{−2βc_{0}+2βε} ds)^{1−ε}
= C(ε)t^{(−2βc_{0}+2βε+1)(1−ε)},

where C(ε) is a positive, ε-dependent constant. Hence for t > T_{5}(ω, ε),

\frac{log R(t, ω)}{log t} ≤ \frac{log C(ε)}{log t} + (−2βc_{0} + 2βε + 1)(1 − ε).

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 27 Hence, for each 0 < ε < 1 ∨ c_{0} ∨ (2βc_{0} − 1)/(2β), there is an Ω_{ε}^{∗} with ℙ[Ω_{ε}^{∗}] = 1 such that ω ∈ Ω_{ε}^{∗} implies

lim sup_{t→∞} \frac{log R(t, ω)}{log t} ≤ (−2βc_{0} + 2βε + 1)(1 − ε).

Now let Ω^{∗} = ∩_{n∈ℕ}Ω_{1/n}^{∗}, so Ω^{∗} is an almost sure set. This implies

lim sup_{t→∞} \frac{log R(t, ω)}{log t} ≤ (−2βc_{0} + 2β \frac{1}{n} + 1)(1 − \frac{1}{n}), ω ∈ Ω^{∗}, n ∈ ℕ.

Letting n → ∞ now yields (74). □

The proof of Theorem 10 now proceeds by applying Lemma 19 part (b) iteratively.

Proof of Theorem 10. If α > (2β + 1)/(2(β − 1)), then α > 1 + (2(β − 1))^{−1} > 1 + (2β)^{−1}. By (71), with c_{0} = α − 1, (72) holds, and c_{0}β − 1/2 − α = \frac{1}{2}(2α(β − 1) − (2β + 1)) > 0. Hence c_{0}β > 1/2 + α > 1/2, so the alternative (i) in part (b) of Lemma 19 holds, and we have (46). Therefore, by Lemma 14 it now follows that (35) holds, and so Lemma 15 enables us to conclude that parts (ii), (iii) of Theorem 10 are also true.

Now, consider the case 1 + (2(β − 1))^{−1} < α ≤ (2β + 1)(2(β − 1))^{−1}. Notice that if α > 1 + (2(β − 1))^{−1}, then immediately α > 1 + (2β)^{−1}. Let c_{0} = α − 1 > (2(β − 1))^{−1} and consider the sequence defined by c_{n+1} = βc_{n} − 1/2, n ≥ 0. Note that c_{0}β ≤ 1/2 + α, and that the iteration for the sequence may be rewritten as c_{n+1} − (2(β − 1))^{−1} = β(c_{n} − (2(β − 1))^{−1}). Since c_{0} > (2(β − 1))^{−1}, and β > 1, the sequence is increasing and lim_{n→∞} c_{n} = ∞. Therefore, there exists a minimal n_{0} > 1 such that c_{n_{0}}β > 1/2 + α, while

c_{n}β ≤ \frac{1}{2} + α, n = 0, 1, …, n_{0} − 1.

Therefore, by Lemma 19, for 1 + (2β)^{−1} < 1 + (2(β − 1))^{−1} < α ≤ (2β + 1)(2(β − 1))^{−1}, we have

lim sup_{t→∞} \frac{log X(t)}{log t} ≤ −c_{0}

and c_{0}β ≤ \frac{1}{2} + α. Hence, by Lemma 19 (b) alternative (ii), we have

lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{1}, a.s.

If n_{0} ≠ 1, since c_{1} > c_{0} > (2β)^{−1}, but c_{1}β ≤ \frac{1}{2} + α, we may apply Lemma 19(b) alternative (ii) again to give

lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{2}, a.s.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 28 Continuing to iterate in this way leads to

lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{n_{0}−1}, a.s.

where c_{n_{0}−1}β ≤ \frac{1}{2} + α, so applying Lemma 19(b) alternative (ii) gives

lim sup_{t→∞} \frac{log |X(t)|}{log t} ≤ −c_{n_{0}}, a.s.

However, because c_{n_{0}}β > \frac{1}{2} + α, the next iteration of Lemma 19 (b) invokes alternative (i), thereby enabling us to conclude that (46) holds. The proof now concludes as in the case α > (2β +1)(2(β −1))^{−1} above, and Theorem 10 is proven in the case α ∈ (1 + (2(β − 1))^{−1}, (2β + 1)(2(β − 1))^{−1}] also. □

Appendix

Here, we prove a result which bounds the decay rate of ∫_{t}^{∞} A(s) dB(s) as t → ∞ for a scalar process A = {A(t), F^{B}(t); 0 ≤ t < ∞} which is in L^{2}(ℝ^{+}) a.s.

If A ∈ L^{2}(ℝ^{+}) a.s. then

lim_{t→∞} ∫_{0}^{t} A(s) dB(s) exists a.s.

by the martingale time change theorem. If we denote this limit by ∫_{0}^{∞} A(s) dB(s), we may define, for each t ≥ 0

∫_{t}^{∞} A(s) dB(s) := ∫_{0}^{∞} A(s) dB(s) − ∫_{0}^{t} A(s) dB(s)

This is a well-defined F^{B}(∞)-measurable random variable, for every t ≥ 0; note, however, that it is not a stochastic process, but rather a one-parameter family of random variables. Observe that ∫_{t}^{∞} A(s) dB(s) is obviously not F^{B}(t)-measurable. However, in the sequel, it will suffice to understand the asymptotic behaviour of the random function

t ↦ (∫_{t}^{∞} A(s) dB(s)) (ω) for all ω in an almost sure set.

Proof of Lemma 13. By the comments preceding this Lemma, it is evident that both numerator and denominator in (45) exist for all t ≥ 0, and, moreover, that both have limit zero, as t → ∞, almost surely. Define for t ≥ 0

W (t) = \{ \begin{matrix}tB(1/t), & t > 0, \\ 0, & t = 0. \end{matrix}

Hence W = {W (t), F^{W} (t); 0 ≤ t < ∞} is a standard one-dimensional Brownian motion. Moreover, its natural filtration can be expressed in

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 29 terms of that of B; indeed F^{W} (t) = F^{B}(1/t) for all t. If we define Y by

Y (t) = \{ \begin{matrix}A(1/t), & t > 0, \\ 0, & t = 0, \end{matrix}

then Y = {Y (t), F^{W} (t); 0 ≤ t < ∞}. Furthermore, the process has continuous sample paths. Next, define the family of F^{B}(∞)- measurable random variables {M(τ); τ ≥ 0} by

M(τ) = \{ \begin{matrix}∫ ∞_{\frac{1}{τ}} A(s) dB(s), & τ > 0, \\ 0, & τ = 0. \end{matrix}

Then the function t ↦ M(t, ω) is continuous for almost all ω ∈ Ω, and lim_{τ→∞} M(τ) exists a.s.. Moreover, for τ > 0, we have

∫_{0}^{τ} (\frac{1}{u}Y (u))^{2} du = ∫ ∞_{\frac{1}{τ}} A(u)^{2} du < ∞, a.s.

Furthermore, we have

lim_{τ↓0} ∫_{0}^{τ} (\frac{1}{u}Y (u))^{2} du = 0, a.s.

Hence, for each τ > 0, ∫_{0}^{τ} \frac{1}{u}Y (u) dW (u) exists. Indeed, as

∫ ∞_{\frac{1}{τ}} A(s) dB(s) = ∫_{0}^{τ} \frac{1}{u}Y (u) dW (u),

we can write

M(τ) = \{ \begin{matrix}∫_{0}^{τ} \frac{1}{u}Y (u) dW (u), & τ > 0, \\ 0, & τ = 0. \end{matrix}

Consequently, M = {M(τ), F^{W} (τ); 0 ≤ τ < ∞} is a continuous local martingale with strictly increasing square variation, which, for τ > 0, is given by

⟨M⟩(τ) = ∫_{0}^{τ} (\frac{1}{u}Y (u))^{2} du = ∫ ∞_{\frac{1}{τ}} A(u)^{2} du.

Then ⟨M⟩ satisfies lim_{τ↓0}⟨M⟩(τ) = 0, a.s., and lim_{τ→∞}⟨M⟩(τ) < ∞ a.s.. By the martingale time change theorem, there exists a standard Brownian motion B̃ such that

M(τ) = B̃(⟨M⟩(τ)), 0 ≤ τ < ∞ a.s.

The law of the iterated logarithm now gives

lim sup_{T ↓0} \frac{|B̃(T)|}{\sqrt{2T log log(1/T)}} = 1, a.s.

But as lim_{τ↓0}⟨M⟩(τ) = 0, a.s., and τ ↦ ⟨M⟩(τ) is continuous and strictly increasing, we have

lim sup_{τ↓0} \frac{|B̃(⟨M⟩(τ)|}{\sqrt{2⟨M⟩(τ) log log(⟨M⟩(τ)^{−1})}} = 1, a.s.

Therefore, as B̃(⟨M⟩(τ)) = M(τ), τ ≥ 0,

lim sup_{τ↓0} \frac{|M(τ)|}{\sqrt{2⟨M⟩(τ) log log(⟨M⟩(τ)^{−1})}} = 1, a.s.

Therefore,

lim sup_{t→∞} \frac{|∫_{t}^{∞} A(s) dB(s)|}{\sqrt{2 ∫_{t}^{∞} A(s)^{2} ds log log (∫_{t}^{∞} A(s)^{2} ds)^{−1}}}

= lim sup_{τ↓0} \frac{| ∫ ∞_{\frac{1}{τ}} A(s) dB(s)|}{\sqrt{2 ∫ ∞_{\frac{1}{τ}} A(s)^{2} ds log log (∫ ∞_{\frac{1}{τ}} A(s)^{2} ds)^{−1}}}

= lim sup_{τ↓0} \frac{|M(τ)|}{\sqrt{2⟨M⟩(τ) log log(⟨M⟩(τ)^{−1})}} = 1, a.s.

which establishes the desired result. □

Acknowledgement The author gratefully acknowledges both David Reynolds, for discussions concerning this problem, and Markus Reiß, who alerted him to the result of Motoo [9].

References

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[3] J. A. D. Appleby and D. W. Reynolds. Subexponential solutions of linear Volterra integro-differential equations and transient renewal equations. Proc. Roy. Soc. Edinburgh. Sect. A, 132A:521–543, 2002.

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[5] J. A. D. Appleby and D. W. Reynolds. Non-exponential stability of scalar stochastic Volterra equations. Statist. Probab. Lett., 62(4):335–343, 2003.

[6] M. A. Berger and V. J. Mizel. Volterra equations with Itô integrals I. J. Integral Equations, 2(3):187–245, 1980.

[7] K. Itô and H. P. McKean. Diffusion processes and their sample paths. Springer, Berlin, 2nd edition, 1974.

[8] I. Karatzas and S. E. Shreve. Brownian Motion and Stochastic Calculus, volume

113 of Graduate Texts in Mathematics. Springer, New York, 1991.

EJQTDE, Proc. 7th Coll. QTDE, 2004 No. 1, p. 31

[9] M. Motoo. Proof of the law of the iterated logarithm through diffusion equation. Ann. Inst. Statis. Math., 10:21–28, 1959.

(Received September 29, 2003)

Centre for Modelling with Differential Equations (CMDE), School of Mathematical Sciences, Dublin City University, Dublin 9, Ireland

E-mail address: john.appleby@dcu.ie
URL: http://webpages.dcu.ie/~applebyj