Asymptotics of a renewal-like recursion and an integral equation
aa r X i v : . [ m a t h . C A ] M a y ASYMPTOTICS OF A RENEWAL-LIKE RECURSIONAND AN INTEGRAL EQUATION ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORIDepartment of Probability Theory and Statistics,E¨otv¨os Lor´and UniversityP´azm´any P. s. 1/C, H-1117 Budapest, Hungary
E-mail address: [email protected], [email protected]
Abstract.
We consider a renewal-like recursion and prove thatthe solution is polynomially decaying under suitable conditions.We prove similar results for the corresponding integral equation.In both cases coefficients and functions are of more general formthan in the classic cases. Introduction
In this paper we examine the asympotics of a renewal-like recursionand a similar integral equation. The motivation comes from probabilitytheory; more precisely, in a random model of publication activity [1]the asymptotic distribution of the weights of the authors satisfy suchequations.The recursion is of the form(1) x n = n − X j =1 w n,j x n − j + r n , w n,j = a j + b j n + c n,j ( n = 1 , , . . . ) , where w n,j ≥
0, and a, b, c are decaying at least exponentially fast. Theprecise assumptions are formulated later. Our goal is to prove that x n is polynomially decaying as n → ∞ under suitable conditions, and todetermine the exponent.Similar recursions are widely examined, see e.g. Milne-Thompson[11], Cooper–Frieze [4]. In those cases either the coefficients are special,or only the last m terms appear on the right-hand side for some fixed m . Now all previous terms are present, and the weights depend bothon n and j . Date : 21 May 2012.2000
Mathematics Subject Classification.
Key words and phrases.
Renewal equation, renewal theory.The European Union and the European Social Fund have provided financialsupport to the project under the grant agreement no. T ´AMOP 4.2.1./B-09/KMR-2010-0003. ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI
On the other hand, omitting ( b n ) and ( c n ) and supposing that ( a n )is a probability distribution, we get the well-known renewal formula [6,Chapter XIII]. The asymptotics were examined in a more subtle wayin [5], for instance, by weaker Tauberian type assumptions. In our casethe coefficients are of more general form; however, we have strongerconditions on them. An example will show that these assumptions cannot be totally omitted (see Remark 4).The continuous counterpart is the following integral equation, whichis a Volterra equation of the second kind.(2) g ( t ) = Z t w t,s g ( t − s ) ds + r ( t )for t > g (0) = 1. The kernel w t,s is supposed to be written inthe following form. 0 ≤ w t,s = a ( s ) + b ( s ) t + d + c t,s , where a is a probability density function, and again, a, b, c are decreas-ing fast. We will show that g ( t ) is between two polynomially decayingfunctions under suitable conditions, and give the exponent. In addi-tion, assuming that g is decreasing, we will prove that g ( t ) is polyno-mially decaying as t → ∞ . We use Laplace transforms and Tauberiantheorems in this part.Omitting b and c we get a classic renewal equation [7, Chapter XI].In Section 2 we formulate the main results for both cases. Sections3 and 4 contain the proofs for the discrete and the continuous cases,respectively. Section 5 contains the Laplace transform methods.2. Main results
The discrete recursion.
Consider the following recursion:(3) x n = n − X j =1 w n,j x n − j + r n , w n,j = a j + b j n + c n,j , ( n = 1 , , . . . ) , where w n,j ≥
0, and a n , b n , c n,j , r n satisfy the following conditions.(r1) a n ≥ n ≥
1, and the greatest common divisor of the set { n : a n > } is 1;(r2) r n ≥
0, and there exists such an n that r n > SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 3 (r3) there exists z > < ∞ X n =1 a n z n < ∞ , ∞ X n =1 | b n | z n < ∞ , ∞ X n =1 n − X j =1 | c n,j | z j < ∞ , ∞ X n =1 r n z n < ∞ . It is clear that x n ≥ n ≥ x n ). It is polynomially de-caying; the exponent is also given. Theorem 1.
Suppose that the sequence ( x n ) satisfies recursion (3) ,conditions (r1)–(r3) hold, and ( x n ) has infinitely many positive terms.Then x n n − γ q n → C as n → ∞ , where C is a positive constant, q is thepositive solution of equation P ∞ n =1 a n q n = 1 , and γ = ∞ P n =1 b n q n ∞ P n =1 na n q n . Remark . The condition on w n,j in recursion (3) can be modified inthe following way. w n,j = a j + b j n − j + c n,j , n = 1 , , . . . . The difference may be added to the remainder term c n,j , because wehave ∞ X n =1 n − X j =1 (cid:12)(cid:12)(cid:12)(cid:12) b j n − j − b j n (cid:12)(cid:12)(cid:12)(cid:12) y j = ∞ X j =1 | b j | y j ∞ X n = j +1 (cid:16) n − j − n (cid:17) = ∞ X j =1 | b j | y j j X n =1 n , which is finite for 0 < y < z . Since the generating function of thesequence ( a n ) is left continuous at point z , there exists y < z such that P ∞ n =1 a n y n > Remark . The condition that the sequence has infinitely many positiveterms is necessary as the following example shows. Let r = 1, r n = 0if n >
1, and w n,j = a n (cid:16) − n − j (cid:17) . Then we get that x = 1, x = x = · · · = 0. ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI The integral equation.
Now we examine an integral equation,which is similar to recursion (3). Namely, let g : R → R be the solutionof the following integral equation; we will explain later why the solutionexists.(4) g ( t ) = Z t w t,s g ( t − s ) ds + r ( t )for t > g (0) = 1. Here0 ≤ w t,s = a ( s ) + b ( s ) t + d + c t,s , and the following conditions hold.(i1) a ∈ L [0 , ∞ ) is a probability density function concentrated onthe set of positive real numbers. That is, a is nonnegative al-most everywhere, and R ∞ a ( s ) ds = 1.(i2) b ∈ L [0 , ∞ ), and d is a positive constant.(i3) r ∈ L [0 , ∞ ) is a nonnegative, continuous function.(i4) c : [0 , ∞ ) → R is (jointly) measurable, c t,s is integrable on[0 , t ] with respect to s for all t >
0, and lim t →∞ c t,s = 0 for a.e. s > z > Z ∞ a ( t ) z t dt < ∞ , Z ∞ | b ( t ) | z t dt < ∞ , and(i6) z t Z t | c t,s | ds and r ( t ) z t are directly Riemann integrable withrespect to t on [0 , ∞ ).Recall that a nonnegative function h is directly Riemann integrableon [0 , ∞ ) (see p. 361 of [7]), if and only if it is (Riemann) integrableon every finite interval, and for all τ > ∞ X n =1 sup nτ ≤ θ ≤ ( n +1) τ h ( θ ) < ∞ , that is, the upper Riemann sum of h with span τ is finite. As usual,we say that a real function h is directly Riemann integrable if both itspositive and negative parts are directly Riemann integrable. This isequivalent to the direct Riemann integrability of | h | .Equation (4) is a nonlinear Volterra-type integral equation of the sec-ond kind. It is easy to check that all conditions of Theorem 3.2. of [10]hold for this equation in a finite interval 0 ≤ t ≤ T . Thus, applying thetheorem we get that the equation has a unique and continuous solutionfor all positive T . Hence g ( t ) is defined on the set of nonnegative realnumbers, and it is continuous. Since the proof of Theorem 3.2. of [10]relies on Picard approximation, and g (0), w , r are all nonnegative, itis clear that g ( t ) is nonnegative for all t ≥ SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 5
Our main results are about the asymptotics of g ( t ) as t → ∞ . Firstwe give the order of g by proving lower and upper bounds. Thenassuming that g is decreasing, we will find the asymptotics of g usingLaplace transforms. Theorem 2.
Let g be the solution of equation (4) . Suppose that w isnonnegative, all conditions (i1)–(i6) hold, and for all T > there exists t > T such that g ( t ) > . Introduce γ = R ∞ b ( s ) ds R ∞ sa ( s ) ds . Then < lim inf t →∞ g ( t ) t − γ ≤ sup t g ( t ) t − γ < ∞ holds. Theorem 3.
Let g be the solution of equation (4) . In addition to theconditions of Theorem 2 suppose that g is decreasing. Then g ( t ) t − γ → C holds for some < C < ∞ as t → ∞ . The discrete case: Proof of Theorem 1.
Preliminaries.
We may assume that P ∞ n =1 a n = 1 and q = 1.Condition (r3) implies that q exists, and q < z . Define ˜ x n = q n x n ,˜ a j = q j a j , ˜ b j = q j b j , ˜ c n,j = q j c n,j , ˜ r n = q n r n for n, j ≥
1. We get that P ∞ n =1 ˜ a n = 1 and˜ x n = n − X j =1 (cid:16) ˜ a j + ˜ b j n − j + ˜ c n,j (cid:17) ˜ x n − j + ˜ r n , n = 1 , , . . . . Moreover, condition (r3) holds with ˜ z = z/q . Thus we may assumethat P ∞ n =1 a n = 1, and z >
1, indeed.
Lemma 1. x n > for every n large enough. Proof. If a k > k , then for every sufficiently large n we have w n,k >
0. To see this, note that lim n →∞ b k n − k = 0, and lim n →∞ c n,k = 0holds for fixed k due to condition (r3). Hence, if x n > n is largeenough, then x n + k >
0. This implies that x n + ℓk > ℓ = 1 , , . . . .Due to condition (r1), every sufficiently large n is a linear combinationof some values of k for which a k >
0. Therefore x n is positive for every n large enough. (cid:3) We need some more notations.Define y n = x n n − γ for n ≥
1. We have(5) y n = n − X j =1 w n,j (cid:16) − jn (cid:17) γ y n − j + r n n − γ , n = 1 , , . . . . ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI From the Taylor expansion of the function f ( x ) = (1 − x ) t for x ≥ | f ( x ) − tx | = (cid:12)(cid:12)(cid:12) t ( t − x (1 − θx ) t − (cid:12)(cid:12)(cid:12) ≤ | t ( t − | x e − θx ( t − ≤ | t ( t − | x e x | t − | ≤ | t ( t − | x e nεx , where 0 ≤ θ ≤
1, and ε > e ε < z holds with z ofcondition (r3), while n is so large that nε > | t − | holds.Therefore(6) (cid:16) − jn (cid:17) γ = 1 − γjn + R n,j , where(7) | R n,j | ≤ | γ ( γ − | j n e jε holds uniformly in j for all n large enough, say n ≥ L . Assuming thatthe coefficients satisfy equation (3) we get that(8) w n,j (cid:16) − jn (cid:17) γ = a j + 1 n (cid:0) b j − γja j (cid:1) − n γjb j + w n,j R n,j + c n,j (cid:16) − γjn (cid:17) . Boundedness of ( y n ) . Our next goal is to prove that the se-quence ( y n ) is bounded from above, and its limes inferior is positive.Before doing so we prove another lemma. Lemma 2.
For every positive integer k we have (9) ∞ X n = k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − k X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) < ∞ . Proof.
Using equation (8) we obtain that ∞ X n = k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − k X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ∞ X n = k (cid:12)(cid:12)(cid:12)(cid:12) n n − k X j =1 (cid:0) b j − γja j (cid:1) − γn n − k X j =1 jb j ++ n − k X j =1 w n,j R n,j + n − k X j =1 c n,j (cid:16) − γjn (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ∞ X n = k n ∞ X j = n − k +1 (cid:12)(cid:12) γja j − b j (cid:12)(cid:12) + | γ | n n − X j =1 j | b j | ++ n − X j =1 | w n,j R n,j | + (cid:0) | γ | (cid:1) n − X j =1 | c n,j | ! . For the first term we used that P ∞ j =1 (cid:0) b j − γja j (cid:1) = 0 holds by thedefinition of γ . SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 7
Let us divide the sum into four parts and examine them separately.By condition (r3) for the first two we have that ∞ X n = k n ∞ X j = n − k +1 (cid:12)(cid:12) γja j − b j (cid:12)(cid:12) ≤ ∞ X n = k ∞ X j = n − k +1 (cid:0) | γ | ja j + | b j | (cid:1) = | γ | ∞ X j =1 j a j + ∞ X j =1 j | b j | < ∞ ; ∞ X n =1 | γ | n n − X j =1 j | b j | ≤ | γ | ∞ X n =1 n ∞ X j =1 j | b j | < ∞ . Moreover, using (7), we obtain that ∞ X n = L n − X j =1 | w n,j R n,j | ≤ | γ ( γ − | ∞ X n = L n − X j =1 (cid:0) a j + | b j | + | c n,j | (cid:1) j n e jε ≤ | γ ( γ − | " ∞ X j =1 ∞ X n = j +1 (cid:0) a j + | b j | (cid:1) j n e jε + ∞ X n =1 n − X j =1 | c n,j | z j < ∞ by condition (r3) and the choice of ε .Similarly, ∞ X n =1 n − X j =1 | c n,j | ≤ ∞ X n =1 n − X j =1 | c n,j | z j < ∞ also holds. Thus we have proved (9). (cid:3) Lemma 3. ( y n ) is bounded from above. Proof.
Let z n = max (cid:8) , y , . . . , y n (cid:9) . Then y n ≤ z n , z n is increasing,and z n ≤ z n − max ( , n − X j =1 w n,j (cid:16) − jn (cid:17) γ + r n n − γ ) ≤ z n − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + r n n − γ ! = z n − (1 + s n ) , where s n = (cid:12)(cid:12)(cid:12)P n − j =1 w n,j (cid:0) − jn (cid:1) γ − a j (cid:12)(cid:12)(cid:12) + r n n − γ .Iterating this we obtain that sup n ≥ z n ≤ Q ∞ n =1 (1 + s n ). In order toshow that this quantity is finite it is sufficient to prove that P ∞ n =1 s n < ∞ holds. The latter is implied by Lemma 2 with k = 1, and by thefact that ∞ X n =1 r n n − γ = O ∞ X n =1 r n z n ! < ∞ . ´AGNES BACKHAUSZ AND TAM ´AS F. M ´ORI Thus the sequence ( y n ) is bounded from above. (cid:3) Lemma 4. lim inf n →∞ y n > . Proof.
This is similar to the upper bound, therefore we only outlinethe proof, omitting the details.Based on Lemma 1, suppose that x n > n ≥ N . By condition(r2) and the definition of y n we have that y n ≥ n − N X j =1 w n,j (cid:16) − jn (cid:17) γ y n − j . We obtain that for z n = min N ≤ j ≤ n y j the following inequality holds. z n ≥ z n − min ( , n − N X j =1 w n,j (cid:16) − jn (cid:17) γ ) ≥ z n − n − N X j =1 a j − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − N X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! = z n − − ∞ X j = n − N +1 a j − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − N X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ! = z n − (1 − s n ) , where s n = ∞ X j = n − N +1 a j + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n − N X j =1 w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . This implies that inf n ≥ N y n ≥ lim n →∞ z n ≥ z N Q ∞ n = N +1 (1 − s n ). For theproof of the positivity of the right-hand side we need to show that P ∞ n = N +1 s n < ∞ . This is a consequence of Lemma 2 with k = N , andthe following estimation. ∞ X n = N +1 ∞ X j = n − N +1 a j ≤ ∞ X j =1 ja j < ∞ . We conclude that inf n ≥ N y n >
0, and hence lim inf n →∞ y n > (cid:3) Final step.
The remaining part of the proof is similar to theproof of the discrete renewal theorem.
SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 9
Equation (8) implies that(10) y n = n − X j =1 (cid:18) a j + b j − γja j n − j − ( b j − γja j ) jn ( n − j ) − n γjb j ++ w n,j R n,j + c n,j (cid:16) − γjn (cid:17)(cid:19) y n − j + r n n − γ . Fix a positive integer N . Then we obtain from (10) by summationthat(11) N X n =1 y n = N X n =1 n − X j =1 (cid:18) a j + b j − γja j n − j (cid:19) y n − j + u N with an appropriately chosen sequence ( u N ). Here u N is convergent as N → ∞ ; let u denote the limit. In order to show this, since ( y n ) isbounded, it is sufficient to prove that ∞ X n =1 n − X j =1 (cid:18) | b j − γja j | jn ( n − j ) + 1 n | γb j | j + | w n,j R n,j | + | c n,j | (cid:0) | γ | (cid:1)(cid:19) < ∞ ; ∞ X n =1 r n n − γ < ∞ . We have almost done it before; the only thing left is to show the con-vergence for the first term in the double sum. In this case ∞ X n =1 n − X j =1 | b j − γja j | jn ( n − j ) ≤ ∞ X n =1 n − X j =1 j | b j | + | γ | j a j n ( n − j )= ∞ X j =1 (cid:0) j | b j | + | γ | j a j (cid:1) ∞ X n = j +1 n ( n − j ) ≤ ∞ X j =1 (cid:0) j | b j | + | γ | j a j (cid:1) ∞ X n =1 n < ∞ . Introduce variable m = n − j instead of n in equation (11). Then N X n =1 y n = N − X j =1 N X n = j +1 (cid:18) a j + b j − γja j n − j (cid:19) y n − j + u N = N − X j =1 N − j X m =1 (cid:18) a j + b j − γja j m (cid:19) y m + u N = N − X m =1 y m N − m X j =1 (cid:18) a j + b j − γja j m (cid:19) + u N . Since ∞ X j =1 (cid:18) a j + b j − γja j m (cid:19) = 1 , we have N X n =1 y n = N X m =1 y m − ∞ X j = N − m +1 (cid:18) a j + b j − γja j m (cid:19)! + u N . These imply that N − X m =0 y N − m ∞ X j = m +1 a j = N X m =1 y m ∞ X j = N − m +1 a j = u N − N X m =1 y m ∞ X j = N − m +1 b j − γja j m . The second term on the right-hand side converges to 0 as N → ∞ ,because N X m =1 y m ∞ X j = N − m +1 | b j − γja j | m ≤ (sup n y n ) N X m =1 m ∞ X j = N − m +1 (cid:0) | b j | + | γ | ja j (cid:1) . The sum inside is estimated in the following way. Let ε > e ε < z holds. Then ∞ X j = N − m +1 (cid:0) | b j | + | γ | ja j (cid:1) ≤ K e − ( N − m ) ε , where K = ∞ X j =1 (cid:0) | b j | + | γ | ja j (cid:1) e jε < ∞ . Now, using notation M = h √ N i , we get that N X m =1 Km e − ( N − m ) ε ≤ N − M X m =1 Km e − ( N − m ) ε + N X m = N − M +1 Km e − ( N − m ) ε ≤ N Ke − Mε + M KN − M , which tends to 0 as N → ∞ . We conclude that(12) N − X m =0 y N − m ∞ X j = m +1 a j → u ( N → ∞ ) . Modifying the proof of Lemma 2, namely, using equation (8), condi-tion (r3) and the fact that the sequence of arithmetic means converges
SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 11 to zero if the original sequence is nonnegative and converges to zero, itis easy to see that n − X j =1 (cid:12)(cid:12)(cid:12)(cid:12) w n,j (cid:16) − jn (cid:17) γ − a j (cid:12)(cid:12)(cid:12)(cid:12) → n → ∞ ) . This and (5), together with the boundedness of ( y n ) imply that(13) y n − n − X j =1 a j y n − j → , as n → ∞ .From now on the argument is the usual one.Let ( n k ) be a subsequence of the natural numbers that satisfieslim k →∞ y n k = lim sup n →∞ y n =: y. From (13) we get that for all ℓ ≤ M the following estimation holds. y = lim k →∞ y n k ≤ a ℓ lim inf k →∞ y ( n k − ℓ ) + lim sup k →∞ X j
0. By iteration we obtain thatlim k →∞ y ( n k − ℓ − ... − ℓ i ) = y, for all positive a ℓ , . . . , a ℓ i . By condition (r1) for all sufficiently large m we have lim k →∞ y n k − m = y . Modifying the subsequence we may assumethat this holds for all m = 0 , , . . . . Hence choosing N = n k in (12) wecan see that y ∞ X m =0 ∞ X j = m +1 a j = u. For y = lim inf n →∞ y n the same argument shows that y ∞ X m =0 ∞ X j = m +1 a j = u. Hence y = y , that is, the limit lim n →∞ y n = C exists. We have alreadyproved that this is finite and positive. This implies Theorem 1. (cid:3) Remark . It is well known that if y n = P n − j =1 a j y n − j holds for all n ,instead of in (13), then y n is convergent. On the other hand, (13) isnot yet sufficient for the convergence of ( y n ). For example, let y n =2 + sin (cid:0) log(1 + n ) (cid:1) . Then | y n − y n − j | ≤ log(1 + n ) − log(1 + n − j ) ≤ j n − j , hence (cid:12)(cid:12)(cid:12)(cid:12) y n − n − X j =1 a j y n − j (cid:12)(cid:12)(cid:12)(cid:12) ≤ y n ∞ X j = n a j + n − X j =1 | y n − y n − j | a j ≤ ∞ X j = n a j + 3 n − X j =1 ja j n − j ≤ ∞ X j = n a j + 3 M − X j =1 ja j n − M + 3 n − X j = M ja j ≤ ∞ X j = n a j + 3 n − M ∞ X j =1 ja j + 3 ∞ X j = M ja j . This converges to zero with M = n/
2, but y n does not converge. Remark . If w k,i = a i , then x k → C by the arithmetic version ofthe renewal theorem. The following example shows that a remainder,though converging to 0, may change this. Let ( a i ) be arbitrary, x k =2 + sin (cid:0) log( k + 1) (cid:1) , and w k,i = a i + (cid:18) x k − k − X j =1 x k − j a j (cid:19)(cid:18) k − X j =1 x j (cid:19) − . Then w k,i = a i + o (1), and x k = k − P i =1 x k − i w k,i + 2 δ k, .4. The continuous case: Proof of Theorem 2.
Preliminaries.Lemma 5.
Under the conditions of Theorem 2 we have g ( t ) > forevery t large enough. Proof.
Choose 0 < s and δ > S ′ = { s ∈ (0 , s ) : a ( s ) > δ } has positive Lebesgue measure.Let ℓ ( s ) = sup { t ≥ s : | c t,s | ≥ δ } . Then ℓ ( s ) < ∞ for a.e. s > ℓ is not necessarily Borel measurable, yet itis Lebesgue measurable by the measurable projection theorem, for thesuperlevel set { ℓ > K } is just the projection of the two dimensionalmeasurable set { ( s, t ) : 0 < s ≤ t, K < t, | c t,s | ≥ δ } onto the first SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 13 coordinate. Hence U t = { ℓ < t } , t > , s ) \ U t tends to 0 as t → ∞ . The same holds for the sequence V t = { s ∈ (0 , t ] : | b ( s ) | ≤ δ ( t + d ) } . Thus we can find a threshold T ≥ s suchthat the Lebesgue measure of S = S ′ ∩ U T ∩ V T is positive. Obviously, w t,s ≥ δ for all t ≥ T and s ∈ S . By the Lebesgue density theorem wemay assume that S only consists of points with density 1.By the continuity of g there exists a whole open interval I above T where g ( t ) is separated from zero. Let ε denote the length of I , and η > g over I . Then for t ∈ S + I we have g ( t ) ≥ Z t w t,s g ( t − s ) ds ≥ Z S ∩ ( t − I ) w t,s g ( t − s ) ds ≥ δη λ (cid:0) S ∩ ( t − I ) (cid:1) > , where λ stands for the Lebesgue measure.Since the set sum S + I is an open set, we can iterate this procedureto obtain that g is positive everywhere on the set I ∪ ( S + I ) ∪ ( S + S + I ) ∪ ( S + S + S + I ) ∪ . . . . The proof can be completed by showing that this set contains everysufficiently large real number. In other words, if t is large enough, thenit can be written in the form t = s + · · · + s n + r , where s , . . . , s n ∈ S , n ∈ N , and 0 < r < ε . In fact, this is true for arbitrary S ⊂ R + thathas two incommensurable elements α and β (hence for every set ofpositive Lebesgue measure). Indeed, by the equidistribution theoremthere exists positive integers k and m such that k < m αβ < k + εβ , that is, kβ < mα , and their distance is less than ε . Consequently, inthe finite sequence nkβ < ( n − kβ + mα < ( n − kβ + 2 mα < · · · < nmα the distance between neighbouring terms is less than ε . If n is largeenough, then ( n + 1) kβ < nmα , i.e., the largest term of the sequenceabove is bigger than the smallest term of the next sequence. Thusevery t ≥ nkβ is sufficiently close to a positive linear combination of α and β . (cid:3) Let us introduce the notation H ( t ) = g ( t ) ( t + d ) − γ ( t ≥ . From (4) we obtain the following integral equation for H .(14) H ( t ) = Z t w t,s (cid:18) t − s + dt + d (cid:19) γ H ( t − s ) ds + r ( t ) ( t + d ) − γ for t >
0, and H (0) = d − γ .Let us choose ε in a similar way as we did in the discrete case.Namely, we have e ε < z with z of condition (i5). In what followsequations (15), (16), and (17) correspond to (6), (7), and (8), resp.Firstly,(15) (cid:18) t − s + dt + d (cid:19) γ = (cid:18) − st + d (cid:19) γ = 1 − γst + d + R t,s , where(16) | R t,s | ≤ | γ ( γ − | s t e sε , if t is large enough, say t ≥ L . Finally, from the decomposition of w t,s we get that(17) w t,s (cid:16) − st + d (cid:17) γ = a ( s ) + b ( s ) − γsa ( s ) t + d − γsb ( s )( t + d ) + w t,s R t,s + c t,s (cid:16) − γst + d (cid:17) . Boundedness of H . The method of proof is discretization; inthis way all we need to do is similar to what we did in the discretecase.Before proving boundedness we need another lemma.
Lemma 6.
For every fixed T ≥ the function A ( t ) = (cid:12)(cid:12)(cid:12)(cid:12)Z t − T (cid:18) w t,s (cid:16) − st + d (cid:17) γ − a ( s ) (cid:19) ds (cid:12)(cid:12)(cid:12)(cid:12) is directly Riemann integrable on [ T, ∞ ) . Proof.
Fix τ >
0. Then(18) ∞ X n =1 sup nτ ≤ θ ≤ ( n +1) τ Z θ | c θ,s | z s ds < ∞ , according to condition (i6). Now we prove that(19) ∞ X n = (cid:6) Tτ (cid:7) sup nτ ≤ θ ≤ ( n +1) τ A ( θ ) = ∞ X n = (cid:6) Tτ (cid:7) sup ≤ θ ≤ τ A ( nτ + θ ) < ∞ . SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 15
From the definition of γ it follows that R ∞ (cid:0) b ( s ) − γsa ( s ) (cid:1) ds = 0. Usingthis and equation (17) we obtain that A ( nτ + θ ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z nτ + θ − T (cid:18) w nτ + θ,s (cid:16) − snτ + θ + d (cid:17) γ − a ( s ) (cid:19) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z nτ + θ − T b ( s ) − γsa ( s ) nτ + θ + d ds − γ ( nτ + θ + d ) Z nτ + θ − T sb ( s ) ds + Z nτ + θ − T w nτ + θ,s R nτ + θ,s ds + Z nτ + θ − T c nτ + θ,s (cid:16) − γsnτ + θ + d (cid:17) ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ Z ∞ nτ + θ − T | b ( s ) − γsa ( s ) | nτ + θ + d ds + | γ | ( nτ + θ + d ) Z nτ + θ s | b ( s ) | ds + Z nτ + θ | w nτ + θ,s R nτ + θ,s | ds + (cid:0) | γ | (cid:1) Z nτ + θ | c nτ + θ,s | ds holds for all θ > ∞ X n = (cid:6) Tτ (cid:7) sup ≤ θ ≤ τ nτ + θ + d Z ∞ nτ + θ − T | b ( s ) − γsa ( s ) | ds ≤ ∞ X n = (cid:6) Tτ (cid:7) nτ + d Z ∞ nτ − T | b ( s ) − γsa ( s ) | ds ≤ ∞ X n =1 nnτ + d + T Z ( n +1) τnτ | b ( s ) − γsa ( s ) | ds ≤ τ Z ∞ (cid:0) | γ | sa ( s ) + | b ( s ) | (cid:1) ds, which is finite by condition (i5).The sum of the second terms is also finite by condition (i5). ∞ X n =1 sup ≤ θ ≤ τ | γ | ( nτ + θ + d ) Z nτ + θ s | b ( s ) | ds ≤ | γ | τ ∞ X n =1 n Z ∞ s | b ( s ) | ds < ∞ . By inequality (16), condition ( i ε in the pre-liminaries, we obtain the following estimation for the sum of the third terms. ∞ X n = (cid:6) Lτ (cid:7) sup ≤ θ ≤ τ Z nτ + θ | w nτ + θ,s R nτ + θ,s | ds ≤ ∞ X n = (cid:6) Lτ (cid:7) sup ≤ θ ≤ τ Z nτ + θ (cid:18) a ( s ) + | b ( s ) | d + | c nτ + θ,s | (cid:19) | γ ( γ − | s nτ + θ ) e sε ≤ | γ ( γ − | ∞ X n =1 Z ( n +1) τ (cid:18) a ( s ) + | b ( s ) | d (cid:19) s τ n e sε ds + | γ ( γ − | ∞ X n =1 sup ≤ θ ≤ τ Z nτ + θ | c nτ + θ,s | z s ds. In the right-hand side the last sum is finite by (18). In the first sumthe integrand is nonnegative, hence by Fubini’s theorem we obtain that ∞ X n =1 Z ( n +1) τ (cid:18) a ( s ) + | b ( s ) | d (cid:19) s τ n e sε ds ≤ τ Z ∞ ∞ X n =1 n ! (cid:18) a ( s ) + | b ( s ) | d (cid:19) s e sε ds < ∞ , by the choice of ε . Thus the sum of the third terms is finite, too.Finally, for the sum of the fourth terms we clearly have ∞ X n =1 sup ≤ θ ≤ τ Z nτ + θ | c nτ + θ,s | ds ≤ ∞ X n =1 sup ≤ θ ≤ τ z nτ + θ Z nτ + θ | c nτ + θ,s | ds < ∞ . Putting these together we obtain that all four parts of A give finitesums, hence (19) holds. The nonnegativity and integrability of A isclear, for it is a continuous function of t . Thus the proof of the lemmais completed. (cid:3) Lemma 7. H ( t ) is bounded from above. We define Z ( t ) = max { , H ( s ) : 0 ≤ s ≤ t } for t ≥
0. This is finite,because H , as well as g , is continuous.First we give an upper bound for sup <θ ≤ τ H ( t + θ ), where t and τ are fixed positive numbers. Introduce w ∗ t,s = w t,s (cid:18) − st + d (cid:19) γ < t, ≤ s ≤ t. SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 17
Using the nonnegativity of w and r , equation (14), and the definitionof Z , we get that(20) H ( t + θ ) = Z θ w ∗ t + θ,s H ( t + θ − s ) ds + Z t + θθ w ∗ t + θ,s H ( t + θ − s ) ds + r ( t + θ )( t + θ + d ) − γ ≤ Z θ w ∗ t + θ,s ds Z ( t + θ ) + (cid:20)Z t + θθ w ∗ t + θ,s ds + r ( t + θ )( t + θ + d ) − γ (cid:21) Z ( t )= Z θ w ∗ t + θ,s ds [ Z ( t + θ ) − Z ( t )]+ (cid:20)Z t + θ w ∗ t + θ,s ds + r ( t + θ )( t + θ + d ) − γ (cid:21) Z ( t ) . Next we want to prove that there exists τ >
0, and for every τ ,0 < τ ≤ τ , a positive integer N ( τ ) such that(21) sup ≤ θ ≤ τ Z θ w ∗ nτ + θ,s ds ≤ , provided n > N ( τ ).To show this we will give an upper bound on(22) w ∗ nτ + θ,s = (cid:18) − snτ + θ + d (cid:19) γ (cid:18) a ( s ) + b ( s ) nτ + θ + d + c nτ + θ,s (cid:19) . We clearly have (cid:18) − snτ + θ + d (cid:19) γ ≤ (cid:18) θd (cid:19) | γ | ≤ exp (cid:16) θ | γ | d (cid:17) , and a ( s ) + b ( s ) nτ + θ + d ≤ a ( s ) + | b ( s ) | d . Hence, for 0 ≤ θ ≤ τ we can write Z θ w ∗ nτ + θ,s ds ≤ exp (cid:16) θ | γ | d (cid:17) (cid:20) Z θ (cid:16) a ( s ) + | b ( s ) | d (cid:17) ds + Z θ | c nτ + θ,s | ds (cid:21) ≤ exp (cid:16) τ | γ | d (cid:17) Z τ (cid:16) a ( s ) + | b ( s ) | d (cid:17) ds + z nτ + θ Z θ | c nτ + θ,s | ds, if n is large enough, namely, n ≥ | γ | / dε will do. The first term in theright-hand side can be arbitrarily small if τ is fixed small enough. Asto the second term, it can be estimated in the following way.sup ≤ θ ≤ τ z nτ + θ Z θ | c nτ + θ,s | ds ≤ sup nτ ≤ θ ≤ ( n +1) τ z θ Z θ | c θ,s | ds, which tends to 0 as n → ∞ by condition (i6). Thus (21) is satisfied if n is greater than a certain threshold N ( τ ).For any 0 < τ ≤ τ and t = nτ , n > N ( τ ) inequality (20) impliesthatsup <θ ≤ τ H ( t + θ ) ≤
12 [ Z ( t + τ ) − Z ( t )] ++ sup <θ ≤ τ (cid:20)Z t w ∗ t + θ,s ds + r ( t + θ + d )( t + θ + d ) − γ (cid:21) Z ( t ) . Here we use that Z is nonnegative and increasing by definition.We clearly have Z ( t + τ ) = max (cid:8) Z ( t ) , sup <θ ≤ τ H ( t + θ ) (cid:9) . There-fore we obtain that Z ( t + τ ) ≤ max ( Z ( t ) ,
12 [ Z ( t + τ ) − Z ( t )] ++ sup <θ ≤ τ (cid:20)Z t w ∗ t + θ,s ds + r ( t + θ )( t + θ + d ) − γ (cid:21) Z ( t ) ) , from which it follows that Z ( t + τ ) − Z ( t ) ≤
12 [ Z ( t + τ ) − Z ( t )] ++ sup <θ ≤ τ (cid:20)Z t w ∗ t + θ,s ds − r ( t + θ )( t + θ + d ) − γ (cid:21) Z ( t ) ! + ;where x + denotes max ( x, Z ( t + τ ) − Z ( t ) ≤ sup <θ ≤ τ (cid:20)Z t w ∗ t + θ,s ds − r ( t + θ )( t + θ + d ) − γ (cid:21)! + Z ( t ) . We continue with deriving an upper bound for the right-hand side.Since a is a probability density function, we havesup <θ ≤ τ (cid:20)Z t + θ w ∗ t + θ,s ds − (cid:21) ≤ sup <θ ≤ τ (cid:20)Z t + θ w ∗ t + θ,s ds − Z t + θ a ( s ) ds (cid:21) ≤ sup <θ ≤ τ (cid:12)(cid:12)(cid:12)(cid:12)Z t + θ (cid:0) w ∗ t + θ,s − a ( s ) (cid:1) ds (cid:12)(cid:12)(cid:12)(cid:12) . SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 19
Therefore Z ( t + τ ) − Z ( t ) ≤ Z ( t ) sup <θ ≤ τ (cid:12)(cid:12)(cid:12)(cid:12)Z t + θ (cid:0) w ∗ t + θ,s − a ( s ) (cid:1) ds (cid:12)(cid:12)(cid:12)(cid:12) ++ sup <θ ≤ τ r ( t + θ )( t + θ + d ) − γ ! for all 0 < τ ≤ τ , t = nτ , n ≥ N ( τ ).Similarly to Lemma 3 of the discrete case, for the boundedness of Z ( nτ ) from above it suffices to prove that(23) ∞ X n =1 sup <θ ≤ τ (cid:12)(cid:12)(cid:12)(cid:12)Z nτ + θ (cid:0) w ∗ nτ + θ,s − a ( s ) (cid:1) ds (cid:12)(cid:12)(cid:12)(cid:12) ++ ∞ X n =1 sup ≤ θ ≤ τ r ( nτ + θ )( nτ + θ + d ) − γ < ∞ . Lemma 6 with T = 0 implies that the first sum is finite. Since bycondition (i6) r ( t ) z t is directly Riemann integrable, it follows that thesecond sum is also finite. Thus we conclude that the sequence Z ( nτ ) = max { , H ( s ) : 0 ≤ s ≤ nτ } is bounded from above if τ is small enough. Hence the function H isalso bounded from above. (cid:3) Lemma 8. lim inf t →∞ H ( t ) > . Proof.
Like in the discrete case, we omit the details that are straight-forward modifications of the previous lemma, and only give a sketch ofthe proof.Based on Lemma 5, we can suppose that H ( t ) > t ≥ T .This time define Z ( t ) = min { H ( s ) : T ≤ s ≤ t } , for t ≥ T .Let us derive a lower bound for H ( t + θ ), where t > T and θ > H ( t + θ ) = Z t + θ w ∗ t + θ,s H ( t + θ − s ) ds + r ( t + θ )( t + θ + d ) − γ ≥ Z θ w ∗ t + θ,s H ( t + θ − s ) ds + Z t − T + θθ w ∗ t + θ,s H ( t + θ − s ) ds ≥ Z ( t + θ ) Z θ w ∗ t + θ,s ds + Z ( t ) Z t − T + θθ w ∗ t + θ,s ds = [ Z ( t + θ ) − Z ( t )] Z θ w ∗ t + θ,s ds + Z ( t ) Z t − T + θ w ∗ t + θ,s ds. Now Z is decreasing. Applying (21) we obtain that H ( t + θ ) ≥
12 [ Z ( t + θ ) − Z ( t )] + Z ( t ) Z t − T + θ w ∗ t + θ,s ds for 0 < θ ≤ τ . Taking infimum, subtracting Z ( t ) and using that a is aprobability density function we get that Z ( t + τ ) − Z ( t ) ≥ min ( ,
12 [ Z ( t + τ ) − Z ( t )] ++ Z ( t ) (cid:20) inf <θ ≤ τ Z t − T + θ w ∗ t + θ,s ds − (cid:21) ) , from which it follows that Z ( t + τ ) − Z ( t ) ≥ ( , Z ( t ) (cid:20) inf <θ ≤ τ Z t − T + θ w ∗ t + θ,s ds − (cid:21) ) ≥ − " sup <θ ≤ τ (cid:12)(cid:12)(cid:12)(cid:12)Z t − T + θ (cid:0) w ∗ t + θ,s − a ( s ) (cid:1) ds (cid:12)(cid:12)(cid:12)(cid:12) + Z ∞ t − T a ( s ) ds Z ( t ) . Similarly to Lemma 4, in order to prove that lim n →∞ Z ( T + nτ ) > ∞ X n =1 Z ∞ ( n − τ a ( s ) ds + ∞ X n =1 sup <θ ≤ τ (cid:12)(cid:12)(cid:12)(cid:12)Z nτ + θ (cid:0) w ∗ T + nτ + θ,s − a ( s ) (cid:1) ds (cid:12)(cid:12)(cid:12)(cid:12) < ∞ . For the first term we have ∞ X n =1 Z ∞ ( n − τ a ( s ) ds ≤ τ Z ∞ sa ( s ) ds < ∞ by condition (i5). The finiteness of the second term follows directlyfrom Lemma 6.Thus we proved that lim n →∞ Z ( T + nτ ) >
0, which immediatelyimplies that lim inf t →∞ H ( t ) >
0, as needed. (cid:3) The monotonic case: Proof of Theorem 3
First note that γ ≤ g is decreas-ing.We consider the following integral equation.(24) g ( x ) = Z x g ( x − u ) a ( u ) du + Z x g ( x − u ) b ( u ) x + d du ++ Z x g ( x − u ) c x,u du + r ( x ) . SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 21
In the sequel we define the Laplace transform of an integrable func-tion f as F ( s ) = lim y →∞ Z y e − sx f ( x ) dx for s ∈ C , provided the limit exists and it is finite.Denote the Laplace transforms of functions g , a , b , r by G , A , B , R , respectively. These Laplace transforms are well defined and holo-morphic on the half-plane H = { s ∈ C : Re s > } ; this follows from theconditions on a , b , r and Theorem 2. Moreover, A , B , and R are alsoholomorphic in a neighbourhood of the origin. Let f be either of thefunctions above, then we have F ′ ( s ) = − Z ∞ e − sx xf ( x ) dx, s ∈ H . Multiplying both sides of equation (24) by ( x + d ) e − sx , then inte-grating, and using the well known properties of the Laplace transformwe obtain that − G ′ ( s ) + d · G ( s ) = − [ G ( s ) A ( s )] ′ ++ d · G ( s ) A ( s ) + G ( s ) B ( s ) + C ( s ) − R ′ ( s ) + d · R ( s )for s ∈ H , where C ( s ) = Z ∞ e − sx ( x + d ) Z x g ( x − u ) c x,u du dx. This is finite and holomorphic in a neighbourhood of 0 by condition(i6) and Theorem 2.After rearranging we have G ′ ( s ) = G ( s ) (cid:20) d − B ( s ) − A ′ ( s )1 − A ( s ) (cid:21) + R ′ ( s ) − d · R ( s ) − C ( s )1 − A ( s ) . This is an inhomogeneous linear differential equation of order one for G . Restricted to the set of positive real numbers we know that thesolution is unique with any condition of type G ( s ) = t , and there isan explicit formula for it. Introducing the notations L ( s ) = d − B ( s ) − A ′ ( s )1 − A ( s ) , R ∗ ( s ) = − R ′ ( s ) − d · R ( s ) − C ( s )1 − A ( s ) , all solutions of the differential equation can be obtained in the form(25) G ( s ) = exp (cid:18) − Z s L ( t ) dt (cid:19) (cid:20) C + Z s R ∗ ( t ) exp (cid:18)Z t L ( u ) du (cid:19) dt (cid:21) for s >
0, with an appropriate constant C .From the results of Theorem 2 it follows that G ( s ) → s goes toinfinity on the real line. On the other hand, L ( t ) → d > t → ∞ ,thus the first exponential tends to infinity as s → ∞ . Hence there can exist at most one C for which equation (25) is satisfied. Conditions (i5)and (i6) imply that R ∗ ( t ) ≤ C /t for some C ; in addition, L ( t ) ≤ C d also holds for t > /
2. Therefore we have Z + ∞ s R ∗ ( t ) exp (cid:18)Z st L ( u ) du (cid:19) dt ≤ Z + ∞ s C t exp (cid:0) C d ( s − t ) (cid:1) dt ≤ C s with some constant C , if s > / C = Z + ∞ R ∗ ( t ) exp (cid:18)Z t L ( u ) du (cid:19) dt, which is finite, in (25) we get that(26) G ( s ) = Z + ∞ s R ∗ ( t ) exp (cid:18)Z st L ( u ) du (cid:19) dt, for s >
0, and this G ( s ) → s → ∞ on the real line. Hence this isthe Laplace transform of g on the set of positive numbers.Since G ( s ) is holomorphic on the half-plane H , it is the unique ex-tension of the solution given above. The right-hand side of (26) iswell-defined on H , giving a holomorphic function on H , which extends G from the set of positive real numbers to H . Thus we have that theLaplace transform of g is given by (26) on the whole half-plane H .Now we examine the behaviour of G around zero. In what follows h , h , . . . will always denote functions that are holomorphic in a neigh-bourhood of the origin. Let us start with L . Using the Taylor expansionof the exponential function we get(27) A ( s ) = 1 − s Z ∞ ta ( t ) dt + s Z ∞ t a ( t ) dt + . . . , s ∈ C , which implies that L ( s ) = d − B ( s ) − A ′ ( s )1 − A ( s ) = B (0) − A ′ (0) s R ∞ ta ( t ) dt + h ( s ) = − γ + 1 s + h ( s ) . Furthermore we have Z s L ( t ) dt = ( γ + 1) log s + h ( s ) , s ∈ H . Here we chose an arbitrary holomorphic branch of the logarithm onthe right half-plane H . Once the logarithm is defined on H , then s γ +1 and s − ( γ +1) are also meaningful there. Thus we obtain that(28) exp (cid:18)Z st L ( u ) du (cid:19) = (cid:18) ts (cid:19) γ +1 exp (cid:0) h ( t ) − h ( s ) (cid:1) , s, t ∈ H . SYMPTOTICS OF RENEWAL-LIKE EQUATIONS 23
One can similarly derive that sR ∗ ( s ) is holomorphic in a neighbour-hood of 0, hence(29) R ∗ ( s ) exp (cid:0) h ( s ) (cid:1) = h ( s ) s , s ∈ H . Finally, from equation (26) we obtain that(30) G ( s ) = Z + ∞ s R ∗ ( t ) exp (cid:18)Z st L ( u ) du (cid:19) dt = s − ( γ +1) exp (cid:0) − h ( s ) (cid:1) Z + ∞ s t γ +1 R ∗ ( t ) exp (cid:0) h ( t ) (cid:1) dt, s ∈ H . Suppose first that γ is not a negative integer, and consider onlypositive values of s . Then, with a sufficiently small positive ε , by (29)we have Z ∞ s t γ +1 R ∗ ( t ) exp (cid:0) h ( t ) (cid:1) dt = C + Z εs t γ +1 R ∗ ( t ) exp (cid:0) h ( t ) (cid:1) dt = C + Z εs t γ h ( t ) dt = C + s γ +1 h ( s )for s ∈ (0 , ε ), where C is a constant. Hence G ( s ) = exp (cid:0) − h ( s ) (cid:1) (cid:0) C s − ( γ +1) + h ( s ) (cid:1) = h ( s ) + s − ( γ +1) h ( s ) , from which the k th derivative of G can be written in the following form. G ( k ) ( s ) = h ( s ) + s − ( γ +1+ k ) h ( s ) . Choose a positive integer k such that 0 < γ + k + 1, then it followsthat(31) s γ + k +1 G ( k ) ( s ) → K as s → +0, with some finite constant K .Before going further, we prove a similar relation for γ = − k , where k is a positive integer. In this case we have Z ∞ s t γ +1 R ∗ ( t ) exp (cid:0) h ( t ) (cid:1) dt = C + Z εs t γ +1 R ∗ ( t ) exp (cid:0) h ( t ) (cid:1) dt = C + Z εs t γ h ( t ) dt = C + s γ +1 h ( s ) + C log s, where C and C are constants, and s ∈ (0 , ε ). The term log s comesfrom the ( k − h . Then G ( s ) = exp (cid:0) − h ( s ) (cid:1) (cid:0) h ( s ) + C s k − log s (cid:1) = h ( s ) + s k − h ( s ) log s, consequently, G ( k − ( s ) = h ( s ) + h ( s ) log s , and finally G ( k ) ( s ) = h ( s ) s + h ( s ) log s, s ∈ (0 , ε ) . This implies that s G ( k ) ( s ) → K as s → +0, with some finite constant K . Thus, (31) remains valid fornegative integer values of γ .Now we apply Karamata’s Tauberian theorem (see e.g. [7, Theo-rem XIII.5.2], [2, Theorem 1.7.1]). We will use the following notation.Functions v and w are asymptotically equal to each other, that is, v ( x ) ∼ w ( x ) as x → ∞ ) if v/w tends to 1 as x → ∞ ). v ( x ) ∼ · w ( x ) means that v/w tends to 0 as x → ∞ ). Thelatter is the same as v = o ( w ). Theorem A.
Let U be a non-decreasing right-continuous function on (0 , ∞ ) such that its Laplace transform ω ( s ) = R ∞ e − sx dU ( x ) exists for s > . If ℓ is slowly varying at infinity, ≤ ρ < ∞ , and ≤ c < ∞ ,then each of the relations ω ( s ) ∼ cs − ρ ℓ (cid:18) s (cid:19) as s → +0 and U ( t ) ∼ c ρ + 1) t ρ ℓ ( t ) as t → ∞ implies the other. We apply this theorem to U ( x ) = R x g ( u ) u k du , for which ω is con-stant times G ( k ) . From equation (31) we get that(32) Z x g ( u ) u k du ∼ A k x γ + k +1 as x → ∞ , for some A k ≥
0. Note that the constant A k depends on k .In order to finish the proof of Theorem 3 we need another Tauberiantype theorem, giving the asymptotics of g ( u ) u k from the asymptoticsof its integral function. We will use the monotonicity of g at this point.We say that a function f is slowly oscillating if for any ε > δ > f ( u ) < f ( x ) (1 + ε ) holds for all x < u
0, any δ > δ ) k < ε would satisfythe condition.Slow oscillation is generally a sufficient condition of Tauberian typetheorems. For example, Theorem 17.2. of [9] states the following. Theorem B.
Let f be defined on an interval ( a, ∞ ) , and suppose that f ( x ) ∼ Ax α as x → ∞ with some real numbers α, A . If f is m timesdifferentiable and lim inf f ( m ) ( y ) − f ( m ) ( x ) x α − m ≥ as x → ∞ and < y/x → , then f ( j ) ( x ) ∼ Aα ( α − . . . ( α − j + 1) x α − j as x → ∞ , for ≤ j ≤ m . Based on equation (32), we can apply this theorem with f ( x ) = − Z x g ( u ) u k du, m = j = 1 , and α = γ + k + 1 . Then we get that g ( x ) x − γ is convergent as x → ∞ .From Theorem 2 it follows that the limit of g ( x ) x − γ is positive andfinite, which is just our Theorem 3. (cid:3) Remark . Since the Laplace transform method usually gives only localresults in the discrete case, and we needed global results there, it isreasonable to use classic renewal techniques. On the other hand, thosemethods rely on convolution in the continuous case, which was notuseful for our integral equation.
Acknowledgement.
Authors are indebted to G´abor Hal´asz for his in-valuable help with Laplace transforms.
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Department of Probability Theory and Statistics, E¨otv¨os Lor´andUniversity, P´azm´any P. s. 1/C, H-1117 Budapest, Hungary
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