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Lemma 4 suppose that q satisfies the difference

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Lemma 4.Suppose thatqsatisfies the difference differential equation(1.1)withβ=1. Suppose further that there exists a positive real numberλsuch thatq(u) = O(eλu),0< u→ ∞.Then there exists a complex numberAsuch that for allu >0,q(u) =Aq(u),whereqis given by Definition 1 withβ=1, i.e.q(u) =Aintegraldisplay0expbraceleftbiggux+msummationdisplayj=1αjintegraldisplayx01evjttdtbracerightbiggdx,u >0.Proof.Partial summation enables us to rewrite the differential difference equa-tion (1.1) in the form(uq(u))=msummationdisplayj=1cj{q(u+vj)q(u+vj1)},u >0,
DIFFERENCE DIFFERENTIAL EQUATIONS7wherecj=msummationdisplayi=jαi,j= 1,2, . . . , m.Now integrate. There exists a complex numberAsuch that for allu >0,uq(u) =A+msummationdisplayj=1cjintegraldisplayvjvj-1q(u+t)dt.(4.3)LetM:=4λmsummationdisplayj=1|cj|eλvj.The exponential growth requirement implies that there existu0MandB >0such that for alluu0,|q(u)| ≤Beλu.Inserting this inequality into (4.3) yields|uq(u)| ≤ |A|+Bmsummationdisplayj=1|cj|integraldisplayvjvj-1eλ(u+t)dt,uu0.It follows that|q(u)| ≤|A|u+Beλuλumsummationdisplayj=1|cj|eλvj|A|u+Beλu4,uu0.(4.4)Now we claim that|q(x)| ≤2|A|/xforxu0. To prove the claim, fixxu0andobserve that ifBeλx/4≤ |A|/x, then (4.4) gives|q(x)| ≤2|A|/x.On the other hand,ifBeλx/4>|A|/x, then we must haveBeλu/4>|A|/ufor allux. Since (4.4)holds for alluu0, we then get|q(u)| ≤Beλu2,ux.We can now insert this latter inequality back into (4.3) and repeat the previousreasoning withBreplaced byB/2. In general, ifnis a positive integer such thatBeλx4·2n1>|A|x,then iterating the previous argumentntimes will yield|q(u)| ≤|A|u+Beλu4·2n,ux.(4.5)Note that the bound (4.5) is valid forux, wherexis the same as above.IfAnegationslash= 0, letnbe the least positive integer such thatBeλx4·2n|A|x.Then (4.5) gives|q(x)| ≤2|A|x.(4.6)Sincexu0is arbitrary, (4.6) must hold for allxu0.On the other hand,ifA= 0, then lettingn→ ∞in (4.5) shows thatq(u) = 0 for allux.Butagain, sincexu0is arbitrary, we must haveq(u) = 0 for alluu0. In otherwords, (4.6) also holds for allxu0whenA= 0.This establishes the claim.Invoking Theorem 3 now completes the proof.square
8DAVID M. BRADLEYWe can now show that any solution to (4.1) withκ >0 must be wildly oscillatory.

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Term
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Prime number, DAVID M BRADLEY

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