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 existu0≥MandB >0such that for allu≥u0,|q(u)| ≤Beλu.Inserting this inequality into (4.3) yields|uq(u)| ≤ |A|+Bmsummationdisplayj=1|cj|integraldisplayvjvj-1eλ(u+t)dt,u≥u0.It follows that|q(u)| ≤|A|u+Beλuλumsummationdisplayj=1|cj|eλvj≤|A|u+Beλu4,u≥u0.(4.4)Now we claim that|q(x)| ≤2|A|/xforx≥u0. To prove the claim, fixx≥u0andobserve 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 allu≥x. Since (4.4)holds for allu≥u0, we then get|q(u)| ≤Beλu2,u≥x.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 thatBeλx4·2n−1>|A|x,then iterating the previous argumentntimes will yield|q(u)| ≤|A|u+Beλu4·2n,u≥x.(4.5)Note that the bound (4.5) is valid foru≥x, 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)Sincex≥u0is arbitrary, (4.6) must hold for allx≥u0.On the other hand,ifA= 0, then lettingn→ ∞in (4.5) shows thatq(u) = 0 for allu≥x.Butagain, sincex≥u0is arbitrary, we must haveq(u) = 0 for allu≥u0. In otherwords, (4.6) also holds for allx≥u0whenA= 0.This establishes the claim.Invoking Theorem 3 now completes the proof.square