Revisiting extensions of regularly varying functions ∗ MeitnerCadena 5 1 Abstract 0 2 RelationshipsamongtheclassesM,M∞,andM−∞andtheclassofO-regularlyvarying functionsareshown. TheseresultsarebasedontwocharacterizationsofM, M∞, and pr M−∞providedbyCadenaandKratzin[7]andanewonegiveninthisnote. A Keywords: regularlyvaryingfunction; slowlyvaryingfunction; O-RV;largedeviations; 7 extremevaluetheory 2 AMSclassification:26A12;60F10 ] A C 1 Introduction . h t ApositiveandmeasurablefunctionU definedonR+isaregularlyvarying(RV)functionif a m U(tx) [ lim <∞ (t>0). (1) x→∞ U(x) 2 v Ifthislimitequals1,U isaslowlyvarying(SW)function. ClassesRV andSV ofregularlyand 8 slowlyvaryingfunctionswereintroducedbyKaramata[10]in1930.Sincethentheoryofthese 8 functionshasbeendevelopedinmanydirections. Systematictreatmentofthistheorycanbe 4 foundine.g.[6]and[14]. 6 0 ExtensionsofRVfunctionshavebeenobtainedbyletting(1)tovary.Anearlyextensionofthis . typewasgivenbyAvakumovic´in1936[4].HeintroducedtheclassO-RV ofO-regularlyvarying 2 0 (O-RV)functionsU whichsatisfythefollowingconditioninsteadof(1): 5 v:1 0<U∗(t):=xl→im∞UU((txx))≤xl→im∞UU((txx))=:U∗(t)<∞ (t≥1). (2) i X RecentlyCadenaandKratz[7]gaveanextensionofRVfunctionsbyalsoletting(1)tovary,but r theydesigneditinadifferentwaytothepreviousone. TheyintroducedtheclassM which a consistsinfunctionsU satisfyingthefollowingconditioninsteadof(1): U(x) U(x) ∃ρ∈R,∀ǫ>0, lim =0 and lim =∞. (3) x→∞ xρ+ǫ x→∞ xρ−ǫ We have clearly RV (O-RV and, for instance using Theorem 1 (see Corollary 1), RV(M. TherearisesthenaturalquestionofhowO-RV andM arerelatedbetweenthem. Weunder- takethisstudyhelpingusofcharacterizationsoftheseclasses: recallingwell-knowncharac- terizationsofO-RV andgivingproofsofthreecharacterizationsofM,twoofthemprovidedin [7]andanewonegiveninthisnote. ∗UPMCParis6&CREAR,ESSECBusinessSchool;E-mail:[email protected],[email protected],or [email protected] 1 CadenaandKratzalsointroducedthefollowingnaturalextensionsofM. U(x) M∞ := U:R+→R+:U ismeasurableandsatisfies∀ρ∈R, lim =0 (4) x→∞ xρ ½ ¾ U(x) M−∞ := U:R+→R+:U ismeasurableandsatisfies∀ρ∈R, lim =∞ . (5) x→∞ xρ ½ ¾ The new characterization givenfor M isextended toM∞ and M−∞. Relationships among M∞andM−∞andO-RV arealsoinvestigatedinthisnote. This noteisorganizedasfollows. Themainresultsarepresented inthenextsection, intro- ducingpreviouslynotationsanddefinitions. First,thenewcharacterizationsofM,M∞,and M−∞basedonlimitsaregiven. Next,analysesofuniformconvergenceinthesecharacteriza- tionsarepresentedand,finally,relationshipsamongO-RV andM,M∞andM−∞areshown. AllproofsarecollectedinSection3.Conclusionispresentedinthelastsection. 2 MainResults ForapositivefunctionU withsupportR+ itslower andupperordersaredefinedby(seee.g. [6]) log(U(x)) log(U(x)) µ(U) := lim , ν(U) := lim . x→∞ log(x) x→∞ log(x) Throughoutthisnotelog(x)representsthenaturallogarithmofx. WenoticethattheclassesM,M∞,andM−∞definedin(3),(4),and(5)areabitweakerthan thecorrespondingclassesgivenin[7],andthateachofthemisdisjointfromeachother.More- over,usingstraightforwardcomputations,onecanprovethatρdefinedin(3)isunique,hence itwillbedenoted byρ , andonecanshowthatǫ>0in(3)canbetakensufficiently small. U Additionally,onecanprovethatM isstrictlylargerthanRV,forinstanceusingTheorem1(see Corollary1),andthatM∞isrelatedtothedomainofattractionofGumbel(see[7]). ThenewcharacterizationsofM,M∞,andM−∞follow. Theorem1. LetU:R+→R+beameasurablefunction.Then (i) U∈M withρ =−τiff U U(tx) ∀r <τ,∃x >1,∀x≥x , lim tr =0 a a t→∞ U(x)  (6)  ∀r >τ,∃x >1,∀x≥x , lim trU(tx)=∞. b b t→∞ U(x) (ii) U∈M∞iff  U(tx) ∀r ∈R,∃x >1,∀x≥x , lim tr =0. (7) 0 0 t→∞ U(x) (iii) U∈M−∞iff U(tx) ∀r ∈R,∃x >1,∀x≥x , lim tr =∞. (8) 0 0 t→∞ U(x) Example1. 1. ConsiderameasurableandpositivefunctionUwithsupportR+suchthat,forx≥x with 0 somex >1,U(x)=x log(x). 0 ± 2 Notingthat,fort,x>1, U(tx) log(x) 0 ifr >1 lim tr = lim tr−1 = t→∞ U(x) t→∞ log(tx) ∞ ifr <1, ½ provides,takingτ=−1andapplyingTheorem1,(i),U∈M withρ =1. U 2. LetU beafunctiondefinedbyU(x):=xsin(x),x>0. Writing U(tx) tr =tr+sin(tx)xsin(tx)−sin(x) U(x) gives,forr ∈R, U(tx) U(tx) lim tr =∞ and lim tr =0. t→∞ U(x) t→∞ U(x) Hencethe necessarycondition ofTheorem1, (i), is not satisfiedand consequently gives U6∈M. It follows a consequence of Theorem 1. This result was proved by Cadena and Kratz in [7] combiningaresultprovidedin[8]andanothercharacterizationofM (seeTheoremCKlater). Corollary1. RV ( M. Notethat,fromCorollary1,RV ⊆M O-RV. TherearenotcommonelementsbetwTeenO-RV andM undertheirdefinitionsgivenin(2)and (3)respectively,butobservingthecharacterizationofM giveninTheorem1oneidentifiesthe quotientU(tx) U(x),whichappearsin(2). Thenextexampleexploitsthislinktoshowafirst relationshipbetweenO-RV andM. ± Example2. M 6⊆ O-RV. LetUbeafunctiondefinedbyU(x):=exp (logx)αcos (logx)β ,x>0,where0<α,β<1such thatα+β>1. © ¡ ¢ª Prof. PhilippeSouliergaverecommendationstocorrectanerrorinanearlyversionofthisex- ample. Ontheonehand,notingthat,forx,t>e,usingthechangesofvariabley=log(x)ands=log(t) andobservingthats→∞ast→∞, U(tx) lim tr = limexp rs+(s+y)αcos (s+y)β −yαcos yβ t→∞ U(x) s→∞ n1 y α ¡ ¢ yα ¡ ¢o 0 ifr >0 = limexp s r+ 1+ cos (s+y)1/3 − cos yβ = s→∞ s1−α s s ∞ ifr <0, ½ µ ³ ´ ¡ ¢ ¡ ¢¶¾ ½ provides,takingτ=0andapplyingTheorem1,U∈M withρ =0. U Ontheotherhand,writing,forx>e andt>0,usingthepreviouschangesofvariables,withx suchthat logtx β=π 2+2kπ,foragivent, ¡ ¢ ± U(tx) =exp − logx αcos (logx)β U(x) n ¡ ¢ ¡ ¢o =exp −yαcos ((π 2+2kπ)1/β−s)β =expn (π 2+2¡kπ)±1/β−s αsin ((π 2¢o+2kπ)1/β−s)β−(π 2+2kπ) . n¡ ± ¢ ¡ ± ± ¢o 3 Since (π 2+2kπ)1/β−s β−(π 2+2kπ)→0ask→∞,wehave ¡ ± ¢ ± lim (π 2+2kπ)1/β−s αsin ((π 2+2kπ)1/β−s)β−(π 2+2kπ) k→∞ h¡ ± ((π 2+2kπ)1¢/β−s)¡β−±(π 2+2kπ) ± ¢i = lim , k→∞ ± (π 2+2kπ)1/β−s±−α ¡ ± ¢ whichisanindeterminationoftype0 0.Then,applyingL’Hopital’srulewehave ± ((π 2+2kπ)1/β−s)β−(π 2+2kπ) lim k→∞ ± (π 2+2kπ)1/β−s±−α ¡ ± ((π 2+2kπ¢)1/β−s)β−(π 2+2kπ) = lim(2π)α/β k→∞ ± k−α/β ± β ((π 2+2kπ)1/β−s)β−1(π 2+2kπ)1/β−1−1 = lim − (2π)α/β+1 , k→∞ α ± k−α/β−1± whichisanindeterminationoftype0 0.Then,applyingagainL’Hopital’srulewehave ± ((π 2+2kπ)1/β−s)β−(π 2+2kπ) lim k→∞ ± (π 2+2kπ)1/β−s±−α = lim s¡β(±1−β)(2π)α/β+2¢((π 2+2kπ)1/β−s)β−2(π 2+2kπ)1/β−2 k→∞ α(α+β) ± k−α/β−2 ± β(1−β) = lim s (2π)(α+β−1)/βk(α+β−1)/β k→∞ α(α+β) ∞ ifs>0 = −∞ ifs<0. ½ Then,weget,fort>1, U(tx) U∗(t)= lim =∞, x→∞ U(x) and,fort<1, U(tx) U∗(t)= lim =0, x→∞ U(x) whichcontradict(2),soU6∈O-RV.Inparticular,U6∈SV. Next,theuniformconvergencesinxoflimitsgivenin(6),(7),and(8)areanalyzed.Tothisaim, wewillusethenexttworesults. Proposition1. LetU:R+→R+beameasurablefunction.Then (i) IfU ∈M withρ =−τ,thenthereexistsx >1suchthat,forx ≤c<d <∞,thereexist U 0 0 0<M <M satisfying,forx∈[c;d],M ≤U(x)≤M . c d c d (ii) IfU ∈M∞,thenthereexistsx0>1suchthat,forc≥x0,thereexistMc >0satisfying,for x∈[c;∞),U(x)≤M . c (iii) IfU∈M−∞,thenthereexistsx0>1suchthat,ford≥x0,thereexistMd>0satisfying,for x∈[d;∞),U(x)≥M . d 4 Proposition2(Givenin[2]). LetµbetheLebesguemeasureonR,Aameasurablesetofpositive measure,and xn n∈Naboundedsequenceofrealnumbers.Then,µ(A)≤µ limn→∞(xn+A) . © ª ¡ ¢ Nowtheresultsonuniformconvergencesarepresented.Theirproofsareinspiredby[3]. Theorem2(UniformConvergenceTheorem(UCT)). LetU:R+→R+beameasurablefunc- tion.Then (i) IfU∈M withρ =−τandr<τ,then,foranyx ≤c<d<∞forsomex >1, U a a U(tx) lim tr sup =0. t→∞ x∈[c;d] U(x) (ii) IfU∈M withρ =−τandr>τ,then,foranyx ≤c<d<∞forsomex >1, U b b U(tx) lim tr inf =∞. t→∞ x∈[c;d] U(x) (iii) IfU ∈M∞satisfying,fors>1,U(x)≥Ms forx∈[1;s]andsomeMs >0,then,forr ∈R andanyconstantsx ≤c<d<∞forsomex >1, 0 0 U(tx) lim tr sup =0. t→∞ x∈[c;d] U(x) (iv) IfU ∈M−∞satisfying,fors>1,U(x)≤Ms forx∈[1;s]andsomeMs >0,then,forr ∈R andanyconstantsx ≤c<d<∞forsomex , 0 0 U(tx) lim tr inf =∞. t→∞ x∈[c;d] U(x) Note that UCT cannot be extended to infinite intervals. For instance, from the functionU giveninExample2wehavethatcomputingthesupremumofthequotientU(tx) U(x)inxon [x ;∞),foranyx >1,givesalways∞,andhenceonecannotdeducethatρ =0. 0 0 U ± ThenextresultsonO-RV,M,M∞,andM−∞willbeusedtogivemorerelationshipsbetween theseclasses.OnO-RV weneed: Proposition3(seee.g. [11],[14],[1],[9],and[6]). LetU :R+→R+beameasurablefunction. Thenthefollowingstatementsareequivalent: (i) U∈O-RV. (ii) Thereexistα,β∈Randx >1,c>0suchthat,forallt≥1andx≥x , 0 0 U(tx) c−1tβ≤ ≤ctα. U(x) (iii) Thereexistfunctionsη(x)andφ(x)boundedon[x ;∞),forsomex ≥1,suchthat 0 0 x dy U(x)=exp η(x)+ φ(y) , x≥1. ½ Z1 y ¾ OnM weneedthenexttwocharacterizationsofM givenbyCadenaandKratzin[7]. Forthe sakeofcompletenessofthisnote,wegivethemasTheoremCKandindicatetheirproofs.Part oftheseproofsarecopiedfrom[7]. TheoremCK. LetU :R+→R+ beameasurablefunction. Thenthefollowingstatementsare equivalent: (i) U∈M withρ =τ. U 5 log(U(x)) (ii) lim =τ. x→∞ log(x) (iii) Thereexistb>1andmeasurablefunctionsα,β,andδsatisfying,asx→∞, α(x) log(x)→0, β(x)→τ, δ(x)→1, ± suchthat x ds U(x)=exp α(x)+δ(x) β(s) , x≥x forsomex ≥b. 1 1 ½ Zb s ¾ Remark1. IfF isthetailofadistributionF associatedtoarandomvariable(rv)X,someau- thors(seee.g.[12]and[13])saythatX isheavy-tailedifthelimit log F(x) η:= lim x→∞ lo³g(x) ´ existsandtakesanegativevalue. Wenoticethatthischaracterizationdoesnotcoverrvswithheavytailssatisfyingη=0orwith heavytailsforwhichsuchlimitdoesnotexist. Indeed,ontheoneside,fromTheoremCKone hasthatη=0impliesthatF ∈M withρ =0,beingaparticularcaseofthesefunctionstheSV F functions,whichareconsideredheavy-tailed.Ontheotherside,CadenaandKratzpresentedin log F(x) [7]familiesoftailsF forwhichthelimit lim doesnotexist,forinstancethenexttail x→∞ lo³g(x) ´ definedby(see[7]) Letα>0,β<−1, x >1,anddefinetheseriesx =x(1+α)n,n≥1,whichsatisfies a n a x →∞asn→∞. ItisnothardtoprovethatthetailF associatedtoarv X and n definedby 1 x∈[0;x ) 1 F(x):=( xnα(1+β) x∈[xn;xn+1),∀n≥1 satisfies log F(x) α(1+β) log F(x) lim =− <−α(1+β)= lim . x→∞ lo³g(x) ´ 1+α x→∞ lo³g(x) ´ Notethatif−α(1+β) (1+α)<1,thentheexpectedvalueofX is∞,whichmeans thatX canbeconsideredasaheavy-tailedrv. ± WenoticefromtherepresentationsofUviaO-RV andM giveninProposition3,(iii),andThe- oremCK,(iii),respectively,thatakeydifferencebetweenthoserepresentationsisthepresence ofaboundedfunctionundertheintegralsymbol. Motivatedbythisobservation,webuiltthe nextfunctionbelonging toO-RV butnottoM. Thisaimisreachedbybuildingabounded xφ(s)ds functionφsuchthatthelimit lim 1 s doesnotexist.Notethatifthislimitexists,then, x→∞Rlog(x) applyingTheoremCK,(iii),U∈M. Example3. O-RV 6⊆ M. x ds LetU:R+→R+beameasurablefunctionsatisfying,forx≥1,U(x)=exp φ(s) ,where ½Z1 s ¾ thefunctionφhassupport[1;∞)andisdefinedby 0 ifx∈[1;e)orx∈I withnodd φ(x)= n 1 ifx∈I withneven, ½ n 6 whereI =[een;een+1),n∈N. n Ontheonehand,applyingProposition3,onehasU∈O-RV. Ontheotherhand,writing,forx>1,usingthechangeofvariabley=log(s) log(x), log(U(x)) xφ(s)ds 1 ± = 1 s = φ eylog(x) dy log(x) Rlog(x) Z0 ³ ´ gives,takingx =een,n=2,3,..., n n−1 (−1)ke−k ifnisodd lolgo(gU(x(xnn)))=nkX=−11Zekek/e+n1/enφ³eyen´dy= nkX=−01(−1)k+1e−k ifniseven, andonethengets  kX=1 1 ifnisodd log(U(x )) 1+e−1 lim n = n→∞ log(xn)  e−1 ifniseven, 1+e−1 whichimpliesν(U)−µ(U)≥(1−e−1) (1+e−1)>0,hencethelimit lim log(U(x))doesnotexist x→∞ log(x) andthus,applyingTheoremCK,U6∈±M. NowwegiveanotherrelationshipbetweenO-RV andM. Proposition4. LetU:R+→R+beameasurablefunction.IfU∈O-RVandthelimit log(U(x)) lim x→∞ log(x) exists,thenU∈M. TherelationshipsofM∞andM−∞withO-RV aresimpler. Proposition5. Forλ∈ ∞,−∞ , M O-RV=;. λ © ª T 3 Proofs ProofofTheorem1. • Proofofthenecessaryconditionof(i) AssumeU∈M withρ =−τ.Letr ∈Rsuchthatr 6=0. U – Ifr <τ Let0<ǫ<τ−r andδ>0.Byhypothesis,thereexistsaconstantx >1suchthat,for 0 x≥x ,U(x)≤δx−τ+ǫ,andthereexistsx >1suchthat,forx≥x ,U(x)≥x−τ−ǫ δ. 0 1 1 Hence,settingx :=max(x ,x ),forx≥x andt>1, a 0 1 a ± U(tx) tr ≤δ2tr(tx)−τ+ǫxτ+ǫ=δ2t−τ+r+ǫx2ǫ, U(x) andtheassertionthenfollowsast→∞since−τ+r+ǫ<0. 7 – Ifr >τ Let0<ǫ<r−τandδ>0.Byhypothesis,thereexistsaconstantx >1suchthat,for 0 x≥x ,U(x)≤δx−τ+ǫ,andthereexistsx >1suchthat,forx≥x ,U(x)≥x−τ−ǫ δ. 0 1 1 Hence,settingx :=max(x ,x ),forx≥x andt>1, a 0 1 a ± U(tx) 1 1 tr ≥ tr(tx)−τ−ǫxτ−ǫ= tr−τ−ǫx−2ǫ, U(x) δ2 δ2 andtheassertionthenfollowsast→∞sincer−τ−ǫ>0. • Proofofthesufficientconditionof(i) Letδ>0andη>0. Onetheonehand,sinceτ−δ 2<τ,byhypothesis,thereexistsaconstantx >1such a U(xt) that, for x ≥x , lim tτ−δ/2 ± =0. Hence, given x ≥x , thereexists t =t (x)>1 a a a a t→∞ U(x) suchthat,fort≥t ,tτ−δ/2U(tx)≤ηU(x),or a U(tx) xτ−δU(x) ≤η . (9) (tx)−τ+δ tδ/2 Onetheotherhand,sinceτ+δ 2>τ,byhypothesis,thereexistsaconstantx >1such b U(xt) that, for x ≥x , lim tτ+δ/2 ± =∞. Hence,given x ≥max(x ,x ),thereexists t = b a b b t→∞ U(x) t (x)>1suchthat,fort≥t ,tτ+δ/2U(tx)≥ηU(x),or b b U(tx) ≥ηxτ+δU(x)tδ/2. (10) (tx)−τ−δ Combining (9) and (10), given x ≥max(x ,x ) and for t ≥max(t ,t ), and using the a b a b changeofvariabley=txwithy→∞ast→∞,provide,forδ>0, U(y) U(y) lim =0 and lim =∞, y→∞y−τ+δ y→∞y−τ−δ whichimpliesthatU∈M withρ =−τ. U • Proofofthenecessaryconditionof(ii) Letr ∈Randη>0. Setr′<−r. SinceU ∈M∞thereexistsaconstantx0>1suchthat, forx≥x ,U(x)≤ηxr′.Hence,fort>1, 0 U(tx) tr+r′xr′ tr ≤η , U(x) U(x) andtheassertionthenfollowsast→∞sincer+r′<0. • Proofofthesufficientconditionof(ii) Let r ∈R. Taking r′ <−r, by hypothesis, there exists a constant x >1such that, for 0 x≥x , lim tr′U(xt) =0. Hence,forη>0,thereexistsaconstant t >1suchthat,for 0 0 t→∞ U(x) t≥t ,tr′U(tx)≤ηU(x),or 0 U(tx) U(x) ≤η . (tx)r xrtr+r′ Usingthechangeofvariabley=txandnotingthaty→∞ast→∞give,forr∈R,being r+r′>0, U(y) lim =0, y→∞ yr whichmeansthatU∈M∞. 8 • Proofofthenecessaryconditionof(iii) Letr ∈Randη>0. Setr′>−r.SinceU∈M−∞thereexistsaconstantx0>1suchthat, forx≥x ,U(x)≥ηxr′.Hence,fort>1, 0 trU(tx)≥η xr′ tr+r′, U(x) U(x) andtheassertionthenfollowsast→∞sincer+r′>0. • Proofofthesufficientconditionof(iii) Let r ∈R. Taking r′ <−r, by hypothesis, there exists a constant x >1such that, for 0 x≥x , lim tr′U(xt) =∞. Hence,forη>0,thereexistsaconstantt >1suchthat,for 0 0 t→∞ U(x) t≥t ,tr′U(tx)≥ηU(x),or 0 U(tx)≥ηU(x)t−r−r′. (tx)r xr Usingthechangeofvariabley=txandnotingthaty→∞ast→∞give,forr∈R,being −r−r′>0, U(y) lim =0, y→∞ yr whichmeansthatU∈M∞. ProofofCorollary1. LetU∈RVwithtailindexρ.Then,fort>1, U(tx) lim tr =tr+ρ, x→∞ U(x) whichimpliesthat,forǫ>0,thereexistsaconstantx >1suchthat,forx≥x , 0 0 U(tx) tr+ρ−ǫ≤tr ≤tr+ρ+ǫ. U(x) Hence,settingτ=−ρ,gives,ontheonehand,forr <τ, U(tx) −ǫ≤ lim tr ≤ǫ, t→∞ U(x) U(tx) whichimplies lim tr =0takingǫarbitrary,and,ontheotherhand,forr >τ, t→∞ U(x) U(tx) lim tr =∞. t→∞ U(x) Thereforeonehas,applyingTheorem1,thatU∈M withρ =ρ. U Finally,afunctionbelongingtoM butnottoRV isforinstancethefunctiongiveninExample 2. ProofofProposition1. • Proofof(i) Letǫ>0.BydefinitionofU∈M withρ =−τ,thereexistconstantsx ,x >1suchthat, U a b forx≥x ,U(x)≤x−τ+ǫ, and, forx≥x ,U(x)≥x−τ−ǫ. a b So, for x ≥ x :=max(x ,x ), x−τ−ǫ ≤U(x)≤ x−τ+ǫ. Hence, for any x ≤c <d <∞, 0 a b 0 onehas, setting M :=min(c−τ−ǫ,d−τ+ǫ)and M :=max(c−τ−ǫ,d−τ+ǫ), thatU satisfies c d M ≤U(x)≤M foranyx∈[c;d]. c d 9 • Proofof(ii) Letǫ>0. BydefinitionofU ∈M∞,thereexistsaconstantx0>1suchthat,forx≥x0, U(x)≤xǫ.Hence,foranyc≥x ,onehas,settingM :=cǫ,thatU satisfiesU(x)≤M for 0 c c anyx∈[c;∞). • Proofof(iii) Letǫ>0. BydefinitionofU ∈M−∞,thereexistsaconstantx0>1suchthat,forx≥x0, U(x)≥xǫ.Hence,foranyd ≥x ,onehas,settingM :=dǫ,thatU satisfiesU(x)≥M 0 d d foranyx∈[d;∞). ProofofTheorem2. LetµbetheLebesguemeasureonR. • Proofof(i) LetU ∈M withρ =−τandletr <τ. ApplyingTheorem1,(i),thereexistsx >1such U a that,forx≥x , a U(tx) lim tr =0. t→∞ U(x) Letx ≤c<d<∞.ThenusingEgoroff’stheorem(seee.g.[5]),thereexistsameasurable a A⊆[c;d]ofapositivemeasuresuchthat U(tx) limsuptr =0. t→∞x∈A U(x) Letusprovebycontradictionthatthepreviouslimitholdson[c;d]. Thensupposethat thereexistǫ>0, xn n∈N⊆[c;d],and tn n∈N⊆R+suchthattn→∞and © ª © ª U(t x ) lim tr n n >ǫ. (11) n→∞ n U(xn) ByProposition2onehas,denotinglog(A)= log(x):x∈A andnotingthatlog(A)hasa positivemeasure, © ª µ lim(log(A)−log(x )) ≥µ logA >0, n n→∞ ³ ´ ¡ ¢ whichimpliesthatthereexistaconstantlog(u)∈Randasubsequence xni i∈N⊆ xn n∈N suchthatlog(x )+log(u)∈log(A),i.e.ux ∈A.Notethatu>0. ni ni © ª © ª ByProposition1,(i),thereexist0<M ≤M <∞suchthatM ≤U(x)≤M ,x∈(c;d). c d c d Hence,onethenhas tr U(tnixni)= tni r U tuni uxni urU(uxni)≤ tni r U tuni uxni urMd. ni U(x ) u U³(ux ) ´ U(x ) u U³(ux ) ´ M ni µ ¶ ni ni µ ¶ ni c Notingthat tni r U((tni u)uxni)→0sinceux ∈Aandt u→∞asn →∞provide u U(ux ) ni ni i µ ¶ ± ni U(t x ) ± tr ni ni →0asn →∞,whichcontradicts(11). ni U(x ) i ni • Proofof(ii) LetU ∈M withρ =−τandletr <τ. ApplyingTheorem1,(i),thereexistsx >1such U b that,forx≥x , b U(tx) lim tr =∞. t→∞ U(x) 10