Candforeachn2INfnE CandassumethatH1limn1fntftuniformlyonEandH2foreachn2INlimtpfntAnexistsThenalimn1AnAexistsandblimtpftAThatislimtplimn1fntlimn1limtpfntcrJoel ID: 185240
Download Pdf The PPT/PDF document "InterchangingtheOrderofSummationCorollar..." is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
InterchangingtheOrderofSummationCorollary(InterchangingtheOrderofSummation)If1Xj=11Xk=1 ajk 1then1Xj=11Xk=1ajk=1Xk=11Xj=1ajkRemark.ThehypothesisP1j=1P1k=1 ajk 1reallymeansthatforeachj2IN,1Xk=1 ajk =Mj1and1Xj=1Mj1Thetwodoublesumsintheconclusionreallymean1Xj=11Xk=1ajk=limn!1nXj=11Xk=1ajk=limn!1nXj=1limm!1mXk=1ajk=limn!1limm!1nXj=1mXk=1ajk1Xk=11Xj=1ajk=limm!1mXk=11Xj=1ajk=limm!1mXk=1limn!1nXj=1ajk=limm!1limn!1nXj=1mXk=1ajkThatalloftheselimitsexistispartoftheconclusionofthecorollary.Thisresultisacorollaryofthefollowingtheorem,whichhasalreadybeenproveninclass.Theorem.LetXbeametricspace,EXandp2E0,thesetoflimitpointsofE.Letf:E! Cand,foreachn2IN,fn:E! Candassumethat(H1)limn!1fn(t)=f(t)uniformlyonEand(H2)foreachn2IN,limt!pfn(t)=AnexistsThen(a)limn!1An=Aexistsand(b)limt!pf(t)=A.Thatis,limt!plimn!1fn(t)=limn!1limt!pfn(t).c\rJoelFeldman.2008.Allrightsreserved.February4,2008InterchangingtheOrderofSummation1 t#p8:n!1z }| {f1(t)f2(t)f3(t)unif !f(t)####(b)A1A2A3(a) !AProofoftheCorollary:SetX=IR,E=1;1 2;1 3;1 m; andp=0.Write1 m=tmandde nefn(tm)=nXj=1mXk=1ajkf(tm)=1Xj=1mXk=1ajkThein nitesuminthede nitionoff(tm)convergesbycomparisonwithP1j=1Mj.Veri cationof(H1):Foreachj2IN, mPk=1ajk Mjforallm2IN.SotheWeierstrassM{testimpliesthatfnconvergestof,uniformlyonE.Veri cationof(H2):Foreachj2IN,1Pk=1ajkconvergesabsolutelybythehypothesisthat1Pk=1 ajk =Mj1.Solimm!1fn(tm)=limm!1nXj=1mXk=1ajk=nXj=1limm!1mXk=1ajk=Anexists.Sothetheoremnowtellsisthatlimn!1An=limn!1limm!1nXj=1mXk=1ajkandlimm!1f(tm)=limm!1limn!1nXj=1mXk=1ajkexistandareequal. Example.Hereisanexamplewhichillustratestheneedforthehypothesisthatthedoublesumconvergesabsolutely.Wechooseajk=8�&#x-5.3;㠔:1ifj=k=11ifk=j+1 1ifk=j 10otherwisec\rJoelFeldman.2008.Allrightsreserved.February4,2008InterchangingtheOrderofSummation2 ThisexampleisriggedtogivethepartialsumsSmn=mXj=1nXk=1ajk=(1ifm=n2ifn�m0ifnmPictoriallyajk k! j 11000# 10100 0 1010 00 101 ...............Smn n! m 1222! 2# 0122! 2 0012! 2 0001 .........0 # ####& 2 0000!0 1Forany xedn,Sm;n=0forallm�nandsoconvergesto0asm!1.Hencelimn!1limm!1mXj=1nXk=1ajk=limn!1limm!1Sm;n=limn!10=0Similarly,foreach xedm,Sm;n=2foralln�mandsoconvergesto2asn!1.Hencelimm!1limn!1mXj=1nXk=1ajk=limm!1limn!1Sm;n=limm!12=2AndthesequenceSm;m=1convergesto1asm!1.Solimm!1mXj=1mXk=1ajk=limm!1Sm;m=limm!11=1c\rJoelFeldman.2008.Allrightsreserved.February4,2008InterchangingtheOrderofSummation3