De64257ne the function IR IR by the requirements that for 1 and that 2 for all real So is periodic of period 2 1 2 3 4 Now de64257ne 0 4 As 1 the series converges uniformly by the Weiersrtrass test with As is a continuous function is a un ID: 76843
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ANowhereDi erentiableContinuousFunctionThesenotescontainastandard(1)exampleofafunctionf:IR!IRthatiscontinuouseverywherebutdi erentiablenowhere.De nethefunction':IR!IRbytherequirementsthat'(x)=xforx22 11]andthat'(x+2)='(x)forallrealx.So'isperiodicofperiod2. 1'(x) 1 2 3 41234xNowde nef(x)=1Xn=0 3 4n'(4nx)As'(x)j1,theseriesconvergesuniformlybytheWeiersrtrassM{testwithMn= 3 4n.As'isacontinuousfunction,f(x)isauniformlimitofcontinuousfunctionsandhenceiscontinuous.Wenow xanyx2IRandprovethatfisnotdi erentiableatxbyexhibitingasequencehm m2INofrealnumbersconvergingto0suchthat1 hmf(x+hm) f(x)divergesasm!1.Infacthm=1 24 mwiththesignchosen(2)sothatthereisnointegerstrictlybetween4mxand4m(x+hm).Wenextcomputethemagnitudeofthenthtermin1 hmf(x+hm) f(x).Thatis,wecompute\rm;nwhere\rm;n=1 hm 3 4n'(4nx+4nhm) '(4nx)=2(3n)4m n'(4nx1 24n m) '(4nx)Casen�m:Inthiscase1 24n misaneveninteger.So\rm;n=0because'(4nx1 24n m)='(4nx)because'hasperiod2.Casen=m:Recallthatthesignofhmwaschosensothatsothatthereisnointegerstrictlybetween4mxand4m(x+hm).So 4mx;'(4mx)and 4m(x+hm);'(4mx+4mhm)lieonthesameramp(i.e.straightlinesegment)inthegraphof',above.Eachofthoserampshasslope 1or+1.So '(4mx+4mhm) '(4mx) =4mhm=1 2and \rm;n =2(3m)4m m1 2=3mCasenm:Since '(y) '(x) jy xforallx;y2IR,wealwayshavethat \rm;n 2(3n)4m n1 24n m=3nPuttingtheseboundstogether 1 hmf(x+hm) f(x) = 1Xn=0\rm;n = mXn=0\rm;n j\rm;mj m 1Xn=0\rm;nj3m m 1Xn=03n=3m 1 3m 1 3=1 2(3m+1)Sureenough,thisdivergesasm!1.Sofisnotdi erentiableatx (1)ThisparticularexampleisduetoJohnMcCarthyandappearedintheAmericanMathematicalMonthly,Vol.LX,No.10,December1953.In1872,Weierstrassgavetheexamplef(x)=P1n=0bncos(anx)forb1andab&#x-2.0;锣1+3 2.ItisdiscussedinACourseinMathematicalAnalysisbyE.Goursat(translatedbyE.R.Hedrick).(2)Toseethatthesignmaybechoseninthisway,observethat4m[x+1 24 m] 4m[x 1 24 m]=1.Either4m[x+1 24 m]and4m[x 1 24 m]arebothintegers,inwhichcasetherearenointegersintheopeninterval(4m[x 1 24 m];4m[x+1 24 m])andwemaychooseeithersignforhm.Orthereisexactlyoneintegerintheopeninterval(4m[x 1 24 m];4m[x+1 24 m]).Thisoneintegeriseither4mx,inwhichcasewemaychooseeithersignforhm,orisin(4m[x 1 24 m];4mx),inwhichcasewechoosehm=+1 24 m,orisin(4mx;4m[x+1 24 m])inwhichcasewechoosehm= 1 24 m.c\rJoelFeldman.2008.Allrightsreserved.February4,2008ANowhereDi erentiableContinuousFunction