胡适耕 实变函数答案 第三章A(3)

?当fn???f,则?fn?有子列fnk几乎处处收敛于f．

??∴

?Xf??limfnk?lim?fnk?const???

XX∴f可积 ∴f?L1. 23．设

??Xfnd???，则?X?fd????nXfnd?.

证 由于fn?M?(X)，故于是

??Xfnd????fnd???

X?fnn?L.设Fn??fk,则limFn??fk．

1k?1n?nk?1∴Fn?又

?fk?1k??fk??fk?L1．

k?1k?1n??Xfn???fn???故fn?L1?fn?L1．

X因为limn?XFn?lim?nX?fk?1Xnk?lim??fk???fk ，由控制收敛定理，有

nk?1Xk?1Xkn?lim?Fn??limFn??nXXn?fk?1?，则

??fd????XnXfnd?.

24．设0?a?b,fn(x)?ae?nax?be?nbx(n?1,2?)，验证

???0fndm???0?fdm，且??n?0fndm??.

证 ∵

??0fndm????01?nbx?nax??nax?nbxae?bedx???n(e?e)0?0

∴

??0fndm?0

?nax??f0?ndm???0?(ae?be?nbx)dx???0ae?axbe?bx(?)dx ?ax?bx1?e1?e 72

???0?11?ax?bxd(1?e)?d(1?e) ?ax?bx?01?e1?e1?e ??limln??0?01?e?ax∴

?bx??1?e?b?b?lnlim?ln

??0?01?e?a?a???0fndm??xn?0?fndm，取xn??xnln(b/a)，则

n(b?a)??0fndm??(?fn)dm??fndm

0??(be?nbx?ae?nax)dm???ae?nax?be?nbx?dm

0xnxn??x1?1?nax(e?e?nbx)n?(e?nbx?e?nax)

xnn0nab??2??a?b?a?a?b?a? ???????n??b?b????ab??b?ab?2??a??a?a?fndm?????????.

n??b??b????∴???0xsin()?ndx． 25．求lim?n0nx??1?()??n??xsin()1n，在0,1上，f(x)?解 令fn(x)??1， ??nnnx?x???1?()1?()????nn????在?1,???上，fn(x)?1x2Cn?()2n?11?

n(n?1)x2(n?1)x2?()2n2n73

?1x2x2?22n?1x2x2?24?4(n?2) x2?1,x?(0,1)?1取g(x)??4，则g(x)?L?0,???．

,x?(1,??)??x2又fn(x)?0,a.e(n??)，由控制收敛定理limn??0fn(x)??0?0.

0?26．求limn??01?nx2dx. 2n(1?x)1?nx21?nx2?1， 解 令fn(x)?，在?0,1?上，fn(x)?22n1?nx(1?x)11?nxnx2?2?4(n?2) 在?1,???上，fn(x)?12?2x2x2Cnx?Cn?x41?n?1x21?2221??1,x?(0,1)?1取g(x)??4，则g(x)?L?0,???．

,x?(1,??)?2x?又?fn(x)?0,a.e(n??)，由控制收敛定理limn??0?1?nx2dx?0?0. 2n?0(1?x)27．求limn??0nxsin5nxdx. 221?nx解 令fn(x)?nx1nx5sinnx，在上，f(x)??0,1??1， n1?n2x22nxx211?3， 3nx2x2在?1,???上，fn(x)??11,x?(0,1)??x21取g(x)??，则g(x)?L?0,???

?13,x?(1,??)??x2

74

又fn(x)?0,a.e(n??)，由控制收敛定理limn??0fn(x)dx??0?0.

0?28．求limn??0x2?n(1?)dx.

nx2?n)，在?0,1?上，fn(x)?1， 解 令fn(x)?(1?n在?1,???上，fn(x)?111??， x2nx2x2(1?)1?n?nn?1,x?(0,1)?1取g(x)??1，则g(x)?L?0,???.

,x?(1,??)??x2又?limfn(x)?limnn1?xx2??(1?)?n????2?nx2?e?x

2由控制收敛定理limn?0fn(x)dx??e?xdx?0?2?2.

29．求limn??01(nx?)?ndx.

x1x解 令fn(x)?(nx?)，在?0,1?上，fn(x)??n11(nx?)nx?1，

xnxn1在?1,???上，fn(x)???(n?2)， 2n2n2(1?nx)(nx)x?1,x?(0,1)?1取g(x)??1，则g(x)?L?0,???.

,x?(1,??)??x2又fn(x)?0,a.e(n??) ∴limn??0fn(x)dx??0?0.

0? 75

30．求limn??0e?xndx.

n解 令fn(x)?e?x，在?0,1?上，fn(x)?在?1,???上，fn(x)?1exn?1?1， 0e1exn?1?x?e． xe取g(x)???1,x?(0,1)?x?e,x?(1,??)1，则g(x)?L?0,???.

?fn(x)?1,a.e,x?(0,1)又∵?

f(x)?0,a.e,x?(1,??)?n∴limn??0fn(x)dx?lim?fn(x)dx?lim?n0n1??1fn(x)dx??1??0?1.

011??31.

??0?sinxan?1dx??2(a?1) ． ex?an?1n?1 证 由??1知，当x?(0,??)时，

?ex?1，故有

?sinxe?x?sinx?x?xn??(esinx)(?e)， ?x?xe??1??en?0令fk(x)??(en?1k?xsinx)(?e?x)n?1(k?1,2,?)，则fk(x)?x， xe?1sinx，且 xe??当0?x?1时，易验知fk(x)?而当1?x??时，易验知fk(x)?

12?，取控制函数 ex?1exx 0?x?1 ex?1F(x)?

2 1?x?? ex76

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