(完整)實變與泛函期末試題答案

經(jīng)典word整理文檔，僅參考，雙擊此處可刪除頁眉頁腳。本資料屬于網(wǎng)絡整理，如有侵權，請聯(lián)系刪除，謝謝?。O()是(,)上的實值連續(xù)函數(shù)，則對于任意常數(shù)，{|()}是一開集,而ExfxafxaE{x|f(x)}{|()}Exfxa設xE(x)a在f(x)(,)fx(x,x)00000()，即fxaU(x,)E,0故為由{|()}x0Ex0Exfxa{|()}ExfxaEE{x|f(x)}設,則是則{x},xEx0EE0nxx(n)n.分0由知xEf(x)a，fnnf(x)f(limx)limf(x)a，0nnnn即分xE0由{|()}分Exfxax0對{|()},{|()}5分ExfxaExfxaE知,7分EEEEE{|()}{|()}ExfxaExfxaxfxa{|()}mE,{f(x)}E是aeae則....f(x)n0對,E{()}mEE(\).E,使fx在且Enn}Ei11n}][n,][|ff,kn],iiiikii1i1f(x)則n}]f(x)。ni10,,nn使i0i001xEn}]E[n,]E[ff,kn],10i0iik0o11f(x)f(x).n6分i00，n},使m(E\n}])En}],0iii1,i(\[,)n.,對mEEni,ni1m(E\[n,，2iii取EE在12分n}],{f(x)}Einm(E\E)m(E\n}])m(E\[n,])m(E\[n,])11iiiiii1i11im(E\E[n,]),15分2iii1i1Ef(x)n(x)F(x)a.e.E在F(x)E于，nfn(x)f(x)，fn則()在fxlimf(x)dxf(x)dx。EnnEE(x)f(x)fx()f(x).fnni于EFxfxFx()()于E()()于EF(x)可積,可得到f(x)F(x)nfxnif(x)Ef(x)在E在分nlimf(x)dxf(x)dxnnEE0在E0mE()Fx當且eEme.分F(x)dx4e(x)f(x)0nNfNn，][ffn0nN2mE分F(x)dx4E[fnf](f(x)f(x))dx＝f(x)dxf(x)dxnnEEEf(x)f(x)nE=f(x)f(x)f(x)f(x)nnE[ff]E[ff]nnfx(()())()()fxfxfxnnE[ff]E[ff]nn22F(x)mE[ff]E[fnf]n2mE4=分22（設mE。limf(x)dxf(xdxn上可積，由非負可測函數(shù)nEEE。因在積分的定義LmEF(x)F(x)dxF(x)dx),0(limF(x)dxF(x)dx,EE,kkEkEEkEkkmEk，F(xiàn)(x)dxF(x)dx4kEEkF(xdx=F(x)dxF(xdxEEEEkk分F(x)dx[F(x)]dx4kEEkE{ff}kn于，f(x)f(x)2F(x).,n1,2,Enkf(x)f(x)f(x)f(x)0nnlimf(x)f(x)dx0Elimf(x)dxf(xdx(x)f(x)0nfknnnEEnEnNN分()()fxfxn2kf(x)dxf(x)dxf(x)f(x)nnEEEf(x)f(x)dxf(x)f(x)dxnnEEEkkF(x)dx22EEk242分由(x)f(x)()().fxffxinnf(x)F(x)a.e.于E，nf(x)F(x)于E,nif(x)F(x)于E。由()Fx(x)和()fxL.2分fnFxdx()在FxE()limFxdx()0k0，kEkEk3F(x)dxF(x)dx,5kEEkF(xdx=F(x)dxF(xdxEEEE(k([F(x)]FxF(x))k)kk分F(x)dxF(x)dx5kEEkeE，0F(x)dx,5e0nNf(x)f(x)EENnk分ffknk當=nNHEffmHnknnkF(x)dx。5HnEH(EH)H(EE)(EH)，nnnkkn時nNf(x)dxf(x)dxf=(x)f(x)dxnnEEE(x)f(x)≤fnE＝f(x)f(x)dx＋nEEf(x)f(x)dx＋nf(x)f(x)dxnEHHknkn（EHEff[])5(knknk≤F(x)dxF(x)dxkEEHkkn22＜55＝分5limf(x)dxf(x)dx。nnEE121278集(,)(,)(,)P339999分4124mPm39分1222133332122223113333231121分3130.n23I,nnn2nn2()0(當時分nn3故2m*P()n分3即0.分m*Pdtlim1。tn1(0,)n1tnn1111當(,n2分tttn1tnn當)t11n12n1t1n11ttt1t2nnn12ntt,n2分22n1t2nn121,t(t),t令F(t)n24,2t1F(t),分tn11tnn且546，F(xiàn)(t)1tt12(0,)0即()在Ft上分1分ett1nn1tnndtlimedt1分tt1ntn1(0,)(0,)nn是TXTX設為Xx0,,,,.分xTxTxnxTxxTxTx2n10210n10是Xxnn1d(x,x)dTx,Tx)m1mmm1(,dxx)(dTxTx,)mm1m1m2(,)(,),dxxmdxx2m1m210mnd(x,x)d(x,x)d(x,x)d(x,x)mnmm1m1m2n1n()(,)dxxmm1n1011nm(,),dxxm10101nmmd(x,x)d(x,x(nm).1mn01m,n,d(x,x)即x是X分mn性nn1，由X的完備知存在xX,使.因xxm分d(x,Tx)d(x,x)d(x,Tx)mmd(x,x)d(x,x)0,mm，m1故d(x,Tx)0，即x所以為T的不動得x分~~~xXTxx~~~d(x,x)dTx,)d(x,x),~~因1,故d(x,x)0，即xx，分設0，(xg(x)f(x)g(x)f，Ef(x)dxg(x)dx.EE6()()0,gxfx[g(x)f(xx0分E若[g(x)f(xx0,E則()()0,分gxf