實(shí)變函數(shù)(曹廣福)1到5章答案

1、第一章習(xí)題參考解答1第一章習(xí)題參考解答第一章習(xí)題參考解答3等式成立的的充要條件是什么？)()(CBACBA?????解：若，則.)()(CBACBA?????ACBACBAC???????)()(即，.AC?反過來假設(shè)因?yàn)?所以.故AC?BCB??)(CBABA????.CBA??)(?)(CBA??最后證，CBACBA?????)()(事實(shí)上，則且。若，則；)(CBAx????Ax?CBx??Cx?CBAx???)(若，則，故.從而.

2、Cx?Bx?CBABAx?????)(CBACBA?????)()(.即.AACBACBAC??????????)()(AC?反過來，若，則因?yàn)樗杂忠驗(yàn)?，AC?BCB??)(CBABA????AC?所以故)(CBAC???)()(CBACBA?????另一方面，且，如果則AxCBAx??????)(CBx??Cx?；如果因?yàn)椋怨?則CBAx?)(??Cx?CBx??Bx?BAx??.從而CBAx???)(CBACBA?????)(

3、)(于是，)()(CBACBA?????4對(duì)于集合A，定義A的特征函數(shù)為，假設(shè)??????AxAxxA01)(?是一集列，證明：??nAAA21（i）)(inflim)(inflimxxnnAnnA???（ii）)(suplim)(suplimxxnnAnnA???證明證明：（i），，時(shí)，.)(inflimnnmNnnnAAx???????N??0n0nm??mAx?所以，所以故1)(?xmA?1)(inf0??xmAnm?1)(inf

4、sup)(inflim????xxmnAnmNbAn??第一章習(xí)題參考解答3反過來，，使?NnnaxfxxExn????????1)(|1?1)(|naxfxEx???即故Exanaxf????且1)()(|axfxEx??所以故)(|1)(|1axfxEnaxfxEn???????1)(|)(|1naxfxEaxfxEn??????7設(shè)是E上的實(shí)函數(shù)列，具有極限，證明對(duì)任意常數(shù)a都有：)(xfn)(xf1)(|inflim1)(|in

5、flim)(|11kaxfxEkaxfxEaxfxEnnknnk?????????????證明：證明：，即，且N?????kaxfxEx)(|kaaxf1)(???Ex?因?yàn)?，使，有，故Nnxfxfnn?????，)()(limnm??kaxfn1)(??所以，)(1)(|nmkaxfxExm??????x1)(|kaxfxEmnm????=，由k的任意性：1)(|kaxfxExmnmNn???????1)(|inflimkaxfxEm

6、n??，反過來，對(duì)于，1)(|inflim1kaxfxExnnk??????1)(|inflim1kaxfxExnnk???????，有=，即Nk??1)(|inflimkaxfxExmn???1)(|kaxfxEmnmNn??????時(shí)，有：且，所以，且nmNn????，kaxfm1)(??Ex?kaxfxfmm1)()(lim???.，故從而Ex???k又令Exaxf??且)()(|axfxEx??故=)(|axfxE?1)(|in