實(shí)變函數(shù)論與泛函分析曹廣福1到5章課后答案

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)()(CBACBA?????成立的的充要條件是什么？解：若)()(CBACBA?????，則ACBACBAC???????)()(.即，AC?.反過來假設(shè)AC?因?yàn)锽CB??.所以)(CBABA????.故CBA??)(?)(CBA??.最后證，CBACBA?????)()(事實(shí)上，)(CBAx????則Ax?且CBx??。若Cx?，則CBAx???)(；若Cx?，則Bx?

2、，故CBABAx?????)(.從而CBACBA?????)()(.AACBACBAC??????????)()(.即AC?.反過來，若AC?，則因?yàn)锽CB??所以)(CBABA????又因?yàn)锳C?，所以)(CBAC???故)()(CBACBA?????另一方面，AxCBAx??????)(且CBx??，如果Cx?則CBAx?)(??；如果Cx?因?yàn)镃Bx??，所以Bx?故BAx??.則CBAx???)(.從而CBACBA?????)(

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

4、sup)(inflim????xxmnAnmNbAn??第一章習(xí)題參考解答3即Exanaxf????且1)(故)(|axfxEx??所以)(|1)(|1axfxEnaxfxEn???????故1)(|)(|1naxfxEaxfxEn??????7設(shè))(xfn是E上的實(shí)函數(shù)列，具有極限)(xf，證明對(duì)任意常數(shù)a都有：1)(|inflim1)(|inflim)(|11kaxfxEkaxfxEaxfxEnnknnk?????????????證

5、明：證明：N?????kaxfxEx)(|，即kaaxf1)(???，且Ex?因?yàn)镹nxfxfnn?????，)()(lim，使nm??，有kaxfn1)(??，故，)(1)(|nmkaxfxExm?????所以?x1)(|kaxfxEmnm????1)(|kaxfxExmnmNn???????=1)(|inflimkaxfxEmn??，由k的任意性：1)(|inflim1kaxfxExnnk??????，反過來，對(duì)于1)(|infli

6、m1kaxfxExnnk???????，Nk??，有1)(|inflimkaxfxExmn???=1)(|kaxfxEmnmNn??????，即nmNn????，時(shí)，有：kaxfm1)(??且Ex?，所以，kaxfxfmm1)()(lim???且Ex?.??k又令，故Exaxf??且)(從而)(|axfxEx??故)(|axfxE?=1)(|inflim1kaxfxEnnk?????8設(shè))(xfn是區(qū)間（a，b）上的單調(diào)遞增的序列，即?