實分析與復(fù)分析JnensenHolder'sinequaliyiPPT課件

1、第1頁/共29頁第2頁/共29頁第3頁/共29頁,),)()()(xfxfVfJxfJ ),)()()(fxfVfJxfJ, 0)(, 0, 0)(),()(xfJxfJxfJxfJ第4頁/共29頁)()(xfJxfJ, )()(fxfVfJ),(),(11LfJLf,dfJdfJdfJ第5頁/共29頁dffdVdfJdfJ ,)(ffVfJ第6頁/共29頁 dfJfJ1)( )( fJdfJ fJ 第7頁/共29頁第8頁/共29頁第9頁/共29頁第10頁/共29頁第11頁/共29頁第12頁/共29頁triviality. In fact,fgfgfg,dfgdfgdfg.dfgfgd第13頁

2、/共29頁If the first inequality in (1) is a equality, then,dfgfgd,Re(),iezzufg 令令則則設(shè)設(shè),arg ,if gdzzz ez 設(shè)設(shè)令令其其中中,ufgfg 因因此此，第14頁/共29頁f gdzzfgdfgd,udfgd,.fgfgalmost every x( ) ( )( ) ( ) ,if x g xef x g x .almost every x 第15頁/共29頁If( ) ( )( ) ( ) ,if x g xef x g xalmost every,thenxiiifgdeefgdef gd,fgd 第1

3、6頁/共29頁We may as well supposetrivial. In fact, if p=，then q=1.)(sup1gfdgxfessdfgx第17頁/共29頁If the second inequality in (1) is a equality, then,1dgfgfdfg1,.fgfgalmost every x 0If ( )then( ),g xf xf ，and( )for.f xfalmost every x In turn is obvious.第18頁/共29頁So we suppose that 1,.p q 第19頁/共29頁第20頁/共29頁.p

4、JFdf d 第21頁/共29頁In fact,11()()pqpJFJFdfgd 11()()ppqqf gg df d 1()ppJ FFFd 第22頁/共29頁11()()ppqqqppfgdfgg d1111,()()pqfgdqqpp (),qdg d 第23頁/共29頁inequality,By Jensen s1()pJ Ffgd ,)(FJFJ1()pJFf d 1111,()()pppfgdf d 第24頁/共29頁 111()pppfgdf d 11ppqqg df d pqfg Our conclusion (1) is then an immediate consequ

5、ence of Jensens inequality asis the condition for equality.第25頁/共29頁If the second inequality in (1) is a equality, then.qpgfdfg.)(FJFJByinequality, is a constant.Jensen sF(forsome constant).qqppFfg 1111,()ppqpgffppqpq 第26頁/共29頁1for.pgfalmost every x In turn is obvious. In fsct,1ifforthenpgfalmost every x ，,pppfgdf df 111qqqq