Page 103 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
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Measure Theory and Functional Analysis




                    Notes          Making use of Hölder’s inequality, we get
                                                                          1         1
                                                                          p         q
                                                        |fg|d      |f| d     |g| d
                                                                                q
                                                                     p
                                                        X         X         X
                                                               =   f      g                                … (3)
                                                                    p   q
                                   From (2) and (3) it follows that
                                                         |F (f)|    f      g  .
                                                                    p   q
                                             F|f|
                                   Hence  sup     : f  Lp(X) and f  0  g
                                              f                        q
                                                p
                                                             F      g            (Using definition of the norm) … (4)
                                                                    q
                                                                    q–1
                                   Further,                let f = |g|  sgn  g                             … (5)
                                   Since                 sgn g = 1, we get

                                                              p
                                                           |f| = |g| p(q–1)  = |g| q              ( p (q – 1) = q)
                                                              1           1
                                                             p            q
                                   Thus,  f   L  (X) and   |f| d  =   |g| d                                … (6)
                                                                      q
                                                         p
                                             p
                                                      X           X
                                                    1
                                                    p
                                                                    q /p
                                                q
                                                          p
                                   But       |g| d     |g| d   =  g  q
                                            X         X
                                   which implies on using (6) that
                                                                    q /p
                                                             f   =  g                                      … (7)
                                                              p     q

                                   Now                     F (f) =  fg d  |g| q 1  g sgng d
                                                                 X       X

                                                                     q
                                                               =  |g| d    g  q
                                                                             q
                                                                 X
                                                          q
                                   Hence                g   g  = F (f)     F      f  .
                                                          q                  p
                                   and this on using (7) yields that

                                                              q              q/p
                                                            g  = F (f)     F     g
                                                              q              q
                                                           q q /p
                                                         g     =   g       F                               … (8)
                                                           q        q
                                                                                                        ( g   0)






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