Page 101 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
P. 101

Measure Theory and Functional Analysis




                    Notes          We next show that T is onto   .
                                                           q
                                                                     *
                                   Let ( )     , we shall show that is a g     such that T maps g into ( ).
                                       k    q                        p                      k
                                   Let x      so that
                                          p
                                                                  n
                                                              x =   x e k
                                                                     k
                                                                  k 1
                                   We shall show that

                                                                  n
                                                           g (x) =   x k  k  is the required g.
                                                                  k 1

                                   Since the representation for x is unique, g is well defined and moreover it is linear on   . To
                                                                                                           p
                                   prove it is bounded, consider

                                                                   n        n
                                                         |g (x)| =    k  x k   k  x k
                                                                  k 1      k 1
                                                                         1         1
                                                                   n     p   n     q
                                                                       p         q
                                                                     x  k       k        (Using Hölder’s inequality)
                                                                   k 1      k 1
                                                                             1
                                                                      n      q
                                                         |g (x)|   x   | | q
                                                                          k
                                                                      k 1
                                          g is bounded linear functional on   .
                                                                       p
                                   Since  e       for k = 1, 2, …, we get
                                           k   p
                                          g (e ) =   for any k so that
                                             k   k
                                          T  = ( ) and T is on   *  onto   .
                                           g   k            p      q
                                   We next show that
                                            Tf   =   f   so that T is an isometry.

                                   Since Tf   ,  we have from (6) and (10) that
                                            q
                                                             1
                                                             q
                                                            q  =   Tf       f
                                                           k
                                                      k 1


                                   Also,                 x       x =   x e k . Hence
                                                             p           k
                                                                     k 1

                                                           f (x) =   x (e )  x  k  k
                                                                     k
                                                                       k
                                                                  k 1      k 1




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