Page 97 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
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Measure Theory and Functional Analysis




                    Notes                                  |y |    f   or max {|y |}                  [   x   = 1]
                                                             i               i
                                                                   f                                      … (10)
                                       From (8) and (10), we obtain
                                                              f  = max. {|y |} so that
                                                                        i
                                                                            n
                                       y    f is an isometric isomorphism of L  to   *.
                                                                            1
                                                    n
                                               n
                                       Hence   *   .
                                               1
                                               n
                                   (iii)  Let L =    with the norm
                                                             x  = max {|x | : 1, 2, 3, …, n}.
                                                                       i
                                       Now f defined in (1) above is continuous as in (1). Let L  represents the set of all continuous
                                                          n
                                       linear functionals on    so that
                                                             L =   n  * .

                                       Now we determine the norm of y’s which makes y    f as isometric isomorphism.

                                                                   n       n
                                                         |f (x)| =   x y i  |x ||y |.
                                                                              i
                                                                                  i
                                                                      i
                                                                  i 1     i 1
                                                      n                    n
                                                 But   |x ||y |    max (|x |)  |y |
                                                         i   i          i      i
                                                     i 1                   i 1
                                       Hence we have

                                                                   n                    n
                                                         |f (x)|    |y | ( x ) so that f  |y |            … (11)
                                                                                            i
                                                                       i
                                                                   i 1                  i 1
                                       Consider the vector x defined by
                                                                 |y |
                                                             x =   i   when y    0 and x  = 0 otherwise.  … (12)
                                                              i   y        i       i
                                                                   i
                                                                      |y |
                                       Hence                 x   =  max  i  1 .
                                                                      |y |
                                                                        i
                                                                   n        n
                                       and               |f (x)| =   |x y |   |y |
                                                                      i  i      i
                                                                  i 1       i 1
                                       Therefore
                                                         n
                                                          |y | = |f (x)|     f    x   =   f  .
                                                             i
                                                         i 1
                                                         n
                                                          |y |     f                                      … (13)
                                                             i
                                                         i 1



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