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p
                                                                     Unit 8: Bounded Linear Functional on the L -spaces



               It follows now from (11) and (13) that                                           Notes
                                          n
                                              i  so that y
                                     f  =   |y |          f is an isometric isomorphism.
                                          i 1
               Hence,    n  *   .
                            n
                            1
               This completes the proof of the theorem.




             Note  We need the signum function for finding the conjugate spaces of some infinite
             dimensional space which we define as follows:
             If   is a complex number, then


                  sgn      if  0
                       | |
                        0 if  0

                 (i) |sgn  | = 0 if   = 0 and | sgn  | = 1 if     0


                 (ii)   sgn   = 0 if   = 0 and   sgn   =    =| |, if     0.
                                               | |

          Theorem 4: The conjugate space of    is   , where
                                       p    q
                  1  1
                        = 1 and 1 < p <  .
                  p  q

          or        * p   .
                      q

          Proof: Let x = (x )      so that   |x | p                              … (1)
                       n    p           n
                                    n 1
                                                th
          Let e  = (0, 0, 0, …, 1, 0, 0, …) where 1 is in the n  place.
              n
                                      e       for n = 1, 2, 3, … .
                                           p
                                       n
                                                                                 *
          We shall first determine the form of f and then establish the isometric isomorphism of    onto
                                                                                 p
           q .
          By using (e ), we can write any sequence
                   n
                                              n
                 (x , x , … x , 0, 0, 0, …) in the form   x e  and
                   1  2   n                      k  k
                                              k 1
                     n
                  x    x e   (0, 0, 0,  , x  , x  ,  ).
                        k  k           n 1  n 2
                     k 1





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