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Measure Theory and Functional Analysis




                    Notes
                                                              1
                                             n                p
                                   Now   x    x e       |x | p                                             … (2)
                                               k  k       k
                                            k 1       k n 1
                                   The R.H.S. of (2) gives the remainder after n terms of a convergent series (1).

                                                              1
                                                             p
                                   Hence                |x | p    0 as n   .                               … (3)
                                                          k
                                                     k n 1
                                   From (2) and (3) if follows that


                                                              x =   x e                                    … (4)
                                                                     k  k
                                                                  k 1
                                                    n
                                   Let f     *  and S  =   x e  then
                                           p    n      k  k
                                                    k 1
                                                             S    x as n                              (Using (4))
                                                              n
                                   Since f is linear, we have

                                                                  n
                                                           f (S ) =   x f(e ) .
                                                             n       k  k
                                                                  k 1
                                   Also f is continuous and S     x, we have
                                                       n
                                                           f (S )   f (x) as n   .
                                                             n
                                                                  n
                                                           f (x) =   x f(e )                               … (5)
                                                                     k  k
                                                                  k 1
                                   which gives the form of the functional on   .
                                                                      p
                                   Now we establish the isomeric isomorphism of   *  onto   , for which proceed as follows:
                                                                                  q
                                                                           p
                                   Let f (e ) =   and show that the mapping
                                        k    k
                                   T :   *        given by
                                            q
                                       p
                                   T (f) = ( ,  , …,  , …) is an isomeric isomorphism of   *  onto   .
                                                                                        q
                                          1  2    k                              p
                                   First, we show that T is well defined.
                                   For let x     , where x = ( ,  , …,  , 0, 0, …) where
                                             p
                                                           2
                                                                 n
                                                         1
                                                                  |  | g 1  sgn  k 1 k  n
                                                                             ,
                                                               =    k
                                                              k     0           n  k
                                                                     q–1
                                                           | | = | |  for 1   k   n.
                                                             k     k
                                                                       p                   1  1
                                                                    (q 1)
                                                          | | = |   |    = | | .                q   p(q 1)  q
                                                                              q
                                                              p
                                                            k      k        k
                                                                                           p  q
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