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Measure Theory and Functional Analysis




                    Notes
                                                            1
                                            n               p
                                                          p
                                   Now  x     x e k     x k                                                … (2)
                                               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
                                               p
                                   Hence     x k     0 as n                                                … (3)
                                         k n 1
                                   From (2) and (3), it follows that


                                       x =   x e k .                                                       … (4)
                                              k
                                           k 1
                                                   n
                                         *
                                   Let  f    and  s  n  x e  then
                                         p
                                                       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 isometric isomorphism of    onto   , for which we proceed as follows:
                                                                           p
                                                                                  q
                                   Let f (e ) =   and show that the mapping
                                        k    k
                                                                 *
                                                             T :         given by                        … (6)
                                                                      q
                                                                 p
                                                                                 *
                                   T (f) = ( ,  , …,  , …) is an isometric isomorphism of    onto   .
                                          1  2    k                              p      q
                                   First, we show that T is well defined.
                                   For let x   , where x = ( ,  , …,  , 0, 0, …)
                                            p
                                                        1
                                                           2
                                                                n
                                                g 1  sgn  , 1 k  n
                                   where   =   k      k
                                         k     0           n  k
                                                           | | = | | q – 1  for 1   k   n.
                                                             k     k
                                                                                            1  1
                                                                      p                   
                                                                    (q 1)
                                                          | | =          = | | .                 q   p(q 1) q
                                                                             q
                                                              p
                                                            k      k        k              p  q
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