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Measure Theory and Functional Analysis




                    Notes
                                                                  x (1/n)     x (1/n)
                                   Also                 |G  (x)| =    f(t) dt  n  |f(t)|dt  M .
                                                          n
                                                                    x           x
                                   Hence |G  (x)|  M,   n   N and   x   [a, b].
                                           n
                                                                             2
                                                                       2
                                                    [G (x) – f (x)] 2   (M + M)  = 4M , x   [a, b].
                                                      n
                                   On applying Lebesgue bounded convergence theorem, we get
                                                      b           b
                                                  Lim (G  n  f) 2  =  Lim (G n  f) 2  0
                                                   n
                                                      a           a
                                                              2
                                                    Lim G n  f  = 0
                                                     n        2
                                                    Lim G n  f
                                                    n         2
                                   or               Lim f G  n  = 0
                                                     n
                                     for given   > 0,  n   N, such that n  n
                                                    o                o
                                                         f – G  <
                                                             n  2
                                   Particularly for n = n .
                                                    o
                                                        f  G   <
                                                            n o 2
                                                          f – g  <                               (Taking  G  = g)
                                                               2                                          n o
                                   Thus there exists a continuous function  G (x) g(x)
                                                                    n o
                                                                   x (1/n )
                                                                       o
                                                               =  n o   f(t) dt, x [a, b],
                                                                     x
                                   which satisfies the given condition.

                                   8.2 Summary


                                      A linear functional f on a normed space N  is said to be bounded if there is a constant
                                                                           1
                                       k > 0 such that
                                                         |f (x)|  k   x  ,    x   N
                                                                             1
                                      If x     and f is bounded linear functional on   , then f has the unique representation of
                                             p                                p
                                       the form as an infinite series.

                                                           f (x) =   x f(e )
                                                                         k
                                                                     k
                                                                  k 1
                                                      *
                                      The norm of f      is given by
                                                      p
                                                                            1
                                                                            q
                                                              f  =   |f(e )| q
                                                                        k
                                                                   k 1



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