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Unit 16: Continuous Linear Transformations




                                                                                                Notes
                                              q 1        q 1
                              Now      =        sgn          sgn
                                   k  k   k  k      k   k   k    k
                                       = | |  = | | p   (Using property of sgn function)  … (7)
                                            q
                                   k  k    k     k
                                                 1
                                           n     p
                                               p
                                    x  =      k
                                          k 1
                                                 1
                                           n     q
                                               q
                                       =      k                                   … (8)
                                          k 1
          Since we can write
                                          n
                                     x =     k e k , we get
                                         k 1

                                          n         n
                                   f (x) =   k  f (e )  k  k
                                                k
                                         k 1        k 1
                                          n   q
                                   f (x) =                         ( Using (7))              … (9)
                                             k
                                         k 1
          We know that for every x    p

                                | f (x)|   f  x ,

          which upon using (8) and (9), gives

                                                            1
                                          n           n     p
                                              q           q
                                 |f (x)|     k    f      k
                                         k 1         k 1
          which yields after simplification.

                                     1
                               n     p
                                   q
                                  k       f                                      … (10)
                              k 1
          since the sequence of partial sums on the L.H.S. of (10) is bounded, monotonic increasing, it
          converges. Hence

                                     1
                               n     q
                                   q
                                  k       f                                      … (11)
                              k 1
          so the sequence ( ) which is the image of f under T belongs to    and hence T is well defined.
                                                              q
                         k
          We next show that T is onto   .
                                  q
                                                 *
          Let ( )     , we shall show that there is a  g   such that T maps g into ( ).
                   q
               k                                 p                     k


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