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Measure Theory and Functional Analysis




                    Notes


                                     Notes  Let L be a linear space. A mapping p : L   R is called a sub-linear functional on L if
                                     it satisfies the following two properties namely,

                                     (i)  p (x + y)   p (x) + p (y)    x, y   L  (sub additivity)
                                     (ii)  p (  x) =   p (x),     0    (positive homogeneity)
                                     Thus p defined on L in the above theorems is a sub-linear functional on L.

                                   Some Applications of the Hahn-Banach Theorem

                                   Theorem: If N is a normed linear space and x    N, x    0 then there exists a functional f    N* such
                                                                     o     o                          o
                                   that
                                                               f  (x ) =   x    and   f    = 1.
                                                               o  o    o      o
                                   Proof: Let M denote the subspace of N spanned by x , i.e.,
                                                                             o
                                                               M = {  x  :   any scalar}.
                                                                      o
                                   Define f : M    F (R or C) by

                                                                  f (  x ) =     x   .
                                                                      o      o
                                   We show that f is a functional on M with   f  = 1.
                                   Let x , x    M so that
                                       1  2
                                   x  =   x  and x  =   x . Then
                                    1  1  o    2  2  o
                                                       f (x  + x ) = f (  x  +   x )
                                                          1  2      1  o  2  o
                                                               = (  +  )   x
                                                                   1  2   o
                                   But              (  +  )   x   =     x  +     x
                                                      1  2   o    1  o    2  o
                                                               = f (x ) +  f (x )
                                                                    1     2
                                   Hence               f (x  + x ) = f (x ) + f (x )                       … (1)
                                                          1  2      1    2
                                   Let k be a scalar (real or complex). Then if x   M, then x =   x .
                                                                                    o
                                   Now f (kx) = f (k   x ) = k     x    = k f (x)                          … (2)
                                                   o       o
                                   If follows from (1) and (2) that f is linear.
                                   Further, we note that since x    M with   = 1, we get
                                                         o
                                                           f (x ) =   x   .
                                                             o     o
                                   For any x   M, we get, | f (x)| = | | || x || =   x  =  x
                                                                    o       o
                                                         |f (x) | =   x

                                          f is bounded and we have

                                             |f(x)|
                                          sup      = 1 for x   M and x   0.
                                               x
                                   So by definition of norm of a functional, we get
                                                              f  = 1.




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