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Unit 15: Banach Space: Definition and Some Examples




                                                                                                Notes
                                           2           2
                                       =  z 1  2 z z  2  z 2      z , z 2  z , z 2  z z  2
                                                                     1
                                                                           1
                                                                                  1
                                                1
                                       = (|z | + |z |) 2
                                           1     2
                               |z  + z |  |z | + |z |
                                 1   2    1     2
          or                     z  + z       z    +   z                   (   x   = |x|)
                                 1  2     1     2
          Also                       x  = | x| = | | |x| = | |   x
          Hence all the conditions of normed linear space are satisfied. Thus both C or R are normed linear
          space. And by Cauchy general principle of convergence, R and C are complete under the matrices
          induced by the norm. So R and C are Banach spaces.

                                                                 n
                                                                       n
                 Example 2: Euclidean and Unitary spaces: The linear space R  and C  of all n-tuples (x ,
                                                                                     1
          x  …, x ) of real and complex numbers are Banach spaces under the norm
           2    n
                                                 1/2
                                           n
                                     x  =   |x | 2
                                              i
                                          i 1
          [Usually called Euclidean and unitary spaces respectively].
          Solution: (i) Since each |x |   0, we have
                              i
                                               x    0

                        n
                            2
          and   x  = 0      |x | = 0    x  = 0, i = 1, 2, …, n
                           i        i
                       i 1
                    (x , x , … x ) = 0
                     1  2   n
                    x = 0
          (ii) Let x = (x , x , …, x )
                    1  2    n
                                                n
                                                     n
          and y = (y , y , … y ) be any two numbers of C  (or R ). Then
                  1  2    n
                                 x + y   2  =    (x , x , …, x ) + (y , y , … y )   2
                                            1  2   n    1  2   n
                                       =    (x + x ), (x  + y ), …, (x + y )   2
                                            1  1  2   2     n   n
                                          n
                                       =   |x  y | 2
                                             i  i
                                         i 1
                                          n
                                           |x  y |(|x | |y |)
                                             i  i   i    i
                                         i 1

                                          n            n
                                           |x  y ||x |  |x   y ||y |
                                             i  i  i      i   i  i
                                         i 1          i 1
          Usually Cauchy inequality for each sum, we get

                                                   2        1             1         1
                                           n          n      2   n         2  n      2
                                 x + y   2  =   |x i  y |  |x | 2  |x i  y | 2  |y | 2
                                                                                 i
                                                         i
                                                 i
                                                                        i
                                          i 1         i 1        i 1         i 1
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