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Unit 24: Inner Product and Inner Product Spaces




                                                                                                Notes
                 Example 13: If F be the real field and V be the set of polynomials, in a variable x over F
          of degree 2 or less. In V we define an inner product by: If p(x), q(x)  V, then
                                           1
                              (p(x), q(x)) =   p ( ) ( ) dx
                                                x
                                             x
                                               q
                                          1
                                              2
          Let us start with the basis   = 1,   = x,   = x  of V and obtain orthogonal set by applying Gram-
                                1    2    3
          Schmidt process. Let
                                              1
                                          1
                                       =
                                      1        2
                                          1
                                           1
                                     2
          as                        1  =   1.dx  2.
                                          1
                                     ’ =    ( ,  )
                                      2   2   2  1  1
                                                                1
                                            1   1          1 x  2
                                       =  x      x .1 dx  x
                                             2  1          2  2
                                                                1
          So the orthonormal    is given by
                            2
                                          x       x        3
                                       =                     x
                                      2    ,     1    1/2  2
                                           2      2
                                                 x dx
                                                1
          Finally
                                     ,  =   ( ,  )    ( ,  )
                                     3    3   3  2  2   3  1  1
                                                 3    3     2 1    1
                                       =  x 2  x 2 ,  x  x  x  ,  ,
                                                 2    2        2   2

          Now
                                                           1
                                   3      3  1         3 x  4
                               x 2 ,  x  =    x  2  ,xdx      0
                                   2      2  1         2 4
                                                           1
          and

                                   1      1  1        1 x 3  1  2
                                x  2  ,  =    x 2  ,1dx
                                    2     2  1         2  3   3
                                                           1
          Thus
                                            1
                                     ,  =  x 2
                                     3
                                            3
          and normalized    is given by
                         3
                                         x  2  1  3  x 2  1 3     10   2
                                           ,             2   1/2    (3x  1)
                                       =          1               4        .
                                     3     3         2  1
                                                    x     dx
                                                  1    3


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