Page 295 - Matrix Analysis & Applied Linear Algebra
P. 295

5.3 Inner-Product Spaces                                                           291


                                    guarantees that  x y  +  x z  =  x y + z  . Now prove that  x αy  = α  x y
                                    for all real α. This is valid for integer values of α by the result just established,
                                    and it holds when α is rational because if β and γ are integers, then

                                                 β                                  β      β
                                            2
                                          γ   x   y   =  γx βy  = βγ  x y  =⇒    x   y   =    x y  .
                                                γ                                   γ      γ
                                    Because  x + αy  and  x − αy  are continuous functions of α (Exercise
                                    5.1.7), equation (5.3.8) insures that  x αy  is a continuous function of α. There-
                                    fore, if α is irrational, and if {α n } is a sequence of rational numbers such that
                                    α n → α, then  x α n y → x αy  and  x α n y  = α n  x y → α  x y  , so
                                     x αy  = α  x y  .

                   Example 5.3.4
                                                                                  n
                                    We already know that the euclidean vector norm on C  is generated by the stan-
                                    dard inner product, so the previous theorem guarantees that the parallelogram
                                    identity must hold for the 2-norm. This is easily corroborated by observing that
                                                    2         2         ∗               ∗
                                              x + y  +  x − y  =(x + y) (x + y)+(x − y) (x − y)
                                                    2         2
                                                                                     2      2
                                                                =2 (x x + y y)=2( x  +  y  ).
                                                                     ∗
                                                                           ∗
                                                                                     2      2
                                    The parallelogram identity is so named because it expresses the fact that the
                                    sum of the squares of the diagonals in a parallelogram is twice the sum of the
                                    squares of the sides. See the following diagram.
                                                                                 x + y

                                                             y  ||x - y||  ||x + y||

                                                         ||y||
                                                                   ||x||     x






                   Example 5.3.5
                                    Problem: Except for the euclidean norm, is any other vector p-norm generated
                                    by an inner product?
                                    Solution: No, because the parallelogram identity (5.3.7) doesn’t hold when
                                                             2          2        2      2
                                    p 
=2. To see that  x + y  +  x − y  =2  x  +  y    p   is not valid for
                                                                        p
                                                                                 p
                                                             p
                                              n
                                    all x, y ∈C  when p 
=2, consider x = e 1 and y = e 2 . It’s apparent that
                                            2    2/p           2
                                     e 1 + e 2   =2  =  e 1 − e 2   , so
                                            p                  p
                                                  2          2    (p+2)/p             2      2
                                          e 1 + e 2   +  e 1 − e 2   =2   and   2  e 1   +  e 2    =4.
                                                  p          p                        p      p
   290   291   292   293   294   295   296   297   298   299   300