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Unit 23: Hilbert Spaces: The Definition and Some Simple Properties




          (i)  ( x + y, z) =   (x, z) +   (y, z),   x, y, z   X and  ,     C.  (Linearity property)  Notes

          (ii)  (x, y)  = (y, x)                                    (Conjugate symmetry)
          (iii)  (x, x)   0, (x, x) = 0    x = 0                         (Non-negativity)

          A complex inner product space X is a linear space over C with an inner product defined on it.




             Notes

             1.  We can also define inner product by replacing C by R in the above definition. In that
                 case, we get a real inner product space.
             2.  It should be noted that in the above definition (x, y) does not denote the ordered pair
                 of the vectors x and y. But it denotes the inner product of the vectors x and y.

          Theorem 1: If X is a complex inner product space then
          (a)  ( x –  y, z) =   (x, z) –   (y, z)

          (b)  (x,  y +  z) =  (x, y) +  (x, z)

          (c)  (x,  y –  z) =   (x, y) –  (x, z)
          (d)  (x, 0) and (0, x) = 0 for every x   X.

          Proof: (a)         ( x –  y, z) = ( x + (– ) y, z)
                                       =   (x, z) + (– ) (y, z)
                                       =   (x, z) –  (y, z).

          (b)                (x,  y +  z) =  ( y  z, x) ( y, x) ( z, x)

                                       =  (y, x)  (z,x)

                                       =  (x, y)  (x,z)

          (c)                 (x,  y –  z) = (x,  y + (–  ) z) =  (x, y) (  )(x,z)

                                       =  (x, y)  (x,z)

          (d)                     (0, x) = (0 , x) = 0 ( , x) = 0, where  is the zero

                                                        element of x and  (x, 0) (0, x)  0  0 .

          Further note that (x, y + z) = (x, |y + 1|z) =  1 (x, y) +  1 (x, z)
          Hence (x, y + z) = (x, y) + (x, z).
          This completes the proof of the theorem.










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