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Measure Theory and Functional Analysis




                    Notes          Theorem 5: In a Hilbert space the inner product is jointly continuous i.e.,
                                                            x    x, y    y   (x , y )   (x, y)
                                                             n     n       n  n
                                   Proof: We have
                                                 |(x , y ) – (x, y)| = |(x , y ) – (x , y) + (x , y) – (x,, y)|
                                                    n  n            n  n   n     n
                                                               = |(x , y  – y) + (x  – x, y)|
                                                                    n  n      n
                                                                               (by linearity property of inner product)
                                                                 |(x , y  – y)| + |(x  – x, y)|  [ |  +  |   | | + | |]
                                                                   n  n         n
                                                                  x  +  y  – y  +   x  – x     y   [By Schwarz inequality]
                                                                   n    n        n
                                   Since x    x and y    y as n     .
                                        n        n
                                   Therefore   y  – y     0 and   x  – x    0 as h   . Also (x ) is a continues sequence, it is bounded
                                             n  n          n                    n
                                   so that   x     M   n.
                                          n
                                   Therefore
                                          | (x , y ) – (x, y) |    0 as n     .
                                             n  n
                                   Hence (x , y )    (x, y) as n     .
                                          n  n
                                   This completes the proof of the theorem.
                                   Theorem 6: A closed convex set E in a Hilbert space H continuous a unique vector of smallest
                                   norm.

                                   Proof: Let   = inf {  e  ; e   E}
                                   To prove the theorem it suffices to show that there exists a unique x   E s.t.   x   =  .
                                   Definition of   yields us a sequence (x ) in E such that
                                                                 n
                                                        Lim x n  =                                         … (1)
                                                        n
                                                          x   x
                                   Convexity of E implies that   m  n  E . Consequently
                                                            2

                                                     x   x
                                                      m   n         x  + x      2                          … (2)
                                                       2            m  n
                                   Using parallelogram law, we get
                                                                      2
                                               x  + x    +   x  – x  2  = 2   x    + 2   x    2
                                                     2
                                               m   n     m   n      n       n
                                                                      2
                                   or                    x  – x  2  = 2   x    + 2   x    –   x  – x    2
                                                                              2
                                                         m   n      m       n     m   n
                                                                              2
                                                                      2
                                                                 2   x    + 2   x    – d  2           (Using (2))
                                                                    m       n
                                                                  0 as m, n                           (Using (1))
                                            x  – x  2    0 as m, n
                                            m   n
                                          (x ) is a CAUCHY sequence in E.
                                           n
                                            x   E such that  Lim x  x , since H is complete and E is a closed subset of H, therefore
                                                        n   n
                                          E is also complete and consequently (x ) is in E is a convergent sequence in E.
                                                                         n


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