Page 227 - DMTH505_MEASURE_THEOREY_AND_FUNCTIONAL_ANALYSIS
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Measure Theory and Functional Analysis




                    Notes          Thus (s ) is a Cauchy sequence in B and since B is complete,   a vector x   B such that
                                        n
                                                             s    x and therefore
                                                              n
                                                             x  =  lim s  n  lim s n  2  3 ,
                                                                  n       n
                                   i.e., x   S .
                                          3
                                   It now follows by the continuity of T that

                                                           T (x) =  T lim s  n
                                                                   n

                                                               =  lim T(s )
                                                                       n
                                                                 n
                                                               =  lim (y  1  y  2   y )
                                                                                n
                                                                 n
                                                               = y
                                   Hence                      y   T (S )
                                                                    3
                                   Thus                       y    S     y   T (S ). Accordingly
                                                                             3
                                                             S   T (S )
                                                                    3
                                   This completes the proof of the lemma.





                                     Note  If B and B  are Banach spaces, the symbol S (x; r) and S  (x; r) will be used to denote
                                     open spheres with centre x and radius r in B and B  respectively. Also S  and  S  will denote
                                                                                             r    r
                                     these spheres when the centre is the origin. It is easy to see that
                                                         S (x; r) = x + S  and S  = r S
                                                                     r    r   1
                                   For, we have
                                                      y   S (x; r)     y – x   < r
                                                                    z   < r and y – x = z, z   S
                                                                                       r
                                                                  y = x + z and   z  < r

                                                                  y   x + S
                                                                         r
                                   Thus                  S (x; r) = x + S
                                                                     r
                                                                     x
                                   and            S  = {x :   x   < r} =  x :  1
                                                   r                 r

                                                               = {r . y   y   < 1}
                                                               = r S
                                                                   1
                                   Thus                      S = r S
                                                              r    1
                                   Now we prove an important lemma which is key to the proof of the open mapping theorem.







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