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Measure Theory and Functional Analysis




                    Notes          If x is any vector in B we can by the Archimedean property of real numbers find a positive
                                   integer n such that n >   x  , i.e., x   S ,
                                                                n
                                   therefore

                                                                 
                                                              B =   S  n
                                                                  n 1
                                   and since t is onto, we have
                                                             B = T (B)

                                                                    
                                                               = T    S  n
                                                                    n 1

                                                                 
                                                               =    T (S )
                                                                       n
                                                                  n 1
                                   Now B  being complete, Baire’s theorem implies that some  T S   possesses an interior point
                                                                                      n 0
                                   Z . This in turn yields a point y     T S   such that y  is also an interior point of  T S  .
                                    0                       0     n 0         0                         n 0
                                   Further, maps j : B    B  and g : B    B
                                   defined respectively by j (y) = y = y – y  and g (y) = 2 n  y
                                                                  0            0
                                   where n  is a non-zero scalars, are homeomorphisms as shown below f is one-to-one and onto.
                                         o
                                            –1
                                   To show f, f  are continuous, let y    B  and y     y in B.
                                                              n        n
                                   Then   f (y ) = y  – y     y – y  = f (y)
                                            n    n  0      0
                                                                –1
                                          –1
                                   and    f  (y ) = y  + y     y + y  = f  (y)
                                              n   n   0      0
                                   Hence f and f  are both continuous so that is a homeomorphism.
                                              –1
                                   Similarly g : B    B  : g (x) = 2n y is a homeomorphism for, g is one-to-one, onto and bicontinuous
                                                           0
                                   for n    0.
                                      0
                                   Therefore we have
                                   (i)  f (y ) = 0 = origin in B  is an interior point of  f T S  .
                                          0                                     n
                                   (ii)  f T S   =  f T S
                                            n 0
                                                       n 0
                                                 =  T S   y
                                                      n 0  0

                                                    T S                 y   T S
                                                      2 n 0                0    n 0

                                   (iii)  T S    =  T 2 S  2n T S
                                           2 n 0      n 0  1  0   1
                                                 =  g (T(S )) g (T(S ))
                                                        1
                                                                 1
                                                 =  2 T(S )
                                                    n 0  1



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