Page 226 - DMTH401_REAL ANALYSIS
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Real Analysis




                    Notes                                   1               1       1
                                                        m
                                                                  m
                                                                        m
                                                                               m
                                                     g (4 x  ±  )- g (4 x ) =  4 x  ±  -  4 x  =
                                                            2               2       2
                                       and finally
                                                            m
                                                         æ  3 ö  1 1  m  n
                                                     a n  = ç ÷  2  2 1  4 m  =  3 =  3
                                                         è
                                                          4 ø
                                             n
                                                n
                                   (c)  a n  £  3 , if  < m .
                                                                    1                    1           1
                                                                                    m
                                                                             n
                                                               n
                                                       n
                                       |g(4 (x + h )) – g(4 x)| = |g(4 x  ±    4 n–m ) – g(4 x)| £ |4 x  ±  4 n–m  – 4 x|=   4 n–m ,
                                           n
                                                                                                m
                                                m                   2                    2           2
                                       and therefore
                                                            n
                                                         æ  3 ö  m n  1  1  n
                                                               -
                                                     a £ ç ÷  4        =  3
                                                         è  4 ø  2  4 m
                                                      n            1
                                                                   2
                                       Putting all things together we get
                                                       x
                                                                x
                                                      f ( + h  ) - f ( )  ¥
                                                           m          a  = a  +  ...+ a  +a  + a  +  ... =
                                                                                     +
                                                                                          2
                                                                                     1
                                                                                    m
                                                                                          +
                                                                                         m
                                                                  = å n    1     m  
                                                          h
                                                           m        n = 0
                                                                                      = 0 by (a)
                                                     = ||a  + a  + ... +a  ³ |a | – |a +... + a  | ³
                                                         1   2     m    m     1     m–1
                                                                                  by(b)
                                                     ... ³|a | – |a | – |a | – ... – |a  |  =  3  –|a | – |a | – ... –|a  | ³
                                                                                      m
                                                          m     1    2        m–1         1     2       m–1
                                                     by(c)                    3 -  3  3
                                                                               m
                                                                           m
                                                                2
                                                             1
                                                         m
                                                      ³  3 – 3 – 3  – ... – 3 m–1  = 3  –   =  (3 m - 1  -  1) ® ¥ .

                                                                               3 1   2
                                                                                -
                                   Remark: In original constructive proof  of this theorem in 1872, Karl Weierstrab used f(x) =
                                     ¥  n    n
                                       a  cos(b  p  ) x  with a a Î (0, 1) and with positive odd integer b both satisfying ab > 1 + 3/4p.
                                   å n  0
                                      =
                                   Interesting  is,  that despite  of  the  differentiability  of  cosine, the  limit function  will not  be
                                   differentiable anywhere.
                                                                    Figure  17.5









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