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Real Analysis




                    Notes          The sequence of derivatives is
                                                                   f‘(x) = cos(nx)
                                   Which does not converge (take for example x = ).


                                          Example:  Find  a sequence of differentiable  functions that converges uniformly to  a
                                   continuous limit function but the limit function is not differentiable

                                   we found a sequence of differentiable functions that converged point wise to the continuous,
                                   non-differentiable function f(x) = |x|. Recall:

                                                      ì     1           1
                                                                    x
                                                                 1
                                                      ï - x -  2n  if -   -  n
                                                      ï
                                                      ï n  2      1     1
                                                 f (x)=  í  x  if -  <  x <
                                                  n     2         n     n
                                                      ï
                                                      ï    1     1
                                                      ï x -   if    x   1
                                                      î   2n     n
                                   That same sequence also converges uniformly, which we will see by looking at ‘|| f  – f|| .
                                                                                                        n     D
                                   We will find the sup in three steps:
                                   If  1  x –1/n:
                                      f (x) – f(x)| = |–x –  /  + x| =  /
                                                              1
                                                      1
                                       n               2n       2n
                                   If  –1/n < x < 1/n:
                                                                       1
                                                                     2
                                                                            3
                                                   n
                                                       2
                                      |f (x) – f(x)| | /  x | + |x|   /   /  +  /  =  /
                                                                n
                                                                   1
                                        n            2            2  n   n   2n
                                   If  1/n x  1:
                                                      1
                                      |f (x) – f(x)| = |x –  /  – x| =  /
                                                              1
                                        n               2n      2n
                                   Thus, || f  – f||  <  /  which implies that  f  converges uniformly to f. Note that all  f  are
                                                    3
                                           n     D    2n                n                                  n
                                   continuous so that the limit function must also be continuous (which it is). But clearly f(x) = |x|
                                   is not differentiable at x = 0.
                                   16.3 Uniform Convergence and Integrability
                                                  ¥
                                   Theorem 2: Let  { }f  n n= 1   be a sequence of real-valued functions on  [ , ].a b  If all  f  are Riemann-
                                                                                                  n
                                   integrable and  f ®  f  uniformly then f is Riemann-integrable and
                                                n
                                                                 b         b
                                                                 a ò  f  n ( )dx ®  a ò  f ( ) .
                                                                               dx
                                                                              x
                                                                    x
                                                                                                      n
                                                      :
                                   Proof: Recall partition  P a t=  0  <  t <  ... t =  . b  The upper Darboux sum  ( , )U f p = å  i=  1 sup [ i t - 1 , i t  ]
                                                                 <
                                                                   n
                                                             1
                                                                                          x
                                            )
                                   f ( )(t -  t i- 1  and the lower Darboux sum  ( , )L f P = å  n i=  1 inf [ i t - 1 , i t  ]  f  ( ) (t - t i-  1 ).  f  is  Riemann-
                                     x
                                        i
                                                                                             i
                                   integrable on  [ , ]a b  if and only if  " > 0 PU  ( , ) L ( , ) <  .
                                                                        f
                                                                               f
                                                                            -
                                                                                 P
                                                                    $
                                                                          P
                                   Claim 1: f is Riemann-integrable.
                                                                                
                                   Let   >  0.  Since  f ®  f  uniformly  n$ s.t.  f -  f  sup <  4(b a ) .  Since  f  is R-integrable  P$  s.t.
                                                                      n
                                                                                           n
                                                 n
                                                                                 -
                                     f
                                               P
                                   U ( , ) L ( , ) <  /2.  We have
                                             f
                                       P
                                         -
                                      n
                                             n
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