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Real Analysis




                    Notes          where the convergence is convergence in measure with respect to . In 1988, Tišer showed that if
                                                                          s  2
                                                                     s 2 i 1  £  i
                                                                       +
                                                                          i  a
                                   for some a > 5/2, then

                                                              1      f(y)d (y) ¾¾¾ f(x)
                                                                          
                                                            (B (x)) ò  r B (x)  r  0
                                                              r
                                                          p
                                   for -almost all x and all f  L (H, ; R), p > 1.
                                   As of 2007, it is still an open question whether there exists an infinite-dimensional Gaussian
                                   measure  on a separable Hilbert space H so that, for all f  L (H, ; R),
                                                                                    1
                                                                  1
                                                                             
                                                            lim       ò  f(y)d (y) =  f(x)
                                                            r   (B (x))  r B (x)
                                                              0
                                                                  r
                                   for -almost all x  H. However, it is conjectured that no such measure exists, since the s  would
                                                                                                         i
                                   have to decay very rapidly.
                                                                  æ  1 ö
                                                         a
                                          Example: If a ¹  0, ( ) arctan ç ÷
                                                            =
                                                        f
                                                                  è ø
                                                                   a
                                                a
                                   The function      is not continuous at the point (x, a) = (0, 0) and the function f(a) has a
                                               2
                                              x + a 2
                                                                         p                       p
                                                                                 +
                                                                                                         -
                                   discontinuity a = 0, because f(a) approaches  +  as a  0  and approaches  -  as a  0 .
                                                                         2                       2
                                                             1  a
                                   If we now differentiate f(a) =   ò 0 x + a 2  dx  with respect to a under the integral sign, we get
                                                               2
                                                              1
                                              2
                                    d       1 x - a 2     x         1
                                       a
                                      f ( ) =  ò   dx = -      = -      which is, of course, true for all values of a except
                                              2
                                                         2
                                   da       0 x + a  2  x +  a  2  1 + a  2
                                                              0
                                   a= 0.
                                          Example: The principle of  differentiating under  the integral sign may sometimes  be
                                   used to evaluate a definite integral.
                                                           p
                                                               -
                                                                 a
                                   Consider integrating  ( )f a =  0 ò  ln(1 2 cos(x) + a 2 )dx (for a >  1)
                                   Now,
                                                                              +
                                                             d       p - 2 cos(x) 2a
                                                               f ( ) =  ò          dx
                                                                 a
                                                             da      0 2 cos(x)- a  + a 2
                                                              1  pæ     (1 - a ) 2  ö
                                                            =   0 ç ò  1 -       2 ÷  dx
                                                              a  è  1 2 cos(x) + a ø
                                                                        a
                                                                      -
                                                                                      p
                                                             p  2 ì     æ 1 + a  æ  x ö ö  ü
                                                           =   -  í arctan ç  ×  tan ç ÷÷  ý
                                                             a  a  î    è 1 - a  è  2 øø  0 þ



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