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Measure Theory and Functional Analysis




                    Notes                                                p          p
                                   or                   |f + g| p   |f||f g|  q  |g||f g|  q
                                   Integrating, we get

                                                                          p           p
                                                           p
                                                      |f g| dx    |f||f g|  q  |g||f g| dx                 … (4)
                                                                                      q
                                   Using (3), relation (4) becomes

                                                                          1          1
                                                           p
                                                      |f g| dx      |f| dx  p  |g| dx  p  |f g| dx  1 q
                                                                      p
                                                                                             p
                                                                                 p
                                                                1
                                                            p
                                   Dividing each term by   |f g| dx  q  , we get
                                                            1  1  q      1  p       1  p
                                                        p
                                                                                p
                                                                     p
                                                   |f g| dx        |f| dx     |g| dx
                                                       1  1          1  1
                                   But                       1    1
                                                       p  q          q  p

                                                             1           1          1
                                                              p           p          p
                                                                     p
                                                         p
                                                                                p
                                   So               |f g| dx       |f| dx     |g| dx
                                   or                     f + g      f  +   g
                                                              p     p    p
                                   Hence the proof.



                                     Note  Equality hold in Minkowski’s inequality if and only if one of the functions f and g
                                     is a multiple of the other.
                                   Theorem 2: Minkowski’s inequality for 0 < p < 1. If 0 < p < 1 and f, g are non-negative functions
                                   in L , then
                                      p
                                                                  f + g       f  +   g  .
                                                                      p    p    p
                                   Proof: For this proceed  as in  theorem Minkowski’s  inequality and  applying the  Hölder’s
                                                                       p
                                                                                      q
                                   inequality for 0 < p < 1 for the functions f   L  and (f + g) p/q   L , we get
                                                                       1 p         1 q
                                                     |f||f g| p/q     |f| p  |f g| p/q q
                                                                                 )
                                                                       1 p       1 q
                                                     |f||f g| p/q     |f| p  |f g| p                       … (i)

                                   Also g   L , proceeding as above, we get
                                           p
                                                                         1 p       1 q
                                                                     p
                                                    |g||f g| p/q     |g| dx  |f g| p                      … (ii)







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