If f = f ( x ) is absolutely continuous on [0,1] such that f ′ ∈ L 2 ( [ 0 , 1 ] ) and f ( 0 ) = 0. Prove then that lim x → 0 + f ( x ) x 1 2 = 0The hint says to use Fundamental theorem of calculus but I don't see where to use that.I know if f ′ ∈ L 2 ( [ 0 , 1 ] ) then ( ∫ [ 0 , 1 ] | f ′ ( x ) | 2 d x ) 1 2 &lt; + ∞ .I don't know what to do from here though. Any hints would be appreciated. Am I merely using L'Hôpital's rule, since I get 0/0?

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If   f  =  f  (  x  ) is absolutely continuous on [0,1] such that       f    ′    ∈      L    2    (  [  0  ,  1  ]  ) and   f  (  0  )  =  0. Prove then that      lim          x      →              0        +                        f      (      x      )              x                        1          2                      =  0The hint says to use Fundamental theorem of calculus but I don&#039;t see where to use that.I know if       f    ′    ∈      L    2    (  [  0  ,  1  ]  ) then            (            ∫          [      0      ,      1      ]        |      f    ′    (  x  )      |          2        d  x                    )                            1        2              &amp;lt;  +  ∞  .I don&#039;t know what to do from here though. Any hints would be appreciated. Am I merely using L&#039;Hôpital&#039;s rule, since I get 0/0?

Lisbonaid · Accepted Answer

For   x  &amp;gt;  0, using the fundamental theorem of calculus:                              |                                    f              (              x              )                                      x                                |                                    =                  |                                                    ∫                0                x                                            f                ′                            (              t              )              d              t                                      x                                |                                                  ≤                              ∫                          |                                                      1                                            [                0                ,                x                ]                                      (            t            )                          f              ′                        (            t            )                          |                        d            t                                x                                                            =                              ‖                          1                              [                0                ,                x                ]                                                    f              ′                                      ‖              1                                            x                                                            ≤                              x            ‖                          f              ′                                      ‖              2                                            x                                                            =                  x                ‖                  f          ′                          ‖          2                                                  →        0            as   x  →      0    +  , where the last inequality is Hölder&#039;s inequality.

If f = f ( x ) is absolutely continuous on [0,1] such that f &#x

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