Let ( X , A ) = ( [ − 1 , 1 ] , B [ − 1 , 1 ] ) be a measurable space equipped with the Lebesgue measure, m.Define ϕ : L 1 ( m ) → R by ϕ ( f ) = ∫ ( x − 1 2 ) f ( x ) d m ( x )To find | | ϕ | | , I think the general strategy is to find an upper bound on ϕ, and then construct a function f such that it is attained.In our case, we have | | ϕ | | = i n f { c ≥ 0 : | ϕ ( f ) | ≤ c | | f | | }. We have that | ϕ ( f ) | ≤ ∫ | ( x − 1 2 ) | | f ( x ) | d m ( x )I can split this up into two intervals: [−1,0] and [0,1]. On [0,1], we have that | x − 1 2 | ≤ 1 2 . On [−1,0], we have that | x − 1 2 | ≤ 3 2 . Hence | ϕ ( f ) | ≤ 2 ∫ | f | . This is my bound. However, I am unable to finish the proof because I am not sure how to construct a function f which attains this bound.

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Let   (  X  ,      A    )  =  (  [  −  1  ,  1  ]  ,            B              [        −        1        ,        1        ]              ) be a measurable space equipped with the Lebesgue measure, m.Define   ϕ  :            L      1        (    m    )    →      R   by  ϕ  (  f  )  =  ∫  (  x  −      1    2    )  f  (  x  )  d  m  (  x  )To find       |        |    ϕ      |        |  , I think the general strategy is to find an upper bound on   ϕ, and then construct a function f such that it is attained.In our case, we have      |        |    ϕ      |        |    =  i  n  f  {  c  ≥  0  :      |    ϕ  (  f  )      |    ≤  c      |        |    f      |        |    }. We have that       |    ϕ  (  f  )      |    ≤  ∫      |    (  x  −      1    2    )      |        |    f  (  x  )      |    d  m  (  x  )I can split this up into two intervals: [−1,0] and [0,1]. On [0,1], we have that       |    x  −      1    2        |    ≤      1    2  . On [−1,0], we have that       |    x  −      1    2        |    ≤      3    2  . Hence       |    ϕ  (  f  )      |    ≤  2  ∫      |    f      |  . This is my bound. However, I am unable to finish the proof because I am not sure how to construct a function f which attains this bound.

Alexia Hart · Accepted Answer

We have that      |    ϕ  (  f  )      |    =      ∫          [      −      1      ,      1      ]            |    (    x    −          1      2        )    f    (    x    )    |    ≤      |        |    x  −      1    2        |              |                      L        ∞            [      −      1      ,      1      ]            |        |    f      |              |                      L        1            [      −      1      ,      1      ]        ≤      3    2        |        |    f      |              |                      L        1            [      −      1      ,      1      ]      and we deduce that      |        |    ϕ      |        |    ≤      3    2        (  1  )Fixing   n  ∈            N              &amp;gt;      0       we consider the functions define as      f    n    (  x  )  :=  n      χ          [      −      1      ,      −      1      +      1              /            n      ]      where   χ is the indicator function. Now we have that      f    n    ∈      L    1    (  [  −  1  ,  1  ]  )    and        |        |    f      |              |                      L        1            (      [      −      1      ,      1      ]      )        =  1Therefore, Hence, developing the calculations easily obtains that  ϕ  (      f    n    )  =  −      3    2    +      1          2      n      Then we have that      |    ϕ  (      f    n    )      |    ≤      |    −          3      2        +          1              2        n              |        |        |        f    n        |              |                      L        1            (      [      −      1      ,      1      ]      )      So,      |        |    ϕ      |        |    ≥      |    −          3      2        +          1              2        n              |      ∀  n  ∈            N              &amp;gt;      0            (  2  )We can now conclude from the the previous equation (number (1) and (2) ) that      |        |    ϕ      |        |    =      3    2

Lucian Maddox · Accepted Answer

The norm of   φ is defined as  ‖  φ  ‖  =      sup          ‖      f              ‖                              L            1                    [          −          1          ,          1          ]                          |          ∫              [        −        1        ,        1                            (              x    −          1      2                      )                  f    (    x    )        d    x        |  Clearly,      |          ∫              [        −        1        ,        1        ]                            (              x    −          1      2                      )                  f    (    x    )        d    x        |    ≤      max          x      ∈      [      −      1      ,      1      ]        ]            |        x  −      1    2              |              |          ∫              [        −        1        ,        1        ]                    |        f    (    x    )          |            d    x        |    =      3    2    ‖  f      ‖                  L        1            [      −      1      ,      1      ]      For   f  (  x  )  =      2    a        χ          [      −      1      ,      −      1      +      a      ]      , where   a  ∈  (  0  ,  1  ), clearly,   ‖  f      ‖                  L        1            [      −      1      ,      1      ]        =  1, while  φ  (  f  )  =      ∫          [      −      1      ,      1      ]                  (        x  −      1    2              )          f  (  x  )    d  x  =      2    a        ∫          −      1              −      1      +      a                  (        x  −      1    2              )          d  x  =  −      3    2    +      a    2    ,and hence      |    φ  (  f  )      |    =      3    2    −      a    2  and hence  ‖  φ  ‖  ≥      3    2    −      a    2    ,for all   a  ∈  (  0  ,  1  ). Consequently   ‖  φ  ‖  =  3      /    2.

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