Suppose that ( A , Σ , m ) is a measure space and H is a linear functional on L ∞ ( A , Σ , m ). If U := { u : A → R | u is measurable, bounded and ∫ A u d m = 1 } and there are functions u 1 , u 2 ∈ U such that H ( u 1 ) ≤ 0 and H ( u 2 ) ≥ 0 ,then my question is:Is there a function u 3 ∈ U such that H ( u 3 ) = 0?

Question

Suppose that   (  A  ,  Σ  ,  m  ) is a measure space and   H is a linear functional on             L              ∞              (  A  ,  Σ  ,  m  ). If      U    :=      {    u    :    A    →          R                           |                                 u             is measurable, bounded and                         ∫                      A                          u                 d                  m                =        1              }  and there are functions       u          1        ,      u          2        ∈      U   such that  H  (      u          1        )  ≤  0    and    H  (      u          2        )  ≥  0  ,then my question is:Is there a function       u          3        ∈      U   such that   H  (      u          3        )  =  0?

Jake Landry · Accepted Answer

For each   t  ∈  [  0  ,  1  ], define       v          t        ∈      U   as follows:      v          t                          =                    def              t  ⋅      u          1        +  (  1  −  t  )  ⋅      u          2        .Clearly, each       v          t       is measurable and bounded, being a linear combination of measurable and bounded functions. Also,                    ∀        t        ∈        [        0        ,        1        ]        :                          ∫                      A                                    v                      t                                  d                  m                                    =                  ∫                      A                          [        t        ⋅                  u                      1                          +        (        1        −        t        )        ⋅                  u                      2                          ]                d                  m                                                  =                  ∫                      A                          t        ⋅                  u                      1                                  d                  m                +                  ∫                      A                          (        1        −        t        )        ⋅                  u                      2                                  d                  m                                                  =        t                  ∫                      A                                    u                      1                                  d                  m                +        (        1        −        t        )                  ∫                      A                                    u                      2                                  d                  m                                                  =        t        ⋅        1        +        (        1        −        t        )        ⋅        1                                          =        1.            Hence, indeed,       v          t        ∈      U   for all   t  ∈  [  0  ,  1  ]. By the linearity of   H, we get  ∀  t  ∈  [  0  ,  1  ]  :    H  (      v          t        )  =  H  (  t  ⋅      u          1        +  (  1  −  t  )  ⋅      u          2        )  =  t  ⋅  H  (      u          1        )  +  (  1  −  t  )  ⋅  H  (      u          2        )  .Notice that   H  (      v          0        )  =  H  (      u          2        )  ≥  0 and   H  (      v          1        )  =  H  (      u          1        )  ≤  0. By applying the Intermediate Value Theorem to the continuous function  t  ⟼  t  ⋅  H  (      u          1        )  +  (  1  −  t  )  ⋅  H  (      u          2        )defined on the closed interval   [  0  ,  1  ], we see that there exists a       t          ∗        ∈  [  0  ,  1  ] for which   H  (      v                  t                  ∗                      )  =  0. We can therefore set       u          3        :=      v                  t                  ∗                    .

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