Let ( X , M , μ ) be a measure space s.t there exists an infinite sequence of disjoints measurable sets of strictly positive finite measure. Show that L 1 ( X , M , μ ) is not reflexive.My attempt: this question followed a question that asked to prove that a TVS V is reflexive ⟺ the closed unit ball in V is weakly compact. I tried showing that { f ∈ L 1 : ‖ f ‖ 1 ≤ 1 } is not weakly compact, but I wasn't able to find an open covering by weak open sets s.t it has no finite subcovering. Any help would be appreciated.

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Let   (  X  ,  M  ,  μ  ) be a measure space s.t there exists an infinite sequence of disjoints measurable sets of strictly positive finite measure. Show that       L    1    (  X  ,  M  ,  μ  ) is not reflexive.My attempt: this question followed a question that asked to prove that a TVS   V is reflexive     ⟺   the closed unit ball in   V is weakly compact. I tried showing that   {  f  ∈      L    1    :  ‖  f      ‖    1    ≤  1  } is not weakly compact, but I wasn&#039;t able to find an open covering by weak open sets s.t it has no finite subcovering. Any help would be appreciated.

Layla Love · Accepted Answer

I&#039;ll turn my comment into an answer. Suppose that       E    i   are pairwise disjoint sets which satisfy   μ  (      E    i    )  ∈  (  0  ,  ∞  ) for all   i and call       e    i    =  μ  (      E    i        )          −      1            χ                  E        i            , where   χ denotes the indicator function of a set.By the Eberlein-Smulian theorem, sequential weak compactness is the same as weak compactness for Banach spaces, so it suffices to argue that the sequence   (      e    i    :  i  =  1  ,  …  ), which lie in the unit ball of       L    1    (  X  ,  M  ,  μ  ), cannot have a weakly convergent subsequence. Suppose that a subsequence       e                  n        k             converges weakly to some             e      ~       and define  g  =      ∑          k      =      1        ∞    (  −  1      )    k        e                  n        k              .Notice that because the       E    i   are pairwise disjoint,   ‖  g      ‖                  L        ∞            (      X      ,      M      ,      μ      )        =  1. Define a bounded linear functional on       L    1    (  X  ,  M  ,  μ  ) by  Λ  (  f  )  =      ∫          X        g  f  d  μso that       |    Λ  (  f  )      |    ≤  ‖  f      ‖                  L        1            (      X      ,      M      ,      μ      )      . By construction, we have   Λ  (      e                  n        k              )  =  (  −  1      )    k  , which does not converge, a contradiction.

Let ( X , M , μ ) be a measure space s.t there exists an in

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