Assume that f ∈ L loc 1 ( R n , R ) is such that ∫ R n f ( x ) g ( x ) d x ≥ 0for all g ∈ C c ( R n , R + ) . Here C c ( R n , R + ) denotes the set of all continuous functions on R n taking non-negative real values and which have compact support. I think it follows that f ( x ) ≥ 0 for a.e. x ∈ R n . However I only managed to prove this implication in the case where f is (essentially) bounded. How to go from this simpler case to the general case ?

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Assume that   f  ∈      L          loc              1        (            R              n        ,      R    ) is such that      ∫                            R                          n                      f  (  x  )  g  (  x  )  d  x  ≥  0for all   g  ∈      C          c        (            R              n        ,            R              +        )  . Here       C          c        (            R              n        ,            R              +        ) denotes the set of all continuous functions on             R              n       taking non-negative real values and which have compact support. I think it follows that   f  (  x  )  ≥  0 for a.e.   x  ∈            R              n      . However I only managed to prove this implication in the case where   f is (essentially) bounded. How to go from this simpler case to the general case ?

Kiana Cantu · Accepted Answer

Fix a family of mollifier   (      g    ϵ        )    ϵ    &amp;gt;  0. For any   R  &amp;gt;  0 let      f    R    =  f      χ                  B                  R          +          1                            ,where       B          R      +      1       is the ball of radius   R  +  1 in             R        n  . Then       f    R    ∗      g    ϵ   converges in       L    1    (            R        n    ) to       f    R  . Restricting to       B    R  ,       f    R    ∗      g    ϵ   converges in       L    1    (      B    R    ) to       f    R    =  f.Also, if   ϵ  &amp;lt;  1, then the condition implies that      f    R    ∗      g    ϵ    (  x  )  ≥  0  ,        ∀  x  ∈      B    R    .Together with the convergence of       f    R    ∗      g    ϵ    →  f in       L    1    (      B    R    ), there is a sequence       ϵ    n    →  0 so that      f    R    ∗      g                  ϵ        n              →  fa.e. in       B    R  . Thus   f  ≥  0 in       B    R  . Since   R  &amp;gt;  0 is arbitrary, choosing   R  =  K. Using  {  x  ∈            R        n    :  f  (  x  )  &amp;lt;  0  }  =      ⋃          K      ∈              N              {  x  ∈      B    K    :  f  (  x  )  &amp;lt;  0  }  ,one also has   f  ≥  0 a.e. in             R        n  .

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