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EnvivyEvoxys6 2022-07-10 Answered
Assume that f L loc 1 ( R n , R ) is such that
R n f ( x ) g ( x ) d x 0
for 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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Answers (1)

Kiana Cantu
Answered 2022-07-11 Author has 22 answers
Fix a family of mollifier ( g ϵ ) ϵ > 0. For any R > 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 < 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 f
a.e. in B R . Thus f 0 in B R . Since R > 0 is arbitrary, choosing R = K. Using
{ x R n : f ( x ) < 0 } = K N { x B K : f ( x ) < 0 } ,
one also has f 0 a.e. in R n .
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