I'm wondering whether the following statement holds:Let f n , f : R → R...

Riya Hansen

Riya Hansen

Answered

2022-07-03

I'm wondering whether the following statement holds:
Let f n , f : R R 0 + be functions with f n ( x ) d x = f ( x ) d x = 1 and for every bounded and countinuous function g : R R the following integral-convergence
g ( x ) f n ( x ) d x n g ( x ) f ( x ) d x
holds. Then it follows that f n f almost everywhere.
Intuitively the statement looks false, but I can't find a counterexample. If it doesn't hold: changes the further assumption that the f n , f have to be continuous anything?
Kind regards

Answer & Explanation

Franco Cohen

Franco Cohen

Expert

2022-07-04Added 8 answers

A classical counterexample is
f n ( x ) := ( 1 + sin ( 2 π n x ) ) 1 { x ( 0 , 1 ) } ,
and
f ( x ) := 1 { x ( 0 , 1 ) } .
We have f n = f = 1 and, for every bounded continuous g : R R , f n g f g as n (e.g., by Riemann–Lebesgue lemma). However, ( f n ) n 1 does not converge at all.
Bruno Pittman

Bruno Pittman

Expert

2022-07-05Added 4 answers

Riemann-Lebesgue lemma gives you directly that 0 1 g ( x ) sin ( 2 π n x ) d x 0 as n

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