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Jan 9, 2023 at 0:26 comment added Yemon Choi As a variant on Vincius Novelli's answer: instead of using the Schwarz class, note that VN's argument actually proves that if $\mu$ is a finite measure on ${\mathbb R}$ then it annihilates every $f\in C_0({\mathbb R})$ which is the (inverse) Fourier transform of an $L^1$-function on $\widehat{\mathbb R}$. The class of all such $f$ is denoted by $A({\mathbb R})$, this is the Fourier algebra of ${\mathbb R}$, which is known to be dense in $C_0({\mathbb R})$ using e.g. Stone-Weierstrass. One advantage of this approach is that it works on any locally compact abelian group.
Jan 8, 2023 at 14:32 vote accept Boby
Jan 7, 2023 at 18:31 review Close votes
Jan 23, 2023 at 3:05
Jan 7, 2023 at 16:09 answer added Vinícius Novelli timeline score: 4
Jan 7, 2023 at 16:05 history edited Daniele Tampieri CC BY-SA 4.0
Minor Math Jaxing + fixed typos
Jan 7, 2023 at 15:19 history asked Boby CC BY-SA 4.0