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Mar 26, 2020 at 13:54 comment added Nik Weaver "Is there a way to see ...?" --- in principle, yes, but I think you'd have to build up a lot of machinery that's immediately available for multiplication operators. "How can one use this?" --- sorry, I don't understand, use it for what?
Mar 26, 2020 at 10:20 comment added Lenard Velasquez Thank you for the clarification. Is there a way to see that $\mathcal{A}'=\mathcal{A}$ without using the Fourier transform? In fact one has that if I apply the Fourier transform to $\mathcal{A}$ this space becomes $L^{\infty}(\mathbb{R})$. How can one use this?
Mar 19, 2020 at 13:48 comment added Nik Weaver (I.e., $\mathcal{A}' = \mathcal{A}$.)
Mar 19, 2020 at 13:05 history answered Nik Weaver CC BY-SA 4.0