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Sep 20, 2018 at 10:37 vote accept Giulia S-A.
Sep 19, 2018 at 15:45 answer added Mizar timeline score: 2
Sep 19, 2018 at 13:16 comment added Nate Eldredge @TaQ: All I meant by "pointwise convergence is for free" was the trivial observation that we have $\widehat{f_n}(x) \to \widehat{f}(x)$ for each $x$, simply because $\widehat{f_n}(x) = \langle f_n, e^{i x \cdot}\rangle$. I did not think as far ahead as you got. It'd be great if you would post it as an answer.
Sep 19, 2018 at 8:11 comment added TaQ @NateEldredge How is "X-wise convergence is for free" defined? If one knows (x) that elements in a bounded set in $\mathscr E'(\mathbb R)$ have "uniformly bounded" supports and orders, then the assertion follows trivially from formula (2) $\ |f(z)|\le\gamma\,(1+|z|)^N{\rm e}^{\,r|\rm{Im}\,z|} $ in the statement of the Paley−Wiener theorem in Rudin's Functional Analysis (Th. 7.23, p. 183 in my edition). Is (x) "for free"?
Sep 19, 2018 at 5:52 history edited Martin Sleziak CC BY-SA 4.0
typo in the title
Sep 18, 2018 at 9:01 vote accept Giulia S-A.
Sep 20, 2018 at 10:37
Sep 18, 2018 at 6:46 answer added Jochen Wengenroth timeline score: 5
Sep 18, 2018 at 0:15 comment added Nate Eldredge Initial thoughts: Pointwise convergence is for free. By the mean value property, it's sufficient to get $L^1$ convergence on compact subsets, so we might try to show there's a dominating function, or uniform integrability. And if I'm not mistaken, $\mathcal{E}(\mathbb{R})$ is a Frechet space, so we have the uniform boundedness principle available, which seems like it ought to help.
Sep 17, 2018 at 9:00 review First posts
Sep 17, 2018 at 9:15
Sep 17, 2018 at 8:56 history asked Giulia S-A. CC BY-SA 4.0