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S Jul 7 at 19:05 history bounty ended CommunityBot
S Jul 7 at 19:05 history notice removed CommunityBot
Jul 4 at 14:55 vote accept NancyBoy
Jul 4 at 14:07 history edited LSpice CC BY-SA 4.0
While this is on the front page, typo; removed "thank you"
Jul 4 at 9:46 answer added Bazin timeline score: 10
Jul 3 at 3:35 comment added RobPratt Cross-posted: math.stackexchange.com/questions/4940919/…
Jul 2 at 20:11 history edited NancyBoy CC BY-SA 4.0
Simplifying conditions
S Jun 29 at 17:46 history bounty started NancyBoy
S Jun 29 at 17:46 history notice added NancyBoy Draw attention
Jun 15 at 8:55 comment added Igor Khavkine @WillieWong Ah yes, a perfectly valid point!
Jun 15 at 0:02 comment added Willie Wong @IgorKhavkine: $\mathbb{R}$ has two "ends", I can easily imagine a situation where $f$ has exponential decay on one end and $g$ on the other.
Jun 14 at 17:37 history edited LSpice CC BY-SA 4.0
Title
Jun 14 at 10:20 comment added Igor Khavkine The precise condition quoted in the answer by Nate Eldredge uses the $L^2$-norm on the shifted real axis. Consult the cited reference for more details. More specialized literature might give analogous results with different norms.
Jun 14 at 10:07 comment added NancyBoy Thank you @IgorKhavkine for the post you mentinned. A quick question, when it is mentionned "the fourier transform should decay exponentially", we are talking about the module of the fourier tranform ?. It is to say that if $|\mathcal{F}[f*g](u)| < e^{-a|u|}$ for some $a>0$, $f*g$ is analytic ?
Jun 14 at 9:07 comment added Igor Khavkine Have a look at the properties of the Fourier transform of analytic functions at MO23679. If you pick your functions to be analytic at least in some strip around the real axis, I suspect that the answer will come out: iff one of $f$ or $g$ is analytic.
Jun 14 at 8:27 history edited Carlo Beenakker CC BY-SA 4.0
fixed latex in title
Jun 14 at 7:56 history asked NancyBoy CC BY-SA 4.0