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Mar 23, 2020 at 16:09 comment added Andy @MarineGalantin No, my indicator example is in $L1 \cap L2$. You are correct that integrable functions go to continous ones
Mar 23, 2020 at 15:55 comment added Marine Galantin @Andy right sorry. Imeant it goes from integable functions to continuous functions, but we don't know anything about the final space. However, what you're saying, is that the Fourier transform of a L1 inter L2 function is going to be in L1 inter L2?
Mar 23, 2020 at 11:22 comment added Bazin Your function $\psi$ is certainly a tempered distribution, then you have a valid inversion formula.
Mar 23, 2020 at 1:44 comment added Andy No, consider the indicator of [0,1], for many $t$ the fourier is $1/t$. However if your function is smoother, it will have rapidly decreasing fourier transform, which will put it back in $L^1$. For instance if you have a function in $C^k$ with the derivative in $L^1$, then integration by parts shows it goes down at least like $1/t^k$.
Mar 22, 2020 at 21:39 comment added Marine Galantin @Andy thanks for the comment. Right I was thinking about Parseval's formula, but Fourier is taking an $L^1$ function to an $L^1$ function right ? Or have I missed something ? In the paper, they also talk about square integrability, and I don't get what s the link between that and having a Fourier transform.
Mar 22, 2020 at 20:32 comment added Andy Can you talk slightly about the domains of your function? Multiplying by $e^{ak}$ will make it explode in one of the directions for instance. I think what you're after anyway is called the laplace transform, and terry tao has good notes (and the inversion) of it. Anyway if you're talking about fourier as $L^2(R) \to L^2(R)$ it is invertible (Parsevals formula shows the $L^2$ norm of the fourier is like that of the original function you started with).
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Mar 22, 2020 at 19:14 history asked Marine Galantin CC BY-SA 4.0