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I have a function $C_T(k)$ that is not $L_1$, because its limit in negative infinity is a constant. So I dampened it by $ e^{\alpha k} $. Let's call the transformed function (of the dampened function) $\psi$.

My aim was to compute a few things on $\psi$, and then take back the function to the time domain, and multiply by $ e^{-\alpha k} $ to get back my original function. However, I am not sure why $\psi$ would be integrable thus Fourier transformable.

My question would be, is the function $\psi$ invertible ?

If the answer to the question is "it depends", does it help to know that $\psi$ can be writen as a bell behaved function of a characteristic function ? Like a constant times a characteristic function ?

I know that if the PDF related to that characteristic function had jump, thus it would mean that $\psi$ is not integrable since a Fourier transform is always continuous. But here, the PDF has no jump. Thus I don't know if this much information would allow me to say anything about the integrability of $\psi$.

In the worst case scenario that the information I am giving is not enough, what should I look at to know if $\psi$ is integrable without having an analytical expression for it ?

Thank you.

Post Scriptum : I am trying to be as precise as I can, without giving you the full details of the domain and the research paper I am dealing with. However, by experience, I know people will complain about my non capacity of being precise enough, so here is, in advance, the precise paper and the precise moment that confuses me. It is here in Carr Madan's paper, page 3 : http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.348.4044&rep=rep1&type=pdf

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  • $\begingroup$ 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). $\endgroup$
    – Andy
    Commented Mar 22, 2020 at 20:32
  • $\begingroup$ @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. $\endgroup$
    – Marine Galantin
    Commented Mar 22, 2020 at 21:39
  • $\begingroup$ 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 $1/t^k$. $\endgroup$
    – Andy
    Commented Mar 23, 2020 at 1:44
  • $\begingroup$ Your function $\psi$ is certainly a tempered distribution, then you have a valid inversion formula. $\endgroup$
    – Bazin
    Commented Mar 23, 2020 at 11:22
  • $\begingroup$ @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? $\endgroup$
    – Marine Galantin
    Commented Mar 23, 2020 at 15:55

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