The original question:

Assume I have a real function $f(t)$ with Fourier transform of $\hat{f}(w)$.

Can one say anything about the inverse Fourier transform of $\frac{1}{1\pm\hat{f}(w)}$?

An answer (edit: the answer was deleted, and is now a comment on the original question):

In summary, if the function $\frac{1}{1\pm\hat{f}(w)}$ belongs to the Schwartz class, then its inverse Fourier transform exists and vise versa. See here.

To quote Ron Gordon's question, under what conditions is $\frac{1}{1\pm\hat{f}(w)}$ Schwartz?

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    $\begingroup$ The big question to answer here is what is meant by the Fourier transform because the comment is a bit misleading in that it gives the impression that the function you mentioned must be Schwartz but that is not necessarily the case. The Fourier transform can be defined in different function spaces and so you need to be more specific. It is defined on $L^1$ as an integral transform but it is defined on $L^2$ by limiting arguments of functions in $L^1 \cap L^2$. $\endgroup$ – Cameron Williams Jul 31 '13 at 17:09

I'm shamefully rusty with this sort of honest analysis, but here's my stab at it.

Suppose that $\tfrac{1}{1\pm \hat{f}(w)}$ is Schwartz. Then, in particular, for every multi-index $\alpha$, $$ C_\alpha := \sup_w \left| \frac{w^\alpha}{1 \pm \hat{f}(w)} \right| < +\infty, $$ so that for all $w$, $$ \left| \frac{w^\alpha}{1 \pm \hat{f}(w)} \right| \leq C_\alpha, $$ and hence $$ \left|\hat{f}(w)\right| \geq \left|1\pm \hat{f}(w)\right|-1 \geq \frac{1}{1+C_\alpha}|w^\alpha| - 1. $$ Thus, $\left|\hat{f}(w)\right|$ grows super-polynomially as $\|w\| \to +\infty$, and hence $f$ cannot even be a tempered distribution, let alone an element of some $L^p$ space for $p \geq 1$.

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  • $\begingroup$ Should the last fraction be $\frac{1}{C_\alpha}$? Everything else you wrote makes sense, thanks! $\endgroup$ – angryavian Jul 31 '13 at 20:45
  • $\begingroup$ $C_\alpha < 1+C_\alpha$, so I can replace $C_\alpha$ with $1+C_\alpha$, just to make sure that I'm not dividing by $0$. On the other hand, if $C_\alpha = 0$, then I guess you're immediately in business for super-polynomial growth? $\endgroup$ – Branimir Ćaćić Jul 31 '13 at 20:50

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