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Given that, for some $f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$, we have the Fourier transform of $f$ given as \begin{equation} F(k):=\int_{-\infty}^\infty e^{-ikx}f(x)\text{d}x~~~(k\in\mathbb{R}),\end{equation} is there any known concise expression for $g$ in terms of $f$, or at least an estimate for $|g|$ in terms of $|f|$, where \begin{equation}g(x):=\frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}|F(k)|\text{d}k~~~(x\in\mathbb{R})?\end{equation} I can't find anything on this problem in the literature I've looked through. Any information would be greatly appreciated!

[Note: I am assuming, indeed, that $g$ is even well-defined; if it is not, can anybody give me a hint/idea as to why?]

Stromael
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  • Even if $f\in L^1\cap L^2$, there is no guarantee that $F\in L^1$. Example: $$f(x)=e^{-x} \begin{cases} 1 & x \ge 0\ 0 & x <0 \end{cases}$$ whose Fourier transform is $$F(k)=\frac{C}{1+ik}$$(the constant $C$ depends on the definition of Fourier transform, with your notation should be $1$ but I am too lazy to check now). – Giuseppe Negro Mar 31 '14 at 19:14
  • Thanks, you are indeed correct. However I don't mind if $F\in L^2\backslash L^1$. I was simply restricting my Fourier transform operator to $L^1\cap L^2$ to ensure the hypotheses of the Plancherel theorem are satisfied so that the Fourier transform is isometric.

    If $F\in L^1$ can we conclude an estimate for $|g|$ in terms of $|f|$?

     – Stromael Mar 31 '14 at 19:43
  • The only thing that I can think of is that, if $f\in H^1(\mathbb{R}^n)$, then $\lvert f \rvert \in H^1(\mathbb{R}^n)$ (this is a theorem of Stampacchia). In Fourier terms this means that the Fourier transform of $\lvert f \rvert$ decays so fast that it stays in $L^2$ even after multiplication by a polynomial term of degree one. – Giuseppe Negro Apr 03 '14 at 20:50

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