I have a function $f\in L^1(\mathbb{R})$. Its Fourier tranfsorm is uniformly continuous and goes to zero at infinity. I know, in addition, that if I assume that $xf(x)$ is integrable, then the Fourier transform is continuously differentiable. My question is: there are weaker assumptions under which one can claim that the Fourier transform is just in $W^{1,1}$ (at least on bounded intervals)?
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1$\begingroup$ Can you make the question more precise? What kind of criteria are you interested in? $\endgroup$– user1688Commented Jan 11, 2017 at 11:29
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$\begingroup$ I would like to know if there exist (in the literature) sufficient conditions on functions $f\in L^1(\mathbb{R})$ that guarantee that the Fourier transform of $f$ is a function in $W_{loc}^{1,1}(\mathbb{R})$. $\endgroup$– user103449Commented Jan 12, 2017 at 11:10
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$\begingroup$ Yes, many such criteria exist, for instance $f$ being a Schwartz function will do. $\endgroup$– user1688Commented Jan 12, 2017 at 11:51
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$\begingroup$ I agree, but the assumption of being a Shwartz function is a very "strong" assumption (even stronger than "xf(x) integrable") and entails much more that $W^{1,1}$-regularity for the Fourier transform, since it is an isometry on the set of the Schwartz functions. I was rather interested in some kind of "minimal" criterium. Precisely, since I already know that assuming "$xf(x)$ integrable" entails that the Fourier transform is $C^1$, I wondered if there were weaker assumptions under which the Fourier transform would be just $W^{1,1}$. $\endgroup$– user103449Commented Jan 16, 2017 at 12:54
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