Does anyone could give me a sufficient condition on $f$ so that the Fourier transform of $f$ (denoted as $\hat{f}$) is in $L^{1}(\mathbb R)$. The Fourier transform here is the linear operator $\mathcal F$ define for an $L^{1}$ function as $$ \hat{f}(\xi)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb R} f(x) e^{—i\xi x} \ dx $$ Thanks !



If $f\in L^{1}$ then $\hat{f}\in C^{1}_{0}$, though not necessarily $L^{1}$. The continuity, however, precludes exclusion in $L^{1}$ due to singularities. Hence, the only obstruction is decay of $\hat{f}$ at infinity.

Now since you are just after a sufficient condition that is fairly general, what can one say about the decay of $|\hat{f}|$ when $f$ has $k$ continuous derivatives? If this sounds completely alien to you, then integrate by parts and see what you get. The answer is $\hat{f}=O(|x|^{-k})$ for large $x$.

Now what can you say about the integrability of $C|x|^{-k}$ away from the origin? Or consider $C(1+|x|^{k})^{-1}$ to consider all of $\mathbb{R}$ instead, if that's easier.

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  • $\begingroup$ thanks ! that is exactly what I was looking for. $\endgroup$ – Eddy May 6 '14 at 13:56

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