Is the Fourier transform of the function $f$ which agrees with $1_{[-1.1]}|x|^\alpha$ on $[-1,1]$ and then decays very fast to zero to become a compactly supported continuous function, is in $L^1(\mathbb R)$, where $\alpha\in(1,2)$? My guess is that the answer is true. If I can show that the identity $\widehat{D^\alpha f}(\zeta)=(2\pi i\zeta)^\alpha\hat{f}(\zeta)$ holds in some sense and we have $D^\alpha(1_{[-1,1]}|x|^\alpha)=\alpha!|x|1_{[-1,1]}$ in some sense. Then $\hat{f}$ would have enough decay at infinity to make it in $L^1(\mathbb R)?$ May be fractional derivatives and their interaction with the Fourier transform might be helpful. But I could not find any good reference (easily readable) which can make the above argument rigorous? For my purpose if one such $f$ exists for which the Fourier transform is integrable is enough?
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asked Jun 27, 2020 at 18:48
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1$\begingroup$ If $\hat{f}\in L^1(\mathbb{R})$ then $f\in C(\mathbb{R})$ so the answer is no. But this question is more appropriate for math.stackexchange. $\endgroup$– Aleksei KulikovCommented Jun 27, 2020 at 18:53
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$\begingroup$ @Aleksei. I have edited the question. The edited question is what I meant originally. $\endgroup$– A beginner mathmaticianCommented Jun 27, 2020 at 19:00
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$\begingroup$ The smoothness of this is better than that of $|x|$, which already has Fourier coefficients $\simeq 1/n^2$, so the answer is yes. $\endgroup$– Christian RemlingCommented Jun 27, 2020 at 20:25
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$\begingroup$ @Christian. Ok. I missed that observation. Clearly, for $|x|$ we have decay $|\zeta|^{-2}$ at $\infty.$ But how to prove your statement rigorously? $\endgroup$– A beginner mathmaticianCommented Jun 28, 2020 at 3:42
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