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I am struggle to prove the following theorem.

If $f$ $\in$ $C^1$ and $f$ , $f'$ $\in$ $L^1$ , then $\mathcal{F}$$f$ $\in$$L^1$.

($\mathcal{F}$$f$ is Fourier transform of $f$ )

I doubt that whether this theorem is right.

Please tell me proof or counterexample.

(following statement is supplement)

I already know that $f$ is bounded , $\mathcal{F}$$f$ $\in$ $L^2$ and $f$ $\in$ $L^2$ under the hypothesis.

And, since $f$, $f'$ $\in$ $L^1$ then $\mathcal{F}$$f$ , $\mathcal{F}$$f'$ is bounded.

Davide Giraudo
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Nikolai
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  • $f' \in L^2$ so $\mathcal{F} [f'] = i\omega \hat{f} \in L^2$ whence $\langle 1/(1+|\omega|),(1+|\omega|) |\hat{f}| \rangle \le ...$ – reuns Nov 29 '18 at 10:27
  • @reuns Why f' is belong to L2 ? – Nikolai Nov 29 '18 at 11:28
  • @reuns If $f'$ is bounded, your inequality is correct. But, only $f'$ in $L^1$, can't lead $f'$ is bounded. – Nikolai Nov 29 '18 at 12:02
  • Sorry I missed the ambiguity : you should read $C^1$ as the Banach space with sup norm of $f$ and $f'$. If you read it as just "continuously differentiable" it is very possible the claim isn't true – reuns Nov 29 '18 at 12:06
  • Otherwise we can look at something like $f' \ast |x|^{-1/2+\epsilon}$ to obtain that $(1+| \omega|^{1/2+\epsilon}) \hat{f} \in L^2$ – reuns Nov 29 '18 at 12:15

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