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Let $\hat{f}(\xi)$ be a smooth function on $\mathbb{R}^n$ that decays like $|D^\alpha_\xi \hat{f}(\xi)| \lesssim (1 + |\xi|^2)^{-\frac{1}{4}(1 + |\alpha|)}$, where $\alpha$ is a multi-index such that $D^\alpha_\xi = D^{\alpha_1}_{\xi_1}...D^{\alpha_n}_{\xi_n}$, and $|\alpha| = \alpha_1 + ... + \alpha_n$. The question is, since $\hat{f}$ has sufficient decay properties, would the inverse Fourier transform $f(x)$ of $\hat{f}(\xi)$ be smooth? Thanks in advance for any help!

Edit: In view of user225318's answer, now I am curious whether $f$ can be said to be smooth away from the origin.

Guest
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  • See this. – science Apr 12 '15 at 13:48
  • @science Thanks for your comment. Can you please explain a bit how this helps? I don't see it right away. The function $\hat{f}(\xi)$ here is not Schwartz class. Also, I would want to infer smoothness of $f$, not integrability properties. – Guest Apr 12 '15 at 14:27
  • Take $\alpha = 0$, you have that $\hat{f}$ decays like $(1 + |\xi|^2)^{-1/4}$. On $\mathbb{R}^n$ does guarantees neither that $\hat{f}$ is in $L^{1}$ or $L^2$. I don't think if you can even guarantee that you have a inverse Fourier transform that is a function. – user225318 Apr 12 '15 at 15:36
  • To be more precise, take $g(x) = 1/ \sqrt{1 + |x|^2}$. We have that $D^\alpha g \lesssim g^{|\alpha|}$. So $g$ satisfies all the assumptions (it decays much faster than you assumed). It is in $L^2(\mathbb{R}^n)$ only in $n = 1$... // Do you perhaps mean to ask a different question? – user225318 Apr 12 '15 at 15:42
  • @user225318 Here $\hat{f}(\xi)$ is different from any $g(\xi)$ in the sense that $\hat{f}(\xi)$ has been obtained already as the Fourier transform of an $L^2$ function. – Guest Apr 12 '15 at 16:08

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In view of this comment clarifying the question, we can give a counter example. Consider

$$ f = \exp( - |x|) $$

which is in any $L^p$. Its Fourier transform is well known to be (up to a constant depending on the dimension)

$$ (1 + |\xi|^2)^{-(n+1)/2} $$

This function is smooth and decays much faster (with all derivatives) than what you supposed. (In fact it is $L^1$ and so the inverse Fourier transform converges everywhere to $f$.)

Clearly $f$ is not differentiable at the origin.

Community
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user225318
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  • You are right. Thanks for the clarification. However, I was wondering: since decay conditions actually have to do with the behavior of a function outside a compact set (including the origin), perhaps we could say that $f$ is smooth away from the origin? – Guest Apr 12 '15 at 17:30
  • No. Spatial translations becomes phase shifts under Fourier transform. – user225318 Apr 12 '15 at 17:33
  • Furthermore, you know that if $f = L^1 \cap C^\infty$ you must have $\hat{f}$ decaying faster than any polynomial rate. And any such function $f$ is also $L^2$ from interpolation. So in fact the statement you asked about is not just false, but so false that there is no hope for any easy fix. – user225318 Apr 12 '15 at 17:35
  • But $f$ does not have to be $L^1$, right? How about the following calculation? Fix $|\eta| = [\frac{n}{2}] + 1 + k$, where $k \in \mathbb{N}$. $|D^\gamma_\xi\hat{f}(\xi)| \leq C(1 + |\xi|^2)^{-\frac{1}{4}(1 + |\gamma|)}$. When $\gamma$ is large, $\xi^\alpha D^\gamma_\xi \hat{f}(\xi) \in L^2 \Rightarrow D^\alpha_x(x^\gamma f(x)) \in L^2$. Since this happens for all $|\alpha| \leq |\eta|$, we can say, $x^\gamma f(x) \in H^{|\eta| + k}$, which is $C^k$ by Sobolev embedding. Now let $k$ vary. – Guest Apr 12 '15 at 19:37
  • Dividing a smooth function by $x^\gamma$ doesn't yield you a smooth function. – user225318 Apr 17 '15 at 20:06