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It is well known that if $\varphi$ is a Schwartz function on $\mathbb{R}$ (i.e. smooth and decaying at infinity faster than polynomials), then its Fourier transform decays faster than polynomials. More precisely, for any $M>0$ there exists a constant $C_M>0$ such that \begin{equation}\tag{1}\label{1} |\widehat{\varphi}(\lambda)|=\left|\int_{\mathbb{R}} \varphi(x)e^{-2\pi i x \lambda}\,dx\right|\le C_M \lambda^{-M}, \end{equation} for any $\lambda>0$. The proof of \eqref{1} uses integration by parts.

My questions regard potential relaxing of the smoothness assumption imposed on $\varphi$:

  1. Is there a function which is non-differentiable at some point, but for which \eqref{1} still holds?
  2. Take $\varphi(x)=|x|e^{-x^2}$. Does \eqref{1} hold for this function, and if not, what is the optimal rate of decay of $|\widehat{\varphi}(\lambda)|$?

I think the answer to the second question should be $\lambda^{-1}$ (same as in van der Corput estimate with a non-smooth cutoff function), but I am not able to work out the details. All hints will be appreciated.

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    $\begingroup$ A few hints far from a complete answer: (1) Let $\psi_r(\gamma)=\gamma^r\hat{\phi}(\gamma)$. $\psi_r$ also decays fast. $\psi_r=\hat{\phi^{(r)}}$ the Fourier transform of the distributional $r$th derivative of $\phi$. Since $\psi_r\in L^2$, so is $\phi^{(r)}$. It is left to show that the pointwise derivatives exist. (2) The Fourier transform of $f(x) = |x|e^{x^2} = sgn(x) xe^{-x^2} $ is the Hilbert transform of $\gamma e^{-\gamma^2}$ up to constants. $\endgroup$
    – Onur Oktay
    Jul 6 at 10:37
  • $\begingroup$ Thanks! About (2), I agree, but I can't see how to use the fact that f is a Hilbert transform of $\gamma e^{-\gamma^2}$ to answer the question? $\endgroup$
    – Tony419
    Jul 7 at 7:57

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I answer question $(1)$, assuming only that $\varphi$ is integrable.

If $(1)$ holds, then $\hat{\varphi}$ is also integrable, so Fourier inversion formula applies. For almost every $x \in \mathbb{R}$, $$\varphi(x) = \int_{\mathbb{R}}\hat{\varphi}(\lambda)e^{i2\pi x\lambda}\mathrm{d}\lambda.$$ Since for every $n \in \mathbb{N}$, the functions $\lambda\mapsto\lambda^n\hat{\varphi}(\lambda)$ are integrable, the right-hand side defines a $\mathcal{C}^\infty$ function. Hence $\varphi$ coincides almost everywere with a $\mathcal{C}^\infty$ function.

Now, I answer question $(2)$. I hope that my computations are right. Set $\varphi(x) = |x|e^{-x^2}$ and $\psi(x) = e^{-|x|}$. For every non null $x$, $$\varphi(x) + \psi(x) = |x|e^{-x^2}+e^{-|x|}.$$ $$\varphi'(x) + \psi'(x) = \mathrm{sign}(x)(e^{-x^2}-2x^2e^{-x^2}-e^{-|x|}).$$ $$\varphi''(x) + \psi''(x) = \mathrm{sign}(x)(-6xe^{-x^2}+4x^3e^{-x^2})+e^{-|x|}.$$
These quantities have the same limits at $0+$ as at $0-$. Therefore, the function $\varphi+\psi$ is $\mathcal{C}^2$ and $(\varphi+\psi)^{(k)}$ for $k=0,1,2$ are integrable. Moreover, $(\varphi+\psi)''$ has bounded variation. Hence $$(\hat{\varphi}+\hat{\psi})(\lambda) = o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$ But $$\hat{\psi}(\lambda) = \int_0^\infty e^{-x}(e^{-i2\pi x\lambda}+e^{i2\pi x\lambda})\mathrm{d}x = \frac{1}{1+i2\pi\lambda}+\frac{1}{1-i2\pi\lambda} = \frac{2}{1+4\pi^2\lambda^2}.$$ Hence $$\hat{\varphi}(\lambda) = \frac{-1}{2\pi^2\lambda^2} + o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$

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  • $\begingroup$ Thank you for a nice answer! I am curious about your solution to the second part. How did you realize that adding $\psi(x)=e^{-|x|}$ will cancel out the singularity of $\varphi$ at 0 (up to order 2)? Did you just guess that or is there some deeper philosophy behind it? $\endgroup$
    – Tony419
    Jul 7 at 7:49
  • $\begingroup$ @Tony419 It was clear for me that $\phi+\psi$ was $\mathcal{C}^1$ (by looking at left and right derivatives at $0$), and that its derivative has finite variation. When I derivated once more, I was surprised to see that $\phi+\psi$ is $\mathcal{C}^2$. There is no deeper philosophy. $\endgroup$ Jul 7 at 12:23
  • $\begingroup$ I see, thanks for clarifying. $\endgroup$
    – Tony419
    Jul 7 at 17:54
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This is a comment, not an answer (I am not entitled). It should be put on record that no computations are required to answer your questions in the negative. This follows from the basic facts that the Fourier transform is a bijection (even an lcs isomorphism) from the Schwartz space onto itself and is its own inverse (up to a factor and a change of sign in the exponent, which are of no relevance here).

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Amir Sagiv
    Jul 6 at 15:35
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    $\begingroup$ I would agree with this comment in the present situation if $\hat{\phi}$ was assumed to be in the Schwarz space. $\endgroup$ Jul 6 at 16:35

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