Decay of the Fourier transform of a non-differentiable function 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 $\varphi$:

*

*Is there a function which is non-differentiable at some point, but for which \eqref{1} still holds?

*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.
 A: 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
$$\hat{\varphi}(\lambda) = \frac{-1}{2\pi^2\lambda^2} + o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$
A: 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).
