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.



Decay of the Fourier transform of a non-differentiable function
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
$$\hat{\varphi}(\lambda) = \frac{-1}{2\pi^2\lambda^2} + o(\lambda^{-3})~\mathrm{as}~\lambda\to\pm\infty.$$


