TITLE: When is the Gaussian function $e^{-|\cdot|}$ in $W^{k,p}$?
QUESTION [1 upvotes]: I want to determine for which $p,k$ do we get that $e^{-|x|}$ belong to $W^{k,p}(\mathbb{R}^{n})$.
I know that 
$$e^{-|x|}\in W^{k,p}(\mathbb{R}^{n})\iff \int_{\mathbb{R}^{n}}|\frac{\partial^{\alpha}}{\partial x^{\alpha}}e^{-|x|}|^{p}\,dx\qquad\text{for all }|\alpha|\le k.$$
However, I am having trouble determining the integral due to the absolute value signs. The integrand appears to be symmetric so I presume that means we can write something like
$$\begin{aligned}\int_{\mathbb{R}^{n}}|\frac{\partial^{\alpha}}{\partial x^{\alpha}}e^{-|x|}|^{p}\,dx&=2\int\left(\frac{\partial^{\alpha}}{\partial x^{\alpha}} e^{-x}\right)^{p}\,dx
\\
&=2\int((-1)^{\alpha\cdot p}e^{-px})\,dx\end{aligned}$$
but I'm not really sure? Do I have to convert to spherical coordinates, perhaps?

REPLY [1 votes]: The function is infinitely smooth outside of one point and is very nice  at infinity, so its membership in $W^{k,p}$ is determined only by the $L^p$-integrability of $k$th derivatives near $0$. Your mistake is in focusing on the radial derivative; the tangential ones matter more for such functions.
The derivatives of order $1$ are bounded. 
The problematic derivative of order $2$ is taken twice in the tangential direction: try computing  $\frac{\partial^2}{\partial y^2}\exp(-\sqrt{x^2+y^2})$ at $(x,y)=(r,0)$. Near the origin this derivative is of size $1/r$ where $r$ is the distance to the origin. This pattern continues, with some of $k$th order derivatives being $\sim 1/r^{k-1}$ near $0$. Hence, 
$$e^{-|\cdot|}\in W^{k,p}(\mathbb{R}^n) \iff (k-1)p<n-1 \quad \text{ or } \ k=1$$ 
(the case  $k=n=1$ does not fit the first inequality, so it's stated separately.)
I don't think I would want to write out such computations in detail, but if it was necessary, I'd compute derivatives in Euclidean coordinates at a point $(r,0,\dots,0)$, proving by induction that they are $P(r)e^{-r}/r^{k-1}$ where $P$ is a polynomial.
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