I am given this function: $$f(x) = e^{- \sqrt{|x|}}$$
and I want to find $k\in \mathbb{N}, \ p \ge 1$ such that $f \in W^{kp} (\mathbb{R})= \{ f \in L^p (\mathbb{R}) \ | \ \forall \alpha \le k: \ D^{\alpha} f_w \in L^p (\mathbb{R}) \}$
Here $D^{\alpha} f_w$ is the $\alpha$-th weak derivative of $f$.
$f'_w = h$ such that $\forall \varphi \in C^{\infty}_0 : \int f(x) \varphi'(x) dx = - \int h(x) \varphi(x) dx$
So what I do is I try to find the first weak derivative of $f$. Then I establish for what $p$ : $f'_w \in L^p$.
Then I would find $f''_w$ and find the proper $p$ by integrating $$\int_{\mathbb{R}} |f''_w(t)|^p dt$$
And I would continue this up until I get a function which isn't weakly differentiable.
And my problem is in integrating:
$$\int_{\mathbb{R}} f(x) \varphi'(x) dx = \int_{\mathbb{R}} e^{- \sqrt{|x|}} \varphi'(x) dx = \int_{-\infty}^0 e^{- \sqrt{-x}} \varphi'(x) dx + \int_0^{\infty} e^{- \sqrt{x}} \varphi'(x) dx$$
$$= -\int_{-\infty}^0 \frac{e^{- \sqrt{-x}}}{2 \sqrt{-x}} \varphi(x) dx - \int_0^{\infty} \frac{-e^{- \sqrt{x}}}{2 \sqrt{x}} \varphi(x) dx$$
So it appears that $$f'_w(x) = -sgn(x) \cdot \frac{e^{- \sqrt{|x|}}}{2 \sqrt{|x|}}$$
Is that correct so far?