a question about generalized derivatives in Sobolev space Let $\delta>-\frac{1}{2}$, $s\in (0,\delta+\frac{1}{2})$, $X_+(x)=x , x>0$  or $X_+(x)=0, x\le 0$.
How to prove that$(1-\lambda^2)_+^{\delta}\in W^{s,2}(R)$?
Where $W^{s,2}(R)$ is a Sobolev space for every Generalized Derivatives not higher than $s$ orders of the element in $W^{s,2}(R)$ belong to $L^2$.

Let $f(t) = (1-t^2)^\delta_+$.  In order to show that $f \in W^{s,2}(\mathbb R)$, I would try to argue that $\hat f(w) =O( |w|^{-\delta-1})$ as $|w|\to\infty$.  So w.l.o.g. $w>0$ and $w$ is large.
$$ \hat f(w) = \int_{-1}^1 (1-t^2)^\delta \cos(2\pi wt) \, dt = \int_{-1}^{-\eta} (1-t^2)^\delta \cos(2\pi wt) \, dt + \int_{-\eta}^{\eta} (1-t^2)^\delta \cos(2\pi wt) \, dt + \int_{\eta}^{1} (1-t^2)^\delta \cos(2\pi wt) \, dt ,$$
where $\eta = \frac1w \lfloor wt \rfloor- \frac1{2w}$.  The first and last terms are dominated by $O(w^{-1-\delta})$ since $t^2 - \eta^2 = O(w^{-1})$, and $\delta > -1$.  For middle term, integrate by parts and get
$$ - \frac1{2\pi w}\int_{-\eta}^{\eta} 2t (1-t^2)^{\delta-1} \sin(2\pi wt) \, dt .$$
If $\delta \le 0$, by analyzing the size of the integrand at $t = \pm1$, we see this is dominated by 
$$ \frac1{2\pi w} \int_{w^{-1}}^1 s^{\delta-1} \, ds \le C w^{-1-\delta} ,$$
and we are done.
If $\delta > 0$, repeat the process: let $\eta_2 = \eta - \frac1{2\omega}$, split the integral as
$$- \frac1{2\pi w}\int_{-\eta}^{\eta_2} 2t (1-t^2)^{\delta-1} \sin(2\pi wt) \, dt
- \frac1{2\pi w}\int_{-\eta_2}^{\eta_2} 2t (1-t^2)^{\delta-1} \sin(2\pi wt) \, dt
- \frac1{2\pi w}\int_{\eta_2}^{\eta} 2t (1-t^2)^{\delta-1} \sin(2\pi wt) \, dt
.$$
Since $\delta > 0$, again the first and last terms are bounded by $O(w^{-1-\delta})$.  Integrate by parts for the middle term and get
$$ - \frac1{2\pi w^2}\int_{-\eta_2}^{\eta_2} [2 (1-t^2)^{\delta-1} + 4t^2 (1-t^2)^{\delta-2}\cos(2\pi wt) \, dt .$$
If $\delta \le 1$, we can again analyze the size of the integrand at $t = \pm1$, and see this is again dominated by $w^{-1-\delta}$.
If $\delta > 1$, repeat the process again and again as necessary.
A final word.  I am self-taught in the subject of Sobolov spaces.  So I don't know if there is a standard argument.  I also waited 12 hours hoping someone else would answer, but that didn't happen.
