Fractional powers of the operator $B: L_2(\mathbb{R}) \mapsto L_2(\mathbb{R})$, $Bf = f-f^{''}$. Consider the linear operator $B: L_2(\mathbb{R}) \mapsto L_2(\mathbb{R})$ defined by the following mapping: $Bf = f-f^{\prime\prime} \equiv (I-\Delta)f$, where $\Delta$ is the Laplace operator that takes two derivatives. 
One way of alternatively defining $B$ is to consider the relationship one has in the Fourier domain: that is to say that if $\mathcal{F}$ is the Fourier transform, then $B$ can be defined as a linear operator such that the following relation holds:
$$
\begin{align*}
\mathcal{F}(Bf)(t) = (1+t^2)(\mathcal{F}f)(t).
\end{align*}
$$
This allows us to naturally define arbitrary powers of $B$, and namely for any $s\in \mathbb{R}$, we can define $B^s$ such that it satisfies the following relation:
$$
\begin{align*}
\mathcal{F}(B^sf)(t) = (1+t^2)^s (\mathcal{F}f)(t).
\end{align*}
$$
Note that when $s$ is an integer, then $B^s f = (I-\Delta)^s f$. 
$\textbf{Question:}$ We know that when $s$ is an integer, then by the above relation, that $B^s f1_{A} = 1_{A}B^s f$ almost everywhere, where $1_{A}$ is the indicator function on some subset $A\subset \mathbb{R}$. I am curious as to whether this same relation holds when $s$ is no longer an integer, as I know that fractional derivatives just in general confuse me.
 A: One way your identity $1_{A}B^{s}f = B^{s}(1_{A}f)$ can fail: $1_{A}f$ may not be in the domain of $B^{s}$ even if $f$ is differentiable. For example, let $f$ be piecewise linear with $f(-2)=0$ and derivative
$$
                   f'=\chi_{[-2,-1]}-\chi_{[1,2]}.
$$
Then $f \in \mathcal{D}(B^{p})$ for $0 \le \rho \le 1$. However, $f\chi_{[-1,1]}=\chi_{[-1,1]}$ is not in the domain of $B^{\rho}$ for any $\rho \ge 1/2$. This is because there is no $\rho \ge 1/2$ for which $(1+s^{2})^{\rho/2}\mathcal{F}(f)|_{s}$ is square-integrable because the Fourier transform of $\chi_{[-1,1]}$ is the sinc function $\sin(s)/s$ which does not decay fast enough for $s^{\rho}\sin(s)/s$ to be square-integrable for $\rho \ge 1/2$.
Another way your identity can fail is that fractional powers of derivatives do not generally preserve support. In $R^{n}$, you can get an explicit formula for the fractional powers of $(-\Delta)$ "provided $f$ is regular enough":
$$
          (-\Delta)^{s}f = c_{n,s}\int_{\mathbb{R}^{n}}\frac{f(x)-f(y)}{|x-y|^{n+2s}}\,dy
$$
http://www.ma.utexas.edu/mediawiki/index.php/Fractional_Laplacian
