I have a question about functions that have bounded support in $\mathbb{R}$. In particular, suppose that I have a function $f$ with support $A\subset \mathbb{R}$ so that $A$ is compact. Without loss of generality, let's suppose that $A$ is a closed interval $[a,b]$.
Define $B$ to be the differential operator, such that $Bf = \frac{d}{dx}f$. If $s$ is an integer, then aside from the set $\{a,b\}$, which is of Lebesgue measure zero, we have that $B^s f = 0$ for any $x \in [a,b]^C$, the complement of $[a,b]$. My question is, aside from the points $\{a,b\}$ is this also true for any $s\in \mathbb{R}$, that is to say that $B^s f = 0$ for $x\in[a,b]^{C}$ almost everywhere?
$\textbf{Edit:}$ The reason why I feel this is not trivial is because as I understand it, fractional derivatives are no longer 'local' functionals, unlike their integer counterparts.
This fails for every fractional $s$. Consider what happens on the Fourier transform side. If $f$ has bounded essential support, then $\hat f$ is real-analytic. Taking derivative of order $s$ amounts to multiplying $\hat f$ by $(i\xi)^s$. If $s$ is fractional, the product will not be even $C^\infty$ smooth in general (unless $\hat f$ happens to vanish near the origin). Hence, the derivative will have unbounded essential support.