If we study C* algebras, at one point or another we are exposed to the idea of the Continuous functional Calculus, i.e. if $\mathcal{Y}$ is a unital C* algebra, $\xi \in \mathcal{Y}$ is normal, that is $\xi^*\xi = \xi\xi^*$, then we can make sense of the expression $f(\xi)$, where $f \in \mathcal{C}(\sigma(\xi))$ (= all complex valued, continuous functions on the spectrum $\sigma(\xi)$ of $\xi$). I just recently stumbled upon a video about fractional derivatives, and shortly after I read the corresponding wikipedia article (https://en.wikipedia.org/wiki/Fractional_calculus). In the wikipedia article it is mentioned that one can use the continuous functional calculus in order to give precise meaning to $D^\alpha$, where $D$ is a differentiation operator and $\alpha >0$. As the article points to pseudo-differential operators, and I unfortunately don't have much of a clue of that, I didn't quite seem to find an explanation of how to define $D^\alpha$ in that way. So my question is as follows:
Which C* algebra $\mathcal{Y}$ are we considering so that $D \in \mathcal{Y}$ and $D$ is normal with respect to the involution on $\mathcal{Y}$, so that we may properly define $D^\alpha$?
I don't know where the Wikipedia article is coming from, but I don't think what you want to do can be done.
First, I'm not aware of a context where a differential operator is bounded as an operator on a Hilbert space. So I don't see how you would put a differential operator inside a C$^*$-algebra.
Second, differential operators are not positive, so doing functional calculus with a function that is only defined for positive values (as are fractional powers) is no-go.