In the following I am talking about the Caputo fractional derivative defined via:
$$D^\alpha f(t):=\frac1{\Gamma(\lceil\alpha\rceil-\alpha)}\int_0^t \frac{d^{\lceil \alpha\rceil}f(\tau)}{d\tau^{\lceil\alpha\rceil}}\frac1{(t-\tau)^{\alpha-\lceil\alpha\rceil+1}}d\tau$$ which can be understood as deriving $f$ a total $\lceil\alpha\rceil$ times and then applying a fractional integral of power $\lceil\alpha\rceil-\alpha$. If you want to answer with any of the others, that is fine.
The following picture on wikipedia shows how the derivative of the identity changes as $\alpha$ varies from $-1$ to $1$. It looks like a very continuous (in $\alpha$) process, I would like to be able to formulate the question of how it could be continuous and also see whether or not it is.
We could consider the vector space $C^\infty([0,1])$ with metric
$$d(f,g)=\sum_n2^{-n} \frac{\|(f-g)^{(n)}\|_\infty}{1+\|(f-g)^{(n)}\|_\infty}$$ On this space the usual derivative operators are all linear and continuous. Also the fractional derivatives appear to be linear (and continuous?). The space of continuous linear maps on $C^\infty$ $\mathcal L(C^\infty([0,1]))$ can be naturally given the topology of pointwise continuity.
Is the map $[0,1]\to \mathcal L(C^\infty([0,1])), \alpha\mapsto D^\alpha$ continuous with this topology? Is there some way of rephrasing this question so that $D^\alpha$ changes continuously with $\alpha$?