Norm in the space of Riemann-Liouville integrals/derivatives? (aka fractional integrals/derivatives) What is the natural norm to consider in the space of Riemann-Liouville integrals/derivatives of a function?
Context: Let $f\in L^2 ([a,b])$. Define the Riemann-Liouville fractional integral of order $\alpha\geq 0$ (respectively fractional derivative) as:
$$I_{a^+}^{\alpha} (f)(x) = \frac{1}{\Gamma (\alpha)} \int_a^x (x-y)^{\alpha-1} f(y)dy$$
respectively,
$$D_{a^+}^{\alpha} (f)(x) = \frac{1}{\Gamma (1-\alpha)} \frac{d}{dx}\int_a^x \frac{f(y)}{(x-y)^{\alpha}}dy.$$
It follows that $D_{a^+}^{\alpha} I_{a^+}^{\alpha} (f) = f$ and viceversa.
Denote by $I_{a^+}^{\alpha} (L^2)$ the imagine of $L^2([a,b])$ by the operator $I_{a^+}^{\alpha}$.
Question: if I have an operator $T_{\alpha}:I_{a^+}^{\alpha} (L^2)\rightarrow L^2([a,b]) $ which is a linear isomorphism. What is the norm considered on the imagine space? That is,
$$\|T_{\alpha}f\|_{L^2([a,b])} \leq \|T_{\alpha}\|_{op} \|f\|_{?}$$
where $\|T_{\alpha}\|_{op}$ denotes the operator norm.
 A: I'll take your word for it that indeed $D^\alpha$ and $I^\alpha$ are mutually inverse since I have not seen that form of the fractional derivative before. It can be shown that $I^\alpha:L^2(a,b)\rightarrow L^2(a,b)$ with
$$\|I^\alpha\|_2\le\frac{(b-a)^{\alpha/2}}{\alpha|\Gamma(\alpha)|}\|f\|_2$$
It follows that $I^\alpha(L^2)=\{f\in L^2(a,b)\,:\,D^\alpha f\in L^2(a,b)\}$, so we can define an operator $A^\alpha$ on $L^2(a,b)$:
$$D(A^\alpha)=I^\alpha(L^2),\quad A^\alpha f=D^\alpha f$$
Since $I^\alpha$ is bounded it quickly follows that $A^\alpha$ is closed. A natural norm is hence the graph norm
$$\|f\|_{I^\alpha}:=\|f\|_2+\|A^\alpha f\|_2=\|f\|_2+\|D^\alpha f\|_2$$
which makes $I^\alpha(L^2)=D(A^\alpha)$ a Banach space. (In fact, it is a Hilbert space under the inner product $(f|g)_{I^\alpha}=(f|g)_{L^2}+(D^\alpha f|D^\alpha g)_{L^2}$.) Since $I^\alpha$ is bounded, it is not hard to show that $\|f\|_{I^\alpha,0}:=\|D^\alpha f\|_2$ is also a norm on $I^\alpha(L^2)$ which is equivalent to the one just given.
Of course, this is simply one choice of (equivalence class of) norm; there are many others. For a rather artificial example we could define
$$\|f\|_{I^\alpha,T_\alpha}:=\|T_\alpha f\|_2$$
which would make the two spaces isometrically isomorphic. The graph norm would probably be more useful as it is analogous to the norm on the Sobolev spaces $H^k(a,b)$.
