is easy to check for any $\varphi\in L_1([0,T])$, $$\begin{aligned}
          \label{e:DI}
D^{\beta}_tI^{\beta}\varphi =\varphi.\end{aligned}$$ Furthermore, if $\varphi$ is absolutely continuous on $[0,T]$ then by $$\begin{aligned}
 \label{e:pbetav}
\partial^{\beta}_t\varphi=I^{1-\beta}\frac{d}{dt}\varphi.\end{aligned}$$ Thus by $$\begin{aligned}
 \label{e:Dbetadt}
D^{1-\beta}_t\partial^{\beta}_t \varphi=\frac{d}{dt}\varphi, \quad \text{a.e.}.\end{aligned}$$

Denote by $$E_{\beta}(z)=\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(\beta k+1)}, \quad z \in {\mathbb{C}}$$ the Mittag-Leffler function. We will also use the generalized Mittag-Leffler function $$E_{\beta,\gamma}(z):=\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(\beta k+\gamma)}, \quad z \in {\mathbb{C}}.$$

We assume $\beta, \gamma \in (0,1)$. It is well known (see e.g. (12) in [@D Theorem 1.3-4]) that, when $-1 < \gamma -\beta < 1$, $E_{\beta,\gamma}(t)$ is bounded on $(-\infty,0]$ and $$\label{mittag}
\lim_{{\mathbb{R}}\ni t \to \infty}  t E_{\beta,\gamma}(-t) =\frac{1}{\Gamma (\gamma-\beta)}.$$ Furthermore, for any constant $\lambda$, $$\label{frac heat}
\varphi(t):=E_{\beta}(\lambda t^{\beta})$$ is a solution of the equation $$\partial^{\beta}_t \varphi=\lambda \varphi, \quad t>0$$ with the initial condition $\varphi(0)=1$.

The following is a classical result. We provide a proof for the readers’ convenience.

\[l:DaE\] Let $a\in (0,1)$ and $b \ge 0$, then $$\label{seattle1}
 D_t^a E_{\beta}(-b
t^{\beta})=t^{-a}E_{\beta,1-a}(-bt^{\beta}),$$ $$\label{seattle2}
I^aE_{\beta}(-b
t^{\beta})=t^{a}E_{\beta,1+a}(-bt^{\beta}).$$

We first prove (\[seattle1\]). One can easily check (see e.g. [@EIK (5.1.2)]) $$\label{seattle3}
D_t^at^{b-1}=\frac{\Gamma(b)}{\Gamma(b-a)}t^{b-a-1}, \quad a,b>0.$$ Thus, $$\begin{aligned}
D_t^a E_{\beta}(-b t^{\beta})&=&D_t^a\left(\sum_{k=0}^\infty
\frac{(-1)^kb^{2k}t^{\beta k}}{\Gamma(\beta k+1)}\right)\\
&=&\sum_{k=0}^\infty\frac{(-1)^kb^{2k}}{\Gamma(\beta
k+1)}\frac{\Gamma(\beta k+1)}{\Gamma(\beta k+1-a)} t^{\beta
k-a}\\
&=&t^{-a}\sum_{k=0}^\infty\frac{(-bt^{\beta})^k}{\Gamma(\beta
k+1-a)}\\
&=&t^{-a}E_{\beta,1-a}(-bt^{\beta}).\end{aligned}$$

To prove (\[seattle2\]), in place of (\[seattle3\]), it is enough to use $$I^at^{b-1}=\frac{\Gamma(b)}{\Gamma(b+a)}t^{b+a-1}, \quad 