\{0\}$ and $m\leq 3$, the functions $p, q, \frac{\partial}{ \partial t}p(t,\cdot), D^m p$ and $D^m q$ are continuous and bounded on $[0,T]\times K$.

\(i) is a consequence of (\[frac heat\]). See (5.2.44) and (A.19) of [@EIK] for (v). Others can be found in Propositions 5,1 and 5.2 of [@EIK]. In particular, (vi) is a consequence of (5.2.6) and (5.2.13) of [@EIK].

We need the following integration by parts formula.

\[L:2.6\] Suppose that $f$ is absolutely continuous on $[0, T]$ with $f(T)=0$ and $g$ is continuous and $I^{1-\beta}g$ is absolutely continuous on $[0, T]$. Then $$\int_0^T f(t)D^\beta_t g(t) dt = \int_0^T G(t) D^\beta_t F(t) dt,$$ where $F(t)=f(T-t)$ and $G(t)=g(T-t)$.

Note that $$\begin{aligned}
 \int_0^T |f'(t)| \left( \int_0^t \theta^{-\beta}
 |g(t-\theta )| d\theta\right) dt  \le \|g\|_{L_\infty([0,T])}
 \int_0^T |f'(t)| dt  \int_0^T \theta^{-\beta}
d\theta <\infty.\end{aligned}$$ Thus, using $f(T)=0$, the integration by part and the Fubini’s Theorem, we get $$\begin{aligned}
\int_0^T f(t) D^\beta_t g(t) dt
&=& \int_0^T f(t) \frac{d}{dt} (I^{1-\beta}  g) (t) dt  \nonumber\\
&=& f(t) (I^{ 1-\beta }  g) (t) \Big|_0^T -\int_0^T f'(t) (I^{1-\beta } g)(t) dt  \nonumber\\
&=& - \frac{1}{\Gamma (1-\beta)} \int_0^T f'(t) \left( \int_0^t \theta^{-\beta}
 g(t-\theta ) d\theta\right) dt \nonumber\\
 &=& -  \frac{1}{\Gamma (1-\beta)} \int_0^T \theta^{-\beta}   \left( \int_\theta^T
f'(t)  g(t-\theta ) dt \right) d\theta 