\label{e:sd} .\end{aligned}$$ As $F(0)=f(T)=0$ and $f$ is absolutely continuous, by and we have $D^\beta_t F = \partial^\beta_t F=(I^{1-\beta} F')$. So by the integration by part and the Fubini’s Theorem $$\begin{aligned}
\int_0^TG(t)  D^\beta_t F(t) dt
&=&  \int_0^T G(t) (I^{1-\beta} F') (t) dt \\
&=& \frac{1}{\Gamma (1-\beta )} \int_0^T G(t) \left( \int_0^t \theta^{-\beta}
F'(t-\theta ) d\theta \right) dt \\
&=&  \frac{1}{\Gamma (1-\beta )} \int_0^T \theta^{-\beta}  \left( \int_\theta^T G(t)
F'(t-\theta ) dt\right) d\theta  \\
&=& - \frac{1}{\Gamma (1-\beta )} \int_0^T \theta^{-\beta}  \left( \int_\theta^T
f' (T-t+\theta ) g(T-t) dt\right) d\theta  \\
&=& - \frac{1}{\Gamma (1-\beta )} \int_0^T \theta^{-\beta}  \left( \int_\theta^T
f' (s) g(s-\theta) ds\right) d\theta,\end{aligned}$$ which is $
\int_0^T f(s) D^\beta_t g(s) ds$ by . This proves the lemma.

\[L:2.3\] For each $(t, x)\in (0,
\infty) \times {\mathbb{R}}^d\setminus\{0\}$, $$\label{eqn lap}
\Delta q (t,x)=\frac{\partial}{\partial t}p (t,x).$$

Since both are continuous it is enough to prove that the equality holds for almost all $(t, x)$. Let $\phi (t)$ and $\psi (x)$ be smooth functions with compact support in $(0, T)$ and ${\mathbb{R}}^d\setminus\{0\}$ respectively. Define $\Phi (s):=\phi (T-s)$. Since $\phi (t)$ is smooth function with compact support in $(0, T)$, using Lemma \[lemma list\](i), Lemma \[L:2.6\] and , for every $x \in {\mathbb{R}}^d$ $$\begin{aligned}
 \int^T_0  D^{1-\beta}_t  \Phi(s)   