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I'd like to prove this lemma since this lemma asserts the regularity of the heat equation by using the cut-off function and mollification.

Let $\Omega\subset\mathbb{R}^n$. Define $\Omega_T=\Omega\times(0,T]$. Let $\phi$ be the fundamental solution for the heat equation, $$\phi=\left\{ \begin{array}{ll}\frac{1}{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{4t}}\quad & \textrm{for}~t>0 \\ ~~~~~~~0\quad & \textrm{for}~t<0\end{array} \right.$$

Assume that $f$ is bounded in $\mathbb{R}^{n+1}$, $f\equiv0$, and $f(x,t)\equiv0$ for $|t|\geq T_1>T$. Further define $u(x,t)=\phi*f=\int_{\mathbb{R}^{n+1}}\phi(x-y,t-s)f(y,s)~dyds$ . Then

1) $u\in C^\infty(\Omega_T)$,

2)$u_t(x,t)-\Delta u(x,t)=0$ in $\Omega_T$, and

3)$D^\alpha_{x,t}u(x,T)$ exist for $x\in\Omega$.

In order to show $u\in C^\infty(\Omega_T)$, it is enough to consider only near points of $(x_0,t_0)$ in $\Omega_T$. It's an important idea that we just find some $\tilde{\phi}$ satisfying $\phi*f=\tilde{\phi}*f$ near $(x_0,t_0)$. Now we take a smooth cut-off function $\zeta_\epsilon$ for $\epsilon>0$, \begin{equation} \zeta_\epsilon(x,t)=\left\{ \begin{array}{ll} 1 & \textrm{if $(x,t)\in B(0,\epsilon/2)$}\times(-\epsilon/2,\epsilon/2) \\ 0 & \textrm{if $(x,t)\in\mathbb{R}^{n+1}\backslash[ B(0,\epsilon)\times(-\epsilon,-\epsilon)]$}. \end{array} \right. \end{equation}

Using above then we could define $\tilde{\phi}$, \begin{equation} \tilde{\phi}(x,t)=\phi(x,t)(1-\zeta_\epsilon(x,t)). \end{equation}

Since $\phi\in C^\infty(\mathbb{R}^{n+1})$ except near $(0,0)\in\mathbb{R}^{n+1}$, $\tilde{\phi}\in C^\infty(\mathbb{R}^{n+1})$.

For any fixed $(x_0,t_0)\in\Omega_T$ and all $(y,s)\in\mathbb{R}^{n+1}$,

showing that $\phi(x-y,t-s)f(y,s)=\tilde{\phi}(x-y,t-s)f(y,s)$ implies $(i)$.

If $(x-y,t-s)\in(-\epsilon,\epsilon)\times B(0,\epsilon)$ then $t_0-2\epsilon<s<t_0+2\epsilon$ and this implies that $y\in B(0,2\epsilon)$, using $f(y,s)=0$ in $\Omega_T$ yields $\phi(x-y,t-s)f(y,s)=\tilde{\phi}(x-y,t-s)f(y,s)$.

If not then $\zeta_\epsilon=0$ yields $\phi(x-y,t-s)f(y,s)=\tilde{\phi}(x-y,t-s)f(y,s)$. $f$ is uniformly bounded in $\Omega_T$ hence $u(x,t)\in C^\infty(\Omega_T)$.

Now a direct evaluation asserts $(ii)$ since $\phi_t-\Delta\phi=0$ and by using $(i)$. Omit the subscript $\epsilon$ of $\zeta_\epsilon$, \begin{equation} \begin{aligned} u_t-\Delta u& =\frac{\partial}{\partial t}\int_{\mathbb{R}^{n+1}}\tilde{\phi}(x-y,t-s)f(y,s)~dyds \\ & \quad-\Delta_x\int_{\mathbb{R}^{n+1}}\tilde{\phi}(x-y,t-s)f(y,s)~dyds \\ & =\int_0^T\int_{\Omega_T\backslash B(0,\epsilon)}(\phi_t(1-\zeta)-\phi\zeta_t-\Delta\phi(1-\zeta)-\phi\Delta\zeta)f(y,s)~dyds \\ & \quad+\int_0^T\int_{B(0,\epsilon)}0\cdot f(y,s)~dyds\quad(\because~\zeta=1) \\ & =\int_0^T\int_{\Omega_T\backslash B(0,\epsilon)}(\phi_t-\Delta\phi)f(y,s)~dyds=0 \end{aligned} \end{equation} This result only validates in $\Omega_T$.

Now I want to prove (3) of this lemma, it's quite difficult to me. I think that $D_xu$ and $D^2_xu$ might exist at $t=T$but how can I control the ball at $t=T$? Moreover, it might be a one-sided derivative $D_tu(x,T)$ if $t\rightarrow T^-$, I cannot figure out how to dominate the distance by $O(\epsilon)$.

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