# Regularity of the heat equation

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 $$\epsilon>0$$, $$$$\zeta_\epsilon(x,t)=\left\{ \begin{array}{ll} 1 & \textrm{if (x,t)\in B(0,\epsilon/2)}\times(-\epsilon/2,\epsilon/2) \\ 0 & \textrm{if (x,t)\in\mathbb{R}^{n+1}\backslash[ B(0,\epsilon)\times(-\epsilon,-\epsilon)]}. \end{array} \right.$$$$

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

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 $$t_0-2\epsilon and this implies that $$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{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} 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 $$O(\epsilon)$$.