Jtdylktuy

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

Assume that $f$ is bounded in $mathbb{R}^{n+1}$, $fequiv0$, 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) $uin C^infty(Omega_T)$,

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

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

In order to show $uin C^infty(Omega_T)$, 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 $(x,t)inmathbb{R}^{n+1}backslash[ B(0,epsilon)times(-epsilon,-epsilon)]$}.
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 $phiin C^infty(mathbb{R}^{n+1})$ except near $(0,0)inmathbb{R}^{n+1}$, $tilde{phi}in C^infty(mathbb{R}^{n+1})$.

For any fixed $(x_0,t_0)inOmega_T$ and all $(y,s)inmathbb{R}^{n+1}$,

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 $t_0-2epsilon<s<t_0+2epsilon$ and this implies that $yin B(0,2epsilon)$, 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 $phi_t-Deltaphi=0$ 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_xint_{mathbb{R}^{n+1}}tilde{phi}(x-y,t-s)f(y,s)~dyds \
& =int_0^Tint_{Omega_Tbackslash B(0,epsilon)}(phi_t(1-zeta)-phizeta_t-Deltaphi(1-zeta)-phiDeltazeta)f(y,s)~dyds \
& quad+int_0^Tint_{B(0,epsilon)}0cdot f(y,s)~dydsquad(because~zeta=1) \
& =int_0^Tint_{Omega_Tbackslash B(0,epsilon)}(phi_t-Deltaphi)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 $trightarrow T^-$, I cannot figure out how to dominate the distance by $O(epsilon)$.