---
abstract: 'Let $P=\partial_{t}+\partial_{i}(a^{ij}\partial_{j})$ be a backward parabolic operator. It is shown that under certain conditions on $\left\{ a^{ij}\right\} $, if $u$ satisfies $\left|Pu\right|\leq C(\left|u\right|+\left|\nabla u\right|)$, $\left|u(x,t)\right|\lesssim e^{C\left|x\right|^{2}}$ in $\mathbb{R}^{n}\times[0,T]$ and $\left|u(x,0)\right|\lesssim e^{-M\left|x\right|^{2}}$ for all $M>0$, then $u$ vanishes identically in $\mathbb{R}^{n}\times[0,T]$.'
author:
- Tu Nguyen
bibliography:
- 'UC.bib'
title:
- On a question of Landis and Oleinik
- 
---

Introduction
============

Let $P$ be a backward parabolic operator on $\mathbb{R}^{n}$,$$Pu=\partial_{t}u+\mbox{div}(A\nabla u)$$ where $A(x,t)=(a^{ij}(x,t))_{i,j=1}^{n}$ is a real, symmetric matrix such that for some $\lambda>0$, $$\lambda\left|\xi\right|^{2}\leq a^{ij}(x,t)\xi_{i}\xi_{j}\leq\lambda^{-1}\left|\xi\right|^{2}\mbox{ for all }\xi\in\mathbb{R}^{n}.\label{eq:ellipticity}$$ It was conjectured by Landis and Oleinik [@MR0402268] that if $Pu=b(x,t)\cdot\nabla u+a(x,t)u$ in $\mathbb{R}^{n}\times[0,T]$ and $\left|u(x,0)\right|\lesssim e^{-\left|x\right|^{2+\epsilon}},\,\forall x\in\mathbb{R}^{n}$ then $u\equiv0$ in $\mathbb{R}^{n}\times[0,T]$, provided $A,b$ and $c$ satisfy appropriate conditions at infinity.

Escauriaza, Kenig, Ponce and Vega [@MR2231129] showed that this is true when $P$ is the backward heat operator (i.e. $A(x,t)\equiv\mbox{Id}$) and $b$ and $c$ are bounded. They also obtained a similar result when the domain is $\mathbb{R}_{+}^{n}\times[0,T]$. The aim of this paper is to extend these results to parabolic operators with variable coefficients.

Suppose that $\{a^{ij}\}$ satisfy the ellipticity condition (\[eq:ellipticity\]), and for some $\epsilon>0$$$\begin{aligned}
 &  & \left|\nabla_{x}a^{ij}(x,t)\right|\lesssim\left\langle x\right\rangle ^{-1-\epsilon},\,\,\,\left|\partial_{t}a^{ij}(x,t)\right|\lesssim1,\\
 &  & \left|a^{ij}(x,t)-a^{ij}(x,s)\right|\lesssim\left\langle x\right\rangle ^{-1}\left|t-s\right|^{1/2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall x\in\mathbb{R}^{n};\,\,\, t,s\in[0,T].\end{aligned}$$ Assume that $u$ satisfies the inequalities $$\left|Pu\right|\leq C(\left|u\right|+\left|\nabla u\right|)\,\,\,\,\,\mbox{ in }\mathbb{R}^{n}\times[0,T]$$ and $$\left|u(x,t)\right|\lesssim e^{C\left|x\right|^{2}}\,\,\,\,\,\,\,\,\forall(x,t)\in\mathbb{R}^{n}\times[0,T],$$ for some $C>0$. Then

1.  If $\left|u(x,0)\right|\lesssim e^{-M\left|x\right|^{2}}$ for all $M>0$, then $u\equiv0$.

2.  If $u(x,0)\not\equiv0$ then there exists $M>0$ such that if $\left|x\right|>M$, $$\int_{B(x,1)}\left|u(y,0)\right|^{2}dy\gtrsim e^{-M\left|x\right|^{2}\log\left|x\right|}\,\,\,\,\,\,\,\mbox{and\,\,\,\,\,\,\,}\int_{B(x,\left|x\right|/2)}\left|u(y,0)\right|^{2}dy\gtrsim e^{-M\left|x\right|^{2}}.$$

 

We also obtain a similar result where the domain is a half-space.

Let $\mathbb{R}_{+}^{n}=\left\{ x\in\mathbb{R}^{n}:x_{1}>0\right\} $. Suppose that $\{a^{ij}\}$ satisfy the ellipticity condition (\[eq:ellipticity\]), and for some $\epsilon>0$$$\begin{aligned}
 &  & \left|\nabla_{x}a^{ij}(x,t)\right|\lesssim\left\langle x\right\rangle ^{-1-\epsilon},\,\,\,\left|\partial_{t}a^{ij}(x,t)\right|\lesssim1,\\
 &  & \left|a^{ij}(x,t)-a^{ij}(x,s)\right|\lesssim\left\langle x\right\rangle ^{-1}\left|t-s\right|^{1/2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\forall x\in\mathbb{R}_{+}^{n};\,\,\, t,s\in[0,T].\end{aligned}$$ Assume that $u$ satisfies the inequalities $$\left|Pu\right|\leq C(\left|u\right|+\left|\nabla u\right|)\,\,\,\,\,\mbox{ in }\mathbb{R}_{+}^{n}\times[0,T]$$ and $$\left|u(x,t)\right|\lesssim e^{C\left|x\right|^{2}}\,\,\,\,\,\,\,\,\forall(x,t)\in\mathbb{R}_{+}^{n}\times[0,T],$$ for some $C>0$. Then

1.  If $\left|u(x,0)\right|\lesssim e^{C\left|x\right|^{2}-Mx_{1}^{2}}$ for all $M>0$, then $u\equiv0$.

2.  If $u(\cdot,0)\not\equiv0$ then there exists $M>0$ such that if $R>M$, $$\int_{B(Re_{1},1)}\left|u(y,0)\right|^{2}dy\gtrsim e^{-MR^{2}\log R}\,\,\,\,\,\,\,\mbox{and\,\,\,\,\,\,\,}\int_{B(Re_{1},R/2)}\left|u(y,0)\right|^{2}dy\gtrsim e^{-MR^{2}}.$$

The proof in [@MR2231129] for the heat operator used a Carleman inequality together with a scaling argument to show $u(x,0)$ has a doubling property which implies that $$\int_{B(x,\left|x\right|/2)}\left|u(y,0)\right|^{2}dy\gtrsim e^{-M\left|x\right|^{2}}$$ for some $M>0$. This argument breaks down in the variable coefficients case, as it requires a uniform bound on $\left\Vert \nabla a_{x_{0},R}^{ij}\right\Vert _{L^{\infty}}$ where $a_{x_{0},R}^{ij}(x,t)=a^{ij}(x_{0}+Rx,R^{2}t)$ and $R>\left|x_{0}\right|$.

To prove the first part of Theorem 1.1, we first show that if $\left|u(x,0)\right|\lesssim e^{-M\left|x\right|^{2}}$ for all $M>0$ then there exists $T_{0}\in[0,T]$ such that for any $M\geq0$, $$\int_{0}^{T_{0}}\int_{B(x,\left|x\right|/2)}u^{2}(y,t)dydt\lesssim e^{-M\left|x\right|^{2}}\mbox{ \,\,\,\,\,\, if }\left|x\right|\geq R_{M}.$$ Then we show that if $u(\cdot,0)\not\equiv0$, for any $T_{0}\in[0,T]$, the following lower bound holds $$\int_{0}^{T_{0}}\int_{B(x,\left|x\right|/2)}u^{2}(y,t)dydt\gtrsim e^{-C_{2}\left|x\right|^{2}}$$ where $C_{2}=C_{2}(T_{0},u)\geq0$. (a similar bound for the Schrödinger equation was proved in [@MR2273975].) Thus, we must have $u(\cdot,0)\equiv0$, which then implies $u\equiv0$ in $\mathbb{R}^{n}\times[0,T]$. The proof of Theorem 1.2 follows the same argument, using anisotropic Carleman inequalities instead, as now $u$ decays in the direction of $x_{1}$ only.

We would like to mention a unique continuation result of [@MR1971939; @MR2001174]. Let $u$ be a solution of the inequality $\left|Pu\right|\leq M(\left|u\right|+\left|\nabla u\right|)$ in $B(0,1)\times[0,T]$ which vanishes to infinite order at $0$, i.e. $\left|u(x,0)\right|\leq C_{k}\left|x\right|^{k}$ for all $k\geq0$, $\forall x\in B(0,1)$. Then $u(\cdot,0)\equiv0$ in $B(0,1)$. We have benefited from the Carleman inequalities and ideas contained in these papers, and also from those of [@MR040; @MR1979722; @MR2005639; @MR2273975; @MR2231129].

Details of the proofs of Theorem 1.1 and 1.2 are in section 2 and 3, respectively. The proofs of the Carleman inequalities used in section 2 and 3 will be gathered in section 4, together with other auxiliary lemmas.

**Acknowledgement.** I would like to thank my thesis advisor Carlos Kenig for suggesting the problem, and for his invaluable guidance and support.

Proof of theorem 1.1
====================

We first remark that by considering $P_{r}=\partial_{t}+\mbox{div}\left(A_{r}\nabla\right)$ where $a_{r}^{ij}(x,t)=a^{ij}(rx,r^{2}t)$ and $u_{r}(x,t)=u(rx,r^{2}t)$ for suitably small $r>0$, we can assume that the constant $C$ in the hypothesis of Theorem 1.1 is as small as we like, say $C\leq\lambda^{5}/100$, and that $\left|\partial_{t}a^{ij}(x,t)\right|\leq C$. Furthermore, we can take $T=1$.

Upper bound
-----------

In this section we will adapt the arguments of [@MR2005639] to show that under the hypothesis of Theorem 1.1, there exists $T_{0}>0$ such that for any $M>0$, if $\left|x\right|\geq R_{M}>0$ $$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{-M\left|x\right|^{2}}\mbox{ for all }t\in[0,T_{0}].\label{eq: upperbound}$$ First, we prove the weaker bound $$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{-M\left|x\right|^{2}}\,\,\,\,\,\mbox{if }0\leq t\lesssim M^{-1}.\label{eq:weak upperbound}$$ (Note that the time interval of this weaker bound shrinks as $M\rightarrow\infty$.) Then we combine this bound with $M=2$ and another Carleman inequality to obtain (\[eq: upperbound\]).

### First step

We will use the following Carleman inequality of [@MR1971939].

\[1st Carleman\]Suppose $a^{ij}(0,0)=\delta_{ij}$ and $\left|a^{ij}(x,t)-a^{ij}(y,s)\right|\leq L\left(\left|x-y\right|+\left|s-t\right|^{1/2}\right)$. Then there is a constant $N=N(n,\lambda,L)>0$ such that for any $\alpha\geq2$ there is a positive function $\sigma:(0,\frac{4}{\alpha})\rightarrow\mathbb{R}_{+}$ satisfying $$N^{-1}\leq\frac{\sigma(t)}{t}\leq1$$ so that if $v\in C_{c}^{\infty}(\mathbb{R}^{n}\times[0,\frac{2}{\alpha}))$ and $0<a<1/\alpha$, then

$$\begin{aligned}
 &  & \int_{\mathbb{R}^{n+1}}(\alpha^{2}v^{2}+\alpha\sigma_{a}\left|\nabla v\right|^{2})\sigma_{a}^{-\alpha}G_{a}dxdt\leq N\int_{\mathbb{R}^{n+1}}\sigma_{a}^{1-\alpha}\left|Pv\right|^{2}G_{a}dxdt\\
 &  & +\sigma(a)^{-\alpha}\left[-\frac{a}{N}\int_{\mathbb{R}^{n}}\left|\nabla v(x,0)\right|^{2}G_{a}(x,0)dx+\alpha N\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}(x,0)dx\right]\\
 &  & +\alpha^{\alpha}N^{\alpha}\sup_{t\geq0}\int_{\mathbb{R}^{n}}(v^{2}+\left|\nabla v\right|^{2})dx\end{aligned}$$

Here $G_{a}(x,t)=(t+a)^{-n/2}e^{-\left|x\right|^{2}/4(t+a)}$ and $\sigma_{a}(t)=\sigma(t+a)$.

Since the hypothesis of the lemma requires $a^{ij}(0,0)=\delta_{ij}$, we first need to make a change of variable. Let $x_{0}\in\mathbb{R}^{n}$ with $\left|x_{0}\right|\gtrsim1$. Let $S=A(x_{0},0)^{1/2}$, $z_{0}=S^{-1}x_{0}$, $\widetilde{u}(x,t)=u(Sx,t)$ and $\widetilde{A}(x,t)=S^{-1}A(Sx,t)S^{-1}$. Then $\widetilde{A}(z_{0},0)=\mbox{Id}$ and $$\partial_{t}\widetilde{u}+\mbox{div}(\widetilde{A}\nabla\widetilde{u})|_{(x,t)}=\partial_{t}u+\mbox{div}(A\nabla u)|_{(Sx,t)}.$$ Let $\widetilde{u}_{R}$ be a rescale of $\widetilde{u}$ centered at $z_{0}$, $\widetilde{u}_{R}(x,t)=\widetilde{u}(z_{0}+Rx,R^{2}t)$ where $R=\lambda\left|x_{0}\right|/4$. Then $\widetilde{u}_{R}$ satisfies $$\left|P_{R}\widetilde{u}_{R}\right|\leq R^{2}\left|\widetilde{u}_{R}\right|+R\left|\nabla\widetilde{u}_{R}\right|$$ where $P_{R}=\partial_{t}+\mbox{div}(\widetilde{A}_{R}\nabla)$, and $\widetilde{A}_{R}(x,t)=\widetilde{A}(z_{0}+Rx,R^{2}t)$.

From the hypothesis of Theorem 1.1 and our choice of $R$, it is easy to see that $$\left|\nabla\widetilde{a}_{R}^{ij}(x,t)\right|\lesssim1,\left|\widetilde{a}_{R}^{ij}(x,t)-\widetilde{a}_{R}^{ij}(x,s)\right|\lesssim\left|t-s\right|^{1/2}$$ in $B(0,2)\times[0,1/R^{2})$. Furthermore, $\widetilde{A}_{R}(0,0)=\widetilde{A}(z_{0},0)=\mbox{Id}$. Thus, we can apply Lemma \[1st Carleman\] to $P_{R}$ and $v=\widetilde{u}_{R}\psi(x)\varphi(t)$, where $\chi_{[0,1/\alpha]}\leq\varphi\leq\chi_{[0,2/\alpha)}$ and $\chi_{B(0,1)}\leq\psi\leq\chi_{B(0,2)}$ are bump functions, $\alpha\geq2R^{2}$ is a positive constant to be chosen. Let $E=B(0,2)\times[0,2/\alpha)\backslash B(0,1)\times[0,1/\alpha]$ then $$\left|P_{R}v\right|\leq R^{2}\left|v\right|+R\left|\nabla v\right|+\alpha(\left|\widetilde{u}_{R}\right|+\left|\nabla\widetilde{u}_{R}\right|)\chi_{E}.$$ Hence, by Lemma \[1st Carleman\], $$\begin{aligned}
\int_{\mathbb{R}^{n+1}}(\alpha^{2}v^{2}+\alpha\sigma_{a}\left|\nabla v\right|^{2})\sigma_{a}^{-\alpha}G_{a}dxdt & \lesssim & N\int_{\mathbb{R}^{n+1}}\sigma_{a}^{1-\alpha}(R^{2}\left|v\right|+R\left|\nabla v\right|)^{2}G_{a}dxdt+\\
 &  & N\int_{E}\sigma_{a}^{1-\alpha}\alpha^{2}(\left|\widetilde{u}_{R}\right|+\left|\nabla\widetilde{u}_{R}\right|)^{2}G_{a}dxdt+\\
 &  & \alpha^{\alpha}N^{\alpha}\sup_{t\geq0}\int_{\mathbb{R}^{n}}(v^{2}+\left|\nabla v\right|^{2})dx+\\
 &  & \alpha N\sigma(a)^{-\alpha}\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}dx.\end{aligned}$$ If $\alpha\geq2NR^{2}$ then the first term on the right hand side can be absorbed by the left hand side. Also, $\sigma_{a}(t)^{-\alpha}G_{a}(x,t)\leq N^{\alpha}\alpha^{\alpha+\frac{n}{2}}$ in $E$, and $\left|\widetilde{u}_{R}\right|+\left|\nabla\widetilde{u}_{R}\right|\lesssim e^{CR^{2}}$ by hypothesis on $u$ and Lemma \[basic L8 bound\] in the Appendix. Thus, we obtain$$\begin{aligned}
\int_{\mathbb{R}^{n+1}}(\alpha^{2}v^{2}+\alpha\sigma_{a}\left|\nabla v\right|^{2})\sigma_{a}^{-\alpha}G_{a}dxdt & \lesssim & N^{\alpha}\alpha^{\alpha+\frac{n}{2}}e^{2CR^{2}}+\alpha N\sigma(a)^{-\alpha}\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}dx.\end{aligned}$$

Let $\rho=\frac{1}{Ne}$, and $a=\frac{\rho^{2}}{2\alpha}$. Then $$\sigma_{a}(t)^{-\alpha+1}G_{a}(x,t)\geq\alpha^{\alpha+\frac{n}{2}-1}N^{2\alpha+n-2}\mbox{ \,\,\,\,\,\,\,\,\,\,\,\, in }B(0,2\rho)\times[0,\frac{\rho^{2}}{2\alpha}]$$ and $$\sigma(a)^{-\alpha}G_{a}(x,0)\leq N^{\alpha}a^{-\alpha-\frac{n}{2}}=(2\alpha e^{2})^{\alpha+\frac{n}{2}}N^{3\alpha+n}\,\,\,\,\mbox{for all }x.$$ Hence, $$\begin{aligned}
 &  & \alpha^{\alpha+\frac{n}{2}}N^{2\alpha+n-2}\int_{B(0,2\rho)\times[0,\frac{\rho^{2}}{2\alpha}]}(v^{2}+\left|\nabla v\right|^{2})dxdt\lesssim N^{\alpha}\alpha^{\alpha+\frac{n}{2}}e^{2CR^{2}}\\
 &  & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\alpha^{\alpha+\frac{n}{2}+1}(2e)^{2\alpha+n}N^{3\alpha+n+1}\int_{B(0,2)}v^{2}(x,0)dx\end{aligned}$$ or $$\int_{B(0,2\rho)\times[0,\frac{\rho^{2}}{2\alpha}]}(v^{2}+\left|\nabla v\right|^{2})dxdt\lesssim N^{2-\alpha-n}e^{2CR^{2}}+\alpha(2e)^{2\alpha+n}N^{\alpha+3}\int_{B(0,2)}v^{2}(x,0)dx.$$ We now choose $\alpha=MR^{2}$ then the first term in the right hand side is bounded by $e^{-MR^{2}}$. The second term is also bounded by $e^{-MR^{2}}$ by the decay hypothesis on $u(\cdot,0)$. Thus, for any $M>2N$, $$\int_{B(0,2\rho)\times[0,\frac{\rho^{2}}{2MR^{2}}]}(v^{2}+\left|\nabla v\right|^{2})dxdt\lesssim e^{-MR^{2}}.$$ By Lemma \[basic L8 bound\], this implies $$\left|v\right|+\left|\nabla v\right|\lesssim e^{-MR^{2}}\,\,\,\,\,\,\mbox{in }B(0,\rho)\times[0,\frac{\rho^{2}}{4MR^{2}}].$$ Undoing the change of variable, we get $$\left|u(x_{0},t)\right|+\left|\nabla u(x_{0},t)\right|\lesssim e^{-MR^{2}}\,\,\,\,\,\,\mbox{if }0\leq t\leq\frac{\rho^{2}}{4M}.$$ This proves (\[eq:weak upperbound\]).

### Second step.

\[lem:upper 1\] Let $\epsilon$ be the constant in the hypothesis of Theorem 1.1. Let $$G(x,t)=\exp\left(c(T-t)\left|x\right|+\left|x\right|^{2}\right)$$ where $0\leq c\leq R^{1+\epsilon/8}$. Then for any $v\in C_{c}^{\infty}(\left\{ R\leq\left|x\right|\leq R^{1+\epsilon/8}\right\} \times[0,T])$, the following inequality holds $$\begin{aligned}
 &  & \frac{\lambda^{2}}{4}\int_{\mathbb{R}^{n+1}}v^{2}Gdxdt+\frac{\lambda^{2}}{4}\int_{\mathbb{R}^{n+1}}\left|\nabla v\right|^{2}Gdxdt\leq\int_{\mathbb{R}^{n+1}}\left|Pv\right|^{2}Gdxdt\\
 &  & +\lambda^{-1}\int_{\mathbb{R}^{n}}\left|\nabla v(x,T)\right|^{2}G(x,T)dx+R^{2+\epsilon/4}\int_{\mathbb{R}^{n}}v^{2}(x,0)G(x,0)dx,\end{aligned}$$ provided $R\gtrsim1$.

This Carleman inequality is an extension of a Carleman inequality in [@MR2005639] to the case of variable coefficients. As $\{a^{ij}\}$ are no longer constants, it is necessary to put a restriction on the support of $v$ (compared with [@MR2005639].) We will prove this inequality in the Appendix. We now deduce (\[eq: upperbound\]) from (\[eq:weak upperbound\]) and Lemma \[lem:upper 1\].

\[pro:upper\] Suppose that $u$ is as in the hypothesis of Theorem 1.1, and $\left|u(x,0)\right|\lesssim e^{-M\left|x\right|^{2}}$ for all $M>0$. Let $T=\rho^{2}/8N$, where $\rho$ and $N$ are as above. Then for all $M>0$,$$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{-M\left|x\right|^{2}}$$ for all $t\in[0,T/4]$.

Fix $M>0$. Let $$v(x,t)=u(x,t)\theta(x)$$ where $$\theta(x)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }\left|x\right|<R-1\mbox{ or }\left|x\right|>MR+1\\
1 & \mbox{ if }R<\left|x\right|<MR\end{array}\end{cases}$$ Since $$Pv=\theta Pu+2\left\langle A\nabla u,\nabla\theta\right\rangle +u\Delta\theta,$$ it follows that$$\begin{aligned}
\left|Pv\right| & \leq & C\theta\left(\left|u\right|+\left|\nabla u\right|\right)+2\lambda^{-1}\left|\nabla u\right|\left|\nabla\theta\right|+\left|u\Delta\theta\right|\\
 & \leq & C\left(\left|v\right|+\left|\nabla v\right|\right)+\left|u\right|\left(C\left|\nabla\theta\right|+\left|\Delta\theta\right|\right)+2\lambda^{-1}\left|\nabla u\right|\left|\nabla\theta\right|\\
 & \leq & C\left(\left|v\right|+\left|\nabla v\right|\right)+C'\left(\left|u\right|+\left|\nabla u\right|\right)\chi_{E}\end{aligned}$$ where $E=\left(\left\{ R-1<\left|x\right|<R\right\} \cup\{MR<\left|x\right|<MR+1\}\right)\times[0,T]$, $C'\leq4\lambda^{-1}$.

Choose $c=MR/T$. Then for large $R$, $c\leq R^{1+\epsilon/8}$ and $\mbox{supp}v\subset\left\{ R\leq\left|x\right|\leq R^{1+\epsilon/8}\right\} $. Thus, we can apply the previous lemma to $v$ to obtain$$\begin{aligned}
\int_{0}^{T}\int_{\mathbb{R}^{n}}(v^{2}+\left|\nabla v\right|^{2})Gdxdt & \lesssim & \int_{\mathbb{R}^{n}}\left|\nabla v(x,T)\right|^{2}G(x,T)dx+R^{2+\epsilon/4}\int_{\mathbb{R}^{n}}v^{2}(x,0)G(x,0)dx\\
 &  & +\int_{E}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt.\end{aligned}$$ In the previous subsection, we have shown that

$$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{-2N\left|x\right|^{2}}\mbox{ for all }t\in[0,T].$$ Hence, as $G(x,T)=e^{\left|x\right|^{2}}$, we have $$\int_{\mathbb{R}^{n}}\left|\nabla v(x,T)\right|^{2}Gdx\lesssim1.$$ Since $\left|u(x,0)\right|\lesssim e^{-2M\left|x\right|^{2}}$ and $G(x,0)\leq e^{(M+1)\left|x\right|^{2}}$ if $\left|x\right|\geq R$ by our choice of $c$, it follows that $$R^{2+\epsilon/4}\int_{\mathbb{R}^{n}}v^{2}(x,0)Gdx\lesssim1.$$ In $\{MR<\left|x\right|<MR+1\}$, $G(x,t)\leq e^{2\left|x\right|^{2}}$, hence, $$\int_{0}^{T}\int_{MR<\left|x\right|<MR+1}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim R^{n-1}.$$ In $\left\{ R-1<\left|x\right|<R\right\} $, $G(x,t)\leq e^{(M+1)R^{2}}$, so

$$\int_{0}^{T}\int_{R-1<\left|x\right|<R}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim e^{(M+2)R^{2}}.$$ Thus,$$\int_{0}^{T}\int_{\mathbb{R}^{n}}(v^{2}+\left|\nabla v\right|^{2})Gdxdt\lesssim e^{(M+2)R^{2}}.$$ As $G(x,t)\geq e^{4MR^{2}}$ in $\{6R\leq\left|x\right|\leq7R\}\times[0,T/2]$, this implies $$\begin{aligned}
\int_{0}^{T/2}\int_{6R\leq\left|x\right|\leq7R}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)dxdt & \lesssim & e^{-MR^{2}},\end{aligned}$$ provided $R\geq R_{M}$. This and Lemma \[basic L8 bound\] prove the proposition.

Lower Bound
-----------

In this subsection, assuming $u(\cdot,0)\not\equiv0$, we will show that the following lower bound holds for any $T\leq1$, $$\int_{0}^{T}\int_{R<\left|x\right|<2R}u^{2}(x,0)dx\gtrsim e^{-C_{2}R^{2}}\label{eq:lower}$$ To prove this, we first adapt arguments of [@MR2198840] and [@MR040] to show that there exists $s>0,$ such that for small $t$, we have $$\int_{R<\left|x\right|<2R}u^{2}(x,t)dx\gtrsim e^{-R^{s}}.\label{eq:weak lower}$$ Then we use this bound together with a bootstrap argument to obtain .

### First step

Since $u(\cdot,0)\not\equiv0$, we can suppose that, $$\int_{B(e_{1},\rho\lambda/4)}u^{2}(x,0)dx\ne0.$$ Here $\rho$ is a positive constant to be chosen. By using Lemma \[Lemma\], and multiplying $u$ by a constant if necessary, we can assume that $$\int_{B(e_{1},\rho\lambda/2)}u^{2}(x,t)dx\geq L\label{eq:trivia}$$ if $t$ is small enough. Here $L$ is a large constant to be chosen.

We will use the doubling property of $u(\cdot,0)$ proved by Escauriaza, Fernández and Vessella. We present their arguments here in the form that we need. Let $x_{0}=\left|x_{0}\right|e_{1}$, and $v$ be as in section 2.1. As before, if $\alpha\geq2NR^{2}$ the following inequality holds$$\begin{aligned}
 &  & \int_{\mathbb{R}^{n+1}}(\alpha^{2}v^{2}+\alpha\sigma_{a}\left|\nabla v\right|^{2})\sigma_{a}^{-\alpha}G_{a}dxdt\leq N^{\alpha}\alpha^{\alpha+\frac{n}{2}}e^{CR^{2}}+\nonumber \\
 &  & \sigma(a)^{-\alpha}\left[-\frac{a}{N}\int_{\mathbb{R}^{n}}\left|\nabla v(x,0)\right|^{2}G_{a}dx+\alpha N\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}dx\right].\label{eq:first 1}\end{aligned}$$ Let $\rho=\frac{1}{Ne}$ and $0<a\leq\rho^{2}/2\alpha$. Then, $$\begin{aligned}
\alpha^{2}\int_{\mathbb{R}^{n+1}}v^{2}\sigma_{a}^{-\alpha}G_{a} & \geq & \alpha^{2}\int_{0}^{\rho^{2}/\alpha}dt\int_{B(0,2\rho)}(t+a)^{-\alpha-\frac{n}{2}}e^{-\rho^{2}/(t+a)}v^{2}(x,t)dx\nonumber \\
 & \geq & N_{\rho}\alpha^{2}\int_{a}^{a+\rho^{2}/\alpha}s^{-\alpha-\frac{n}{2}}e^{-\rho^{2}/s}ds\int_{B(0,\rho)}v^{2}(x,0)dx\label{eq:first 2}\\
 & \geq & N_{\rho}\alpha^{2}\int_{\rho^{2}/2\alpha}^{\rho^{2}/\alpha}s^{-\alpha-\frac{n}{2}}e^{-\rho^{2}/s}ds\int_{B(0,\rho)}v^{2}(x,0)dx\nonumber \\
 & \geq & \frac{N_{\rho}\alpha^{\alpha+\frac{n}{2}+1}N^{2\alpha}}{2}\int_{B(0,\rho)}v^{2}(x,0)dx.\nonumber \end{aligned}$$ (we have used Lemma \[Lemma\] in the second inequality. $N_{\rho}$ is the constant appears in that lemma.) Here, $\alpha$ has to satisfy $$\rho^{2}/\alpha\leq N_{\rho}^{-1}\min\left\{ R^{-2},1/\log\left(\frac{N_{\rho}\int_{B(0,1)\times[0,R^{-2}]}v^{2}(x,t)dxdt}{\int_{B(0,\rho)}v^{2}(x,0)dx}\right)\right\}$$ As $\left|v(x,t)\right|\lesssim e^{CR^{2}}$, we can take $$\alpha=\rho^{2}N_{\rho}\left(2R^{2}+\log\frac{N_{\rho}}{\int_{B(0,\rho)}v^{2}(x,0)dx}\right).$$ For this value of $\alpha$, $$\frac{N_{\rho}\alpha^{\alpha+\frac{n}{2}+1}N^{2\alpha}}{2}\int_{B(0,\rho)}v^{2}(x,0)dx\geq N^{\alpha}\alpha^{\alpha+\frac{n}{2}}e^{CR^{2}}.$$ This together with (\[eq:first 1\]) and (\[eq:first 2\]) show that $$-\frac{a}{N}\int_{\mathbb{R}^{n}}\left|\nabla v(x,0)\right|^{2}G_{a}(x,0)dx+\alpha N\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}(x,0)dx\geq0$$ or,$$2a\int_{\mathbb{R}^{n}}\left|\nabla v(x,0)\right|^{2}G_{a}(x,0)dx+\frac{n}{2}\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}(x,0)dx\leq4\alpha N^{2}\int_{\mathbb{R}^{n}}v^{2}(x,0)G_{a}(x,0)dx$$ for all $a\leq\rho^{2}/2\alpha$.

By Lemma \[Lmm doubl\], this implies that $$\int_{B(0,2r)}v^{2}(x,0)\leq e^{128\alpha N^{2}}\int_{B(0,r)}v^{2}(x,0)$$ for all $0\leq r\leq1/2$. It follows that there exists positive constant $C_{1}$ and $C_{2}$ such that if $r\leq\rho/2$, $$\left(\int_{B(0,\rho)}v^{2}(x,0)\right)^{1+C_{1}\log\frac{\rho}{r}}\leq e^{C_{2}R^{2}\log\frac{\rho}{r}}\int_{B(0,r)}v^{2}(x,0).\label{eq:true doubling}$$ Taking $r=\rho\lambda^{2}/2$, we see that there are constants $J$ and $K$ so that $$\left(\int_{B(0,\rho)}v^{2}(x,0)\right)^{K}\leq e^{JR^{2}/2}\int_{B(0,\rho\lambda^{2}/2)}v^{2}(x,0).$$ This implies, after undoing the changes of variable, $$\left(\int_{B(x_{0},\rho\lambda R)}u^{2}(x,0)\right)^{K}\leq\lambda^{-K}e^{JR^{2}}\int_{B(x_{0},\rho\lambda R/2)}u^{2}(x,0).\label{eq:doubling}$$ We now use a chain-of-balls argument similar to that of [@MR040]. Let $x_{k+1}=(1-\frac{\rho\lambda^{2}}{8})x_{k}$ for $k=0,1,2,\ldots$. Then by (\[eq:doubling\]),

$$\left(\int_{B(x_{k+1},\rho\lambda^{2}\left|x_{k+1}\right|/8)}u^{2}(x,0)\right)^{K}\leq\lambda^{-K}e^{J\left|x_{k}\right|^{2}}\int_{B(x_{k},\rho\lambda^{2}\left|x_{k}\right|/8)}u^{2}(x,0)\,\,\,\,\,\,\,\, k=0,1,\ldots$$ Let $m=[\log\left|x_{0}\right|/\log\frac{8}{8-\rho\lambda^{2}}]$ then $\left|x_{m}\right|\sim1$, hence $$\int_{B(x_{m},\rho\lambda^{2}\left|x_{m}\right|/8)}u^{2}(x,0)\geq\lambda^{K}e^{-J\left|x_{m}\right|^{2}}\left(\int_{B(e_{1},\rho\lambda^{2}/8)}u^{2}(x,0)\right)^{K}\geq1.$$ (we have used (\[eq:trivia\]) in the last inequality.) It follows that $$\int_{B(x_{0},\rho\lambda^{2}\left|x_{0}\right|/8)}u^{2}(x,0)\geq\lambda^{\frac{K^{m+1}-K}{K-1}}e^{-\frac{J(K^{m}-1)}{K-1}\left|x_{0}\right|^{2}},$$ which, by the choice of $m$, implies $$\int_{B(x_{0},\rho\lambda\left|x_{0}\right|/2)}u^{2}(x,0)\geq e^{-C_{s}\left|x_{0}\right|^{s}}$$ for some positive constants $s$ and $C_{s}$. The same inequality holds for $u(\cdot,t)$ if $t$ is small so that (\[eq:trivia\]) holds.

### Second step.

We now use (\[eq:weak lower\]) and another Carleman inequality to prove (\[eq:lower\]). Let $\psi\in C_{c}^{\infty}(0,T)$ be a positive bump function satisfying $$\psi(t)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }t\in[0,\frac{T}{8}]\cup[\frac{7T}{8},T]\\
4 & t\in[\frac{T}{4},\frac{3T}{4}]\end{array}.\end{cases}$$ Let $\delta\in(1,1+\epsilon/2)$, where $\epsilon$ is the constant in the hypothesis of Theorem 1.1. Let $$S_{R,T}:=\left\{ (x,t):R^{1/\delta}\leq\left|x\right|\leq R,T/8\leq t\leq7T/8\right\} .$$

\[lem:lower 1\]Let $G(x,t)=e^{\varphi(x,t)}$ where $\varphi(x,t)=E_{1}R(T-t)\left|x\right|+E_{2}\left|x-R\psi(t)e_{1}\right|^{2}$. Here $E_{1}\gtrsim T^{-2},E_{2}\gtrsim1$ are constants that may depend on $R$, but $E_{1}/E_{2}\geq100/T$ is a fixed constant. Then if $R\geq R_{0}=R_{0}(E_{1}/E_{2},T,\lambda)$, $$\begin{aligned}
 &  & E_{1}^{3}R^{2}\int_{\mathbb{R}_{+}^{n+1}}v^{2}Gdxdt+E_{2}\int_{\mathbb{R}_{+}^{n+1}}\left|\nabla v\right|^{2}Gdxdt\lesssim\int_{\mathbb{R}_{+}^{n+1}}\left|Pv\right|^{2}Gdxdt,\end{aligned}$$ for any $v\in C_{c}^{\infty}(S_{R,T})$. The implicit constant depends only on $T$ and $\lambda$.

We give a proof of this lemma in the Appendix. Note that in contrast to Lemma \[lem:upper 1\], here the main term in $\varphi$ is $E_{1}R(T-t)\left|x\right|$, as $E_{1}\gg E_{2}$. The use of the shift $x-R\psi(t)e_{1}$ originates in a Carleman inequality for Schrödinger equations proved in [@MR2273975] (see their Lemma 3.1)

The next proposition, a corollary of this lemma, is the basis of our bootstrap argument.

\[pro:iteration\]Let $u$ be as in Theorem 1.1. Suppose that for some $s\geq2$, there exist $C_{s}>0$ such that$$\int_{T/4}^{3T/4}\int_{R\leq\left|x\right|\leq2R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{s}R^{s})$$ for all $R\geq C_{s}$. Let $s_{1}=\max\left\{ 2,\frac{s-1}{\delta}+1\right\} $, where $1<\delta<1+\frac{\epsilon}{2}$. Then there is $C_{s_{1}}>0$ such that

$$\int_{0}^{T}\int_{R-1\leq\left|x\right|\leq R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{s_{1}}R^{s_{1}})$$ for all $R\geq C_{s_{1}}$.

Let $v(x,t)=u(x,t)\theta(x,t)$ where $\theta(x,t)=\theta_{1}(x)\theta_{2}(x-R\psi(t)e_{1})$, with $\psi$ defined as above, and $$\theta_{1}(x)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }\left|x\right|<R^{1/\delta}\mbox{ or }\left|x\right|>cR\\
1 & \mbox{ if }R^{1/\delta}+1\leq\left|x\right|\leq cR-1\end{array}\end{cases}$$ $$\theta_{2}(x)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }\left|x\right|<2R\\
1 & \mbox{ if }\left|x\right|>3R,\end{array}\end{cases}$$ with $c=2^{-11}$. Clearly, $\mbox{supp}(v)\subset S_{R,T}$.

We have$$\begin{aligned}
\left|Pv\right| & \leq & C\left(\left|v\right|+\left|\nabla v\right|\right)+\left|u\right|\left(C\left|\nabla\theta\right|+\left|\partial_{t}\theta\right|+\left|\Delta\theta\right|\right)+2\lambda^{-1}\left|\nabla u\right|\left|\nabla\theta\right|\\
 & \leq & C\left(\left|v\right|+\left|\nabla v\right|\right)+C'\left(\left|u\right|+\left|\nabla u\right|\right)\chi_{E}\end{aligned}$$ where $E=\mbox{supp}\nabla\theta$ and $C'\leq\lambda^{4}/T$.

Applying the previous Carleman inequality to $v$, we get$$\int_{0}^{T}\int_{\mathbb{R}^{n}}(v^{2}+\left|\nabla v\right|^{2})Gdxdt\lesssim\int_{E}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt.$$ Since$$\begin{aligned}
\inf_{\begin{array}{cc}
16R^{1/\delta}\leq\left|x\right|\leq32R^{1/\delta}\\
T/4\leq t\leq3T/4\end{array}}\left\{ G(x,t)\right\}  & \geq & \exp\left(4E_{1}TR^{1+\frac{1}{\delta}}+E_{2}\left(4R-32R^{1/\delta}\right)^{2}\right),\end{aligned}$$ if $E_{1}\geq2\cdot16^{s}C_{s}R^{\frac{s-1}{\delta}-1}$ and $E_{1}/E_{2}=256/T$ then $$\begin{aligned}
\int_{0}^{T}\int_{\mathbb{R}^{n}}(\left|v\right|^{2}+\left|\nabla v\right|^{2})Gdxdt & \gtrsim & \exp\left(4E_{1}TR^{1+\frac{1}{\delta}}+E_{2}\left(4R-32R^{1/\delta}\right)^{2}-16^{s}C_{s}R^{s/\delta}\right)\\
 & \geq & \exp\left(3E_{1}TR^{1+\frac{1}{\delta}}+16E_{2}R^{2}\right)=:\Sigma\end{aligned}$$ The set $E$ is contained in the union of $\{R^{1/\delta}\leq\left|x\right|\leq R^{1/\delta}+1\}$, $\{2R\leq\left|x-R\psi(t)e_{1}\right|\leq3R\}\cap\{\left|x\right|\leq cR\}$ and $\{cR-1\leq\left|x\right|\leq cR\}$. In $\{R^{1/\delta}\leq\left|x\right|\leq R^{1/\delta}+1\}$, $$G(x,t)\leq\exp\left(2E_{1}TR^{1+\frac{1}{\delta}}+E_{2}\left(4R+2R^{1/\delta}\right)^{2}\right)$$ and $\left|u\right|+\left|\nabla u\right|\lesssim e^{CR^{2/\delta}}$, hence$$\begin{aligned}
 &  & \int_{0}^{T}\int_{R^{1/\delta}\leq\left|x\right|\leq2R^{1/\delta}}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim\\
 &  & \exp\left(2E_{1}TR^{1+\frac{1}{\delta}}+16E_{2}R^{2}+20E_{2}R^{1+\frac{1}{\delta}}+CR^{2/\delta}\right)\ll\Sigma/4.\end{aligned}$$ In $\{2R\leq\left|x-R\psi(t)e_{1}\right|\leq3R\}\cap\{\left|x\right|\leq cR\}$, $$G(x,t)\leq\exp\left(c^{2}E_{1}TR^{2}+9E_{2}R^{2}\right)\leq\exp(10E_{2}R^{2})$$ hence $$\int_{0}^{T}\int_{2R\leq\left|x-R\psi(t)e_{1}\right|\leq3R,\left|x\right|\leq cR}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\ll\Sigma/4.$$ Thus, we conclude that $$\Sigma/4\leq\int_{0}^{T}\int_{cR-1\leq\left|x\right|\leq cR}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt.$$ Since in $\{cR-1\leq\left|x\right|\leq cR\}$, $G\leq\exp(25E_{2}R^{2})$, we obtain $$\int_{0}^{T}\int_{cR-1\leq\left|x\right|\leq cR}(\left|u\right|^{2}+\left|\nabla u\right|^{2})dxdt\geq\exp(-9E_{2}R^{2}).$$ Recall that we need $E_{1}\geq2\cdot16^{s}C_{s}R^{\frac{s-1}{\delta}-1}$ and $E_{1}\gtrsim T^{-2}$. With the minimum choice $E_{1}\sim\max\{1,R^{\frac{s-1}{\delta}-1}\}$, we obtain$$\int_{0}^{T}\int_{cR-1\leq\left|x\right|\leq cR}(\left|u\right|^{2}+\left|\nabla u\right|^{2})dxdt\geq\exp(-C_{s_{1}}R^{s_{1}}).$$ for large $R$. The proposition follows from this.

\[pro:lower\]Suppose $u$ satisfies the assumption of Theorem 1.1. If $u(\cdot,0)\not\equiv0$ then for any $T\leq1$, there exist $C_{2}=C_{2}(T,u)>0$ such that

$$\int_{0}^{T}\int_{R-1\leq\left|x\right|\leq R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{2}R^{2})$$ for all $R\geq C_{2}$.

This is a consequence of repeatedly applying the previous proposition. Let $s_{0}=s$ where $s$ is the exponent appeared in (\[eq:weak lower\]), and$$s_{k+1}=2+\left(\frac{s_{k}-1}{\delta}-1\right)_{+},\,\,\,\,\, k=1,2,3,\ldots$$ It is simple to check that there is $k_{0}$ such that $s_{k}=2$ for all $k\geq k_{0}$. Clearly, we can assume that on $[0,T]$, (\[eq:weak lower\]) holds. Let $a_{k}=T\left(\frac{1}{2}-2^{k-k_{0}-1}\right)$ and $b_{k}=T\left(\frac{1}{2}+2^{k-k_{0}-1}\right)$. Since $$\int_{a_{0}}^{b_{0}}\int_{R<\left|x\right|<2R}\left|u(x,t)\right|^{2}dxdt\gtrsim e^{-R^{s}},$$ the previous proposition (applied to the time interval $[a_{1},b_{1}]$) shows that $$\int_{a_{1}}^{b_{1}}\int_{R-1\leq\left|x\right|\leq R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{s_{1}}R^{s_{1}})\,\,\,\,\,\,\,\mbox{if }R\geq C_{s_{1}}$$ for some positive $C_{s_{1}}$. Induction then shows that for any $k$, there is $C_{s_{k}}>0$ such that $$\int_{a_{k}}^{b_{k}}\int_{R-1\leq\left|x\right|\leq R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{s_{k}}R^{s_{k}})\,\,\,\,\,\,\,\mbox{if }R\geq C_{s_{1}}.$$ In particular when $k=k_{0}$ we obtain $$\int_{0}^{T}\int_{R-1\leq\left|x\right|\leq R}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim e^{-C_{2}R^{2}}.$$

Proof of Theorem 1.1
--------------------

1\. Suppose otherwise $u\not\equiv0$. We can assume without loss of generality that $u(\cdot,0)\not\equiv0$. (if not, we can translate to a time $0<s<1$ such that $u(\cdot,s)\not\equiv0$. The bounds $\left|u(x,s)\right|\lesssim e^{-M\left|x\right|^{2}}$ for all $M$, follows from (\[eq: upperbound\])). But then we are in position to apply Proposition \[pro:lower\], and obtain a lower bound that contradicts the upper bound of Proposition \[pro:upper\]. Thus, we must have $u\equiv0$.

2\. Let $T=\rho^{2}/8N$. Inspecting the proof of Proposition \[pro:upper\], we see that to obtain the upper bound$$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{-M\left|x\right|^{2}}\mbox{ in }(B_{7R}\backslash B_{6R})\times[0,T/4],$$ for some $M\geq2N$, it suffices to have $$\int_{B(x,1)}u^{2}(y,0)\leq e^{-2M\left|x\right|^{2}}$$ for all $x\in B_{2MR}\backslash B_{R/2}$. Hence, in order to avoid contradiction with the lower bound (\[eq:lower\]), we must have $$\sup_{x\in B_{2MR}\backslash B_{R/2}}\int_{B(x,1)}u^{2}(y,0)\geq e^{-4M^{2}R^{2}},$$ if $M\geq2\max\{N,C_{2}\}$ (here $C_{2}$ is the constant appears in Proposition \[pro:lower\]). This and (\[eq:doubling\]) together with a chain-of-balls argument shows that $$\inf_{x\in B_{MR}\backslash B_{R}}\int_{B(x,\rho\lambda R)}u^{2}(y,0)\geq e^{-M_{1}R^{2}},$$ for some $M_{1}>0$. Combining this with the doubling inequality (\[eq:true doubling\]), we obtain

$$\inf_{x\in B_{MR}\backslash B_{R}}\int_{B(x,1)}u^{2}(y,0)\geq e^{-M_{2}R^{2}\log R}.$$ These estimates prove the second part of the theorem.

As the cutoff functions used in the proof of Theorem 1.1 are radial, the same results and proofs apply to solutions of $\left|Pu\right|\lesssim\left|u\right|+\left|\nabla u\right|$ in $\left(\mathbb{R}^{n}\backslash B_{R}\right)\times[0,1].$

Proof of theorem 1.2
====================

The proof of Theorem 1.2 is very similar to that of Theorem 1.1, using anisotropic Carleman inequalities. We use the notation $x=(x_{1},x')$.

Upper bound
-----------

For the first step, the same argument as in section 2.1.1 shows that for all $M>0$ $$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{C\left|x\right|^{2}-Mx_{1}^{2}}\mbox{ for all }x\in\mathbb{R}_{+}^{n},\label{eq:thm 2 upper 1}$$ if $0\leq t\lesssim M^{-1}$(here $C$ is the constant in the statement of Theorem 1.2. Now we can only rescale with $R\sim x_{1}$, resulting in the weaker bound.)

For the second step, we will need the next lemma, which is inspired by a Carleman inequality in [@MR1979722]. To ease notations, we will assume that $a_{\infty}^{1j}=0$ for $j\ne1$ where $a_{\infty}^{ij}=\lim_{x\rightarrow\infty}a^{ij}(x,t)$. Otherwise, we will need to replace $\varphi$ below by$$\tilde{\varphi}(x,t)=\varphi(x_{1},Bx',t)$$ where $B$ is a positively definite, symmetric $(n-1)\times(n-1)$-matrix, satisfying $\sum_{j\ne1}B^{ij}a_{\infty}^{1j}=0$ for all $i=2,3,\ldots,n$. The reader can check that the conclusion of the lemma holds with such a modification of $\varphi$. (we only use $a_{\infty}^{1j}=0$ to control the term $I_{4}$ in the proof.)

Let $\epsilon$ be the constant in the hypothesis of Theorem 1.2. Let $G(x,t)=e^{\varphi(x,t)}$ where $$\varphi(x,t)=-\frac{\lambda\left|x'\right|^{2}}{8s}+\frac{c(S^{\alpha}-s^{\alpha})}{s^{\alpha}}x_{1}+bs.$$ Here $0\leq c\leq R^{1+\epsilon/8}$, $\alpha$ and $b\leq\alpha/4$ are large fixed constants, $s$ is the translated time variable $s=t+1$, and $S=T+1$. Then for large $R$, for any $v\in C_{c}^{\infty}(\left\{ R\leq x_{1}\leq R^{1+\epsilon/8}\right\} \times[0,T])$, $$\begin{aligned}
 &  & \frac{1}{16}\int_{0}^{T}\int_{\mathbb{R}_{+}^{n}}(cRv^{2}+b\left|\nabla v\right|^{2})Gdxdt\leq\int_{0}^{T}\int_{\mathbb{R}_{+}^{n}}\left|Pv\right|^{2}Gdxdt+\int_{\mathbb{R}_{+}^{n}}\left\Vert \nabla v(x,T)\right\Vert ^{2}Gdx\\
 &  & +\int_{\mathbb{R}_{+}^{n}}(\left|x'\right|^{2}+R^{2+\epsilon})v^{2}(x,0)G(x,0)dx+\int_{\mathbb{R}_{+}^{n}}(\left|x'\right|^{2}+R^{2+\epsilon})v^{2}(x,T)G(x,T)dx\end{aligned}$$

We give a proof of this lemma in the Appendix.

\[pro:upper 1.2\]Suppose that $$\left|u(x,t)\right|+\left|\nabla u(x,t)\right|\lesssim e^{C\left|x\right|^{2}-2^{\alpha}x_{1}^{2}}\mbox{ \,\,\,\,\,\,\, }\forall(x,t)\in\mathbb{R}_{+}^{n}\times[0,T],\label{eq: thm 2 upper}$$ Let $d=\frac{2^{\alpha+1}(T+2)}{\alpha T}$, where $\alpha$ is as in the previous lemma. Then for any $M>0$ we have

$$\int_{0}^{T/2}\int_{dR<x_{1}<2dR,\left|x'\right|<R}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)dxdt\lesssim e^{-MR^{2}}$$

Let $$v(x,t)=u(x,t)\theta(x)\mbox{ where }\theta(x)=\theta_{1}(x)\theta_{2}(x),$$ and $$\theta_{1}(x)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }x_{1}<R-1\mbox{ or }x_{1}>MR+1\\
1 & \mbox{ if }R<x_{1}<MR\end{array}\end{cases}$$ $$\theta_{2}(x)=\begin{cases}
\begin{array}{cc}
1 & \mbox{ if }\left|x'\right|<r\\
0 & \mbox{ if }\left|x'\right|>r+1\end{array}\end{cases}$$ We will now apply the previous lemma with $c=MR$, to the function $v$ and get$$\begin{aligned}
 &  & \int_{0}^{T}\int_{\mathbb{R}_{+}^{n}}(v^{2}+\left|\nabla v\right|^{2})Gdxdt\lesssim\int_{E}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt+\int_{\mathbb{R}_{+}^{n}}\left\Vert \nabla v(x,T)\right\Vert ^{2}Gdx\\
 &  & +\int_{\mathbb{R}_{+}^{n}}(\left|x'\right|^{2}+R^{2+\epsilon})v^{2}(x,0)G(x,0)dx+\int_{\mathbb{R}_{+}^{n}}(\left|x'\right|^{2}+R^{2+\epsilon})v^{2}(x,T)G(x,T)dx\end{aligned}$$ where $E=\mbox{supp}\nabla\theta\times[0,T]$.

Using (\[eq: thm 2 upper\]) and the decay of $u(\cdot,0)$, we can check easily that the last three integrals in the right hand side are bounded by $R^{2+\epsilon}$, and $$\int_{\left\{ r<\left|x'\right|<r+1,R<x_{1}<MR\right\} \times[0,T]}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\rightarrow0\,\,\,\mbox{as }r\rightarrow\infty.$$ In $MR<x_{1}<MR+1$, $G(x,t)\leq e^{-\frac{\lambda\left|x'\right|^{2}}{8}+2^{\alpha}x_{1}^{2}+2b}$. Hence, because of the bound (\[eq: thm 2 upper\])

$$\int_{0}^{T}\int_{MR<x_{1}<MR+1}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim1.$$ Furthermore,$$\int_{0}^{T}\int_{R-1<x_{1}<R}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim e^{2^{\alpha}MR^{2}}.$$ Thus, we conclude that $$\begin{aligned}
\int_{0}^{T}\int_{\mathbb{R}_{+}^{n}}\left(\left|v\right|^{2}+\left|\nabla v\right|^{2}\right)Gdxdt & \lesssim & e^{2^{\alpha}MR^{2}}.\end{aligned}$$ As in $\{x:dR<x_{1}<2dR,\left|x'\right|<R\}$, $u=v$ and $G(x,t)\geq e^{(2^{\alpha}+1)MR^{2}}$, it follows that$$\int_{0}^{T/2}\int_{dR<x_{1}<2dR,\left|x'\right|<R}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)dxdt\lesssim e^{-MR^{2}}.$$

Using the inequality (\[eq:doubling\]) and a chain-of-balls argument, we can actually take $d$ to be any positive number.

Lower bound
-----------

Assuming $u(\cdot,0)\not\equiv0$, the same argument as in section 2.2 gives the lower bound$$\int_{B(Re_{1},\rho\lambda R)}\left|u(x,t)\right|^{2}dxdt\gtrsim e^{-R^{s}}\,\,\,\,\,\forall t\in[0,T],\,\,\,\forall R\gtrsim1$$ for some $T\leq1$.

For the second step, we will need another Carleman inequality. Let $\delta\in(1,1+\epsilon/2)$ where $\epsilon$ is the constant in the hypothesis of Theorem 1.2, and $$S_{R,T}:=\left\{ (x,t)\in\mathbb{R}_{+}^{n}:R^{1/\delta}\leq x_{1}\leq R,T/8\leq t\leq7T/8\right\} .$$

Let $G(x,t)=e^{\varphi(x,t)}$ where $$\varphi(x,t)=-\frac{\lambda\left|x'\right|^{2}}{8t}+E_{1}R\frac{(T^{\alpha}-t^{\alpha})}{t^{\alpha}}x_{1}+E_{2}\left(x_{1}-R\psi(t)\right)^{2}+bE_{2}t$$ where $\alpha$ and $b$ are suitable fixed positive constants, $E_{1},E_{2}\gtrsim1$ are large constants that may depend on $R$, but $E_{1}/E_{2}$ is a large fixed constant independent of $R$. Then for large $R$, $$\begin{aligned}
 &  & \int_{\mathbb{R}_{+}^{n+1}}v^{2}Gdxdt+\int_{\mathbb{R}_{+}^{n+1}}\left|\nabla v\right|^{2}Gdxdt\leq\int_{\mathbb{R}_{+}^{n+1}}\left|Pv\right|^{2}Gdxdt,\end{aligned}$$ for any $v\in C_{c}^{\infty}(S_{R,T})$. Here and $\psi$ is as in section 2.2.

We will omit the proof of this lemma as it is almost the same as that of Lemma 4.1, except for the important fact that $E_{1}R\frac{(T^{\alpha}-t^{\alpha})}{t^{\alpha}}x_{1}$ is now the dominating term. (This is similar to the relationship between Lemma \[lem:upper 1\] and Lemma \[lem:lower 1\])

\[pro:iteration 2\]Let $u$ and $P$ be as in Theorem 1.2. Suppose that for some $s\geq2$, there are constants $R_{s},C_{s}>0$ such that$$\begin{aligned}
\int_{T/4}^{3T/4}\int_{\begin{array}{c}
R\leq x_{1}\leq2R\\
\left|x'\right|\leq C_{s}R^{s/2}\end{array}}(u^{2}+\left|\nabla u\right|^{2})dxdt & \gtrsim & \exp(-C_{s}R^{s})\end{aligned}$$ for all $R\geq R_{s}$. Let $s_{1}=\max\left\{ 2,\frac{s-1}{\delta}+1\right\} $ for some $1<\delta<1+\frac{\epsilon}{2}$. Then there is $R_{s_{1}},C_{s_{1}}$ such that

$$\int_{0}^{T}\int_{\begin{array}{c}
R\leq x_{1}\leq2R\\
\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}\end{array}}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{s_{1}}R^{s_{1}})$$ for all $R\geq R_{s_{1}}$.

Let $$v(x,t)=u(x,t)\theta(x,t)\mbox{ where }\theta(x,t)=\theta_{1}(x)\theta_{2}(x_{1}-R\psi(t))\theta_{3}(x'),$$ where $\psi$ is defined as before, and $$\theta_{1}(x)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }x_{1}<R^{1/\delta}\mbox{ or }x_{1}>cR\\
1 & \mbox{ if }R^{1/\delta}+1\leq x_{1}\leq cR-1\end{array}\end{cases}$$ $$\theta_{2}(r)=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }r<2R\\
1 & \mbox{ if }r>3R,\end{array}\end{cases}$$

$$\theta_{3}(x')=\begin{cases}
\begin{array}{cc}
0 & \mbox{ if }\left|x'\right|>C_{s_{1}}R^{s_{1}/2}+1\\
1 & \mbox{ if }\left|x'\right|<C_{s_{1}}R^{s_{1}/2},\end{array}\end{cases}$$ where $c$ and $C_{s_{1}}$ are positive constants to be choosen. It is clear that $\mbox{supp}(v)\subset S_{R,T}$.

Applying the previous Carleman inequality to $v$, as before we get$$\int_{\mathbb{R}_{+}^{n+1}}(v^{2}+\left|\nabla v\right|^{2})Gdxdt\lesssim\int_{E}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt,$$ where $E=\mbox{supp}\nabla\theta$.

Because in the set $\{x:10^{\alpha}R^{1/\delta}\leq x_{1}\leq2\cdot10^{\alpha}R^{1/\delta},\left|x'\right|\leq C_{s}(10^{\alpha}R^{1/\delta})^{s/2}\}\times[\frac{T}{4},\frac{3T}{4}]$,$$\begin{aligned}
G(x,t) & \geq & \exp\left(-D_{s}^{'}R^{s/\delta}+10^{\alpha}E_{1}R^{1+\frac{1}{\delta}}+E_{2}\left(4R-2\cdot10^{\alpha}R^{1/\delta}\right)^{2}\right)\end{aligned}$$ we have, $$\begin{aligned}
\int_{\mathbb{R}_{+}^{n+1}}(\left|v\right|^{2}+\left|\nabla v\right|^{2})Gdxdt & \gtrsim & \exp\left(10^{\alpha}E_{1}R^{1+\frac{1}{\delta}}+E_{2}\left(4R-2\cdot10^{\alpha}R^{1/\delta}\right)^{2}-C_{s}R^{s/\delta}-D_{s}^{'}R^{s/\delta}\right)\\
 & \gtrsim & \exp\left(9^{\alpha}E_{1}R^{1+\frac{1}{\delta}}+16E_{2}R^{2}\right)=:\Sigma\end{aligned}$$ if $E_{1}/R^{\frac{s-1}{\delta}-1}$ and $E_{1}/E_{2}$ are large enough.

In $\{R^{1/\delta}\leq x_{1}\leq R^{1/\delta}+1\}$, $$G(x,t)\leq\exp\left(-\frac{\lambda\left|x'\right|^{2}}{8}+8^{\alpha}E_{1}R^{1+\frac{1}{\delta}}+16E_{2}R^{2}\right)$$ so using the bound (\[eq: thm 2 upper\]) we get $$\int_{0}^{T}\int_{R^{1/\delta}\leq x_{1}\leq2R^{1/\delta}}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim\exp\left(8^{\alpha}E_{1}R^{1+\frac{1}{\delta}}+16E_{2}R^{2}\right)\ll\Sigma.$$ In $\{2R\leq\left|x_{1}-R\psi(t)\right|\leq3R\}\cap\{x_{1}\leq cR\}$, $$G(x,t)\leq\exp\left(-\frac{\lambda\left|x'\right|^{2}}{8}+c8^{\alpha}E_{1}R^{2}+9E_{2}R^{2}\right)$$ Hence, if $c$ is chosen to be small enough, $$\int_{0}^{T}\int_{2R\leq\left|x_{1}-R\psi(t)\right|\leq3R,x_{1}\leq cR}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim\exp\left(10E_{2}R^{2}\right)\ll\Sigma.$$ In $\{C_{s_{1}}R^{s_{1}/2}\leq\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}+1\}$, $$G(x,t)\leq\exp(-\lambda C_{s_{1}}^{2}R^{s_{1}}/8+c8^{\alpha}E_{1}R^{2}+16E_{2}R^{2})$$ Note that by our choice of $E_{1}$ and $E_{2}$, $E_{1}R^{2}\sim E_{2}R^{2}\sim R^{s_{1}}$, so if we choose $C_{s_{1}}$ big enough, $$\int_{0}^{T}\int_{x_{1}<cR,C_{s_{1}}R^{s_{1}/2}\leq\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}+1}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt\lesssim1.$$ Thus, we conclude that $$\Sigma\lesssim\int_{0}^{T}\int_{cR<x_{1}<cR+1,\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}}\left(\left|u\right|^{2}+\left|\nabla u\right|^{2}\right)Gdxdt.$$ Since in $\{cR<x_{1}<cR+1,\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}\}$, $G\leq\exp(KR^{s_{1}})$, we obtain $$\int_{0}^{T}\int_{\begin{array}{c}
cR\leq x_{1}\leq cR+1\\
\left|x'\right|\leq C_{s_{1}}R^{s_{1}/2}\end{array}}(\left|u\right|^{2}+\left|\nabla u\right|^{2})dxdt\geq\exp(-KR^{s_{1}}).$$ The proposition follows immediately from this.

\[pro:lower 1.2\]Let $u$ and $P$ be as in Theorem 1.2. If $u(\cdot,0)\not\equiv0$ then then for any $T\leq1$, there exist $C_{2}=C_{2}(T,u)>0$ such that

$$\int_{0}^{T}\int_{\begin{array}{c}
R\leq x_{1}\leq2R\\
\left|x'\right|\leq C_{2}R\end{array}}(u^{2}+\left|\nabla u\right|^{2})dxdt\gtrsim\exp(-C_{2}R^{2})$$ for all $R\geq C_{2}$.

The proof is similar to that of Proposition \[pro:lower\], using Proposition \[pro:iteration 2\] instead of Proposition \[pro:iteration\]. We omit the details.

Using Proposition \[pro:upper 1.2\] (see also the remark after it) and Proposition \[pro:lower 1.2\], the proof of Theorem 1.2 is identical to that of Theorem 1.1. We omit the details.

appendix
========

Some auxiliary lemmas
---------------------

The first lemma is a standard estimate for solutions of parabolic inequalities, we refer to [@MR001].

\[basic L8 bound\]Suppose that in $\Omega^{*}:=B(0,2)\times[0,2R^{-2}]$ , the following inequality holds $$\left|Pv\right|\leq R^{2}\left|v\right|+R\left|\nabla v\right|.$$ Then $$\left\Vert v\right\Vert _{L^{\infty}(\Omega)}+\left\Vert \nabla v\right\Vert _{L^{\infty}(\Omega)}\leq C_{n}R^{c}\left\Vert v\right\Vert _{L^{2}(\Omega^{*})}$$ where $\Omega=B(0,1)\times[0,R^{-2}]$ and $c$ is a constant depending only on $n$.

The next two lemmas are from [@MR2198840] (see also [@MR2231129]).

\[Lemma\]For $\rho\in(0,1/2)$, there is constant $N_{\rho}>0$ such that if $$\left|Pv\right|\leq R^{2}\left|v\right|+R\left|\nabla v\right|$$ in $\Omega^{*}:=B(0,2)\times[0,2R^{-2}]$ then $$\int_{B(0,\rho)}v^{2}(x,0)dx\leq N_{\rho}\int_{B(0,2\rho)}v^{2}(x,t)dx$$ for all $$0\leq t\leq N_{\rho}^{-1}\min\left\{ R^{-2},1/\log\left(\frac{N_{\rho}\int_{\Omega^{*}}v^{2}(x,0)dxdt}{\int_{B(0,\rho)}v^{2}(x,0)dx}\right)\right\} .$$

 

\[Lmm doubl\]Suppose $v\in C_{c}^{\infty}(\mathbb{R}^{n})$ such that for some $C>1$, $$2a\int_{\mathbb{R}^{n}}\left|\nabla v\right|^{2}e^{-\left|x\right|^{2}/4a}dx+\frac{n}{2}\int_{\mathbb{R}^{n}}v^{2}e^{-\left|x\right|^{2}/4a}dx\leq C\int_{\mathbb{R}^{n}}v^{2}e^{-\left|x\right|^{2}/4a}dx,$$ for all $0<a\leq1/(12C)$. Then$$\int_{B(0,2r)}v^{2}dx\leq e^{32C}\int_{B(0,r)}v^{2}dx$$ for all $0\leq r\leq1/2$.

Proof of the Carleman inequalities
----------------------------------

In this section we will prove the Carleman inequalities that were used in the proofs of Theorems 1.1 and 1.2. We will use the following notations$$\begin{aligned}
 & \Delta v=\mbox{div}(A\nabla v)\\
 & \left\Vert \nabla v(x,t)\right\Vert =\left\langle A(x,t)\nabla v(x,t),\nabla v(x,t)\right\rangle ^{1/2}.\end{aligned}$$ We recall the following lemma of [@MR1971939] (see also [@MR2001174], [@MR2198840]).

\[Basic lmm\]Suppose $\sigma(t):\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ is a smooth function, $\alpha$ is a real number, $F$ and $G$ are differentible functions, $G$ positive. Then the following identity holds for $v\in C_{c}^{2}(\mathbb{R}^{n}\times[0,T])$$$2\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}w^{2}Gdxdt+\frac{1}{2}\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}v^{2}MGdxdt-\frac{\alpha}{2}\int_{\mathbb{R}_{+}^{n+1}}\sigma^{-\alpha}v^{2}\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)Gdxdt$$ $$+\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}\left[\left(\log\frac{\sigma}{\sigma'}\right)'+\frac{\partial_{t}G-\Delta G}{G}-F\right]\left\Vert \nabla v\right\Vert ^{2}Gdxdt+2\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}\left\langle D_{G}\nabla v,\nabla v\right\rangle Gdxdt$$ $$-\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}v\left\langle A\nabla v,\nabla F\right\rangle Gdxdt=2\int_{\mathbb{R}_{+}^{n+1}}\frac{\sigma^{1-\alpha}}{\sigma'}wPvGdxdt+\int_{\mathbb{R}^{n}\times\{T\}}\frac{\sigma^{1-\alpha}}{\sigma'}\left\Vert \nabla v\right\Vert ^{2}Gdx-$$ $$-\int_{\mathbb{R}^{n}\times\{0\}}\frac{\sigma^{1-\alpha}}{\sigma'}\left\Vert \nabla v\right\Vert ^{2}Gdx+\frac{1}{2}\int_{\mathbb{R}^{n}\times\{T\}}\frac{\sigma^{1-\alpha}}{\sigma'}v^{2}(F-\frac{\alpha\sigma'}{\sigma})Gdx-$$ $$-\frac{1}{2}\int_{\mathbb{R}^{n}\times\{0\}}\frac{\sigma^{1-\alpha}}{\sigma'}v^{2}(F-\frac{\alpha\sigma'}{\sigma})Gdx.$$ where$$w=\partial_{t}v-\left\langle A\nabla\log G,\nabla v\right\rangle +\frac{Fv}{2}-\frac{\alpha\sigma'}{2\sigma}v,$$ $$M=\left(\log\frac{\sigma}{\sigma'}\right)'F+\partial_{t}F+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)-\left\langle A\nabla\log G,\nabla F\right\rangle ,$$

and $$D_{G}^{ij}=a^{il}\partial_{kl}(\log G)a^{kj}+\frac{\partial_{l}(\log G)}{2}\left[a^{kj}\partial_{k}a^{il}+a^{ki}\partial_{k}a^{jl}-a^{kl}\partial_{k}a^{ij}\right]+\frac{1}{2}\partial_{t}a^{ij}.$$

We will first derive a corollary of this lemma which will be used to prove all of our Carleman inequalities. Letting $\alpha=0$ and $\sigma(t)=e^{t}$ in Lemma \[Basic lmm\], we obtain the following identity for $v\in C_{c}^{2}(\mathbb{R}^{n}\times[0,T])$,$$\begin{aligned}
 &  & 2\int_{\mathbb{R}_{+}^{n+1}}w^{2}Gdxdt+\frac{1}{2}\int_{\mathbb{R}_{+}^{n+1}}v^{2}MGdxdt-\int_{\mathbb{R}_{+}^{n+1}}v\left\langle A\nabla v,\nabla F\right\rangle Gdxdt\nonumber \\
 &  & +\int_{\mathbb{R}_{+}^{n+1}}\left\Vert \nabla v\right\Vert ^{2}\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)Gdxdt+2\int_{\mathbb{R}_{+}^{n+1}}\left\langle D_{G}\nabla v,\nabla v\right\rangle Gdxdt\nonumber \\
 & = & 2\int_{\mathbb{R}_{+}^{n+1}}wPvGdxdt+\int_{\mathbb{R}^{n}}\left\Vert \nabla v(x,T)\right\Vert ^{2}Gdx-\int_{\mathbb{R}^{n}}\left\Vert \nabla v(x,0)\right\Vert ^{2}Gdx\label{eq: iden1}\\
 &  & +\frac{1}{2}\int_{\mathbb{R}^{n}}v^{2}(x,T)FGdx-\frac{1}{2}\int_{\mathbb{R}^{n}}v^{2}(x,0)FGdx.\nonumber \end{aligned}$$ where $$M=\partial_{t}F+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)-\left\langle A\nabla F,\nabla\log G\right\rangle .$$ Note that if $\nabla F$ is differentiable, we can integrate by parts to obtain $$-\int_{\mathbb{R}_{+}^{n+1}}v\left\langle A\nabla v,\nabla F\right\rangle Gdxdt=\frac{1}{2}\int_{\mathbb{R}_{+}^{n+1}}v^{2}\Delta FGdxdt+\frac{1}{2}\int_{\mathbb{R}_{+}^{n+1}}v^{2}\left\langle A\nabla F,\nabla\log G\right\rangle Gdxdt.$$ Then this term can be combined with the second term of the left hand side. However, in our applications, $\nabla F$ might not be differentiable, so we approximate $F$ by some $C^{2}$ function $F_{0}$ and use the above identity with $F_{0}$ in place of $F$. Then, using Cauchy-Schwarz, we arrive at the following lemma.

\[lem:basic 2\]Suppose $v\in C_{c}^{2}(\mathbb{R}^{n}\times[0,T])$, then$$\begin{aligned}
 &  & \frac{1}{2}\int_{\mathbb{R}_{+}^{n+1}}v^{2}M_{0}Gdxdt+\int_{\mathbb{R}_{+}^{n+1}}\left[2\left\langle D_{G}\nabla v,\nabla v\right\rangle +\left\Vert \nabla v\right\Vert ^{2}\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)\right]Gdxdt\nonumber \\
 &  & -\int_{\mathbb{R}_{+}^{n+1}}v\left\langle A\nabla v,\nabla(F-F_{0})\right\rangle Gdxdt\leq\int_{\mathbb{R}_{+}^{n+1}}\left|Pv\right|^{2}Gdxdt+\int_{\mathbb{R}^{n}}\left\Vert \nabla v(x,T)\right\Vert ^{2}Gdx\nonumber \\
 &  & -\int_{\mathbb{R}^{n}}\left\Vert \nabla v(x,0)\right\Vert ^{2}Gdx+\frac{1}{2}\int_{\mathbb{R}^{n}}v^{2}(x,T)FGdx-\frac{1}{2}\int_{\mathbb{R}^{n}}v^{2}(x,0)FGdx.\label{eq:inequa 1}\end{aligned}$$ where $$M_{0}=\partial_{t}F+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)+\Delta F_{0}-\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle$$ and $$D_{G}^{ij}=a^{il}\partial_{kl}(\log G)a^{kj}+\frac{\partial_{l}(\log G)}{2}\left[a^{kj}\partial_{k}a^{il}+a^{ki}\partial_{k}a^{jl}-a^{kl}\partial_{k}a^{ij}\right]+\frac{1}{2}\partial_{t}a^{ij}.$$

We will now prove our Carleman inequalities using Lemma \[lem:basic 2\].

\[Proof of lemma 2.2\]As $\mbox{supp}v\subset\{R\leq\left|x\right|\leq R^{1+\epsilon/8}\}\times[0,T]$, we will assume that $R\leq\left|x\right|\leq R^{1+\epsilon/8}$ in all the computation below.

Since $\nabla^{2}\varphi\geq\mbox{Id}$, $\left|\nabla\log G\right|\leq R^{1+\epsilon/8},$ and $\left|\nabla a^{ij}(x,t)\right|\leq R^{-1-\epsilon}$, it follows that $D_{G}\geq\frac{\lambda^{2}}{2}\mbox{Id}$ for large $R$. To make the gradient term (i.e. the second term in (\[eq:inequa 1\])) positive, we will choose $F$ satisfying$$\left|\frac{\partial_{t}G-\Delta G}{G}-F\right|\leq\lambda^{4}/2,\label{eq:proof 1 eq 1}$$ so that $$2\left\langle D_{G}\nabla v,\nabla v\right\rangle +\left\Vert \nabla v\right\Vert ^{2}\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)\geq\frac{\lambda^{2}}{2}\left|\nabla v\right|^{2}.$$ Let $\varphi(x,t)=c(T-t)\left|x\right|+\left|x\right|^{2}$, then $$\begin{aligned}
\frac{\partial_{t}G-\Delta G}{G} & = & \partial_{t}\varphi-\Delta\varphi-a^{ij}\partial_{i}\varphi\partial_{j}\varphi\\
 & = & -c\left|x\right|-x_{j}\partial_{i}a^{ij}(x,t)\left(c(T-t)\left|x\right|^{-1}+2\right)-a^{ij}(x,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}\\
 &  & -a^{ij}(x,t)\left[\delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}\right],\end{aligned}$$ As the second term is of order $O(R^{-7\epsilon/8})$, if we let $$\begin{aligned}
F(x,t) & = & -c\left|x\right|+\frac{\lambda^{4}}{3}-a^{ij}(x,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}\\
 &  & -a^{ij}(x,t)\left[\delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}\right].\end{aligned}$$ then (\[eq:proof 1 eq 1\]) is satisfied. Moreover, $$-R^{2+\epsilon/4}\lesssim F\lesssim-R^{2},\,\,\,\,-\frac{\lambda^{4}}{2}\leq\frac{\partial_{t}G-\Delta G}{G}-F\leq-\frac{\lambda^{4}}{4}.$$ We have$$\begin{aligned}
\partial_{t}F(x,t) & = & -\partial_{t}a^{ij}(x,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}+2ca^{ij}(x,t)x_{i}x_{j}\left|x\right|^{-1}\left(c(T-t)\left|x\right|^{-1}+2\right)\\
 &  & -\partial_{t}\left\{ a^{ij}(x,t)\left[\delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}\right]\right\} .\end{aligned}$$ The second term on the right hand side is positive by ellipticity of $\{a^{ij}\}$. Noting that the last terms of $F$ and $\partial_{t}F$ are $O(R^{\epsilon/8})$, we get $$\begin{aligned}
\partial_{t}F+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right) & \geq & -\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)a^{ij}(x,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}\\
 &  & -\partial_{t}a^{ij}(x,t)a^{ij}(x,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}+O(R^{\epsilon/8})\\
 & \gtrsim & R^{2}.\end{aligned}$$ (note that $\left|\partial_{t}a^{ij}(x,t)\right|\leq C\leq\lambda^{5}/100$.)

For the approximation $F_{0}$ of $F$, we choose $$\begin{aligned}
F_{0}(x,t) & = & -c\left|x\right|+\frac{\lambda^{4}}{3}-a^{ij}(X,t)x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}\\
 &  & -a^{ij}(X,t)\left[\delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}\right],\end{aligned}$$ where $X=(2R,0,\ldots,0)$.

As $$\left|a^{ij}(x,t)-a^{ij}(X,t)\right|=O(R^{-7\epsilon/8}),\,\,\,\left|\nabla a^{ij}(x,t)\right|=O(R^{-1-\epsilon})$$ and $$\begin{aligned}
 &  & x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}=O(R^{2+\epsilon/4}),\,\,\,\,\,\nabla\left(x_{i}x_{j}\left(c(T-t)\left|x\right|^{-1}+2\right)^{2}\right)=O(R^{1+\epsilon/4})\\
 &  & \delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}=O(R^{\epsilon/8}),\,\,\,\,\,\\
 &  & \nabla\left(\delta_{ij}\left(c(T-t)\left|x\right|^{-1}+2\right)-c(T-t)x_{i}x_{j}\left|x\right|^{-3}\right)=O(R^{-1+\epsilon/8}),\end{aligned}$$ we have$$\nabla(F-F_{0})=O(R^{1-5\epsilon/8}).$$ Easy computation shows $$\Delta F_{0}=O(R^{\epsilon/4}).$$ Thus, $$M_{0}=\partial_{t}F+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)+\Delta F_{0}-\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle \gtrsim R^{2}.$$ Finally, we can use Cauchy-Schwarz to control the remaining term as follows $$\left|\int_{0}^{T}\int_{\mathbb{R}^{n}}v\left\langle A\nabla v,\nabla(F-F_{0})\right\rangle Gdxdt\right|\leq\frac{M_{0}}{4}\int_{\mathbb{R}^{n}}v^{2}Gdxdt+\frac{\lambda^{2}}{4}\int_{\mathbb{R}^{n}}\left|\nabla v\right|^{2}Gdxdt.$$ This show that the left hand side of (\[lem:basic 2\]) is greater than $$\frac{R^{2}}{4}\int_{\mathbb{R}^{n}}v^{2}Gdxdt+\frac{\lambda^{2}}{4}\int_{\mathbb{R}^{n}}\left|\nabla v\right|^{2}Gdxdt.$$ In our case, $F<0$ so the third and fourth terms in the right hand side of (\[lem:basic 2\]) are negative. Thus, the lemma is proved.

 

\[Proof of lemma 2.4\]As $\nabla^{2}\varphi\geq2E_{2}\mbox{Id}$, the first term in $D_{G}$ is at least $2\lambda^{2}E_{2}\mbox{Id}$. The middle three terms of $D_{G}$ are $O(E_{1}R^{1-\frac{1+\epsilon}{\delta}})$, and the last term is bounded by $C\leq\lambda^{4}$. Thus, $D_{G}\geq\lambda^{2}E_{2}\mbox{Id}$. To make the gradient term positive, we will chose $F$ satisfying$$\left|\frac{\partial_{t}G-\Delta G}{G}-F\right|\leq\lambda^{4}E_{2},$$ so that then $$2\left\langle D_{G}\nabla v,\nabla v\right\rangle +\left\Vert \nabla v\right\Vert ^{2}\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)\geq\lambda^{2}E_{2}\left|\nabla v\right|^{2}.$$ Let $\widetilde{x}=x-R\psi(t)e_{1}$. Then we have

$$\begin{aligned}
\frac{\partial_{t}G-\Delta G}{G} & = & \partial_{t}\varphi-\Delta\varphi-a^{ij}\partial_{i}\varphi\partial_{j}\varphi\\
 & = & -E_{1}R\left|x\right|-2E_{2}R\psi'(t)\left(x_{1}-R\psi(t)\right)\\
 &  & -\partial_{i}a^{ij}(x,t)\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\\
 &  & -a^{ij}(x,t)\left(E_{1}R(T-t)\frac{x_{i}}{\left|x\right|}+2E_{2}\widetilde{x_{i}}\right)\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\\
 &  & -a^{ij}(x,t)\left[-E_{1}R(T-t)\frac{x_{i}x_{j}}{\left|x\right|^{3}}+\delta_{ij}\left(E_{1}R(T-t)\left|x\right|^{-1}+2E_{2}\right)\right].\end{aligned}$$

Note that in $S_{R,T}$ we have$$\left|\nabla a^{ij}(x,t)\right|\leq\left\langle x\right\rangle ^{-1-\epsilon}\leq R^{-(1+\epsilon)/\delta},$$ hence$$\left|\partial_{i}a^{ij}(x,t)\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\right|\lesssim E_{1}R^{1-\frac{1+\epsilon}{\delta}}.$$ Thus, we choose$$\begin{aligned}
F(x,t) & = & -E_{1}R\left|x\right|-2E_{2}R\psi'(t)\left(x_{1}-R\psi(t)\right)+\lambda^{4}E_{2}/2\\
 &  & -a^{ij}(x,t)\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\left(E_{1}R(T-t)\frac{x_{i}}{\left|x\right|}+2E_{2}\widetilde{x_{i}}\right)\\
 &  & -a^{ij}(x,t)\left[-E_{1}R(T-t)\frac{x_{i}x_{j}}{\left|x\right|^{3}}+\delta_{ij}\left(E_{1}R(T-t)\left|x\right|^{-1}+2E_{2}\right)\right]\end{aligned}$$ Also, let

$$\begin{aligned}
F_{0}(x,t) & = & -E_{1}R\left|x\right|-2E_{2}R\psi'(t)\left(x_{1}-R\psi(t)\right)+\lambda^{4}E_{2}/2\\
 &  & -a^{ij}(X,t)\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\left(E_{1}R(T-t)\frac{x_{i}}{\left|x\right|}+2E_{2}\widetilde{x_{i}}\right)\\
 &  & -a^{ij}(X,t)\left[-E_{1}R(T-t)\frac{x_{i}x_{j}}{\left|x\right|^{3}}+\delta_{ij}\left(E_{1}R(T-t)\left|x\right|^{-1}+2E_{2}\right)\right]\end{aligned}$$

where $X=(2R^{1/\delta},0,\ldots,0)$.

In the support of $v$, $T-t\geq T/8$, and $\left|\tilde{x}\right|\leq5R$, so by ellipticity of $\left\{ a^{ij}\right\} $, $$\begin{aligned}
 &  & -a^{ij}\left(E_{1}R(T-t)\frac{x_{j}}{\left|x\right|}+2E_{2}\widetilde{x_{j}}\right)\left(E_{1}R(T-t)\frac{x_{i}}{\left|x\right|}+2E_{2}\widetilde{x_{i}}\right)\\
 &  & \leq-\lambda\left|E_{1}R(T-t)\frac{x}{\left|x\right|}-2E_{2}\tilde{x}\right|^{2}\lesssim-T^{2}E_{1}^{2}R^{2}.\end{aligned}$$ The other terms in $F$ are bounded by $E_{1}R^{2}$, $E_{2}R^{2}/T$, $E_{2}$, and $E_{1}TR^{1-\frac{1}{\delta}}$. Hence, for large $R$, $$F\lesssim-E_{1}^{2}T^{2}R^{2}\mbox{ and }F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)\gtrsim E_{1}^{2}E_{2}R^{2}$$ It is easy to check that $$\begin{aligned}
\partial_{t}F & = & O(E_{1}^{2}R^{2})\\
\Delta F_{0} & = & O(E_{1}^{2}R^{2-\frac{2}{\delta}})\end{aligned}$$ which is smaller than $F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)$ provided $E_{2}\gg1$. Using $\left|a^{ij}(x,t)-a^{ij}(X,t)\right|\lesssim R^{1-\frac{1+\epsilon}{\delta}}$ and $\left|\nabla\left(a^{ij}(x,t)-a^{ij}(X,t)\right)\right|=R^{-\frac{1+\epsilon}{\delta}}$ in $S_{R,T}$, we get $$\left|\nabla(F-F_{0})\right|\lesssim R^{3-\frac{2+\epsilon}{\delta}}E_{1}^{2}$$ hence $$\left|\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle \right|\lesssim R^{4-\frac{2+\epsilon}{\delta}}E_{1}^{3}\ll E_{1}^{2}E_{2}R^{2}$$ for large $R$, as $\delta<1+\frac{\epsilon}{2}$.

Putting together these estimates, we obtain $$M_{0}=\partial_{t}F+\Delta F_{0}+F\left(\frac{\partial_{t}G-\Delta G}{G}-F\right)-\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle \geq E_{1}^{2}E_{2}R^{2}.$$ Finally, since $$M_{0}E_{2}\gg R^{6-\frac{2(2+\epsilon)}{\delta}}E_{1}^{4}\gtrsim\left|\nabla(F-F_{0})\right|^{2},$$ we can control the remaining term by Cauchy-Schwarz, $$\left|\int_{\mathbb{R}_{+}^{n+1}}u\left\langle A\nabla u,\nabla(F-F_{0})\right\rangle Gdxdt\right|\leq\frac{M_{0}}{2}\int u^{2}Gdxdt+\frac{\lambda^{2}E_{2}}{8}\int\left|\nabla u\right|^{2}Gdxdt$$ Thus, the lemma is proved.

 

\[Proof of lemma 3.1\]As $\nabla^{2}\varphi\geq-\frac{\lambda}{8}\mbox{Id}$ and$$\left|\nabla\log G\right|\left|\nabla a^{ij}\right|=O(\left|x\right|^{-7\epsilon/8}),\,\,\,\,\left|\partial_{t}a^{ij}\right|\leq C\leq\lambda^{4}/100,$$ it follows that if $$H:=\frac{\partial_{t}G-\Delta G}{G}-F\geq\frac{1}{\lambda}$$ then the gradient term is positive. We have $$\begin{aligned}
\frac{\partial_{t}G-\Delta G}{G} & = & \frac{\lambda}{16s^{2}}\sum_{i,j\ne1}\left(2\delta_{ij}-\lambda a^{ij}(x,t)\right)x_{i}x_{j}-\frac{c\alpha S^{\alpha}x_{1}}{s^{\alpha+1}}+b\\
 &  & -a^{11}(x,t)\frac{c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}}-\sum_{j\ne1}a^{1j}(x,t)\frac{\lambda x_{j}}{2s}\frac{c(S^{\alpha}-s^{\alpha})}{s^{\alpha}}\\
 &  & -\partial_{i}(a^{ij}\partial_{j}\varphi).\end{aligned}$$ Since $c\leq x_{1}^{1+\epsilon/8}$, from the decay of $\nabla a^{ij}$ it follows that $\left|\partial_{i}(a^{ij}\partial_{j}\varphi)\right|\lesssim1.$ If we choose$$\begin{aligned}
F(x,t) & = & \frac{\lambda}{16s^{2}}\sum_{i,j\ne1}\left(2\delta_{ij}-\lambda a^{ij}(x,t)\right)x_{i}x_{j}-\frac{c\alpha S^{\alpha}x_{1}}{s^{\alpha+1}}\\
 &  & -a^{11}(x,t)\frac{c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}}-\sum_{j\ne1}a^{1j}(x,t)\frac{\lambda x_{j}}{2s}\frac{c(S^{\alpha}-s^{\alpha})}{s^{\alpha}},\end{aligned}$$ then for large $b$, $$2b\geq H\geq b/2$$ implying the positivity of the gradient term.

Consider four terms of $(H+\partial_{t})F$ corresponding to four terms of $F$.

1.  $I_{1}:=\left(\frac{1}{2}Hs-1\right)\frac{\lambda}{8s^{3}}\sum_{i,j\ne1}\left(2\delta_{ij}-\lambda a^{ij}(X,t)\right)x_{i}x_{j}-\sum_{i,j\ne1}\partial_{t}a^{ij}x_{i}x_{j}\geq b\left|x'\right|^{2}$\
    for large $b$.

2.  $$I_{2}:=\left(\frac{\alpha+1}{s}-H\right)\frac{c\alpha S^{\alpha}x_{1}}{s^{\alpha+1}}\geq\frac{c\alpha^{2}S^{\alpha}x_{1}}{4s^{\alpha+1}}\geq\frac{1}{8}c\alpha^{2}x_{1}$$ if $\alpha\geq4b$.

3.  $$\begin{aligned}
    I_{3}: & = & \left(\frac{2\alpha}{s}-H\right)\frac{a^{11}(x,t)c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}}+\frac{2a^{11}\alpha(S^{\alpha}-s^{\alpha})}{s^{\alpha+1}}-\\
     &  & -\partial_{t}a^{11}(x,t)\frac{c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}}\geq\frac{\lambda\alpha c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}},\end{aligned}$$

    \
    again if $\alpha\geq4b$.

4.  $$\begin{aligned}
    I_{4}: & = & -a^{1j}(x,t)\frac{\lambda cx_{j}}{2}\left(-\frac{(\alpha+1)S^{\alpha}}{s^{\alpha+2}}+\frac{1}{s^{2}}+\frac{H(S^{\alpha}-s^{\alpha})}{s^{\alpha+1}}\right)\\
     &  & -\partial_{t}a^{1j}(x,t)\frac{\lambda cx_{j}}{2}\frac{S^{\alpha}-s^{\alpha}}{s^{\alpha+1}}.\end{aligned}$$

    \
    Since we are assuming $a_{\infty}^{1j}=0$, $\left|a^{1j}(x,t)\right|\lesssim\left\langle x\right\rangle ^{-\epsilon}$, hence $$\left|I_{4}\right|\leq2^{\alpha}\alpha c\left|x'\right|\left\langle x\right\rangle ^{-\epsilon}+\left|\partial_{t}a^{1j}(x,t)\right|\frac{\lambda c\left|x'\right|}{2}\frac{S^{\alpha}-s^{\alpha}}{s^{\alpha+1}}$$ Recall that $c\leq R^{1+\epsilon/8}\leq x_{1}^{1+\epsilon/8}$, hence the first term is bounded by $\frac{1}{4}(I_{1}+I_{2})$. Also, by dilation, we can assume $\left|\partial_{t}a^{ij}\right|\leq C\ll1$, so that the second term is bounded by $\frac{1}{4}(I_{1}+I_{3})$. Thus, $$\left|I_{4}\right|\leq\frac{I_{1}}{2}+\frac{I_{2}+I_{3}}{4}.$$

From these estimates, we obtain $$(H+\partial_{t})F\geq\frac{I_{1}+I_{2}}{2}\geq\frac{1}{32}(b\left|x'\right|^{2}+c\alpha^{2}x_{1}).$$ As an approximation of $F$, we choose $$\begin{aligned}
F_{0}(x,t) & = & \frac{\lambda}{16s^{2}}\sum_{i,j\ne1}\left(2\delta_{ij}-\lambda a^{ij}(X,t)\right)x_{i}x_{j}-\frac{c\alpha S^{\alpha}x_{1}}{s^{\alpha+1}}\\
 &  & -a^{11}(X,t)\frac{c^{2}(S^{\alpha}-s^{\alpha})^{2}}{s^{2\alpha}}-\sum_{j\ne1}a^{1j}(X,t)\frac{\lambda x_{j}}{2s}\frac{c(S^{\alpha}-s^{\alpha})}{s^{\alpha}},\end{aligned}$$ where $X=(R,0,\ldots,0)$. Simple calculation shows that $$\left|\Delta F_{0}\right|\lesssim\left\langle x\right\rangle ^{-1-\epsilon}(\left|x'\right|+c)+1\lesssim1$$ and

$$\left|\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle \right|\lesssim\left\langle x\right\rangle ^{-1-\epsilon}\left(\left|x'\right|+c\right)^{3}+R^{-\epsilon}\left(\left|x'\right|+c\right)^{2}\lesssim R^{-\epsilon/2}(b\left|x'\right|^{2}+c\alpha^{2}x_{1}).$$ (the implicit constants depend on $\lambda$ but not on $R$).

It follows that for large $R$,

$$M_{0}=(\partial_{t}+H)F+\Delta F_{0}-\left\langle A\nabla(F-F_{0}),\nabla\log G\right\rangle \gtrsim b\left|x'\right|^{2}+c\alpha^{2}x_{1}.$$ We use Cauchy-Schwarz to control the remaining term $$\left|\int_{\mathbb{R}_{+}^{n+1}}u\left\langle A\nabla u,\nabla(F-F_{0})\right\rangle Gdxdt\right|\leq\frac{1}{4}\int u^{2}M_{0}Gdxdt+\frac{b}{4}\int\left|\nabla u\right|^{2}Gdxdt.$$ As $\left|F\right|\lesssim\left|x'\right|^{2}+R^{2+\epsilon}$, the lemma is proved.

[<span style="font-variant:small-caps;"> </span>]{}

[<span style="font-variant:small-caps;">Department of Mathematics, University of Chicago, 5734 S. University Ave., Chicago, IL 60637, USA</span>]{}

*E-mail address*: `tu@math.uchicago.edu`
