\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 189, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/189\hfil Backward uniqueness]
{Backward uniqueness for heat equations with coefficients of bounded
 variation in time}

\author[S. Tarama \hfil EJDE-2013/189\hfilneg]
{Shigeo Tarama} 

\address{Shigeo Tarama \newline
Lab. of Applied Mathematics, Osaka City University, Osaka 558-8585, Japan}
\email{starama@mech.eng.osaka-cu.ac.jp}

\thanks{Submitted February 27, 2013. Published August 28, 2013.}
\subjclass[2000]{35K05, 16D10}
\keywords{Heat equation; backward Cauchy problem; uniqueness}

\begin{abstract}
 Uniqueness of solutions to the backward
 Cauchy problem for heat equations with coefficients of bounded variation
 in time is shown through the Carleman estimate.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

We consider a heat operator in the time backward form
\begin{equation}\label{heat}
Lu=\partial_t+\sum_{j,k=1}^d\partial_{x_j}(a_{jk}(x,t)\partial_{x_j}u),
\end{equation}
with real bounded and measurable  coefficients $a_{jk}(x,t)$ on
$\mathbb{R}^d\times [0,T]$, for some $T>0$, satisfying
$a_{jk}(x,t)=a_{kj}(x,t)$ ($j,k=1,2,\dots,d$) and
\begin{equation}\label{elliptic}
 \sum_{j,k=1}^d a_{jk}(x,t)\xi_j\xi_k \ge D_0|\xi|^2
\end{equation}
for any $\xi\in \mathbb{R}^d$ with some positive $D_0$.

The Cauchy problem for $Lu=f$ on $\mathbb{R}^d\times [0,T]$  with Cauchy data on
$t=0$ is not well-posed. But the uniqueness of solutions to the Cauchy
problem is valid under  some conditions on the coefficients.
Since the work of Mizohata \cite{mz}, there are many works on this problem.
See for example the survey paper of  Vessella \cite{vess} and the papers
cited therein. But it seems that the backward uniqueness for heat
operators with  discontinuous coefficients  is not well studied.

We consider an operator whose coefficients
$a_{jk}(x,t)$ ($j,k=1,2,\dots,d$) are of bounded variation in $t$
uniformly with respect to $x\in \mathbb{R}^d$. That is, there exists a constant
$M\ge 0$ such that we have
\begin{equation}\label{hyp-2}
\sum_{l=1}^L\sup_{x \in \mathbb{R}^d}|a_{jk}(x,t_l)-a_{jk}(x,t_{l-1})|\le M
\end{equation}
for any partition of $[0,T]$, $t_0=0<t_1<\dots<t_L=T$,
which means that $a_{jk}(x,t)$ is a $C_b^0(\mathbb{R}^d)$-valued function on
$[0,T]$ with  bounded variation. Here $C_b^0(\mathbb{R}^d)$ is a space of bounded
and continuous functions on $\mathbb{R}^d$.
While we assume that $a_{jk}(x,t)$ are Lipschitz continuous in $x$ uniformly
with respect to  $t$, that is, we assume that we have, with some $L\ge 0$,
\begin{equation}\label{hyp-3}
\sum_{j,k=1}^d|a_{jk}(x,t)-a_{jk}(y,t)|\le L|x-y|
\end{equation}
for any $x,y\in \mathbb{R}^d$ and any $t\in[0.T]$.
 Under these conditions we show the following.

\begin{theorem}\label{thm1.1}
Assume that the coefficients  $a_{jk}(x,t)$ ($j,k=1,2,\dots,d$)
of  the operator \eqref{heat} are real, bounded and  symmetric and
satisfy \eqref{elliptic}, \eqref{hyp-2} and \eqref{hyp-3}.
Let $u(x,t)\in L^2([0,T],H^1(\mathbb{R}^d _x)) \cap C^0([0,T],L^2(\mathbb{R}^d _x))$ satisfy
$Lu\in L^2([0,T],L^2(\mathbb{R}^d _x))$,
\[
\|Lu(\cdot,t)\|\le C\|u(\cdot,t)\|_1\quad \text{almost all $t\in[0,T]$},
\]
and $u(x,0)=0$. Then we have $u(x,t)=0$ on $\mathbb{R}^d\times [0,T]$.
\end{theorem}

Here the spaces $L^2$ and $H^1$ and their norm $\|\cdot\|$ and
$\|\cdot\|_1$ are standard ones whose  definitions are given below.
For a Banach space $X$, we denote by $L^2([0,T],X)$ and $C^0([0,T],X)$
the space of $X$-valued square integrable functions and  the space of $X$-valued
continuous functions respectively.

\begin{remark} \label{rmk1} \rm
We note that $Lu\in L^2([0,T],L^2(\mathbb{R}_x ^d))$ implies
$\partial_t u(x,t)$ being in $L^2([0,T],H^{-1}(\mathbb{R}_x ^d))$.
 While $\partial_t u(x,t)$ being in $L^2([0,T],H^{-1}(\mathbb{R}_x ^d))$ and
$u(x,t)$ in $L^2([0,T],H^1(\mathbb{R}_x ^d))$ imply
 $u(x,t)\in  C^0([0,T],L^2(\mathbb{R}_x ^d))$
(see for example \cite[Theorem 1. \S 1.1 Ch. XVIII]{dl}).
Then the assumption $u(x,t)\in C^0([0,T],L^2(\mathbb{R}^d _x))$ follows from the
other assumptions.
\end{remark}

Theorem \ref{thm1.1} is shown by using the Carleman estimate. Here, in order
to indicate the principal idea of proof,  we show how to obtain the Carleman
estimate for a simple operator $\partial_t u+a(t)\partial_x^2u$.
We assume that the coefficient $a(t)$ satisfies that
$C_1\le a(t)\le C_2$
with positive $C_1,C_2$, and  that for any positive $\varepsilon$ there
exists $T_{\varepsilon}\in(0,\varepsilon]$ such that we have, for any
$h\in [0,T_{\varepsilon}]$,
\[
\int_0^{T_{\varepsilon}}|a(t+h)-a(t)|\,dt\le \varepsilon h.
\]
We remark that the second assumption is satisfied if $a(t)$ is of bounded
variation  and continuous at $t=0$.

We define the weight function $\psi_{1,\gamma}(t)$ by
$\psi_{1,\gamma}(t)=\gamma\int_t^{T_{\varepsilon}}e^{\psi_{\gamma}(t)}\,d\tau$,
 where
\[
\psi_{\gamma}(t)=\int_t^{T_{\varepsilon}}\frac{1}{\varepsilon}(1+
\gamma |a(\tau+1/\gamma)-a(\tau)|+\gamma\int_0^1|a(\tau+s/\gamma)
-a(\tau)|\,ds)\,d\tau.
\]

We note that $0\le \psi_{\gamma}(t)\le 5/2$  when $\gamma\ge 1/T_{\varepsilon}$.
Under these conditions, we show that there exist positive $\varepsilon$,
$\gamma_0$ and $C$ such that we have the estimate
\[
\int_0^{T_{\varepsilon}}(\gamma\|e^{2\psi_{1,\gamma}(t)}u\|^2
+\|e^{2\psi_{1,\gamma}(t)}\partial_xu\|^2)\,dt
\le C\int_0^{T_{\varepsilon}}\|e^{2\psi_{1,\gamma}(t)}
(\partial_tu+a(t)\partial_x^2u)\|^2\,dt
\]
for any $\gamma\ge \gamma_0$ and $u(x,t)$ satisfying $u(x,0)=0$ and
$u(x,T_{\varepsilon})=0$.
Plancherel's theorem implies that we have only to show the estimate
\begin{equation} \label{C-1}
(\gamma +\xi^2)\int_0^{T_{\varepsilon}}|e^{2\psi_{1,\gamma}(t)}u(t)|^2\,dt
\le C\int_0^{T_{\varepsilon}}|e^{2\psi_{1,\gamma}(t)}\hat{L}u(t)|^2\,dt
\end{equation}
for any $\xi\in\mathbb{R}$, $\gamma\ge \gamma_0$ and $u(t)$ satisfying $u(0)=0$ and
$u(T_{\varepsilon})=0$. Here $\hat{L}u=\dfrac{d}{dt}u-a(t)\xi^2u$.
By setting $u(t)=e^{-\psi_{1,\gamma}(t)}v(t)$, we see that the estimate
above is equivalent to
\begin{equation} \label{C-2}
(\gamma +\xi^2)\int_0^{T_{\varepsilon}}|v(t)|^2\,dt
\le C\int_0^{T_{\varepsilon}}|\tilde{L}v(t)|^2\,dt
\end{equation}
where $\tilde{L}=\frac{d}{dt}+\gamma e^{\psi_{\gamma}(t)}-a(t)\xi^2$.
In the following, we show \eqref{C-2} for a real valued $v(t)$ satisfying
$v(0)=0$ and $v(T_{\varepsilon})=0$.

We see from $v(0)=v(T_{\varepsilon})=0$ that
\[
-\int_0^{T_{\varepsilon}}v(t)\tilde{L}v(t)\,dt
=\int_0^{T_{\varepsilon}}(a(t)\xi^2-\gamma e^{\psi_{\gamma}(t)})(v(t))^2\,dt.
\]
Then, when  $C_1\xi^2\ge 2e^{5/2}\gamma$, we have
\[
\int_0^{T_{\varepsilon}}|v(t)||\tilde{L}v(t)|\,dt
\ge (C_1\xi^2/2)\int_0^{T_{\varepsilon}}|v(t)|^2\,dt.
\]
Hence we have
\[
C_3\int_0^{T_{\varepsilon}}|\tilde{L}v(t)|^2\,dt
\ge(\xi^2+\gamma)\int_0^{T_{\varepsilon}}|v(t)|^2\,dt
\]
with some $C_3$, when $C_1\xi^2e^{-5/2}/2\ge \gamma\geq 1$.

For the case where $C_1\xi^2e^{-5/2}/2\le \gamma$, we first  remark that
\[
\int_0^{T_{\varepsilon}}(\tilde{L}v(t))^2\,dt
=\int_0^{T_{\varepsilon}}\Bigl((v'(t))^2
+(\gamma e^{\psi_{\gamma}(t)}-a(t)\xi^2)^2(v(t))^2\Bigr)\,dt+I
\]
where\[
I=2\int_0^{T_{\varepsilon}}v'(t)(\gamma e^{\psi_{\gamma}(t)}-a(t)\xi^2)v(t)\,dt.
\]
To estimate $I$, we regularize $a(t)$ by $a_{\gamma}(t)
=\int_0^1a(t+s/\gamma)\,ds$.
Since $a_{\gamma}(t)=\gamma\int_t^{t+1/\gamma}a(s)\,ds$, we see that
\[
|a_\gamma(t)-a(t)|\le \int_0^1|a(t+s/\gamma)-a(t)|\,ds, \quad
|a'_\gamma(t)|\le \gamma|a(t+1/\gamma)-a(t)|.
\]
Note that $|a_\gamma(t)|\le C_2$.
We set $I=I_1+I_2$, where
\begin{gather*}
I_1=2\int_0^{T_{\varepsilon}}v'(t)(\gamma e^{\psi_{\gamma}(t)}
-a_\gamma (t)\xi^2)(v(t))\,dt,\\
I_2=2\int_0^{T_{\varepsilon}}v'(t)(a_\gamma (t)-a(t))\xi^2v(t))\,dt.
\end{gather*}
From $|a_\gamma(t)-a(t)|^2\le 2 C_2|a_\gamma(t)-a(t)|$ and Schwarz's inequality,
 we obtain
\[
|I_2|\le \int_0^{T_{\varepsilon}}|v'(t)|^2\,dt+2
C_2\int_0^{T_{\varepsilon}}\int_0^1|a(t+s/\gamma)-a(t)|\,ds\xi^4
|v(t)|^2\,dt.
\]
Note
$$
I_1=\int_0^{T_{\varepsilon}}(\gamma e^{\psi_{\gamma}(t)}-a_\gamma (t)\xi^2)
\frac{d}{dt}(v(t))^2\,dt,
$$
$v(0)=0$, $v(T_{\varepsilon})=0$, and
\[
-\frac{d}{dt} e^{\psi_{\gamma}(t)}=\frac{e^{\psi_{\gamma}(t)}}{\varepsilon}
(1+
\gamma |a(t+1/\gamma)-a(t)|+\gamma\int_0^1|a(t+s/\gamma)-a(t)|\,ds).
\]
By integrating by parts and $e^{\psi_{\gamma}(t)}\ge 1$,
\[
I_1\ge \int_0^{T_{\varepsilon}}\Bigl(\frac{\gamma}{\varepsilon}
\bigl(1+\gamma |a(t+1/\gamma)-a(t)|+\gamma\int_0^1|a(t+s/\gamma)-a(t)|\,ds\bigr)
+a'_\gamma(t)\xi^2 \Bigr)(v(t))^2\,dt.
\]
Hence, noting $|a'_\gamma(t)|\le \gamma|a(t+1/\gamma)-a(t)|$, we see that,
if $\gamma/\varepsilon\ge \xi^2$,
\[
I_1\ge \int_0^{T_{\varepsilon}} (\frac{\gamma}{\varepsilon}+
\frac{\gamma^2}{\varepsilon}\int_0^1|a(\tau+s/\gamma)-a(\tau)|\,ds)(v(t))^2\,dt.
\]
Then  from $I\ge I_1-|I_2|$ it follows that, if $\gamma/\varepsilon\ge \xi^2$,
\[
I \ge \int_0^{T_{\varepsilon}}\Bigl(
(\frac{\gamma}{\varepsilon}+(\frac{\gamma^2}{\varepsilon}
-2C_2\xi^4)\int_0^1|a(t+s/\gamma)-a(t)|\,ds\Bigr)
(v(t))^2\,dt -\int_0^{T_{\varepsilon}}|v'(t)|^2\,dt,
\]
from which we see that
\[
I \ge \int_0^{T_{\varepsilon}}\frac{\gamma}{\varepsilon}
(v(t))^2\,dt -\int_0^{T_{\varepsilon}}|v'(t)|^2\,dt,
\]
when $\gamma/\varepsilon\ge \xi^2$ and $\gamma^2/\varepsilon\ge 2C_2\xi^4$.

We have some positive $\varepsilon$ not depending on $\xi$ or on $\gamma$
such that, if $C_1\xi^2e^{-5/2}/2\le \gamma$, we have
$\gamma/\varepsilon\ge \xi^2$ and $\gamma^2/\varepsilon\ge 2C_2\xi^4$.
Therefore, with such $\varepsilon$, we see that
\[
\int_0^{T_{\varepsilon}}(\tilde{L}v(t))^2\,dt
\ge \int_0^{T_{\varepsilon}}\frac{\gamma}{\varepsilon}
(v(t))^2\,dt
\]
if   $\gamma\ge 1/T_{\varepsilon}$  and $\gamma\ge C_1\xi^2e^{-5/2}/2 $.
Then we have with some positive $C_4$,
\[
C_4\int_0^{T_{\varepsilon}}(\tilde{L}v(t))^2\,dt
\ge
(\gamma+\xi^2)\int_0^{T_{\varepsilon}}
(v(t))^2\,dt
\]
if $\gamma\ge 1/T_\varepsilon$ and $\gamma\ge C_1\xi^2e^{-5/2}/2 $.
Hence we obtain the  estimate \eqref{C-2}.

We remark that we need a more precise estimate than the estimate
above for the proof of Theorem \ref{thm1.1}.

In the next section we recall the properties of the Hardy-Littlewood
decomposition and  the properties of functions of bounded variation
for the preliminaries. We draw the Carleman estimate in the section 3.
Finally we give the proof of Theorem \ref{thm1.1} in the section 4.
In this study the author is inspired by the paper of
Del Santo and Pruzzi \cite{del}.

We denote the space of square integrable functions on
$\mathbb{R}^d$ by $L^2(\mathbb{R}^d)$.  The inner product in $L^2(\mathbb{R}^d)$ is given by
\[
(u,v)=\int_{\mathbb{R}^d}u(x)\overline{v(x)}\,dx
\]
and the norm by
 $\|v(\cdot)\|=\Big(\int_{\mathbb{R}^d}|v(x)|^2\,dx\Big)^{1/2}$.

The space $H^1(\mathbb{R}^d)$ consists of $u(x)\in L^2(\mathbb{R}^d)$ whose derivatives
 $\partial_{x_j}u(x)$ ($j=1,2,\dots,d$) belong also to $L^2(\mathbb{R}^d)$.
The norm $\|\cdot\|_1$ of   $H^1(\mathbb{R}^d)$ is given by
$ \|u(\cdot)\|_1=\sqrt{\|u(\cdot)\|^2+\sum_{j=1}^d\|\partial_{x_j}u(\cdot)\|^2}$.
We set $\|\nabla u\|^2=\sum_{j=1}^d\|\partial_{x_j}u\|^2$.

Let $C^\infty(\Omega)$ be the space of infinitely differentiable functions
on $\Omega$,  $W^{1,\infty}(\mathbb{R}^d)$ the space of bounded and Lipschitz
continuous functions on $\mathbb{R}^d$ with the norm
\[
\|u(\cdot)\|_{W^{1,\infty}}=\|u(\cdot)\|_{L^{\infty}}+\sum_{j=1}^d
\|\partial_{x_j}u(\cdot)\|_{L^{\infty}}.
\]
Here we denote by $\|u(\cdot)\|_{L^{\infty}}$ the essential supremum of $|u(x)|$ on $\mathbb{R}^d$.

We denote by $\hat{v}(\xi)$ the Fourier transform of $v(x)$ given by
\[
\int_{\mathbb{R}^d}e^{-i x\xi}v(x)\,dx,
\]
while the inverse Fourier transform of $w(\xi)$ is defined by
\[
\frac{1}{(2\pi)^d}\int_{\mathbb{R}^d}e^{i x\xi}w(\xi)\,d\xi.
\]
In the following, we use $C$ or $C$ with some suffix in order to denote
positive constants that may be different line by line.


\section{Preliminaries}
\subsection{The Littlewood-Payley decomposition}

We recall some properties of the Littlewood-Payley decomposition
and the related results referring to \cite{met}.

Let $\phi_0(\xi)\in C^{\infty}(\mathbb{R}^d)$ satisfy
$0\le \phi_0(\xi) \le 1$, $ \phi_0(\xi)=1$ for $|\xi|\le 11/10$ and
$ \phi_0(\xi)=0$ for $|\xi|\ge 19/10$.
We define $\phi_n(\xi)$ with $n=1,2,3,\dots$ by
\[
\phi_n(\xi)=\phi_0(\frac{\xi}{2^{n}})-\phi_0(\frac{\xi}{2^{n-1}}).
\]


For a function $\phi(\xi)$, we denote the Fourier multiplier with
$\phi(\xi)$ by $\phi$; that is,
$\phi v$ is the inverse Fourier transform of $\phi(\xi)\hat{v}(\xi)$.
We remark that
\begin{equation}\label{l-2}
C^{-1}\|u\|^2\le \sum_{n=0}^\infty \|\phi_n u\|^2\le C\|u\|^2.
\end{equation}

\begin{lemma} \label{lem2.1}
For $a(x)\in W^{1,\infty}(\mathbb{R}^d)$, we have
\begin{equation}\label{commutator-1}
\sum_{n=0}^{\infty}\|\phi_n au-a\phi_n u\|_1^2
\le C(\|a\|_{W^{1,\infty}}\|u\|)^2.
\end{equation}
\end{lemma}

\begin{proof}
We define the paraproduct $T_au$ by $\sum_{l=0}^{\infty}a_{l-3} \phi_l u$.
 Here $a_l$ is  the inverse Fourier transform of
$\phi_0(2^{-l}\xi)\hat{a}(\xi)$. It is well known that we have
\[
\|au -T_a u\|_1\le C\|a\|_{W^{1,\infty}}\|u\|.
\]
See for example \cite[Theorem 5.2.8]{met}.
This estimate and \eqref{l-2} imply
\begin{gather*}
\sum_{n=0}^{\infty}\|( a\phi_nu-T_a \phi_n u)\|_1^2
\le C(\|a\|_{W^{1,\infty}}\|u\|)^2,
\\
\sum_{n=0}^{\infty}\|\phi_n( au-T_a u)\|_1^2\le C(\|a\|_{W^{1,\infty}}\|u\|)^2.
\end{gather*}
Then we have  to show only that
\[
\sum_{n=0}^{\infty}\|\phi_nT_a u-T_a \phi_n u)\|_1^2
\le C(\|a\|_{W^{1,\infty}}\|u\|)^2.
\]

In the following, we assume that $l$ and $n$ are non-negative integers.
Note  that  the spectrum of $a_{l-3}\phi_l u$, that is, the support of
the Fourier transform of  $a_{l-3}\phi_l u$ is contained in
$2^{l-2}<|\xi|<2^{l+2}$ if $l\ge 1$, while the spectrum of $a_{-3}\phi_0 u$
is contained in $|\xi|<2$. Then we see that
\[
\phi_na_{l-3}\phi_l u=0\quad |l-n|\ge 3.
\]
We remark also that $\phi_l(\xi)\phi_n(\xi)=0$ if $|l-n|>1$.
Then $\phi_nT_a u-T_a \phi_n u$ is equal to
\[
\sum_{l;|l-n|\le 2}\phi_na_{l-3}\phi_l u-a_{l-3}\phi_l \phi_n u,
\]
which, using the symbol of commutator, is equal to
\[
\sum_{l;|l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u .
\]
Since $a_l(x)=\int_{\mathbb{R}^d}a(y)2^{ld}P(2^l(x-y))\,dy$ where $P(x)$
is the inverse Fourier transform of $\phi_0(\xi)$, then we have
$|a_l(x)-a_l(y)|\le C\|a\|_{W^{1,\infty}} |x-y|$.
Note that $[\phi_n,a_{l-3}]u$ is equal to
\[
2^{nd}\int_{\mathbb{R}^d}Q(2^n(x-y))(a_l(y)-a_l(x))u(y)\,dy
\]
where $Q(x)=P(x)-2^{-d}P(x/2)$ if $n\ge 1$ and $Q(x)=P(x)$ if $n=0$.
Then
\[
|[\phi_n,a_{l-3}]\phi_l u(x)|\le C\|a\|_{W^{1,\infty}}2^{-n}2^{nd}
\int_{\mathbb{R}^d}P_1(2^n(x-y))|u(y)|\,dy
\]
where $P_1(x)=|P(x)||x|$.
Since $P_1(x)$ is integrable, we have
\[
\|[\phi_n,a_{l-3}]\phi_l u\|\le C\|a\|_{W^{1,\infty}}2^{-n}\|u\|.
\]

If $|l-n|\le 2$, the spectrum of $[\phi_n,a_{l-3}]\phi_l u$ is contained
in $|\xi|\le2^{ n+2}$, then
\[
\|[\phi_n,a_{l-3}]\phi_l u\|_1\le
(2^{n+4}+1)\|[\phi_n,a_{l-3}]\phi_l u\|.
\]
Hence we obtain
\[
\|\sum_{l; |l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u\|_1\le
\sum_{l; |l-n|\le 2}C\|a\|_{W^{1,\infty}}\|\phi_l u\|.
\]
Since $\phi_nT_a u-T_a \phi_n u=\sum_{l; |l-n|\le 2}[\phi_n,a_{l-3}]\phi_l u$,
we have
\[
\|\phi_nT_a u-T_a \phi_n u\|_1\le
\sum_{l; |l-n|\le 2}C\|a\|_{W^{1,\infty}}\|\phi_l u\|,
\]
from which we obtain \eqref{commutator-1}.
\end{proof}

\subsection{Bounded variation}

Next we recall the properties of functions with  bounded variation.
(See, for example,  the appendix of \cite{Br} for the detail.)
Let $X$ be a Banach space with a norm $\|\cdot\|_X$  and let $f(t)$ be
a $X$-valued function on $[0,T]$ with  bounded variation; that is,
whose  total variation $V(f,[0,T])$ given by
\[
V(f,[0,T])=\sup_{\substack{\text{any partition of $[0,T]$ }\\
t_0=0<t_1<\cdots<t_L=T}}\sum_{l=1}^L\|f(t_l)-f(t_{l-1})\|_X
\]
is finite.
Then, setting $V_f(t)=V(f,[0,t])$, we have
\[
\|f(t)-f(s)\|_X\le V_f(t)-V_f(s)
\]
for any $0\le  s\le  t\le T$, which implies that  $f(t)$ has at most
countably many discontinuous points and there exists
$f(t+0)=\lim_{h\searrow 0}f(t+h)$ for any $t\in [0,T)$
and that we have
\begin{equation}\label{bv-int0}
\int_0^{T-h}\|f(t+h)-f(t)\|_X\,dt\le hV_f(T)
\end{equation}
for any $0\le h\le T$.

We see that \eqref{bv-int0} implies
\begin{equation}\label{bv-int}
\int_0^{T/2}\|f(t+h)-f(t)\|_X\,dt\le hV_f(T)
\end{equation}
for any $0\le h\le T/2$.

Furthermore we have $\|f(t+0)-f(t)\|_X=V_f(t+0)-V_f(t)$.
Then we see from \eqref{bv-int} that, when $f(t)$ is right continuous at $t=0$,
 that is $f(0+0)=f(0)$, for any $\varepsilon>0$ there exists a positive
$T_{\varepsilon}$ such that we have
\[
\int_0^{T_{\varepsilon}}\|f(t+h)-f(t)\|_X\,dt\le h\varepsilon
\]
for any $0\le h\le T_{\varepsilon}$.

\begin{remark}\label{rem-bv} \rm
It follows from the argument above and  assumption \eqref{hyp-2} that
the right limit $\lim_{h\searrow 0}a_{jk}(x,t+h)$ converges uniformly
on $\mathbb{R}^d_x$ for any $t\in [0,T)$. Then we see that  \eqref{elliptic}
and \eqref{hyp-3} still hold for $a_{jk}(x,t+0)=\lim_{h\searrow 0}a_{jk}(x,t+h)$.
Furthermore, we have $a_{jk}(x,t+0)=a_{jk}(x,t)$ except for at most countably
 many $t$. Then, in Theorem \ref{thm1.1}, we may assume that $a_{jk}(x,t+0)=a_{jk}(x,t)$
on $[0,T]$ uniformly with respect to $x\in \mathbb{R}^d$.
\end{remark}

\section{Carleman estimate}

Noting Remark \ref{rem-bv},
we  may assume that for any positive $\varepsilon$, there exist
$T_{\varepsilon}>0$  such that we have
\begin{equation}\label{3-1}
\int_0^{T_{\varepsilon}}\sum_{j,k=1}^d\|a_{jk}(\cdot,t+h)-a_{jk}
(\cdot,t)\|_{L^\infty}\,dt\le \varepsilon h
\end{equation}
for any $h\in [0,T_{\varepsilon}]$.
Here we may assume that $T_{\varepsilon}\le \varepsilon$.

We define $\psi_{\gamma}(t)$ and $\psi_{1,\gamma}(t)$ with
$\gamma \ge 1/T_{\varepsilon}$ by
\begin{gather*}
\psi_{\gamma}(t)=\int_t^{T_{\varepsilon}}\Bigl(\frac{1}{\varepsilon}
+\frac{1}{\varepsilon}\sum_{j,k=1}^d\gamma
\int_0^1\|a_{jk}(x,\tau+\frac{s}{\gamma})-a_{jk}(x,\tau)\|_{L^\infty}\,ds\Bigr)
\,d\tau, \\
\psi_{1,\gamma}(t)=\gamma\int_t^{T_{\varepsilon}} e^{\psi_{\gamma}(\tau)}\,d\tau.
\end{gather*}
We note that, since
\[
\sum_{j,k=1}^d\int_0^{T_{}\varepsilon}
\|a_{jk}(x,t+\frac{s}{\gamma})-a_{jk}(x,t)\|_{L^\infty}\,dt\le
\frac{\varepsilon s}{\gamma}
\]
for $s\in[0,1]$ and $\gamma\ge 1/T_\varepsilon$,
 we have, on $[0,T_{\varepsilon}]$,
\begin{equation}\label{est-phi}
0\le \psi_{\gamma}(t) \le \frac{3}{2}.
\end{equation}

In this section we show the following Carleman estimate.

\begin{proposition} \label{calreman-prop}
There exists a positive constant $\varepsilon_0$
so that, for any $\varepsilon\in(0,\varepsilon_0)$ we have, with a
positive $\gamma_{\varepsilon}$,
 \begin{equation}\label{calreman}
\begin{aligned}
&\frac{\gamma}{\varepsilon}\int_0^{T_{\varepsilon}}
 e^{2\psi_{1,\gamma}(t)}\|u(\cdot,t)\|^2\,dt+
\frac{1}{\varepsilon}\int_0^{T_{\varepsilon}}
 e^{2\psi_{1,\gamma}(t)}\|\nabla u(\cdot,t)\|^2\,dt\\
&\le C\int_0^{T_{\varepsilon}}e^{2\psi_{1,\gamma}(t)}\|Lu(\cdot,t)\|^2\,dt
\end{aligned}
 \end{equation}
for any $\gamma\ge \gamma_{\varepsilon}$ and any
$u(x,t)\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$ satisfying
$\partial_{x_j} u(x,t)\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$
($j=1,2,\dots,d$) and  $Lu\in L^2(\mathbb{R}^d \times [0,T_{\varepsilon}])$,
 $u(x,0)=0$  and $u(x,T_{\varepsilon})=0$.
 Here the constant $C$ is independent of $\varepsilon$ and of $\gamma$.
\end{proposition}

We define the operator $L_\gamma$ by
$L_\gamma=e^{\psi_{1,\gamma}(t)}L e^{-\psi_{1,\gamma}(t)}$; that is,
\[
L_\gamma u=\partial_tu+\gamma e^{\psi_{\gamma}(t)}u
+\sum_{j,k=1}^d\partial_{x_j}( a_{jk}(x,t)\partial_{x_j}u).
\]
Then, by replacing $u$ by $e^{\psi_{1,\gamma}(t)}u$, \eqref{calreman}
is equivalent to
\begin{equation}\label{calreman-1}
 \frac{\gamma}{\varepsilon}\int_0^{T_{\varepsilon}}\|u(\cdot,t)\|^2\,dt+
\frac{1}{\varepsilon}\int_0^{T_{\varepsilon}}\|\nabla u(\cdot,t)\|^2\,dt
 \le C\int_0^{T_{\varepsilon}}\|L_\gamma u(\cdot,t)\|^2\,dt.
 \end{equation}
We remark that  from \eqref{commutator-1} it follows that
\[
\sum_{n=0}^\infty \|\phi_n \partial_{x_j}(a_{jk}\partial_ku)
-\partial_{x_j}(a_{jk}\partial_k\phi_n u)\|^2\le
 C\|\partial_{x_{k}}u\| ^2,
\]
from which we obtain
\[
\sum_{n=0}^\infty \|\phi_n L_{\gamma}u-L_{\gamma}\phi_n u\|^2\le
 C\|u\|_1 ^2.
\]
 Then, by \eqref{l-2} we get
 \begin{equation}\label{sum}
\sum_{n=0}^\infty \|L_{\gamma}\phi_n u\|^2 \le
 C( \| L_{\gamma}u\|^2  +\|u\|_1 ^2).
\end{equation}
Therefore, we consider the estimate of $\|L_{\gamma}\phi_n u\|$.
Note that  \eqref{elliptic} implies
 \begin{equation}\label{Garding}
\sum_{j,k=1}^d(a_{jk}(x,t)\partial_{x_{k}}v,\partial_{x_{j}}v)\geq
D_0\|\nabla v\|^2,
\end{equation}
from which and from \eqref{est-phi}
we obtain the following: for $u(x,t)$ satisfying $u(x,0)=0$ and
$u(x,T_{\varepsilon_0})=0$,
\[
- \int_0^{T_\varepsilon}(L_{\gamma}\phi_n u,\phi_n u)
\ge D_0\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
-\gamma e^{3/2}\int_0^{T_\varepsilon}\|\phi_n u\|^2\,dt.
\]
When $\frac{D_0}{4}2^{2(n-1)}\ge \gamma e^{3/2}$ and $n\ge 1$, we see,
noting $ \|\nabla (\phi_n u)\|^2\ge 2^{2(n-1)}\|\phi_n u\|^2$, that
\[
- \int_0^{T_\varepsilon}(L_{\gamma}\phi_n u,\phi_n u)
\ge \frac{D_0}{2}\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
+\gamma e^{3/2}\int_0^{T_\varepsilon}\|\phi_n u\|^2\,dt.
\]
Hence, by $|(L_{\gamma}\phi_n u,\phi_n u)|
\le \frac{\varepsilon}{2}\|L_{\gamma}u\|^2+\frac{1}{2\varepsilon}\|u\|^2$
 we get
\begin{equation}\label{carleman-00}
\varepsilon\int_0^{T_\varepsilon}\|L_{\gamma}u\|^2\,dt\ge
D_0\int_0^{T_\varepsilon}\|\nabla (\phi_n u)\|^2\,dt
+\int_0^{T_\varepsilon}( 2\gamma-\frac{1}{\varepsilon})\|\phi_n u\|^2\,dt.
\end{equation}
For the case where  $\frac{D_0}{4}2^{2(n-1)}\le \gamma e^{3/2}$ with
$\gamma\ge 1/T_{\varepsilon}$,
we have the following lemma.

\begin{lemma}\label{carleman-lemma}
There exists a positive $\varepsilon_0$ such  that
under the condition that \eqref{3-1} is valid for $0<\varepsilon<\varepsilon_0$,
we  have the following estimates.
 When $0<\varepsilon<\varepsilon_0$ and
$\frac{D_0}{4}2^{2(n-1)}\le \gamma e^{3/2}$ with $\gamma\ge 1/T_{\varepsilon}$,
we have
 \begin{equation}\label{carleman-01}
 C\int_0^{T_{\varepsilon}}\| L_{\gamma}\phi_n u\|^2\, dt
 \geq
  \frac{1}{\varepsilon}\int_0^{T_{\varepsilon}}\|\nabla\phi_n u\|^2\,dt+
  \frac{\gamma}{\varepsilon}\int_0^{T_{\varepsilon}}\|\phi_n u\|^2\,dt
  \end{equation}
  for any $u(x,t)$ satisfying $u(x,0)=0$ and $u(x,T_{\varepsilon_0})=0$
 \end{lemma}

 \begin{proof}
Note that
 \begin{align*}
 \| L_{\gamma}\phi_n u\|^2
&=\|\partial_t(\phi_n u)\|^2+
 \|\gamma e^{\psi_{\gamma}}\phi_n u+\sum_{j,k=1}^d\partial_{x_j}(a_{jk}
\partial_{x_k}\phi_n u)\|^2 \\
&\quad +
 2\Re (\partial_t(\phi_n u),\gamma e^{\psi_{\gamma}}\phi_n u)
+  \sum_{j,k=1}^d 2\Re (\partial_t(\phi_n u),\partial_{x_j}(a_{jk}
\partial_{x_k}\phi_n u)).
 \end{align*}
Let $\chi(s)\in C^{\infty}(\mathbb{R})$ satisfy $\chi(s)\ge 0$ on $\mathbb{R}$, $\chi(s)=0$
on $(-\infty,0]\cup [1,\infty)$ and $\int_{-\infty}^{\infty}\chi(s)\,ds=1$.
 Set $D_1=\sup_{s\in\mathbb{R}}|\chi(s)|+|\chi'(s)|$. We define the regularization of
$a_{jk}$,  $a^{\gamma}_{jk}(x,t)$, by
 \[
 a_{jk}^{\gamma}(x,t)=\gamma\int_{-\infty}^{\infty}\chi(\gamma (s-t))
a_{jk}(x,s)\,ds.
\]
 We see that
\[
  a_{jk}^{\gamma}(x,t)=\int_{-\infty}^{\infty}\chi(s)a_{jk}(x,t+s/\gamma)\,ds
\]
from which and from $\int_{-\infty}^{\infty}\chi(s)\,ds=1$ we see
  \[
   a_{jk}^{\gamma}(x,t)-a_{jk}(x,t)=\int_{-\infty}^{\infty}\chi(s)(a_{jk}
(x,t+s/\gamma) -a_{jk}(x,t))\,ds,
\]
while  from
$\partial_t a^{\gamma}(x,t)=-\gamma\int_{-\infty}^{\infty}\chi'(s)a_{jk}
(x,t+s/\gamma)\,ds$ and  $\int_{-\infty}^{\infty}\chi'(s)\,ds=0$,
it follows that
\[
   \partial_ta_{jk}^{\gamma}(x,t)=-\gamma\int_{-\infty}^{\infty}\chi'(s)
(a_{jk}(x,t+s/\gamma)
-a_{jk}(x,t))\,ds.
\]
Then we have
\begin{gather*}
| a_{jk}^{\gamma}(x,t)-a_{jk}(x,t)|\le D_1\int_0^1|a_{jk}(x,t+s/\gamma)
-a_{jk}(x,t)|\,ds,\\
| \partial_ta_{jk}^{\gamma}(x,t)|\le D_1\gamma\int_0^1|a_{jk}(x,t+s/\gamma)
-a_{jk}(x,t)|\,ds.
\end{gather*}
Furthermore we note  that
\[
| a_{jk}^{\gamma}(x,t)|\le \|a_{jk}(\cdot,t)\|_{L^\infty}
\]
implies
\[
| a_{jk}^{\gamma}(x,t)-a_{jk}(x,t)|\le 2\|a_{jk}(\cdot,t)\|_{L^\infty}.
\]
Then we have
\[
| a_{jk}^{\gamma}(x,t)-a_{jk}(x,t)|
\le \sqrt{2\|a_{jk}(\cdot,t)\|_{L^\infty}}
\Big(D_1\gamma\int_0^1|a_{jk}(x,t+s/\gamma)
-a_{jk}(x,t)|\,ds\Big)^{1/2}.
\]
Using the estimates above, we  estimate the term
$ (\partial_t(\phi_n u),\partial_{x_j}(a_{jk}\partial_{x_k}\phi_n u))$.
Note that
\begin{equation}
\begin{aligned}
&(\partial_t(\phi_n u),\partial_{x_j}(a_{jk}\partial_{x_k}\phi_n u))\\
&= (\partial_t(\phi_n u),\partial_{x_j}(a_{jk}^{\gamma}\partial_{x_k}\phi_n u))
+(\partial_t(\phi_n u),\partial_{x_j}((a_{jk}-a_{jk}^{\gamma})
 \partial_{x_k}\phi_n u)).
\end{aligned}
\end{equation}
 Note that
$|(\partial_t(\phi_n u),\partial_{x_j}((a_{jk}-a_{jk}^{\gamma})\partial_{x_k}
\phi_n u))|=
|(\partial_t\partial_{x_j}(\phi_n u),((a_{jk}-a_{jk}^{\gamma})\partial_{x_k}
\phi_n u))|$ which is dominated by
\[
2^{n+1}\|\partial_t(\phi_n u)\|\sqrt{2\|a_{jk}(\cdot,t)\|_{L^\infty}}
\Big(D_1\int_0^1\|a_{jk}(\cdot,t+s/\gamma)
-a_{jk}(\cdot,t)\|_{L^\infty}\,ds\Big)^{1/2}\|\nabla \phi_n u\|.
\]
Here we used $\phi_n(\xi)=0$ for $|\xi|\ge 2^{n+1}$.
Setting $K=\sum_{j,k=1}^d\sup_{t\in[0,T_{\varepsilon}]}
\|a_{jk}(\cdot,t)\|_{L^\infty}$, we obtain, from Schwarz's inequality,
\begin{align*}
&\sum_{j,k=1}^d|(\partial_t(\phi_n u),\partial_{x_j}((a_{jk}-a_{jk}^{\gamma})
 \partial_{x_k}\phi_n u))|
\\
&\leq 2^{n+1}\|\partial_t(\phi_n u)\|\sqrt{2K}
\Big(D_1\sum_{j,k=1}^d \int_0^1\|a_{jk}(\cdot,t+s/\gamma)
-a_{jk}(\cdot,t)\|_{L^\infty}\,ds\Big)^{1/2}
\|\nabla \phi_n u\|.
\end{align*} % 3.11
Then we get
\begin{align*}
&2\sum_{j,k=1}^d|(\partial_t(\phi_n u),\partial_{x_j}((a_{jk}
-a_{jk}^{\gamma})\partial_{x_k}\phi_n u))|\\
&\le \|\partial_t(\phi_n u)\|^2+ 2^{2(n+1)+1}KD_1\sum_{j,k=1}^d
\int_0^1\|a_{jk}(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds
\|\nabla \phi_n u\|^2.
\end{align*} % 3.12
Hence
\begin{align*}
&\|\partial_t(\phi_n u)\|^2+2\sum_{j,k=1}^d \Re (\partial_t(\phi_n u),
 \partial_{x_j}((a_{jk}-a_{jk}^{\gamma})\partial_{x_k}\phi_n u))
\\
&\geq - 2^{2(n+1)+1}KD_1\sum_{j,k=1}^d\int_0^1\|a_{jk}(\cdot,t+s/\gamma)
-a_{jk}(\cdot,t)\|_{L^\infty}\,ds
\|\nabla \phi_n u\|^2.
\end{align*}
On the other hand, noting that
\begin{align*}
&\sum_{j,k=1}^d 2\Re (\partial_t(\phi_n u),\partial_{x_j}(a_{jk}^{\gamma}
 \partial_{x_k}\phi_n u))\\
&=\sum_{j,k=1}^d(\partial_{x_j}\phi_n u,(\partial_ta_{jk}^{\gamma})
 \partial_{x_k}\phi_n u)-
\sum_{j,k=1}^d\partial_t(\partial_{x_j}\phi_n u,a_{jk}^{\gamma}
 \partial_{x_k}\phi_n u),
\end{align*}
 we see that, when $u(x,0)=0$ and $u(x,T_{\varepsilon_0})=0$ are satisfied,
\begin{align*}
&|\int_0^{T_{\varepsilon}}2\Re (\partial_t(\phi_n u),
\partial_{x_j}(a^{\gamma}_{jk}\partial_{x_k}\phi_n u))\,dt|\\
&\leq \int_0^{T_{\varepsilon}}(\sum_{j,k=1}^d D_1\gamma\int_0^1\|a_{jk}
(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds)\|\nabla \phi_n u\|^2\,dt.
\end{align*}
Therefore, we see that, when $u(x,0)=0$ and $u(x,T_{\varepsilon_0})=0$
are satisfied,
 \begin{align*}
& \int_0^{T_{\varepsilon}}
 \Bigl( \|\partial_t(\phi_n u)\|^2+ 2\Re (\partial_t(\phi_n u),
\sum_{j,k=1}^d\partial_{x_j}(a_{jk}\partial_{x_k}\phi_n u))
  \Bigr)\,dt\\
&\geq
-( 2^{2(n+1)+1}K+\gamma)D_1\sum_{j,k=1}^d\int_0^{T_\varepsilon}
\int_0^1\|a_{jk}(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds
\|\nabla \phi_n u\|^2\,dt.
\end{align*}
Similarly, we have
\[
2\Re (\partial_t(\phi_n u),\gamma e^{\psi_{\gamma}}\phi_n u)=
\partial_t(\phi_n u,\gamma e^{\psi_{\gamma}}\phi_n u)-\gamma
(\frac{d}{dt}e^{\psi_{\gamma}})\|\phi_n u\|^2.
\]
We note that
\[
-\gamma (\frac{d}{dt}e^{\psi_{\gamma}})=\gamma e^{\psi_{\gamma}}\times
(\frac{1}{\varepsilon}+\frac{\gamma}{\varepsilon} \sum_{j,k=1}^d
\int_0^1\|a_{jk}(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds)
\]
from which and from $e^{\psi_{\gamma}(t)}\ge 1$, we obtain
\[
-\gamma (\frac{d}{dt}e^{\psi_{\gamma}})\ge
\frac{\gamma}{\varepsilon}(1+\gamma \sum_{j,k=1}^d
\int_0^1\|a_{jk}(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds)\,.
\]
Then we see that, when $u(x,0)=0$ and $u(x,T_{\varepsilon})=0$ are satisfied,
\begin{align*}
&\int_0^{T_{\varepsilon}}2\Re (\partial_t(\phi_n u),\gamma e^{\psi_{\gamma}}
\phi_n u)\,dt \\
&\ge  \int_0^{T_{\varepsilon}}\frac{\gamma}{\varepsilon}(1+\gamma \sum_{j,k=1}^d
\int_0^1\|a_{jk}(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds)
\|\phi_n u\|^2\,dt
\end{align*}
Therefore,
\begin{align*}
\int_0^{T_{\varepsilon}}\| L_{\gamma}\phi_n u\|^2\,dt
&\ge \int_0^{T_{\varepsilon}}\frac{\gamma}{\varepsilon}\|\phi_n u\|^2\,dt
\\
&\quad +\int_0^{T_{\varepsilon}}(\frac{\gamma^2}{\varepsilon} \|\phi_n u\|^2
-(2^{2(n+1)+1}KD_1+D_1\gamma)\|\nabla \phi_n u\|^2
)\\
&\quad \times \Big(\sum_{j,k=1}^d \int_0^1\|a_{jk}
(\cdot,t+s/\gamma)-a_{jk}(\cdot,t)\|_{L^\infty}\,ds \Big)\,dt.
\end{align*}
Since $\|\nabla \phi_n u\|^2\le 2^{2(n+1)}\|\phi_n u\|^2$,
  when $\frac{D_0}{4}2^{2(n-1)}\leq \gamma e^{3/2}$, we have
 \begin{gather*}
 (2^{2(n+1)+1}KD_1+D_1\gamma)\|\nabla \phi_n u\|^2
 \le C \gamma^2 \| \phi_n u\|^2, \\
\|\nabla \phi_n u\|^2\le C\gamma \| \phi_n u\|^2.
\end{gather*}
Choosing  $\varepsilon_0$  small, we have,
for $0<\varepsilon<\varepsilon_0$, 
\[
(\frac{\gamma^2}{\varepsilon} \|\phi_n u\|^2
-(2^{2(n+1)+1}KD_1+D_1\gamma)\|\nabla \phi_n u\|^2)\ge 0.
\]
Hence
\[
\int_0^{T_{\varepsilon}}\| L_{\gamma}\phi_n u\|^2\,dt
\ge \int_0^{T_{\varepsilon}}\frac{\gamma}{\varepsilon}\|\phi_n u\|^2\,dt.
\]
Using $\gamma \| \phi_n u\|^2\ge \frac{1}{C}\|\nabla \phi_n u\|^2$, we obtain
\[
\int_0^{T_{\varepsilon}}\| L_{\gamma}\phi_n u\|^2\,dt
\ge
\int_0^{T_{\varepsilon}}(\frac{\gamma}{2\varepsilon}\|\phi_n u\|^2+
\frac{1}{2C\varepsilon}\|\nabla\phi_n u\|^2)
\,dt.
\]
\end{proof}

 Now we complete the proof of Proposition \ref{calreman-prop}.
We choose $\varepsilon_0$ and $\gamma_{\varepsilon}$ so that the assertion
of Lemma  \ref{carleman-lemma} is valid.
Furthermore we choose $\gamma_{\varepsilon}$ so large that we have
$\gamma_{\varepsilon}> 2/\varepsilon$. Then we
obtain from \eqref{carleman-00}
\[
C\int_0^{T_\varepsilon}\|L_{\gamma}\phi_nu\|^2\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla \phi_nu\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| \phi_nu\|^2\,dt
\]
for $n$ satisfying $\frac{D_0}{4}2^{2(n-1)}\geq \gamma e^{3/2}$ with
$\gamma\ge \gamma_\varepsilon$.
Hence  it follows  from the estimate above, \eqref{carleman-01} and
\eqref{l-2} that
\[
C\sum_{n=0}^{\infty}\int_0^{T_\varepsilon}\|L_{\gamma}\phi_nu\|^2\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla u\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| u\|^2\,dt.
\]
Noting \eqref{sum}, we have
\[
C\int_0^{T_\varepsilon}(\|L_{\gamma}u\|^2+\|u\|_1^2)\,dt\ge
\frac{1}{\varepsilon}\int_0^{T_\varepsilon}\|\nabla u\|^2\,dt+
\frac{\gamma}{\varepsilon}\int_0^{T_\varepsilon}\| u\|^2\,dt.
\]
Then, by choosing $\varepsilon_0$ so small, we obtain the desired estimate
\eqref{calreman-1}. The proof of Proposition \ref{calreman-prop}
is complete.

\section{Proof of Theorem \ref{thm1.1}}

 First, we show  time local uniqueness under the assumptions of
Theorem \ref{thm1.1}.
Then, using the well-known continuity argument, we show that the assertion
 of Theorem \ref{thm1.1} is valid.

 \begin{proposition}\label{local-u1}
 Under the assumptions of Theorem \ref{thm1.1}, there exists  $t_0\in(0,T]$
such that we have $u(x,t)=0$ for $t\in[0,t_0]$.
 \end{proposition}

\begin{proof}
Set $f=Lu$. Then we see from the assumptions of Theorem \ref{thm1.1}, that
$f\in L^2([0,T],L^2(\mathbb{R}^d ))$ and
$\|f(\cdot,t)\|^2\le C_0(\|\nabla u(\cdot,t)\|^2+\| u(\cdot,t)\|^2)$
for almost all $t\in[0,T]$.
 Let the  non-negative function $\chi_0(t)\in C^{\infty}(\mathbb{R})$ satisfy
\[
\chi_0(t)= \begin{cases}
1& t<3/4\\
0& t>7/8\,.
\end{cases}
\]
Set $u_{\varepsilon}(x,t)=\chi_0(t/T_{\varepsilon})u(x,t)$.
Here we use the notation of Proposition \ref{calreman-prop}.
Then we see that $u_{\varepsilon}(x,0)=0$,
$u_{\varepsilon}(x,T_\varepsilon)=0$ and that
\[
Lu_{\varepsilon}=\chi_0(t/T_{\varepsilon})f+\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon}u.
\]
Then  from \eqref{calreman} we obtain
\begin{align*}
&\int_0^{T_\varepsilon}(\frac{1}{\varepsilon}\|\nabla u_\varepsilon \|^2
+\frac{\gamma}{\varepsilon}\|u_\varepsilon \|^2)e^{2\psi_{1,\gamma}(t)}\,dt\\
&\le C \int_0^{T_\varepsilon}(\|\chi_0(t/T_{\varepsilon})f\|^2
+(\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon})^2\|u\|^2)
e^{2\psi_{1,\gamma}(t)}\,dt.
\end{align*}
 Since
 \[
 \|\chi_0(t/T_{\varepsilon})f\|^2\le C_0(\|\nabla u_{\varepsilon}\|^2+
 \|u_\varepsilon\|^2),
 \]
 by choosing $\varepsilon$  small, we have
 \begin{equation}\label{carleman-03}
 \int_0^{T_\varepsilon}(\frac{1}{\varepsilon}\|\nabla u_{\varepsilon}\|^2
+\frac{\gamma}{\varepsilon}\|u_{\varepsilon}\|^2)e^{2\psi_{1,\gamma}(t)}\,dt
\le C  \int_0^{T_\varepsilon}(\frac{\chi_0'(t/T_{\varepsilon})}{T_\varepsilon})^2
\|u_{\varepsilon}\|^2e^{2\psi_{1,\gamma}(t)}\,dt
 \end{equation}
for any $\gamma\ge \gamma_{\varepsilon}$.
Since $\chi_0'(t/T_{\varepsilon})=0$ for $t\le 3T_\varepsilon/4$ and
$\psi_{1,\gamma}(t)$ is decreasing, we note that the right-hand side
of \eqref{carleman-03} can be dominated by
\[
C_{\varepsilon}e^{2\psi_{1,\gamma}(3T_{\varepsilon}/4)}.
\]
Since $\psi_{\gamma,1}=\gamma\int_t^{T_\varepsilon}e^{\psi_{\gamma}(\tau)}\,d\tau$
 and $e^{\psi_{\gamma}(\tau)}\ge 1$ for  $\tau\ge0$, we see that
for $t\in[0,T_\varepsilon/4]$,
\[
\psi_{\gamma,1}(t)\geq \psi_{\gamma,1}(T_\varepsilon/4)
\geq \psi_{\gamma,1}(3T_\varepsilon/4)+\gamma T_\varepsilon/2.
\]
Then, noting that $u_{\varepsilon}(x,t)=u(x,t)$ on $[0,3T_{\varepsilon}/4]$,
 we see that
 \[
\frac{\gamma}{\varepsilon} e^{ 2\psi_{\gamma,1}(3T_\varepsilon/4)
+\gamma T_\varepsilon}\int_0^{T_\varepsilon/4}  \|u\|^2\,dt
  \]
is not greater than the left hand side of  \eqref{carleman-03}.
Then we have
\[
\int_0^{T_\varepsilon/4}
 \|u\|^2\,dt\le \frac{ \varepsilon C_{\varepsilon}}{\gamma}
e^{ -\gamma T_\varepsilon}.
\]

As $\gamma$ tends to infinity, the right hand side converges to zero.
Then we see that $\int_0^{T_\varepsilon/4}
 \|u\|^2\,dt=0$, which implies $u(x,t)=0$ on $[0,T_\varepsilon/4]$.
\end{proof}

Using the same argument we have the following proposition.

 \begin{proposition}\label{local-u2}
 We assume that  the assumptions of Theorem \ref{thm1.1} except for
$u(x,0)=0$ are  satisfied. For any $t_0\in[0,T)$ there exists
$t_1\in(t_0,T]$ such that, if $u(x,t_0)=0$, then  we have
 $u(x,t)=0$ for $t\in[t_0,t_1]$.
 \end{proposition}

 Now we prove Theorem \ref{thm1.1}.
 Let $S$ be the subset  of $(0,T)$ that consists of $t_0\in(0,T)$  satisfying
 $u(x,t)=0$ on $[0,t_0]$. From Proposition \ref{local-u1} and Proposition
\ref{local-u2} we see that the set $S$ is not empty and open set.
Since $u(x,t)\in C^0([0,T],L^2(\mathbb{R}^d))$, we see that $S$ is  closed subset
of $(0,T)$. Then the connectedness of $(0,T)$ implies $S=(0,T)$.
Then we see that $u(x,t)=0$ on $[0,T)$. Hence
 $u(x,t)=0$ on $[0,T]$. The proof of Theorem \ref{thm1.1} is complete.

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\end{document}