In the book **Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I** page186. The authors proved such a Theorem. In the following $P=\sum_{i,j=1}^{n}\partial_{i}(p_{ij}\partial_{j}\cdot)$ is a second order partial differential operator and the coefficient $p_{ij}$ is smooth. Theorem5.2. Let $\Omega$ be a connected open subset in $\mathbb{R}^{d}$ and let $\omega\subset\Omega$, with $\omega\neq\emptyset$. If $u\in H^{2}(\Omega)$ satisfies$$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\omega$, then $u$ vanishes in $\Omega$. Sketch of Proof: The authors set $F=supp(u)$(so $F$ is a closed set according the definition of support set of function), and then they prove that $F$ is also open to get $u=0$. They take a point $x^{(1)}$ in $F\setminus \mathrm{int}(F)$ and to get a contradiction by the following Proposition(Proposition5.1 in Lebeau book). Their proof of contradiction is by constructing a family of Ball $\mathscr{B}_{t}=B(x^{(0)},(1-t)r_{1}+tr_{2})$ where $x^{(0)}\in\Omega\setminus F$, and they say according to the following proposition, we can get **if u vanishes in $\mathscr{B}_{t}$ with $0\leq t\leq1$, then there exists $\epsilon>0$ such that $u$ vanishes in $\mathscr{B}_{t+\epsilon}$**, here, we just know $B(x^{(0)};r_{1})\subset\Omega\setminus F$ in which $u=0$, the question is **how can we push the radius from $r_{1}$ to $r_{2}$**, the $\epsilon$ in each step is different, does there exist such a situation, that in each step, the increase of radius is$\frac{1}{10},\frac{1}{10^{2}},\frac{1}{10^{3}},\dots$ so that we can not get $B(x^{(0)};r_{2})$? Just in ordinary differential equation, when we study the extension of classical solution of $\dot{x}(t)=f(x(t))$? Everytime we extend a little, but the total step is finite? Proposition5,1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\quad,a.e.\quad in\Omega$$ for some $C>0$ and $u(x)=0$ in $\{x\in V;\phi(x)\geq\phi(x^{(0)})\}$. Then $u$ vanishes in a neighborhood of $x^{(0)}$. **Remark, the author didn't tell us how large of the neighborhood**