A question on the proof of unique continuation for the case $u\in H^{2}$ in Le Rousseau, Lebeau and Robbiano book on Carleman estimates

Question

In the book Elliptic Carleman Estimates and Applications to Stabilization and Controllability, Volume I, page 186 (MR4436025, Zbl 1497.35005), the authors proves a unique continuation theorem which is the argument of this question.
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.

Theorem 5.2. 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 $F=\operatorname{supp}(u)$ (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 $F=\emptyset$ and thus $u=0$.
They take a point $x^{(1)}$ in $F\setminus \operatorname{int}(F)$ and work in order to get a contradiction by the following proposition (Proposition 5.1 in the book). Precisely, their proof of by contradiction is worked out by constructing a family of balls $$ \mathscr{B}_{t}=B\big(x^{(0)},(1-t)r_{1}+tr_{2}\big) $$ where $x^{(0)}\in\Omega\setminus F$, and then proving (according to the said proposition) that 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 $u=0$: now the question is how can we push the radius from $r_{1}$ to $r_{2}$?
The $\epsilon$ in each step is different, so how 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 intersect $B(x^{(0)};r_{2})$?
Is it just as 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 how the total step is finite?

Proposition 5.1. Let $u\in H^{2}_{\rm loc}(\Omega)$ and $$|Pu(x)|\leq C\Big(|u(x)|+|Du(x)|\Big),\text{ a.e. in }\Omega$$ 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 is.

Answer 1

I think one can answer your question (specifically the bold parts of your text) without having any knowledge of the precise argument of Lebeau; if you find the answer lacking, please let me know and also include a bit more detail (precise Theorem statements etc) since I don't have access to Lebeau's book at the moment.

With that in mind, I shall assume you accept the statement

If $u$ vanishes in $\mathscr{B}_t$, then $u$ vanishes in $\mathscr{B}_{t+\epsilon}$.

In this case, let $t_0 \in [0,1]$ be defined to be $$ \sup \big\{ t\in [0,1] : u \text{ vanishes on } \mathscr{B}_t \big\} $$ The supremum exists as the set is non-empty (you've accepted that $0$ belongs to said set), and is upper bounded (by $1$). I claim that $u$ vanishes on $\mathscr{B}_{t_0}$: this follows from the fact that $\mathscr{B}_{t_0} = \cup_{t\in [0,t_0)} \mathscr{B}_{t}$.

Now suppose $t_0 < 1$, then by the Proposition which you accept you must have $u$ also vanishes $\mathscr{B}_{t_0+\epsilon}$ which contradicts the definition of $t_0$.

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The reason that this differs from the local existence theorem for ODEs is that in that case, the extension condition is

If $u$ is a solution on $[0,t]$, then $u$ is a solution on $[0,t+\epsilon)$.

(Note the difference between open and closed domains.) And we can analogously set

$$ t_0 = \sup \big\{ t : u \text{ is a solution on } [0,t] \big\} $$

But now we cannot claim $u$ is a solution on $[0,t_0]$ to use the local existence theorem, since the definition of $t_0$ only guarantees that $u$ is a solution on $[0,t]$ for every $t < t_0$ (note the strictly less symbol). And $\cup_{t < t_0} [0,t] = [0,t_0)$ and not $[0,t_0]$.