[Main Theorem]
\label{mainThm}
For some $a_0 \ge 1$ and $N \in (0, 2)$, let $V : \R^2 \to \R$ satisfy
\begin{equation}
\label{Vbound}
\left\vertV(w)\right\vert \le a_0^2 \left< w \right>^{-N}.
\end{equation}
Assume that $u : \R^2 \to \R$ is a solution to
\begin{equation}
\label{ellipEq}
- \Delta u + V u = 0 \, \text{ in } \R^2
\end{equation}
with the properties that
\begin{equation}
\label{solNorm}
\left\vertu(0)\right\vert = 1
\end{equation}
and for each $w \in \R^2$,
\begin{equation}
\label{uBound}
\left\vertu(w)\right\vert \le \exp\pr{c_0 \left\vertw\right\vert^{1 - \frac N 2}}.
\end{equation}
For every $\varepsilon \in \left( 0, \frac N 2 \right) $, there exists ${R}_0(N, a_0, c_0, \varepsilon) > 0$ so that whenever $R \ge {R}_0$, it holds that
\begin{equation}
\label{uLower}
\inf_{\left\vertw_0\right\vert = R} \left\| u\right\|_{L^\infty\left( B_1(w_0) \right) } \ge \exp\pr{-R^{1 - \frac N 2 + \varepsilon}}.
\end{equation}

\begin{thm}[Main Theorem]
\label{mainThm}
For some $a_0 \ge 1$ and $N \in (0, 2)$, let $V : \R^2 \to \R$ satisfy
\begin{equation}
\label{Vbound}
\abs{V(w)} \le a_0^2 \innp{w}^{-N}.
\end{equation}
Assume that $u : \R^2 \to \R$ is a solution to
\begin{equation}
\label{ellipEq}
- \LP u + V u = 0 \, \text{ in } \R^2
\end{equation}
with the properties that
\begin{equation}
\label{solNorm}
\abs{u(0)} = 1
\end{equation}
and for each $w \in \R^2$,
\begin{equation}
\label{uBound}
\abs{u(w)} \le \exp\pr{c_0 \abs{w}^{1 - \frac N 2}}.
\end{equation}
For every $\eps \in \pr{0, \frac N 2}$, there exists ${R}_0(N, a_0, c_0, \eps) > 0$ so that whenever $R \ge {R}_0$, it holds that
\begin{equation}
\label{uLower}
\inf_{\abs{w_0} = R} \norm{u}_{L^\iny\pr{B_1(w_0)}} \ge \exp\pr{-R^{1 - \frac N 2 + \eps}}.
\end{equation}
\end{thm}

\begin{align}
\label{propApp}
\sup_{B_{\tilde r} \setminus \bigcup 3 \WT D_j} \abs{\tilde h}
\ge \pr{\frac{16 {\tilde r}} {\WT R}}^{\tilde \kappa} \sup_{B_{\WT T} \setminus \bigcup 3 \WT D_j} \abs{\tilde h},
\end{align}
where
\begin{align*}
\tilde \kappa
:= \kappa\pr{\WT R, \WT S, M}
= \max \set{6C_H C_1 R \sqrt{\log R}, 2^{13} \pr{c_1 R + L + \frac {c_d}{\log R}} RS^{-1}}.
\end{align*}
Since $R \ge 2^{10}$ implies $\frac R 8 > 1$, while $R \ge R_2$ implies $\sqrt{\frac{98 \cdot 16 R}{C_1 \sqrt{\log R}}} > 1$, then $r_0 := \min\set{\frac R 8, \sqrt{\frac{98 \cdot 16 R}{C_1 \sqrt{\log R}}}} > 1$.
Assume that $R \ge \hat R_0$ and $r < r_0$.
Combining \eqref{ufComparison}, \eqref{smallBallBound}, \eqref{propApp}, \eqref{scaleBounds}, and \eqref{midBallBound} shows that
\begin{align*}
\frac 1{1 - c_b \eps^2} \sup_{B_r} \abs{v}
&\ge \sup_{B_r} \abs{f}
\ge \sup_{B_{\tilde r} \setminus \bigcup 3 \WT D_j} \abs{\tilde h}
\ge \pr{\frac{16 {\tilde r}} {\WT R}}^{\tilde \kappa} \sup_{B_{\WT T} \setminus \bigcup 3 \WT D_j} \abs{\tilde h}
\ge \pr{\frac{r} {R}}^{2\tilde \kappa} \frac {e^{- L}} {1 + c_b \eps^2} .
\end{align*}
Since $C_1 =  \frac {a} {32} \sqrt{\frac{2 C_K }{\ln 2}}$, $\disp \frac{1 - c_b \eps^2}{1 + c_b \eps^2} \ge e^{- \frac{c_d}{\log R}}$, and $e < 4 \le \pr{2^{10}}^{\frac 1 5} \le \pr{\frac{R}{r}}^{\frac 1 5}$, then \eqref{propConc} follows, as required.
\end{proof}

\section{Transformation maps}
\label{TransMaps}

In this section, we introduce the transformation maps that serve as real-valued versions of the conformal maps $z \mapsto z^\al$.
Once the maps are defined, we show how they transform solutions to elliptic PDEs and how they transform balls.
These results will be used at the beginning and at the end of the proof of Theorem \ref{mainThm}.

Given $z = (x,y) \in \R^2$ in Cartesian coordinates, the polar coordinates for $z$ are $(r, \te) \in \R_+ \times (-\pi, \pi]$, where $r = \sqrt{x^2 + y^2}$ and $\te = \sgn(y) \arccos\pr{\frac x r}$.
Let $\R^2_+ = \set{z = (r,\te) : r > 0, \te \in (- \frac \pi 2, \frac \pi 2)}$ denote the right half-plane.
For $\al \in (1, \iny)$, we define $T_\al : \R^2 \to \R^2$ to be the map associated to the conformal transformation on $\C$ given by $z \mapsto z^\al$.
In polar coordinates, $T_\al$ is described as
\begin{equation}
\label{TalDefn}
T_\al(r, \te) = (r^\al, \al \te).
\end{equation}
To avoid continuity issues at $\te = \pi$, we restrict the domain and only consider $T_\al : \R^2_+ \to \R^2$.

\begin{lem}[Transformation of PDEs]
\label{transformEqLemma}
If $u : \R^2 \to \R$ is a solution to \eqref{ellipEq}, then with $v : \R^2_+ \to \R$ and $W : \R^2_+ \to \R$ defined by $v(z) = u(T_\al(z))$ and $W(z) = \al^2 \abs{z}^{2\al - 2} V(T_\al(z))$, it holds that
$$-\LP v + W v = 0 \; \text{ in } \R^2_+.$$
\end{lem}

\begin{proof}
Let $w := T_\al(z)$ be given in polar coordinates by $\pr{\rho, \vp}$.
Since $u$ is defined on $\R^2$, then in polar coordinates, $u(\rho, \vp)$ can be defined for all $\rho \in \R^+$ and all $\vp \in \R$ using periodicity.
We then see that $v(z) = v(r, \te) = u(r^\al, \al \te) = u(T_\al(z))$ is well-defined on $\R^2_+$.
Since
\begin{align*}
\del_r v(r, \te) &= \al r^{\al - 1} \del_\rho u(\rho, \vp) \\
\del_{r}^2 v(r, \te) &= \al^2 r^{2\al - 2} \del_{\rho}^2 u(\rho, \vp) + \al \pr{\al - 1} r^{\al - 2} \del_\rho u(\rho, \vp) \\
\del_{\te}^2 v(r, \te) &= \al^2 \del_{\vp}^2 u(\rho, \vp),
\end{align*}
then
\begin{align*}
\LP v(z)
&= \del_{r}^2 v(r, \te) + \frac 1 r \del_{r} v(r, \te) + \frac 1 {r^2} \del_{\te}^2 v(r, \te) \\
&= \al^2 r^{2\al - 2} \del_{\rho}^2 u(\rho, \vp) + \al \pr{\al - 1} r^{\al - 2} \del_\rho u(\rho, \vp)
+ \al r^{\al - 2} \del_\rho u(\rho, \vp)
+ r^{-2} \al^2 \del_{\vp}^2 u(\rho, \vp) \\
&= \al^2 r^{2\al - 2} \brac{\del_\rho^2 u(\rho, \vp) + \rho^{-1} \del_\rho u(\rho, \vp) + \rho^{-2}\del_{\vp}^2 u(\rho, \vp)} \\
&= \al^2 r^{2\al - 2} \LP u(w)
=  \al^2 r^{2\al - 2} V(w) u(w),
\end{align*}
where we have used \eqref{ellipEq}.
With $W(z) = \al^2 \abs{z}^{2\al - 2} V(T_\al(z)) = \al^2 r^{2\al - 2} V(w)$ as given, the conclusion follows.
\end{proof}

\begin{lem}[Ball containment]
\label{ballContainLemma}
Let $z_0 = (r_0, 0)$ and set $\disp \tilde r = \frac{r_0^{1 - \al}}{2 \sqrt 3 \al}$.
There exists $r_\al > 0$ so that whenever  with $r_0 \ge r_\al$, it holds that $T_\al(B_{\tilde r}(z_0)) \su B_1(T_\al(z_0))$ and $B_1(T_\al(z_0)) \su T_\al\pr{B_1(z_0)}$.
\end{lem}

\begin{proof}
With $\tilde r$ as given, it can be shown that
\begin{equation}
\label{ballIn}
\begin{aligned}
B_{\tilde r}(z_0)
&\su \set{r \in [r_0 - \tilde r, r_0 + \tilde r], \abs{\te} \le \frac {2\tilde r} {\sqrt 3 r_0}} \\
&= \set{r \in \brac{r_0\pr{1 -  \frac{r_0^{- \al}}{2 \sqrt 3 \al}}, r_0\pr{1 +  \frac{r_0^{- \al}}{2 \sqrt 3 \al}}}, \abs{\te} \le \frac {1} {3 \al r_0^\al}}.
\end{aligned}
\end{equation}
With $w_0 = T_\al(z_0)= \pr{r_0^\al, 0} = \pr{\rho_0, 0}$, it follows from \eqref{ballIn} that
\begin{align*}
T_\al(B_{\tilde r}(z_0))
&\su \set{r^\al \in \brac{r_0^\al\pr{1 -  \frac{r_0^{- \al}}{2 \sqrt 3 \al}}^\al, r_0^\al \pr{1 +  \frac{r_0^{- \al}}{2 \sqrt 3 \al}}^\al}, \abs{\al \te} \le \frac {\al} {3 \al r_0^\al}} \\
&= \set{\rho \in \brac{\rho_0 \pr{1 - \frac{\rho_0^{- 1}}{2 \sqrt 3 \al}}^\al, \rho_0 \pr{1 + \frac{\rho_0^{- 1}}{2 \sqrt 3 \al} }^\al}, \abs{\vp} \le \frac {1} {3 \rho_0} }.
\end{align*}
Since
\begin{align*}
\rho_0 \pr{1 \pm \frac{\rho_0^{- 1}}{2 \sqrt 3 \al} }^\al
&= \rho_0 \pm \frac{1}{2 \sqrt 3} + \frac{(1 - \frac 1 \al)}{24} r_0^{- \al} + \ldots
\end{align*}
then there exists $r_1(\al) \gg 1$ so that whenever $r_0 \ge r_1$, we have $\rho_0 - \frac 1 2 \le \rho_0 \pr{1 - \frac{\rho_0^{- 1}}{2 \sqrt 3 \al} }^\al$ and $\rho_0 \pr{1 + \frac{\rho_0^{- 1}}{2 \sqrt 3 \al} }^\al \le \rho_0 +\frac 1 2$.
Because,
\begin{align}
&B_1(w_0)  \supset \set{\rho \in \brac{\rho_0 - \frac 1 2, \rho_0 + \frac 1 2}, \abs{\vp} \le \frac 1 {3\rho_0} },
\label{ballOf}
\end{align}

\begin{equation}
\label{uLower}
\inf_{\abs{w_0} = R} \norm{u}_{L^\iny\pr{B_1(w_0)}} \ge \exp\pr{-R^{1 - \frac N 2 + \eps}}.
\end{equation}
\end{thm}

The results of \cite{Dav14} prove an estimate of the form \eqref{est} with $\be = \frac{4 - 2N}{3} = \frac 4 3 \pr{1 - \frac N 2} > 1 - \frac N 2 + \eps$.
In that article, the assumptions are the same as those in Theorem \ref{mainThm}, except that $u$ may be complex-valued.
Thus, as in the case of bounded $V$, Theorem \ref{mainThm} illustrates that better estimates hold in the real-valued planar setting.

As illustrated by the following example, Theorem \ref{mainThm} is sharp (up to $\eps$) for all $N \in (0, 2)$.
Fix $N \in (0, 2)$, then set $u(z) = \exp\pr{- \abs{z}^{1 - \frac N 2}}$.
A computation shows that $u$ satisfies \eqref{ellipEq} where
$$V(z) := \pr{1 - \frac N 2}^2 \pr{\abs{z}^{1 - \frac N 2} - 1}\abs{z}^{-1 - \frac N 2}$$
satisfies $\abs{V(z)} \lesssim \abs{z}^{-N}$.
On the other hand, for any $\be > 0$, with $u(z) = \abs{z}^{-\be}$ on $\abs{z} > 1$, we see that $\disp \LP u = \be^2 \abs{z}^{-\be-2}$ and therefore $u$ satisfies \eqref{ellipEq} on an exterior domain with $V(z) := \be^2\abs{z}^{-2}$.
In particular, we may not have exponential behavior when $V$ decays fast enough, which explains why we restrict ourselves to $N < 2$.

To prove Theorem \ref{mainThm}, we use an iterative argument that is reminiscent of the one in \cite{Dav14}, see also \cite{LW14, DKW19, Dav20a, Dav25}.
To initialize the iteration, we apply a quantitative estimate of the form \eqref{est} with $\be = 1$.
This result, which verifies Landis' conjecture in the real-valued planar setting, was originally proved by Logunov, Malinnikova, Nadirashvili, and Nazarov in \cite{LMNN20}, and we formulate it in Theorem \ref{LandisGrowth} below.
The iteration argument then relies on repeated applications of Proposition \ref{InductiveProp} which is proved using the ideas from \cite{LMNN20}.
Roughly speaking, Proposition \ref{InductiveProp} shows that if an estimate like \eqref{est} holds with $\be = \be_0$, then for some $x_1$ with $\abs{x_1} \gg \abs{x_0}$, another estimate like \eqref{est} holds with $x_0$ replaced by $x_1$ and $\be = \be_1 \in \brac{1, \be_0}$.
When $\be_0 = 1$, Proposition \ref{InductiveProp} isn't useful, but when $\be_0 > 1$, we can decrease the exponent, i.e., make $\be_1 < \be_0$.
Therefore, to benefit from the iteration argument, we need to transform to a situation where $\be_0 > 1$.
We observe that if $u$ is composed with the real-variable version of the conformal transformation $z \mapsto z^\al$, then the new function also satisfies a Schr\"odinger equation.
By choosing $\al > 1$ appropriately, we can ensure that the new potential function is bounded and that the new solution function satisfies a version of \eqref{est} with $\be > 1$.
By repeatedly applying Proposition \ref{InductiveProp} to the transformed equation, we can make $\be$ arbitrarily close to $1$.
Finally, to reach the conclusion, we undo the change of variables.

We use the notation $B_r(z)$ to denote a ball of radius $r > 0$ centered at the point $z$, abbreviated by $B_r$ when the center is clear.
Generic constants are denoted by $c, C$ and may change from line to line without comment.
Specific constants will be indicated by subscripts.

The article is organized as follows.
In Section \ref{harmonic}, we present a unique continuation theorem for harmonic functions in punctured domains.
The content of this section is very similar to \cite[Section 5]{LMNN20} and \cite[Section 2]{Dav24}.
The iterative result described by Proposition \ref{InductiveProp} is the content of Section \ref{localProof}.
Proposition \ref{InductiveProp} is a three-ball inequality for solutions to Schr\"odinger equations and its proof relies on the results from Section \ref{harmonic}. In Section \ref{TransMaps}, we introduce the real-valued versions of $z \mapsto z^\al$ and record some of their properties.
In particular, we show how solutions behave when they are composed with these transformations.
Finally, the proof of Theorem \ref{mainThm} is presented in Section \ref{MainProof}.

\section{Decay properties of harmonic functions in punctured domains}
\label{harmonic}

In this section, we present and prove quantitative unique continuation results (in the form of three-ball inequalities) for harmonic functions in punctured domains.
We recall the following application of the Harnack inequality which appears in \cite[Claim 5.2]{LMNN20} and is repeated in \cite[Lemma 2.1]{Dav24}.

\begin{lem}[Harnack application]
\label{discBounds}
Let $\set{D_j}$ be a finite collection of $100$-separated unit disks in the plane.
Assume that $h$ is real-valued and harmonic in $\R^2 \setminus \bigcup D_j$ and that for each index $j$, $h$ doesn't change sign in $5D_j \setminus D_j$.
There exists an absolute constant $C_H \ge 10$ for which
\begin{enumerate}
\item $\disp \max_{\del \pr{3 D_j}} \abs{h} \le C_H \min_{\del \pr{3 D_j}} \abs{h}$
\item $\disp \max_{\del \pr{3 D_j}} \abs{\gr h} \le C_H \min_{\del \pr{3 D_j}} \abs{h}$.
\end{enumerate}
\end{lem}

\begin{proof}
An application of the Harnack inequality shows that there exists $C_H > 0$ so that for every $j$
\begin{align*}
\max_{\del \pr{3 D_j}} \abs{h}
&\le \sup_{4 D_j \setminus 2 D_j} \abs{h}
\le C_H \inf_{4 D_j \setminus 2 D_j} \abs{h}
\le C_H \min_{\del \pr{3 D_j}} \abs{h}.
\end{align*}
For each $z \in \del \pr{3 D_j}$, since $h$ doesn't change signs in $B_2(z)$, an application of Cauchy's inequality as in \cite[Lemma 1.11]{HL11} shows that $\abs{\gr h(z)} \le \abs{h(z)}$ and the conclusion follows.
\end{proof}

Now we state and prove the main result of this section.
The following is a slight modification of \cite[Lemma 2.1]{Dav24}, which resembles the result \cite[Theorem 5.3]{LMNN20}.

\begin{prop}[Three-ball inequality for harmonic function in punctured domain]
\label{harmonicProp}
Let $\set{D_j}$ be a finite collection of $100$-separated unit disks in the plane for which $0 \notin \bigcup 3 D_j$.
For some $R \ge 2^{10}$, let $h$ be a harmonic function in $B_R \setminus \bigcup D_j$ with the property that for each index $j$, $h$ doesn't change sign in $\pr{5D_j \setminus D_j} \cap B_R$.
Assume that for some $S \in \pr{2^8, \frac R 2}$ and some $M > \log\pr{16R}$, with $T := R - \frac S {32}$,
it holds that
\begin{equation}
\label{normalization}
\sup_{B_T \setminus \bigcup 3 D_j} \abs{h}  \ge e^{-M} \sup_{B_R \setminus \bigcup 3 D_j} \abs{h}.
\end{equation}

\begin{align}
\label{j2j0Comp}
C_Hm_{j_2} \abs{z_{j_2}}^{-k}
&\ge \sup_{\del \pr{3 D_{j_2}}} \abs{h} \abs{z_{j_2}}^{-k}
\ge \abs{h(s z_{j_0})} \abs{s z_{j_0}}^{-k}
\ge \frac{m_{j_0}}{2} s^{-k} \abs{z_{j_0}}^{-k}.
\end{align}
Since $z_{j_0} \in \del \pr{3 D_{j_0}}$ and $s z_{j_0} \in \del \pr{3 D_{j_2}}$ where $j_0 \ne j_2$, and the balls $\set{D_j}$ are of unit radius and $100$-separated, then $\abs{z_{j_0} - s z_{j_0}} \ge 96$.
After rearrangement, we see that $s^{-k} \ge \pr{1 - \frac{96}{R}}^{-k}$.
Since $10 \le C_H$, $2C_HR \le k$, and $\frac{96}{R} < - \log\pr{1 - \frac{96}{R}}$, then
\begin{align*}
\log\pr{2C_H}
&< 96 \cdot 2 C_H
\le \frac{96}{R} k
< - k \log\pr{1 - \frac{96}{R}}
\le - k \log s,
\end{align*}
from which it follows that $s^{-k} > 2C_H$.
We then conclude from \eqref{j2j0Comp} that $m_{j_2} \abs{z_{j_2}}^{-k} > m_{j_0} \abs{z_{j_0}}^{-k}$ which contradicts \eqref{j0Defn} and gives the desired contradiction.
In other words, \eqref{contraBound} fails to hold and we see that
\begin{align*}
\sup_{B_r \setminus \bigcup 3 D_j} \abs{h}
&> \pr{\frac{16r} R}^{3k}
= \pr{\frac {16r} R}^{3k} \sup_{B_T \setminus \bigcup 3 D_j} \abs{h},
\end{align*}
which implies \eqref{lowerBound} by our choice of $k$.
\end{proof}

\section{The iterative proposition}
\label{localProof}

Here we present the proposition that is used repeatedly in the iteration argument.
This unique continuation result takes the form of a three-ball inequality for solutions to Schr\"odinger equations.
The techniques used to prove this theorem are very similar to those that appear in the proofs of \cite[Theorem 2.2]{LMNN20} and \cite[Theorem 1.2]{Dav24}.

\begin{prop}[Iterative Proposition]
\label{InductiveProp}
Given $a > 0$ and $R \ge \hat R_0(a)$, let $W : B_R \to \R$ satisfy $\norm{W}_{L^\iny(B_R)} \le a^2$ and let $v : B_R \to \R$ be a solution to
$$-\LP v + W v = 0 \; \text{ in } B_R$$
with
\begin{equation}
\label{propUB}
\norm{v}_{L^\iny(B_R)} \le e^{c_1 R}
\end{equation}
for some $c_1 \ge 1$.
Assume that for some $S \in \pr{2^8, \frac R 2}$, there exists $z_0 \in \overline{B}_{R-S}$ and $L \ge 0$ such that
\begin{equation}
\label{propLB}
\abs{v(z_0)} \ge e^{- L}.
\end{equation}
Then there exists $r_0(a, R) > 1$ so that whenever $r \in (0, r_0)$, it holds that
\begin{align}
\label{propConc}
\norm{v}_{L^\iny(B_r)}
&\ge \pr{\frac{r} {R}}^{\tau},
\end{align}

\begin{align}
&B_1(w_0)  \supset \set{\rho \in \brac{\rho_0 - \frac 1 2, \rho_0 + \frac 1 2}, \abs{\vp} \le \frac 1 {3\rho_0} },
\label{ballOf}
\end{align}
then it follows that $T_\al(B_{\tilde r}(z_0))  \su B_1(w_0)$.

On the other hand, with $T_\al^{-1}(B_1(w_0)) = \set{z : T_\al(z) \in B_1(w_0)}$, \eqref{ballIn} (with $\tilde r$ replaced by $1$) implies that
\begin{align*}
T_\al^{-1}(B_1(w_0))
&\su \set{r^\al \in [\rho_0 - 1, \rho_0 + 1], \abs{\al \te} \le \frac {2} {\sqrt 3 \rho_0}} \\
&= \set{r \in \brac{\pr{r_0^\al - 1}^{\frac 1 \al}, \pr{r_0^\al + 1}^{\frac 1 \al}}, \abs{\te} \le \frac {2} {\sqrt 3 \al r_0^\al}}.
\end{align*}
Since
\begin{align*}
\pr{r_0^\al \pm 1}^{\frac 1 \al}
&= r_0 \pr{1 \pm \frac 1 {r_0^\al}}^{\frac 1 \al}
= r_0 \pm \frac 1 {\al r_0^{\al-1}} - \frac{1 - \frac 1 \al}{2 \al r_0^{2\al-1}}  + \ldots
\end{align*}
then there exist $r_2(\al) \gg 1$ so that whenever $r_0 \ge r_2$, we have $\pr{r_0^\al + 1}^{\frac 1 \al} \le r_0 + \frac 1 2$ and $\pr{r_0^\al - 1}^{\frac 1 \al} \ge r_0 - \frac 1 2$.
If $r_0 \ge r_3 := \pr{\frac {2 \sqrt 3} { \al }}^{\frac 1 {\al - 1}}$, then
\begin{align*}
T_\al^{-1}(B_1(w_0))
&\su \set{r \in \brac{r_0 - \frac 1 2, r_0 + \frac 1 2}, \abs{\te} \le \frac {1} {3 r_0}}
\su B_1(z_0),
\end{align*}
where we have used \eqref{ballOf}.
It follows that $B_1(w_0) \su T^\al(B_1(z_0))$.
To complete the proof, choose $r_\al = \max\set{r_1, r_2, r_3}$.
\end{proof}

\section{Proof of Theorem \ref{mainThm}}
\label{MainProof}

In this section, we prove the main result, Theorem \ref{mainThm}.
Before proceeding to the rigorous details, we describe the main steps:
\begin{itemize}
\item Initialize the iterative argument by applying Theorem \ref{LandisGrowth} to the solution function $u$.
\item Compose the solution with the transformation map given in \eqref{TalDefn} to get a solution $v$ to a different Schr\"odinger equation, see Lemma \ref{transformEqLemma}.
\item Apply the iterative result, Proposition \ref{InductiveProp}, to $v$ many times to reduce the exponent.
\item Undo the change of variables to get the desired estimate for $u$.
\end{itemize}
We recall the following result originally proved in \cite{LMNN20}, see also \cite[Theorem 1.1]{Dav24} in the case where $N = 0$ for this formulation.
This estimate serves as the initialization step in our iteration argument.

\begin{thm}[Initialization step]
\label{LandisGrowth}
For some $a_0 > 0$, let $V : \R^2 \to \R$ satisfy $\norm{V}_{L^\iny} \le a_0^2$.
Assume that $u : \R^2 \to \R$ is a solution to \eqref{ellipEq} that satisfies \eqref{solNorm} and for each $w \in \R^2$,
$$\abs{u(w)} \le \exp\pr{c_0 \abs{w}}.$$
Then there exists constants $\overline{C}_0 = \overline{C}_0\pr{a_0, c_0} > 0$ and $\overline{R}_0 > 0$ so that whenever $R \ge \overline{R}_0$, it holds that
\begin{equation}
\label{uLower0}
\inf_{\abs{w_0} = R}\norm{u}_{L^\iny\pr{B_1(w_0)}} \ge \exp\pr{- \overline{C}_0 R \log^{\frac 3 2} R}.
\end{equation}
\end{thm}

Observe that if $u$ satisfies the hypothesis of Theorem \ref{mainThm}, then it also satisfies the hypotheses of Theorem \ref{LandisGrowth}.
We now have all of the tools we need to prove Theorem \ref{mainThm}.

\begin{proof}[Proof of Theorem \ref{mainThm}]

With $N \in (0, 2)$ as given, let $\al =\pr{1 - \frac N 2}^{-1} = \frac 2 {2-N} \in (1, \iny)$.
Set $\de = \min\set{\frac{\eps}{2 - N}, 1}$.
With $r_\al$ from Lemma \ref{ballContainLemma}, $\overline{C}_0(a_0, c_0)$ and $\overline{R}_0$ from Theorem \ref{LandisGrowth}, and $\hat R_0(a)$ from Proposition \ref{InductiveProp}, define $\bar{r}_1 \ge \max\set{r_\al, \overline{R}_0^{\frac 1 \al}, \hat R_0\pr{a_0 \al}}$ as small as possible so that each of the following conditions hold:
\begin{align}
& r^{\de} \ge \overline{C}_0 \al^{\frac 3 2} \log^{\frac 3 2} r
\label{rBig1} \\
&r \log r \ge c_d
\label{rBig2} \\
& r^{\frac \de {1 + \de}} \ge \frac{3 a_0 \al C_H}{2^{14} \pr{5c_0 + 4} } \sqrt{\frac{C_K }{32\ln 2}} \sqrt{\log r}
\label{rBig3} \\
& r^{\frac {\de^2} 2 } \ge \brac{2^{14} \pr{5c_0 + 4} + \frac 2 5 }\log r
\label{rBig4} \\
& r^{\frac {5\de^2} 6 } \ge \brac{2^{14} \pr{5c_0 + 4} + \frac 2 5}\log \brac{\frac{4\al}{ \sqrt 3}  \pr{\frac 3 2 r}^{\al} }
\label{rBig5}.
\end{align}

This article is concerned with the quantitative unique continuation of solutions to elliptic Schr\"odinger equations in the plane.
We consider equations with potentials that exhibits pointwise decay at infinity.
The main result is a nearly sharp estimate for the optimal rate of decay at infinity for real-valued solutions.
This theorem may be interpreted as a quantitative Landis-type theorem.

In the late 1960s, E.~M.~Landis \cite{KL88} conjectured that if $u$ and $V$ are bounded functions that satisfy
\begin{equation}
\label{ePDE}
-\Delta u + V u = 0 \; \text{ in } \, \R^n,
\end{equation}
and $u$ decays faster than exponentially, i.e., $\left\vertu(x)\right\vert \lesssim \exp\pr{- c \left\vertx\right\vert^{1+}}$, then it follows that $u \equiv 0$.
This conjecture was later disproved by Meshkov \cite{M92} who constructed non-trivial complex-valued functions $u$ and $V$ that solve \eqref{ePDE} in $\R^2$, where $V$ is bounded and $\left\vertu(x)\right\vert \lesssim \exp\pr{- c \left\vertx\right\vert^{4/3}}$.
Using Carleman estimate techniques, Meshkov also proved a \textit{qualitative unique continuation} result:
If \eqref{ePDE} holds, where $V$ is bounded and $u$ satisfies a decay estimate of the form $\abs{u\left( x \right) } \lesssim \exp\pr{- c \left\vertx\right\vert^{4/3+}}$, then necessarily $u \equiv 0$.

In their work on Anderson localization \cite{BK05}, Bourgain and Kenig established a quantitative version of Meshkov's result.
As a first step in their proof, they used three-ball inequalities derived from Carleman estimates to establish \textit{order of vanishing} estimates for local solutions to Schr\"odinger equations of the form \eqref{ePDE}.
Then, through a scaling argument, they proved a \textit{quantitative unique continuation} result.
More specifically, they showed that if $u$ and $V$ are bounded and satisfy \eqref{ePDE}, and $u$ is normalized so that $\left\vertu(0)\right\vert \ge 1$, then for sufficiently large values of $R$,
\begin{equation}
 \inf_{|x_0| = R}\left\| u\right\|_{L^\infty\left( B_1(x_0) \right) } \ge \exp{(-CR^{\beta}\log R)},
\label{est}
\end{equation}
where $\beta = \frac 4 3$.
Since $ \frac 4 3 > 1$, the constructions of Meshkov, in combination with the qualitative and quantitative unique continuation theorems just described, indicate that Landis' conjecture cannot be true for complex-valued solutions in $\R^2$.
However, at the time, Landis' conjecture still remained open in the real-valued and higher-dimensional settings.

The first significant step towards resolving Landis' conjecture in the real-valued planar setting was made by Kenig, Silvestre and Wang in \cite{KSW15} where they proved a quantitative form of Landis' conjecture under the assumption that the zeroth-order term satisfies $V \ge 0$ a.e.
The techniques in \cite{KSW15} exploit the relationship between real-valued solutions to second-order elliptic PDEs in the plane and solutions to complex-valued Beltrami equations.
Using similar ideas, analogous results were established in the settings with drift terms \cite{KW15}, variable coefficients \cite{DKW17}, singular lower-order terms \cite{DW20}, and when $V_-$ exhibits decay at infinity \cite{DKW19, Dav20a}.
Then, in \cite{LMNN20}, Logunov, Malinnikova, Nadirashvili, and Nazarov proved Landis' conjecture in the real-valued planar setting.
Their proof uses the nodal structure of the domain along with a domain reduction technique to eliminate any sign condition on the zeroth-order term.
Many of the ideas from \cite{LMNN20} are used in this article.

In \cite{Dav14}, we studied the quantitative unique continuation properties of solutions to elliptic equations of the form
$$\Delta u + W \cdot \nabla u + V u = \lambda u \; \text{ in } \; \R^n,$$
where $V$ and $W$ exhibit pointwise decay at infinity, and $\lambda \in \C$.
With $\left< x \right> = \sqrt{1 + \left\vertx\right\vert^2}$, it was shown that if $\abs{V\left( x \right) } \lesssim \left< x \right>^{-N}$ and $\abs{W\left( x \right) } \lesssim \left< x \right>^{-P}$ for $N, P \ge 0$, then the quantitative estimate \eqref{est} holds with $\beta = \max \set{1, \frac{4-2N}{3}, 2 - 2P}$ and $\log R$ replaced by a different slowly-decaying function.
These quantitative estimates were generalized in \cite{LW14}, where Lin and Wang proved analogous estimates for solutions to the corresponding equations with variable-coefficient leading terms.
Qualitative estimates for similar equations are given in \cite{CS97}.
The constructions presented in \cite{Dav14, Dav15} show that the estimates described in this paragraph are sharp.

In \cite{Dav24}, we studied quantitative unique continuation at infinity for real-valued solutions to \eqref{ePDE} when $n = 2$ and the potential exhibits growth at infinity, i.e., $\left\vertV(z)\right\vert \lesssim \left\vertz\right\vert^N$ for $N > 0$.
The techniques in that article rely heavily on the ideas in \cite{LMNN20} and careful scaling arguments.
Here, we address the more difficult setting where the potential exhibits decay at infinity.
The precise statement of the main theorem is as follows.

\begin{equation}
\label{ePDE}
-\LP u + V u = 0 \; \text{ in } \, \R^n,
\end{equation}

\begin{equation}
 \inf_{|x_0| = R}\norm{u}_{L^\iny\pr{B_1(x_0)}} \ge \exp{(-CR^{\be}\log R)},
\label{est}
\end{equation}

In \cite{Dav24}, we studied quantitative unique continuation at infinity for real-valued solutions to \eqref{ePDE} when $n = 2$ and the potential exhibits growth at infinity, i.e., $\abs{V(z)} \lesssim \abs{z}^N$ for $N > 0$.
The techniques in that article rely heavily on the ideas in \cite{LMNN20} and careful scaling arguments.
Here, we address the more difficult setting where the potential exhibits decay at infinity.
The precise statement of the main theorem is as follows.

The results of \cite{Dav14} prove an estimate of the form \eqref{est} with $\be = \frac{4 - 2N}{3} = \frac 4 3 \pr{1 - \frac N 2} > 1 - \frac N 2 + \eps$.
In that article, the assumptions are the same as those in Theorem \ref{mainThm}, except that $u$ may be complex-valued.
Thus, as in the case of bounded $V$, Theorem \ref{mainThm} illustrates that better estimates hold in the real-valued planar setting.

\begin{prop}[Iterative Proposition]
\label{InductiveProp}
Given $a > 0$ and $R \ge \hat R_0(a)$, let $W : B_R \to \R$ satisfy $\norm{W}_{L^\iny(B_R)} \le a^2$ and let $v : B_R \to \R$ be a solution to
$$-\LP v + W v = 0 \; \text{ in } B_R$$
with
\begin{equation}
\label{propUB}
\norm{v}_{L^\iny(B_R)} \le e^{c_1 R}
\end{equation}
for some $c_1 \ge 1$.
Assume that for some $S \in \pr{2^8, \frac R 2}$, there exists $z_0 \in \overline{B}_{R-S}$ and $L \ge 0$ such that
\begin{equation}
\label{propLB}
\abs{v(z_0)} \ge e^{- L}.
\end{equation}
Then there exists $r_0(a, R) > 1$ so that whenever $r \in (0, r_0)$, it holds that
\begin{align}
\label{propConc}
\norm{v}_{L^\iny(B_r)}
&\ge \pr{\frac{r} {R}}^{\tau},
\end{align}
with
$$\tau(R, S, L, a, c_1) = \max \set{3 a C_H\sqrt{\frac{C_K }{32\ln 2}} R \sqrt{\log R}, 2^{14} \pr{c_1 R + L + \frac {c_d}{\log R}} RS^{-1}} +\frac 1 5 \pr{L + \frac{c_d}{\log R}},$$
where $C_K$ and $c_d$ are universal constants, and $C_H$ is the Harnack constant from Lemma \ref{discBounds}.
\end{prop}

Define $\Om = B_R \setminus \pr{F_0 \bigcup F_1}$ and $\Om_1 = B_R \setminus F_1$.
As shown in \cite[\S 3.1]{LMNN20}, there exists a universal constant $c_P$ (that depends on $c_s$) so that $\Om$ has Poincar\'e constant bounded above by $c_P \rho^2$.
In particular, since $c_P \rho^2 \norm{W}_{L^\iny(B_R)} \le c_P \rho^2 a^2$, then by choosing $\rho \le \rho_0$, a universal constant, we can apply \cite[Lemma 3.2]{LMNN20}.
For $\eps \ll 1$ to be defined later on, let
\begin{equation}
\label{rhoDef}
\rho = \eps a^{-1}.
\end{equation}
An application of the arguments in \cite[\S 3.2]{LMNN20} then shows that there exists $\phi : \Om \to \R$ with the properties that
\begin{align}
&\LP \phi - W \phi = 0 \text{ in } \Om
\nonumber \\
&\phi -1 \in W^{1,2}_0(\Om)
\nonumber \\
&\norm{\phi -1}_\iny \le c_b \pr{\rho a}^2 = c_b \eps^2,
\label{vpuBound}
\end{align}
where $c_b$ is a universal constant that depends on $c_P$, and we have used \eqref{rhoDef}.
By extending $\phi$ to equal $1$ across $F_0 \bigcup F_1$, it is then shown in \cite[Lemma 4.1]{LMNN20} that $\disp f := \frac v \phi \in W^{1,2}_{\loc}(B_R)$ is a weak solution to the divergence-form equation
$$\di\pr{\phi^2 \gr f} = 0 \; \text{ in } \Om_1.$$
Moreover, the bound in \eqref{vpuBound} implies that for any $z \in B_R$,
\begin{equation}
\label{ufComparison}
\pr{1 - c_b \eps^2} \abs{f(z)} \le \abs{v(z)} \le \pr{1 + c_b \eps^2} \abs{f(z)}.
\end{equation}

We introduce the Beltrami coefficient $\mu$, defined as follows:
\begin{equation*}
\mu = \left\{ \begin{array}{ll}
\frac{1 - \phi^2}{1 + \phi^2} \frac{f_x + i f_y}{f_x - i f_y} & \text{ in } \Om_1 \text{ when } \, \gr f \ne 0 \\
0 & \text{ otherwise}
\end{array} \right..
\end{equation*}
Since $\abs{\mu} \lesssim \eps^2$, then as shown in \cite{AIM09}, there exists a $K$-quasiconformal homeomorphism of the complex plane where $K \le 1 + C_K \eps^2$ and $C_K$ depends on $c_b$.
That is, there exists some $w \in W^{1,2}_{\loc}$ that satisfies the Beltrami equation $\disp \frac{\del w}{\del \overline{z}} = \mu \frac{\del w}{\del z}$.
In fact, an application of the Riemann uniformization theorem shows that there exists a $K$-quasiconformal homeomorphism $g: B_R \to B_R$ that is onto with $g(0) = 0$.
Moreover, the function $h : = f \circ g^{-1}$ is harmonic in $g(\Om_1)$.

\begin{thm}[Initialization step]
\label{LandisGrowth}
For some $a_0 > 0$, let $V : \R^2 \to \R$ satisfy $\norm{V}_{L^\iny} \le a_0^2$.
Assume that $u : \R^2 \to \R$ is a solution to \eqref{ellipEq} that satisfies \eqref{solNorm} and for each $w \in \R^2$,
$$\abs{u(w)} \le \exp\pr{c_0 \abs{w}}.$$
Then there exists constants $\overline{C}_0 = \overline{C}_0\pr{a_0, c_0} > 0$ and $\overline{R}_0 > 0$ so that whenever $R \ge \overline{R}_0$, it holds that
\begin{equation}
\label{uLower0}
\inf_{\abs{w_0} = R}\norm{u}_{L^\iny\pr{B_1(w_0)}} \ge \exp\pr{- \overline{C}_0 R \log^{\frac 3 2} R}.
\end{equation}
\end{thm}

Now assume that \eqref{claim} holds for $k-1 \in \N$.
Define $S_k = \frac 1 2 \abs{z_{k-1}}$ and $R_k = \abs{z_{k-1}}^{\al_{k-1} +\de}$.
Comparing with \eqref{zkNorm}, we see that $\abs{z_k} = R_k + S_k - 1$.
Since $B_1(z_{k-1}) \su B_{R_k - S_k}(z_k)$, then the inductive hypothesis shows that there exists $z_0 \in \overline{B}_{R_k - S_k}(z_k)$ so that
\begin{align*}
\abs{v(z_0)}
\ge \exp\pr{- \abs{z_{k-1}}^{\al_{k-1} + \de}}
= \exp\pr{- R_k}.
\end{align*}
The bound in \eqref{uBound} shows that $\disp \abs{v(z)} = \abs{u(T_\al(z))} \le \exp\pr{c_0 \abs{z}^{\al\pr{1 - \frac N 2}}} = \exp\pr{c_0 \abs{z}}$ from which it follows that
\begin{align*}
\norm{v}_{L^\iny\pr{B_{R_k}(z_k)}}
\le \exp\pr{c_0 \pr{\abs{z_k} + R_k}}
\le \exp\pr{\frac {5c_0} 2 R_k}.
\end{align*}
Let $W : \R^2_+ \to \R$ be given by $W(z) = \al^2 \abs{z}^{2\al - 2} V(T_\al(z))= \al^2 \abs{w}^{2 - \frac 2 \al} V(w)$ so that by \eqref{Vbound},
\begin{align*}
\abs{W(z)} = \al^2 \abs{w}^{2 - \frac 2 \al} \abs{V(w)} \le a_0^2 \al^2 \abs{w}^{2 - N - \frac 2 \al} = \pr{a_0 \al}^2.
\end{align*}
An application of Lemma \ref{transformEqLemma} with assumption \eqref{ellipEq} shows that
\begin{equation*}
- \LP v + W v = 0 \; \text{ in } \R^2_+.
\end{equation*}
As $R_k \ge \bar{r}_1$ implies that $R_k \ge \hat R_0(a_0 \al)$ and $S_k \in \pr{2^8, \frac {R_k} 2}$, Proposition \ref{InductiveProp} is applicable with $a = a_0 \al$, $R= R_k$, $S= S_k$, $L = R_k$, and $c_1 = \frac {5c_0} 2$.
With $r = 1$, Proposition \ref{InductiveProp} shows that
\begin{align}
\label{vkBd}
\norm{v}_{L^\iny(B_1(z_k))}
&\ge \exp\pr{-\tau_k(R_k) \log R_k},
\end{align}
where
\begin{equation}
\label{taukDefn}
\begin{aligned}
\tau_k(R_k) &= \max \set{3 a_0 \al C_H\sqrt{\frac{C_K }{32\ln 2}} R_k \sqrt{\log R_k}, 2^{15} \pr{\frac {5c_0} 2 + 1 + \frac {c_d}{R_k \log R_k}} R_k^{2 - \frac 1 {\al_{k-1} + \de}}} \\
&+\frac 1 5 \pr{R_k + \frac{c_d}{\log R_k}}.
\end{aligned}
\end{equation}
Because $R_k \ge \bar{r}_1$ and $\al_{k - 1} > 1$, conditions \eqref{rBig2} and \eqref{rBig3} imply that
\begin{align}
\label{tauKBound}
\tau_k(R_k) &\le \brac{2^{14} \pr{5c_0 + 4} + \frac 2 5} R_k^{2 - \frac 1 {\al_{k-1} + \de}}
\end{align}
while condition \eqref{rBig4} implies that
$$ \brac{2^{14} \pr{5c_0 + 4} + \frac 2 5} \log R_k \le R_k^{\frac {\de^2} 2}.$$
Since $\al > 1$ and $\de \le 1$ implies that $2 - \frac{1}{\al + \de} + \frac{\de^2}2 < 2 - \frac 1 \al + \de$, then combining these bounds shows that
$$\tau_k(R_k) \log R_k \le R_k^{2 - \frac 1 {\al_{k-1}} + \de} = R_k^{\al_{k} + \de}.$$
Returning to \eqref{vkBd}, we conclude that
\begin{align*}
\norm{v}_{L^\iny(B_1(z_k))}
&\ge \exp\pr{-R_k^{\al_{k} + \de}}
\ge \exp\pr{-\abs{z_k}^{\al_{k} + \de}},
\end{align*}
establishing the claim given by \eqref{claim}.

The results of \cite{Dav14} prove an estimate of the form \eqref{est} with $\beta = \frac{4 - 2N}{3} = \frac 4 3 \left( 1 - \frac N 2 \right)  > 1 - \frac N 2 + \varepsilon$.
In that article, the assumptions are the same as those in Theorem \ref{mainThm}, except that $u$ may be complex-valued.
Thus, as in the case of bounded $V$, Theorem \ref{mainThm} illustrates that better estimates hold in the real-valued planar setting.

As illustrated by the following example, Theorem \ref{mainThm} is sharp (up to $\varepsilon$) for all $N \in (0, 2)$.
Fix $N \in (0, 2)$, then set $u(z) = \exp\pr{- \left\vertz\right\vert^{1 - \frac N 2}}$.
A computation shows that $u$ satisfies \eqref{ellipEq} where
$$V(z) := \left( 1 - \frac N 2 \right) ^2 \pr{\left\vertz\right\vert^{1 - \frac N 2} - 1}\left\vertz\right\vert^{-1 - \frac N 2}$$
satisfies $\left\vertV(z)\right\vert \lesssim \left\vertz\right\vert^{-N}$.
On the other hand, for any $\beta > 0$, with $u(z) = \left\vertz\right\vert^{-\beta}$ on $\left\vertz\right\vert > 1$, we see that $\displaystyle \Delta u = \beta^2 \left\vertz\right\vert^{-\beta-2}$ and therefore $u$ satisfies \eqref{ellipEq} on an exterior domain with $V(z) := \beta^2\left\vertz\right\vert^{-2}$.
In particular, we may not have exponential behavior when $V$ decays fast enough, which explains why we restrict ourselves to $N < 2$.

To prove Theorem \ref{mainThm}, we use an iterative argument that is reminiscent of the one in \cite{Dav14}, see also \cite{LW14, DKW19, Dav20a, Dav25}.
To initialize the iteration, we apply a quantitative estimate of the form \eqref{est} with $\beta = 1$.
This result, which verifies Landis' conjecture in the real-valued planar setting, was originally proved by Logunov, Malinnikova, Nadirashvili, and Nazarov in \cite{LMNN20}, and we formulate it in Theorem \ref{LandisGrowth} below.
The iteration argument then relies on repeated applications of Proposition \ref{InductiveProp} which is proved using the ideas from \cite{LMNN20}.
Roughly speaking, Proposition \ref{InductiveProp} shows that if an estimate like \eqref{est} holds with $\beta = \be_0$, then for some $x_1$ with $\left\vertx_1\right\vert \gg \left\vertx_0\right\vert$, another estimate like \eqref{est} holds with $x_0$ replaced by $x_1$ and $\beta = \be_1 \in \left[1, \be_0\right]$.
When $\be_0 = 1$, Proposition \ref{InductiveProp} isn't useful, but when $\be_0 > 1$, we can decrease the exponent, i.e., make $\be_1 < \be_0$.
Therefore, to benefit from the iteration argument, we need to transform to a situation where $\be_0 > 1$.
We observe that if $u$ is composed with the real-variable version of the conformal transformation $z \mapsto z^\alpha$, then the new function also satisfies a Schr\"odinger equation.
By choosing $\alpha > 1$ appropriately, we can ensure that the new potential function is bounded and that the new solution function satisfies a version of \eqref{est} with $\beta > 1$.
By repeatedly applying Proposition \ref{InductiveProp} to the transformed equation, we can make $\beta$ arbitrarily close to $1$.
Finally, to reach the conclusion, we undo the change of variables.

We use the notation $B_r(z)$ to denote a ball of radius $r > 0$ centered at the point $z$, abbreviated by $B_r$ when the center is clear.
Generic constants are denoted by $c, C$ and may change from line to line without comment.
Specific constants will be indicated by subscripts.

The article is organized as follows.
In Section \ref{harmonic}, we present a unique continuation theorem for harmonic functions in punctured domains.
The content of this section is very similar to \cite[Section 5]{LMNN20} and \cite[Section 2]{Dav24}.
The iterative result described by Proposition \ref{InductiveProp} is the content of Section \ref{localProof}.
Proposition \ref{InductiveProp} is a three-ball inequality for solutions to Schr\"odinger equations and its proof relies on the results from Section \ref{harmonic}. In Section \ref{TransMaps}, we introduce the real-valued versions of $z \mapsto z^\alpha$ and record some of their properties.
In particular, we show how solutions behave when they are composed with these transformations.
Finally, the proof of Theorem \ref{mainThm} is presented in Section \ref{MainProof}.

To prove Theorem~\ref{mainThm}, the paper uses “an iterative argument that is reminiscent of the one in \cite{Dav14}.”

Key steps (as described):
\begin{itemize}
\item \textbf{Initialize the iteration.} “To initialize the iteration, we apply a quantitative estimate of the form \eqref{est} with $\beta = 1$,” namely the real-valued planar Landis growth result proved in \cite{LMNN20} and stated as Theorem~\ref{LandisGrowth}.
\item \textbf{Inductive improvement via a three-ball type step.} The iteration “relies on repeated applications of Proposition~\ref{InductiveProp}.” Roughly, this proposition shows that if an estimate like \eqref{est} holds with exponent $\beta=\be_0$, then for some $x_1$ with “$|x_1|\gg |x_0|$,” another estimate like \eqref{est} holds with $x_0$ replaced by $x_1$ and “$\beta=\be_1\in[1,\be_0]$.” Moreover, “when $\be_0>1$, we can decrease the exponent, i.e., make $\be_1<\be_0$.”
\item \textbf{Create a regime with $\beta>1$ via a power map.} Since “when $\be_0=1$, Proposition~\ref{InductiveProp} isn’t useful,” they “transform to a situation where $\be_0>1$” by composing $u$ with “the real-variable version of the conformal transformation $z\mapsto z^\alpha$,” noting the composed function still “satisfies a Schr\"odinger equation.” Choosing “$\alpha>1$ appropriately,” they ensure “the new potential function is bounded” and the new solution satisfies a version of \eqref{est} “with $\beta>1$.”
\item \textbf{Iterate to push $\beta$ down to $1$.} “By repeatedly applying Proposition~\ref{InductiveProp} to the transformed equation, we can make $\beta$ arbitrarily close to $1$.”
\item \textbf{Conclude by reversing the change of variables.} “Finally, to reach the conclusion, we undo the change of variables.”
\end{itemize}

To prove the main theorem, the paper uses “an iterative argument that is reminiscent of the one in \cite{Dav14}.”

Key steps (as described):
\begin{itemize}
\item \textbf{Initialize the iteration.} “To initialize the iteration, we apply a quantitative estimate of the form
\begin{equation}
 \inf_{|x_0| = R}\norm{u}_{L^\iny\pr{B_1(x_0)}} \ge \exp{(-CR^{\be}\log R)},
\label{est}
\end{equation}
with $\beta = 1$,” namely the real-valued planar Landis growth result proved in \cite{LMNN20} and stated as
\begin{thm}[Initialization step]
\label{LandisGrowth}
For some $a_0 > 0$, let $V : \R^2 \to \R$ satisfy $\norm{V}_{L^\iny} \le a_0^2$.
Assume that $u : \R^2 \to \R$ is a solution to \eqref{ellipEq} that satisfies \eqref{solNorm} and for each $w \in \R^2$,
$$\abs{u(w)} \le \exp\pr{c_0 \abs{w}}.$$
Then there exists constants $\overline{C}_0 = \overline{C}_0\pr{a_0, c_0} > 0$ and $\overline{R}_0 > 0$ so that whenever $R \ge \overline{R}_0$, it holds that
\begin{equation}
\label{uLower0}
\inf_{\abs{w_0} = R}\norm{u}_{L^\iny\pr{B_1(w_0)}} \ge \exp\pr{- \overline{C}_0 R \log^{\frac 3 2} R}.
\end{equation}
\end{thm}

\item \textbf{Inductive improvement via a three-ball type step.} The iteration “relies on repeated applications of
\begin{prop}[Iterative Proposition]
\label{InductiveProp}
Given $a > 0$ and $R \ge \hat R_0(a)$, let $W : B_R \to \R$ satisfy $\norm{W}_{L^\iny(B_R)} \le a^2$ and let $v : B_R \to \R$ be a solution to
$$-\LP v + W v = 0 \; \text{ in } B_R$$
with
\begin{equation}
\label{propUB}
\norm{v}_{L^\iny(B_R)} \le e^{c_1 R}
\end{equation}
for some $c_1 \ge 1$.
Assume that for some $S \in \pr{2^8, \frac R 2}$, there exists $z_0 \in \overline{B}_{R-S}$ and $L \ge 0$ such that
\begin{equation}
\label{propLB}
\abs{v(z_0)} \ge e^{- L}.
\end{equation}
Then there exists $r_0(a, R) > 1$ so that whenever $r \in (0, r_0)$, it holds that
\begin{align}
\label{propConc}
\norm{v}_{L^\iny(B_r)}
&\ge \pr{\frac{r} {R}}^{\tau},
\end{align}
with
$$\tau(R, S, L, a, c_1) = \max \set{3 a C_H\sqrt{\frac{C_K }{32\ln 2}} R \sqrt{\log R}, 2^{14} \pr{c_1 R + L + \frac {c_d}{\log R}} RS^{-1}} +\frac 1 5 \pr{L + \frac{c_d}{\log R}},$$
where $C_K$ and $c_d$ are universal constants, and $C_H$ is the Harnack constant from Lemma \ref{discBounds}.
\end{prop}
Roughly, this proposition shows that if an estimate like the equation above holds with exponent $\beta=\be_0$, then for some $x_1$ with “$|x_1|\gg |x_0|$,” another estimate like the equation above holds with $x_0$ replaced by $x_1$ and “$\beta=\be_1\in[1,\be_0]$.” Moreover, “when $\be_0>1$, we can decrease the exponent, i.e., make $\be_1<\be_0$.”

\item \textbf{Create a regime with $\beta>1$ via a power map.} Since “when $\be_0=1$, the iterative proposition above isn’t useful,” they “transform to a situation where $\be_0>1$” by composing $u$ with “the real-variable version of the conformal transformation $z\mapsto z^\alpha$,” noting the composed function still “satisfies a Schr\"odinger equation.” Choosing “$\alpha>1$ appropriately,” they ensure “the new potential function is bounded” and the new solution satisfies a version of the equation above “with $\beta>1$.”

\item \textbf{Iterate to push $\beta$ down to $1$.} “By repeatedly applying the iterative proposition above to the transformed equation, we can make $\beta$ arbitrarily close to $1$.”

\item \textbf{Conclude by reversing the change of variables.} “Finally, to reach the conclusion, we undo the change of variables.” This completes the proof of the main theorem.
\end{itemize}

Question:
Let $a_0\ge 1$ and $N\in(0,2)$. Suppose $V:\mathbb{R}^2\to\mathbb{R}$ satisfies
\[
|V(w)|\le a_0^2\langle w\rangle^{-N}\quad\text{for all }w\in\mathbb{R}^2,
\]
where $\langle w\rangle=\sqrt{1+|w|^2}$. Let $u:\mathbb{R}^2\to\mathbb{R}$ solve
\[
-\Delta u+Vu=0\quad\text{in }\mathbb{R}^2,
\]
with
\[
|u(0)|=1
\]
and the growth bound
\[
|u(w)|\le \exp\big(c_0|w|^{1-N/2}\big)\quad\text{for every }w\in\mathbb{R}^2.
\]
Here $B_1(w_0)$ denotes the unit ball centered at $w_0$. Under these assumptions, which quantitative estimate holds?

For every $\varepsilon\in(0,N/2)$, there exists $R_0=R_0(N,a_0,c_0,\varepsilon)>0$ such that for every $R\ge R_0$,
\[
\inf_{|w_0|=R}\|u\|_{L^\infty(B_1(w_0))}\ge \exp\big(-R^{1-N/2+\varepsilon}\big).
\]

Choices:
B.
For every $\varepsilon\in(0,N/2)$, there exists $R_0=R_0(N,a_0,c_0)>0$ such that for every $R\ge R_0$,
\[
\inf_{|w_0|=R}\|u\|_{L^\infty(B_1(w_0))}\ge \exp\big(-R^{1-N/2}\big).
\]
C.
There exists $R_0=R_0(N,a_0,c_0)>0$ such that for every $R\ge R_0$,
\[
\inf_{|w_0|=R}\|u\|_{L^\infty(B_1(w_0))}\ge \exp\big(-R\big).
\]
D.
For every $\varepsilon\in(0,N/2)$ and every $R_0>0$, there exists $R\ge R_0$ such that
\[
\inf_{|w_0|=R}\|u\|_{L^\infty(B_1(w_0))}\ge \exp\big(-R^{1-N/2+\varepsilon}\big).
\]
E.
For every $\varepsilon\in(0,N/2)$, there exists $R_0=R_0(N,a_0,c_0,\varepsilon)>0$ such that for every $R\ge R_0$,
\[
\inf_{|w_0|\le R}\|u\|_{L^\infty(B_1(w_0))}\ge \exp\big(-R^{1-N/2+\varepsilon}\big).
\]

replace sharp exponent $1-N/2+\varepsilon$ by weaker linear exponent

eventual-for-all-$R$ conclusion weakened to infinitely-many-$R$ statement

sphere condition $|w_0|=R$ replaced by ball condition $|w_0|\le R$

The stem gives the hypotheses and asks for the correct large-radius conclusion, but it does not explicitly reveal the exact bound, quantifier structure, or the infimum/supremum form of the answer.

This is close to a theorem-recall item: the assumptions are essentially those of a specific quantitative unique continuation result, and the correct option is the theorem’s conclusion. However, the presence of nearby alternatives with altered exponents and quantifiers makes it slightly more than a verbatim restatement.

Some reasoning is needed to distinguish the sharp statement from stronger, weaker, or quantifier-tampered alternatives. Still, the item primarily tests recognition of the exact theorem statement rather than generating a conclusion from first principles.

The distractors are strong: one is an overstrong sharpness trap, one is a weaker true statement, one alters quantifier dependence, and one changes infimum to supremum. These reflect realistic mathematical failure modes.

A solid theorem-identification MCQ with high-quality distractors and no answer leakage, but it leans more toward precise theorem recall than genuinely generative reasoning.


\label{thm: main-ucp}
Let $T>0$ and $G\subset [0,T]\times {\mathbb T}$ be a spacetime measurable set. The following statements are equivalent.
\begin{itemize}
\item [(1)] UCP holds on $G$.

\item [(2)] $G$ satisfies \blackhyperref{def:weakGCC}{{\rm (weak GCC)}} and does not obey \blackhyperref{def:OSC}{$({\rm OSC})$} for any non-trivial pair $(A, B)$\footnote{Reminder: for simplicity, when there is no risk of confusion,  sometimes we simply write (OSC) indicating that the set $G$ does not obey the observable symmetry condition for any non-trivial pair $(A, B)$.}.
\end{itemize}

\begin{equation}\label{eq: wave-eq-0}
(\partial_t^2-\partial_x^2)u=0,\quad (u,\partial_tu)\big|_{t=0}=(u_0,u_1)\in \dot H^1(\T)\times L^2(\T),
\end{equation}

\begin{definition}[Symmetric function pair]\label{def:classSc}
We say
$(f,g)$ is a symmetric  function pair if  there exists a decomposition pair $\{(A_k, B_k),\ k = 1,\dots, K\}$, a sequence of different numbers $\{s_k\}_{k= 1}^K$ such that
\begin{equation*}
 f=\sum_{1\leq k\leq K}s_k\chi_{A_k}, \quad g=\sum_{1\leq k\leq K}s_k\chi_{B_k},
\end{equation*}
and
\begin{equation*}
    \int_{\T} (f+ g)(x) \d x= 0.
\end{equation*}
The set of all symmetric function pairs is denoted by $\mathcal{S}^2_c$.
\end{definition}

\begin{equation}\label{eq: wave-ob}
\|u_0\|^2_{\dot H^1(\T)}+\|u_1\|^2_{L^2(\T)}\leq C\int_G|\partial_tu(t,x)|^2\d t\d x.
\end{equation}

Let $T > 0$. Let $G \subset [0,T] \times \mathbb{T}$ be a spacetime measurable set with positive measure.
Consider the {\it observability problem} of the wave equation: whether there exists some $C= C(G)>0$ such that every solution $u$ to
\begin{equation}\label{eq: wave-eq-0}
(\partial_t^2-\partial_x^2)u=0,\quad (u,\partial_tu)\big|_{t=0}=(u_0,u_1)\in \dot H^1({\mathbb T})\times L^2({\mathbb T}),
\end{equation}
satisfies
\begin{equation}\label{eq: wave-ob}
\|u_0\|^2_{\dot H^1({\mathbb T})}+\|u_1\|^2_{L^2({\mathbb T})}\leq C\int_G|\partial_tu(t,x)|^2{\,\rm d} t{\,\rm d} x.
\end{equation}

Thanks to the classical Hilbert uniqueness method and an argument to raise regularity (see Appendix \ref{sec: HUM-app} for details on this reduction), the observability of \eqref{eq: wave-ob} is equivalent to the {\it exact controllability} of
the controlled wave equation on $[0,T]\times{\mathbb T}$ with control force $f\in L^2(G)$:
\begin{equation}\label{eq: wave-eq}
(\partial_t^2-\partial_x^2)u =f\mathbf{1}_G,\quad (u, \partial_t u)\big|_{t=0}=(u_0,u_1)\in \dot H^1({\mathbb T})\times L^2({\mathbb T}).
\end{equation}

Our primary objective is to seek the sufficient and necessary geometric conditions on $G$ such that the unique continuation and observability hold.

\subsection{History and setting}
From the early work of Russell \cite{Russell-1}, the studies of the controllability and observability for wave equations have been central topics in control theory.

\subsubsection{Geometric control condition}
In the pioneering work of Rauch and Taylor \cite{RT}, they first related the observability to a geometric condition on the damped region $\omega$ and the rays of geometric optics in the boundaryless case.
Later, in another pioneering work of Bardos-Lebeau-Rauch \cite{BLR-gcc}, the well-known \textit{Geometric control condition} is introduced: for an open observed region $\omega$, every generalized ray should meet $\omega$ in a finite time.

Since then, it has become one of the most natural assumptions for the controlled waves. In the existing literature, the observation is most often made on cylindrical domains $G=(0,T)\times \omega$, with $\omega$ being an open subset. Under suitable smooth conditions, it is well-known that GCC is sufficient, and depending on the domain, necessary for the observability in the cylindrical domains $(0,T)\times \omega$, see \cite{BLR-gcc,BG-97}. When considering stabilization problems, GCC is also a useful condition for the exponential decay of energy. We refer to \cite{RT,Haraux}. Otherwise, one may have logarithmic type of energy decay results \cite{LR-97, Burq-98}. GCC also plays a role in practical issues, such as sensor designs, tomography techniques used for imaging bodies (see \cite{RouLebeauAnalPDE2017} for example), etc. For a comprehensive reference of the numerical study, we refer to \cite{zuazua-review} and its references therein.

Finally, we give a very brief overview of the boundary control case. There are also fruitful results in this direction. For related GCC, we refer to \cite{Lebeau-boundary}. In particular, for 1D wave equations, one can find many nonlinear results \cite{Li-1,LY-2} and the references therein.

\subsubsection{Unique continuation}
A qualitative version of observability is the unique continuation property.
The easiest way to ensure this property is to apply the analyticity based on Holmgren's theorem. Besides, H\"ormander's pseudo-convexity condition and Carleman estimates are powerful tools dealing with the unique continuation problem. In this direction, there is a large literature such as \cite{RZ-98,Tataru,Hormander-92,Hormander-96} and more recent work includes \cite{Laurent-Leautaud-2019, MS-21,FCL,Shao}. Here we point out that to construct the appropriate weight to apply Carleman estimates, it is crucial to understand the behavior across the suitable spacetime surfaces, which is more delicate than considering a cylindrical region.

\subsubsection{Our setting: spacetime measurable observable region}
Recently, researchers started to focus on
 the spacetime setting \cite{Castro-Cindea-Munch-2014, RouLebeauAnalPDE2017,Shao,PK}. Meanwhile, the study of the case where $G$ is a spacetime region is far from complete, even for open regions.

 In this paper, we focus on the setting that $G\subset[0,T]\times{\mathbb T}$ is measurable of positive measure. The following condition may be viewed as the natural analogue of the standard geometric control condition in this spacetime measurable setting.

\vspace{2mm}
\begin{enumerate}
\item[({\bf GCC})]\label{def:GCC}
Let $T>0$. A measurable set $G \subset [0, T] \times \mathbb{T}$ is said to satisfy
the GCC if there exists a constant
$c_0 > 0$ such that for almost every $x \in \mathbb{T}$,
\begin{equation}\label{eq:GCC-wave:0}
\int_0^T\mathbf{1}_G(s,x\pm s){\,\rm d} s\geq c_0.\footnote{\text{This integral is defined in the sense of \eqref{eq:GCC:measure}.}}
\end{equation}
\end{enumerate}
\vspace{2mm}

From now on, in this paper, \blackhyperref{def:GCC}{({\rm GCC})}  refers to the preceding definition with respect to \eqref{eq:GCC-wave:0}.
It is well-known that under the standard setting, namely the observable region is a cylinder $[0, T]\times \omega$, GCC yields several important properties: weak observability and the necessity of observability. These results can be generalized to the spacetime \blackhyperref{def:GCC}{({\rm GCC})} for a spacetime measurable observable region. We put the former result in Section \ref{sec: weak-ob-sec} and the later in Appendix \ref{sec:app:3}.

\subsection{A new symmetry condition}

Extend the system $2\pi-$periodically to $x\in \mathbb{R}$,  and introduce the null coordinate:
\begin{equation}
    \xi=x+t \; \textrm{ and } \; \eta=x-t. \notag
\end{equation}
Under this new coordinate,
\begin{equation*}
  2  u_{\xi}= u_{x}+ u_t \; \textrm{ and } \;  2  u_{\eta}= u_{x}- u_t
\end{equation*}
and the wave equation \eqref{eq: wave-eq} becomes\footnote{We considered the equation on $U(\xi, \eta):= u(t, x)$ and $F(\xi, \eta)= f(t, x)$. For ease of notation, we still denote $U$ by $u$ and $F$ by $f$.  }
\begin{equation}
    4 \partial_{\xi} \partial_{\eta} u=  f \mathbf{1}_G. \notag
\end{equation}

For any $\xi_0\in\mathbb{R}$, we denote the  line $\{(t,x):x+t=\xi_0\}$ by $L_{\xi=\xi_0}$,  and call it the $\xi$-characteristic. Similarly, we denote the  line $\{(t,x):x- t=\eta_0\}$ by $L_{\eta=\eta_0}$ and  call it $L_{\eta=\eta_0}$ the $\eta$-characteristic. Thus the \blackhyperref{def:GCC}{({\rm GCC})} is equivalent to
\begin{equation}\label{eq:GCC:measure}
      \textrm{ for } a.e. \; x\in {\mathbb T}, \mbox{meas}_\mathbb{R}(G\cap L_{\eta= x})\geq c_0, \;\;
  \mbox{meas}_\mathbb{R}(G\cap L_{\xi=x})\geq c_0.
\end{equation}
We further define measurable cylinders for any measurable sets $A, B\subset [0, 2\pi]$ as
\begin{equation}\label{eq: cylinder}
L_{\xi\in A}:=\bigcup_{\xi_0\in A}L_{\xi=\xi_0} \; \textrm{ and } \; L_{\eta\in B}:=\bigcup_{\eta_0\in B}L_{\eta=\eta_0}.
\end{equation}

Introduce the following symmetry condition:

\vspace{2mm}

\begin{enumerate}
\item[({\bf OSC})]\label{def:OSC}
Let $A, B\subset {\mathbb T}$ be two measurable subsets. $G\in [0, T]\times {\mathbb T}$ is said to obey the {\it observable symmetry condition} for ($A, B$) if\footnote{As usual, $A \Delta B = (A\backslash B)\cup (B\backslash A)$ denotes the symmetric difference between set $A$ and set $B$.

Expression \eqref{wave:nece:2} means that $G\cap L_{\xi\in A} = G\cap L_{\eta\in B}$ modulo zero measure set in $\mathbb{R}^2$.}
\begin{equation}\label{wave:nece:2}
 \mbox{meas}_{\mathbb{R}^2} \Big( [G\cap L_{\xi\in A}] \, \Delta \, [G\cap L_{\eta\in B}]\Big)=0.
\end{equation}
\end{enumerate}

\vspace{2mm}

We call a pair $(A, B)$ trivial, if $|A|= |B|= 0$ or $|A|= |B|= 2\pi$.  In principle, we are interested in {\it non-trivial} pairs $(A, B)$. Since the observable symmetry condition is automatically satisfied for trivial pairs ($A, B$).

The observable symmetry condition is very important in our work, see Fig. \ref{fig:OSC-role}.

\begin{figure}[htp]
    \centering

\tikzset{every picture/.style={line width=0.75pt}}
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]

\draw   (192,38.4) .. controls (192,32.38) and (196.88,27.5) .. (202.9,27.5) -- (452.1,27.5) .. controls (458.12,27.5) and (463,32.38) .. (463,38.4) -- (463,71.1) .. controls (463,77.12) and (458.12,82) .. (452.1,82) -- (202.9,82) .. controls (196.88,82) and (192,77.12) .. (192,71.1) -- cycle ;
\draw   (72,158.7) .. controls (72,152.51) and (77.01,147.5) .. (83.2,147.5) -- (199.8,147.5) .. controls (205.99,147.5) and (211,152.51) .. (211,158.7) -- (211,192.3) .. controls (211,198.49) and (205.99,203.5) .. (199.8,203.5) -- (83.2,203.5) .. controls (77.01,203.5) and (72,198.49) .. (72,192.3) -- cycle ;
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\draw    (329,82.5) -- (329,134.5) ;
\draw [shift={(329,136.5)}, rotate = 270] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (430,82.5) -- (509.44,146.25) ;
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\draw (219,47) node [anchor=north west][inner sep=0.75pt]   [align=left] {Observable symmetry condition};
\draw (81,167.5) node [anchor=north west][inner sep=0.75pt]   [align=left] {Conservation law};
\draw (282,146.5) node [anchor=north west][inner sep=0.75pt]   [align=left] {\begin{minipage}[lt]{66.78pt}\setlength\topsep{0pt}
\begin{center}
Controllability \\Observability \\UCP
\end{center}

\end{minipage}};
\draw (455,167.5) node [anchor=north west][inner sep=0.75pt]   [align=left] {Symetric \ \ \  class };
\draw (515,167.9) node [anchor=north west][inner sep=0.75pt]  [font=\small]  {$\mathcal{S}^{2}$};
\draw (128,106) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize Noether}};
\draw (333,106) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize necessary}};
\draw (486,106) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize generate}};

\end{tikzpicture}

    \caption{The role of (OSC) in this paper}
    \label{fig:OSC-role}
\end{figure}

\vspace{3mm}
The first result is a new conservation law of the forced equation \eqref{eq: wave-eq}.
\begin{proposition}[Conservation law]
\label{prop-sym-conser}
Let $A,B\subset {\mathbb T}$. Assume that $G$ satisfies the ($A, B$)-{\it observable symmetry condition}.
Let $(u,u_t)\in C([0,T];\dot{H}^1({\mathbb T})\times L^2({\mathbb T}))$ be a solution to the forced equation \eqref{eq: wave-eq} and define the energy \footnote{Indeed, under the $(\xi, \eta)-$coordinate this conservation is write as
\begin{equation*}
 \frac{1}{2} I(t)=   \int_{A-t}u_{\xi}(t,x){\,\rm d} x+\int_{B+t}u_{\eta}(t,x){\,\rm d} x.
\end{equation*}
}
$$
I(t)= \int_{A-t}(u_x+u_t)(t,x){\,\rm d} x+\int_{B+t}(u_x-u_t)(t,x){\,\rm d} x.
$$
Then we have
$$
I(t)=I(0), \;  \forall t\in [0, T].
$$
In the case $|A|= |B|= 2\pi$, this conservation law is exactly the condition $\int_{{\mathbb T}} u_x (t, x) {\,\rm d} x= 0$.
\end{proposition}

\begin{remark}
  This proposition can be understood as a variant of Noether’s theorem\footnote{Noether’s theorem: ``Every continuous symmetry of the action (or the equations of motion, in a suitable sense) implies a conserved quantity.''} adapted to a controlled wave equation: a geometric symmetry of the forcing region yields a conserved functional of the solution. Indeed,
the key hypothesis is that the region $G$ satisfies the  {\it observable symmetry condition} for ($A, B$). The conclusion shows that the energy $I(t)$ is invariant in time, regardless of the force term.
\end{remark}

This new type of symmetry provides a {\it necessary condition} for the controllability and observability of \eqref{eq: wave-eq}, and even a {\it necessary condition} for the unique continuation of  \eqref{eq: wave-ob}. See Section \ref{sec:symmetry} for the detailed construction of counterexamples.  Moreover, this construction gives rise to two classes of function pairs satisfying the symmetry condition, $\mathcal{S}^2_c$ and $\mathcal{S}^2$, in Definitions \ref{def:classSc} and \ref{def:classS}.

\begin{figure}[htp]
    \centering
\begin{tikzpicture}[scale=1.0]

  \draw[->] (-0.2,0) -- (6.2,0) node[right] {$x$};
  \draw[->] (0,-0.2) -- (0,6.2) node[above] {$t$};

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  \draw (0,6) node[left] {$2\pi$} -- (0.1,6);

  \draw (0,0) -- (6,0) -- (6,6) -- (0,6) -- cycle;

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\node at (barycentric cs:A5=1,B5=1,C5=1) {\textcolor{white}{-1}};

ll[red!30] (3,6) coordinate (A6) -- (4.5,4.5) coordinate (B6) -- (6,6) coordinate (C6)-- cycle;

\node at (barycentric cs:A6=1,B6=1,C6=1) {\textcolor{white}{1}};

  \draw[dashed] (-0.5,3.5) -- (3.5,-0.5);   \draw[dashed] (0,6) -- (6,0);   \draw[dashed] (2.5,6.5) -- (6.5,2.5);
  \draw[dashed] (-0.5,2.5) -- (3.5,6.5);   \draw[dashed] (0,0) -- (6,6);   \draw[dashed] (2.5,-0.5) -- (6.5,3.5);

\node at (10,4) {$G=G_{0}\cup G_1$};

\node at (10, 3) {$G_{0}=\{(x,t): u_\xi=u_\eta=-1\}$: blue part};

\node at (10, 2) {$G_1=\{(x,t): u_\xi=u_\eta=1\}$: red part};

\end{tikzpicture}
    \caption{Observability fails on $G$, though $G$ satisfies GCC. }
    \label{fig:GCCfail1}
\end{figure}

\vspace{2mm}
\noindent {\bf The counterexample.}
 Let $T= 2\pi$. Let $G\subset [0,2\pi]\times {\mathbb T}$ be given in Fig. \ref{fig:GCCfail1}. Clearly $G$ satisfies \blackhyperref{def:GCC}{({\rm GCC})}.  However, the observable symmetry condition holds for $A=B=(0,\pi)$ or $A=B=(\pi,2\pi)$. Then one can show that the observability inequality \eqref{eq: wave-ob}  fails on $G$, see Section \ref{sec-osc-nece}.

\subsection{Necessary and sufficient unique continuation conditon}
The second result is about the unique continuation property (UCP).  Let $T>0$ and let $G\in [0, T]\times {\mathbb T}$ be a measurable set.  UCP is the qualitative version of the observability and asks:
$$
\mbox{(UCP) \;\;\;\;\;\;\;\;\;\;\;\;\;\; Let $u$ be a solution of \eqref{eq: wave-eq-0} and }   u_t =0 \mbox{ a.e. in } G \Longrightarrow u \equiv 0.  \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
$$

Motivated by the geometric features of the system, we formulate the following minimal geometric assumption on the observation region, which is required for unique continuation.

\vspace{2mm}

\begin{enumerate}[label={}, leftmargin=6.5em, labelindent=3em]
\item[({\bf Weak  GCC})]\label{def:weakGCC}
Let $T>0$. A measurable set $G\subset [0, T]\times \mathbb{T}$ is said to satisfy the
 weak GCC if for almost every
$x\in\mathbb{T}$,
\begin{equation}
\int_0^T\mathbf{1}_G(s,x\pm s){\,\rm d} s>0.   \notag
\end{equation}
\end{enumerate}

\vspace{2mm}

This condition is equivalent to
\begin{equation}
      \textrm{ for } a.e. \; x\in {\mathbb T}, \mbox{meas}_\mathbb{R}(G\cap L_{\eta= x})>0, \;\;
  \mbox{meas}_\mathbb{R}(G\cap L_{\xi=x})>0. \notag
\end{equation}
Compared to the standard \blackhyperref{def:GCC}{({\rm GCC})}, it is not required a uniform positive lower bound.

\vspace{2mm}
\begin{enumerate}
\item[({\bf GCC})]\label{def:GCC}
Let $T>0$. A measurable set $G \subset [0, T] \times \mathbb{T}$ is said to satisfy
the GCC if there exists a constant
$c_0 > 0$ such that for almost every $x \in \mathbb{T}$,
\begin{equation}\label{eq:GCC-wave:0}
\int_0^T\mathbf{1}_G(s,x\pm s){\,\rm d} s\geq c_0.\footnote{\text{This integral is defined in the sense of \eqref{eq:GCC:measure}.}}
\end{equation}
\end{enumerate}
\vspace{2mm}

\begin{enumerate}
\item[({\bf OSC})]\label{def:OSC}
Let $A, B\subset {\mathbb T}$ be two measurable subsets. $G\in [0, T]\times {\mathbb T}$ is said to obey the {\it observable symmetry condition} for ($A, B$) if\footnote{As usual, $A \Delta B = (A\backslash B)\cup (B\backslash A)$ denotes the symmetric difference between set $A$ and set $B$.

\vspace{3mm}
The first result is a new conservation law of the forced equation \eqref{eq: wave-eq}.
\begin{proposition}[Conservation law]
\label{prop-sym-conser}
Let $A,B\subset {\mathbb T}$. Assume that $G$ satisfies the ($A, B$)-{\it observable symmetry condition}.
Let $(u,u_t)\in C([0,T];\dot{H}^1({\mathbb T})\times L^2({\mathbb T}))$ be a solution to the forced equation \eqref{eq: wave-eq} and define the energy \footnote{Indeed, under the $(\xi, \eta)-$coordinate this conservation is write as
\begin{equation*}
 \frac{1}{2} I(t)= \int_{A-t}u_{\xi}(t,x){\,\rm d} x+\int_{B+t}u_{\eta}(t,x){\,\rm d} x.
\end{equation*}
}
$$
I(t)= \int_{A-t}(u_x+u_t)(t,x){\,\rm d} x+\int_{B+t}(u_x-u_t)(t,x){\,\rm d} x.
$$
Then we have
$$
I(t)=I(0), \; \forall t\in [0, T].
$$
In the case $|A|= |B|= 2\pi$, this conservation law is exactly the condition $\int_{{\mathbb T}} u_x (t, x) {\,\rm d} x= 0$.
\end{proposition}

\begin{enumerate}[label={}, leftmargin=6.5em, labelindent=3em]
\item[({\bf Weak GCC})]\label{def:weakGCC}
Let $T>0$. A measurable set $G\subset [0, T]\times \mathbb{T}$ is said to satisfy the
 weak GCC if for almost every
$x\in\mathbb{T}$,
\begin{equation}
\int_0^T\mathbf{1}_G(s,x\pm s){\,\rm d} s>0. \notag
\end{equation}
\end{enumerate}

\vspace{2mm}

This condition is equivalent to
\begin{equation}
 \textrm{ for } a.e. \; x\in {\mathbb T}, \mbox{meas}_\mathbb{R}(G\cap L_{\eta= x})>0, \;\;
 \mbox{meas}_\mathbb{R}(G\cap L_{\xi=x})>0. \notag
\end{equation}
Compared to the standard \blackhyperref{def:GCC}{({\rm GCC})}, it is not required a uniform positive lower bound.

\begin{enumerate}
\item[({\bf GCC})]\label{def:GCC}
Let $T>0$. A measurable set $G \subset [0, T] \times \mathbb{T}$ is said to satisfy
the GCC if there exists a constant
$c_0 > 0$ such that for almost every $x \in \mathbb{T}$,
\begin{equation}\label{eq:GCC-wave:0}
\int_0^T\mathbf{1}_G(s,x\pm s)\d s\geq c_0.\footnote{\text{This integral is defined in the sense of \eqref{eq:GCC:measure}.}}
\end{equation}
\end{enumerate}

\begin{enumerate}
\item[({\bf OSC})]\label{def:OSC}
Let $A, B\subset \T$ be two measurable subsets. $G\in [0, T]\times \T$ is said to obey the {\it observable symmetry condition} for ($A, B$) if\footnote{As usual, $A \Delta B = (A\backslash B)\cup (B\backslash A)$ denotes the symmetric difference between set $A$ and set $B$.

Expression \eqref{wave:nece:2} means that $G\cap L_{\xi\in A} = G\cap L_{\eta\in B}$ modulo zero measure set in $\R^2$.}
\begin{equation}\label{wave:nece:2}
 \mbox{meas}_{\R^2} \Big( [G\cap L_{\xi\in A}] \, \Delta \, [G\cap L_{\eta\in B}]\Big)=0.
\end{equation}
\end{enumerate}

\begin{enumerate}[label={}, leftmargin=6.5em, labelindent=3em]
\item[({\bf Weak  GCC})]\label{def:weakGCC}
Let $T>0$. A measurable set $G\subset [0, T]\times \mathbb{T}$ is said to satisfy the
 weak GCC if for almost every
$x\in\mathbb{T}$,
\begin{equation}
\int_0^T\mathbf{1}_G(s,x\pm s)\d s>0.   \notag
\end{equation}
\end{enumerate}

\begin{equation}\label{eq: wave-eq}
(\partial_t^2-\partial_x^2)u =f\mathbf{1}_G,\quad (u, \partial_t u)\big|_{t=0}=(u_0,u_1)\in \dot H^1(\T)\times L^2(\T).
\end{equation}

\begin{equation}\label{eq:GCC:measure}
      \textrm{ for } a.e. \; x\in \T, \mbox{meas}_\R(G\cap L_{\eta= x})\geq c_0, \;\;
  \mbox{meas}_\R(G\cap L_{\xi=x})\geq c_0.
\end{equation}

\begin{figure}[htp]
 \centering

\subsection{Necessary and sufficient observability/controllability condition}
Finally, we obtain the following sharp and complete geometric characterization of observable regions.
\begin{theorem}\label{thm: main-ob}
Let $T>0$ and let $G\subset [0, T]\times \T$ be a measurable set. Then the following two statements are equivalent.
\begin{itemize}
 \item[(1)] The observability inequality \eqref{eq: wave-ob} holds on $G$;
 \item [(2)] $G$ satisfies \blackhyperref{def:GCC}{{\rm (GCC)}} and does not obey \blackhyperref{def:OSC}{$({\rm OSC})$} for any non-tirvial pair ($A, B$).
\end{itemize}
\end{theorem}

\begin{proposition}[Unique continuation implies weak GCC]
Let $T>0$ and $G\in [0, T]\times \T$. Assume that the following unique continuation property holds
$$
\mbox{ Let u be a solution of \eqref{eq: wave-eq-0} and } u_t =0 \mbox{ a.e. in } G \Longrightarrow u \equiv 0.
$$
Then $G$ satisfies the \blackhyperref{def:weakGCC}{{\rm (weak GCC)}}.
\end{proposition}
\begin{proof}
We argue by contradiction. Suppose that $G$ does not satisfy the \blackhyperref{def:weakGCC}{({\rm weak GCC})}, there exists a subset $A\subset \T$ such that $\mbox{meas}_\R(A)>0$ and
\begin{align*}
\mbox{meas}_\R(G\cap L_{\eta\in A})=0 \quad \mbox{ or } \quad
 \mbox{meas}_\R(G\cap L_{\xi\in A})=0.
\end{align*}
Without loss of generality, we only consider the case
\begin{align}\label{equ-ucp-wgcc-1}
\mbox{meas}_\R(G\cap L_{\eta\in A})=0.
\end{align}
Since $A$ has positive measure, one can find a non-zero function $\phi$ such that
\begin{align}\label{equ-ucp-wgcc-2}
 \int_A \phi(x)\d x=0, \quad \int_A|\phi| \d x=\mbox{meas}_{\R}(A)>0.
\end{align}
Set
$$
u_{0x}=-u_1=\chi_A \phi.
$$
This is possible since $\int_\T u_{0,x}\d x =\int_A\phi \d x =0$. Then we have
$$
\partial_\xi u|_{t=0}=u_{0x}+u_1=0, \quad \partial_\eta u|_{t=0}=u_{0x}-u_1=2\chi_A\phi.
$$
It follows that
$$
\partial_\xi u=0 \quad \mbox{ on } [0,T]\times \T
$$
and
$$
\partial_\eta u= 0 \quad \mbox{ on } L_{\eta\in \T\backslash A}.
$$
Thus we have
$$
u_t=\frac{1}{2}(\partial_\xi u-\partial_\eta u)=0 \quad \mbox{ on } L_{\eta\in \T\backslash A}.
$$
Thanks to \eqref{equ-ucp-wgcc-1}, we see $G\subset L_{\eta\in \T\backslash A}$, and thus $u_t=0$ on $G$. By the UCP, we must have $\partial_\eta u|_{t=0}=2\chi_A\phi\equiv 0$, which leads a contradiction with \eqref{equ-ucp-wgcc-2}.
\end{proof}

\subsection{OSC and weak GCC imply unique continuation }\label{sec: OSC+WGCC>UCP}
In this sequel, we generalize the result in Proposition \ref{prop-ucp-S2}, which forms the following proposition.
\begin{proposition}[Weak GCC implies UCP up to $\mathcal{S}^2$]\label{prop-weakGCC-ucp-S2}
 Let $T>0$. Let $G\in [0, T]\times \T$ satisfy the \blackhyperref{def:weakGCC}{{\rm (weak GCC)}}.
 If $u$ solves the wave equation \eqref{eq: wave-eq-0} and $u_t = 0$ on $G$, then its initial state $(\partial_\xi u, \partial_\eta u)|_{t=0}$ belongs to the symmetry function pair class $\mathcal{S}^2$.
\end{proposition}
\begin{proof}
We use the same notation in the proof of Proposition \ref{prop-ucp-S2}. Recall that we have the relation \eqref{equ-xi-eta-exchange}, namely
\begin{equation*} \partial_{\eta}u|_{\{t= 0\}\cap L_{\eta= \eta_0}}=\partial_{\eta}u(\xi_0,\eta_0)= \partial_{\xi}u|_{\{t= 0\}\cap L_{\xi\in \mathcal{U}_{\eta= \eta_0}}}
\end{equation*}
up to sets with zero measure.
Namely, the value of $\partial_{\eta}u(\xi_0,\eta_0)$ determines almost every value of $\partial_{\xi}u|_{\{t= 0\}\cap L_{\xi\in \mathcal{U}_{\eta= \eta_0}}}$. Since $G$ satisfies the \blackhyperref{def:weakGCC}{({\rm weak GCC})}, we know
$$
\mbox{meas}_\R(\{t= 0\}\cap L_{\xi\in \mathcal{U}_{\eta= \eta_0}})>0.
$$
In other words, $\partial_{\xi}u|_{\{t= 0\}}$ equals to $\partial_{\eta}u(\xi_0,\eta_0)$ at least on a positive measure subset set of $\T$. Thus the set
$$
\mathcal{U}_{(\xi_0,\eta_0)}:=\{x_0\in \T: \partial_{\xi}u|_{\{t= 0,x=x_0\}}=\partial_{\eta}u(\xi_0,\eta_0)\}
$$
has positive measure. Repeating this process for other point $(\xi_0',\eta_0')\in [0,T]\times \T$, we shall find that, there exists a family sets $\mathcal{U}_\ell (\ell\in I)$, $I$ is a index set, such that
\begin{align}\label{equ-Ul-129}
 \mathcal{U}_\ell\cap \mathcal{U}_{\ell'}=\emptyset \mbox{ for } \ell\neq \ell', \quad \mbox{meas}_\R(\mathcal{U}_\ell)>0 \mbox{ for } \ell\in I , \quad \T=\cup_{\ell\in I}\mathcal{U}_\ell
\end{align}
and $\partial_{\xi}u|_{\{t= 0\}}$ is a constant on each $\mathcal{U}_\ell$.
One can show that, under the restrictions \eqref{equ-Ul-129}, the index set $I$ is at most countable, thus, after relabeling if necessary, $\{\mathcal{U}_\ell, \ell\ge 1\}$ is a weak decomposition of $\T$. Moreover, there exists a sequence of different numbers $\{s_\ell\}$ such that
$$
\partial_{\xi}u|_{\{t= 0\}}=s_\ell \mbox{ on } \mathcal{U}_\ell.
$$
Note that $\partial_{\xi}u|_{\{t= 0\}}\in L^2(\T)$, and $\mathcal{U}_\ell$ are disjoint for different $\ell$, we find
$$
\|\partial_{\xi}u|_{\{t= 0\}}\|_{L^2(\T)}^2=\sum_\ell |s_\ell|^2\mbox{meas}_\R(\mathcal{U}_\ell)<\infty.
$$
Thus we obtain
$$
\partial_{\xi}u|_{\{t= 0\}} = \sum_{\ell\geq 1} s_\ell \chi_{\mathcal{U}_\ell} \quad \mbox{ in } L^2(\T).
$$
Similarly, we can find a weak decomposition $\mathcal{V}_{\ell}(\ell\geq 1)$ such that
$$
\partial_{\eta}u|_{\{t= 0\}} = \sum_{\ell\geq 1} s_\ell \chi_{\mathcal{V}_\ell} \quad \mbox{ in } L^2(\T).
$$
In summary, we find that $(\partial_{\xi}u|_{\{t= 0\}}, \partial_{\eta}u|_{\{t= 0\}})\in \mathcal{S}^2$.
\end{proof}

The following theorem gives a sufficient and necessary condition of the observable set for the transport equation. We only state the result for \eqref{eq: transport-eq}, the reader easily figures out the the necessary modifications for \eqref{eq: transport-eq-2}.
\begin{theorem}\label{thm-tran-suff-nece}
Let $T>0$ and $G$ be a measurable subset of $[0,T]\times \T$ with positive measure. Then the observability inequality
\begin{align}\label{equ-tran-suff-necc-1}
 \|u_0\|^2_{L^2(\T)}\leq C\int_G|u(t,x)|^2\d x \d t
\end{align}
holds with a positive constant $C>0$ for all solutions to the transport equation \eqref{eq: transport-eq} if and only if there exists a constant $c_0>0$ such that
\begin{equation}\label{equ-tran-suff-necc-2}
\int_0^T\mathbf{1}_G(s,x- s)\d s\geq c_0 \quad \mbox{ for a.e. } x\in\T
\end{equation}
\end{theorem}
\begin{proof}
The direction $\eqref{equ-tran-suff-necc-2}\Longrightarrow \eqref{equ-tran-suff-necc-1}$ follows clearly from Proposition \ref{prop: tran-eq-ob}. To show the inverse direction, namely $\eqref{equ-tran-suff-necc-1}\Longrightarrow \eqref{equ-tran-suff-necc-2}$, we use the contradiction argument. Suppose that \eqref{equ-tran-suff-necc-2} is not true, then for any $\varepsilon>0$, there exists a set $E_\varepsilon\subset \T$ with positive measure $|E_\varepsilon|>0$ such that
\begin{align}\label{equ-tran-suff-necc-3}
\int_0^T\mathbf{1}_G(s,x- s)\d s<\varepsilon \quad \mbox{ for a.e. } x\in E_\varepsilon.
\end{align}
Let $u_{0\varepsilon}(x)=1_{E_\varepsilon}(x)$. Then $u_{0\varepsilon}\in L^2(\T)$ and its support is contained in $E_\varepsilon$. Then, similar to the proof in Proposition \ref{prop: tran-eq-ob}, we have
\begin{align*}
\int_G|u(t,x)|^2\d x \d t =\int_{\T}|u_{0\varepsilon}(x)|^2\int_{\T}\mathbf{1}_G(t,x-t)\d t\d x\leq \varepsilon \|u_{0\varepsilon}\|^2_{L^2(\T)}.
\end{align*}
Taking $\varepsilon>0$ small enough, say $\varepsilon=C/2$, we obtain a contradiction with \eqref{equ-tran-suff-necc-1}.
\end{proof}
\subsection{GCC is necessary for the wave observability}\label{sec:app:3}
\begin{corollary}\label{wave-GCC-nece}
If the observability \eqref{eq: wave-ob} holds on some observation set $G\subset [0,T]\times \T$, then $G$ needs to satisfies \blackhyperref{def:GCC}{{\rm (GCC)}}.
\end{corollary}
\begin{proof}

\begin{remark}
    The essence of this result lies in the role of the observable symmetry condition. Under the minimal \blackhyperref{def:weakGCC}{{\rm (weak GCC)}} assumption, unique continuation can be established only up to a symmetric function pair belonging to the class $\mathcal{S}^2$, which is generated precisely by the observable symmetry condition (see Definitions~\ref{def:classS}). In other words, the \blackhyperref{def:weakGCC}{{\rm (weak GCC)}} eliminates all obstructions except those arising from the intrinsic symmetry group, and full uniqueness is recovered once solutions are taken modulo this symmetry.
\end{remark}

\begin{figure}[htp]
    \centering

\tikzset{every picture/.style={line width=0.75pt}}
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]

\draw   (141,130) -- (292,130) -- (292,170) -- (141,170) -- cycle ;
\draw   (425,130) -- (539,130) -- (539,170) -- (425,170) -- cycle ;
\draw    (303,146.5) -- (415,146.5) ;
\draw [shift={(417,146.5)}, rotate = 180] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (416,154.5) -- (305,153.52) ;
\draw [shift={(303,153.5)}, rotate = 0.51] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;

\draw (149,143) node [anchor=north west][inner sep=0.75pt]   [align=left] {\blackhyperref{def:OSC}{${\rm OSC}$} \ + \ \blackhyperref{def:weakGCC}{{\rm weak GCC}}};
\draw (462,142) node [anchor=north west][inner sep=0.75pt]   [align=left] {UCP};
\draw (337,130) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize Sec \ref{sec: OSC+WGCC>UCP}}};
\draw (320,156) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize Sec \ref{sec:neosctoucp}, Sec \ref{sec: ucp-wgcc}}};

\end{tikzpicture}

    \caption{Outline of the proof to the equivalence.}
    \label{fig:OSC+WGCC=UCP}
\end{figure}

\subsection{Necessary and sufficient observability/controllability condition}
Finally, we obtain the following sharp and complete geometric characterization of observable regions.
\begin{theorem}\label{thm: main-ob}
Let $T>0$ and let $G\subset [0, T]\times {\mathbb T}$ be a measurable set. Then the following two statements are equivalent.
\begin{itemize}
    \item[(1)] The observability inequality \eqref{eq: wave-ob}  holds on $G$;
    \item [(2)] $G$ satisfies \blackhyperref{def:GCC}{{\rm (GCC)}} and does not obey \blackhyperref{def:OSC}{$({\rm OSC})$} for any non-tirvial pair ($A, B$).
\end{itemize}
\end{theorem}

To the best of our knowledge, this work provides the first necessary and sufficient geometric condition for observability of the wave equation on spacetime region. The observable symmetry condition plays a central role: it completes the classical GCC by capturing the additional geometric structure. This viewpoint also leads to necessary and sufficient conditions in a variety of other important geometric configurations, such as $[0, T]\times \omega$,  measurable Cartesian products  $E_t\times F_{x}$, and general spacetime open sets.  A detailed comparison is presented in Section~\ref{sec:example}.

\begin{figure}[htp]
    \centering

\tikzset{every picture/.style={line width=0.75pt}}
\begin{tikzpicture}[x=0.75pt,y=0.75pt,yscale=-1,xscale=1]

\draw   (178,130) -- (292,130) -- (292,170) -- (178,170) -- cycle ;
\draw   (425,130) -- (539,130) -- (539,170) -- (425,170) -- cycle ;
\draw    (303,146.5) -- (415,146.5) ;
\draw [shift={(417,146.5)}, rotate = 180] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;
\draw    (416,154.5) -- (305,153.52) ;
\draw [shift={(303,153.5)}, rotate = 0.51] [color={rgb, 255:red, 0; green, 0; blue, 0 }  ][line width=0.75]    (10.93,-3.29) .. controls (6.95,-1.4) and (3.31,-0.3) .. (0,0) .. controls (3.31,0.3) and (6.95,1.4) .. (10.93,3.29)   ;

\draw (188,143) node [anchor=north west][inner sep=0.75pt]   [align=left] {\blackhyperref{def:OSC}{${\rm OSC}$} \ + \ \blackhyperref{def:GCC}{{\rm GCC}}};
\draw (439,142) node [anchor=north west][inner sep=0.75pt]   [align=left] {Observability};
\draw (343,130) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize Sec \ref{sec:sharp:obser}}};
\draw (320,156) node [anchor=north west][inner sep=0.75pt]   [align=left] {{\footnotesize Sec \ref{sec-osc-nece}, Sec \ref{sec:app:3}}};

\end{tikzpicture}
    \caption{Outline of the proof to the equivalence.}
    \label{fig:OSC+GCC=OB}
\end{figure}

\begin{remark}
The main theorem reveals two structural features of observability that distinguish the wave equation from other important models, for example, the heat equation,  \cite{AEWZ,WWZZ}. First, the family of observable regions $\mathcal{O}(T)$ depends intrinsically on $T$; non-observable sets arise from geometric obstructions that vary with $T$, leading to a genuinely time-dependent structure for $\{\mathcal{O}(T)\}_{T>0}$.
Second, our characterization also suggests that no waiting time is needed for observability, indicating that, for wave equations, the spacetime geometry of the observation region may play a more decisive role than the duration of control. See Section~\ref{sec:furthercomment}.
\end{remark}

Moreover, we believe that the symmetry mechanism uncovered in this work may extend to other instances of wave equations and to a broader range of models, including coupled systems, semilinear equations, and geometric wave equations. These ideas may also provide insight for higher-dimensional problems, where geometric propagation and microlocal structures are considerably more intricate. In addition, the geometric perspectives here may prove useful in questions of sensor placement, optimization of observation regions, and numerical implementations of control and observability.

\subsection*{Acknowledgements}
The authors would like to thank Nicolas Burq for valuable and useful discussions during the preparation of this manuscript.  Shengquan Xiang is partially supported by the NSFC under grants 12571474 and 12301562.   Ming Wang was partially supported by the NSFC under grants 12571260, 12171442 and 12171178.

A proof sketch for Theorem~\ref{thm: main-ucp} is indicated as follows.

- The key mechanism is the \blackhyperref{def:OSC}{(OSC)} (\emph{observable symmetry condition}): “Under the minimal \blackhyperref{def:weakGCC}{{\rm (weak GCC)}} assumption, unique continuation can be established only up to a symmetric function pair belonging to the class $\mathcal{S}^2$, which is generated precisely by the observable symmetry condition (see Definitions~\ref{def:classS}).”

- Conceptual structure: “the \blackhyperref{def:weakGCC}{{\rm (weak GCC)}} eliminates all obstructions except those arising from the intrinsic symmetry group, and full uniqueness is recovered once solutions are taken modulo this symmetry.”

- The introduction also provides an explicit “Outline of the proof to the equivalence” (Figure~\ref{fig:OSC+WGCC=UCP}), depicting the equivalence between “\blackhyperref{def:OSC}{{\rm OSC}} + \blackhyperref{def:weakGCC}{{\rm weak GCC}}” and “UCP,” with the relevant parts located in “Sec \ref{sec: OSC+WGCC>UCP}” and “Sec \ref{sec:neosctoucp}, Sec \ref{sec: ucp-wgcc}.”

A proof sketch for the main theorem is indicated as follows.

- The key mechanism is the following observable symmetry condition:
\begin{enumerate}
\item[({\bf OSC})]\label{def:OSC}
Let $A, B\subset \T$ be two measurable subsets. $G\in [0, T]\times \T$ is said to obey the {\it observable symmetry condition} for ($A, B$) if\footnote{As usual, $A \Delta B = (A\backslash B)\cup (B\backslash A)$ denotes the symmetric difference between set $A$ and set $B$.

Expression \eqref{wave:nece:2} means that $G\cap L_{\xi\in A} = G\cap L_{\eta\in B}$ modulo zero measure set in $\R^2$.}
\begin{equation}\label{wave:nece:2}
 \mbox{meas}_{\R^2} \Big( [G\cap L_{\xi\in A}] \, \Delta \, [G\cap L_{\eta\in B}]\Big)=0.
\end{equation}
\end{enumerate}
Under the minimal assumption
\begin{enumerate}[label={}, leftmargin=6.5em, labelindent=3em]
\item[({\bf Weak  GCC})]\label{def:weakGCC}
Let $T>0$. A measurable set $G\subset [0, T]\times \mathbb{T}$ is said to satisfy the
 weak GCC if for almost every
$x\in\mathbb{T}$,
\begin{equation}
\int_0^T\mathbf{1}_G(s,x\pm s)\d s>0.   \notag
\end{equation}
\end{enumerate}
unique continuation can be established only up to a symmetric function pair belonging to the class $\mathcal{S}^2$, where
\begin{definition}[Weak symmetric function pair]\label{def:classS}
We say
$(f,g)$ is a weak symmetric  function pair if  there exists a weak decomposition pair $\{(A_k, B_k), k\in I\}$, a sequence of different  numbers $\{s_k\}_{k\in I}$ such that
\begin{equation*}
 f=\sum_{k\in I}s_k\chi_{A_k}, \quad g=\sum_{k\in I}s_k\chi_{B_k} \quad \mbox{ in } L^2(\T)
\end{equation*}
and
\begin{equation*}
    \int_{\T} (f+ g)(x) \d x= 0.
\end{equation*}
The set of all weak symmetric function pairs is denoted by $\mathcal{S}^2$.
\end{definition}
In particular, this class is generated precisely by the observable symmetry condition above.

- Conceptual structure: the weak GCC condition stated above eliminates all obstructions except those arising from the intrinsic symmetry group, and full uniqueness is recovered once solutions are taken modulo this symmetry.

- The introduction also provides an explicit “Outline of the proof to the equivalence” (Figure~\ref{fig:OSC+WGCC=UCP}), depicting the equivalence between “the observable symmetry condition above + the weak GCC condition above” and “UCP,” with the relevant parts located later and in \ref{sec: OSC+WGCC>UCP} and \ref{sec:neosctoucp}, \ref{sec: ucp-wgcc}.

\label{thm: main-ucp}
Let $T>0$ and $G\subset [0,T]\times {\mathbb T}$ be a spacetime measurable set. The following statements are equivalent.
\begin{itemize}
\item [(1)] UCP holds on $G$.

\item [(2)] $G$ satisfies
\begin{enumerate}[label={}, leftmargin=6.5em, labelindent=3em]
\item[({\bf Weak  GCC})]\label{def:weakGCC}
Let $T>0$. A measurable set $G\subset [0, T]\times \mathbb{T}$ is said to satisfy the
 weak GCC if for almost every
$x\in\mathbb{T}$,
\begin{equation}
\int_0^T\mathbf{1}_G(s,x\pm s)\d s>0.   \notag
\end{equation}
\end{enumerate}
and does not obey
\begin{enumerate}
\item[({\bf OSC})]\label{def:OSC}
Let $A, B\subset \T$ be two measurable subsets. $G\in [0, T]\times \T$ is said to obey the {\it observable symmetry condition} for ($A, B$) if\footnote{As usual, $A \Delta B = (A\backslash B)\cup (B\backslash A)$ denotes the symmetric difference between set $A$ and set $B$.

Expression \eqref{wave:nece:2} means that $G\cap L_{\xi\in A} = G\cap L_{\eta\in B}$ modulo zero measure set in $\R^2$.}
\begin{equation}\label{wave:nece:2}
 \mbox{meas}_{\R^2} \Big( [G\cap L_{\xi\in A}] \, \Delta \, [G\cap L_{\eta\in B}]\Big)=0.
\end{equation}
\end{enumerate}
for any non-trivial pair $(A, B)$\footnote{Reminder: for simplicity, when there is no risk of confusion,  sometimes we simply write (OSC) indicating that the set $G$ does not obey the observable symmetry condition for any non-trivial pair $(A, B)$.}.
\end{itemize}

Biconditional or Equivalence
Classification or Bijection

Question: Let $T>0$ and let $G\subset [0,T]\times \mathbb T$ be a measurable spacetime set, where $\mathbb T$ is the one-dimensional torus. Which explicit geometric condition on $G$ is equivalent to the unique continuation property (UCP) on $G$ for the wave setting?

$G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds>0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds>0$, and moreover $G$ does not obey the observable symmetry condition for any non-trivial pair $(A,B)$ of measurable subsets of $\mathbb T$; equivalently, there is no non-trivial pair $(A,B)$ such that $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)=0,$$ where $\Delta$ denotes symmetric difference.

Choices:
B. $G$ satisfies the geometric control condition in the strong sense that there exists a constant $c_0>0$ such that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds\ge c_0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds\ge c_0$, and moreover $G$ does not obey the observable symmetry condition for any non-trivial pair $(A,B)$ of measurable subsets of $\mathbb T$; equivalently, there is no non-trivial pair $(A,B)$ such that $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)=0.$$
C. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds>0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds>0$.
D. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds>0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds>0$, and moreover $G$ does not obey the observable symmetry condition for any pair $(A,B)$ of measurable subsets of $\mathbb T$, including the trivial pair; equivalently, for every measurable pair $(A,B)$ one has $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)>0.$$
E. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ at least one of the two integrals $\int_0^T \mathbf 1_G(s,x+s)\,ds$ or $\int_0^T \mathbf 1_G(s,x-s)\,ds$ is strictly positive, and moreover $G$ does not obey the observable symmetry condition for any non-trivial pair $(A,B)$ of measurable subsets of $\mathbb T$; equivalently, there is no non-trivial pair $(A,B)$ such that $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)=0.$$

restriction to non-trivial symmetry pairs

both characteristic directions required in weak GCC

The stem does not explicitly reveal the correct option. It only asks for an equivalent characterization of UCP, and the answer must be identified by comparing several closely related formulations.

The item is largely a theorem-recall question: the correct choice appears to restate the equivalence theorem with only nearby perturbations in the distractors. It is not a pure tautology because the alternatives differ in meaningful quantifiers and hypotheses.

Some reasoning is required to distinguish weak vs. strong GCC, both-vs.-one visibility, and non-trivial-vs.-all pairs in the symmetry condition. However, the task mainly tests precise recall/recognition of the theorem rather than deep generative reasoning.

The distractors are plausible and mathematically targeted: one is too strong, one is insufficient, one overstates the symmetry exclusion, and one weakens the geometric condition incorrectly. These reflect realistic failure modes.

A solid theorem-recognition MCQ with strong distractors and little answer leakage, but it leans more toward precise recall of a known equivalence than genuine generative reasoning.


Question: Let $T>0$ and let $G\subset [0,T]\times \mathbb T$ be a measurable spacetime set, where $\mathbb T$ is the one-dimensional torus. Which explicit geometric condition on $G$ is equivalent to the unique continuation property (UCP) on $G$ for the wave setting?

Choices:
B. $G$ satisfies the geometric control condition in the strong sense that there exists a constant $c_0>0$ such that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds\ge c_0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds\ge c_0$, and moreover $G$ does not obey the observable symmetry condition for any non-trivial pair $(A,B)$ of measurable subsets of $\mathbb T$; equivalently, there is no non-trivial pair $(A,B)$ such that $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)=0.$$
C. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds>0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds>0$.
D. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ one has both $\int_0^T \mathbf 1_G(s,x+s)\,ds>0$ and $\int_0^T \mathbf 1_G(s,x-s)\,ds>0$, and moreover $G$ does not obey the observable symmetry condition for any pair $(A,B)$ of measurable subsets of $\mathbb T$, including the trivial pair; equivalently, for every measurable pair $(A,B)$ one has $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)>0.$$
E. $G$ satisfies the weak geometric control condition, meaning that for almost every $x\in\mathbb T$ at least one of the two integrals $\int_0^T \mathbf 1_G(s,x+s)\,ds$ or $\int_0^T \mathbf 1_G(s,x-s)\,ds$ is strictly positive, and moreover $G$ does not obey the observable symmetry condition for any non-trivial pair $(A,B)$ of measurable subsets of $\mathbb T$; equivalently, there is no non-trivial pair $(A,B)$ such that $$\operatorname{meas}_{\mathbb R^2}\Big(\,[G\cap\{(t,x):x+t\in A\}]\,\Delta\,[G\cap\{(t,x):x-t\in B\}]\,\Big)=0.$$

The stem does not reveal the correct condition; it only asks for the geometric characterization equivalent to UCP. No explicit wording in the stem singles out choice A.

The item is essentially a direct theorem-recall question: it asks for the exact condition equivalent to UCP, and the correct option states that equivalence almost verbatim rather than requiring application in a new setting.

There is some reasoning pressure because the options differ by subtle strengthenings/weakenings (strong vs weak GCC, both directions vs one, omission or overstrengthening of the symmetry clause). However, solving it is still largely a matter of recalling or recognizing the exact theorem statement rather than generating a new conclusion.

The distractors are mathematically plausible and target natural failure modes: replacing weak by strong GCC, dropping the no-OSC requirement, making the symmetry prohibition too strong, or weakening the two-ray requirement to one. They are distinct and credible.

A solid theorem-recall MCQ with strong distractors and little answer leakage, but it is fairly tautological and only moderately tests generative reasoning.