Weak unique continuation for bounded potentials in three dimensions

Localization at a zero point. Fix $z_0\in U_0$ and choose $r_0>0$ with $B_{r_0}(z_0)\subseteq U_0$ and $R>0$ with ${\bar{B}}_R(z_0)\subseteq \mathrm{\Omega }$ (both possible by [L11]); translate so that $z_0=0$. Choose $\chi \in C_c^{\mathrm{\infty }}(B_R(0);[0,1])$ with $\chi =1$ on $B_{R\mathrm{/}2}(0)$ [L2] and set $u:=\chi \psi$. Since $\psi \in H_{\mathrm{l}\mathrm{o}\mathrm{c}}^2(\mathrm{\Omega })$ and $\chi$ is smooth with compact support in $\mathrm{\Omega }$, the weak Leibniz rule [L3] through order two shows that $u$, extended by zero, lies in $H^2({\mathbb{R}}^3)$ with weak Laplacian the square integrable function $\mathrm{\Delta }u=\chi \mathrm{\Delta }\psi +2⟨\mathrm{\nabla }\chi ,\mathrm{\nabla }\psi ⟩+(\mathrm{\Delta }\chi )\psi$; the hypothesis that $\mathrm{\Delta }\psi =V\psi$ holds distributionally says exactly that the distribution $\mathrm{\Delta }\psi$ is the regular distribution of the locally integrable function $V\psi$, so testing this identity against $\chi \phi$ for arbitrary $\phi \in C_c^{\mathrm{\infty }}(\mathrm{\Omega })$ gives $\chi \mathrm{\Delta }\psi =\chi V\psi$, and $\chi \in C_c^{\mathrm{\infty }}$ is $C^2$ so its classical Laplacian is defined [L10]. Thus $\mathrm{\Delta }u=Vu+e$ as $L^2$ classes with $e:=2⟨\mathrm{\nabla }\chi ,\mathrm{\nabla }\psi ⟩+(\mathrm{\Delta }\chi )\psi$, and $e$ is supported in the annulus $\{R\mathrm{/}2\le \mathrm{\mid }x\mathrm{\mid }\le R\}$ because $\chi =1$, hence $\mathrm{\nabla }\chi =\mathrm{\Delta }\chi =0$, on $B_{R\mathrm{/}2}(0)$. Finally $u=0$ almost everywhere on $B_{r_0}(0)\subseteq U_0$, so $u$ is compactly supported in the annulus $\{0<r_0\le \mathrm{\mid }x\mathrm{\mid }\le R\}$.

Carleman estimate with the potential split. Let $M$ be the essential supremum of $\mathrm{\mid }V\mathrm{\mid }$ over $B_R(0)$ with respect to Lebesgue measure [L5]; it is finite because $B_R(0)$ is compactly contained in $\mathrm{\Omega }$ and $V\in L_{\mathrm{l}\mathrm{o}\mathrm{c}}^{\mathrm{\infty }}(\mathrm{\Omega })$, and by [L5] one has $\mathrm{\mid }V\mathrm{\mid }\le M$ almost everywhere on $B_R(0)$. Applying [L1] to $u$ with parameter $s\ge 1$, $\tau =s+1$, and using $\mathrm{\mid }V\mathrm{\mid }\le M$ together with $r\le R$ on the support of $u$, one has $\mathrm{\parallel }{r}^{-\tau +2}Vu{\mathrm{\parallel }}_2\le M{R}^2\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2$ and therefore $\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2\le \frac{2}{s}(M{R}^2\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2+\mathrm{\parallel }{r}^{-\tau +2}e{\mathrm{\parallel }}_2)$.

Absorption and the inner-ball bound. For every integer $s\ge \max \{1,4M{R}^2\}$ the factor satisfies $2M{R}^2\mathrm{/}s\le 1\mathrm{/}2$, so absorbing the first term in the estimate of step 2.1 gives $\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2\le \frac{4}{s}\mathrm{\parallel }{r}^{-\tau +2}e{\mathrm{\parallel }}_2$; since $e$ is supported in $\{R\mathrm{/}2\le \mathrm{\mid }x\mathrm{\mid }\le R\}$ where $r^{2-\tau }\le (R\mathrm{/}2{)}^{2-\tau }$ (the exponent $2-\tau \le 0$), this yields $\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2\le \frac{4}{s}(R\mathrm{/}2{)}^{2-\tau }\mathrm{\parallel }e{\mathrm{\parallel }}_2$. For every $0<\sigma \le R\mathrm{/}2$ one has $u=\psi$ almost everywhere on $B_{\sigma }(0)$ (because $\chi =1$ on $B_{R\mathrm{/}2}(0)$) and $r\le \sigma$ there, so $\mathrm{\parallel }\psi {\mathrm{\parallel }}_{L^2(B_{\sigma }(0))}=\mathrm{\parallel }u{\mathrm{\parallel }}_{L^2(B_{\sigma }(0))}\le {\sigma }^{\tau }\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2\le \frac{4(R\mathrm{/}2{)}^2}{s}(\frac{2\sigma }{R}{)}^{\tau }\mathrm{\parallel }e{\mathrm{\parallel }}_2$.

Limit and local propagation. For fixed $\sigma <R\mathrm{/}2$ the bound of step 3.1 tends to $0$ as the integer $s\to \mathrm{\infty }$, because $\mathrm{\parallel }e{\mathrm{\parallel }}_2$ is fixed by the localization of step 1.1, $2\sigma \mathrm{/}R<1$ and $\tau =s+1$; hence $\psi =0$ almost everywhere on $B_{\sigma }(0)$ for every $\sigma <R\mathrm{/}2$. Since $B_{R\mathrm{/}2}(0)$ is the countable union of the positive-radius balls $B_{(R\mathrm{/}2)(1-1\mathrm{/}n)}(0)$, $n\ge 2$, and a countable union of null sets is null by [L8] and [L9], it follows that $\psi =0$ almost everywhere on $B_{R\mathrm{/}2}(0)$. Restated without the translation: if $z\in \mathrm{\Omega }$ has a neighbourhood on which $\psi$ vanishes almost everywhere and $R>0$ satisfies ${\bar{B}}_R(z)\subseteq \mathrm{\Omega }$, then $\psi =0$ almost everywhere on $B_{R\mathrm{/}2}(z)$.

The zero set is open in $\mathrm{\Omega }$ and carries no mass. Let $Z:=\bigcup \{U\subseteq \mathrm{\Omega }:U\text{ open},\psi =0\text{ a.e. on }U\}$. Then $Z$ is open and nonempty (it contains $U_0$), and $\psi =0$ almost everywhere on $Z$: for each point of $Z$ some open zero neighbourhood exists by the definition of $Z$, and [L7] supplies a rational box contained in it and containing that point. Consider the entire subfamily $\mathcal{Q}$ of the fixed countable rational-box basis consisting of boxes $Q\subseteq \mathrm{\Omega }$ on which $\psi =0$ almost everywhere. Its union equals $Z$: every $Q\in \mathcal{Q}$ is an open zero neighbourhood, and every point of $Z$ lies in such a box. For a fixed measurable representative, each exceptional set $\{x\in Q:\psi (x)\mathrm{\ne }0\}$ is null; the union over this countable subfamily is null by [L8] and [L9]. Thus $\psi =0$ almost everywhere on $Z$, without selecting a box or radius simultaneously for an uncountable family of points.

$\mathrm{\Omega }\setminus Z$ is open. Let $x\in \mathrm{\Omega }\setminus Z$ and choose $\epsilon >0$ with $\bar{B}(x,\epsilon )\subseteq \mathrm{\Omega }$: [L11] gives $\delta >0$ with $B(x,2\delta )\subseteq \mathrm{\Omega }$, and $\bar{B}(x,\delta )\subseteq B(x,2\delta )\subseteq \mathrm{\Omega }$, so $\epsilon :=\delta$ qualifies. If some $y\in Z\cap B(x,\epsilon \mathrm{/}3)$ existed, then $R:=\epsilon -\mathrm{\mid }x-y\mathrm{\mid }>2\epsilon \mathrm{/}3$; every $z\in {\bar{B}}_R(y)$ satisfies $\mathrm{\mid }z-x\mathrm{\mid }\le \mathrm{\mid }z-y\mathrm{\mid }+\mathrm{\mid }y-x\mathrm{\mid }\le R+\mathrm{\mid }x-y\mathrm{\mid }=\epsilon$, so ${\bar{B}}_R(y)\subseteq \bar{B}(x,\epsilon )\subseteq \mathrm{\Omega }$, and step 4.1 would give $\psi =0$ almost everywhere on $B_{R\mathrm{/}2}(y)$; since $\mathrm{\mid }x-y\mathrm{\mid }<\epsilon \mathrm{/}3<R\mathrm{/}2$, the point $x$ lies in that ball, so $x$ would have a neighbourhood on which $\psi$ vanishes almost everywhere, contradicting $x\mathrm{\notin }Z$. Hence $B(x,\epsilon \mathrm{/}3)\cap Z=\mathrm{\varnothing }$, and $\mathrm{\Omega }\setminus Z$ is open.

Conclusion. The sets $Z$ and $\mathrm{\Omega }\setminus Z$ are disjoint, open, and their union is $\mathrm{\Omega }$, and $Z$ is nonempty; since $\mathrm{\Omega }$ is connected [L6], no separation of $\mathrm{\Omega }$ exists, so $\mathrm{\Omega }\setminus Z=\mathrm{\varnothing }$ and $Z=\mathrm{\Omega }$. By step 5.1, $\psi =0$ almost everywhere on $\mathrm{\Omega }$, which is the class statement $\psi =0$ on $\mathrm{\Omega }$.