Em Boundary And Spectral Prerequisites

Change of variables. Assume first $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\})$. The map $(t,n)\mapsto e^{-t}n$ is a diffeomorphism of $\mathbb{R}\times S^2$ onto ${\mathbb{R}}^3\setminus \{0\}$; since $u$ is smooth and compactly supported away from the origin, the functions $v(t,n):=e^{-t\mathrm{/}2}u(e^{-t}n)$ and $w(t,n):=e^{st}v(t,n)$ are smooth and $w$ is compactly supported in $t$ uniformly in $n$. By [L1] with the substitution $r=e^{-t}$, whose derivative $\mathrm{\mid }\mathrm{d}r\mathrm{/}\mathrm{d}t\mathrm{\mid }=e^{-t}=r$ is supplied by [L2], the measure becomes $dx=e^{-3t}dtd\sigma (n)$; writing $u(x)=e^{(1\mathrm{/}2-s)t}w(t,n)$ gives $r^{-2\tau }\mathrm{\mid }u{\mathrm{\mid }}^{2}dx=e^{2\tau t}{e}^{(1-2s)t}\mathrm{\mid }w{\mathrm{\mid }}^2{e}^{-3t}dtd\sigma =\mathrm{\mid }w{\mathrm{\mid }}^{2}dtd\sigma ,$ because $2\tau -2s-2=0$. Hence ${\int }_{{\mathbb{R}}^3}{r}^{-2\tau }\mathrm{\mid }u{\mathrm{\mid }}^{2}dx={\int }_{\mathbb{R}}{\int }_{S^2}\mathrm{\mid }w{\mathrm{\mid }}^{2}d\sigma dt$.

Spherical Laplacian and the polar form of $\mathrm{\Delta }$. For $g\in C^2(S^2)$ let $G(x):=g(x\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$ be its degree-zero homogeneous extension to ${\mathbb{R}}^3\setminus \{0\}$ and define ${\mathrm{\Delta }}_{S^2}g(n):=\mathrm{\Delta }G(n)$ for $n\in S^2$. This is well defined: by the chain rule, $\mathrm{\Delta }(G\circ m_t)=t^2(\mathrm{\Delta }G)\circ m_t$ for every $t>0$ and every $C^2$ function on ${\mathbb{R}}^3\setminus \{0\}$, while $G\circ m_t=G$; hence $(\mathrm{\Delta }G)(tn)=t^{-2}\mathrm{\Delta }G(n)$, and $r^2\mathrm{\Delta }G(rn)=\mathrm{\Delta }G(n)$ does not depend on $r$, so the value at $r=1$ defines a function of $n$ alone. Define ${\mathrm{\nabla }}_{S^2}g(n):=\mathrm{\nabla }G(n)$, which is tangent because $n\cdot \mathrm{\nabla }G(n)=0$; for a smooth tangential field $Y$ on $S^2$, define ${\mathrm{div}}_{S^2}Y(n):=\mathrm{div}\tilde{Y}(n)$ with $\tilde{Y}(x)=Y(x\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$. For $\psi \in C^2((0,\mathrm{\infty }))$ the product rule gives $\mathrm{\nabla }(\psi (r)G(x))={\psi }^{\mathrm{\prime }}(r)G(x)x\mathrm{/}r+\psi (r)\mathrm{\nabla }G(x)$, and $⟨\mathrm{\nabla }G(x),x⟩\mathrm{/}r={\mathrm{\partial }}_{r}G=0$; using [L3] twice and the direct computation $\mathrm{div}(x\mathrm{/}r)={\sum }_i{\mathrm{\partial }}_i(x_i\mathrm{/}r)=3\mathrm{/}r-r^2\mathrm{/}{r}^3=2\mathrm{/}r$, $\mathrm{div}\left({\psi }^{\mathrm{\prime }}G\frac{x}{r}\right)={\psi }^{\mathrm{\prime }\mathrm{\prime }}G+{\psi }^{\mathrm{\prime }}G\frac{2}{r},\mathrm{div}(\psi \mathrm{\nabla }G)={\psi }^{\mathrm{\prime }}⟨\frac{x}{r},\mathrm{\nabla }G⟩+\psi \mathrm{\Delta }G=\psi \mathrm{\Delta }G,$ so that $\mathrm{\Delta }(\psi (r)G)=({\psi }^{\mathrm{\prime }\mathrm{\prime }}(r)+\frac{2}{r}{\psi }^{\mathrm{\prime }}(r))G+r^{-2}\psi (r){\mathrm{\Delta }}_{S^2}g(n)$, the last step using $\mathrm{\Delta }G(rn)=r^{-2}{\mathrm{\Delta }}_{S^2}g(n)$. In particular, for a harmonic homogeneous polynomial $H_l$ of degree $l$ and $h:=H_l{\mathrm{\mid }}_{S^2}$, the identity with $\psi (r)=r^l$ and $G(x)=H_l(x)\mathrm{/}\mathrm{\mid }x{\mathrm{\mid }}^l$ gives $0=\mathrm{\Delta }{H}_l=r^{l-2}(l(l+1)h+{\mathrm{\Delta }}_{S^2}h)$, hence ${\mathrm{\Delta }}_{S^2}h=-l(l+1)h$. The same computation applied to a general $F\in C^2({\mathbb{R}}^3\setminus \{0\})$ with $f(r,n):=F(rn)$, whose gradient splits as $\mathrm{\nabla }F=({\mathrm{\partial }}_{r}f)n+r^{-1}{\mathrm{\nabla }}_{S^2,n}f$ into radial and tangential parts, gives the general polar form $\mathrm{\Delta }F(rn)={\mathrm{\partial }}_r^{2}f+\frac{2}{r}{\mathrm{\partial }}_{r}f+r^{-2}{\mathrm{\Delta }}_{S^2,n}f$; the divergence of a tangential field $Y$ with $Y(x)=\tilde{Y}(x\mathrm{/}\mathrm{\mid }x\mathrm{\mid })$ is $r^{-1}{\mathrm{div}}_{S^2}\tilde{Y}$ in these coordinates, and ${\mathrm{div}}_{S^2}{\mathrm{\nabla }}_{S^2}={\mathrm{\Delta }}_{S^2}$.

The conjugated operator. Apply the polar form of step 1.2 to $u(x)=r^{-1\mathrm{/}2}v(t,n)$, $t=-\ln r$, whose radial part is $f(r,n)=r^{-1\mathrm{/}2}v(-\ln r,n)$. Coordinate differentiation by [L4] gives ${\mathrm{\partial }}_{r}u=-\frac{1}{2}{r}^{-3\mathrm{/}2}v-r^{-3\mathrm{/}2}{v}_t$ and ${\mathrm{\partial }}_r^{2}u=r^{-5\mathrm{/}2}(v_{tt}+2v_t+\frac{3}{4}v)$, so ${\mathrm{\partial }}_r^{2}u+\frac{2}{r}{\mathrm{\partial }}_{r}u=r^{-5\mathrm{/}2}(v_{tt}-\frac{1}{4}v),$ while $r^{-2}{\mathrm{\Delta }}_{S^2,n}u=r^{-5\mathrm{/}2}{\mathrm{\Delta }}_{S^2}v$. Hence $\mathrm{\Delta }u=r^{-5\mathrm{/}2}(v_{tt}+{\mathrm{\Delta }}_{S^2}v-\frac{1}{4}v)\mathrm{.}$ Substituting $v=e^{-st}w$ gives $v_{tt}=e^{-st}({\mathrm{\partial }}_t-s{)}^{2}w$ and ${\mathrm{\Delta }}_{S^2}v=e^{-st}{\mathrm{\Delta }}_{S^2}w$, so with $L:=({\mathrm{\partial }}_t-s{)}^2+{\mathrm{\Delta }}_{S^2}-\frac{1}{4}$ one has $\mathrm{\Delta }u=r^{-5\mathrm{/}2}{e}^{-st}Lw\mathrm{.}$

Modal decomposition. For almost every $t$ the function $w(t,\cdot )$ is smooth on $S^2$, hence in $L^2(S^2,d\sigma )$, and by [L5] it expands as $w(t,\cdot )={\sum }_{l,m}{w}_{lm}(t)Y_{lm}$ in a complete orthonormal basis $\{Y_{lm}\}$ of degree-$l$ spherical harmonics, the coefficients $w_{lm}(t)=⟨w(t,\cdot ),Y_{lm}⟩$ being smooth in $t$. For complex-valued $w$, apply the real harmonic basis expansion to its real and imaginary parts and combine the coefficients; orthogonality then gives the same complex $L^2$ expansion and absolute-square Parseval identity. By step 1.2, each $Y_{lm}$ satisfies ${\mathrm{\Delta }}_{S^2}{Y}_{lm}=-l(l+1)Y_{lm}$, and ${\mathrm{\Delta }}_{S^2}$ is symmetric for the surface pairing: extend two smooth sphere functions $g,h$ homogeneously with degree zero to the annulus $1<\mathrm{\mid }x\mathrm{\mid }<2$. Their normal derivatives vanish on both boundary spheres. Subtracting the two first Green identities [L13] gives ${\int }_{1<\mathrm{\mid }x\mathrm{\mid }<2}(H\mathrm{\Delta }G-G\mathrm{\Delta }H)=0$; using $\mathrm{\Delta }G(rn)=r^{-2}{\mathrm{\Delta }}_{S^2}g(n)$ and polar coordinates [L1], the radial factor is ${\int }_1^{2}dr=1$, so ${\int }_{S^2}({\mathrm{\Delta }}_{S^2}g)hd\sigma ={\int }_{S^2}g({\mathrm{\Delta }}_{S^2}h)d\sigma$. Applying this to the real and imaginary parts of $w(t,\cdot )$ and to $Y_{lm}$ gives the modal identity without any unproved surface integration formula, so the coefficients of $Lw$ against $Y_{lm}$ are exactly $(L_l{w}_{lm})(t)$; hence $Lw(t,\cdot )={\sum }_{l,m}(L_l{w}_{lm})(t)Y_{lm},L_l:=({\mathrm{\partial }}_t-s{)}^2-(l+\frac{1}{2}{)}^2,$ since $-l(l+1)-\frac{1}{4}=-(l+\frac{1}{2}{)}^2$. Parseval for the complete family at each $t$ [L6] and Tonelli's theorem against the counting measure on the index set [L8] give ${\int }_{\mathbb{R}}{\int }_{S^2}\mathrm{\mid }w{\mathrm{\mid }}^{2}d\sigma dt={\sum }_{l,m}{\int }_{\mathbb{R}}\mathrm{\mid }{w}_{lm}{\mathrm{\mid }}^{2}dt,{\int }_{\mathbb{R}}{\int }_{S^2}\mathrm{\mid }Lw{\mathrm{\mid }}^{2}d\sigma dt={\sum }_{l,m}{\int }_{\mathbb{R}}\mathrm{\mid }{L}_l{w}_{lm}{\mathrm{\mid }}^{2}dt,$ all terms being nonnegative; each $w_{lm}$ is smooth and compactly supported on $\mathbb{R}$, and $L_l{w}_{lm}$ is its second derivative combination, again compactly supported and square integrable.

Fourier multiplier bound, mode by mode. Fix $(l,m)$. In the $2\pi$ convention of [L7], the distributional Fourier transform of $L_l{w}_{lm}$ is the bounded continuous function $m_l{\widehat{w}}_{lm}$ with symbol $m_l(\xi )=(2\pi i\xi -s{)}^2-(l+\frac{1}{2}{)}^2$, and Plancherel gives $\mathrm{\parallel }{L}_l{w}_{lm}{\mathrm{\parallel }}_2^2={\int }_{\mathbb{R}}\mathrm{\mid }{m}_l(\xi ){\mathrm{\mid }}^2\mathrm{\mid }{\widehat{w}}_{lm}(\xi ){\mathrm{\mid }}^{2}d\xi$. Put $x=4{\pi }^2{\xi }^2\ge 0$, $A=s-l-\frac{1}{2}$ and $B=s+l+\frac{1}{2}$. Then $s^2-(l+\frac{1}{2}{)}^2=AB$ and $A^2+B^2+2AB=(A+B{)}^2=4s^2$, so $\mathrm{\mid }{m}_l(\xi ){\mathrm{\mid }}^2=(AB-x{)}^2+4s^{2}x=(x+A^2\left)\right(x+B^2)\ge A^2{B}^2\mathrm{.}$ For integers $l\ge 0$ and $s\ge 1$ one has $\mathrm{\mid }AB\mathrm{\mid }=\mathrm{\mid }{s}^2-(l+\frac{1}{2}{)}^2\mathrm{\mid }\ge s-\frac{1}{4}$: on the range $l\le s-2$, nonempty only for $s\ge 2$, the function $l\mapsto s^2-(l+\frac{1}{2}{)}^2$ is decreasing, so $AB\ge s^2-(s-\frac{3}{2}{)}^2=3s-\frac{9}{4}\ge s-\frac{1}{4}$; at $l=s-1$ and $l=s$ the values are $s-\frac{1}{4}$ and $s+\frac{1}{4}$; and on the range $l\ge s+1$ one has $\mathrm{\mid }AB\mathrm{\mid }=(l+\frac{1}{2}{)}^2-s^2\ge (s+\frac{3}{2}{)}^2-s^2=3s+\frac{9}{4}\ge s-\frac{1}{4}$. Since $s-\frac{1}{4}\ge \frac{s}{2}$ for every $s\ge 1$, conclude $\mathrm{\mid }{m}_l(\xi ){\mathrm{\mid }}^2\ge s^2\mathrm{/}4$ for all $\xi$, and therefore $\mathrm{\parallel }{w}_{lm}{\mathrm{\parallel }}_2\le \frac{2}{s}\mathrm{\parallel }{L}_l{w}_{lm}{\mathrm{\parallel }}_2\mathrm{.}$

The weighted right-hand side. Combining step 2.1 with the measure identity of step 1.1 gives $r^{-2\tau +4}\mathrm{\mid }\mathrm{\Delta }u{\mathrm{\mid }}^{2}dx=r^{-2\tau +4}\cdot r^{-5}{e}^{-2st}\mathrm{\mid }Lw{\mathrm{\mid }}^2\cdot r^{3}dtd\sigma =r^{2-2\tau }{e}^{-2st}\mathrm{\mid }Lw{\mathrm{\mid }}^{2}dtd\sigma =\mathrm{\mid }Lw{\mathrm{\mid }}^{2}dtd\sigma ,$ because $r^{2-2\tau }=e^{(2\tau -2)t}=e^{2st}$; hence ${\int }_{{\mathbb{R}}^3}{r}^{-2\tau +4}\mathrm{\mid }\mathrm{\Delta }u{\mathrm{\mid }}^{2}dx={\int }_{\mathbb{R}}{\int }_{S^2}\mathrm{\mid }Lw{\mathrm{\mid }}^{2}d\sigma dt\mathrm{.}$

Summation. Squaring the mode-wise inequality of step 2.3, summing over all $(l,m)$ and inserting the two identities of step 2.2 and step 3.1 yields ${\int }_{{\mathbb{R}}^3}{r}^{-2\tau }\mathrm{\mid }u{\mathrm{\mid }}^{2}dx={\sum }_{l,m}\int \mathrm{\mid }{w}_{lm}{\mathrm{\mid }}^2\le \frac{4}{s^2}{\sum }_{l,m}\int \mathrm{\mid }{L}_l{w}_{lm}{\mathrm{\mid }}^2=\frac{4}{s^2}{\int }_{{\mathbb{R}}^3}{r}^{-2\tau +4}\mathrm{\mid }\mathrm{\Delta }u{\mathrm{\mid }}^{2}dx,$ which is the asserted inequality after taking square roots. Both sides are finite because $u$ is compactly supported in an annulus on which the two weights are bounded.

Density extension to $H^2$ functions supported in an annulus. Now let $u\in H^2({\mathbb{R}}^3)$ be compactly supported in $\{0<a\le \mathrm{\mid }x\mathrm{\mid }\le b\}$, so its weak derivatives through order two exist as locally square integrable functions and $\mathrm{\Delta }u$ is their weak sum [L12]. By [L9] fix ${\chi }_0\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3;[0,1])$ with ${\chi }_0=1$ on ${\bar{B}}_{a\mathrm{/}4}(0)$ and $\mathrm{supp}{\chi }_0\subseteq B_{a\mathrm{/}2}(0)$, and put $\chi :=1-{\chi }_0$; then $\chi$ is smooth, $\chi =0$ on ${\bar{B}}_{a\mathrm{/}4}(0)$, $\chi =1$ outside $B_{a\mathrm{/}2}(0)$, and $\chi u=u$ because $\chi =1$ on a neighbourhood of $\mathrm{supp}u$. By [L10] choose $u_k\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$ with $u_k\to u$ in $W^{2,2}$; then $\chi u_k\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3\setminus \{0\})$ with $\chi u_k\to \chi u=u$ in $W^{2,2}$, because the weak Leibniz expansion of [L11] through order two bounds $\mathrm{\parallel }\chi (u_k-u){\mathrm{\parallel }}_{W^{2,2}}$ by a constant times $\mathrm{\parallel }{u}_k-u{\mathrm{\parallel }}_{W^{2,2}}$, and likewise $\mathrm{\Delta }(\chi u_k)\to \mathrm{\Delta }(\chi u)=\mathrm{\Delta }u$ in $L^2$. Every $\chi u_k$ and every $\mathrm{\Delta }(\chi u_k)=\chi \mathrm{\Delta }{u}_k+2⟨\mathrm{\nabla }\chi ,\mathrm{\nabla }{u}_k⟩+(\mathrm{\Delta }\chi )u_k$ is supported in $\{\mathrm{\mid }x\mathrm{\mid }\ge a\mathrm{/}4\}$, as is $\mathrm{\Delta }u$ because weak derivatives of a class supported in $\{a\le \mathrm{\mid }x\mathrm{\mid }\le b\}$ are supported there, and on $\{\mathrm{\mid }x\mathrm{\mid }\ge a\mathrm{/}4\}$ the weights $r^{-\tau }$ and $r^{-\tau +2}$ are bounded by $(a\mathrm{/}4{)}^{-\tau }$ and by $(a\mathrm{/}4{)}^{-\tau +2}$ (the exponent $2-\tau \le 0$). Applying step 4.1 to each $\chi u_k$ and passing to the limit in $\mathrm{\parallel }{r}^{-\tau }\chi u_k{\mathrm{\parallel }}_2\le \frac{2}{s}\mathrm{\parallel }{r}^{-\tau +2}\mathrm{\Delta }(\chi u_k){\mathrm{\parallel }}_2$ gives $\mathrm{\parallel }{r}^{-\tau }u{\mathrm{\parallel }}_2\le \frac{2}{s}\mathrm{\parallel }{r}^{-\tau +2}\mathrm{\Delta }u{\mathrm{\parallel }}_2$, and the weak derivatives and weak Laplacian of $u$ are those named in the statement.

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.