Proof. A Radon measure on Euclidean space is concentrated on its support. If \(D_E<\infty\), choose a maximal \(D_E/4\)-separated subset of \(E\). The disjoint ambient balls of radius \(D_E/8\) around its points lie in a ball of radius \(9D_E/8\) around any fixed point of \(E\). Comparison of their \(d\)-dimensional volumes bounds their number by \(9^d\). Maximality gives a cover of \(E\) by closed balls of radius \(D_E/4\), hence by open balls of radius \(D_E/2\). The asserted total-mass bound follows from AD upper regularity at their centers.

If \(\mu(B(a,r))>0\), choose \(x\in E\cap B(a,r)\). When \(2r\le D_E\), \(B(a,r)\subset B(x,2r)\) and the AD upper bound gives \(\mu(B(a,r))\le 2^n C_{\rm AD}r^n\). When \(2r>D_E\), the preceding total-mass estimate gives \(\mu(B(a,r))\le9^dC_{\rm AD}r^n\). For \(D_E=\infty\) only the first case is needed. Since \(n\le d\), \(2^n\le9^d\).

Finally, dyadic annuli and (3) give \[\int_{|a-y|>\varepsilon}|K_n(a-y)|^2\,d\mu(y) \le 2^nG\varepsilon^{-n}\sum_{j=0}^{\infty}2^{-jn}<\infty.\] Cauchy–Schwarz proves the last assertion. ◻

Proof. The kernel is odd and satisfies \[|K_n(z)|=|z|^{-n},\qquad |\nabla K_n(z)|\le C_n|z|^{-n-1}.\] For \(h>0\), upper growth and a dyadic decomposition toward the origin give \[ \int_{0<|x-y|<h}|x-y|^{1-n}\,d\sigma(y)\le C_n G h. \tag{6}\] Multiplication by \(\mathop{\mathrm{Lip}}\varphi\) bounds the diagonal singularity in the first integral of (5). The second integral is separated from the support of \(\varphi\) at its finite boundary, and for \(|y-a|\ge2R\) the kernel difference is bounded by \(C_nR|y-a|^{-n-1}\). Summation over exterior annuli proves absolute convergence and the stated tail bound.

After enlarging to a common localization ball containing both subtraction points, changing that point adds a vector independent of \(x\), whose integral against \(\varphi\) is zero. Changing the localization ball merely moves a finite cross term between the two integrals: expanding it, using oddness, and again using \(\int\varphi\,d\sigma=0\) shows that their sum is unchanged. This calculation can first be made with both an inner and an outer cutoff; the estimates just proved justify removing them.

The difference between a hard cutoff at \(s\) and a bounded smooth cutoff that vanishes on \([0,s]\) and equals one on \([2s,\infty)\) has kernel bounded by \(C s^{-n}\mathbf1_{\{s<|x-y|<2s\}}\). Both its row and column integrals are bounded by \(CG\). The Schur test therefore gives the asserted \(L^2\) bound. For an indicator input that equals one on a neighborhood of the test support, antisymmetrization on a smaller neighborhood has the integrable majorant already established. Dominated convergence identifies its limit with the local term in (5); separated exterior terms have ordinary absolute convergence. ◻

Proof. Uniform mass bounds on compact balls give sequential compactness by diagonal weak compactness on an exhaustion of \(\mathbb R^d\). The upper growth bound passes to the limit by testing slightly larger balls. The lower bound at zero ensures nonzeroness. If \(x\in\mathop{\mathrm{supp}}\sigma\), every fixed neighborhood of \(x\) has positive limiting mass and hence meets the approximating supports eventually. If \(x_j\to x\) and \(x_j\in\mathop{\mathrm{supp}}\sigma_j\), a fixed sufficiently small support ball at \(x_j\) has mass at least a fixed positive multiple of its radius to the power \(n\). A compactly supported bump near \(x\), or the Portmanteau inequality on a closed ball, transfers this positive mass to the limit. This proves both support assertions and, by shrinking and enlarging radii before taking limits, the lower growth bound. Only fixed radii are used at this step, so they are less than \(L_j\) eventually. Harmless dimensional changes in \(c,G\) would also suffice.

We give the operator-bound argument to account for hard boundaries. On a fixed pair of compact supports, the product measures \(\sigma_j\times\sigma_j\) converge weakly. For all but countably many \(\varepsilon>0\), the set \(|x-y|=\varepsilon\) has zero product measure under \(\sigma\times\sigma\). At such a radius the truncated bilinear form on a compact continuous scalar input and a compact continuous \(\mathbb R^d\)-valued dual test passes to the limit. Their \(L^2\) norms pass to the limit as well, so its bound is the original operator bound. Exhausting space leaves only a countable exceptional set of radii. For any specified \(\varepsilon>0\), choose good radii decreasing to \(\varepsilon\). Dominated convergence on compact supports, away from the diagonal, gives the same bilinear estimate for the hard cutoff \(|x-y|>\varepsilon\). Density in \(L^2\), together with the \(L^2\) integrability of the exterior kernel from Lemma 4, identifies the resulting bounded operator with the actual integral for every \(L^2\) input.

Choose a compact Lipschitz bump \(\eta\) with \(\int\eta\,d\sigma>0\). For large \(j\) put \[c_j=\frac{\int\varphi\,d\sigma_j}{\int\eta\,d\sigma_j}, \qquad \varphi_j=\varphi-c_j\eta.\] Then \(c_j\to0\). In (5), first restrict both variables to a fixed large ball with null boundary for the limit measure, and replace \(K_n(x-y)\) by the continuous capped kernel \((x-y)/\max\{h,|x-y|\}^{n+1}\). Weak convergence of product measures gives convergence of the capped pairing on this bounded product. The change on \(|x-y|\le h\) is uniformly \(O(h)\) by (6) and upper growth; the exterior is uniformly \(O(T^{-1})\) by Lemma 5. The same estimates apply to the uniformly bounded corrected tests. Letting \(h\downarrow0\) and \(T\to\infty\) proves the pairing assertion. ◻

Proof. Put \(r_Q=\ell(Q)\), \(w_Q=r_Q^{n/2}\) and \(f_Q=\varphi_Q/w_Q\). The support and Lipschitz conditions give \(|\varphi_Q|\le2J\): compare a point of its support with a point on the sphere of radius \(Jr_Q\) where the function vanishes. The tests are in \(L^1\cap L^2\) by compact support. If \(r_P\le r_Q\), cancellation in \(f_P\) gives \[\begin{align*} |\langle f_Q,f_P\rangle| &=\left|\int(f_Q(x)-f_Q(z_P))f_P(x)\,d\mu(x)\right|\\ &\le H_J\frac{r_P}{r_Q}\frac{w_P}{w_Q}, \qquad H_J=GJ^{n+1}(2J+1)^2. \tag{12}\end{align*}\] Indeed, on the support of \(f_P\) the difference is at most \(Jr_P/(r_Qw_Q)\), and \[\int|f_P|\,d\mu\le \frac{2JG(Jr_P)^n}{w_P}.\] The symmetric estimate follows by reversing \(P,Q\).

A nonzero inner product requires the two support balls to meet. Fix \(Q\) and a finer level \(r_P=2^{-k}r_Q\). Every interacting cube \(P\) lies in \(B(z_Q,(2J+C_0)r_Q)\), up to its null boundary. Same-level cells are pairwise disjoint, so (10) and global upper growth imply \[ \sum_{\substack{P\text{ at this level}\\ \langle f_Q,f_P\rangle\ne0}}r_P^n \le H'_J r_Q^n, \qquad H'_J=c_1^{-1}G(2J+C_0)^n. \tag{13}\] At a coarser level \(r_P=2^kr_Q\), the same argument in \(B(z_Q,(2J+C_0)r_P)\) bounds the number of interacting cubes by \(H'_J\). These are estimates in the measure dimension \(n\).

Multiply (12) by \(w_P\). Summing first over a finer level using (13), and then over a coarser level using its counting bound, gives respectively \[\sum_{r_P=2^{-k}r_Q}|\langle f_Q,f_P\rangle|w_P \le H_JH'_J2^{-k}w_Q, \qquad \sum_{r_P=2^kr_Q}|\langle f_Q,f_P\rangle|w_P \le H_JH'_J2^{-k}w_Q.\] Zero inner products may be discarded. Summing the geometric series therefore yields \[\sum_{P\in\mathcal A}|\langle f_Q,f_P\rangle|w_P \le4H_JH'_Jw_Q.\] For completeness, a symmetric matrix \((a_{QP})\) of nonnegative entries with \(\sum_P a_{QP}w_P\le Bw_Q\) satisfies \[\sum_{Q,P}a_{QP}|t_Qt_P|\le B\sum_Q|t_Q|^2.\] This follows by applying \(2|t_Qt_P|\le |t_Q|^2w_P/w_Q+|t_P|^2w_Q/w_P\) and using symmetry. Apply it to \(a_{QP}=|\langle f_Q,f_P\rangle|\). The resulting bound for \(\|\sum_Qt_Qf_Q\|_2^2\) gives (11) by Hilbert-space duality, with \(B_J=4H_JH'_J\).

For the vector assertion, write \(v_Q=\int f_Qu\,d\mu\). Then \(|\langle e_Q,v_Q\rangle|^2\le|v_Q|^2\). Expand the latter in an orthonormal basis, apply the scalar estimate to each coordinate of \(u\), and sum by Parseval’s identity. The constant is unchanged. ◻

Proof. Fix \(Q_0\), put \(L=\ell(Q_0)\), and fix the single outer ball \[S=B(z_{Q_0},(C_0+2)L).\] Let \(\mathcal A\subset\mathcal O(J,v)\cap\mathcal D(Q_0)\) be finite. For each selected cube write \(r_Q=\ell(Q)\) and choose a witnessing test \(\varphi_Q\) and a unit-bounded direction \(e_Q\), so that \[|\langle e_Q,\mathcal R_\mu(\varphi_Q)\rangle|>vr_Q^n.\] For \(r_Q\le L/(2J)\) the support of the test lies in \(S\), and its distance from \(S^c\) is at least \(L\). With \(\nu=\mu|S\), cancellation gives \[\begin{align*} \mathcal R_\mu(\varphi_Q)-\mathcal R_\nu(\varphi_Q) =\int\varphi_Q(x)\int_{S^c} [K_n(x-y)-K_n(z_Q-y)]\,d\mu(y)\,d\mu(x). \end{align*}\] This integral is absolutely convergent. The kernel derivative bound and upper growth on dyadic annuli give \[\int_{|y-z_Q|\ge L}|y-z_Q|^{-n-1}\,d\mu(y) \le C(n)G/L.\] Since \(|x-z_Q|\le Jr_Q\) on the support and \(\int|\varphi_Q|\,d\mu\le2JG(Jr_Q)^n\), it follows that \[ |\mathcal R_\mu(\varphi_Q)-\mathcal R_\nu(\varphi_Q)| \le T_J(r_Q/L)r_Q^n, \qquad T_J=C(n)G^2J^{n+2}. \tag{14}\] Choose an integer \(N\ge1\), depending only on \(J,v\) and the fixed data, with \[2^{-N}\le\min\{(2J)^{-1},\ v/[2(T_J+1)]\}.\] If \(Q\) lies at least \(N\) generations below \(Q_0\), its restricted pairing therefore has magnitude at least \((v/2)r_Q^n\) in the direction \(e_Q\).

The same hard-truncated function \(R_{\mu,\varepsilon}1_S\) can be used for all these fine cubes. For each such cube, \[\int\varphi_Q(x)R_{\mu,\varepsilon}1_S(x)\,d\mu(x) =\frac12\iint_{S\times S,\ |x-y|>\varepsilon} (\varphi_Q(x)-\varphi_Q(y))K_n(x-y)\,d\mu(x)d\mu(y).\] As \(\varepsilon\downarrow0\) this converges to \(\mathcal R_\nu(\varphi_Q)\). Antisymmetry gives the displayed identity; Lipschitz cancellation and upper growth make its near-diagonal majorant integrable. Also \(\int\varphi_Q\,d\nu=0\) because the support lies in \(S\). Lemma 9 and the hard-truncation bound imply \[\sum_{Q\in\mathcal A} \frac{\left|\int\varphi_Q\langle e_Q, R_{\mu,\varepsilon}1_S\rangle\,d\mu\right|^2}{r_Q^n} \le B_J C_{\rm R}^2\mu(S).\] Restrict the sum to the fine cubes and pass to the limit in this finite sum. We obtain \[\sum_{\substack{Q\in\mathcal A\\r_Q\le2^{-N}L}}r_Q^n \le4v^{-2}B_JC_{\rm R}^2\mu(S).\] Now \(\mu(S)\le G(C_0+2)^nL^n\le G(C_0+2)^nc_1^{-1}\mu(Q_0)\), so (10) bounds the total mass of the fine cubes by a fixed multiple of \(\mu(Q_0)\). At each of the fewer than \(N\) remaining levels, disjointness and containment in \(Q_0\) bound the selected masses by \(\mu(Q_0)\). This proves the desired estimate for every finite \(\mathcal A\), hence for the entire family. ◻

Proof. Every integral in (25) is finite, since its integrand is bounded on the ball and upper growth implies local finiteness. Markov’s inequality shows that \[F=\{x\in B(0,R):\mathop{\mathrm{dist}}(x,S)^2+\mathop{\mathrm{dist}}(x,W)^2<\eta^2\} \quad\hbox{satisfies}\quad \sigma(F)\ge(c/2)R^n.\] In particular, every point of \(F\) is within \(\eta\) of both planes.

Here is the quantitative nonconcentration needed to choose points in \(F\). If \(V\) is an affine \(k\)-plane, \(k<n\), and \(0<\tau\le1\), then \[ \sigma\{x\in\overline B(0,R):\mathop{\mathrm{dist}}(x,V)<\tau R\} \le G2^n3^k\tau R^n. \tag{28}\] Indeed, projection onto \(V\), centered at the projection of \(0\), sends \(\overline B(0,R)\) into its intrinsic closed \(R\)-ball. A maximal \(\tau R\)-separated subset of this intrinsic ball has at most \((1+2/\tau)^k\) points: its disjoint open \(\tau R/2\)-balls lie in the \((R+\tau R/2)\)-ball, and comparison of their \(k\)-dimensional volumes gives this bound. Maximality gives a \(\tau R\)-net. Ambient balls of radius \(2\tau R\) about its points cover the indicated tube. Upper growth gives \(G2^n3^k\tau^{n-k}R^n\), which implies (28).

Choose \[\tau=\min\{1,c/[4(G2^n3^n+1)]\}.\] The tube mass in (28) is strictly less than \((c/2)R^n\). Thus we can successively choose \(p_0,\ldots,p_n\in F\cap\mathop{\mathrm{supp}}\sigma\) such that, for \(1\le j\le n\), \[\mathop{\mathrm{dist}}\bigl(p_j,p_0+\operatorname{span} \{p_i-p_0:1\le i<j\}\bigr)\ge\tau R.\] At each step the affine span has dimension less than \(n\), and \(\mathop{\mathrm{supp}}\sigma\) has full measure. Put \(v_j=p_j-p_0\). Then \(|v_j|\le2R\). These vectors satisfy \[ \sum_{j=1}^n|t_j|\le \frac KR\left|\sum_{j=1}^nt_jv_j\right|, \tag{29}\] where \(K\) depends only on \(n,\tau\). For completeness, define \(K_0=0\) and \(K_{j+1}=K_j(1+2/\tau)+1/\tau\). Projection perpendicular to the span of the first \(j\) vectors bounds the last coefficient by the norm of the complete linear combination divided by \(\tau R\). Subtracting its last term enlarges that norm by at most \(1+2/\tau\). Induction proves (29) with \(K=K_n\). It also proves independence of the \(v_j\).

Let \(a=\pi_Sp_0\) and \(w_j=\pi_Sp_j-a\), where \(\pi_S\) is orthogonal projection. We have \(|w_j-v_j|\le2\eta\). If \(\eta/R\le1/[4(K+1)]\), (29) and absorption give \[\sum|t_j|\le\frac{2K}{R}\left|\sum t_jw_j\right|.\] The \(n\) vectors \(w_j\) consequently form a basis of the direction of \(S\). Every projected anchor \(a,a+w_1,\ldots,a+w_n\) lies within \(2\eta\) of \(W\). The orthogonal normal-coordinate map for \(W\) is affine and has norm \(\mathop{\mathrm{dist}}(\cdot,W)\). Applying this affine map to \(x=a+\sum t_jw_j\) gives \[\mathop{\mathrm{dist}}(x,W)\le2\eta+4\eta\sum|t_j| \le2\eta+\frac{8K\eta}{R}|x-a|.\] Since \(|a|\le R+\eta\le2R\), this proves (26), for example with \(A=2(1+8K)\) and \(\varepsilon_0=\min\{1,1/[4(K+1)]\}\). All choices preceded \(\sigma,R,S,W,\eta\).

Finally, for arbitrary \(x\), let \(p=\pi_Sx\). Since \(|\pi_S0|\le|a|\le2R\) and the linear part of \(\pi_S\) is a contraction, \(|p|\le|x|+2R\). Hence \[\mathop{\mathrm{dist}}(x,W)\le|x-p|+\mathop{\mathrm{dist}}(p,W) \le\mathop{\mathrm{dist}}(x,S)+A\eta(3+|x|/R).\] Interchanging the planes proves (27). ◻

Proof. Write \(b=2^\gamma>1\) and \(R_i=2^i\). The definition of the infimum in \(e_\sigma\) and the positivity of \(\delta b^i\) give affine planes \(P_i\), \(0\le i\le K\), with \[\int_{B(0,R_i)}\mathop{\mathrm{dist}}(x,P_i)^2\,d\sigma \le2(\delta b^i)^2R_i^{n+2}.\] No attainment of a least-squares infimum is required. For \(0\le i<K\), restricting the error for \(P_{i+1}\) to \(B(0,R_i)\) shows that the sum of neighboring errors is at most \(D(\delta b^i)^2R_i^{n+2}\), where \(D=2+2\cdot2^{n+2}b^2\). Choose \(L>0\) with \(D\le L^2c/2\), and put \(\eta_i=L\delta R_i b^i\). Taking \(\kappa=\varepsilon_0/L\) ensures \(\eta_i\le\varepsilon_0R_i\) through the horizon. Lemma 14 gives, at the same arbitrary point \(x\) in each step, \[\mathop{\mathrm{dist}}(x,P_i)\le\mathop{\mathrm{dist}}(x,P_{i+1}) +AL\delta b^i(3R_i+|x|).\] For \(|x|\le R_k\), the sum of these errors for \(i<k\) is bounded by \(B\delta R_kb^k\), where \[B=AL\left(\frac3{2b-1}+\frac1{b-1}\right).\] Indeed, sum the geometric series with ratios \(2b\) and \(b\) separately. Consequently \(\mathop{\mathrm{dist}}(x,P_0)\le\mathop{\mathrm{dist}}(x,P_k)+B\delta R_kb^k\) on that ball. Squaring, integrating, and using upper growth proves (30) with \(S=P_0\) and \(M=4+2GB^2\). ◻

Proof. An affine isometry identifies \(L\) with \(\mathbb R^n\). Under this identification write \(\mu\) for \(\sigma\). The kernel and its scalar projections commute with this isometry. Reflectionlessness passes to \(\mu\) with its actual measure: extend an intrinsic compact Lipschitz test by composing with orthogonal projection and multiplying by an ambient compact cutoff equal to one on its trace support. Its integral is unchanged. Choose a localization ball containing the extension and use the localization independence of the mean-zero pairing from Lemma 5. Thus no density or full-support assertion is needed to make this identification.

Full support. We first derive an annular bound using only growth and reflectionlessness. For a test supported in \(\overline B(a,R/2)\), denote by \(P_{a,R}(\varphi)\) the vector expression \[\begin{align*} &\frac12\iint_{B(a,R)^2} K_n(x-y)\bigl(\varphi(x)-\varphi(y)\bigr)\,d\mu(x)d\mu(y)\\ &\qquad+\int_{B(a,R)}\varphi(x) \int_{B(a,R)^c}\bigl[K_n(x-y)-K_n(a-y)\bigr]\,d\mu(y)d\mu(x). \end{align*}\] Both terms are absolutely integrable by the Lipschitz cancellation near the diagonal and the extra power of decay in the exterior difference. This expression is defined without a mean-zero assumption; on mean-zero tests it equals \(\mathcal R_\mu\). Splitting the larger ball into the smaller ball and its complement, and using oddness, gives \[ P_{a,R}(\varphi)-P_{a,r}(\varphi) =\left(\int\varphi\,d\mu\right) \int_{B(a,R)\setminus B(a,r)}K_n(a-y)\,d\mu(y) \tag{32}\] when \(0<r\le R\) and \(\mathop{\mathrm{supp}}\varphi\subset\overline B(a,r/2)\). One may first perform the splitting with a finite outer cutoff; the integrable exterior differences justify its removal.

Choose a nonnegative Lipschitz bump equal to one on \(B(a,s)\) and zero outside \(B(a,2s)\), and divide it by its \(\mu\)-integral. For \(a\in\mathop{\mathrm{supp}}\mu\) the resulting function \(\eta_s\) has integral one and satisfies \[\|\eta_s\|_\infty\le (cs^n)^{-1},\qquad \mathop{\mathrm{Lip}}(\eta_s)\le (cs^{n+1})^{-1},\qquad |P_{a,4s}(\eta_s)|\le C_n(G^2/c+G)=:A.\] For the last estimate the interior term is at most a constant times \((cs^{n+1})^{-1}(Gs^n)(Gs)\); the exterior term is at most a constant times \(Gs/s\), since \(\int\eta_s\,d\mu=1\). Apply reflectionlessness to \(\eta_{r/4}-\eta_{R/4}\) and then use (32). It follows that \[\left|\int_{B(a,R)\setminus B(a,r)}K_n(a-y)\,d\mu(y)\right| \le 2A\qquad(0<r\le R).\] Letting inner radii decrease strictly to \(\varepsilon\) gives the same bound on \(\{y:\varepsilon<|y-a|<R\}\). At fixed \(\varepsilon>0\) the kernel is integrable there, so this passage does not require zero mass on spheres.

Suppose now that \(z\notin\mathop{\mathrm{supp}}\mu\), and let \(a\) be a nearest support point to \(z\). Translate \(a\) to zero and put \(e=a-z\ne0\). The nearest-point property gives \[2\langle e,x\rangle+|x|^2\ge0\qquad(x\in\mathop{\mathrm{supp}}\mu).\] Define \(p_e(x)=\max(0,\langle e,x\rangle)/|x|^{n+1}\) for \(x\ne0\), and set \(p_e(0)=0\). Then \[p_{-e}(x)\le\tfrac12|x|^{1-n},\qquad p_e(x)=\langle e,K_n(x)\rangle+p_{-e}(x).\] Upper growth makes the first majorant integrable on bounded balls: its integral on \(B(0,R)\) is at most \(C_nGR\), as follows by dyadic annuli. The signed annular bound therefore bounds the nonnegative truncated integrals of \(p_e\) on \(B(0,1)\) uniformly. Monotone convergence proves \(p_e\in L^1(\mu|_{B(0,1)})\).

Take a weakly convergent subsequence of the normalized blowups \(\mu_r(A)=r^{-n}\mu(rA)\) as \(r\downarrow0\). Upper growth supplies compactness, and the lower bound at zero makes the limit \(\tau\) nonzero. These are the conclusions of Lemma 6, applied to the blowups. The rescaled quadratic inequality is \(2\langle e,x\rangle+r|x|^2\ge0\); hence \(\mathop{\mathrm{supp}}\tau\subset\{\langle e,x\rangle\ge0\}\). For every fixed \(T>0\), \[\int_{B(0,T)}\max(0,\langle e,x\rangle)\,d\mu_r(x) \le T^{n+1}\int_{B(0,rT)}p_e(y)\,d\mu(y)\longrightarrow0.\] Testing with compact nonnegative cutoffs shows that \(\tau\) is also supported in \(\{\langle e,x\rangle\le0\}\). It is consequently supported on a proper hyperplane in \(\mathbb R^n\). This is impossible for a nonzero measure with upper \(n\)-growth: covering open sets by disjoint dyadic cubes and enclosing each cube in a ball shows that such a measure is bounded above by \(C_nG'\,dx\), where \(G'\) is its growth constant. A proper hyperplane has Lebesgue measure zero. This contradiction proves \(\mathop{\mathrm{supp}}\mu=\mathbb R^n\).

A positive bounded density. The same cube comparison gives \(\mu=f\,dx\) with \(f\le C_nG\) almost everywhere. Full support makes the lower ball bound valid at every point. Lebesgue differentiation then gives \(f\ge c/\omega_n\) almost everywhere. Choose a measurable representative with \[ 0<c_1\le f(x)\le C_1<\infty\qquad(x\in\mathbb R^n). \tag{33}\] Changing \(f\) on a null set has not changed \(\mu\).

A local representative for the pairing. Fix a unit direction \(u\in\mathbb R^n\), a center \(a\), and radii \(0<H\) and \(2H\le R\). Put \(B=B(a,R)\) and \[T_{\varepsilon,B}1(x)=\int_B \frac{\langle u,x-y\rangle}{\max(\varepsilon,|x-y|)^{n+1}}\,d\mu(y).\] There is a constant \(D_1\), independent of \(B\) and \(\varepsilon>0\), such that \[ \|T_{\varepsilon,B}1\|_{L^2(\mu|_B)} \le D_1\mu(B)^{1/2}. \tag{34}\] For the direct operator hypothesis, subtract the hard truncation. The error kernel is bounded by \(\varepsilon^{-n}\mathbf1_{\{|x-y|\le\varepsilon\}}\). Its row and column integrals are at most \(2^nG\), so Schur’s inequality gives (34) from the original operator bound. Restricting the hard operator to a ball preserves that bound, by extending the input by zero.

Here is also the justification under the source assumption. Work first in \(\mathbb R^d\) on the ambient ball corresponding to \(B\). Its boundary has \(\sigma\)-measure zero by (33). Restricted source measures therefore converge weakly as finite measures. The same Schur estimate gives uniformly bounded capped bilinear pairings on every source restriction. For fixed \(\varepsilon>0\) these pairings pass to the limit against two bounded continuous tests: the capped kernel is bounded continuous, as are the squared tests giving their \(L^2\) norms. Take the input test to be one and the other test to be the capped output for the limiting finite measure. That output is bounded continuous, by dominated convergence at this fixed cap. The resulting bilinear estimate gives its genuine \(L^2\) bound (34). Isometric pullback gives precisely the displayed intrinsic transform. This argument passes no singular kernel through weak convergence.

Choose a subsequence \(\varepsilon_j\downarrow0\) along which \(T_{\varepsilon_j,B}1\) converges weakly in \(L^2(\mu|_B)\) to \(v\). This one subsequence works against every test in that Hilbert space. Oddness gives, for bounded Lipschitz \(\varphi\), \[\int_B T_{\varepsilon,B}1\,\varphi\,d\mu =\frac12\iint_{B^2} \frac{\langle u,x-y\rangle(\varphi(x)-\varphi(y))} {\max(\varepsilon,|x-y|)^{n+1}}\,d\mu(x)d\mu(y).\] Removing the cap on the right changes the integral by at most \(C_nG\mathop{\mathrm{Lip}}(\varphi)\mu(B)\varepsilon\). Thus \(v\) represents the actual interior singular pairing. On \(B(a,H)\) the function \[F(x)=\int_{B(a,R)^c} \langle u,K_n(x-y)-K_n(a-y)\rangle\,d\mu(y)\] is well defined and bounded: the kernel difference is at most \(C_n|x-a||y-a|^{-n-1}\), whose exterior integral is at most \(C_nG|x-a|/R\). Fubini now identifies the scalar component of \(\mathcal R_\mu(\varphi)\) with \(\int(v+F)\varphi\,d\mu\) for mean-zero compact Lipschitz tests supported in \(B(a,H)\).

Choose a compact Lipschitz test \(\eta\) supported in \(B(a,H)\) with \(\int\eta\,d\mu=1\), using the positive mass of that ball, and put \(b=\int(v+F)\eta\,d\mu\). For any smooth compact test \(q\) in this smaller ball, reflectionlessness applied to \(q-(\int q\,d\mu)\eta\) gives \(\int(v+F-b)q\,d\mu=0\). Since \(\mu=f\,dx\), the fundamental lemma for locally integrable functions applied to \((v+F-b)f\) proves \[ v+F=b\quad\text{$\mu$-almost everywhere on }B(a,H). \tag{35}\] This is a statement about the transform; constancy of \(f\) has not yet been used or proved.

An unweighted equation and density constancy. Let \(g\in C_c^\infty(B(a,H))\) be real with \(\int g\,dx=0\). By (33), \(g/f\) belongs to \(L^2(\mu|_B)\). It is therefore an allowed test for the weak convergence, even though it need not be Lipschitz. Moreover, \(\mu\) and Lebesgue measure have the same null sets. Testing with \(g/f\) and using (35) yields \[ \lim_j\left[\int g(x)T_{\varepsilon_j,B}1(x)\,dx +\int g(x)F(x)\,dx\right]=0. \tag{36}\] The subsequence and \(b\) were selected before \(g\).

For a compact smooth function \(g\), define the scalar test transform \[R_ug(x)=-\frac12\int_{\mathbb R^n} \frac{\langle u,h\rangle[g(x+h)-g(x-h)]}{|h|^{n+1}}\,dh.\] The integral is absolutely convergent: the first difference is \(O(|h|)\) near zero, and the translated compact supports control infinity. The analogous capped expression equals the capped transform of \(g\) by translation and reflection. It converges pointwise to \(R_ug\) and is uniformly bounded on bounded \(x\)-sets. If \(\mathop{\mathrm{supp}}g\subset B(a,H)\) and \(\int g=0\), then for \(|x-a|\ge2H\) and \(\varepsilon\le H\) the cap is inactive and mean-zero cancellation gives \[|R_{u,\varepsilon}^{\rm cap}g(x)| \le C_n H\|g\|_1|x-a|^{-n-1}.\] Together these bounds form an integrable majorant independent of small \(\varepsilon\). Since \(f\) is bounded, dominated convergence applies to \(\int f R_{u,\varepsilon}^{\rm cap}g\).

For \(\varepsilon\le H\), oddness and Fubini identify the expression in brackets in (36) with \(-\int f R_{u,\varepsilon}^{\rm cap}g\,dx\). The local capped product is absolutely integrable. For the exterior term, use the centered kernel difference, which is absolutely integrable against \(|g(x)|\,dx\,d\mu(y)\); cancel the center term using \(\int g=0\) inside the \(x\)-integral before integrating over the infinite exterior. The open ball and its closed exterior are exact complements. Thus no separate divergent exterior integral or omitted boundary term occurs. It follows that \[ \int_{\mathbb R^n}f(x)R_ug(x)\,dx=0 \qquad\left(g\in C_c^\infty(\mathbb R^n),\ \int g=0\right). \tag{37}\] The ball may be chosen after the test, so this statement has no fixed support restriction.

Let \(u_1,\ldots,u_n\) be an orthonormal basis and \(\psi\in C_c^\infty(\mathbb R^n)\) be real. Each \(g_i=\partial_{u_i}\psi\) has integral zero. We claim \[ \sum_{i=1}^nR_{u_i}(\partial_{u_i}\psi)(x)=\mathcal L\psi(x). \tag{38}\] Indeed, set \[Q_x(t)=\int\frac{\psi(x+th)+\psi(x-th)-2\psi(x)}{|h|^{n+1}}\,dh.\] Scaling gives \(Q_x(t)=tQ_x(1)\) for \(t>0\). Differentiation under the integral at \(t=1\) is justified by second-derivative cancellation near zero; for \(1/2\le t\le2\) the derivative integrand vanishes for all sufficiently large \(|h|\) at fixed \(x\). Consequently \[Q_x(1)=\int\frac{D\psi(x+h)h-D\psi(x-h)h}{|h|^{n+1}}\,dh.\] Expand \(h\) in the orthonormal basis and use the definition of \(R_u\) to obtain (38). Every pairing with bounded \(f\) is integrable by the preceding estimates. Summing (37) therefore gives \[\int f\mathcal L\psi\,dx=0\qquad(\psi\in C_c^\infty(\mathbb R^n)).\] Real and imaginary parts give the same identity for complex tests. Boundedness supplies \(\int|f(x)|(1+|x|)^{-n-2}\,dx<\infty\). Lemma 21 now makes \(f\) affine almost everywhere. A bounded affine function is constant: any nonzero linear part is unbounded along a line, and continuity promotes an almost-everywhere bound to an everywhere bound. Thus \(f=\theta\) almost everywhere, with \(\theta>0\) by (33). The original ball bounds give \(c\le\theta\omega_n\le G\), completing the proof. ◻

Proof. After discarding finitely many indices, the smallness threshold in Lemma 15 holds. That lemma gives \(S_j\) and the uniform moment bounds. Each fixed-radius squared-distance integral tends to zero: choose a larger fixed dyadic radius first, and then use \(K_j\to\infty\) and \(\delta_j\to0\).

This integral conclusion also controls all support points in bounded balls. If \(x_j\in E_j\cap B(0,R)\) and \(\mathop{\mathrm{dist}}(x_j,S_j)\ge t>0\), the \(1\)-Lipschitz distance function is at least \(t/2\) on \(B(x_j,\min\{1,t/2\})\). This ball lies in \(B(0,R+1)\), and its radius is admissible for all large \(j\). The lower growth bound therefore forces a fixed positive lower bound on the squared-distance integral there, a contradiction. Thus \[ \sup_{x\in E_j\cap B(0,R)}\mathop{\mathrm{dist}}(x,S_j)\longrightarrow0 \quad\text{for every fixed }R. \tag{45}\] Because \(0\in E_j\), the nearest points \(a_j\) tend to \(0\). Choose orthonormal tangent and complementary normal frames \(L_j,N_j\). The compactness of the finite-dimensional set of tangent isometries gives a subsequence with \(L_j\to L_\infty\); norm preservation shows that the limit is still an isometry. The normal frames are retained along this subsequence; their convergence is not required.

Lemma 6, followed if necessary by a further subsequence, gives a nonzero limit \(\nu\) with \(0\in\mathop{\mathrm{supp}}\nu\), hard truncation bound \(D_0\), global upper growth \(2^nG\), and \[ \nu(B(x,r))\ge4^{-n}cr^n \qquad(x\in\mathop{\mathrm{supp}}\nu, r>0). \tag{46}\] All frames and moment bounds are kept along the composition of the two subsequences. If \(x\in\mathop{\mathrm{supp}}\nu\), positivity of compact tests in every neighborhood of \(x\) supplies nearby points of \(E_j\) for all large \(j\). Together with (45), this gives \(\mathop{\mathrm{dist}}(x,S_j)\to0\). On the other hand the affine projections \[a_j+L_jL_j^*(x-a_j)\longrightarrow L_\infty L_\infty^*x.\] It follows that \(x=L_\infty L_\infty^*x\). At this stage we have proved only \(\mathop{\mathrm{supp}}\nu\subset L_\infty(\mathbb R^n)\).

We next verify that this same \(\nu\) is reflectionless. Let \(\varphi\) be a compactly supported Lipschitz function with \(\int\varphi\,d\nu=0\). The pairing-convergence assertion of Lemma 6 provides compact Lipschitz tests \(\varphi_j\) of zero \(\mu_j\)-mean, with uniformly bounded supports and Lipschitz constants, such that \[\mathcal R_{\mu_j}(\varphi_j)\longrightarrow \mathcal R_\nu(\varphi).\] Choose \(C\ge1\) bounding their Lipschitz constants. For all sufficiently large \(j\), their supports lie compactly inside \(B(0,A_j)\), so (41), applied to \(\varphi_j/C\), gives \(|\mathcal R_{\mu_j}(\varphi_j)|\le Cv_j\to0\). Thus \(\mathcal R_\nu(\varphi)=0\). This argument applies to every such test on the measure subsequence already selected above.

Apply Lemma 16 to this plane-supported limit, using the inherited hard-truncation bound \(D_0\). It gives \(\nu=q(L_\infty)_\#\mathcal L^n\) with \(q>0\). Its upper growth and (46), evaluated on a ball centered on the plane, give the two bounds for \(q\) in (42). Since \(|u_{j,i}(x)|\le|N_j^*(x-a_j)|=\mathop{\mathrm{dist}}(x,S_j)\), the normal moment bounds follow from those of \(S_j\).

Finally, convergence of the affine projections is uniform on bounded sets, so (45) gives uniform closeness of source support points to the limiting plane. Conversely, if points \(y_j\) of that plane in a fixed bounded ball stayed a positive distance from \(E_j\), a subsequence would converge to a point \(y\) of the plane. A sufficiently small ball about \(y\) would then miss \(E_j\) for all large \(j\). A nonnegative compact test in that ball with positive integral against \(q(L_\infty)_\#\mathcal L^n\) contradicts weak convergence. This proves the stated bilateral convergence. ◻

Proof. It suffices to consider \(H\ge1\). Fix a Lipschitz function \(\chi\) with \(0\le\chi\le1\), \(\mathop{\mathrm{Lip}}(\chi)\le2/H\), equal to one on \(B(0,H)\), and zero outside \(B(0,3H/2)\). Its support lies in \(B(0,2H)\). Choose the smallest dyadic radius \(R\ge4H\), and put \(D=B(0,R)\). All estimates below concern sufficiently large \(j\), so that \(K_j\ge\log_2R\), \(\delta_j\le1\), \(A_j>2H\), and the lower estimate in (48) is available at radius \(H\). Constants may depend on the fixed quantities displayed in the lemma and on \(H,R\), but not on \(j\). We omit the index \(j\) temporarily.

The moment bound at radius \(R\) gives \[ \int_D u^2\,d\mu\le C\delta^2, \qquad \int\chi^2\,d\mu\ge\mu(B(0,H))\ge cH^n>0. \tag{54}\] Define the weighted center and its centered test by \[ b=\frac{\int\chi^2u\,d\mu}{\int\chi^2\,d\mu}, \qquad U=u-b,\qquad \phi=\chi^2U. \tag{55}\] Cauchy–Schwarz, followed by (54), proves \[ |b|\le C\delta,\qquad \int_DU^2\,d\mu\le C\delta^2,\qquad \int|\phi|\,d\mu\le C\delta, \qquad \int|\phi U|\,d\mu\le C\delta^2. \tag{56}\] In particular, these estimates precede any energy bound. The function \(\phi\) has mean zero. It also has a Lipschitz constant bounded independently of \(j\): on the support of \(\chi\), \(|U(x)|\le W+2H+C\), and the product rule for Lipschitz functions gives the bound. The same bound across the complement follows because \(\chi\) vanishes there. Consequently (52) implies \[ |\langle e,\mathcal R_\mu(\phi)\rangle|\le C\delta^3. \tag{57}\]

The positive part of this pairing comes from the exact identity \[\begin{align*} &(U(x)-U(y))(\chi(x)^2U(x)-\chi(y)^2U(y))\\ &\hspace{12mm}= |\chi(x)U(x)-\chi(y)U(y)|^2 -U(x)U(y)(\chi(x)-\chi(y))^2. \tag{58}\end{align*}\] All terms in its integral over \(D\times D\) are integrable before a quantitative energy estimate is made. Indeed, a Lipschitz difference cancels one power of the distance in each factor. For every \(x\in D\), upper growth and dyadic annuli imply \[ \int_D |x-y|^{1-n}\,d\mu(y)\le C G R. \tag{59}\] The integrability of local affine and cutoff energies follows from this bound and the finiteness of \(\mu(D)\). For the signed cutoff error, symmetry and \(2|U(x)U(y)|\le U(x)^2+U(y)^2\) give the sharper estimate \[\begin{align*} &\iint_{D\times D} |U(x)U(y)|\frac{|\chi(x)-\chi(y)|^2}{|x-y|^{n+1}} \,d\mu(x)d\mu(y)\\ &\hspace{12mm}\le C G R\mathop{\mathrm{Lip}}(\chi)^2\int_D U^2\,d\mu \le C\delta^2. \tag{60}\end{align*}\] This is an estimate from the already known second moment; it assumes no bound on the unknown fractional energy.

We next separate the local pairing from its exterior remainder. Set \(q(x,y)=|x-y|^{-n-1}\). The affine identity in (49), and the pairing formula in Lemma 5, give \[\begin{align*} \langle e,\mathcal R_\mu(\phi)\rangle &=\frac12\iint_{D\times D} (U(x)-U(y))(\phi(x)-\phi(y))q(x,y) \,d\mu(x)d\mu(y)+\mathrm{Ext}, \tag{61}\\ \mathrm{Ext} &=\int_D\phi(x)\int_{D^c} \bigl[U(x)q(x,y) -U(y)\{q(x,y)-q(0,y)\}\bigr]\,d\mu(y)d\mu(x). \tag{62}\end{align*}\] To obtain the second identity, the base-point subtraction initially also contributes \(-U(0)q(0,y)\). That term is separately integrable on \(D^c\) and disappears because \(\int\phi\,d\mu=0\). The difference multiplying \(U(y)\) in (62) is kept together.

For \(|x|\le2H\) and \(|y|\ge R\), \[ q(x,y)\le C|y|^{-n-1}, \qquad |q(x,y)-q(0,y)|\le CH|y|^{-n-2}. \tag{63}\] The global affine bound \(|U(y)|\le |y|+W+C\), together with upper growth and (63), makes the exterior expression absolutely integrable. We may therefore integrate (58) and use (61) and (60) to obtain \[ \mathcal E_{\mu,D}(\chi U) \le 2|\langle e,\mathcal R_\mu(\phi)\rangle| +2|\mathrm{Ext}|+C\delta^2. \tag{64}\] It remains to bound the exterior remainder at the scale of the known second moment. Put \(T=2^K\). Upper growth and dyadic annuli give \(\int_{D^c}|y|^{-n-1}\,d\mu(y)\le C/R\). On a controlled shell \(t\le |y|<2t\), where \(t=2^k\) and \(k<K\), Cauchy–Schwarz and the moment at radius \(2t\) give \[ \int_{t\le|y|<2t}|U(y)|\,d\mu(y) \le C\delta t^{n+1+\gamma} \qquad(t\ge R). \tag{65}\] Here the contribution of \(b\) is absorbed using \(t\ge1\). Since \(\gamma<1\), summing the controlled shells gives \[ \int_{R\le|y|<T}\frac{|U(y)|}{|y|^{n+2}}\,d\mu(y) \le C\delta\sum_{\substack{t\ge R\\t\text{ dyadic}}}t^{\gamma-1} \le C\delta. \tag{66}\] Beyond the actual last controlled radius, use the global affine bound \(|U(y)|\le |y|+W+C\). Upper growth then gives \[ \int_{|y|\ge T}\frac{|U(y)|}{|y|^{n+2}}\,d\mu(y) \le C(T^{-1}+T^{-2})\le C T^{-1}. \tag{67}\] The half-open shells and the closed final exterior cover every point, including their boundary spheres. Combining (56) with the kernel and tail bounds (63)–(67) gives the quantitative estimate \[ |\mathrm{Ext}|\le C(\delta^2+\delta T^{-1}). \tag{68}\] Thus the remote part retains the factor \(\delta\) supplied by the centered test. Replacing that factor by a constant would lose the required normalization.

The reduction (64), the tested-pairing bound (57), and (68) now yield \[ \mathcal E_{\mu,D}(\chi U) \le C(\delta^3+\delta^2+\delta T^{-1}). \tag{69}\] On \(B(0,H)\), \(\chi=1\) and the centered difference \(U(x)-U(y)\) is \(u(x)-u(y)\). Restriction to this smaller product set and division by \(\delta^2>0\) give \[ \mathcal E_{\mu,B(0,H)}(u/\delta) \le C\left(1+\delta+\frac{1}{\delta T}\right). \tag{70}\] The last term is bounded by (50). The local \(L^2\) estimate follows directly from the moment bound at a fixed dyadic radius containing \(H\).

For each of the finitely many omitted indices, \(u_j/\delta_j\) is an ordinary Lipschitz function. Upper growth gives its finite \(L^2\) norm on every fixed ball, and (59) gives its finite local energy. Enlarging the bound to cover those indices completes the proof. ◻

Proof. We give the construction, including the norm convergence. On each bounded projection region, fractional energy controls the variance inside small cells. A common subsequence of their finitely many averages will recover both first and second moments, not just weak convergence of the heights. First, (72) and compact support cutoffs imply finite-measure weak convergence on each \(\Omega_H\). Indeed, multiply by a compact continuous cutoff equal to one on \(\overline\Omega_H\); weak convergence of these finite measures passes to restriction to the continuity set \(\Omega_H\). In particular their masses are bounded.

We will also use the consequence \[ \sup\{\mathop{\mathrm{dist}}(x,L):x\in\mathop{\mathrm{supp}}\mu_j\cap B(0,R)\} \longrightarrow0 \quad(R<\infty). \tag{78}\] If this failed, some bounded sequence of support points would stay a positive distance from \(L\). Passing to a subsequence, a fixed small ball about its limit would be disjoint from \(L\) and would contain a smaller ball about each of those support points. The lower growth bound would give the smaller ball a fixed positive mass; its radius is admissible for large \(j\). A continuous compact cutoff, equal to one on all the smaller balls and supported away from \(L\), would then contradict (73). The cutoff is taken on a larger ambient ball, so the smaller support balls retain their full mass even if they cross a localization boundary.

Fix \(H\), and partition the half-open box \((-(H+1),H+1]^n\) into equal dyadic boxes of side length \(h=2(H+1)2^{-k}\). For such a box \(D\), set \[C_D=\Omega_H\cap\pi^{-1}(D).\] These finitely many disjoint measurable cells cover \(\Omega_H\). Each cell boundary has \(\nu\)-measure zero, by (72) and the nullity of coordinate faces. The same holds for intersections of cells from different partition levels. Hence \[ \mu_j(C_D)\longrightarrow\nu(C_D)=q h^n, \qquad \mu_j(C_D\cap C_{D'})\longrightarrow\nu(C_D\cap C_{D'}). \tag{79}\] For each fixed partition level, all its cell masses are therefore at least \((q/2)h^n\) for sufficiently large \(j\). This uses positivity on every full-plane cell, rather than just nonzero total mass. By (78), at that same fixed level the actual support points in one cell have mutual distance at most \((\sqrt n+2)h\) for sufficiently large \(j\). The entire projection cylinder need not have bounded diameter; only pairs seen by the restricted measure are used.

For \(w_j=w_j^a\), let \(m_{j,D}\) be its average on \(C_D\) (zero if the cell has zero mass), and let \(P_k^jw_j\) be the corresponding step function. The exact variance identity gives \[\begin{align*} \int_{C_D}|w_j-m_{j,D}|^2\,d\mu_j &=\frac{1}{2\mu_j(C_D)} \iint_{C_D\times C_D}|w_j(x)-w_j(y)|^2\,d\mu_j(x)d\mu_j(y) \\ &\le C_n q^{-1}h \iint_{C_D\times C_D} \frac{|w_j(x)-w_j(y)|^2}{|x-y|^{n+1}} \,d\mu_j(x)d\mu_j(y). \tag{80}\end{align*}\] All averages are defined for these sufficiently large indices, and their \(L^2\) bounds follow from (74). The sets \(C_D\times C_D\) are disjoint subsets of \(\Omega_H\times\Omega_H\). Summing (80) yields \[ \int_{\Omega_H}|w_j-P_k^jw_j|^2\,d\mu_j \le C_H h. \tag{81}\] There is no factor counting the cells in this estimate.

For fixed \(H,k,D,a\), Cauchy–Schwarz and the positive limiting cell mass bound \(m_{j,D}^a\). A diagonal subsequence therefore makes all these averages converge simultaneously: the collection of indices is countable, including all \(H,k\) and the finite set of coordinates \(a\). Write \(m_D^a\) for their limits and \(P_{H,k}^a\) for the corresponding step function on the limiting plane box.

At fixed levels \(k,l\), expand the squared difference of the two step functions. Convergence of the finitely many coefficients, cell masses, and intersection masses in (79) gives \[\int_{\Omega_H}|P_k^jw_j^a-P_l^jw_j^a|^2\,d\mu_j \longrightarrow \int_{\Omega_H}|P_{H,k}^a-P_{H,l}^a|^2\,d\nu.\] The triangle inequality and (81) show that the last integral is at most \(2C_H(h_k+h_l)\). The step functions thus converge in \(L^2(\nu|_{\Omega_H})\) to some \(v_H^a\).

The reverse triangle inequality also gives \[\left|\|w_j^a\|_{L^2(\mu_j|_{\Omega_H})} -\|P_k^jw_j^a\|_{L^2(\mu_j|_{\Omega_H})}\right| \le (C_H h_k)^{1/2}.\] For fixed \(k\), the step-function norm converges, since its square is the finite sum \(\sum_D(m_{j,D}^a)^2\mu_j(C_D)\). First let \(j\to\infty\), then \(k\to\infty\). This proves (77) with \(v_H^a\) in place of \(v^a\).

For a continuous \(\psi\) on \(\overline\Omega_H\), the signed testing error from replacing \(w_j^a\) by \(P_k^jw_j^a\) is at most \[(C_Hh_k)^{1/2}\|\psi\|_\infty\mu_j(\Omega_H)^{1/2}.\] For fixed \(k\), weak convergence on the continuity cells gives \(\int_{C_D}\psi\,d\mu_j\to\int_{C_D}\psi\,d\nu\). Passing first in \(j\), then in \(k\), proves (75) for \(v_H^a\).

The local limits agree on overlaps. To see this without choosing new subsequences, let \(U=\Omega_H\cap\Omega_J\) and \(\eta(x)=\min\{1,\mathop{\mathrm{dist}}(x,U^c)\}\). This bounded Lipschitz function is positive exactly on \(U\). For a compactly supported Lipschitz \(\psi\), use \(\eta\psi\) in the two signed-moment limits. The source integrals are identical, since \(\eta\psi\) vanishes outside the overlap. Their limits imply \[\int_U\eta(v_H^a-v_J^a)\psi\,d\nu=0.\] Compact Lipschitz tests are dense in \(L^2(\nu|_U)\), so \(\eta(v_H^a-v_J^a)=0\) almost everywhere. Since \(\eta>0\) on \(U\), the local limits agree there. Choosing measurable representatives and, on the plane, using the first region containing a point now produces one measurable \(v^a\) agreeing with every \(v_H^a\) almost everywhere. Countability of the cover ensures this simultaneous agreement.

It remains to prove the weighted squared-moment conclusion. Fix \(H,a\), and abbreviate \(w_j=w_j^a\), \(v=v^a\). There are bounded compactly supported ambient Lipschitz functions \(g\) whose restrictions approximate \(v\) in \(L^2(\nu|_{\Omega_H})\). For completeness, simple functions are dense in this space; regularity of finite plane volume approximates their sets by compact and open sets, and distance cutoffs approximate their indicators. Extend the resulting plane Lipschitz functions by \(\pi\), and multiply by a fixed ambient compact cutoff equal to one on \(\overline\Omega_H\).

For each such fixed \(g\), the norm limit, the signed limit tested against \(g\), and ordinary weak convergence tested against \(g^2\) give \[ \lim_j\int_{\Omega_H}|w_j-g|^2\,d\mu_j =\int_{\Omega_H}|v-g|^2\,d\nu. \tag{82}\] This identity involves \(g\), an actual continuous function, on the moving supports. For any continuous weight \(\psi\), \[\begin{align*} \left|\int_{\Omega_H}\psi(w_j^2-g^2)\,d\mu_j\right| &\le \|\psi\|_\infty \|w_j-g\|_{L^2(\mu_j|_{\Omega_H})}\\ &\hspace{8mm}\cdot \left(\|w_j\|_{L^2(\mu_j|_{\Omega_H})} +\|g\|_{L^2(\mu_j|_{\Omega_H})}\right). \end{align*}\] The last factor remains bounded for an \(L^2\)-approximating family \(g\), after taking \(j\to\infty\) for each fixed \(g\). Ordinary weak convergence handles \(\psi g^2\). Equation (82) and Cauchy–Schwarz for the limiting measure then allow \(g\to v\) in \(L^2(\nu|_{\Omega_H})\). This proves (76) in the stated order of limits. ◻

Proof. We retain the subsequence already selected in Lemma 19. In the proof below every bounded region and every test is chosen after this subsequence and its height \(v\).

Local energy and singular pairings. Let \(\Omega\) be one of the bounded continuity regions used in that lemma. On its common compact closure, the three measures \[\mu_j|_\Omega,\qquad w_j\mu_j|_\Omega, \qquad w_j^2\mu_j|_\Omega\] converge against continuous functions to the corresponding measures \(\nu|_\Omega\), \(v\nu|_\Omega\), and \(v^2\nu|_\Omega\). Their total variations are uniformly bounded. Finite sums of products of continuous functions approximate a continuous function on the product of this compact set uniformly. Thus, after replacing \(k(x-y)\) by \((\max(h,|x-y|))^{-n-1}\), both the quadratic energy and every bilinear pairing against a fixed bounded Lipschitz test converge. For the energy, expand \((w_j(x)-w_j(y))^2\); its three terms use exactly the three measures above. Monotone convergence as \(h\downarrow0\) shows that the limiting energy is no larger than the common bound from Lemma 18. Positive-dimensional upper growth makes the diagonal null.

Estimate (86) applies uniformly to the source and limiting heights on \(\Omega\). The error caused by capping the bilinear kernel is bounded by its absolute integral on \(|x-y|\le h\). Consequently \[\begin{align*} &\iint_{\Omega^2}(w_j(x)-w_j(y))(\phi(x)-\phi(y))k(x-y) \,d\mu_j(x)d\mu_j(y)\tag{94}\\ &\qquad\longrightarrow \iint_{\Omega^2}(v(x)-v(y))(\phi(x)-\phi(y))k(x-y) \,d\nu(x)d\nu(y). \end{align*}\] This argument first fixes \(h\), then takes \(j\to\infty\), and finally takes \(h\downarrow0\).

Inherited moments and the two exterior errors. The weak convergence of \(w_j^2\mu_j\) on an enclosing continuity region, tested with nonnegative cutoffs, passes (90) to the limit at every fixed scale. Changing the radius by at most a factor two gives \[ \int_{L\cap B(0,t)}|v|^2\,d\nu\le C t^{n+2+2\gamma} \qquad(t\ge1). \tag{95}\] Each fixed radius eventually lies below \(2^{K_j}\); no estimate past that horizon for a fixed sample has been used. Cauchy–Schwarz on dyadic shells, followed by summation of \(2^{k(\gamma-1)}\), now gives \[ \int_{L\cap\{|y|>T\}} |v(y)|\,|y|^{-n-2}\,d\nu(y) \le C T^{\gamma-1}\qquad(T\ge2). \tag{96}\] Together with local \(L^2\) integrability and \(q>0\), this proves (92).

Fix \(H\) containing the supports of the tests to be considered. Their uniform supremum bounds and the local second-moment bound imply uniform bounds on \(\int|\phi|\,d\mu_j\) and \(\int|\phi w_j|\,d\mu_j\). For \(2H<T\le 2^{K_j}\), the same shell calculation using (90) controls the renormalized exterior integral outside \(B(0,T)\) by \[ C_\phi\bigl(T^{-1}+T^{\gamma-1} +\delta_j^{-1}2^{-K_j}\bigr). \tag{97}\] Here is the bound beyond the last controlled radius. The affine functions \(u_j\) are \(1\)-Lipschitz. The mass of \(B(0,1)\) is bounded below eventually, by convergence to the plane measure, so (90) implies \(|u_j(0)|\le1+C\delta_j\). Hence \(|u_j(y)|\le C+|y|\). Upper growth gives \[\int_{|y|\ge2^{K_j}}|w_j(y)||y|^{-n-2}\,d\mu_j(y) \le C\delta_j^{-1}2^{-K_j}.\] This bound uses the unnormalized height at \(0\); the normalized value \(w_j(0)\) may diverge. The \(T^{-1}\) term in (97) comes from the \(w_j(x)k(x-y)\) term, whose \(x\)-integral stays on the fixed test support. The kernel difference in (87) gives the remaining terms. Thus the order is to choose \(T\) large and then choose \(j\) large. The analogous limiting tail tends to zero by (96).

A common normalization and mean correction. Take two fixed compact ambient Lipschitz functions \(\phi,\eta\) such that \[\int\phi\,d\nu=0,\qquad \int\eta\,d\nu\ne0.\] Choose one continuity region \(A\) containing \(B(0,R)\) with a positive margin beyond both supports. For each of these tests separately, \[ \mathcal B_{\mu_j,A}(w_j,\phi) \longrightarrow\mathcal B_{\nu,A}(v,\phi), \qquad \mathcal B_{\mu_j,A}(w_j,\eta) \longrightarrow\mathcal B_{\nu,A}(v,\eta). \tag{98}\] For the interior terms this is (94). For the exterior terms first restrict \(y\) to a larger fixed continuity region \(B\). The kernels are separated from their singularities on the nonzero test support. To justify the restrictions, let \(F_h\) be the closed \(h\)-neighborhood of \(\partial A\cup\partial B\), for small \(h>0\), and choose one bounded continuity region containing all these collars. The local second-moment bound, Cauchy–Schwarz, and Portmanteau give \[\limsup_{j\to\infty}\int_{F_h}|w_j|\,d\mu_j \le C\bigl(\limsup_{j\to\infty}\mu_j(F_h)\bigr)^{1/2} \le C\nu(F_h)^{1/2}\longrightarrow0 \qquad(h\downarrow0).\] The final limit uses the \(\nu\)-nullity of both boundaries; the same conclusion holds for ordinary mass. Thus continuous cutoffs approximate both restrictions, with product errors controlled by these collar masses and the other factor’s bounded total variation. Convergence of \(\mu_j\) and \(w_j\mu_j\), followed by uniform approximation of the continuous kernels by finite sums of product functions, now gives the exterior limit on \(B\). Finally, (97) and (96) remove the outer restriction. Both convergences in (98) concern the normalization (85); no Riesz pairing is assigned to a test of nonzero mean.

Set \[c_j=\frac{\int\phi\,d\mu_j}{\int\eta\,d\mu_j}, \qquad \xi_j=\phi-c_j\eta.\] The denominator is nonzero eventually, and compact-test convergence gives \(c_j\to0\). Eventually \(|c_j|\le1\), so \(\xi_j\) has a common compact support, a uniform Lipschitz bound, and zero mean against the whole \(\mu_j\). By linearity with the same \(A\) and the same subtraction point, \[\mathcal B_{\mu_j,A}(w_j,\xi_j) =\mathcal B_{\mu_j,A}(w_j,\phi) -c_j\mathcal B_{\mu_j,A}(w_j,\eta) \longrightarrow\mathcal B_{\nu,A}(v,\phi).\] The bump pairing on the right is already normalized and converges; the only required fact about its coefficient is \(c_j\to0\). On the other hand, (89) and (91) give \(|\mathcal B_{\mu_j,A}(w_j,\xi_j)|\le C_{\phi,\eta}\delta_j^2\to0\). We conclude that \[ \mathcal B_{\nu,A}(v,\phi)=0 \qquad\left(\int\phi\,d\nu=0\right). \tag{99}\]

Intrinsic tests and the fractional operator. Let \(U:\mathbb R^n\to L\) be a linear isometry. For a compactly supported intrinsic Lipschitz function \(g\), choose an ambient radial cutoff \(\chi\) equal to one on a ball containing \(U(\mathop{\mathrm{supp}}g)\), and put \(\phi(x)=\chi(x)g(U^*x)\). This is a compact ambient Lipschitz test and \(\phi(Ut)=g(t)\) for every \(t\). Its plane integral is \(q\int g\). A nonnegative compact radial bump positive at \(0\) supplies the required \(\eta\). Pulling (99) back by \(U\) and canceling the positive factor \(q^2\) therefore gives the same renormalized equation for \(v(Ut)\) against every compact intrinsic mean-zero test. Localization independence allows us to take \(A=B(0,R)\) in the intrinsic plane. We henceforth write this intrinsic height simply as \(v\).

For completeness, we identify this equation with (93). Let \(g\) be compactly supported, smooth and of integral zero, first real-valued, with support in \(B(0,H)\); choose \(R>2H\). For \(\varepsilon>0\), define \(\mathcal L_\varepsilon\) by restricting the \(h\)-integral in (93) to \(|h|\ge\varepsilon\). Reflection and translation give the absolutely convergent single-increment formula \[\mathcal L_\varepsilon g(x) =\int_{|x-y|\ge\varepsilon}(g(x)-g(y))k(x-y)\,dy.\] When \(\varepsilon<R-H\), symmetrization on \(B(0,R)^2\) yields \[\begin{align*} \int v\mathcal L_\varepsilon g ={}&\frac12\iint_{B(0,R)^2\cap\{|x-y|\ge\varepsilon\}} (v(x)-v(y))(g(x)-g(y))k(x-y)\,dx\,dy\\ &+\iint_{B(0,R)\times B(0,R)^c}g(x) [v(x)k(x-y)-v(y)\{k(x-y)-k(-y)\}]\,dx\,dy. \end{align*}\] To justify the exterior equality, integrate the test variable first: \(\int g=0\) permits the subtraction of \(k(-y)\), after which (87) makes the product integrable. The direct term involving \(v(x)\) is separately integrable. This proves the displayed identity without applying Fubini to the uncancelled exterior height term.

The interior limit as \(\varepsilon\downarrow0\) follows from (86). On bounded \(x\), the second difference of \(g\) is bounded by \(C|h|^2\) near \(h=0\), giving an \(\varepsilon\)-independent bound for \(\mathcal L_\varepsilon g\). Outside a fixed ball containing the support, the cutoff is irrelevant and \(\int g=0\) gives \(|\mathcal L_\varepsilon g(x)|\le C_g(1+|x|)^{-n-2}\) for \(0<\varepsilon\le1\). Dominated convergence using (92) therefore applies also to the left-hand side. The resulting identity is \(\int v\mathcal Lg=\mathcal B_{dx,B(0,R)}(v,g)=0\). Real and imaginary parts give the complex statement. All steps use the same \(v\) and the same preselected subsequence, so they apply to every coordinate in a finite normal frame simultaneously. ◻

Proof. We first justify the extension of the test class in (102). We will then compute the positive Fourier multiplier away from zero, deduce that the height is a polynomial, and use the weighted tail bound to exclude degrees at least two. For a Schwartz function \(g\) put \[p(g)=\sup_{y\in\mathbb R^n}(1+|y|)^{n+3} \bigl(|g(y)|+\|D^2g(y)\|\bigr), \qquad M_1(g)=\int_{\mathbb R^n}|y|\,|g(y)|\,dy.\] Here \(D^2g\) is the second real derivative, with its operator norm. The quantity \(p\) is a continuous Schwartz seminorm, and \(M_1(g)\le C_n p(g)\), since \(\int |y|(1+|y|)^{-n-3}\,dy<\infty\). We claim that \[ |\mathcal Lg(x)|\le C_n p(g)(1+|x|)^{-n-2} \qquad\text{if }\int g=0. \tag{103}\] Splitting the increment integral at radius one gives \(|\mathcal Lg(x)|\le C_n p(g)\) for every \(x\). For the spatial decay take \(r=|x|\ge2\) and \(R=r/2\). If \(|h|<R\), the segments from \(x\) to \(x\pm h\) stay outside \(B(0,r/2)\). Taylor’s formula therefore bounds the near part of (100) in absolute value by \[C_n p(g)r^{-n-3} \int_{|h|<R}|h|^{1-n}\,dh \le C_n p(g)r^{-n-2}.\]

For the far part let \(E_x=\{y:|x-y|\ge R\}\) and \(k(z)=|z|^{-n-1}\). Translation and reflection give exactly \[g(x)\int_{|h|\ge R}k(h)\,dh -\int_{E_x}k(x-y)g(y)\,dy.\] These integrals are separately absolutely convergent. The first term is bounded by \(C_n|g(x)|/r\). Using the mean-zero condition in the second term gives the identity \[ \int_{E_x}k(x-y)g(y)\,dy =\int_{E_x}\bigl(k(x-y)-k(x)\bigr)g(y)\,dy -k(x)\int_{E_x^c}g(y)\,dy. \tag{104}\] On \(|y|\le r/2\), the mean value theorem gives \[|k(x-y)-k(x)|\le C_n|y|r^{-n-2}.\] On \(E_x\cap\{|y|>r/2\}\), the kernel difference is bounded by \(C_n r^{-n-1}\), whereas \[\int_{|y|>r/2}|g(y)|\,dy\le\frac{2}{r}M_1(g).\] Finally \(E_x^c\subset\{|y|>r/2\}\), so the last term in (104) has the same bound \(C_n r^{-n-2}M_1(g)\). This includes the correction for the omitted ball around \(x\). The near and far estimates prove (103).

It follows from (101) and (103) that \(v\mathcal Lg\) is integrable for every mean-zero Schwartz function, and \[ \left|\int v\mathcal Lg\right| \le C_n p(g)\int |v(x)|(1+|x|)^{-n-2}\,dx. \tag{105}\] Choose \(\chi,\rho\in C_c^\infty(\mathbb R^n)\) with \(\chi=1\) near zero and \(\int\rho=1\). For a mean-zero Schwartz function \(g\) set \[g_j(x)=\chi(x/j)g(x) -\left(\int\chi(y/j)g(y)\,dy\right)\rho(x).\] Each \(g_j\) is compactly supported and has mean zero. The product rule and rapid decrease show \(\chi(\cdot/j)g\to g\) in every Schwartz seminorm; also \(\int\chi(y/j)g(y)\,dy\to\int g=0\). Thus \(g_j\to g\) in Schwartz space. Applying (105) to \(g_j-g\) extends (102) to every mean-zero \(g\in\mathcal S(\mathbb R^n)\).

We next compute the Fourier multiplier needed for localization away from zero. Use the convention \[\widehat g(\xi)=\int_{\mathbb R^n}e^{-2\pi i x\cdot\xi}g(x)\,dx, \qquad \mathcal F^{-1}g(x)=\int_{\mathbb R^n}e^{2\pi i x\cdot\xi}g(\xi)\,d\xi.\] Define \[a_n(\xi)=\int_{\mathbb R^n} \frac{1-\cos(2\pi\xi\cdot h)}{|h|^{n+1}}\,dh.\] This is an absolutely convergent nonnegative integral: quadratic cancellation controls the origin, and boundedness of \(1-\cos\) controls infinity. If \(\xi\ne0\), the numerator equals \(2\) at \(h=\xi/(2|\xi|^2)\) and is positive on a neighborhood of that point. Consequently \(a_n(\xi)>0\). Orthogonal changes of variables show that \(a_n\) is radial, and the substitution \(u=th\) gives \(a_n(t\xi)=t a_n(\xi)\) for \(t>0\). Hence \[ a_n(\xi)=c_n|\xi|,\qquad c_n=a_n(e_1)>0. \tag{106}\]

For \(g\in\mathcal S(\mathbb R^n)\) the full absolute integral required to interchange frequency and increments is finite: \[\int_{\mathbb R^n}\int_{\mathbb R^n} \frac{1-\cos(2\pi\xi\cdot h)}{|h|^{n+1}}|g(\xi)|\,dh\,d\xi =c_n\int_{\mathbb R^n}|\xi|\,|g(\xi)|\,d\xi<\infty.\] Fubini’s theorem and the second difference of the exponential therefore give, pointwise in \(x\), \[ \mathcal L(\mathcal F^{-1}g)(x) =c_n\int_{\mathbb R^n}e^{2\pi i x\cdot\xi}|\xi|g(\xi)\,d\xi. \tag{107}\] The factor \(-2\) in that second difference cancels the factor \(-1/2\) in (100).

The weighted hypothesis implies \(v\in L^1_{\mathrm{loc}}\) and defines a tempered distribution \[T_v(g)=\int_{\mathbb R^n}v(x)g(x)\,dx.\] Indeed this integral is bounded by \(\sup_x(1+|x|)^{n+2}|g(x)|\) times the integral in (101). All distributional pairings here are complex bilinear, without complex conjugation. Let \(S=\mathcal F^{-1}T_v\), so \(S(g)=T_v(\mathcal F^{-1}g)\). Take \(\psi\in C_c^\infty(\mathbb R^n\setminus\{0\})\). The function \(q(\xi)=\psi(\xi)/|\xi|\), extended by zero near the origin, is smooth and compactly supported. Its inverse Fourier transform is Schwartz and has integral \(q(0)=0\). The extended weak equation and (107) give \[0=T_v\bigl(\mathcal L(\mathcal F^{-1}q)\bigr) =c_nT_v(\mathcal F^{-1}\psi)=c_n S(\psi).\] Thus \(S\) is supported at the origin.

For completeness we prove the point-support consequence; compare (Hörmander 2003, Theorem 2.3.4). Continuity of \(S\) on Schwartz space supplies \(N\in\mathbb N\) and \(C<\infty\) such that \[ |S(g)|\le C\sum_{|\alpha|\le N} \sup_x(1+|x|)^N|\partial^\alpha g(x)|. \tag{108}\] A Schwartz function vanishing near zero is annihilated by \(S\): multiply it by expanding compact cutoffs, use the support property, and pass to the Schwartz limit. Consequently \(S\) has the same value on two Schwartz functions agreeing near zero.

Suppose now that \(\partial^\alpha g(0)=0\) for every \(|\alpha|\le N\). Choose \(\eta\in C_c^\infty(B(0,2))\) equal to one near zero and put \(\eta_\varepsilon(x)=\eta(x/\varepsilon)\). For \(0<\varepsilon<1/2\), Taylor’s formula and the product rule give \[\sup_x(1+|x|)^N |\partial^\alpha(\eta_\varepsilon g)(x)| \le C_g\varepsilon^{N+1-|\alpha|} \le C_g\varepsilon \qquad (|\alpha|\le N).\] To see the power, a derivative of order \(j\) falling on the cutoff costs \(\varepsilon^{-j}\), while the remaining derivative of \(g\), of order \(|\alpha|-j\), is \(O(\varepsilon^{N+1-|\alpha|+j})\) on its support. Since \(S(g)=S(\eta_\varepsilon g)\), letting \(\varepsilon\downarrow0\) in (108) proves \(S(g)=0\).

It follows that \(S\) depends only on the derivatives through order \(N\) at zero. More explicitly, subtract from \(g\) the function \[\eta(x)\sum_{|\alpha|\le N} \frac{\partial^\alpha g(0)}{\alpha!}x^\alpha.\] The remainder has vanishing jet through order \(N\) at zero, and therefore there are constants \(c_\alpha\in\mathbb C\) such that \[S(g)=\sum_{|\alpha|\le N}c_\alpha\partial^\alpha g(0).\] Using \(T_v(g)=S(\widehat g)\) and differentiating the Fourier integral of a Schwartz function at zero gives \[T_v(g)=\int_{\mathbb R^n}P(x)g(x)\,dx, \qquad P(x)=\sum_{|\alpha|\le N}c_\alpha(-2\pi i x)^\alpha.\] Every polynomially weighted Schwartz function is integrable, so the finite sum and these integrals may be interchanged. Since \(v\) is locally integrable, equality on compact smooth tests implies \(v=P\) almost everywhere.

Finally, (101) implies \[\int_{|x|\ge1}|P(x)|\,|x|^{-n-2}\,dx<\infty.\] If \(P\) had degree \(k\ge2\), its nonzero leading homogeneous part would have absolute value bounded below on a nonempty open subset of the unit sphere. On the corresponding cone and for all sufficiently large \(r\), \(|P(r\theta)|\ge c r^k\). The last integral would then dominate a positive constant times \[\int_{R_0}^{\infty}r^{k-3}\,dr,\] which diverges for \(k\ge2\). (For \(n=1\), one uses either of the two rays.) Hence \(P\) has degree at most one. Taking real parts gives \(v=b+\ell\) almost everywhere. If \(|v|\le M\) almost everywhere, continuity gives \(|b+\ell(x)|\le M\) for every \(x\), since any open set violating this bound would have positive measure. Evaluating along rays shows \(\ell=0\). ◻

Proof. Choose a compactly supported Lipschitz cutoff equal to one on a ball containing the common bounded set. Its product with \(v\) is a compactly supported Lipschitz function \(\widetilde v\) agreeing with \(v\) on that set. Equation (113), with \(\psi=\widetilde v\), gives \[\int w_jv\,d\mu_j\longrightarrow \int fv\,d\nu=\int f^2\,d\nu.\] Weak convergence tested against the bounded continuous function \(\widetilde v^2\) gives \[\int v^2\,d\mu_j\longrightarrow \int v^2\,d\nu=\int f^2\,d\nu.\] The restrictions of \(v\) belong to the respective \(L^2\) spaces because they are bounded on the common set. Thus all terms in the identity \[\int|w_j-v|^2\,d\mu_j =\int w_j^2\,d\mu_j-2\int w_jv\,d\mu_j +\int v^2\,d\mu_j\] are integrable. Their limits prove the assertion. ◻

Proof. The orthogonal decomposition gives \[ x-a_j=L_jL_j^*(x-a_j)+\delta_jN_jw_j(x), \qquad |x-P_jx|=\delta_j\|w_j(x)\|. \tag{118}\] Since \(v\) has Lipschitz constant at most \(\|B\|\), \[ \int\|v(x)-v(P_jx)\|^2\,d\mu_j(x) \le\|B\|^2\delta_j^2\int\|w_j\|^2\,d\mu_j \longrightarrow0. \tag{119}\] This estimate also proves integrability of the left side. The inequality \(\|u+z\|^2\le2\|u\|^2+2\|z\|^2\) now yields \[ \int\|w_j(x)-v(P_jx)\|^2\,d\mu_j(x)\longrightarrow0. \tag{120}\]

Set \(c_j=b+Ba_j\), \(A_j'=BL_j\), and define \[ S'_j=\bigl\{a_j+L_jt+\delta_jN_j(c_j+A_j't):t\in\mathbb R^n\bigr\}. \tag{121}\] Its direction is the range of \(L_j+\delta_jN_jA_j'\). By (115), \[L_j^*(L_j+\delta_jN_jA_j')=I_{\mathbb R^n}.\] The parametrizing linear map is therefore injective. Its range has dimension \(n\), and (121) is a nonempty affine \(n\)-plane. This conclusion needs no bound on its slope.

For a given \(x\), choose \(t=L_j^*(x-a_j)\) in (121). Since \(v(P_jx)=c_j+A_j't\), (118) gives \[\frac{\mathop{\mathrm{dist}}(x,S'_j)}{\delta_j} \le\|w_j(x)-v(P_jx)\|.\] The squared left side is measurable and dominated by an integrable function. Integrating and using (120) proves (117). ◻

Proof. Admissibility of the growing horizons implies \(\mathop{\mathrm{diam}}(\mathop{\mathrm{supp}}\mu_j)\to\infty\). Lemma 14 and Proposition 17, applied to (123), give, after discarding finitely many indices and passing to a subsequence, a full constant-density plane limit \[\mu_j\rightharpoonup\nu=q\mathcal H^n|_{L(\mathbb R^n)},\qquad q>0,\] where \(L\) is a linear isometry. They also give approximate fitting planes \(S_j=a_j+L_j(\mathbb R^n)\), with \(a_j\to0\) and \(L_j\to L\), and fixed-plane moment bounds \[ \int_{B(0,2^\ell)}\mathop{\mathrm{dist}}(x,S_j)^2\,d\mu_j(x) \le C_*\delta_j^2(2^\ell)^{n+2+2\gamma} \qquad(0\le\ell\le K_j). \tag{126}\] The constant \(C_*\) is independent of \(j\) and \(\ell\). The fitting planes are approximate minimizers; no existence of an exact minimizer is used.

Choose complementary normal isometries \(N_j\) so that (115) holds. For \(i=1,\dots,q_0\), write \[u_{j,i}(x)=\langle N_je_i,x-a_j\rangle, \qquad w_{j,i}(x)=\delta_j^{-1}u_{j,i}(x),\] where \((e_i)\) is the standard orthonormal basis of \(\mathbb R^{q_0}\). The identity \(\sum_i u_{j,i}(x)^2=\mathop{\mathrm{dist}}(x,S_j)^2\) supplies the normal-coordinate moment bounds required by Lemma 18. That lemma and Lemma 19, with a common subsequence as described above, give measurable limiting heights \(f_i\) on the limiting plane. On every region \(\Omega_m\), they satisfy \[\begin{align*} \int_{\Omega_m}w_{j,i}^2\,d\mu_j &\longrightarrow\int_{\Omega_m}f_i^2\,d\nu, \tag{127}\\ \int_{\Omega_m}w_{j,i}\psi\,d\mu_j &\longrightarrow\int_{\Omega_m}f_i\psi\,d\nu \tag{128}\end{align*}\] for every compactly supported Lipschitz \(\psi\). All these statements concern the same heights and the same subsequence.

Proposition 20 gives the mean-zero fractional equation and the weighted tail integrability of each \(f_i\). Lemma 21 therefore gives a scalar \(b_i\) and a linear functional \(M_i\) on \(\mathbb R^n\) such that \[f_i(Lt)=b_i+M_it\quad\text{for almost every }t\in\mathbb R^n.\] Use the ambient affine representative \(v_i(x)=b_i+M_iL^*x\). Applying Lemma 22 to the finite restrictions on \(\Omega_m\) yields \[\int_{\Omega_m}|w_{j,i}-v_i|^2\,d\mu_j\longrightarrow0.\] Assemble \(v=(v_1,\dots,v_{q_0})=b+Bx\). The Euclidean squared norm is the sum of the squared coordinates, so \[ \int_{\Omega_m}\|w_j-v\|^2\,d\mu_j\longrightarrow0, \qquad \sup_j\int_{\Omega_m}\|w_j\|^2\,d\mu_j<\infty. \tag{129}\] Integrability of each coordinate square justifies the finite summation.

Fix \(r>0\) and choose \(m\) with \(B(0,r)\subset\Omega_m\). The integrands in (129) are nonnegative. Restricting them to \(B(0,r)\) preserves the uniform bound and the zero limit. In particular, this step requires no assertion about mass on the sphere \(\partial B(0,r)\). Lemma 23, applied to the restrictions on this ball, supplies affine \(n\)-planes \(S'_j\) with \[\int_{B(0,r)}\left(\frac{\mathop{\mathrm{dist}}(x,S'_j)}{\delta_j}\right)^2 d\mu_j(x)\longrightarrow0.\] By the definition of the excess, \[0\le\left(\frac{e_{\mu_j}(0,r)}{\delta_j}\right)^2 \le r^{-n-2}\int_{B(0,r)} \left(\frac{\mathop{\mathrm{dist}}(x,S'_j)}{\delta_j}\right)^2d\mu_j(x).\] This proves the first limit in (125) for every fixed \(r\), without another extraction.

Proposition 17 also gives bilateral convergence of the supports to \(L(\mathbb R^n)\) on every bounded ball. This convergence persists on the subsequence used for the height limits. For every fixed \(r>0\), both directed deviations in the definition of \(b_{\mu_j}(0,r)\) therefore tend to zero when evaluated at the plane \(L(\mathbb R^n)\). Their sum divided by \(r\) bounds \(b_{\mu_j}(0,r)\) from above, which proves the second limit in (125). ◻

Proof. We first work at center zero and scale one. If the assertion failed, then for each \(j\ge0\) we could choose a measure \(\mu_j\) satisfying the hypotheses at the index \(j+J_0\), for which at least one conclusion in (134) fails. Thus, with \[\delta_j=2^{-(j+J_0)},\qquad K_j=2(j+J_0)+3,\qquad A_j=8\,2^{K_j},\] we would have either \(e_{\mu_j}(0,1/2)>\delta_j\) or \(b_{\mu_j}(0,1/2)\ge\varepsilon\) at every index. The constants in the growth and operator hypotheses remain \(c,G,D_0\) for this entire sequence. Moreover, \[\begin{align*} \delta_j2^{\gamma K_j} &=2^{3\gamma}\,2^{-(1-2\gamma)(j+J_0)}\longrightarrow0,\\ \delta_j^{-1}2^{-K_j} &=2^{-(j+J_0)-3}\longrightarrow0. \end{align*}\] Thus every hypothesis of Proposition 24 holds. On its common subsequence, the excess divided by \(\delta_j\) is eventually less than one and the bilateral beta number is eventually less than \(\varepsilon\). This contradicts the selected failure of their conjunction. We have obtained one \(J\ge J_0\) that works for every source measure at center zero and scale one.

Fix this \(J\). For arbitrary \(\mu,a,r\) in the statement, define the normalized measure \[\sigma=r^{-n}(T_{a,r})_\#\mu, \qquad T_{a,r}(x)=\frac{x-a}{r}.\] The upper and admissible lower growth constants are unchanged, the origin belongs to its support, and \(2^K\le\mathop{\mathrm{diam}}(\mathop{\mathrm{supp}}\sigma)\). The conjugation of hard truncations under \(T_{a,r}\) preserves the bound \(D_0\); the truncation parameter \(\eta\) for \(\sigma\) corresponds exactly to \(r\eta\) for \(\mu\). Transporting all affine \(n\)-planes by this affine bijection gives \[ e_\sigma(0,s)=e_\mu(a,rs),\qquad b_\sigma(0,s)=b_\mu(a,rs). \tag{135}\] For the first equality, the squared distance contributes \(r^{-2}\), the measure contributes \(r^{-n}\), and the normalizing radius contributes \(s^{-n-2}\); the result is exactly the expression normalized by \((rs)^{n+2}\). The second equality follows directly by scaling both directed distances and the radius by \(r\).

If \(\psi\) is a compactly supported 1-Lipschitz test in \(B(0,A)\) with zero \(\sigma\)-mean, then \(\varphi(x)=\psi(T_{a,r}x)\) has Lipschitz constant at most \(r^{-1}\), is supported in \(B(a,Ar)\), and has zero \(\mu\)-mean. The homogeneity of the kernel, applied to the renormalized pairing of Lemma 5, gives \[\mathcal R_\sigma(\psi)=r^{-n}\mathcal R_\mu(\varphi).\] Hence (133) supplies the normalized bound \(|\mathcal R_\sigma(\psi)|\le\delta^3\). The scale-one assertion applies to \(\sigma\), and (135) gives (134) for \(\mu\). Finally, multiplying or dividing a test by \(r\) proves the equivalence of the two oscillation normalizations. ◻

Proof. For every eligible ordered pair \((Q,Q')\), choose one such \(T\), and let \(\mathcal O_0\) be the resulting finite set. Make these choices before fixing a point. Outside the common null set of lattice boundaries, fix \(x\in R\) and list the parents whose seeds contain \(x\) in increasing order of generation. Their generations are distinct, because their parents also contain \(x\).

Starting with the first parent \(Q_1\), group together every listed parent of generation at most \(k(S(Q_1))\). This group has at most \(N+1\) members. If another parent remains, let \(Q_2\) be the first one. Nesting gives \(Q_2\subset S(Q_1)\), so the chosen charge for \((Q_1,Q_2)\) contains \(x\) and has generation in \([k(S(Q_1)),k(Q_2))\). Repeat this procedure. The intervals supplying successive charges are disjoint: the next interval starts at \(k(S(Q_2))\geq k(Q_2)\). Thus the number of groups is at most \(1+\sum_{T\in\mathcal O_0}\mathbf1_T(x)\), and \[\sum_{Q\in\mathcal F}\mathbf1_{S(Q)}(x) \leq (N+1)\left(\mathbf1_R(x) +\sum_{T\in\mathcal O_0}\mathbf1_T(x)\right).\] Integrating this finite inequality and using the two mass bounds proves the result. In particular, overlapping seeds and arbitrarily long gaps between selected generations cause no difficulty. ◻

Proof. We first fix all numerical parameters. Apply Theorem 25 with \(\gamma=1/4\) and output tolerance \(\varepsilon\). Write its resulting constants as \[\delta=2^{-j_*},\qquad K_*=2j_*+3,\qquad A_*=8\cdot2^{K_*}.\] Choose \[A_{\rm f}>H2^{K_*}+2C_0+1,\qquad 0<\alpha<\min\left\{\frac{\varepsilon H}{4}, \frac{\delta H}{2\sqrt{G+1}}\right\}, \qquad J_{\rm o}>A_*H+2C_0+1, \qquad v=\frac{\delta^3}{2}.\] Here \(G\leq9^dC_{\rm AD}\) is supplied by Lemma 4. Proposition 10 shows that \[\mathcal O=\{T\in\mathcal D: W_{J_{\rm o}}(T)>v\ell(T)^n\}\] is Carleson; denote its constant by \(C_{\rm o}\). Proposition 11, with aperture \(A_{\rm f}\) and tolerance \(\alpha/(2A_{\rm f})\), supplies a Carleson family \(\mathcal C\), a fixed depth \(N\), and, for every \(Q\notin\mathcal C\), a descendant \(S(Q)\subset Q\) at depth at most \(N\) with \[b_\mu(z_{S(Q)},A_{\rm f}\ell(S(Q))) <\frac{\alpha}{2A_{\rm f}}.\] A near-minimizing plane \(L_S\) consequently satisfies both directed bounds \[ \sup_{E\cap B(z_S,A_{\rm f}\ell(S))}\mathop{\mathrm{dist}}(\cdot,L_S) \leq\alpha\ell(S),\qquad \sup_{L_S\cap B(z_S,A_{\rm f}\ell(S))}\mathop{\mathrm{dist}}(\cdot,E) \leq\alpha\ell(S). \tag{136}\] The lattice mass bounds give a uniform \(M\geq1\) such that \[ \mu(Q)\leq M\mu(S(Q)),\qquad M\leq C(d,n,C_{\rm AD})2^{nN}. \tag{137}\]

Before iterating propagation, we ensure its admissible horizon. The required condition is \(2^{K_*}H\ell(S)\leq D_E\). If \(D_E<\infty\), restrict for now to parents satisfying \(\ell(Q)\leq D_E/B_*\), where \(B_*=H2^{K_*}\). Every selected seed then satisfies the condition, as do all later radii in its chain. If \(D_E=\infty\), there is no cutoff. This restriction costs only finitely many generations: the largest lattice scale lies in \([D_E,2D_E)\), so there are at most \(2+\lceil\log_2 B_*\rceil\) omitted generations. At each generation, subcubes of a given \(R\) are disjoint and have total mass at most \(\mu(R)\). We will restore all omitted cubes at this fixed Carleson cost.

We prove the implication needed by Lemma 26. Let \(S\) be one of these flat seeds and let \(Q'\subset S\) be a proper descendant, with \(m=k(Q')-k(S)>0\). Suppose that every proper ancestor of \(Q'\) between \(S\) and \(Q'\) lies outside \(\mathcal O\). We claim that \(\beta_H(Q')<\varepsilon\).

Keep the center \(a=z_{Q'}\) fixed throughout the argument. Let \(T_j\) be the ancestor of \(Q'\) of generation \(k(S)+j\), and set \[r_j=H2^{-j}\ell(S)=H\ell(T_j),\qquad 0\leq j\leq m.\] The center belongs to the closure of every \(T_j\), so \(|a-z_{T_j}|\leq2C_0\ell(T_j)\). In particular, \(B(a,2^{K_*}r_0)\subset B(z_S,A_{\rm f}\ell(S))\). The first bound in (136) and global upper growth give, for \(0\leq i\leq K_*\), \[ e_\mu(a,2^ir_0) \leq\frac{\sqrt G\,\alpha\ell(S)}{2^ir_0} \leq\delta\,2^{i/4}. \tag{138}\]

For \(j<m\), a compactly supported mean-zero test \(\varphi\) with \(\mathop{\mathrm{Lip}}\varphi\leq r_j^{-1}\) and \(\mathop{\mathrm{supp}}\varphi\subset B(a,A_*r_j)\) is admissible in the definition of \(W_{J_{\rm o}}(T_j)\): its support lies in the required enlarged ball, and \(r_j\geq\ell(T_j)\). Therefore \[ |\mathcal R_\mu(\varphi)| \leq v\ell(T_j)^n\leq\delta^3r_j^n. \tag{139}\] We may now iterate Theorem 25 at the radii \(r_0,r_1,\ldots,r_{m-1}\). To check the complete horizon at the \(j\)th step, observe that \[2^ir_j= \begin{cases} r_{j-i},&i\leq j,\\ 2^{i-j}r_0,&i>j. \end{cases}\] In the first case the preceding induction gives excess at most \(\delta\); in the second, (138) gives at most \(\delta2^{(i-j)/4}\leq\delta2^{i/4}\). Thus all horizon inequalities hold at every step. Together with (139), they prove \[e_\mu(a,r_j)\leq\delta\quad(0\leq j\leq m),\qquad b_\mu(a,r_j)<\varepsilon\quad(1\leq j\leq m).\] Since \(r_m=H\ell(Q')\), the claim follows. Notice that no test at the terminal cube \(Q'\) was used.

Fix \(R\in\mathcal D\). Take any finite family \(\mathcal F\) of remaining subcubes of \(R\) with \(\beta_H(Q)\geq\varepsilon\), and remove the members of \(\mathcal C\). For each remaining parent use the seed selected above. If \(Q'\subset S(Q)\) and \(k(Q')>k(S(Q))\), then the just-proved implication, applied contrapositively, supplies \(T\in\mathcal O\) with \[S(Q)\supset T\supset Q',\qquad k(S(Q))\leq k(T)<k(Q').\] Lemma 26 and (137) give \[\sum_{Q\in\mathcal F\setminus\mathcal C}\mu(Q) \leq M(N+1)(1+C_{\rm o})\mu(R).\] Add the Carleson bound for \(\mathcal C\) and the finite coarse-generation cost. Taking the supremum over finite families proves the asserted countable sum. Every parameter was fixed before \(R\) or any seed was chosen. ◻

Proof of Theorem 3. If \(E\) is empty or a singleton, there is no positive admissible radius, and the asserted ball-map conclusion is vacuous. Otherwise Proposition 27 gives (140) for every threshold. The comparison (141) permits application of Theorem 28; its positive image-mass bound also holds with \(\mu\), after changing the positive constant.

For any \(a\in E\) and \(0<r\leq D_E\), apply this conclusion at \(r/2\) and write its map as \(f:B^n(0,r/2)\to\mathbb R^d\). The map \[g:B^n(0,r)\longrightarrow\mathbb R^d,\qquad g(u)=f(u/2),\] has the same image and Lipschitz constant at most \(L/2\). Since \(B(a,r/2)\subset B(a,r)\), it captures at least \(c\theta_0 2^{-n}r^n\) of the original measure in \(B(a,r)\). This also covers \(r=D_E\) when the diameter is finite. All constants were chosen before \(a\) and \(r\), and have only the dependencies asserted in the theorem. Thus the uniform ball-map condition in Definition 2 holds. ◻