Riesz transforms and uniform rectifiability in higher codimension

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 (1.3) give

Cauchy–Schwarz proves the last assertion.□

Proof. The kernel is odd and satisfies

For h>0, upper growth and a dyadic decomposition toward the origin give

Multiplication by \Lip\varphi bounds the diagonal singularity in the first integral of (1.5). 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 (1.5); 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 \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\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\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 \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 1.4, 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

Then c_j\to0. In (1.5), 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 (1.6) and upper growth; the exterior is uniformly O(T^{-1}) by Lemma 1.5. 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

Indeed, on the support of f_P the difference is at most Jr_P/(r_Qw_Q), and

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 (2.1) and global upper growth imply

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 (2.3) by w_P. Summing first over a finer level using (2.4), and then over a coarser level using its counting bound, gives respectively

Zero inner products may be discarded. Summing the geometric series therefore yields

For completeness, a symmetric matrix (a_{QP}) of nonnegative entries with \sum_P a_{QP}w_P\le Bw_Q satisfies

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 (2.2) 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

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

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

This integral is absolutely convergent. The kernel derivative bound and upper growth on dyadic annuli give

Since |x-z_Q|\le Jr_Q on the support and \int|\varphi_Q|\,d\mu\le2JG(Jr_Q)^n, it follows that

Choose an integer N\ge1, depending only on J,v and the fixed data, with

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,

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 2.1 and the hard-truncation bound imply

Restrict the sum to the fine cubes and pass to the limit in this finite sum. We obtain

Now \mu(S)\le G(C_0+2)^nL^n\le G(C_0+2)^nc_1^{-1}\mu(Q_0), so (2.1) 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. We first prove a qualitative assertion for measures with global upper and lower growth. We then localize it by taking a tangent and obtain the uniform radius floor by compactness.

Support convergence and annular bounds. We record the elementary convergence facts needed at both stages. Suppose \sigma_j converge weakly on compact sets to \sigma, have a common upper growth bound, and satisfy

where c,h>0 are fixed. On every bounded region their supports converge in both directions, with an arbitrarily small enlargement of that region. Indeed, if bounded support points x_j stayed a positive distance from \supp\sigma, a subsequence would converge to such a point x. A continuous cutoff near x, together with (2.8) at a sufficiently small fixed radius, would have uniformly positive \sigma_j integrals and zero \sigma integral. Conversely, a ball about a point of \supp\sigma contains a nonnegative compactly supported continuous test with positive \sigma integral. Its \sigma_j integral is eventually positive. Finite coverings make both assertions uniform on bounded sets. The limit also has lower growth: for x\in\supp\sigma and 0<t\leq h, choose x_j\to x and a continuous cutoff equal to one on B(x,t/2) and supported in B(x,t). Eventually B(x_j,t/4)\subset B(x,t/2), so passing the cutoff integrals to the limit gives \sigma(B(x,t))\geq c(t/4)^n.

In particular, if a_j\in\supp\sigma_j tend to a\in\supp\sigma and the limit is sufficiently flat on B(a,2t), then the sources are flat on B(a_j,t). For example, for 0<\eta\leq1,

To check the numerical slack, if the last inequality failed, choose an n-plane whose sum of deviations on B(a,2t) is less than \eta t/32. Approximate a by a_j and both supports within \eta t/32. Points of either set in B(a_j,t) and all the points used to approximate them then lie in B(a,2t). Each of the two deviations for the sources is less than \eta t/8, contradicting b_{\sigma_j}(a_j,t)\geq\eta. This reasoning uses only the lower growth at small fixed radii; the tested radius t need not be below h.

For fixed r,R>0, the kernel in (2.7) is continuous with compact support. If its center varies from a_j to a, the kernels converge uniformly and have a common bounded support. The upper growth bound therefore gives

Here and below support centers in a limit can be approximated by source support centers, as just proved.

For u>0 define the normalized dilation

A change of variables gives the exact identities

Thus a bound |\mathcal S_\sigma(x;r,R)|\leq T at all support centers and all positive radii survives dilations and compact limits. A global lower bound b_\sigma\geq\eta survives a compact limit with the loss \eta\mapsto\eta/64. The compactness and nonvanishing of normalized dilations follow from Lemma 1.6: upper growth controls the mass on each compact set, and lower growth preserves positive mass in every ball about the origin.

Two tangents reduce the containing dimension. Suppose, towards a contradiction, that a nonzero measure \sigma satisfies global bounds

that |\mathcal S_\sigma(x;r,R)|\leq T at every support center and all 0<r\leq R, and that b_\sigma(x,t)\geq\eta>0 everywhere. Decreasing \eta if necessary, assume \eta\leq1. Let P be a linear subspace containing the support, initially P=\R^d.

The support cannot fill P. If it did, upper and lower growth would force \dim P=n. To see this, write k=\dim P and cover its unit ball by O(u^{-k}) balls of radius u. For k<n, upper growth would make its mass tend to zero. For k>n, a family of \gtrsim u^{-k} disjoint balls of radius comparable to u, centered in the unit ball of P, would have total mass \gtrsim u^{n-k}, contradicting the upper bound on a fixed larger ball. When k=n, the support is an n-plane and all its bilateral beta numbers are zero. Each possibility contradicts the assumptions.

Choose c_1\in P\setminus\supp\sigma and a nearest support point a to c_1. Such a point exists because the support is nonempty and closed in finite-dimensional Euclidean space. Set e=a-c_1\in P\setminus\{0\}. The nearest-point inequality implies

After dilation at a by scales u_j\downarrow0 this becomes 2\langle e,x\rangle+u_j|x|^2\geq0. Pass to a nonzero limit \tau. The support-convergence argument shows that

This first tangent retains global upper and lower growth, the annular bound T, and positive global nonflatness.

We next use the sign of the normal kernel to take a second tangent. Define

For 2r\leq R_0, the weight \chi_{0,R_0}-\chi_{0,r} is nonnegative everywhere and equals one when 2r\leq|x|\leq R_0. In fact, inside the R_0-ball the outer cutoff is one, and outside it the inner cutoff is zero. This argument requires no monotonicity of the cutoff. On the support of \tau,

Given v>0, take r=\min(v,R_0)/4 to obtain

Monotone convergence, using p_e(0)=0, proves p_e\in L^1(\tau|_{B(0,R_0)}). This establishes an actual integrability statement before taking the next limit.

For any fixed R>0, exact scaling gives

The convergence follows from absolute continuity of the integral: upper growth and n>0 give \tau(B(0,uR))\to0. Take a nonzero compact limit \nu of these second dilations. Testing with nonnegative compact continuous cutoffs times \langle e,x\rangle_+ shows that this continuous nonnegative function vanishes throughout \supp\nu. The negative normal part already vanishes on every dilated support. Hence

The last inequality uses e\in P\setminus\{0\}: e itself is not in e^\perp. Both tangents retain support in P because their centers lie in P.

The new measure is nonzero, satisfies global upper and lower growth, has annular bound T, and has global beta lower bound at least \eta/4096. For definiteness, the compactness estimates may multiply the upper constant by 2^n and divide the lower constant by 4^n at each tangent; their precise values are immaterial. Strong induction on \dim P, allowing these constants and the positive beta threshold to vary, gives a contradiction. We have proved that a globally regular, globally annular-bounded measure has a flat ball at every prescribed positive tolerance.

Localization and a uniform radius floor. Suppose a measure has global upper growth, lower growth only for 0<t\leq h, and annular bound T for support centers in an open set U with U\cap\supp\sigma\ne\varnothing, and outer radii below a positive cap. If there were no \eta-flat ball in U below another positive cap, take a tangent at any point of U\cap\supp\sigma. Every fixed normalized radius eventually lies below the physical caps, and bounded normalized centers dilate into U. Equations (2.10)– (2.11) and (2.9) therefore make the annular bound and a positive beta lower bound global on the tangent. The lower growth also becomes global, since its normalized cap is h/u_j\to\infty. This contradicts the preceding paragraph. Thus the local qualitative alternative holds with no radius floor.

Normalize the statement of the lemma by a_0=0, s=1. The resulting measures have common global upper growth, origin in their supports, and lower growth at all support centers for 0<t\leq1. If no common \rho worked, choose counterexamples \mu_j with floors \rho_j\downarrow0. The uniform mass bounds on compact sets give a locally weakly convergent subsequence, by the compactness argument in Lemma 1.6. The fixed lower bound at zero keeps its limit \nu nonzero and puts 0\in\supp\nu. The support-convergence argument above transfers the lower bound at radii below a fixed positive cap to \nu. Every fixed annulus with center in B(0,\kappa) and 0<r\leq R<\kappa eventually lies above the floor \rho_j. Its transform on the limit is therefore bounded by T. For every fixed 0<t<\kappa, the failure of flat balls at radius t/2 and (2.9) give

The local qualitative alternative contradicts these two properties. This proves the existence of \rho before the measure is chosen. Finally, (2.11) transports the conclusion back to a_0,s. Only the original lower growth below s\leq D_E has been used.□

Proof. Decompose the one fixed input as

Here |g|,|h|\le1, g vanishes on B(a,r), and h vanishes on B(a,R). All three functions belong to L^2(\mu). Upper growth gives

Cauchy–Schwarz and the operator bound consequently bound each of

by C_{\rm R}(G2^nM)^{1/2}; use r\le R for the middle term.

If u\in L^2(\mu), |u|\le1, and u vanishes on B(a,t), then for |x-a|\le t/2 and 0<\varepsilon<t/2 there is no truncation boundary on its support. The kernel derivative estimate therefore gives

Apply this once to g on A, and twice to h on A and B. Since R_{\mu,\varepsilon}g(a)=\mathcal S_\mu(a;r,R), expanding the two averages of R_{\mu,\varepsilon}(f+g+h) proves (2.13). Finally choose a unit-bounded e with \langle e,\mathcal S_\mu(a;r,R)\rangle=|\mathcal S_\mu(a;r,R)|. This choice precedes \varepsilon; projecting (2.13) onto e proves the last assertion.□

Proof of Proposition 2.3. Put L_0=\max\{1,2C_0\} and M=(8L_0)^n/c_1. Fix the comparison constant E_0 in Lemma 2.5, with this value of M. Apply Lemma 2.4 with

Let \rho>0 be its uniform relative radius. These choices, including \rho, are made before the cube Q is chosen. For now assume \ell(Q)\le D_E and apply that lemma at a_0=z_Q, s=\ell(Q).

The flat branch produces a seed. Suppose the lemma gives a\in E\cap B(z_Q,\kappa\ell(Q)) and \rho\ell(Q)\le t<\kappa\ell(Q) with b_\mu(a,t)<\epsilon_{\rm F}. Choose b\in E^*\cap B(a,t/8), and choose the dyadic scale of P so that

Let P be the cell at that scale containing b. Both b\in Q and \ell(P)<\ell(Q) follow from the core containment and t<\kappa\ell(Q), so exact nesting on E^* gives P\subset Q. Also |z_P-b|\le C_0\ell(P), whence

Choose an affine plane for which the sum of the two deviations in B(a,t) is less than \epsilon_{\rm F}t. Restricting both suprema to the smaller ball gives

The lower bound for t bounds the depth of P uniformly. For example, any integer N_{\rm F}>\log_2(16(A_{\rm F}+C_0)/\rho) suffices.

Cubes without a flat seed have a large annulus. Let \mathcal F_1 be the cubes with \ell(Q)\le D_E for which no seed as in (2.6), with the additional ball containment, exists at depth at most N_{\rm F}. They must lie in the annular branch. Thus each has radii

and a center a_Q in the same core with |\mathcal S_\mu(a_Q;r_Q,R_Q)|>E_0+2. For these fixed positive radii the annular transform is continuous in its center. Indeed the annular kernel is zero near its center, is continuous there after extension by zero, and has bounded compact support locally uniformly in the center; dominated convergence applies using local finiteness. Density of E^* therefore permits replacing a_Q by a'_Q\in E^* so close that

This perturbation uses the strict core and norm margins, and does not change either radius.

Choose dyadic scales s_r,s_R with

using the same monotone rounding rule, so s_r\le s_R. Let A_Q and B_Q be the cells at those scales containing a'_Q. They are descendants of Q with A_Q\subset B_Q\subset Q. Since a cell at scale s has diameter at most L_0s,

Fix an integer I\ge1 larger than \log_2(8L_0/\rho). The depth of A_Q below Q is at most I, and

Neither M nor E_0 depends on \rho; the depth cost H_I is allowed to depend on the now fixed \rho.

Haar Bessel gives the Carleson estimate. Set

Direct integration gives \int h_Q\,d\mu=0 and \|h_Q\|_2^2=1-\mu(A_Q)/\mu(B_Q)\le1. This function is supported in Q and is constant almost everywhere on every cell at least I generations below Q. Partition a finite collection of these functions according to the level of Q modulo I. Within one class, distinct equal-level parents are disjoint. For unequal levels, the gap is at least I: the coarser function is constant on the finer parent, or their supports are disjoint. The finer function’s mean zero gives orthogonality. Bessel’s inequality in each class and summation give

for vector u\in L^2(\mu) and independently chosen |e_Q|\le1. The vector extension follows by the same Parseval argument as in Lemma 2.1.

Fix a top cube Q_0 and a finite \mathcal A\subset\mathcal F_1\cap\mathcal D(Q_0). Use, for every selected cube, the same set S=B(z_{Q_0},(C_0+2)\ell(Q_0)). The annular center and radius bounds ensure B(a'_Q,2R_Q)\subset S. Choose e_Q from the annular vector as in Lemma 2.5; then choose a single

when the family is nonempty. The average gap is greater than one for every selected Q at this same truncation. Consequently

By (2.14), (2.15), and the operator bound,

The empty family is immediate. Taking the supremum over finite subfamilies proves that \mathcal F_1 is Carleson. Finally add the possible top level with \ell(Q)>D_E to obtain \mathcal F. Its mass below any top cube is at most the mass of that top cube. The seed conclusion and its uniform depth bound hold for every cube outside this family.□

Proof. Every integral in (3.1) is finite, since its integrand is bounded on the ball and upper growth implies local finiteness. Markov’s inequality shows that

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

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 (3.4).

Choose

The tube mass in (3.4) is strictly less than (c/2)R^n. Thus we can successively choose p_0,\ldots,p_n\in F\cap\supp\sigma such that, for 1\le j\le n,

At each step the affine span has dimension less than n, and \supp\sigma has full measure. Put v_j=p_j-p_0. Then |v_j|\le2R. These vectors satisfy

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 (3.5) 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)], (3.5) and absorption give

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 \dist(\cdot,W). Applying this affine map to x=a+\sum t_jw_j gives

Since |a|\le R+\eta\le2R, this proves (3.2), 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

Interchanging the planes proves (3.3).□

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

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 3.1 gives, at the same arbitrary point x in each step,

For |x|\le R_k, the sum of these errors for i<k is bounded by B\delta R_kb^k, where

Indeed, sum the geometric series with ratios 2b and b separately. Consequently \dist(x,P_0)\le\dist(x,P_k)+B\delta R_kb^k on that ball. Squaring, integrating, and using upper growth proves (3.6) 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 1.5. 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

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

when 0<r\le R and \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\supp\mu the resulting function \eta_s has integral one and satisfies

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 (3.8). It follows that

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\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

Define p_e(x)=\max(0,\langle e,x\rangle)/|x|^{n+1} for x\ne0, and set p_e(0)=0. Then

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 1.6, applied to the blowups. The rescaled quadratic inequality is 2\langle e,x\rangle+r|x|^2\ge0; hence \supp\tau\subset\{\langle e,x\rangle\ge0\}. For every fixed T>0,

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 \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

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

There is a constant D_1, independent of B and \varepsilon>0, such that

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 (3.10) 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 (3.9). 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 (3.10). 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,

Removing the cap on the right changes the integral by at most C_nG\Lip(\varphi)\mu(B)\varepsilon. Thus v represents the actual interior singular pairing. On B(a,H) the function

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

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 (3.9), 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 (3.11) yields

The subsequence and b were selected before g.

For a compact smooth function g, define the scalar test transform

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 \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

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 (3.12) 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

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

Indeed, set

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

Expand h in the orthonormal basis and use the definition of R_u to obtain (3.14). Every pairing with bounded f is integrable by the preceding estimates. Summing (3.13) therefore gives

Real and imaginary parts give the same identity for complex tests. Boundedness supplies \int|f(x)|(1+|x|)^{-n-2}\,dx<\infty. Lemma 5.2 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 (3.9). 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 3.2 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 \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

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 1.6, followed if necessary by a further subsequence, gives a nonzero limit \nu with 0\in\supp\nu, hard truncation bound D_0, global upper growth 2^nG, and

All frames and moment bounds are kept along the composition of the two subsequences. If x\in\supp\nu, positivity of compact tests in every neighborhood of x supplies nearby points of E_j for all large j. Together with (3.21), this gives \dist(x,S_j)\to0. On the other hand the affine projections

It follows that x=L_\infty L_\infty^*x. At this stage we have proved only \supp\nu\subset L_\infty(\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 1.6 provides compact Lipschitz tests \varphi_j of zero \mu_j-mean, with uniformly bounded supports and Lipschitz constants, such that

Choose C\ge1 bounding their Lipschitz constants. For all sufficiently large j, their supports lie compactly inside B(0,A_j), so (3.17), 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 3.3 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 (3.22), evaluated on a ball centered on the plane, give the two bounds for q in (3.18). Since |u_{j,i}(x)|\le|N_j^*(x-a_j)|=\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 (3.21) 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, \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 (4.2) 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

Define the weighted center and its centered test by

Cauchy–Schwarz, followed by (4.8), proves

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 (4.6) implies

The positive part of this pairing comes from the exact identity

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

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

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 (4.3), and the pairing formula in Lemma 1.5, give

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 (4.16) is kept together.

For |x|\le2H and |y|\ge R,

The global affine bound |U(y)|\le |y|+W+C, together with upper growth and (4.17), makes the exterior expression absolutely integrable. We may therefore integrate (4.12) and use (4.15) and (4.14) to obtain

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

Here the contribution of b is absorbed using t\ge1. Since \gamma<1, summing the controlled shells gives

Beyond the actual last controlled radius, use the global affine bound |U(y)|\le |y|+W+C. Upper growth then gives

The half-open shells and the closed final exterior cover every point, including their boundary spheres. Combining (4.10) with the kernel and tail bounds (4.17)–(4.21) gives the quantitative estimate

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 (4.18), the tested-pairing bound (4.11), and (4.22) now yield

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

The last term is bounded by (4.4). 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 (4.13) 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, (4.26) 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

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 (4.27). 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

These finitely many disjoint measurable cells cover \Omega_H. Each cell boundary has \nu-measure zero, by (4.26) and the nullity of coordinate faces. The same holds for intersections of cells from different partition levels. Hence

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 (4.32), 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

All averages are defined for these sufficiently large indices, and their L^2 bounds follow from (4.28). The sets C_D\times C_D are disjoint subsets of \Omega_H\times\Omega_H. Summing (4.34) yields

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 (4.33) gives

The triangle inequality and (4.35) 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

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 (4.31) 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

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 (4.29) 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,\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

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

This identity involves g, an actual continuous function, on the moving supports. For any continuous weight \psi,

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 (4.36) and Cauchy–Schwarz for the limiting measure then allow g\to v in L^2(\nu|_{\Omega_H}). This proves (4.30) in the stated order of limits.□

Proof. We retain the subsequence already selected in Lemma 4.2. 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

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 4.1. Positive-dimensional upper growth makes the diagonal null.

Estimate (5.3) 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

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 (5.7) to the limit at every fixed scale. Changing the radius by at most a factor two gives

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

Together with local L^2 integrability and q>0, this proves (5.9).

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 (5.7) controls the renormalized exterior integral outside B(0,T) by

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 (5.7) implies |u_j(0)|\le1+C\delta_j. Hence |u_j(y)|\le C+|y|. Upper growth gives

This bound uses the unnormalized height at 0; the normalized value w_j(0) may diverge. The T^{-1} term in (5.14) comes from the w_j(x)k(x-y) term, whose x-integral stays on the fixed test support. The kernel difference in (5.4) 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 (5.13).

A common normalization and mean correction. Take two fixed compact ambient Lipschitz functions \phi,\eta such that

Choose one continuity region A containing B(0,R) with a positive margin beyond both supports. For each of these tests separately,

For the interior terms this is (5.11). 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

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, (5.14) and (5.13) remove the outer restriction. Both convergences in (5.15) concern the normalization (5.2); no Riesz pairing is assigned to a test of nonzero mean.

Set

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,

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, (5.6) and (5.8) give |\mathcal B_{\mu_j,A}(w_j,\xi_j)|\le C_{\phi,\eta}\delta_j^2\to0. We conclude that

Intrinsic tests and the fractional operator. Let U:\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(\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 (5.16) 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 (5.10). 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 (5.10) to |h|\ge\varepsilon. Reflection and translation give the absolutely convergent single-increment formula

When \varepsilon<R-H, symmetrization on B(0,R)^2 yields

To justify the exterior equality, integrate the test variable first: \int g=0 permits the subtraction of k(-y), after which (5.4) 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 (5.3). 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 (5.9) 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 (5.19). 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

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

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 (5.17) in absolute value by

For the far part let E_x=\{y:|x-y|\ge R\} and k(z)=|z|^{-n-1}. Translation and reflection give exactly

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

On |y|\le r/2, the mean value theorem gives

On E_x\cap\{|y|>r/2\}, the kernel difference is bounded by C_n r^{-n-1}, whereas

Finally E_x^c\subset\{|y|>r/2\}, so the last term in (5.21) 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 (5.20).

It follows from (5.18) and (5.20) that v\mathcal Lg is integrable for every mean-zero Schwartz function, and

Choose \chi,\rho\in C_c^\infty(\R^n) with \chi=1 near zero and \int\rho=1. For a mean-zero Schwartz function g set

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 (5.22) to g_j-g extends (5.19) to every mean-zero g\in\mathcal S(\R^n).

We next compute the Fourier multiplier needed for localization away from zero. Use the convention

Define

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

For g\in\mathcal S(\R^n) the full absolute integral required to interchange frequency and increments is finite:

Fubini’s theorem and the second difference of the exponential therefore give, pointwise in x,

The factor -2 in that second difference cancels the factor -1/2 in (5.17).

The weighted hypothesis implies v\in L^1_{\mathrm{loc}} and defines a tempered distribution

Indeed this integral is bounded by \sup_x(1+|x|)^{n+2}|g(x)| times the integral in (5.18). 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(\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 (5.24) give

Thus S is supported at the origin.

For completeness we prove the point-support consequence; compare [7, Theorem 2.3.4]. Continuity of S on Schwartz space supplies N\in\mathbb N and C<\infty such that

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

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 (5.25) 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

The remainder has vanishing jet through order N at zero, and therefore there are constants c_\alpha\in\mathbb C such that

Using T_v(g)=S(\widehat g) and differentiating the Fourier integral of a Schwartz function at zero gives

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, (5.18) implies

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

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 (6.5), with \psi=\widetilde v, gives

Weak convergence tested against the bounded continuous function \widetilde v^2 gives

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

are integrable. Their limits prove the assertion.□

Proof. The orthogonal decomposition gives

Since v has Lipschitz constant at most \|B\|,

This estimate also proves integrability of the left side. The inequality \|u+z\|^2\le2\|u\|^2+2\|z\|^2 now yields

Set c_j=b+Ba_j, A_j'=BL_j, and define

Its direction is the range of L_j+\delta_jN_jA_j'. By (6.7),

The parametrizing linear map is therefore injective. Its range has dimension n, and (6.13) 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 (6.13). Since v(P_jx)=c_j+A_j't, (6.10) gives

The squared left side is measurable and dominated by an integrable function. Integrating and using (6.12) proves (6.9).□

Proof. Admissibility of the growing horizons implies \diam(\supp\mu_j)\to\infty. Lemma 3.1 and Proposition 3.4, applied to (6.15), give, after discarding finitely many indices and passing to a subsequence, a full constant-density plane limit

where L is a linear isometry. They also give approximate fitting planes S_j=a_j+L_j(\R^n), with a_j\to0 and L_j\to L, and fixed-plane moment bounds

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 (6.7) holds. For i=1,\dots,q_0, write

where (e_i) is the standard orthonormal basis of \R^{q_0}. The identity \sum_i u_{j,i}(x)^2=\dist(x,S_j)^2 supplies the normal-coordinate moment bounds required by Lemma 4.1. That lemma and Lemma 4.2, with a common subsequence as described above, give measurable limiting heights f_i on the limiting plane. On every region \Omega_m, they satisfy

for every compactly supported Lipschitz \psi. All these statements concern the same heights and the same subsequence.

Proposition 5.1 gives the mean-zero fractional equation and the weighted tail integrability of each f_i. Lemma 5.2 therefore gives a scalar b_i and a linear functional M_i on \R^n such that

Use the ambient affine representative v_i(x)=b_i+M_iL^*x. Applying Lemma 6.1 to the finite restrictions on \Omega_m yields

Assemble v=(v_1,\dots,v_{q_0})=b+Bx. The Euclidean squared norm is the sum of the squared coordinates, so

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 (6.21) 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 6.2, applied to the restrictions on this ball, supplies affine n-planes S'_j with

By the definition of the excess,

This proves the first limit in (6.17) for every fixed r, without another extraction.

Proposition 3.4 also gives bilateral convergence of the supports to L(\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(\R^n). Their sum divided by r bounds b_{\mu_j}(0,r) from above, which proves the second limit in (6.17).□

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 (6.26) fails. Thus, with

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,

Thus every hypothesis of Proposition 6.3 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

The upper and admissible lower growth constants are unchanged, the origin belongs to its support, and 2^K\le\diam(\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

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 1.5, gives

Hence (6.25) supplies the normalized bound |\mathcal R_\sigma(\psi)|\le\delta^3. The scale-one assertion applies to \sigma, and (6.27) gives (6.26) 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

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 6.4 with \gamma=1/4 and output tolerance \varepsilon. Write its resulting constants as

Choose

Here G\leq9^dC_{\rm AD} is supplied by Lemma 1.4. Proposition 2.2 shows that

is Carleson; denote its constant by C_{\rm o}. Proposition 2.3, 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

A near-minimizing plane L_S consequently satisfies both directed bounds

The lattice mass bounds give a uniform M\geq1 such that

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 7.1. 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

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 (7.1) and global upper growth give, for 0\leq i\leq K_*,

For j<m, a compactly supported mean-zero test \varphi with \Lip\varphi\leq r_j^{-1} and \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

We may now iterate Theorem 6.4 at the radii r_0,r_1,\ldots,r_{m-1}. To check the complete horizon at the jth step, observe that

In the first case the preceding induction gives excess at most \delta; in the second, (7.3) gives at most \delta2^{(i-j)/4}\leq\delta2^{i/4}. Thus all horizon inequalities hold at every step. Together with (7.4), they prove

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

Lemma 7.1 and (7.2) give

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 1.3. If E is empty or a singleton, there is no positive admissible radius, and the asserted ball-map conclusion is vacuous. Otherwise Proposition 7.2 gives (7.5) for every threshold. The comparison (7.6) permits application of Theorem 7.3; 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\R^d. The map

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 1.2 holds.□

Proof. If D_E=0, the support is empty or a singleton. All the truncated operators above vanish \mu-almost everywhere, so the assertions are immediate. Suppose D_E>0. Theorem 1.3 gives uniform rectifiability, with constants \theta,M and admissible radii when the support is bounded. For unbounded support, the forward clauses (a)\Rightarrow(c) and (a)\Rightarrow(b) of Mas–Tolsa’s Theorem 2.3 give, respectively, the hard L^2 and smooth L^p/weak-(1,1) estimates; their Definition 2.1 is precisely the fixed cutoff above, and their Theorem 1.3 includes the vector Riesz family [10]. Their range 1\le n<d contains the present one. We reduce bounded support to this unbounded setting.

Suppose D=D_E<\infty, and fix a\in E. Since n<d, choose an affine n-plane P with \dist(a,P)=4D, and set

For x\in E, the diameter bound gives 3D\le\dist(x,P)\le5D. Thus E and P are disjoint and \supp\sigma=E\cup P is unbounded. Lemma 1.4 and the Euclidean plane-volume bound give

where \omega_n is the volume of the unit ball in \R^n.

We verify lower growth and Lipschitz pieces at every support center and radius. For centers on P, the plane supplies both properties at every radius. For x\in E and r\le D, use the corresponding properties of \mu. If D<r<12D, the ball B(x,D/2) lies in B(x,r) and has mass at least

The uniform-rectifiability map at scale D/2 captures at least \theta(D/2)^n\ge\theta24^{-n}r^n in the same ball. Composing that map with u\mapsto Du/(2r) gives a map on B_{\R^n}(0,r) with the same image and Lipschitz constant at most M. If r\ge12D, the disk in P centered at \pi_Px with radius r/2 lies in B(x,r), since every point z of that disk satisfies

This disk has \sigma-mass at least \omega_n2^{-n}r^n and is the image of B_{\R^n}(0,r) under a 1/2-Lipschitz affine map. These estimates prove that \sigma is globally n-AD regular and uniformly n-rectifiable, with constants controlled by the original dimensions, AD data, \theta, and M.

For any input f in one of the stated L^p(\mu) spaces, extend it by zero on P to obtain \widetilde f. Disjointness gives \|\widetilde f\|_{L^p(\sigma)}=\|f\|_{L^p(\mu)}. For every positive truncation and every x\in E,

The vector Riesz truncations have the same identity. Apply the forward estimates to \sigma and restrict the output to E. Since \sigma|_E=\mu, the identities transfer both the strong bounds and the weak-(1,1) bound to \mu, with the required constant dependence.

In either diameter regime, for a real f\in L^2(\mu), Lemma 1.4 and the size bound in (8.1) give

Consequently the displayed hard integrals converge absolutely for every x and tend to zero as \varepsilon\to\infty. Approximate f in L^2 by compactly supported L^1\cap L^2 inputs f_j. The same kernel estimate gives pointwise convergence of each positive truncation. Taking the supremum over finite sequences gives

so Fatou’s lemma transfers the cited variation bound to the actual integrals for f. For the smooth formulas, the analogous exterior L^{p'}(\mu) kernel estimate, with p'=p/(p-1)>1, and Hölder’s inequality give the same identification for L^p inputs. For L^1 inputs the cutoff kernel is bounded at each fixed positive scale.

Let V=V_\rho((T^{K,\mu}_\varepsilon f)_{\varepsilon>0}). By (8.3), V is finite almost everywhere. Finite variation forces the truncations to be Cauchy as \varepsilon\downarrow0: otherwise a decreasing sequence with infinitely many jumps bounded below would make V infinite. This proves existence of the finite principal value. The definition gives |T^{K,\mu}_\varepsilon f-T^{K,\mu}_\delta f|\le V for every \varepsilon,\delta>0. Sending \delta\downarrow0 gives |T^{K,\mu}_\varepsilon f-T^{K,\mu}f|\le V, while sending \delta\to\infty gives |T^{K,\mu}_\varepsilon f|\le V. Thus the limit belongs to L^2(\mu), and dominated convergence with the majorant V^2\in L^1(\mu) proves strong L^2 convergence. The same argument with Euclidean norms applies to the vector Riesz family.□

Proof of Theorem 7.3. The formulation of Bate, Hyde and Schul uses closed balls and maps defined on all of \R^n [2, Theorem 1.0.4]. If \overline b_E denotes its closed-ball bilateral coefficient, then

Indeed, both closed-ball suprema at radius t are bounded by the open-ball suprema at radius 2t for the same plane. For R<\diam(E)/2, enlarging the spatial ball to B(a,2R) and substituting s=2t therefore transfers the Carleson hypothesis of Theorem 7.3 to closed balls. If D=\diam(E)<\infty and D/2\leq R<D, cover E by a dimensionally bounded number of support-centered open balls of radius D/8. On each enclosing radius-D/4 ball, the open-ball estimate controls the transformed integral for t<D/8. For the remaining D/8\leq t<R, the integral is at most (\log8)\H^n(E)\leq C D^n\leq C2^nR^n. Thus the closed-ball criterion applies with controlled constants in both diameter regimes. AD regularity has the same open/closed-ball equivalence, by doubling the radius and using the total-mass bound when that doubled radius exceeds a finite diameter.

We next pass from Euclidean bilateral flatness to the metric coefficient used in that source’s corona decomposition. Its coefficient \xi_E(a,t) [2, Definition 3.1.3] minimizes, over maps from E\cap\overline B(a,t) to closed radius-t balls in n-dimensional normed spaces, the sum of two quantities: the largest pairwise distance distortion divided by t, and the largest distance from a target-ball point to the image divided by t. We claim

Choose an almost-fitting affine plane L whose summed directed error at radius 2t is at most 2\lambda t, with 0<\lambda<1/12. Orthogonal projection, followed by translation, gives f(x)=\pi_L(x)-\pi_L(a) in the Euclidean closed radius-t ball in the direction of L. Its distance distortion is at most 4\lambda t. For a target vector v with |v|\leq t, the point y=\pi_L(a)+(1-6\lambda)v satisfies |y-a|\leq(1-4\lambda)t. The bilateral estimate supplies z\in E with |z-y|\leq2\lambda t, so z\in\overline B(a,t) and |f(z)-v|\leq8\lambda t. Thus \xi_E(a,t)\leq12\lambda. Letting \lambda decrease to the infimum proves the claim when that infimum is below 1/12. Otherwise the constant map gives \xi_E(a,t)\leq3\leq36\overline b_E(a,2t). The same doubling of scales and finite-diameter coarse-scale bound proved above now transfer the Carleson estimate to \xi_E. By [2, Theorem B, implication (3)\Rightarrow(5)], E therefore has the corona decomposition by normed spaces required in that source’s positive-mass construction, with controlled constants.

This corona decomposition supplies the bi-Lipschitz pieces in [2, Proposition 9.0.2 and Lemma 9.0.4]. Choosing the omitted mass below half the lower AD bound gives a piece F\subset E\cap\overline B(a,s) with \H^n(F)\geq c s^n>0 and a uniformly K-bi-Lipschitz map \phi:F\to\R^n. To see directly that a full parameter-ball image suffices, fix p\in F and put V=(\phi(F)-\phi(p))/(4K). Then V\subset B^n(0,s), and h(u)=\phi^{-1}(4Ku+\phi(p)) on V is 4K^2-Lipschitz. Coordinatewise Lipschitz extension gives a map on all of \R^n with constant at most 4\sqrt d K^2 and image containing F. Choose the preceding bi-Lipschitz piece at s=r/2 and restrict this map to the open ball B^n(0,r). Its image contains F, while \overline B(a,r/2)\subset B(a,r); the mass bound loses only the factor 2^{-n}. This proves precisely the open-ball formulation of Theorem 7.3.□