Uniform excess charge for Coulomb molecules and the outer radius of neutral atoms

Proof. Approximate an n-electron trial state by one with compact support. Wedge it with a normalized spin orbital in a disjoint distant ball. Dilate the orbital to make its kinetic energy small and separate the supports far enough to make the added repulsion small. Disjoint spatial supports make the energy of the normalized wedge the sum of the group energies and their direct cross repulsion. Dropping the added nuclear attraction proves E_{n+1}\leq E_n. A normalized orbital dilated about one nucleus has kinetic energy O(s^{-2}) and attraction at most -c z_m s^{-1}, proving E_1<0.

Suppose \Delta=E_{n-1}-E_n>0, and take a normalized minimizing sequence \psi_j. The form bound above gives a uniform H^1 bound. Choose nonnegative Lipschitz functions g_S,h_S with

They may be constructed as the cosine and sine of a cutoff angle. For each subset I of the particle labels, multiply \psi_j by \prod_{i\in I}g_S(x_i)\prod_{i\notin I}h_S(x_i). The squared norms of these localized states sum to one. The product partition identity for the kinetic energy adds at most Cn/S^2 to the sum of their forms.

The all-g_S component has energy at least E_n times its squared norm. If k<n coordinates carry g_S, slice in the remaining coordinates. Antisymmetry in the k retained coordinates gives their energy lower bound E_k\geq E_{n-1}. Drop the other kinetic terms and all pair terms involving an h_S coordinate. For S>2\max_m|R_m|, their total nuclear attraction is bounded below by -2Zn/S times the component’s squared norm. If p_j(S) is the total squared mass of components other than all-g_S, it follows that

The probability that any |x_i|\geq2S is at most p_j(S), because the all-g_S product vanishes there. Taking first j\to\infty, then S\to\infty, proves tightness.

Local Rellich compactness and tightness give, along a subsequence, strong L^2 convergence to a normalized antisymmetric \Psi, weak H^1 convergence, and almost-everywhere convergence. The attraction converges: the infinitesimal form bound applied to \psi_j-\Psi, with its relative-bound parameter subsequently tending to zero, gives

Cauchy–Schwarz with this weight handles the difference of the attractive expectations. The kinetic energy is weakly lower semicontinuous, and the nonnegative repulsion is lower semicontinuous by Fatou’s lemma. Hence \Psi minimizes. Variation in the form domain gives the weak eigen-equation at E_n.□

Proof. Let \Psi be the minimizer from Lemma 2.1. For l,\varepsilon>0 put

For these fixed parameters, w and its first derivatives are bounded. Test the weak eigen-equation with (\sum_iw(x_i))\Psi and take real parts. This summed multiplier is symmetric; after taking this admissible test we may expand its individual summands in the ambient form.

In the ith summand, the Hamiltonian on the other n-1 coordinates, minus E_n, has nonnegative pairing after weighting by w(x_i). This follows by slicing in x_i and its spin and using E_{n-1}\geq E_n. The remaining kinetic pairing is also nonnegative. Indeed, with \phi=w(x_i)\Psi,

The integration by parts is justified by large-radius cutoffs and H^1 approximation; all displayed coefficients and their needed derivatives are bounded for fixed l,\varepsilon. We obtain

Let l\downarrow0 at fixed \varepsilon. Dominated convergence applies, using V/\varepsilon for attraction and 2K/\varepsilon for a pair. These functions have finite expectations on the form domain. Hence (2.2) holds with w=(V+\varepsilon)^{-1}, and its right side is at most n.

For a configuration outside the null set of nuclear collisions, set

The triangle inequality gives

In fact, multiplying the right side by |x_i-x_j| bounds it above by w(x_i)w(x_j)(V(x_i)+V(x_j))\leq w(x_i)+w(x_j). Weighted Cauchy–Schwarz and z_m\geq1 now imply

Here \sum_mq_{im}^2/z_m\leq\sum_mq_{im}^2\leq s_i^2\leq1. As \varepsilon\downarrow0, each s_i tends to one almost surely, and the squared sum is bounded by n^2. Taking expectations in the last inequality and using (2.2) gives n\geq n^2/(2Z)-n/2, as asserted.□

Proof. The set of strict sectors is nonempty since E_1<0, and finite by Lemma 2.2. After its largest member there are no strict drops. Monotonicity therefore gives E_n=E_N for n>N, whereas E_n\geq E_N for n<N. Existence of \Psi follows from Lemma 2.1. Every other strict sector has particle number at most N.□

Proof. Both F\Psi and F^2\Psi belong to \mathcal Q_n. Test the weak eigen-equation with F^2\Psi and expand the gradients. Subtraction from q_n[F\Psi] cancels all multiplication potentials and mixed-gradient terms, leaving (2.4).□

Proof. Strict binding follows inductively beginning with the vacuum. Suppose \Phi is a ground state in sector n-1. With the radial core cutoff g_S used in the compactness proof, put F_S=\prod_i g_S(x_i). Its norm tends to one, and the multiplier identity gives normalized energy E_{n-1}+O_Z(S^{-2}). Wedge this state with a normalized radial spatial orbital of fixed spin, supported in 3S<|x|<4S, with kinetic energy O(S^{-2}). The supports are separated. Newton’s spherical mean identity is

It follows directly by integrating (s^2+|x|^2-2s|x|t)^{-1/2}/2 over -1<t<1. Every core radius is at most 2S, so the added electron’s attraction plus direct repulsion is at most -(Z-n+1)/(4S). For n\le Z this strictly dominates the kinetic and cutoff errors when S is large. Lemma 2.1 supplies the next ground state. Its tightness argument gives the asserted compactness for every minimizing sequence, completing the induction.□

Proof. Let \Phi be a ground state in a strict sector, supplied by Lemma 2.1. For l,\eps>0 set

For fixed parameters w and its first derivatives are bounded. Test the weak eigen-equation by (\sum_iw(x_i))\Phi and take real parts. This is a symmetric admissible multiplier. In its ith summand, the form on the other n-1 coordinates is at least E_{n-1}\E_\Phi w(x_i), by slicing in x_i and its spin. The selected-coordinate kinetic pairing is nonnegative. Indeed, writing \phi=w(x_i)\Phi and hence \Phi=U(x_i)\phi, integration by parts gives

The identity follows first for smooth compactly supported functions; the bounded coefficients for fixed l,\eps, approximation in H^1, and large-radius cutoffs give it in the present domain. Since E_{n-1}-E_n>0, discarding that additional nonnegative term leaves

Let l\downarrow0 at fixed \eps. Domination by 2K(x_i-x_j)/\eps for pairs and V(x_i)/\eps for attraction, which are integrable on the form domain, justifies the limit. Then w(x)=|x|/(1+\eps|x|). As \eps\downarrow0, both sides increase and the right side tends to nZ. The left side tends to the expectation of \sum_{i<j}(|x_i|+|x_j|)/|x_i-x_j|, which is at least n(n-1)/2 by the triangle inequality. Thus n\leq2Z+1.

The strict indices form a nonempty finite set by Lemma 2.5. After its largest element N_* there can be no further strict drop; monotonicity implies E_n=E_{N_*} for every n>N_*. For n<N_* monotonicity gives E_n\geq E_{N_*}. This proves (2.6), without any assumption on possible plateaus before N_*. Attainment follows once more from Lemma 2.1.□

Proof. For compactly supported f\in L^{5/3} we also have f\in L^{6/5}. Riesz representation on \dot H^1 gives a unique solution u of -\Delta u=4\pi f. The convolution K*f is continuous: near the singularity use K\in L^{5/2}_{\mathrm{loc}} and continuity of translations in that space; away from it use dominated convergence. It is bounded locally and decays like O(|x|^{-1}) at infinity, hence belongs to L^6. It has the same distributional Laplacian as u. A harmonic L^6 function is zero, by mollification and the mean-value property on arbitrarily large balls. Thus u=K*f. Testing its weak equation with u gives the first identity in (3.1); testing with v and using Cauchy–Schwarz gives the second.

For the noncompact nonnegative case, positivity first gives

To obtain this estimate, approximate the absolute value by smooth convex truncations with derivative bounded by one. Completion and an almost-everywhere convergent subsequence extend the estimate to \dot H^1 and identify its dual pairing with actual integration. Let u be the Riesz solution, and let u_j=K*(\mathbf1_{B(0,j)}f). Testing the difference equation with (u_j-u)_+ gives u_j\leq u. These positive-part tests belong to \dot H^1: an L^6 function with gradient in L^2 is in the completion by large-radius cutoff and mollification. For the cutoff error, Hölder’s inequality bounds the gradient of the cutoff times the function by its L^6 norm on the escaping annulus. Now u_j\uparrow K*f and u_j\leq u\in L^6, so dominated convergence in L^6, the distributional equation, and harmonic uniqueness give K*f=u. Its energy identity follows by testing with u. The convolution is continuous also in this case. Choose a point x_0 where its value is finite. On any fixed bounded set of evaluation points, the sufficiently distant kernel tails are dominated by a constant times K(x_0-\cdot)f, which is integrable. Dominated convergence handles that tail, and the local L^{5/3}–L^{5/2} argument handles the remaining compact part.

The spherical average of K at distance d from a sphere of radius s is 1/\max(d,s), by direct angular integration. Decomposing a radial probability into spherical averages proves the smearing assertion. The difference of two independent radial displacements is again radial; applying the same assertion to its distribution and integrating against \rho(x)\rho(y) proves the energy inequality.□

Proof. For a unit spin orbital v, the number operator \sum_{i=1}^k|v\rangle\langle v|_i is a projection on the antisymmetric space: in a Slater basis containing v, its eigenvalue is zero or one. Its expectation is at most \|f\|_2^2. This proves the occupation assertion and its versions with auxiliary integration and mixtures.

For the kinetic bound, take an eigen-expansion \gamma=\sum_\alpha\lambda_\alpha |u_\alpha\rangle\langle u_\alpha|, and let P_s be Fourier projection onto |p|^2/2\leq s. The density of its low-frequency parts is at most Cs^{3/2}: Bessel’s inequality bounds each spin evaluation against the corresponding Fourier vector, and there are two spins. The triangle inequality in the weighted orbital-index and spin norm gives, almost everywhere,

Integrating the left side in x and then in s>0 gives the kinetic trace by Plancherel and the layer-cake formula. The integral of the right side in s is a positive constant times \rho_\gamma(x)^{5/3}. Truncation proves the assertion also when the kinetic trace is infinite.□

Proof. Drop the nonnegative pair repulsion, use E\le0, and split attraction at R=Z^{-1/3}. Outside that ball it is at most Zn/R\le3Z^{7/3}. Inside, Young’s inequality gives

where c is the constant in Lemma 3.2. Combine this with that kinetic lower bound and the energy bound q_n[\psi]\le E+\delta\le\delta.□

Proof. Successive squared-gradient splitting gives

The norm-map inequality bounds |\nabla o|^2 by \sum_h|\nabla p_h|^2. Expanding the product partition in the kinetic form, the mixed derivatives cancel because o\nabla o+c\nabla c=0. This proves the stated IMS bound. Multiplication potentials split exactly by the squared partition.

Fix the recorded subset I, and write u for its out positions and spins and \xi for the remaining positions and spins. Put

where the integral includes the core spin sum. Thus \chi_X is the normalized core vector, with its phase retained. The stage decoration depends only on u and does not change this normalized slice. Fubini gives its H^1 regularity almost surely; permutations of core variables preserve the multiplying factors and therefore its antisymmetry. The conditional core–out repulsion is exactly

Grouping the core form, out kinetic form, and the remaining multiplication potentials proves (3.6). All absolute Coulomb terms and kinetic terms are integrable in this calculation by the form bounds and the finite partition. The slice ratios and their integrals give the stated measurable versions; assign arbitrary values on null slices.

For each I, apply the occupation assertion of Lemma 3.2 to the out group, integrating the other variables. Its one-body matrix \gamma_I^o obeys \gamma_I^o\leq\|\psi_I\|_2^2. Thus \gamma_o=\sum_I\gamma_I^o\leq1, even though the out particle number varies with I. Its kinetic trace is T_o, and its density is o^2\rho_\psi, by summing the squared partition with one label required to be out. Equation (3.2) proves the kinetic assertion.

Finally, the entire construction has a single sampling description. Conditional on the raw configuration, independently label each coordinate by its first-out stage or by final core, with probabilities p_h^2 and c^2. An appended cut only splits the previous core probability. Forgetting its new-stage labels and their variables recovers the old observation. This is the required nesting and proves the tower assertion.□

Proof. For each center y, momentum p, and spin \sigma, use the orbital

Integrate its projector over both spins and |p|\leq(3\pi^2\rho(y))^{1/3} with measure \mathrm dy\,\mathrm dp/(2\pi)^3, obtaining an operator \gamma. The unrestricted packet resolution is the identity: apply Plancherel in p and then integrate g_b^2 in y. Restriction therefore gives 0\leq\gamma\leq1. Momentum-ball integration gives

The reality of g eliminates the kinetic cross term. These traces are finite. Every positive-eigenvalue orbital of \gamma belongs to H^1 and is supported in the packet region.

Take a finite eigen-truncation of \gamma. Occupy each retained orbital independently with Bernoulli probability equal to its eigenvalue. Each realization is a normalized Slater determinant, including the possible vacuum. The average one-body energy is the trace energy. Its average repulsion is the direct energy of the truncated density minus

where here \gamma denotes the finite truncation. In particular the average repulsion is at most the direct energy. Equal-orbital direct and exchange contributions cancel, as required by the Bernoulli occupation rule.

Wedge each realization with the normalized core slice. Because their spatial supports are separated, the terms assigning the two groups to different sets of particle labels have disjoint supports. Normalized antisymmetrization consequently preserves the sum of the group energies and adds exactly their direct cross repulsion. Each resulting vector belongs to its own antisymmetric particle-number sector and has energy at least E. Averaging gives the desired inequality with the finite-truncation density and kinetic energy.

For completeness, passage to the full operator needs no selection of infinite Slater states. The truncated densities increase almost everywhere to g_b^2*\rho, and their kinetic energies increase to (3.8). The full density is bounded and compactly supported, so nuclear integrals converge. The core potential is bounded on the packet region by support separation, so its integrals converge as well. Direct energies converge monotonically. Lemma 3.1 gives D(g_b^2*\rho)\leq D(\rho), proving (3.7). The same argument with no inserted orbitals proves E\leq e_X.□

Proof. On a cube of side b, the spin-two Neumann eigenvalues are \pi^2|m|^2/(2b^2), for m\in\mathbb N_0^3, each with two spin states. The number of lattice points of radius at most s, with this multiplicity, is bounded by

attach a disjoint unit cube to each octant lattice point and enclose their union in the octant ball of radius s+\sqrt3. Thus the jth ordered radius is at least ((3j/\pi)^{1/3}-\sqrt3)_+. Summing their squared radii and using \sum_{j\leq n}j^{2/3}\geq(3/5)n^{5/3} gives the lower bound

for the sum of the first n eigenvalues. The leading coefficient is exactly \cT.

Partition the out coordinates into assignments to the cubes of a grid. Restriction to a cube is admissible for its Neumann form; no zero extension across its boundary is involved. Within each cube the corresponding variables remain antisymmetric. The occupation bound, now in the Neumann basis, gives (3.12) after slicing all auxiliary variables. If n_C denotes the out count in a cube, the leading term is

The error is at most Cb^{-2}(n_o^{4/3}+n_o), since \sum_Cn_C^{4/3}\leq n_o^{4/3}.

Average the grid over a uniform translation in a fundamental cube and over rotations. At a fixed out configuration, its averaged cell density is \sum_{i\ \mathrm{out}}\vartheta_b(x-x_i). Before rotation, the probability that two points are in the same translated cube gives precisely the triangular product in (3.9). Jensen’s inequality bounds the power integral from below by the power of this averaged density. Dropping any chosen particles can only decrease that density and its nonnegative power. This proves (3.10).

For the pair bound, first discard all pairs not both retained. Smearing each retained point by \vartheta_b decreases every distinct pair interaction by Lemma 3.1, because the difference of two independent radial displacements is radial. Each self energy is bounded by C/b, using the kernel’s support and L^\infty bound. Subtracting these self energies from D(\sigma) proves (3.11).□

Proof. The set of admissible s in (4.1) is downward closed. It contains Z^{-1/3} and is bounded above, since every sufficiently large ball contains all nuclei. Increasing open balls show that the condition also holds at the supremum. The radius 2a_y is not admissible, proving the third inequality in (4.2).

If s=a_x-|x-y|/L>0, then B(y,Ls)\subset B(x,La_x), and hence \nu(B(y,Ls))\le a_x^{-3}\le s^{-3}. Thus a_y\ge s. Interchanging x and y proves the Lipschitz bound. If d_y>La_y, a slightly larger ball in the defining family would still contain no nucleus, a contradiction. If d_y/L>1, every radius below d_y/L is admissible, whereas every strictly larger radius contains a charge at least one and has s^{-3}<1. This proves the asserted equality. Finally, the Lipschitz bound makes all scales in B(y,ca_y) comparable to a_y; a mesh of spacing a sufficiently small multiple of a_y gives the stated cover.□

Proof. We first give the upper trial at a target y. Append a cut that is full out on B(y,Aa_y), supported in B(y,2Aa_y), and has gradients bounded by C/a_y. Write X' for the enlarged data, a=a_y, and h=\E[J_y\mid X']. The sliced core vanishes on B(y,Aa). Take uniform packet centers of total mass M_{\rm tr}\ge0 in B(y,a), with packet width a. The packet support lies in B(y,2a) and is separated from the core. The radial mean of the core field and of each far nuclear field is its value at y. Near nuclei contribute positively, and J_y subtracts far out electrons as well as the core. Therefore Lemma 3.5 gives

Here M_{\rm tr} is the expected occupation of the packet trial; it need not be an integer. This use of an arbitrary trial mass is justified by E=\inf_k E_k.

For ah>C_1m_{0,y} choose M_{\rm tr}=\lambda ah. The three positive terms in (4.7), divided by M_{\rm tr}h, are at most

The inequalities am_{0,y}^{1/3}\ge1 and am_{0,y}\ge1 allow us to choose \lambda>0 small and then C_1 large so that their sum is less than a fixed number below one. In the other case use the vacuum trial. For all data this proves

The insertion estimate makes a large positive field expensive. We bound the energy of the removed electrons in terms of positive fields at nearby centers. If the present field exceeded a universal multiple of its proposed bound, this comparison would force a larger normalized field after one further deterministic cut. Repetition gives doubly exponential growth, whereas the elementary nuclear bound and the Lipschitz scale permit only exponential growth. We now quantify this argument, retaining the original e_* throughout.

Start with target v_0=y and the given data X_0=X. At stage j, append the cut just constructed at the current deterministic target v_j; denote the enlarged data by X_{j+1}. The upper estimate (4.8) applies with y=v_j and X'=X_{j+1}. Cover the supports of all cuts, including this new cut, by at most C(j+1) local cells, and let \mathcal U_j be the union of their doubled balls 2B_z. These are the possible next targets: we estimate the removed energy using their fields under the enlarged data, writing X'=X_{j+1} in this calculation. Each newly added scale is comparable to a preceding covering scale by the common local scale properties above; the original target and initial covering scales need not be comparable to one another. Let a_{\min}>0 be the minimum scale of the original target and initial covering cells. The Lipschitz property and the bounded local meshes give, for every later target or covering cell at iteration j,

Indeed each new scale is comparable to a preceding scale; the quantities m(a)=\max(a^{-3},1)+\sqrt{\delta a} and e(a)=m(a)^2/a change by a bounded factor when a does, uniformly in \delta. The total out count obeys

The localization error is at most C^{j+1}Pe_*: for an initial cell this follows from

and for a later cell it follows from a_zm_z\ge1 and \E n_z\le\sqrt Pm_z.

For a covering cell B_z, let V_z be the potential of nuclei in B(z,4a_z). Let h_z(v) be the conditional raw signed field, given the present X', of all sources at distance at least 4a_z from z. It has a jointly measurable harmonic version on 3B_z: all sources are separated from compact subballs, so their kernel derivatives are bounded for the fixed system and may be integrated against the conditional law. On B_z the field of the conditional core satisfies

For v\in2B_z, the exclusion in J_v contains B(z,4a_z), since A=40 and nearby scales are comparable. Extra electrons in h_z can only lower the field. Extra nuclei have total charge at most a_z^{-3} and distance at least 2a_z from v. Hence

Set

Positive parts of harmonic functions are subharmonic. Their ball submean inequality, a fixed covering of B_z, and Minkowski’s inequality give

where the supremum is over v\in\mathcal U_j. The term a_z^{-4} was absorbed using a_z^{-7}\le e_z\le C^{j+1}e_*.

The out addback in (4.11) is at most

Indeed the far exclusion gives distance at least 3a_z; if a_l\gg a_z, Lipschitzness of the scale forces every point of B_l to be a distance at least a fixed multiple of a_l from B_z. Using \|n_l\|_2\le\sqrt{Pe_la_l} and \sqrt{a_l}/\max(a_l,a_z)\le a_z^{-1/2} bounds this addback in L^2 by C^{j+1}\sqrt{Pe_*/a_z}.

For completeness, the nuclear singularities have the required energy scale. If Q_z=\nu(B(z,4a_z))\le a_z^{-3}, weighted convexity gives

In the sum over cells, the power of the number of cells arising from convexity is bounded by C^{j+1}. Lemma 3.2 and Young’s inequality therefore absorb this attraction at cost C^{j+1}e_*. Multiplying the other bounds by \|n_z\|_2\le\sqrt{Pe_za_z} and summing cells, the conditional localization identity, with out-pair repulsion dropped, yields

This calculation used the actual list of cuts; it did not assume the conclusion of the lemma for that list.

Combine (4.8) and (4.15), use \delta\le e_*, and apply conditional Jensen from the enlarged data back to the preceding data. With G(v_j,X_j) defined by the same normalization,

Write g_j=G(v_j,X_j)/\sqrt P. Whenever g_j^2\ge2C^{j+1}, the last inequality permits a deterministic next target v_{j+1}\in\mathcal U_j with

The series

converges. Choose M_0>0 so that \exp(2^{j+1}M_0)\ge2C^{j+1} for every j. If \log g_0>S+M_0, induction gives \log g_j\ge2^jM_0; thus the condition for selecting each next target always holds. All supremums here are of numerical norms for deterministic cuts; choosing a target close to one of them never chooses a target from sampled data. On the other hand, J_v\le Z/(Aa_v) pointwise, and (4.9) gives

For the fixed system and initial cuts the last factor is finite. Its logarithm grows at most linearly in j, contradicting the positive multiple of 2^j forced above. This argument works in both geometries and requires no global lower bound on the scale. It proves (4.6).□

Proof. We divide the proof into the choice of cut, the two energy comparisons, and the deterministic interior bound.

Choice of cut and the conditional field. For a candidate t\in[5a,6a], let the cut transition in a shell of width b, using sine and cosine of an angle. Let n_{\rm del} count the out particles not retained. Such a particle lies in t-7b\le|x-y|\le t+b. For each raw pair of particles the set of t for which both lie in this shell has length at most 8b. Averaging over t therefore bounds its expected squared count by C\beta Pm^2, since B(y,7a) has a bounded comparable-scale cover. The same averaging, applied to the nuclear integral, uses (4.14) to give a bound C\beta a^{-7}. Averaging the sum of the two normalized nonnegative quantities selects one t_0 for which both bounds hold:

The full out-count and IMS bounds in the same display follow from the comparable-scale cover and \E n_z\le\sqrt Pm_z.

On the cut support and on \Omega we have

To see this, the raw sources outside B(y,20a) give a harmonic conditional field on B(y,10a); these electrons are all core particles. Its value at v\in B(y,10a) is at most \E[J_v\mid X]+Ca^{-4}, by the same exclusion comparison as (4.12). Take F to be the positive supremum on B(y,7a). Harmonic means and Lemma 4.2 prove (4.27). The cells of this initial cut have scale comparable to a and D_z\le C\beta^{-1}, so (4.17) verifies that lemma’s hypothesis.

The unconstrained minimizer. The core is at distance at least 4b from \Omega, so its potential is bounded and smooth there. Hence \Phi_X\in L^{5/2}(\Omega). Young’s inequality makes (4.18) coercive in L^{5/3}. The positive Coulomb form is weakly lower semicontinuous: by Lemma 3.1 it is a squared Hilbert norm, or one may express it as the supremum of its affine dual forms. The direct method on the closed positive cone gives a minimizer, and strict convexity of the kinetic term gives uniqueness. Nonnegative variations, followed by signed compact multiplicative variations, give

which is (4.19).

This minimizer depends continuously in L^{5/3} on its field in L^{5/2}. Indeed convergent fields give uniformly bounded minimizers; weak lower semicontinuity and comparison with the limit minimizer identify each weak limit and give convergence of the minimum values. Lower semicontinuity of D then forces convergence of the kinetic norms. Uniform convexity of L^{5/3} gives strong convergence. The conditional field, being jointly measurable and taking values in the separable space L^{5/2}(\Omega), is measurable as a function with values in that space. The continuity just proved establishes the stated measurability of \rho.

The diameter of \Omega is at most 12a, so D(\rho)\ge(\int\rho)^2/(24a). The minimum is nonpositive because zero is admissible. Absorbing V_{\rm loc} with Young’s inequality in (4.27) gives

It follows that

When V_{\rm loc}=0, the same nonpositive-minimum argument uses only \Phi_X\le F and gives the sharper estimates

In particular all trial errors below are integrable in X.

Upper quantum comparison. The centers \rho lie in \Omega=B(y,t_0-4b); packets of width b remain at distance at least 3b from the core. Thus Lemma 3.5 applies, and radial smearing leaves the core potential unchanged. The nuclear loss is bounded at centers by

Weighted convexity, the total local charge bound, and integration of |x|^{-5/2} over a ball of radius 2b give

Nuclei not in the indicated local ball are harmonic on all packets. Young’s inequality now gives, pathwise,

Lower quantum comparison. Reserve a fraction \epsilon T_o of the out kinetic energy. The coarse kinetic inequality absorbs both V_b^{\rm loss} and the nuclear attraction of deleted particles, at cost C\epsilon^{-3/2}\beta^{1/2}a^{-7}, using (4.26). The deleted F contribution costs at most

The retained fine density \sigma is supported in \Omega, because its centers have radius less than t_0-7b and its kernel radius is \sqrt3b. The core potential is harmonic on these kernels, and the loss of nuclear attraction is again covered by V_b^{\rm loss} at the actual particle positions. Applying the remaining fraction of Lemma 3.6 and its pair-repulsion bound in the localization identity gives

Here \E n_o^{4/3}\le(\E n_o^2)^{2/3}; no higher out-count moment is required.

The Euler equation gives the exact expansion

Besides being nonnegative, the convex remainder satisfies

by Young’s inequality applied to the tangent term. For small \epsilon this absorbs \epsilon\cT\int\sigma^{5/3} into half of \mathcal L, with error C\epsilon\int\rho^{5/3}. Average (4.30), compare it with (4.31), and use (4.28), (4.27), and the IMS bound. The terms \epsilon\E\int\rho^{5/3} cost C\epsilon Pe; the packet kinetic term costs Cb^{-2}(\sqrt Pm+a^{-3})\le Cb^{-2}\sqrt Pm. This proves (4.21), after multiplying its constant to restore the full nonnegative remainder, and also the asserted bound for \E\int\sigma^{5/3}.

The source-free improvement. If \nu(B(y,20a))=0, every nucleus and the conditional core are outside the packet supports. The field \Phi_X is harmonic on those supports, so radial averaging is exact and the upper comparison is

For the lower comparison, use all the out kinetic energy. Harmonicity and the deleted-count bound give

The full sharp kinetic and pair bounds of Lemma 3.6 therefore give

Subtract the average upper inequality and use the exact expansion (4.32), the mass estimate (4.29), and the IMS bound. This proves (4.24), including its full convex remainder, without the nuclear loss or an \epsilon error.

There is a useful separate kinetic bound. In the localization identity use only e_X\ge E, drop pair repulsion, and use (4.27). Half the coarse kinetic inequality absorbs V_{\rm loc} at cost Ca^{-7}; also \E(Fn_o)\le CPe. The result is (4.22).

Interior field bound. The core vanishes in B(y,4a) and this ball lies in \Omega. There

Put s=4a and

The Laplacian of the bump is at most 80Cs^6/(s^2-|x-y|^2)^6. For a sufficiently large universal C this is at most \cF times its 3/2 power. Thus U is a supersolution for the displayed equation. The difference W-V_{\rm loc} is bounded above: the far nuclear field is bounded on this ball, the core potential is nonnegative, and K*\rho\ge0. Consequently (W-U)_+ has compact support inside the ball. Nuclear singularities cancel in W-U, which is locally H^1; testing the difference inequality by its positive part yields

The test is valid by compact H^1 approximation: the densities are locally in L^{5/3}, and the nuclear point terms have already canceled. Hence W\le U. On B(y,3a) the bump is Ca^{-4}, proving the first part of (4.23). If V_{\rm loc}=0, this yields W\le Ca^{-4}; the Euler equation then gives \rho\le Ca^{-6} and proves (4.25). In this source-free case,

Finally, weighted powers and \nu(B(y,20a))\le a^{-3} give

□

Proof. Fix small \epsilon>0 and 0<\beta<1/100, to be chosen in that order. We need only consider P\ge\beta^{-4}, since \beta will be fixed universally. Then \beta^{-2}/(am)\le\sqrt P at every target, and the patch of Lemma 4.3 is available. Every raw particle in B_y is retained, and its fine smear is contained in B(y,1.1a) because \beta<1/100. Choose a nonnegative smooth \chi, equal to one there, supported in B(y,2a), with \|\nabla\chi\|_2\le C\sqrt a. Lemma 3.1 and (4.23) give

Divide by Pm^2 and use (4.21). Since \delta a\le m^2, the result is at most

In the atomic distance geometry, using (4.24) instead gives the sharper intermediate inequality

For the common bound it suffices to use (4.37) in both geometries. First choose \epsilon small, then \beta small, and finally a universal threshold P_0 large. The inequalities (4.4) show that the expression is at most 1/2 whenever P>P_0. Enlarge P_0 so that P_0\ge\beta^{-4}; this verifies (4.17) at every target. If P>P_0, we would obtain \E n_y^2/m_y^2\le P/2 for every y, contradicting the definition of P. Thus P\le P_0. Since m_y^2\le2\max(a_y^{-6},1)+2\delta a_y, this proves (4.36).□

Proof. The count assertion is Lemma 4.4. Comparable scales have comparable m and e=m^2/a, uniformly in \delta, so e_*\le C(C_0)e_y in Lemma 4.2. Substitute P\le C into that lemma to obtain the field assertion.□

Proof. Take \beta=a^{0.2} and \epsilon=a^{0.02}. The energy assumption gives m\le Ca^{-3}, while \beta^{-2}/(am)\le a^{1.6}, so the patch hypothesis holds for small a. By Lemma 4.4, P\le C. The terms on the right of (4.21) are bounded respectively by constant multiples of

All are at most Ca^{-7+0.02} for a<1, proving (4.40). Equation (4.22) gives T_o\le Ca^{-7}. Since the cut is full out on B(y,4a), the out density equals \rho_\psi there. Applying Lemma 3.2 proves (4.41).□

Proof. The expectation is \int_{B(y,u)}|x-y|^{-1}\rho_\psi(x)\,dx. Hölder’s inequality, the raw density bound, and direct radial integration give

This is the stated bound in either geometry.□

Proof. By (4.35) and (4.25), 0\le\Delta W\le Ca^{-6} on B(y,2a). Write

For x\in B(y,2a), direct integration of |x-z|^{-1} and |x-z|^{-2} gives

Since \Delta Q=-\Delta W inside that ball, H=W+Q is harmonic there. By (4.25), H\le C_1a^{-4}. Choose C_2>C_1 universally and put h=C_2a^{-4}-H, a positive harmonic function on B(y,2a). The Poisson formula on a sphere of radius 3a/2 gives

Indeed the differentiated Poisson kernel on B(y,a) is bounded by C/a times its value at the center; integrating the nonnegative boundary values gives precisely this inequality. Also h(y)\le Ca^{-4}+W(y)_-, since Q(y)\ge0. Integrate the gradient along segments in B(y,s'), and add the bound for \nabla Q. This proves (4.44).□

Proof. The compact annulus \{\lambda\le|x|\le\Lambda\} has a finite cover by cells B_z whose radii are comparable to one, with constants allowed to depend on \lambda,\Lambda,L. Dilate this cover by u. Because a_{uz}=ua_z, the dilated balls are again cells, now with radii comparable to u. The annular count is bounded by their sum of counts. Minkowski’s inequality and (4.39) bound its L^2 norm by C\sum_z m_z, which is the right side of (4.45). Boundary spheres carry zero raw probability, since the one-electron density is absolutely continuous. Thus the same estimate holds with any choice of open or closed endpoints.□

Proof. On \mathcal C the exact likelihood of an array is

Set it equal to zero outside the chamber. Integration of the permutation sum over \mathcal C equals integration of the labeled product over all arrays, and is therefore one. Define

with the empty product and p_{\ge m} both equal to one. At a tuple with p_{\ge j}>0, use the probability density |\Psi|^2\Lambda_{\ge j}/p_{\ge j} as the conditional raw law. Equivalently, the conditional mean of a nonnegative Borel raw observable G is the ratio

On zero-density tuples use the unconditional raw law. All these versions are jointly Borel, including their dependence on an additional spatial variable. At every parent tuple with p_{\ge j+1}>0, the conditional density of the next array is p_{\ge j}/p_{\ge j+1}. Tonelli’s theorem and \int L_j\dd Y_j=1 show that it integrates to one. Substitution of the ratios gives the tower property, also after nonnegative spatial integration; positive and negative parts give the integrable signed case.

For each multi-index \alpha, |\partial_y^\alpha\mathcal K_z(y)| \le C_\alpha t_{\min}^{-3-|\alpha|} uniformly in z,y. Differentiation under the conditional integral therefore gives all claimed derivatives and bounds. Tonelli gives the mass n. For the potential and its gradient, split the kernels |u|^{-1} and |u|^{-2} at unit distance. The near parts are integrable and are controlled by \|\mu_j\|_\infty; the far parts are controlled by n. Splitting again at a small radius, then using dominated convergence on its complement, proves continuity of the potential and gradient. Finally -\Delta K=4\pi\delta_0 gives (5.1). In the atomic case this argument does not remove the nuclear singularity of V_s.□

Proof. In labeled observation coordinates place the sorting map inside the indicator of A. The translated product density is continuously differentiable in L^1, since \zeta is smooth and compactly supported. Hence p_A is continuously differentiable even for a Borel event. The scalar translation score is

Its law is symmetric and centered. It satisfies \|S(U)\|_q\le Cq^2 for q\ge2: indeed, T=(1-U^2)^{-1} has a finite exponential moment of every order less than one, so integration of its exponential tail gives \|T\|_q\le Cq, while |S(U)|\le2T^2.

For a unit vector v\in\R^{3n} the directional derivative of the likelihood pairs the event indicator with

Multiplication by independent random signs preserves the joint law of the independent symmetric scores. For deterministic coefficients,

gives the Gaussian tail estimate and consequently \|\sum_\alpha\epsilon_\alpha c_\alpha\|_q \le C\sqrt q(\sum_\alpha c_\alpha^2)^{1/2}. Conditioning on the scores and then using the triangle inequality in L^{q/2} yields

The identity \sum_{i,b}v_{i,b}^2=1 and the geometric square-sum bound are precisely what avoid factors depending on dimension or on the number of observations.

Hölder’s inequality at fixed x gives |D_vp_A(x)|\le Cq^{5/2}\ell_j^{-1}p_A(x)^{1-1/q}. For p_A(x)>0 choose q=\max\{2,\log(e/p_A(x))\} and then take the supremum over unit vectors. At a zero the gradient of the nonnegative differentiable function p_A vanishes. Therefore

with right side zero at p_A=0.

Regularize the square root by \sqrt{p_A+\eta}. Since \sqrt u\log^{5/2}(e/u) is bounded on [0,1], these functions have a common Lipschitz bound. Their uniform limit \sqrt{p_A} is Lipschitz. Its gradient vanishes almost everywhere on its zero set, and the ordinary chain rule applies elsewhere. Thus

The permutation sum in (5.3) shows that p_A is invariant under raw particle permutations and is spin independent. The bounded Lipschitz multiplier therefore preserves antisymmetry and the form domain. Its squared norm and its raw law are as claimed by the definition of p_A.

Lemma 2.4, applied with eigenvalue e_\Psi, now gives

Under this conditional law,

Integration of this tail, split at \log(e/p), bounds the fifth moment by C\log^5(e/p) and proves (5.4). The proof for a single array is identical with the square-sum constant equal to one.□

Proof. Write

The first identity in (5.6) follows by multiplying the original P-conditional expectation defining \mu_j by \1_A, using A\in\mathcal F_j, and dividing by P(A). The fresh labels do not alter this expectation. For the second identity, condition the raw-plus-label marginal of Q on X. That marginal is exactly the localization law of \Psi_A; the original observations can be integrated out. Tonelli justifies both identities initially for nonnegative potentials. In particular the first tower is under the original observation law, while the second uses the fresh tilted patch law.

Subtracting the definitions of H_j and h_X, and then using these towers, gives the exact algebraic identity

The estimates below also establish finiteness of all terms, so this subtraction involves no difference of infinite expectations.

For 0<u\le4a, the local density bound and Hölder’s inequality give

The complementary raw potential is bounded by n/(4a). By radial smearing, P^{\rm raw}-P^{\rm ms}\ge0 and only centers with |x_i-y|\le t_{x_i} can contribute. Lipschitz continuity of the width implies |x_i-y|\le t_y+|x_i-y|/2, hence these centers lie in B(y,2t). Equation (5.9) bounds its mean by Ca^{-4}\tau^{1/5}. The same Newton comparison for each fine packet gives

The distance and count assumptions for the deleted particles give \E_Q P^{\rm del}(y)\le Ca^{-4}\sqrt\beta.

For the signed density difference take v_t(z)=\min\{|y-z|^{-1},t^{-1}\}. This is in \dot H^1 and

Lemma 3.1 and Cauchy–Schwarz in probability give

It remains to control the two nonnegative remainders of the truncation inside B(y,t). In the source-free case,

For the nuclear alternative, weighted convexity gives, with Q_{\rm loc}=\sum_\alpha q_\alpha,

The assertion is read as zero if Q_{\rm loc}=0. Uniformly in R\in\R^3,

For completeness, apply Hölder to the second factor with any exponent q\in(3/2,2) and to the first with conjugate exponent q'\in(2,3). Integrals of these radial decreasing powers over a ball are no larger than their centered-ball integrals; the powers of t combine to t^{1/2}. Since \rho_X\le C(a^{-6}+V_{\rm loc}^{3/2}), its remainder is at most

This proves the asserted nuclear bound even when a nucleus lies at y; only V_s(y), not the unsmoothed V(y), was required finite.

For \sigma, only packets centered within t+\sqrt3b of y meet the truncation ball. Extend each such packet’s kernel integral to all space and use radial smearing to bound it by its raw point potential. Equation (5.9) bounds the mean remainder by Ca^{-4}(\tau+\sqrt3\beta)^{1/5}. The truncated \sigma potential is integrable, being at most n/t. Together with Coulomb duality and the just obtained remainder bound, this also proves integrability of the full \rho_X potential. The raw and core potentials were already integrable by (5.9). Hence all the preceding subtractions are valid. Adding the four estimates in (5.8) proves (5.7).□

Proof. It suffices to consider two functions. For a nonnegative compactly supported smooth test \varphi, choose an increasing Lipschitz function \theta_\eta that is zero on (-\infty,0] and one on [\eta,\infty). Test the first inequality with \theta_\eta(v_1-v_2)\varphi and the second with (1-\theta_\eta(v_1-v_2))\varphi. These nonnegative compactly supported H^1 tests are legitimate by mollification, since the right sides are bounded on their support. The sum of the left sides is

The second term is nonpositive. Deleting it and letting \eta\downarrow0 gives the gradient of \max(v_1,v_2) by the Sobolev positive-part formula. Dominated convergence gives the reaction at the maximum; the reactions agree at ties. Induction proves the finite version.

For the last assertion subtract V in each inequality. The singular measures cancel, and the reaction for the offsets becomes F(y,V(y)+a_i(y)). It is locally bounded: it vanishes near every nuclear point, and V is smooth off \operatorname{supp}\nu. Apply the first assertion to the offsets and then add V back.□

Proof. Translate O to the origin. Put u_i=|x_i|, s_k=2^kt and

For s_k/4\le |v|\le8s_k, the nuclear assumptions give c s_k\le d(v)\le C s_k, where d(v) is the distance to the nearest nucleus. Indeed, the inner nucleus supplies the upper bound; the inner and outer alternatives supply the lower bound. Choose t_0 so large that d(v)/L>1 on all these annuli. Lemma 4.1 then gives a_v=d(v)/L\asymp s_k/L.

Each enlarged annulus is covered by a universally bounded number of local balls B(v,a_v) with comparable scales. For example, take a maximal set of centers in the annulus separated by c s_k/L, with c small; the disjoint balls of half that radius give the bound on their number. Lemma 4.4, applied to any \delta-state \omega in the fixed sector N, consequently yields

This proves the result when J is bounded. We may therefore assume that J exceeds a fixed sufficiently large integer.

A weighted eigenstate inequality. Choose 0\le\chi\le1 supported in (32t,T/32), equal to one on [64t,T/64], and satisfying |\chi'(u)|\le C/u, |\chi''(u)|\le C/u^2. Such a function is obtained by multiplying a fixed cutoff at scale t by a fixed cutoff at scale T. Set w(u)=u\chi(u)^2. It is a bounded smooth multiplier that vanishes near the origin. For each weighted index, integration by parts gives

These identities hold first after form-domain approximation and then by continuity, since w and its derivatives are bounded for fixed t,T. Summing the last bound costs at most C\sum_k s_k^{-1}\E n_k\le C/t\le C by (6.2) with \delta=0. After fixing x_i and its spin, the remaining Hamiltonian has energy at least E_{N-1}\ge E. Its contribution against the nonnegative weight w(u_i) is therefore nonnegative. Testing the eigen-equation with \sum_iw(u_i)\Psi shows that weighted repulsion minus attraction is at most C.

For an electron in a shell meeting the support of \chi, keep only its repulsion against electrons at radius <t, at radius \ge T, and in shells l\le k-5. Dropping the other repulsions preserves the upper bound. Write

and define the exterior signed potential at the origin by

For each such k, use a cut that is full out on [s_k/2,4s_k], supported in [s_k/4,8s_k], and has gradients bounded by C/s_k. Let X_k be its out data as in Lemma 3.4; a bar will mean conditional expectation given X_k in this paragraph and below. Set

Every supported shell satisfies 32t\le s_k\le T/64. All sources defining \psi_k are outside \{s_k/4<|v|<8s_k\}: prefix electrons have radius <s_k/16, inner electrons have radius <t\le s_k/32, and exterior electrons have radius \ge T\ge64s_k. Thus \psi_k has a continuous harmonic version on this annulus. Newton’s sphere-average formula gives

where \langle F\rangle_u=(4\pi u^2)^{-1}\int_{|v|=u}F(v)\,dS(v). The shell-k positions are all recorded in X_k. We may therefore condition the selected interaction terms while retaining their weights. The preceding eigenstate inequality becomes

where k ranges over the supported shells. Conditional harmonic versions and sphere averages are justified directly by the positive distance to these sources and their fixed finite total masses.

A size-biased law for one shell. Fix a supported k. Choose a smooth function b_k, equal to one on [s_k,2s_k], supported in (s_k/2,4s_k), with 0\le b_k\le1 and |b_k'|\le C/s_k. Put

Then n_k\le S_k\le n_{\mathrm{neigh},k}. If m_k=0, every shell-k contribution vanishes and can be omitted. Otherwise define

and denote its expectation and L^2 norm by \E_k and \|\cdot\|_{2,k}. The square root of a sum of squares is Lipschitz, and, off its zero set,

Its gradient is zero almost everywhere on its zero set. The multiplier identity in Lemma 2.4 shows that \Psi_k is a \delta_k-state, with

The random variable S_k is X_k-measurable, because its support is inside the full-cut zone. The joint position and cut-label law changes by the factor S_k/m_k. Its conditional position law given X_k is therefore unchanged on S_k>0. In particular all the barred fields and counts above have the same conditional versions under this new law. This permits us to use the local estimates for \Psi_k without changing the fields in (6.4).

A positive bound and a signed harmonic comparison. On the broad annulus of harmonicity, the base and prefix sources lie outside the exclusion defining J_v. Its radius Aa_v is a fixed small fraction of |v|, since A/L is small. The additional electrons subtracted by J_v give the upper bound

For comparable shells the exclusion supplies the denominator c s_k; for sufficiently larger shells radial separation supplies c s_l. These are precisely the electrons absent from the definition of \psi_k.

The cut defining X_k is covered by a universally bounded number of cells at scales comparable to s_k/L. Its gradients are C/s_k and a_v\ge1 there, so it meets the initial-cut hypothesis of Lemma 4.2. For targets in the same broad annulus, that lemma and (6.2), now applied under \Psi_k, give

Conditional Jensen’s inequality bounds \|\bar n_l\|_{2,k} by \|n_l\|_{2,k}. Thus the added count terms are bounded by the convergent series

Positive parts of harmonic functions are subharmonic. Covering the compact annulus \{s_k/2\le|v|\le3s_k\} by a fixed number of interior balls in the broad annulus, and using the ball submean inequality and Minkowski’s inequality, consequently gives

The function A_k/s_k-\psi_k is nonnegative and harmonic in the interior of this latter annulus. Its value at any point of a sphere s_k\le u\le2s_k is at least c times its sphere mean, for one universal c>0. Indeed, after scaling, a bounded chain of interior balls joins any two points of these spheres; on a ball, this comparison follows by comparing its Poisson kernels on a smaller concentric ball. Applying this comparison and (6.3) gives

The negative prefix term retains a fixed positive coefficient.

The endpoint fields. Put s_-=2t and s_+=T/2. For a supported shell and u\in[s_k,2s_k],

At either endpoint all base sources are outside the J_v exclusion. Sphere averaging \bar\varphi and adding back the annular electrons therefore yields

Indeed, the omitted contribution of an annular electron is nonnegative and at most its full Coulomb kernel. The latter has sphere average 1/\max(s,u_i)\le1/\max(s,s_l) for an electron in shell l. For these targets a_v\asymp s/L. The initial cut is still at scale s_k/L, so the energy parameter in Lemma 4.2 satisfies e_*\le C(\min(s,s_k)^{-1}+\delta_k). The precise consequence is

At s=s_- this is at most C(1+\sqrt{\delta_k s_k}). At s=s_+ the interpolation weight gives instead

The inner endpoint count term in (6.9) has norm at most

For the outer endpoint, s_+/\max(s_l,s_+)\le1; we retain its contribution as C(s_k/T)\bar L_{\mathrm{ann}}. Thus (6.8) supplies a nonnegative, X_k-measurable random variable D_k, the same for every electron in shell k, such that

Summation over shells. Changing to the size-biased law and using (6.5), s_k\ge1, and m_k\le\E n_{\mathrm{neigh},k}, gives

The neighbor annuli for supported k lie inside \{t\le|x|<T\} and have bounded overlap, so these costs sum to C\E L_{\mathrm{ann}}. For the outer term, n_k is X_k-measurable and the tower property gives

Here the unconditioned norms use (6.2) with \delta=0, and \sum_k s_k/T\le1. Insert (6.11) into (6.4). The weights W_k=\sum_{i\in k}\chi(u_i)^2 are X_k-measurable and satisfy 0\le W_k\le n_k. Removing the bar on P_k by the tower property, we obtain

Let I_{\mathrm{mid}} consist of the shells entirely contained in [64t,T/64], and set L_{\mathrm{mid}}=\sum_{k\in I_{\mathrm{mid}}}n_k. On these shells W_k=n_k. The sum on the left of (6.13) therefore includes all products n_kn_l with k,l\in I_{\mathrm{mid}} and l\le k-5. For every omitted close pair |k-l|<5, including k=l, use the tilt at k and (6.2) to obtain

The last step uses s_l\asymp s_k and the same calculation as (6.12). Each k has only a bounded number of such neighbors. Thus the close-pair costs sum to C\E L_{\mathrm{ann}}, rather than a cost proportional to J. Expanding L_{\mathrm{mid}}^2 now gives

Only a fixed number of endpoint shells were omitted, so \|L_{\mathrm{ann}}-L_{\mathrm{mid}}\|_2\le C by (6.2). Consequently X=\|L_{\mathrm{ann}}\|_2 satisfies X^2\le C(1+X), proving the lemma.□

Proof. Write \mathcal R=\{R_1,\ldots,R_M\} and d(x)=\operatorname{dist}(x,\mathcal R). We use d(x)\le L a_x from Lemma 4.1. For a Borel set B, write n(B)=\#\{i:x_i\in B\}. The local count bound (4.36), applied with \delta=0, gives

We first prove that

for a sufficiently large universal constant U. A direct summation of local bounds over all large distance scales would give an infinite series. We instead group scales into a bounded number of annuli to which Lemma 6.1 applies.

Nested sets of nuclear representatives. Fix

Thus

Choose an integer j_0 such that 2^{j_0}/L\ge1 and 2^{j_0-1} exceeds the lower threshold for the inner radius in Lemma 6.1. Set U=2^{j_0}. Since the nuclei are distinct, there is an integer l_0<j_0-h such that \mathcal R is 2^{l_0}-separated. Put P_{l_0}=\mathcal R. For each l>l_0, choose a maximal 2^l-separated subset P_l\subset P_{l-1}, where separated means that distinct points have distance at least 2^l.

For each p\in P_{l-1} choose a parent \pi_l(p)\in P_l, keeping \pi_l(p)=p for p\in P_l. Maximality permits this choice with |p-\pi_l(p)|<2^l whenever p is removed. Composing these maps gives an ancestor \alpha_l(R)\in P_l for every original nucleus R, and

Indeed, the accumulated displacement is less than \sum_{k=l_0+1}^l2^k<2^{l+1}; at l=l_0 it is zero. Every representative is an original nucleus, and a removed nucleus never reappears in the nested sets.

For j\ge j_0, put u_j=2^j. Choose a nearest nucleus R(x) measurably, breaking ties by the least index. Assign a point of the dyad u_j\le d(x)<2u_j to c=\alpha_{j-h}(R(x))\in P_{j-h}, and denote its assigned set by A_{j,c}. These are Borel sets, disjoint at each scale, and

Only finitely many crowded pairs. Call (j,c) crowded if there is a distinct w\in P_{j-h} with |w-c|\le D_0u_j; otherwise call it isolated. We claim that the number of crowded pairs is at most CM. For each crowded pair choose one witness w. The points c and w cannot both survive to P_{j+q}, since that set is 2^{j+q}-separated and 2^{j+q}>D_0u_j. Let r be the first level at which either endpoint is removed, and charge (j,c) to one such removal event (r,p), where p\in P_{r-1}\setminus P_r. Then

A fixed removal event can therefore receive charges from at most q+h levels j. At any one of these levels, a center c charging (r,p) is either p itself or has p as its witness. In either case, c\in P_{j-h}\cap\overline B(p,D_0u_j). Such centers are 2^{j-h}-separated. The disjoint balls of radius 2^{j-h-1} around them lie in the ball of radius D_0u_j+2^{j-h-1} around p, so their number is at most (1+2D_0 2^h)^3. This is a fixed constant. There are at most M-1 removal events, since the nested sets eventually contain one nucleus. Consequently

Simultaneous removals cause no difficulty: the charge uses just one of the two removed endpoints.

Each crowded assigned set has bounded expected particle count. To see this, choose a maximal u_j/(2L)-separated subset of A_{j,c}. By (6.18), a volume comparison bounds its cardinality by CL^3. Its balls of radius u_j/(2L) cover A_{j,c}. At every center y we have a_y\ge d(y)/L\ge u_j/L\ge1, so these covering balls are contained in B(y,a_y). Equation (6.15) therefore gives

Together, (6.20) and (6.21) show that all crowded assigned sets cost at most CM expected particles.

Isolated runs give empty annuli. For a fixed nucleus c, the levels j\ge j_0 at which c\in P_{j-h} form an initial integer interval, possibly empty or infinite. Removing its crowded levels leaves consecutive runs of isolated levels. The number of these runs is at most one plus the number of crowded levels for c. Summing over c and using (6.20), the total number of isolated runs is at most CM.

At an isolated level j for c, every original nucleus R satisfies one of the two alternatives

Indeed, if \alpha_{j-h}(R)=c, the first alternative follows from (6.17). Otherwise isolation gives |\alpha_{j-h}(R)-c|>D_0u_j, and the ancestor offset gives the second alternative.

The alternative for a fixed nucleus cannot change between consecutive isolated levels j and j+1 for the same center c. A nucleus near c at level j remains too close to satisfy the far alternative at j+1. Conversely, a nucleus far at j could be near at j+1 only if its fixed distance from c were both greater than (D_0-\varepsilon)u_j and less than 2\varepsilon u_j, which is impossible because D_0>3\varepsilon. Thus the set of near nuclei is constant throughout each isolated run. This argument permits ancestor maps to change and uses only the gap in (6.22).

Consider first a finite run j=a,\ldots,b and put

The near nuclei have distance less than \varepsilon u_a<t/100 from c. All other nuclei have distance greater than (D_0-\varepsilon)u_b>100T. There is a nucleus at c, and T/t=2^{b-a+3} is a power of two. By the choice of j_0, Lemma 6.1 applies. Every assigned point from this run belongs to its annulus, since c is itself a nucleus and

It follows that the union of the assigned sets over this run has expected particle count at most C, independently of its length. For an infinite run, apply the same argument to each finite initial segment and then use monotone convergence for their increasing assigned unions. There are at most CM runs, so the total contribution of all isolated assigned sets is at most CM. Combined with the crowded sets, this proves (6.16).

The remaining distances. It remains to count particles with a_x\ge s and d(x)<U. For the part with d(x)<s, choose a maximal s/2-separated subset of \{d<s,\ a\ge s\}. Its centers lie in \bigcup_{m=1}^M B(R_m,s). Disjoint balls of radius s/4 around them show, by volume comparison, that there are at most CM centers. Their balls of radius s/2 cover the set and are contained in the corresponding local balls B(y,a_y), since a_y\ge s. Equation (6.15) bounds this contribution by CMs^{-3}.

For a distance dyad u\le d(x)<2u, choose a maximal u/(2L)-separated subset. Its centers lie in \bigcup_m B(R_m,2u), so there are at most CL^3M of them. Their local balls cover the dyad, because a_y\ge d(y)/L\ge u/L. Thus (6.15) bounds its expected count by

Take u=2^ks for 0\le k\le K, where K is the least nonnegative integer such that 2^Ks\ge U. These dyads cover s\le d(x)<U. The sum of the inverse-cube terms is bounded by Cs^{-3}. For the constant terms with u<1, use 1\le u^{-3}; the number of remaining dyads, with 1\le u\le2U, is bounded by a universal constant. Their total contribution is therefore at most CMs^{-3}. Adding (6.16) proves (6.14).□

Proof. Smear the point charges at any fixed positive widths and leave the smooth charges unchanged. This decomposes V_\zeta=V_s+V_c with V_s\in\dot H^1 and compactly supported V_c\in L^{5/2}. Thus

For the fixed source and smearing widths, both displayed potential norms are finite. Young’s inequality proves coercivity in both spaces defining \mathcal X_+. A bounded minimizing sequence has weakly convergent subsequences in these two reflexive spaces. Their limits agree as distributions, since both converge against compact smooth tests; nonnegativity passes to this limit. The two linear terms are weakly continuous in their respective spaces, and both positive terms are weakly lower semicontinuous. This proves existence. The strictly convex kinetic term proves uniqueness.

Nonnegative smooth compact variations give

For bounded smooth compactly supported \varphi, the variations (1+t\varphi)\rho are admissible for sufficiently small positive and negative t. Indeed |\varphi\rho|\le\|\varphi\|_\infty\rho, and positivity of the Coulomb kernel bounds its Coulomb energy. The nonnegative pairing property in Lemma 3.1 therefore justifies differentiating these variations and gives \int d\varphi\rho=0. Hence d\rho=0, proving the Euler equation in (7.1) and its distributional Laplacian.

To establish finite mass without assuming it, let the source components have centers R_m and masses Q_m, and put

Newton’s theorem gives V_\zeta\le U. Where \rho>0, the Euler equation gives K*\rho\le V_\zeta. For any ball B, multiply this inequality by 1_Bw\rho and restrict the screening integral below to B. Write q_m(x)=w(x)Q_m/|x-R_m|, so that \sum_mq_m=1. The triangle inequality gives

The resulting symmetrization and Cauchy–Schwarz yield, with m_B=\int_B\rho,

All integrals in this calculation are local and finite; in particular w is bounded on B, and the local Coulomb energy is finite. Increasing B proves \int\rho\le2Q.

For completeness, L^1\cap L^{5/3} implies continuity and decay of the screening potential. The kernel restricted to a fixed ball belongs to L^{5/2}, so its pairing with translates of \rho is continuous. The complementary kernel contributes at most \|\rho\|_1/R when the cutoff radius is R. For fixed R, the local L^{5/3} norm of \rho on a ball whose center tends to infinity tends to zero. First letting the center tend to infinity and then R\to\infty proves decay.

Finally, (W+\varepsilon)_- is compactly supported away from point charges for every \varepsilon>0: W\to0 at infinity, and its positive Coulomb singularities dominate the continuous screening potential. On the support of this test \rho=0, and hence \Delta W\le0. Testing the equation gives \int|\nabla(W+\varepsilon)_-|^2\le0. The test is justified by compact H^1 approximation away from the singularities. Letting \varepsilon\downarrow0 proves W\ge0.□

Proof. For a single source, Newton’s theorem and the Euler equation give \rho(z)\le k(Q/|z-R|)^{3/2}. For every x,R and s>0,

To see the uniformity in R, apply Hölder with exponent q_1\in(3/2,2) to the second factor and its conjugate q_1'\in(2,3) to the first. The integrals of both radial powers over B(x,s) are bounded by their integrals over centered balls of the same radius. Their powers of s combine to s^{1/2}. Split K*\rho(x) at s=Q^{-1/3}. The near part is at most CQ^{3/2}s^{1/2} by (7.4), and the far part at most 2Q/s by Lemma 7.1. Both are CQ^{4/3}.

For a point charge this gives W\ge Q/(2d) for d=|x-R|\le cQ^{-1/3}. Outside that sphere set v=\lambda d^{-4}. For sufficiently small universal \lambda>0,

and its boundary value is below the preceding lower bound. On the exterior domain, testing the equation for v-W with (v-W-\varepsilon)_+ gives a nonpositive gradient square bounded below by zero. This test has compact support since both fields decay. Thus v\le W. Together with the inner estimate this proves (7.2).

For the mixture F=\sum_i(q_i/q)W_i, convexity gives

Compare with the equation for W and test with (F-W-\varepsilon)_+. The support is compact. Near a point charge whose mass was strictly decreased, F-W tends to -\infty, so the test vanishes there. If the masses agree, the singular terms cancel and the difference is locally H^1. The remaining density terms are locally in L^{5/3}, so compact H^1 approximation is valid. Monotonicity of the power gives F\le W.□

Proof. By the scale definition, \nu(B(x,2La_x))>(2a_x)^{-3}. Truncate charges in this ball to total q=(2a_x)^{-3} and apply Lemma 7.2. Their distances from x are at most 2La_x, so (7.2) gives the first bound in (7.7) with a fixed c_0>0.

If some charges contribute to V_c(x), denote their total by q\le a_x^{-3}. The screening estimate for each single field of charge q, followed by the mixture comparison, gives

If there are no contributors, the same conclusion follows from W_0\ge0.

For the upper bound, set a=a_x, s=4a, and let V_{\rm loc} be the potential of nuclei within 20a of x. On B(x,s) the function

is a supersolution of the field equation when C is large: the Laplacian of the second term is at most 80Cs^6/(s^2-|z-x|^2)^6, whereas its pth power has the same spatial dependence and coefficient C^{3/2}. The boundary blowup dominates W_0-V_{\rm loc}. The positive-difference test is legitimate because the point singularities cancel. Thus W_0(x)\le V_{\rm loc}(x)+Ca^{-4}. Subtracting V_c(x) and using the smooth local potential bound proves the upper offset estimate. The expression defining h_0 proves its continuity.

For W_s, the same bump without V_{\rm loc} is a supersolution after its coefficient is enlarged to absorb the locally bounded source 4\pi\theta\nu_s. This proves W_s\le Ca^{-4}. Its Laplacian is bounded by Ca^{-6} on a local ball. Subtract the Newton potential of this source restricted to that ball. Its gradient is bounded by Ca^{-5}, since the integral of |z|^{-2} on a ball of radius a is Ca. The remaining harmonic function is bounded by Ca^{-4} on a smaller ball; its interior gradient bound proves the stated estimate and local C^1 regularity.

For the lower bound, select and truncate the same charges as in the first paragraph, before multiplying them by \theta. Their smearing widths are at most 3c_1a, because their centers are within 2La. A single such blob of charge \theta(2a)^{-3} has bare potential at x at least c_L\theta a^{-4}: every source point is within (2L+3c_1)a of x. Its screening potential is at most C\theta^{4/3}a^{-4}. Choose \theta sufficiently small and apply the mixture comparison to obtain W_s\ge c_\theta a^{-4}.

Decrease \theta further if necessary. The source bound and the lower bound for W_0 imply

Consequently 1.01W_0 is a supersolution for the field of \theta\nu_s; its extra negative point sources have the required sign. The positive-difference test for W_s-1.01W_0, supported away from those point singularities, proves the final comparison. Decay follows from Lemma 7.1.□

Proof. In either case the local upper field estimate in (4.23) or Lemma 7.3 gives h\le Ca^{-4} on a fixed ball about y. The patch core is absent there. The Euler equation, the bound on \nu_s, and the weighted power inequality give

Here all centers in the sum belong to one fixed local ball and their total charge is at most a^{-3}. The same estimate holds for h_0 by Lemma 7.3.

Take the Newton particular solution for this Laplacian restricted to a ball of radius a. Equation (7.4) bounds it by Ca^{-4} on a smaller ball. Its variation over distance s is at most Ca^{-4}(s/a)^{1/2}. In detail, the part within distance Cs of either evaluation point is at most

On a dyadic ring of radius l\ge Cs, the difference of the two Coulomb kernels is at most Cs/l^2, and the source mass in that ring is at most C(a^{-6}l^3+a^{-9/2}l^{3/2}). Summing the rings up to radius a gives the asserted variation.

The harmonic remainder is bounded above by Ca^{-4}. Harnack’s inequality and the harmonic gradient estimate, applied to this upper constant minus the remainder, bound its variation by C(s/a)(a^{-4}+h(y)_-). Combining the estimates gives (7.9), as well as continuity. For h_0, Lemma 7.3 bounds its negative part by Ca^{-4}, which is absorbed by the first term.□

Proof. The scale function is 1/L-Lipschitz. The function a\mapsto c_1a\min(a,r_*)^w has Lipschitz constant O(r_*^w). Thus |t_z-t_y|\le O(r_*^w)|z-y|. If |z-y|\le t_z, this proves the support and comparability assertions. Replacing t_z by t_y changes the kernel by at most Cr_*^wt_y^{-3} on a ball of radius 2t_y, giving the first assertion of (7.12). The chain rule for a Lipschitz width gives the gradient bounds, first pointwise almost everywhere and then in L^2 by the support bound. Since \int\mathcal K_z(y)\,dy=1, Fubini proves preservation of integrals. Weighted Jensen followed by Fubini, using the bound on R1, proves the L^{5/3} estimate.

Newton’s theorem gives the nonnegative representation

Only |z-y|\le2t_y contributes. The local field bound gives

Equation (7.4) therefore bounds the displayed integral by C(a_y^{-6}t_y^2+a_y^{-9/2}t_y^{1/2}), proving (7.14). Both \rho_0 and R\rho_0 belong to L^1\cap L^{5/3}, hence also to L^{6/5}. Their potentials are continuous and decay by the proof of Lemma 7.1; the Sobolev inequality and the Coulomb energy identity put their difference in \dot H^1.□

Proof. First fix \theta as in Lemma 7.3. That lemma implies

with a fixed universal bound. Since \Delta W_s\le \cF W_s^{3/2}, the chain rule gives

Choose \alpha>0 so small that (1+\alpha)(1+\alpha D_*/\cF)\le1.1. Consequently

All chain rules are local: W_s is positive and locally C^1 with locally bounded Laplacian, so multiplication in the weak equation justifies them.

Choose c_b>0 smaller than a sufficiently small fixed fraction of c_0. On a\ge r, r^{4\alpha}W_s^\alpha\le C^\alpha. We may therefore choose positive \epsilon_0,C'_0 so small that, throughout this region, \Gamma_r is a tiny fixed fraction of W_s, is at most 0.1c_0a^{-4}, and is at most (c_b/2)a^{-4}. At the same time,

These choices precede every smallness requirement on r_*.

The equations for h_0 and q_0 give the exact cancellation

Fix y with a_y\ge r. For a contributing center z, the offset oscillation estimate and (7.14) imply

By (7.21), the final error is o(\Gamma_r(y)) uniformly as r_*\to0. Furthermore,

using the gradient and lower bounds for W_s and |z-y|\le2t_y. The chosen small ratio \Gamma_r/W_s and the derivative of the power therefore give

for sufficiently small damping fractions and then sufficiently small r_*. Indeed the limiting coefficient is p/\sqrt{1.01}>1.49. After integration in z, the factor R1(y)=1+o(1) leaves the lower bound 1.3\sqrt{W_s(y)}\Gamma_r(y). Combine this with (7.22) and (7.20) to prove (7.17). All terms represent locally integrable densities, so the assertion includes points of a level set of a up to a Lebesgue-null set; no regularity of those level sets is required.

Because q_0\ge0 and \Gamma_r\le0.1W_0, we have B_r\ge0.9W_0. The upper offset bound follows from b_r\le h_0+q_0. Also

so one may take c'=c_b/2. On the kernel support,

by the lower bound for W_0, scale comparability, and the offset oscillation estimate, after decreasing r_* further. This proves (7.19).

Finally, h_0,q_0 are continuous and locally H^1, and the same is true of \Gamma_r. Their decay proves the remaining assertions.□

Proof. The difference is exactly

The lower bound for W_s gives the assertion with

□

Proof. Write a=a_y and tilt the event in question, if it has positive probability P_A. By the tower property, its average of K*\mu_j is the tilted raw average of the potential of the master smears. Nuclei within a fixed multiple of a contribute at most Ca^{-4} to V_s. More distant electron smears have exactly their bare potential at y: a contributing smear near y has center within 2t_y\ll a. Dropping the remaining negative electron terms therefore gives

Lemmas 4.2 and 4.4, with no initial cut, bound the last expression by Ca^{-4}+C\sqrt{\delta/a}. Indeed, for small a, their scale quantity is bounded by C(a^{-7}+\delta). Lemma 5.2 and \ell_j\ge c_{L_1}a^{1+.01} imply

Solving for P_A proves the assertion. The deterministic threshold comes solely from the universal local field estimates.□

Proof. Since \Delta H_j\ge-Ca_Q^{-6} on Q', adding a quadratic and applying the ball submean inequality gives, for any fixed large C' and a universal C'',

The same inequality follows for distributional Laplacians by mollification. Choose C' above the threshold in Lemma 8.1. Because a_z/a_Q is bounded and a_z\le5L_1r on Q', integration of that tail yields

for arbitrary fixed m,A'. Jensen’s inequality for the spatial average gives the same bound for the integral term. Taking C'_{\rm cap}>C''+1, Markov’s inequality gives the bad-cube probability. On a bad cube its supremum is bounded by the integral term plus C''a_Q^{-4}; the probability just proved controls the latter contribution to every moment. This proves (8.3).

If a point lies in a bad cube, the corresponding term of \Pi dominates (H_j)_+ there. Otherwise the good-cube bound gives the asserted cap, since a_Q and the local scale are comparable by a universal factor. At a fixed point only a bounded number of enlarged cubes occur, proving the estimate for \mathcal E. Every participating or enlarged cube lies within distance C_{L_1}r of some nucleus, by d_y\le La_y. Counting grid cubes in this union proves finiteness and (8.4). Cube suprema are measurable through a countable dense subset, because H_j is continuous. The bounds from Lemma 5.1 also give the last assertion.□

Proof. The argument has three steps. Observation matching first relates the conditional density to the particles retained by a fresh patch. The patch comparison then determines the local offset except for a small event, whose negative-field cost must also be controlled. Finally, two separate tower identities transfer the offset comparison back to the master field.

A count-good event and the fresh patch. Write a=a_y, t=t_y, \tau=t/a. Recall that Rf(y)=\int\mathcal K_z(y)f(z)\,dz: the averaging and the kernel gradient estimates below concern the center variable z. There is a universal K_0 such that

Otherwise tilt this event. Lemma 5.2 gives \delta\le Cr^{-2.02}\log^5(e/r^{42})\le a^{-7+.02} for small r_*. The last inequality is uniform: after multiplication by a^{6.98} its left side is at most C_{L_1}r^{4.96}\log^5(e/r^{42}). Every compatible raw configuration has the same excessive count in B(y,4a), because \sqrt3\ell_j=o(a). A fixed cell cover and Lemma 4.4 contradict this for large enough K_0.

Call the complement of this excessive-count event count-good. On such an observation path, match its entries to any compatible raw configuration. The kernel Lipschitz bound gives

Only observations in B(y,3a) can contribute to this difference. The conditional ratios in Lemma 5.1 are supported on compatible configurations. Averaging (8.10) therefore gives the same bound with \mu_j(y) replacing the raw sum.

If an implication in (8.8) fails with probability at least r^{40}, intersect its failure event with count-good data and call the result A. Then P_A:=\Prob(A)\ge r^{41} for small r. Write P for the original joint law of raw variables and all noisy arrays, and Q=P(\cdot\mid A). Its raw marginal is the law of \psi_A=\sqrt{p_A/P_A}\Psi, with \delta\le a^{-7+.02}. Apply Lemma 4.3 to this state, choosing b=a^{1+.2} and \beta=b/a=a^{.2}. Once selected by shell averaging, the patch radius is deterministic. Extend Q by sampling the fresh cut labels conditionally on the raw variables with the cut’s squared partition probabilities. The raw-label marginal is exactly the patch construction for \psi_A. Write X for its cut data, \rho_X^c for its conditional core density, \rho for its local Thomas–Fermi minimizer, and \sigma for its retained fine density. These are conditional objects under this tilted patch law. In contrast, \mu_j remains the conditional density defined under P.

The fine smearing moves retained centers by at most \sqrt3b. Every particle relevant to the master kernel is in the retained central region. Thus (8.10) implies, Q-almost surely,

Let G=\{D(\sigma-\rho)\le a^{-7+.01}\}. Lemma 4.6 gives

In particular Q(G^c)\le Ca^{.01}. On G, Lemma 3.1 gives

For clarity, after division by a^{-6} these two errors are bounded by

The lower bound \tau\ge c_{L_1}a^w, w=10^{-5}, makes every power positive.

The local inverse comparison and rare negative fields. Put h_X=W-V_c. On G, in the high-density case one must have

Indeed, the opposite inequality and Lemma 7.4 imply on the master kernel h_X(z)\le I_y(\mu_j(y))-\delta_0a^{-4}/8. To cover arbitrarily negative h_X(y), note that h\mapsto h+C\tau h_- is increasing for small \tau; evaluate it at the proposed upper threshold, which is bounded in a^{-4} units. The oscillation error there is o(a^{-4}). The Thomas–Fermi equation, (8.6), and the first part of (8.7) then yield R\rho(y)\le\mu_j(y)-c\delta_0a^{-6}, contradicting (8.13). In the low-density case the lower oscillation estimate and the increasing function h\mapsto h-C\tau h_- similarly give, on G,

Here the second clipping inequality is used.

The deterministic upper bound h_X(y)\le Ca^{-4} controls the positive error on G^c in the low-density case. The high-density case requires a bound on negative fields there. By kernel positivity, \mu_j\ge T_y(h_b)\ge ca^{-6}. Equation (8.11) and \mathcal K_z(y)\le Ct^{-3} force, on every configuration,

The upper oscillation bound implies on this ball

Integrate against \sigma, divide by m_\sigma, and average over G^c. The patch bound controls the W_-\sigma numerator by Ca^{-7+.02}. For the nuclear term, weighted convexity and the uniform integral of |z-R_m|^{-5/2} on a radius-2t ball give

Hölder’s inequality on probability times space therefore gives

We conclude that

Since |I_y|\le Ca^{-4}, the bounds on G now imply the corresponding one-sided comparisons in expectation, with error \delta_0a^{-4}/4+o(a^{-4}).

Transfer of averaged fields. The remaining step compares the expectation of this fresh patch offset with that of the master offset under the event law Q. Apply Lemma 5.3 with the nuclear splitting V=V_s+V_c, master width t=t_y, and fine width b=a^{1.2}. Lemma 4.6 gives the raw local density and Coulomb error bounds; the deleted shell estimate is (4.26). The pointwise density bound required for the nuclear remainder follows from \rho=kW_+^{3/2} and (4.23); its potential on a radius-t ball is Ca^{-4}(t/a)^{1/2} by (7.4). Thus the explicit transfer estimate is

Here \beta=a^{.2}, c_{L_1}a^w\le\tau\le c_1r_*^w, and .01-w/2>0. The two expectations in the common lemma use different conditional laws; this verification identifies neither posterior pathwise.

On the high failure event, every path has H_j(y)<I_y(\mu_j(y))-\delta_0a^{-4}. Equations (8.14) and (8.16), together with (8.17), instead give

a contradiction. The low failure event is contradicted in the same way by (8.15), the deterministic upper bound on h_X, and (8.17). All remainders are uniform: a\asymp_{L_1}r, \tau\ge c_{L_1}a^w, and the smallest power in (8.16) is .004-2.8w>0. The constants were fixed before letting r_*\to0.□

Proof. Conditional Jensen, the master-kernel support, and Lemma 4.4 give

The tail bound implies \|(H_j(y)_+)^{3/2}\|_2\le Ca_y^{-6}. For the nuclear part of the reaction, weighted convexity and the uniform integral of |z-R_m|^{-3/2} on a radius-2t_y ball give

This proves (8.18). Cauchy–Schwarz against \Prob(F_y)<r^{40} contributes r^{20}. Use a_y\asymp_{L_1}r, \tau_y\ge c_{L_1}r^w, and (8.4) to obtain (8.19). The arbitrarily high polynomial bounds in Lemma 8.2 give the other assertions, choosing their exponent larger if necessary. All integrations are licensed by the joint measurable versions already constructed.□

Proof. Define two shifted offsets

Use the following open cover to define u':

The definitions agree on overlaps. On the lower overlap, Lemma 7.7 and (9.3) give

On the upper overlap the shifted cap inequalities, ensured by the choice of L_1, put both u_1 and u_2 below b_{R_+}. Thus the local maxima glue to a continuous locally H^1 function. No assertion about the measure or regularity of a level set of a is required. The new exterior bounds follow from the same comparisons: for R_+\le a<(1+\kappa)R_+ the lower bound is supplied by u_1, and elsewhere the barrier is an included branch.

We verify the differential inequality locally. At a fixed point in one of the open regions, discard branches more than \eta below the maximum. There is a neighborhood on which these discarded branches cannot attain the maximum, while every retained branch is within 2\eta of it. We show that each retained branch satisfies the common inequality

on that neighborhood.

For u_1, the old core inequality suffices on a<r, and monotonicity of f_y gives the required reaction on a\ge R_+. On the gain region r\le a<R_+, the error in () pays \mu_j at points of \mathcal E\cup\mathcal B^{\mathrm h}. At every other gain point, \Pi=0 and near activity yields

Here h_b\le u\le h_{\mathrm{top}}. If \mu_j>T_y(u), then the high comparison applies and I_y(\mu_j)\ge u, including the case where the inverse is clipped at its upper endpoint. It would follow that

contrary to (9.3). Thus 4\pi\mu_j\le f_y(u). The old reaction supplies precisely the new core term on the gain region.

For u_2, the identity \Delta H=-4\pi\nu_s+4\pi\mu_j gives the required core term on all of a<R_+, while the first increment in () pays the Laplacian of \Pi. Consider the remaining points where u_2 is retained. The lower splice or the included barrier gives

The last inequality uses a\le2L_1R_+ and (9.3). On \mathcal E\cup\mathcal B^{\mathrm l} the last error term pays the reaction because u_2\le H and f_y is nondecreasing. Otherwise \Pi=0 and the cap gives H<h_{\mathrm{top}}. If \mu_j<T_y(u_2), the low comparison applies and I_y(\mu_j)\le u_2, also when the inverse is clipped at its lower endpoint. We would obtain

contradicting H=u_2+\lambda_2R_+^{-4}. Therefore 4\pi\mu_j\ge f_y(u_2), as required.

The barrier branch appears only where a>R_+, and Lemma 7.6 gives (9.6) there. The old error density and every additional error are nonnegative, so they can only weaken the required lower bound.

These comparisons cover the exact level sets as well: a=r belongs to the gain region and uses the old reaction, whereas a=R_+ belongs to the new exterior and uses its reaction. All right-side comparisons hold almost everywhere on the chosen neighborhood and therefore upgrade each retained branch’s distributional inequality to (9.6). The source \nu_s is a smooth bounded density for the fixed system, and all branches here are continuous and locally H^1. The reaction is also bounded on bounded ranges of its argument: for a fixed system, t_{\min}:=\inf_z t_z>0 and V_c\in L_c^{5/2} give, for every A>0,

Thus the first, nonsingular part of Lemma 5.4, followed by a partition of unity, proves the inequality for u' on all of \R^3. No vanishing of the reaction near the original nuclei is needed in this step.□

Proof. For the fixed system there are finitely many scales and cubes, and all master smoothing widths have a positive minimum. Lemma 5.1 therefore gives deterministic bounds for the conditional densities, their potentials and their gradients. The offsets and error densities in the assertion are jointly measurable: maxima, clipped inverses and failure flags preserve measurability, and each cube supremum may be taken over a fixed countable dense subset.

Inductively the branches have deterministic continuity moduli and local H^1 bounds on each compact set. Their open definitions agree on overlaps; a finite subcover of a compact set supplies a Lebesgue number and hence a modulus for the glued function. The gradient of a finite maximum is almost everywhere one of the branch gradients. Conditional averaging consequently preserves continuity. It preserves the local H^1 bound by testing weak derivatives against compactly supported vector fields, using Fubini and Cauchy–Schwarz. The error terms have deterministic bounded size and support for the fixed system and step, so the same is true after averaging.

Integrate the distributional inequality of Proposition 9.1 against a fixed nonnegative compact test and then against the explicit positive-parent conditional law of Lemma 5.1. The preceding bounds justify Fubini. The tower property gives \E[\mu_j\mid\mathcal F_{j+1}]=\mu_{j+1}, while convexity gives

This is the required lower bound for the reaction. The exterior bounds and the exterior equality are deterministic and pass to the average.□

Proof. Since r_0<Z^{-1/3}\le a, Lemma 7.6 gives the initial invariant everywhere; there is no initial core region.

At scale r=r_j, apply Lemma 8.4 to the three increments in (). The high-failure estimate pays \mu_j on the gain region; the low-failure estimate pays T_y(H)=f_y(H)/(4\pi) on the remaining participating region. The same Lemma pays membership in \mathcal E and the penalty Laplacian. For \mathcal B\in\{\mathcal B^{\mathrm h},\mathcal B^{\mathrm l}\}, use \1_{\mathcal E\cup\mathcal B}\le\1_{\mathcal E}+\1_{\mathcal B}. All these integrands are nonnegative, so restriction to the indicated subregions only improves their bounds. We obtain

Conditional averaging preserves the unconditional expected integral of a nonnegative density. Because \beta=17-3w>0,

At the last step the data are trivial, so d_n is deterministic. This proves (9.7), uniformly in n. The argument has used the failure probability separately at each spatial point and then integrated it; it has imposed no simultaneous good-data event in space.□

Proof of Theorem 1.1. Use the contradiction sequence and sector eigenstates of Section 8. Its terminal scale is r_*=(B_0M/K_{\mathrm{exc}})^{1/3}. The universal B_0 will exceed the fixed count constant below. All deterministic parameters and tolerances are chosen before taking the sequence limit. The smallness requirements of the preceding lemmas then hold for all its sufficiently late members, uniformly over their scales.

At the final step put

The signed measure S has compact support. In fact \{a<r_*\} lies in a finite union of balls around the nuclei, the error density has compact support, and the nuclei are finite. Since \Delta V_c=-4\pi\nu+4\pi\nu_s, the final invariant implies

The nuclear singularities cancel in the explicit identity

The right-hand side is continuous and locally H^1 on all of \R^3: the last two potentials have bounded compactly supported densities, and the other terms have the stated regularity from the construction.

Far from the nuclei, (9.2) gives U=B_{r_*}=W_0+q_0-\Gamma_{r_*}, which tends to zero by the deterministic field estimates. The compact source potential K*S also tends to zero. For every \epsilon>0, the function (U-K*S-\epsilon)_+ consequently has compact support and is an admissible H^1 test in the subharmonic inequality. Testing gives a nonpositive squared gradient norm, so this positive part vanishes. Letting \epsilon\downarrow0 proves

On every sufficiently large sphere U=B_{r_*}\ge .9W_0>0. The spherical mean of the potential of a compactly supported signed measure, on a sphere centered at the origin and enclosing its support, equals its total mass divided by the sphere radius. Hence

Only positivity on a far sphere is used here; no asymptotic charge formula for W_0 is needed.

At the last scale the data have been completely averaged away, so

If a kernel centered at x contributes to a point with a_y\ge r_*, then its support and the Lipschitz scale estimate imply

For small r_* this forces a_x\ge r_*/2. Each kernel has integral one, and Lemma 6.2 therefore gives

Combining this with (9.7) and (9.9) yields

Choose B_0>2C, increasing it also to ensure the initial scale choice in Section 8. Since K_{\mathrm{exc}}/M\to\infty, the last inequality is impossible. This proves the existence of one universal finite constant bounding the excess per nucleus in every strictly bound sector.

For an arbitrary strictly bound sector, the largest strictly bound sector of the same system has at least as many electrons, so the same bound applies. For every integer N>Z+CM, strict adjacent binding is therefore impossible. Sector monotonicity gives E_N\le E_{N-1}, and exclusion of the strict inequality gives E_N=E_{N-1}. Taking M=1 proves both atomic assertions for every real nuclear charge Z\ge1.□

Proof. We first control the observed count and the positive field throughout the band. At a fixed point and height, a hypothetical failure then defines a tilted state. Its fresh patch compares the observed density with the patch field, including the contribution of rare negative fields; the two expectation towers turn this into a contradiction. Finally, spatial and height nets give the simultaneous statement. Every occurrence of o(1) below tends to zero as s\downarrow0, uniformly for r_0\le r=r_j\le s and the stated bands, after the fixed parameters in the proposition have been chosen.

Kernel geometry. Put a=a_y=d_y/L, t=t_y and \tau=t/a. If \mathcal K_z(y)\ne0, the Lipschitz bound on t gives |z-y|\le t_z\le t_y+Cs^w|z-y|, hence |z-y|\le2t_y. On this support, t_z/t_y=1+O(s^w). Comparison with the fixed-width kernel g_t(y-z)^2 therefore gives

The center derivative is a weak derivative where t_z has a corner; its stated bound also gives the needed Lipschitz estimate. For r/2\le d_y\le8L_1r one has a\asymp_{L_1}r and

For example, when d_y\ge r_0, the formula is \tau=c_1L\min(d_y,s)^w; on the indicated band \min(d_y,s)\ge c_{L_1}d_y. The case d_y<r_0 is comparable, since d_y\ge r_0/2.

An observed-count event. Let M_j be the number of entries of Y_j with radii in [r/4,12L_1r]. For a sufficiently large fixed C_{L_1},

Indeed, if the event had probability p>r^{40}, its tilted raw state from Lemma 5.2 would have offset

Each configuration compatible with the event has more than C_{L_1}r^{-3} raw points in [r/8,13L_1r], because observation displacements are at most \sqrt3\ell_j=o(r). But (4.45) bounds its expected count by C'_{L_1}(r^{-3}+\sqrt{\delta r})\le2C'_{L_1}r^{-3} for small s, a contradiction if C_{L_1}>2C'_{L_1}. Call the complementary event in (10.8) \mathcal C_j. For data in \mathcal C_j, every compatible configuration has at most C_{L_1}r^{-3} centers relevant to the master kernels on r/2\le d_y\le8L_1r. Conditional averaging and (10.6) imply there

A universal spatial field cap. Fix y in the same enlarged band and tilt an event A'=\{H_j(y)>b\,d_y^{-4}\} of positive probability p. The original-law tower gives

To justify the last inequality, a master smear whose potential at y differs from the point potential has |x_i-y|<t_{x_i} and hence |x_i-y|\le2t_y<Aa_y for small s. All terms outside that exclusion therefore have their exact point potential, and dropping the remaining electron terms gives J_y. By (4.39) without an initial cut and Lemma 5.2,

Here C_2 is universal: the local field term in these units is C d_y^4a_y^{-4}=CL^4, and the contribution C d_y^{7/2}\sqrt{D_0} is at most one after decreasing s. Solving (10.10) for p shows that for b\ge2C_2,

Tail integration consequently gives, for each fixed M>0,

Subharmonicity permits a simultaneous conclusion without a Z-dependent derivative estimate. Apply the ball submean inequality with radius d_y/10 for 3r/4\le d_y\le6L_1r. The averaging ball is contained in \{r/2\le d_z\le8L_1r\} and d_z\asymp d_y there, so

The constant C_3 is universal. The integral term has expectation O_{M,L_1}(r^M), since its integration volume is O_{L_1}(r^3). Markov’s inequality proves that, outside an event of probability C_{M,L_1}r^M,

This establishes the claimed order of dependence of the cap.

A fixed spatial point and height. Fix now r\le d_y\le4L_1r and h\in[h_l/2,2h_h]. Consider either failure of (10.5) with tolerance \xi/4, intersected with \mathcal C_j. If its probability were at least r^{42}, call that event A' and write Q=\Prob(\,\cdot\mid A'). The raw marginal is the state \Psi_{A'} of Lemma 5.2, with

for small s. Apply the refined patch (4.42) to this raw state at y, with \beta=a^{.2} and b=\beta a. Extend Q by fresh cut labels sampled conditional on the raw positions, exactly as in that patch. Let X be these patch data, let \rho_X^c be its conditional core density, and let \rho,W,\sigma be its minimizing density, screened field, and fine-smear density. Thus

The roles of these densities and their laws are distinct. The pair \mu_j,H_j keeps its original-observation definition in (10.2). Under the event law Q, the fresh patch gives the conditional core density \rho_X^c and its field \Phi_X=V-K*\rho_X^c; \sigma is the fine smear of retained out particles, whereas \rho is the auxiliary minimizing density and W=\Phi_X-K*\rho. The field comparison is made after taking Q-expectations in (10.19) below.

Each compatible raw configuration matches the observed array Y_j after a permutation, with displacement at most \sqrt3\ell_j per position. Both configurations and their conditional averages at those same data therefore differ, when tested against x\mapsto\mathcal K_x(y), by at most Cr^{-3}\ell_jt^{-4}. This follows from the count bound and the center Lipschitz estimate in (10.6); only entries within 2t+O(\ell_j) of y can contribute. All such raw particles are retained by the fresh patch. Its radial fine smear moves a retained point by at most \sqrt3b, adding at most Cr^{-3}bt^{-4} to the error. Consequently, on every compatible path, including paths where the Coulomb error is large,

In the comparison with the conditional average, one may first compare each raw sum with \sum_i\mathcal K_{Y_{j,i}}(y), then average this same bound using the original posterior law. The array is fixed in both comparisons, so no equality of posterior laws is being used.

Let G=\{D(\sigma-\rho)\le a^{-7+.01}\}. Then Q(G^c)\le Ca^{.01}. The test function z\mapsto\mathcal K_z(y) has support in B(y,2t) and gradient norm at most Ct^{-5/2}. Coulomb duality gives on G

The positivity of the two powers used so far is explicit: .01-4w=.00996 and .005-(5/2)w=.004975.

On G, a high-density antecedent forces

For if the reverse strict inequality held, the oscillation bound (4.44) on B(y,2t) would give W(z)\le(h-\xi/32)d_y^{-4} there for sufficiently small s. This remains uniform when W(y) is arbitrarily negative: the upper estimate is W(y)+C\tau(a^{-4}+W(y)_-), and v\mapsto v+C\tau v_- is increasing for C\tau<1. Using the patch equation and \int\mathcal K_z(y)\,\dd z=1+O(s^w) would then give

contradicting (10.13)–(10.14). Similarly, on G a low-density antecedent forces

Indeed the contrary is positive, so the lower oscillation estimate gives W(z)\ge(h+\xi/32)d_y^{-4} on the contributing ball; its reaction density contradicts the low antecedent with the same strict margin. Constants from d_y=La are fixed before s is decreased.

The negative field on the exceptional patch event. The upper bound (10.16) passes to expectation with an o(a^{-4}) error, by the patch bound W(y)\le Ca^{-4} and Q(G^c)\le Ca^{.01}. To pass the lower bound to expectation we need more than this small probability. Under the high-density antecedent, (10.13) holds on all paths and gives \mathcal R\sigma(y)\ge ca^{-6}, since h\ge h_l/2>0. The support and upper bound of the master kernel imply the deterministic mass bound

For Y=W(y)_-, (4.44) gives throughout this ball W(z)_-\ge(1-C\tau)Y-C\tau a^{-4}. Taking C\tau\le1/2, integrating against \sigma, and using the mass bound yields

Multiply by \1_{G^c} and use the refined patch budget under the same law Q. We obtain

because .02-3w=.01997>0. Thus the high and low antecedents give, respectively,

Transfer between the two conditioning procedures. Apply Lemma 5.3 with V_s=V, V_c=0 and h_X=W. Its two towers are taken under the original observation law and the fresh tilted patch law, respectively. The refined patch provides all its inputs: the raw density bound and Coulomb budget are (4.42), the deleted count bound is (4.26), and \rho\le Ca^{-6} on the contributing ball by (4.25). Therefore

Here \beta=a^{.2} and (10.7) makes every positive-power error vanish. The source-free potential remainder actually has the sharper power \tau^2, so the common upper bound is sufficient. No equality of posterior laws is used in this transfer.

On a high-failure event, its defining inequality is d_y^4H_j(y)<h-\xi/4, whereas (10.18) and (10.19) give d_y^4\E_QH_j(y)\ge h-\xi/16-o(1), a contradiction for small s. On a low-failure event the inequalities are reversed and give the same contradiction. Thus each such failure intersected with \mathcal C_j has probability less than r^{42}. Including (10.8), the two implications at a fixed point and height, with tolerance \xi/4, fail with probability at most Cr^{40}.

Spatial and height nets. Choose a net of the closed band \{r\le d_y\le4L_1r\} with spacing at most r^2 and at most C_{L_1}r^{-3} points. For example, select one band point in each intersecting cube of side r^2/2; the cube count follows from the volume of its O(r^2) enlargement. Choose a finite height net in [h_l/2,2h_h] with mesh at most \xi/64, including its endpoints. Exclude every point-and-height failure, the count failure, and the cap failure with M=40 in (10.11). A union bound gives exceptional probability Cr^{37}, and hence at most Cr^{25} for r<1.

On the retained event, (10.9) shows that d_y^6\mu_j(y) varies by at most Cr^{1-4w}=o(1) between a band point and a net point within Cr^2. To control the field, work in a ball of radius cr around a band point, contained in the enlarged cap band. On this ball B, let Q=K*(\1_B\mu_j). Then H_j+Q is harmonic on B. The value and gradient of Q are bounded by Cr^{-4-3w} and Cr^{-5-3w}, respectively, so H_j+Q is bounded above by Cr^{-4-3w}. Its nonnegative distance from that upper bound has, by the Poisson formula on a smaller concentric ball, oscillation bounded by its central value times the relative distance. It follows that

This estimate also controls the change of the normalized field d_y^4H_j(y), since d_z/d_y=1+O(r). In particular, if its value at y is below h-\xi, its value at z is below h-3\xi/4 for small s. Arbitrarily negative values cause no problem: the relevant upper estimate has the form v+Crv_-+O(r^{1-3w}), increasing in v when Cr<1. If the value at y exceeds h+\xi, it is positive and the cap and (10.20) make its value at z greater than h+3\xi/4.

Suppose now a high-density implication failed at some (y,h). At a nearby spatial net point choose a height-net value h'\in[h-\xi/4,h-\xi/8]. The positive lower height bound and the o(1) density variation preserve the high antecedent at h', while the field is below h-3\xi/4<h'-\xi/4. This is an excluded net failure. For a low-density failure choose h'\in[h+\xi/8,h+\xi/4]; its antecedent is preserved and the field exceeds h+3\xi/4>h'+\xi/4. The height intervals lie inside [h_l/2,2h_h] by the stipulated smallness of \xi. This proves simultaneity and finishes the proposition.□

Proof of Propositions 11.2 and 11.1. Use the observations and master packets (10.1)–(10.2), with \eps fixed during initialization below and Z large enough that m\ge1. The resulting densities and fields satisfy

Write \kappa=1/8. The barrier b_r in the proposition is positive on d\ge r and, for sufficiently small fixed B>0, satisfies

For example, 4\pi k\sqrt B\le19/2 suffices. If R=2r, then

Initialization. With probability at least 1-Cr_0^{35}, the initial conditional density satisfies

To prove this, fix a point y in this annulus and let A_y=\{P_0(y)>0.05Z/r_0\}, an event of \mathcal F_0. Suppose that p=\Prob(A_y)>r_0^{40}. By Lemma 5.2, the raw marginal conditioned on this event is the squared law of an antisymmetric state of offset at most

For the fixed \eps this is at most Z^{7/3} when Z is large enough. The global estimate (3.3) therefore bounds the tilted density \rho_{A_y} by \int\rho_{A_y}^{5/3}\le C Z^{7/3}, with a universal C.

Only packet centers with |x|<2r_0 can contribute to the truncated density in P_0: a center outside this ball has t_x\le c_1s^w|x|<|x|/2, so its packet misses B(0,r_0). A truncated packet has potential at most that of the full packet, which by radial smearing is at most K(y-x). The conditional tower identity and Hölder’s inequality consequently give

Choose \eps small enough that the last coefficient is less than 0.05. This contradicts the definition of A_y, so \Prob(A_y)\le r_0^{40}.

The minimum packet width is t_{\min}=c_1r_0^{1+w}. Integrating |y-z|^{-2} against a bounded packet gives the deterministic bound

A net in the annulus of spacing O(r_0^2) has O(r_0^{-3}) points. The union bound gives failure probability O(r_0^{37}), and the variation between a point and its net point, divided by Z/r_0, is O(r_0^{1-2w})=o(1). For large Z this proves (11.7), with the stated weaker exponent 35.

Let G_0 denote the event in (11.7), and set

On d<2r_0 use the branch

Choose C_3 universally so that 6C_3\ge4\pi k2^{3/2}, and then choose \eps sufficiently small to meet the preceding tilt condition and 4C_3\eps^{3/2}\le0.05. The quadratic term on this ball is at most 0.05Z/r_0, since Zr_0^3=\eps^3. For r_0\le d<2r_0 we thus have v\le2Z/r_0 and

On r_0\le d<1.06r_0, the good-event bound (and the easier bound \mu^{\mathrm{in}}=0 on its complement) give

On 1.9r_0<d<2r_0 we instead have v\le(1/1.9-0.7+0.05)Z/r_0<0. Fix B satisfying the barrier condition and 0<B\le0.1\eps^3. Then v>b_{r_0} on the lower overlap, whereas v<0<b_{r_0} on the upper overlap. Define

The definitions agree on their open overlaps. Lemma 5.4 proves (11.2) initially; near d=r_0 the branch v alone attains the maximum, and its displayed Laplacian supplies both required densities. The exterior upper bound holds, for example, when C_{\mathrm{inv}}\ge\max(B,32\eps^3). Also, using n\le3Z=3\eps^3r_0^{-3},

Order of the remaining constants. The constants C_3,\eps,B have now been fixed universally. Increase C_{\mathrm{inv}} to exceed also the universal cap C_{\mathrm{cap}} in Proposition 10.1. After fixing \gamma in (11.6), choose 0<\lambda_2<\lambda_1<\gamma/2. Choose L_1>2 so large that C_{\mathrm{inv}}L_1^{-4}<\lambda_2. This ensures

Next fix 0<h_l<B/4 and h_h>C_{\mathrm{inv}}, and then choose e_0,\xi>0 sufficiently small that

with \xi also in the range of Proposition 10.1. Only now restrict s according to that proposition, D_0, and the other smallness conditions in this proof. Finally take Z large according to D_0,s and the initialization requirements. Thus no constant is chosen in terms of a later scale or a particular state.

One outward step. Suppose that (11.2) holds at r=r_j, write R=2r, and abbreviate u=u_j. On the good band event of Proposition 10.1, for the band r\le d\le4L_1r=2L_1R, introduce

Define a new function on a fixed open cover by

On bad band data omit v_2 in every row. In both cases set

On the lower overlap, (11.6) gives v_1\ge b_r-\lambda_1R^{-4}>b_R. On the upper overlap, the caps on u and, on good data, H_j put both shifted branches below b_R. The definitions therefore agree. For R\le d<1.08R the lower bound u'\ge b_R follows already from v_1; farther out the barrier is included explicitly. The upper bound u'\le C_{\mathrm{inv}}d^{-4} and the equality u'=b_R for d>L_1R follow from the same caps.

We prove the distributional inequality at radius R before averaging. The open-neighborhood formulation is as follows. At a point y_0 in one of the open sets of (11.10), let M be the maximum of its candidate family and put \eta=e_0R^{-4}. Keep precisely those candidates v with M(y_0)-v(y_0)\le\eta, and choose a maximizer v_* at y_0. Continuity and finiteness give an open neighborhood U of y_0, inside the chosen open set, on which every discarded candidate is strictly below v_* and every retained candidate satisfies

The retained family has maximum M throughout U. At the origin make this construction with the continuous offsets by V; the family there contains only branches having that common singularity.

Each retained branch satisfies on the whole of this neighborhood the common inequality

For v_1, retain its old distributional inequality and compare its right side with the new one almost everywhere. On d<r the old source suffices. On d\ge R, use f(u)\ge f(v_1). On the gain region r\le d<R, bad data are paid by the added error in (11.11). On good data, v_2 is a candidate there, so (11.12) gives

Moreover h_l<7B/8\le d^4u\le C_{\mathrm{inv}}<h_h. If 4\pi\mu_j>f(u), the high implication in (10.5), at height h=d^4u, would give

contrary to (11.9). Hence 4\pi\mu_j\le f(u), and the old reaction supplies the added source.

For v_2, which occurs only on good data, start from its identity on all of U,

This supplies (11.13) on d<R, since \widetilde p\ge0. At any remaining point where v_2 is retained, the lower splice or the included barrier gives M\ge b_R, and d\le2L_1R. Thus

The cap also gives d^4v_2\le C_{\mathrm{cap}}<h_h. If 4\pi\mu_j<f(v_2), the low implication in (10.5), with h=d^4v_2, would imply

which is impossible. Consequently 4\pi\mu_j\ge f(v_2) there. The barrier branch occurs only where d>R and satisfies the common inequality by (11.5).

These verifications compare zero-order densities almost everywhere on the same open set U; they do not restrict or differentiate a weak inequality across an activity set or a sphere. In particular they apply on a neighborhood crossing d=R. The common reaction is F(y,t)=\1_{\{d\ge R\}}f(t), which meets Lemma 5.4’s hypotheses after canceling V near the origin. That lemma proves (11.13) for the local maximum. These local conclusions give the inequality globally by a partition of unity. We have proved the desired invariant at radius R with density \mu_j, function u', and error \widetilde p.

Forgetting the current observations. Define the next offset and error by conditional integration,

We give the regularity justification as well as the formal calculation. For the fixed system t_{\min}>0, and integration of each packet gives deterministic bounds

The same potential bounds hold for the initially truncated density. The explicit initial branches therefore have deterministic local Lipschitz bounds for their offsets. Subtracting a constant and taking a finite maximum preserve such bounds. Where b_R occurs, its offset b_R-V is smooth and separated from the origin. The fixed open cover and agreement on overlaps preserve local bounds after gluing. The errors are bounded, nonnegative, and supported in a deterministic ball. Induction supplies all the claimed bounds at every step, without requiring them to be uniform in Z.

All functions are jointly measurable in data and position. Indeed the conditional densities have the likelihood versions described in Section 10; the initialization flag is a supremum of a continuous function on a fixed compact annulus, testable on a countable dense set; and the band flags and finite maxima are measurable. Conditional integration of functions with deterministic local Lipschitz bounds preserves those bounds. It also commutes with weak derivatives by Fubini. Nonnegative smooth tests in a countable dense family first give simultaneous almost-sure weak inequalities; local bounds extend them to every test. Thus the following conditional integration is legitimate as an inequality of distributions.

Integrate (11.13) conditionally on \mathcal F_{j+1}. The tower property supplies \mu_{j+1}, while convexity gives

The deterministic bounds b_R\le u'\le C_{\mathrm{inv}}d^{-4} and the equality outside L_1R pass to this average. Positivity, boundedness, and support of the error do as well. This proves (11.2) at step j+1.

Summing the lost source. For r\le d_y<2r, every master packet contributing to \mu_j(y) has its center in a fixed annulus comparable to r, and its width is comparable to t_y. Conditional Jensen and the annular count estimate (4.45), applied with offset at most D_0, yield

Here r\le s\le1 and s is small according to D_0, so \max(r^{-3},1)+\sqrt{D_0r}\le Cr^{-3}. The bad band probability is at most Cr^{25} by Proposition 10.1. Cauchy–Schwarz in data and integration over the gain annulus give

Conditional averaging preserves the expected integral. Stopping the sum before any index j and using (11.8) proves (11.3), with the same master kernels and without replacing an expectation by a pathwise bound. At the terminal index the geometric progression of radii gives

The last error is deterministic because \mathcal F_m is trivial. The constants are independent of the number of scales and of Z.

From the potential to the raw electron count. Put r=r_m, u=u_m, and introduce the compactly supported signed measure

The invariant implies \Delta(u-K*S)\ge0. Its singularities cancel in the identity

whose right side is continuous and locally H^1 on all of \R^3. The difference tends to zero at infinity: u=b_r outside L_1r, and the potential of the finite compact source S tends to zero. For each a>0, the positive part (u-K*S-a)_+ has compact support and is an admissible H^1 test in its subharmonic inequality. Testing gives a nonpositive squared gradient integral, hence this positive part vanishes. Letting a\downarrow0 proves u\le K*S.

On the sphere d=2r, the invariant gives u\ge b_r(2r)=15B r^{-4}/256. The spherical mean of K*S there equals the total mass of S divided by 2r, since S is supported in \{d\le r\}. Consequently

No sign condition on the measure S is needed for this mean identity.

Finally \mu_m is the unconditional master-smoothed density. If |x|<s/4, then r_0\le s/2 and D_x^0\le s/2, so t_x\le c_1(s/2)s^w<s/4. The packet at x is therefore wholly contained in B(0,s/2)\subset B(0,r_m). Since every packet has mass one,

Combining this with (11.15), (11.14), and r_m\le s gives

Take c_{\mathrm{def}}=15B/256 and decrease s_*(D_0) so that the error is at most c_{\mathrm{def}}s^{-3} and c_{\mathrm{def}}s^{-3}>1 for every 0<s\le s_*(D_0). All remaining requirements are met by increasing Z_*(D_0,s). This proves the proposition.□

Proof. For x in the broad annulus, every source with |x_i|<h lies outside B(x,Aa_x): indeed h\le|x|/2 and A/L<1/2. Consequently

If 2^jh\le|x_i|<2^{j+1}h and the summand occurs, then

For radii comparable to or below u this follows from the exclusion |x_i-x|\ge Aa_x\ge A\alpha u/L; for radii greater than 2\beta u it follows from |x_i-x|\ge|x_i|/2.

The initial-cut hypotheses of Proposition 4.5 hold uniformly here. Cover its support by a fixed finite number of cells B_z with a_z\asymp u, the constants being allowed to depend on L and the annular ratios. For each target x in the broad annulus these cell scales are comparable to a_x. The gradient bound can be written as \sum_zD_z^2a_z^{-2}\1_{B_z}, with D_z^2\le C(a_z/u)^2. Since a_zm_z\ge1 and the screening parameter P\ge1,

Enlarging the fixed cover constant C_0 suffices for every finite \delta. Thus (4.39), conditional Jensen, the triangle inequality in L^2, and (4.45) applied to (12.3) give, pointwise in the spatial variable,

The series converges: its large-j terms are bounded by a constant times 2^{-j}h^{-1}+\sqrt\delta\,2^{-j/2}h^{-1/2}.

For completeness, conditional fields here have simultaneous spatial versions. On every compact subset of \{|x|>h\}, the kernels in (12.1) and each fixed number of their derivatives have deterministic bounds for the fixed finite system. Conditional integration therefore gives continuous harmonic functions, with measurable dependence on the data. One may first use the conditional law of configurations supplied by the cut and then differentiate under that integral. For a harmonic function H, its positive part is subharmonic. Balls of radius c u about points of the smaller annulus remain in the broad annulus, so their mean-value inequalities imply

Minkowski’s integral inequality and the pointwise bound prove (12.2). Suprema can be taken on countable dense sets, which also proves their measurability.□

Proof. Use Lemma 12.1 with no cut, h=t/4, and u=t. Its convergent series is bounded by C_t(1+\sqrt\delta). The spherical average of the expected field B_{t/4} on |x|=t equals

by the spherical mean of the Coulomb kernel. Bounding this average from above by the positive supremum proves the assertion, with the fixed factor t absorbed into C_t.□

Proof. Take a normalized ground state \Psi_* in sector N_*, where Z\le N_*\le3Z and E_{N_*}=E, as provided by Lemma 2.6. Proposition 11.1, with D_0=0, supplies a fixed sufficiently small t>0 such that

for all sufficiently large Z. Choose a smooth square partition c^2+o^2=1, with c=1 on |x|\le t, c=0 on |x|\ge2t, and |\nabla c|^2+|\nabla o|^2\le C t^{-2} supported in the intervening annulus. Let X be its full out data and set q^c=Z-n_c, where n_c is the number of core particles. Since n_c\le\#\{i:|x_i|<2t\}, we have \E q^c>1.

We first show that the favorable event G=\{n_c\le Z-1\} has probability bounded below independently of Z. Independently of the cut labels, observe one unordered array with the smooth compact noise law of Lemma 5.2, choosing its fixed width so that each spatial displacement is less than t/8. Put

Every entry counted here comes from a particle of radius less than 5t/8, which must be core. Thus q^c\le q^{\rm obs} on the enlarged probability space. Conversely every particle of radius less than t/4 produces an entry in B(0,t/2).

For an event A_\lambda=\{q^{\rm obs}>\lambda\} of probability p>0, the single-array version of (5.4) produces a normalized state \Psi_{A_\lambda} whose raw configuration law is the law conditioned on that event and whose offset satisfies

Every compatible configuration obeys Z-\#\{|x_i|<t/4\}>\lambda. Hence Lemma 12.2 gives

For sufficiently large \lambda this implies p\le e\exp(-c_t\lambda^{2/5}); the inequality is also harmless when p=0. Integrating the tail yields

Here the equality of events uses the integer-valued core count, and \E(q^c)_+\ge\E q^c>1.

We next estimate the average energy of the core slices. The localization error is O_t(1) by (4.45) at offset zero. In the attractive term of (3.6), retain only the core repulsion from particles with radius less than t/4, which are core with probability one. Conditional integration followed by averaging the original cut labels gives

This comparison of expectations uses the raw inner particles; it does not identify fields from different core slices.

Divide the nonzero terms on the right into shells u\le|x_i|<2u, u=2^\ell t, \ell\ge0. For each shell use an auxiliary cut of the original state, full out on that shell, supported in u/2<|x|<4u, with gradient cost C u^{-2}, and call its data Y. All weighted shell positions and their weights o(x_i)^2 are then recorded. Conditional integration of the raw field, with the harmonic version of Lemma 12.1, bounds the shell expectation by \E[n_u M(Y)], where n_u=\#\{u\le|x_i|<2u\}. Equations (4.45) and (12.4) give

The constants are independent of the shell index and this series is summable. For each fixed system the original attraction and repulsion expectations are finite by the form bounds, so the shell decomposition is legitimate; alternatively one can first use finitely many shells and pass to the limit. Dropping the nonnegative out kinetic energy and out–out repulsion from (3.6), and using (12.9), now proves

Every slice satisfies e_X\ge E, and on G sector monotonicity gives e_X\ge E_{n_c}\ge E_{Z-1}. Therefore

which proves (12.7). This argument has used only the zero-offset ground state and the general-offset screening estimates; no neutral offset or ionization gap was assumed.□

Proof. For Z\ge\max(Z_0,2), take a normalized ground state \Phi of H_{Z-1,Z}. It exists by Lemma 2.5, and its offset from E is bounded by Proposition 12.3. In this proof S\ge1 will be a large universal constant, fixed at the end, and b=S^{3/5}. Angle brackets denote normalized Lebesgue averaging over the centers \mathcal A_S=\{y:S\le|y|\le2S\}.

Use the real normalized radial packet g_b from Section 3. Choose smooth real functions c_y,o_y with c_y^2+o_y^2=1, c_y=0 on B(y,2b), c_y=1 outside B(y,3b), and

Set

The unnormalized core F_y\Phi has no particles in B(y,2b). Wedge this core with the normalized one-spin orbital g_b(\cdot-y); the resulting unnormalized Z-electron vector has squared norm A_y. Spatial separation makes the terms assigning different particle labels to the two groups orthogonal. It also makes its energy equal to the core energy, plus the orbital energy times A_y, plus the direct cross repulsion. In particular, with the raw-configuration field

the multiplier identity (2.4) and the variational principle give

This inequality also holds if A_y=0, since its derivation uses the unnormalized trial and its zero norm.

For every raw configuration, averaging the full insertion field gives

To verify this without a symmetry assumption on \Phi, fix the center radius r\in[S,2S]. The nuclear packet average is Z/r since b<r. For any fixed electron position z, the spherical average in y of its packet potential is

This follows by first averaging K(y+v-z) over |y|=r and then integrating v. There are Z-1 electrons, so the averaged field is at least 1/r\ge1/(2S). Radial averaging proves (13.3).

The mean number of particles near a proposed center is small. For large S, 3b<S/2, and a particle can lie in B(y,3b) for some y\in\mathcal A_S only if S/2\le|x_i|\le3S. The fraction of such centers for any particle is at most Cb^3/S^3. Fubini and (4.45), with the bounded offset of \Phi, yield

Writing p_y=1-A_y=\E_\Phi(1-F_y^2), the product inequality 1-\prod_i c_y(x_i)^2\le\sum_i(1-c_y(x_i)^2) gives p_y\le N_y.

It remains to bound the potentially positive field lost on the discarded configurations. The symmetric multiplier G_y=\sqrt{1-F_y^2} is Lipschitz, even where it vanishes, and

For a direct justification, take all products of c_y and o_y in the particle variables except the all-core product. Their Euclidean norm is G_y. The squared gradient of a norm is bounded by the sum of the squared gradients of its components; the full product partition has squared gradient sum \sum_i(|\nabla c_y|^2+|\nabla o_y|^2)(x_i). This proves (13.5) and the stated Lipschitz property.

If p_y>0, put \Phi_y^{\rm loss}=G_y\Phi/\sqrt{p_y}. This is a normalized antisymmetric form-domain vector. By (2.4), (12.7), and (13.5), its offset satisfies

Choose S large enough also that b<A S/L. At each center y in the annulus, the nucleus and every electron included in J_y lie outside the radial packet. Their packet averages equal their point potentials at y. All omitted electron terms enter with a negative sign, whence H_y^{\rm add}\le J_y pointwise in the configuration. Using (4.39) with no initial cut, now in the state \Phi_y^{\rm loss}, gives

Here we may assume S/L\ge1; L is fixed. If p_y=0, the left side is zero and the same bound holds. All expectations in this calculation are finite for a fixed system by the form bounds (the packet potential is bounded as well).

Since p_y\le N_y, (13.4) implies

Also

These estimates explain the width choice: any b=S^a with 1/2<a<2/3 would make the kinetic term and the largest averaged field loss smaller than S^{-1}. All these limits are uniform in Z and in the ground state \Phi. Average (13.2) and write F_y^2H_y^{\rm add}=H_y^{\rm add}-(1-F_y^2)H_y^{\rm add}. Equations (13.3) and (13.7) then show, for one sufficiently large universal choice of S,

In particular \langle A_y\rangle>0. As it is at most one, dividing gives E_{Z-1}-E_Z\ge1/(4S). This proves (13.1).□

Proof of Corollary 1.3. In the notation E_Z(n)=E(n,Z) of the corollary, Proposition 12.3 gives universal C_0,Z_0<\infty such that, for Z\ge Z_0,

The first inequality uses \inf_{n\ge0}E_Z(n)\le E_Z(Z). Proposition 13.1 gives universal c_0>0 and Z_2<\infty such that I_1(Z)\ge c_0 for Z\ge Z_2.

Choose an integer Z_*\ge\max\{2,Z_0,Z_2\}. For each integer 1\le Z<Z_*, the finite-sector Coulomb forms are bounded below and admit finite-energy trial states, so E_Z(Z-1) and E_Z(Z) are finite; the vacuum energy is E_Z(0)=0. Lemma 2.5, applied with n=Z, gives I_1(Z)>0. Thus all the finitely many remaining gaps lie in (0,\infty). Set

These constants satisfy 0<c\le C<\infty. They are universal because Z_* is universal and the exceptional gaps belong to finitely many atoms in the fixed normalization. The large-Z bounds and these choices prove the assertion for every integer Z\ge1.□

Proof. Fix any such ground state \Psi=\Psi_Z and write \E for its raw configuration expectation. By Proposition 12.3 its offset is bounded by a universal constant, and by Proposition 13.1 its ionization gap is at least c>0. For u\ge1, choose a smooth radial cutoff \chi_u(r)=\chi(r/u), where 0\le\chi\le1, \chi=1 on [1,2], and \chi is supported in [1/2,4]. Put

In particular m_u\le N_u. Define

Only its values on the support of \chi_u will be used, so the selected electron never occurs among these inner sources.

Testing the weak eigen-equation by S_u\Psi, and decomposing the result into its i-th weighted terms, gives a useful lower bound. For fixed x_i and spin, the other Z-1 variables are antisymmetric; their energy is at least E_{Z-1} times their squared norm. The selected electron’s kinetic pairing is

The integration by parts follows first for smooth form approximants; w_i and its first two derivatives are bounded, so it passes to the form domain. Keeping only the selected electron’s repulsion from inner sources, and discarding its other nonnegative pair terms, therefore gives

Suppose first that m_u>0. The normalized antisymmetric state

is form admissible. Indeed \sqrt{S_u} is the Euclidean norm of the vector (\chi_u(|x_i|))_i, hence is Lipschitz. Where S_u>0,

its gradient vanishes almost everywhere on its zero set. In particular this is bounded by Cu^{-2}n_{\rm adj}. Identity (2.4) shows that its offset satisfies

There is no assumed positive lower bound on m_u.

We reveal every position carrying a positive shell weight so that S_u is determined by the data. Use an auxiliary cut which is full out on u/2\le|x|\le4u, supported in u/4<|x|<8u, and has squared gradient cost at most Cu^{-2}. Let \omega denote the raw configuration, let \ell denote these cut labels, and let Y=Y(\omega,\ell) be all recorded out data. If Q(\dd\ell\mid\omega) is the conditional label kernel, use exactly this kernel for both states. Their joint laws are then

Integration in \omega includes the spin sum. Every particle with positive w_i is recorded by this cut, so

This equality includes the random labels in the transition regions: particles there have weight zero. It follows from (14.4) that

for every bounded measurable F, and then for integrable F by truncation wherever the conditional expectations are defined. For example, multiply the asserted equality by s(Y) and integrate over an arbitrary data event to check it directly. At the slice level the same fact says that \sqrt{S_u/m_u} is constant in the surviving core variables and cancels when their wavefunction is normalized. The set s=0 has zero P^*-mass and contributes nothing to the original weighted sum.

Let

On the broad annulus u/4<|x|<6u all the inner sources are separated from the evaluation point. Lemma 12.1 supplies continuous harmonic versions of these fields. Applying (14.5) first at a countable dense set of points and then using continuity makes the conditional fields for P and P^* equal on the entire broad annulus, outside one data-null set on \{s>0\}. The two data laws are equivalent on that set. Thus the equality applies at recorded random positions as well as to the positive supremum.

Use (12.5) in the state \Psi_u^*. It gives

This is an application of the screening estimate for arbitrary finite offsets, not of its small-offset refinement. In particular the cut condition is uniform even if m_u is very small: as checked in Lemma 12.1, its cell coefficients obey D_z^2/(a_zm_z)\le C for every \delta_u, because a_zm_z\ge1.

To see explicitly how the weighted count is used, introduce the data-measurable measure

Conditional integration and (14.4) now yield

The quotient may be defined as zero on s=0. This exact change of measure accounts for all shell weights, so there is no further particle-count factor in the last bound.

Substituting (14.3) into (14.7), and then into (14.2), gives

The second line uses m_u\le N_u. When m_u=0 it holds trivially without introducing a biased law. For all sufficiently large u, absorb the last term into the left side. Applying (4.45) in the original state, whose offset is uniformly bounded, gives N_u\le C(1+\sqrt u) and hence

Finally, sum over u=2^jR, j\ge0. The density gives zero mass to the intervening spheres, and \sum_{j\ge0}(2^jR)^{-1}=2/R. This proves (14.1).□

Completion of the proof of Theorem 1.2. For all sufficiently large Z, (12.11) supplies a fixed s_*>0 for which the exterior mass exceeds 1/2. Monotonicity of that mass as a function of the radius implies R(\Psi_Z)\ge s_*. Proposition 14.1 supplies a fixed large R_* for which the exterior mass is at most 1/2, and hence R(\Psi_Z)\le R_*.

For clarity we give uniform arguments over the entire ground-state space for each of the finitely many remaining charges. Fix such a Z. The strict gap E_Z<E_{Z-1} and the compactness conclusion in Lemma 2.5 show that every sequence of normalized neutral ground states has a strongly L^2 convergent subsequence. In particular their configuration laws are uniformly tight. To see directly why compactness gives the required uniform tail, suppose otherwise and choose radii R_k\to\infty and normalized ground states \Psi_k with

After a subsequence \Psi_k\to\Psi strongly in L^2. The bounded multiplication operator \sum_i\1_{\{|x_i|>R_k\}} has norm at most Z, so its expectations in \Psi_k and \Psi differ by at most 2Z\|\Psi_k-\Psi\|_2. Its expectation in the fixed \Psi tends to zero by dominated convergence, a contradiction. Thus a finite upper radius works uniformly for this Z.

For the lower bound, the kinetic estimate (3.3), with the fixed finite offset E_Z-E, gives

uniformly over its normalized ground states. Hölder’s inequality therefore yields

Choose a positive r_Z making this less than Z-1/2. The exterior mass then exceeds 1/2, uniformly over the ground states, so R(\Psi_Z)\ge r_Z.

Taking the minimum of s_* and these finitely many positive lower radii, and the maximum of R_* and the finitely many upper radii, gives universal 0<c\le C<\infty. All density measures used above are absolutely continuous. Their exterior masses are consequently continuous in the radius, start at Z, and tend to zero. Thus the infimum defining the half-electron radius has the stated bounds for every normalized neutral ground state, as required.□