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Intervals and source layers

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

The estimate (4.1) of A localized virial identity compares the boundary energies \(W_f,W_g\) with the radial pressure \(\int w(x)D(x,y)\,d\pi\). OpenAI's proof [OpenAI-LE, Sections 4 and 5] slices space into lines parallel to a direction \(e\), and splits the sources on each line into the intervals where they exceed a level \(t\) (the layers). This lesson prepares the three ingredients: an inequality for the Newton kernel on two parallel lines (Section 1), weights that restrict the direction \(e\) to a small cap around the radial direction (Section 2), and weights on the layer intervals together with an estimate of what they lose (Section 3). The next lesson, Pressure and the energy bound, combines them.

We keep the notation of the two previous lessons: a solution with \(n\ge3\), \(A,B>-2\), \(pq>1\); its sources \(f,g\); \(u=K*g\) and \(v=K*f\); the interaction measure \(\pi\); the functions \(d,\eta,w\) and the exponents \(m=n+3\), \(k=n(m+1)\), \(N=k+n+2\); the energies \(E=E_f+E_g\); for close pairs, \(\ell(z)=\delta d(z)^m\). We write \(\mathbb E_e\) and \(\mathbb P_e\) for the average and the probability over \(e\) with respect to the normalized surface measure \(\sigma/|S|\) on the unit sphere \(S\).

1. Two parallel lines

Fix a unit vector \(e\) and two points \(z\neq z'\) of the hyperplane \(e^\perp\), and write points of the two lines as \(x=z+se\), \(y=z'+s'e\). Let \(\lambda=|z-z'|>0\) and \(h=s-s'\). The Newton kernel restricted to the two lines and its directional factor are \[ \begin{aligned} &\mathcal K(s,s')=K(x-y)\\ &\quad =c_n\bigl(\lambda^2+h^2\bigr)^{(2-n)/2},\\ &\mathcal D(s,s')=D_e(x,y)=\frac{h^2}{\lambda^2+h^2}. \end{aligned} \] For a bounded open interval \(I=(s_-,s_+)\) let \(\overline{\mathcal K}_I(s')=\frac12\bigl(\mathcal K(s_-,s')+\mathcal K(s_+,s')\bigr)\), the average of the kernel at the two endpoints, and define \(\overline{\mathcal K}_J(s)\) for a bounded interval \(J\) of the second line in the same way.

Lemma 1.1 (interval pairs; OpenAI). Let \(I,J\subseteq\mathbb R\) be nonempty open intervals, possibly unbounded, and let \(c_I,c_J\ge0\), with \(I\) bounded if \(c_I>0\) and \(J\) bounded if \(c_J>0\). Then the inequality (1.1) holds: \[ \begin{aligned} &\int_{I\times J}(c_I+c_J)\,\mathcal K\,ds\,ds'\\ &\quad\le\int_{I\times J}\Bigl[c_I\overline{\mathcal K}_I+c_J\overline{\mathcal K}_J\\ &\qquad+\bigl((n-2)\min(c_I,c_J)\,\mathcal D\\ &\qquad\quad+|c_I-c_J|\bigr)\mathcal K\Bigr]ds\,ds', \end{aligned} \] where all terms are nonnegative, the integrals may be infinite, and a term with coefficient \(0\) is \(0\).

Proof. If at most one coefficient is positive, then \(|c_I-c_J|=c_I+c_J\), and (1.1) holds pointwise. Let both be positive; then \(I\) and \(J\) are bounded and all integrals are finite. Let \(m_I,m_J\) be the midpoints, \(d_0=m_I-m_J\), and \(L=|I|\), \(M=|J|\). Integration by parts in \(s\) gives \[ \int_I(\mathcal K-\overline{\mathcal K}_I)\,ds=-\int_I(s-m_I)\,\partial_s\mathcal K\,ds, \] because \[ \begin{aligned} &\int_I(s-m_I)\partial_s\mathcal K\,ds\\ &\quad=\frac L2\bigl(\mathcal K(s_+,s')+\mathcal K(s_-,s')\bigr)\\ &\qquad-\int_I\mathcal K\,ds. \end{aligned} \] With \(\partial_s\mathcal K=-(n-2)\frac h{\lambda^2+h^2}\mathcal K=-\partial_{s'}\mathcal K\) and the same identity in \(s'\), the identity (1.2) holds: \[ \begin{aligned} &\int_{I\times J}\bigl(2\mathcal K-\overline{\mathcal K}_I-\overline{\mathcal K}_J\bigr)\\ &\quad=(n-2)\int_{I\times J}\frac{h^2-d_0h}{\lambda^2+h^2}\,\mathcal K . \end{aligned} \] The image of \(ds\,ds'\) on \(I\times J\) under \(h=s-s'\) has density \(\rho(h)=|I\cap(J+h)|\), the overlap of two intervals of lengths \(L,M\) whose midpoints are \(|h-d_0|\) apart; so \(\rho(h)=F(|h-d_0|)\) with \(F\) nonincreasing. If \(d_0\ge0\) and \(h>0\), then \(|h-d_0|\le|h+d_0|\), so \(\rho(h)\ge\rho(-h)\); the reverse holds if \(d_0\le0\). Since \(h\mapsto h\mathcal K/(\lambda^2+h^2)\) is odd, pairing \(h\) with \(-h\) gives \(d_0\int_{I\times J}\frac h{\lambda^2+h^2}\mathcal K\ge0\). Hence the left side of (1.2) is at most \((n-2)\int_{I\times J}\mathcal D\mathcal K\).

Let \(c=\min(c_I,c_J)\) and multiply this bound by \(c\). Since the endpoint averages are nonnegative, \(\int_{I\times J}(\mathcal K-\overline{\mathcal K}_I)\le\int_{I\times J}\mathcal K\), and likewise for \(J\); multiply these by \(c_I-c\) and \(c_J-c\), whose sum is \(|c_I-c_J|\). Adding the three inequalities and moving the finite endpoint terms to the right gives (1.1). \(\square\)

The formulation without subtraction matters for \(n=3\): there, \(\int\mathcal K\) over an unbounded interval of the other line diverges, and Lemma 1.1 never subtracts such an integral.

2. Weights on caps of directions

For \(\theta\in S\) and \(r>0\) let \(C_r(\theta)=\{e\in S:|e-\theta|<r\}\). The surface measure is invariant under orthogonal maps (as Lebesgue measure is), so \(\sigma(C_r(\theta))\) does not depend on \(\theta\).

Lemma 2.1 (caps). There are \(0<c\le C\), depending only on \(n\), with \(c\,r^{n-1}\le\sigma(C_r(\theta))\le C\,r^{n-1}\) for \(0<r\le1\).

Proof. By polar coordinates, the truncated cone \(T_r=\{x:0<|x|<1,\ x/|x|\in C_r(\theta)\}\) has volume \(\sigma(C_r(\theta))/n\). If \(x\in T_r\), the distance from \(x\) to the line \(\mathbb R\theta\) is at most \(|x|\,|x/|x|-\theta|<r\), so \(T_r\) lies in a cylinder of radius \(r\) and length \(2\) around that line, of volume \(Cr^{n-1}\). Conversely, every \(x=s\theta+y\) with \(\frac12\le s\le\frac34\), \(y\perp\theta\) and \(|y|\le r/4\) has \(|x|<1\) and \(|x/|x|-\theta|^2=2-2s/\sqrt{s^2+|y|^2}\), which is at most \(|y|^2/s^2\le r^2/4\); these points form a set of volume \(cr^{n-1}\) inside \(T_r\). \(\square\)

Fix a smooth \(\psi\colon[0,\infty)\to[0,1]\), equal to \(1\) on \([0,\frac12]\) and to \(0\) on \([1,\infty)\), and for \(0<l\le1\) let \[ Z(l)=\mathbb E_e\,\psi\bigl(|e-\theta|/l\bigr), \] which does not depend on \(\theta\). By Lemma 2.1, \(cl^{n-1}\le Z(l)\le Cl^{n-1}\); differentiating under the integral sign, \(|Z'(l)|\le\sup|\psi'|\,l^{-1}\,\mathbb P_e\bigl(C_l(\theta)\bigr)\le Cl^{n-2}\). Fix \(0<\kappa<1\) and define the cap weights \[ \begin{gathered} h_e(x)=w(x)\,\frac{\psi\bigl(|e-\omega_x|/(\kappa d(x))\bigr)}{Z(\kappa d(x))}\\ (1<|x|<2),\\ h_e(x)=0\quad\text{otherwise}. \end{gathered} \]

Lemma 2.2 (cap weights). For every unit vector \(e\):

  1. \(\mathbb E_eh_e(x)=w(x)\) for every \(x\);
  2. \(0\le h_e(x)\le C_\kappa d(x)^{N-n}\), and \(h_e(x)>0\) only if \(1<|x|<2\) and \(|e-\omega_x|<\kappa d(x)\);
  3. \(h_e\) is continuously differentiable on \(\mathbb R^n\), jointly continuous in \((e,x)\), and \(|\nabla h_e(x)|\le C_\kappa d(x)^{N-n-1}\).

Proof. 1 follows from the definition of \(Z\). 2: \(w\le Cd^{N-1}\) and \(Z(\kappa d)\ge c\kappa^{n-1}d^{n-1}\). 3: On the annulus, differentiating the profile costs at most \(C_\kappa/d\) on its support, since \(|\nabla\omega_x|\le1\), \(|\nabla d|=1\), \(|e-\omega_x|\le\kappa d\) there, and the profile is constant where \(|e-\omega_x|<\kappa d/2\); differentiating \(1/Z(\kappa d)\) costs the factor \(\kappa|Z'(\kappa d)|/Z(\kappa d)\le C/d\); and \(|\nabla w|\le Cd^{N-2}\). This gives the bound. Near \(|x|=1\), \(d\) is smooth and \(w\) vanishes to second order, so \(h_e\) and \(\nabla h_e\) tend to \(0\); near \(|x|=2\), \(d^{N-n}\) and \(d^{N-n-1}\) tend to \(0\). So the extension by \(0\) is continuously differentiable. \(\square\)

Lemma 2.3 (oscillation of the cap weights). Let \(0<\delta<\frac14\). For a close pair \((x,y)\) with endpoint \(z\) (as in Lemma 3.1 of A localized virial identity), \[ \begin{aligned} &|h_e(x)-h_e(y)|\\ &\quad \le C_\kappa\,\delta\,d(z)^{N+2}\\ &\quad \le C_\kappa\,\delta\,d(z)^N, \end{aligned}\tag{2.1} \] and for every solution \[ \begin{aligned} &\int|h_e(x)-h_e(y)|\,d\pi(x,y)\\ &\quad \le C_\kappa\,\delta\,E+C_{\kappa,\delta}. \end{aligned}\tag{2.2} \]

Proof. On the segment from \(x\) to \(y\), \(d\) is comparable to \(d(z)\), so Lemma 2.2(3) gives \(|h_e(x)-h_e(y)|\le C_\kappa d(z)^{N-n-1}\ell(z)\), which is \(C_\kappa\delta d(z)^{N-n-1+m}\), and \(N-n-1+m=N+2\). Integrating against \(\pi\) and using \(d(z)^N\le\eta(x)+\eta(y)\), the close pairs contribute at most \(C_\kappa\delta E\). For far pairs, \(|h_e(x)-h_e(y)|\le h_e(x)+h_e(y)\); the term \(h_e(x)\) is nonzero only if \(d(x)>0\), when \(|x-y|>\ell(x)\), and the tail estimate (4.2) of Positive solutions and their potentials bounds its integral by \[ \begin{aligned} &C_\kappa\int_{B_2}f(x)\,d(x)^{N-n}\,\ell(x)^{2-n}\,dx\\ &\quad \le C_\kappa\delta^{2-n}\int_{B_2}f\,d^{\,N-n-m(n-2)}\le C_{\kappa,\delta}, \end{aligned} \] since \(N-n-m(n-2)=3n+8>0\). The term \(h_e(y)\) is treated in the same way with \(g\). \(\square\)

Lemma 2.4 (cap pressure). For every solution, the bound (2.3) holds: \[ \begin{aligned} &\mathbb E_e\int h_e(x)D_e(x,y)\,d\pi(x,y)\\ &\quad\le\int w(x)D(x,y)\,d\pi(x,y)+C\kappa E, \end{aligned} \] with \(C\) independent of \(\kappa\).

Proof. For a unit vector \(\nu\), \(|(\nu\cdot e)^2-(\nu\cdot\omega_x)^2|\le2|e-\omega_x|\), so \(|D_e-D|\le2\kappa d(x)\) where \(h_e(x)>0\). Averaging over \(e\) with Tonelli's theorem and Lemma 2.2(1), the left side of (2.3) is at most \(\int w(x)\bigl(D+2\kappa d(x)\bigr)\,d\pi\), and \(dw\le C\eta\) gives \(2\kappa\int w(x)d(x)\,d\pi\le C\kappa E_f\). \(\square\)

3. Layer components and their weights

Fix a unit vector \(e\). With \(f(0)=0\) and \(f\) continuous on \(\mathbb R^n\setminus\{0\}\), the set \(\{f>t\}\) is open for every \(t>0\) and does not contain \(0\). For \(x\neq0\) and \(0<t<f(x)\), let \(I=I_f(e,x,t)\) be the connected component containing \(x\) of the intersection of \(\{f>t\}\) with the line \(x+\mathbb Re\); it is an open interval of that line. Give it the weight \(\chi_I=\inf_Ih_e\) if \(I\) is bounded and \(|I|\le2\delta\inf_Id\), and \(\chi_I=0\) otherwise, and write \(\chi_f(e,x,t)=\chi_I\). Define \(J=I_g(e,y,t')\) and \(\chi_g(e,y,t')=\chi_J\) in the same way from \(\{g>t'\}\). Since \(x\in I\), we have \(0\le\chi_f(e,x,t)\le h_e(x)\), and likewise \(0\le\chi_g(e,y,t')\le h_e(y)\).

Lemma 3.1 (measurability). On \(\{(e,x,t):x\neq0,\ 0<t<f(x)\}\), the two distances from \(x\) to the ends of \(I_f(e,x,t)\) (possibly infinite), the ends themselves when finite, and \(\chi_f\) are Borel functions of \((e,x,t)\). The same holds for \(g\).

Proof. The set \(\mathcal L=\{(x,t):x\neq0,\ 0<t<f(x)\}\) is open. The forward distance \(\tau_+(e,x,t)\), the supremum of the \(r\ge0\) with \(x+se\in\{f>t\}\) for \(0\le s\le r\), exceeds \(r\) exactly when, for some rational \(r'>r\), the compact segment from \(x\) to \(x+r'e\) lies in the open layer; that is an open condition in \((e,x,t)\). So \(\tau_+\) is lower semicontinuous, hence Borel, and so is the backward distance \(\tau_-\). The infima of the continuous functions \(h_e\) and \(d\) over the open interval \(I\) equal their infima over the points \(x+se\) with rational \(s\in(-\tau_-,\tau_+)\), which are countable infima of Borel functions. \(\square\)

For each direction define the losses \[ \begin{aligned} &L_f(e)=\int u(x)\int_0^{f(x)}\bigl(h_e(x)\\ &\qquad-\chi_f(e,x,t)\bigr)\,dt\,dx,\\ &L_g(e)=\int v(y)\int_0^{g(y)}\bigl(h_e(y)\\ &\qquad-\chi_g(e,y,t')\bigr)\,dt'\,dy . \end{aligned} \] They are nonnegative, and finite because \(h_e\) is bounded and vanishes outside the annulus \(1\le|x|\le2\).

Lemma 3.2 (averaged loss; OpenAI). For every \(P\ge1\) and every solution, \[ \begin{aligned} &\mathbb E_e\bigl(L_f(e)+L_g(e)\bigr)\\ &\quad \le\bigl(C_\kappa\delta+C_{\kappa,\delta}P^{-1}\bigr)E\\ &\quad +CP, \end{aligned}\tag{3.1} \] where \(C_\kappa\) does not depend on \(\delta\) or \(P\), \(C_{\kappa,\delta}\) does not depend on \(P\), and \(C\) does not depend on \(\kappa,\delta,P\).

Proof. We treat \(f\); the proof for \(g\) is the same. Only \(1<|x|<2\) contributes. Fix such an \(x\), and let \(d=d(x)\) and \(\ell=\delta d^m\).

Short components. If \(|I|\le\ell\), every point of \(I\) is within \(\ell\le\delta d\) of \(x\), so \(\inf_Id\ge(1-\delta)d\) and \(|I|\le\delta d\le2\delta\inf_Id\); thus \(\chi_I=\inf_Ih_e\), and by (2.1) the loss \(h_e(x)-\chi_I\) is at most \(C_\kappa\delta d^N\).

Long components. If \(|I|>\ell\), then one of the two parts of \(I\) on either side of \(x\) contains a segment of length \(\ell/2\) starting at \(x\). Let \(F\) be the set of \(\theta\in S\) for which the segment from \(x\) to \(x+(\ell/2)\theta\) lies in \(\{f>t\}\). By polar coordinates about \(x\), \[ \begin{aligned} \int_{B_{\ell/2}(x)}f&\ge\int_0^{\ell/2}r^{n-1}\int_Ft\,d\sigma\,dr\\ &=\frac{|S|}n\,t\Bigl(\frac\ell2\Bigr)^n\,\mathbb P_e(F). \end{aligned} \] The event \(|I|>\ell\) is contained in \(\{e\in F\}\cup\{-e\in F\}\), and \(B_{\ell/2}(x)\subseteq B_4\), so by (4.2) of Positive solutions and their potentials \[ \begin{aligned} &\mathbb P_e\bigl(|I|>\ell\bigr)\le2\,\mathbb P_e(F)\\ &\quad \le\frac C{t\,\ell^n}\int_{B_4}f\le\frac C{t\,\ell^n}. \end{aligned}\tag{3.2} \]

High levels. Let \(t\ge Pd^{-k}\). On long components the loss is at most \(h_e(x)\le C_\kappa d^{N-n}\), so by (3.2) its average over \(e\) is at most \(C_\kappa d^{N-n}\cdot C\delta^{-n}d^{-mn}P^{-1}d^k\), which is \(C_{\kappa,\delta}P^{-1}d^N\), because \(k=mn+n\). Adding the short components, the average loss at level \(t\) is at most \((C_\kappa\delta+C_{\kappa,\delta}P^{-1})d^N\). Integrating over these levels, then against \(u(x)\,dx\), and using \(d^N\le\eta\), gives at most \((C_\kappa\delta+C_{\kappa,\delta}P^{-1})E_f\).

Low levels. Since \(\mathbb E_eh_e(x)=w(x)\le Cd^{N-1}\), \[ \begin{aligned} &\int_0^{\min\{f(x),Pd^{-k}\}}\mathbb E_e\bigl(h_e(x)\\ &\qquad-\chi_f(e,x,t)\bigr)\,dt\\ &\quad\le Pd^{-k}w(x)\le CPd^{N-1-k}\\ &\quad=CPd^{n+1}\le CP, \end{aligned} \] and integrating against \(u\) over \(B_2\) gives at most \(CP\) by (4.2) of Positive solutions and their potentials. Averages over \(e\) and integrals over \((x,t)\) were exchanged by Tonelli's theorem, which applies by Lemma 3.1. \(\square\)

4. Exercises

Exercise 4.1 (easy). Check the identity \(\int_I(s-m_I)\,\partial_s\mathcal K\,ds=L\,\overline{\mathcal K}_I-\int_I\mathcal K\,ds\) used in the proof of Lemma 1.1.

Exercise 4.2 (easy). Show that \(|I\cap(J+h)|\) is the largest of \(0\) and \(\min\{L,M,\frac{L+M}2-|h-d_0|\}\) for intervals of lengths \(L,M\) with midpoint difference \(d_0\).

Exercise 4.3 (medium). In Lemma 1.1 take \(n=3\), \(c_I=c_J=1\) and \(I=J=(0,1)\). Show that the left side of (1.2) is at most \((n-2)\int\mathcal D\mathcal K\) by computing \(d_0\).

Exercise 4.4 (medium). Show that \(|(\nu\cdot e)^2-(\nu\cdot\omega)^2|\le2|e-\omega|\) for unit vectors \(\nu,e,\omega\).

5. Solutions

4.1. Integrate by parts: \[ \begin{aligned} &\int_{s_-}^{s_+}(s-m_I)\partial_s\mathcal K\,ds\\ &\quad=\bigl[(s-m_I)\mathcal K\bigr]_{s_-}^{s_+}-\int_I\mathcal K\,ds, \end{aligned} \] and \(s_\pm-m_I=\pm L/2\), so the bracket is \(\frac L2\bigl(\mathcal K(s_+,s')+\mathcal K(s_-,s')\bigr)=L\,\overline{\mathcal K}_I\).

4.2. \(I\cap(J+h)\) is the intersection of two intervals of lengths \(L\) and \(M\) whose midpoints are \(m_I\) and \(m_J+h\), at distance \(|d_0-h|\). The intersection is empty when the distance is at least \(\frac{L+M}2\), and otherwise has length \(\frac{L+M}2-|d_0-h|\), capped by the length of the shorter interval.

4.3. Here \(d_0=0\), so the term \(d_0h\) in (1.2) vanishes and (1.2) is the claimed bound with equality.

4.4. \((\nu\cdot e)^2-(\nu\cdot\omega)^2\) is the product of \(\nu\cdot(e-\omega)\) and \(\nu\cdot(e+\omega)\), and \(|\nu\cdot(e-\omega)|\le|e-\omega|\), \(|\nu\cdot(e+\omega)|\le2\).

References