Proof. Let (\ell_i,r_i) be that component, set s=d_i-\ell_i, u=r_i-d_i, T=s+u, and w=s/T. After removing the self-pole, the background H_i(x)=k\log x+\sum_{j\ne i}q_j\log|x-d_j| is well defined and strictly concave on (\ell_i,r_i). To justify the first assertion, for 0<x<1 we have directly W(x)\le k\log x+\log(2-x)<0. Thus the main component contains (0,1), while atomization puts the residual cluster d_i in a different component; hence \ell_i\ge1>0. The second assertion follows by differentiating twice after the self-pole has been removed.

The boundary equations and (2.3) are H_i(\ell_i)=-q_i\log s,\quad H_i(r_i)=-q_i\log u,\quad H_i(d_i)=-q_i\log R_i. Since d_i=(1-w)\ell_i+wr_i, concavity implies -q_i\log R_i \ge-(1-w)q_i\log s-wq_i\log u. Division by -q_i<0 reverses the inequality, and exponentiation gives R_i\le s^{1-w}u^w =T w^{1-w}(1-w)^w\le T/2. For completeness, if h(w)=(1-w)\log w+w\log(1-w), then h'(w)=\log((1-w)/w)+(1-2w)/(w(1-w)) is positive on (0,1/2) and changes sign by symmetry at 1/2; hence w^{1-w}(1-w)^w\le1/2. Thus T\ge2R_i. The main component and all residual components are pairwise disjoint, so adding their lengths proves (2.4). ◻

Proof. For fixed k\ge1, the function r\mapsto r^k(2+r) is continuous and strictly increasing on [0,1], with values 0 and 3 at the endpoints. This proves the existence and uniqueness of \rho(k). For 0<x<1, since d_i-x\le2-x, W(x)\le k\log x+\log(2-x)\le\log(x(2-x))<0. For 0<r<\rho(k), W(-r)\le k\log r+\log(2+r)<0. Thus (-\rho(k),1) lies in the main component. At k=1, \rho(1)=\sqrt2-1; for fixed 0<\rho<1, the left side of the defining equation for \rho(k) decreases as k increases, so its solution increases. ◻

Proof. For 0<y<d_1, logarithmic differentiation of (4.8) gives \frac{\Psi_T'(y)}{\Psi_T(y)} =\frac1y-\frac1k\sum_i\frac{q_i}{d_i-y}=:G(y). Now G'(y)=-\frac1{y^2}-\frac1k\sum_i\frac{q_i}{(d_i-y)^2}<0, while G(y)\to+\infty as y\downarrow0 and G(y)\to-\infty as y\uparrow d_1. Hence G has exactly one zero y_c; the positive function \Psi_T increases on (0,y_c) and decreases on (y_c,d_1), tending to zero at both ends.

For y<0, the same logarithmic derivative is strictly negative, so \Psi_T'(y)>0 because \Psi_T(y)<0. Moreover \Psi_T(y)\to-\infty as y\to-\infty and \Psi_T(y)\to0 as y\uparrow0. Thus \Psi_T(y)=-R(T) has exactly one negative solution.

By atomization, the component containing zero and the component containing d_1 are distinct. Therefore some point of (0,d_1) has W_T\ge0. The displayed relation between W_T, \Psi_T, and R(T) forces R(T)\le\max_{(0,d_1)}\Psi_T=\Psi_T(y_c). If the inequality is strict, \Psi_T=R(T) has two positive solutions and W_T>0 precisely between them; the first is the right endpoint of the main component. If equality holds, the solutions coalesce at y_c and W_T<0 on both adjacent punctured intervals, so y_c is their common boundary point. Together with the unique negative solution, this proves the assertions about the endpoints and the inverse branch. ◻

Proof. We prove the global supporting inequality, rather than assume a formal first variation. For a finite residual measure \sum_iq_i\delta_{d_i} put \alpha_i=q_i/k, so \sum_i\alpha_i=1/k. Formula (4.8) becomes \Psi(y)=y\prod_i(1-y/d_i)^{\alpha_i},\qquad R=\prod_i d_i^{-\alpha_i}. For real y<d_1, all factors 1-y/d_i are positive, and \frac1kW(y)=\log\frac{|\Psi(y)|}{R}. By Lemma 4.1, the two main endpoints are \Phi(-R) and \Phi(R), with the stated contact convention. Lagrange inversion gives \begin{equation} \Phi(z)=\sum_{n\ge1}a_nz^n,\qquad a_n=\frac1n[y^{n-1}]\prod_i(1-y/d_i)^{-n\alpha_i}. \label{eq:4.14} \tag{4.10} \end{equation} Consequently \begin{equation} M_k=\Phi(R)-\Phi(-R)=2\sum_{m\ge0}a_{2m+1}R^{2m+1}. \label{eq:4.15} \tag{4.11} \end{equation} More explicitly, \begin{equation} a_nR^n=\frac1n \sum_{r_1+\cdots+r_N=n-1} \prod_i\frac{(n\alpha_i)_{r_i}}{r_i!} d_i^{-(n\alpha_i+r_i)}. \label{eq:4.16} \tag{4.12} \end{equation} Here (z)_0=1 and (z)_m=z(z+1)\cdots(z+m-1) for m\ge1 is the rising Pochhammer symbol. Every coefficient is nonnegative. If m(d)=\prod_i d_i^{-\gamma_i} with \gamma_i>0, then for every vector v, \frac{v^T(\nabla^2m)v}{m} =\left(\sum_i\frac{\gamma_iv_i}{d_i}\right)^2 +\sum_i\frac{\gamma_iv_i^2}{d_i^2}\ge0. Thus every finite odd partial sum in (4.11) is convex in the positive coordinates. Repeating coordinates according to multiplicity gives the same statement on every common equal-mass quantile partition.

For completeness, we now pass from finite coordinate vectors to the quantiles used below. The masses q_i of an empirical target are rational. Choose a common equal-mass partition which refines all of its constant blocks, sample the reference quantile T_0 on the same cells, and repeat each target value on the corresponding cells. To see explicitly that the changing dimension causes no problem, define m_\ell(T)=\int_0^1T(u)^{-\ell}\,du, \qquad R(T)=\exp\left(-\frac1k\int_0^1\log T(u)\,du\right). The product in the coefficient formula (4.10) has the identity \prod_i(1-y/d_i)^{-nq_i/k} =\exp\left(\frac nk\sum_{\ell\ge1} \frac{m_\ell(T)}\ell y^\ell\right). Thus, for fixed n, a_n is a finite polynomial in m_1,\ldots,m_{n-1}. Under common-partition approximation these moments, R, and their directional derivatives converge, because T\ge1 and Dm_\ell(T)[v]=-\ell\int_0^1v(u)T(u)^{-\ell-1}\,du, \qquad DR(T)[v]=-\frac{R(T)}k\int_0^1\frac{v(u)}{T(u)}\,du. It follows that every fixed odd truncation and its directional derivative converge as the mesh tends to zero.

Let S_m denote the sum of the terms in (4.11) through degree 2m+1. Choose equal-mass step approximations T_{0,N} to T_0, with N a multiple of the common denominator of the target masses. Repeat each target value on the same N cells. Finite-dimensional convexity gives S_m(T)\ge S_m(T_{0,N})+DS_m(T_{0,N})[T-T_{0,N}]. The displayed moment and derivative formulas are dominated uniformly by T_{0,N}\ge1 and |T-T_{0,N}|\le1. Hence, for fixed m, dominated convergence as N\to\infty yields S_m(T)\ge S_m(T_0)+DS_m(T_0)[T-T_0]. At a strictly separated reference, R lies strictly inside the first positive critical value of the inverse branch. To justify that no nearer complex singularity intervenes, recall Pringsheim’s theorem [8, Theorem IV.6]: a power series with nonnegative coefficients and finite radius of convergence is singular at the positive point equal to that radius. The coefficients a_n in (4.10) are nonnegative, while the positive real inverse branch is analytic up to its first critical value. Thus the reference level is strictly inside the disk of convergence. On a smaller closed disk, the inverse series and its material derivative converge locally uniformly by the analytic implicit-function theorem. Hence the series for both S_m(T_0) and DS_m(T_0) converge absolutely. Letting m\to\infty in the preceding finite-truncation inequality gives the result when the target is strictly separated. Indeed, the same Pringsheim argument applies to the target: Lemma 4.1 puts R(T) strictly below its first positive critical value, so its inverse series converges at \pm R(T). Thus its nonnegative odd series converges to the exact inverse-branch width, while the reference value and derivative converge absolutely.

For a target at contact and 0<\lambda<1, define the Abel width M_{k,\lambda}(T)=2\sum_{m\ge0}a_{2m+1}(T) [\lambda R(T)]^{2m+1}. For each finite odd truncation, convexity follows because its monomials are precisely those in (4.12), multiplied by positive constants \lambda^{2m+1}. Apply the common-partition argument to that truncation and then let its degree tend to infinity. Monotone convergence applies on the target side; at the strictly separated reference, both the value and directional derivative converge absolutely. Therefore M_{k,\lambda}(T)\ge M_{k,\lambda}(T_0) +DM_{k,\lambda}(T_0)[T-T_0]. As \lambda\uparrow1, the target value increases to M_k(T): for \lambda<1 it equals \Phi(\lambda R)-\Phi(-\lambda R), and the two real inverse branches converge monotonically to the two level-R endpoints. This is precisely the monotone form of Abel’s boundary theorem, and it also follows directly from monotone convergence of the nonnegative odd series. Strict separation gives absolute convergence of the reference value and material derivative, so their limits may be taken term by term. Letting \lambda\uparrow1 proves (4.9) in the contact case. ◻

Proof. Write d=c-r\cos\theta, 0\le\theta\le\pi, so \,de_I=\,d\theta/\pi. Let A(d)=k+1-\frac{k\sqrt{2a}}d,\qquad a_\pi=A(2)>0. For j\in\{-,+\} there is a unique 0<\rho_j<1 with x_j=c-\frac r2(\rho_j+\rho_j^{-1}). Then K_j=\frac{r(1-\rho_j^2)}{2\rho_j},\qquad P_\rho(\theta)=\frac{1-\rho^2}{1-2\rho\cos\theta+\rho^2} =1+2\sum_{n\ge1}\rho^n\cos(n\theta), and K_j/(d-x_j)=P_{\rho_j}(\theta). Thus (4.7) is \begin{equation} d\xi=B(\theta)\frac{d\theta}{\pi},\qquad B(\theta)=D-\sum_j\sigma_jP_{\rho_j}(\theta), \label{eq:4.18} \tag{4.13} \end{equation} with D=\sum_j\sigma_jP_{\rho_j}(0). This also proves B(0)=0 and B(\theta)>0 for 0<\theta\le\pi.

First assume that the spatial material velocity is smooth and put F(d)=A(d)v(d). Differentiating the preceding definition of g under the integral gives g(d)=\frac{k v(d)}d+\operatorname {pv}\int_I \frac{v(d)-v(e)}{d-e}\,d\eta(e). Differentiating (4.5) with respect to its spatial variable gives \frac kd+\operatorname {pv}\int_I\frac1{d-e}\,d\eta(e)=0. Multiplying this identity by v(d) and subtracting it from the preceding display cancels every term containing v(d) and gives g(d)=\operatorname {pv}\int_I\frac{v(e)}{e-d}\,d\eta(e) =\operatorname {pv}\int_I\frac{F(e)}{e-d}\,de_I(e). At the two exterior crossings, ordinary differentiation and implicit differentiation give \dot M_k=-\sum_j\sigma_j\int_I\frac{F(d)}{d-x_j}\,de_I(d).

Expand F(c-r\cos\theta)=f_0+2\sum_{n\ge1}f_n\cos(n\theta), first as a finite cosine polynomial. Here U_n denotes the Chebyshev polynomial of the second kind, characterized by U_n(\cos\theta)=\sin((n+1)\theta)/\sin\theta. The finite Hilbert transform identity \operatorname {pv}\frac1\pi\int_0^\pi \frac{\cos(n\phi)}{\cos\theta-\cos\phi}\,d\phi =-U_{n-1}(\cos\theta) follows from the zero transform of the constant function, the case n=1, and the cosine recurrence. Moreover \frac1\pi\int_0^\pi P_\rho(\theta)U_{n-1}(\cos\theta)\,d\theta =\begin{cases} \dfrac{1+\rho^2-2\rho^{n+1}}{1-\rho^2},&n\text{ odd},\\[2mm] \dfrac{2\rho(1-\rho^n)}{1-\rho^2},&n\text{ even}. \end{cases} This is obtained by inserting U_{2m}=1+2\sum_{\ell=1}^m\cos(2\ell\theta) and U_{2m-1}=2\sum_{\ell=1}^m\cos((2\ell-1)\theta). We spell out the remaining coefficient calculation. Put \varepsilon_n=\frac1\pi\int_0^\pi U_{n-1}(\cos\theta)\,d\theta =\begin{cases}1,&n\text{ odd},\\0,&n\text{ even},\end{cases} and let S_n(\rho) denote the preceding Poisson integral. The angular density for \xi, the Hilbert-transform representation of g, and the preceding finite-transform identity give \int_Ig\,d\xi=-\frac2r\sum_{n\ge1}f_n \left(D\varepsilon_n-\sum_j\sigma_jS_n(\rho_j)\right). Poisson orthogonality and the displayed first-variation formula give, on the other hand, \dot M_k=-\sum_j\frac{\sigma_j}{K_j} \left(f_0+2\sum_{n\ge1}f_n\rho_j^n\right). Set \Gamma=\sum_j\frac{\sigma_j}{K_j} =\frac2r\sum_j\frac{\sigma_j\rho_j}{1-\rho_j^2}. The coefficient of f_0 in the difference is -\Gamma. If n is even, its coefficient is \begin{align*} &-\frac4r\sum_j\frac{\sigma_j\rho_j^{n+1}}{1-\rho_j^2} -\frac4r\sum_j\frac{\sigma_j\rho_j(1-\rho_j^n)}{1-\rho_j^2}\\ &\hspace{35mm}=-\frac4r\sum_j \frac{\sigma_j\rho_j}{1-\rho_j^2}=-2\Gamma. \end{align*} If n is odd, use D=\sum_j\sigma_jP_{\rho_j}(0) =\sum_j\sigma_j\frac{1+2\rho_j+\rho_j^2}{1-\rho_j^2} and the odd case of the Poisson integral above. The coefficient becomes \frac{2D}{r}-\frac2r\sum_j\sigma_j \frac{1+\rho_j^2}{1-\rho_j^2} =\frac4r\sum_j\frac{\sigma_j\rho_j}{1-\rho_j^2}=2\Gamma. Finally d(\pi)=2, and therefore F(2)=f_0+2\sum_{n\ge1}(-1)^nf_n. The three coefficient formulas are exactly those of -\Gamma F(2). Thus \begin{equation} \boxed{\dot M_k-\int_Ig\,d\xi =-F(2)\sum_j\frac{\sigma_j}{K_j} =-a_\pi v(2)\left(\frac{\sigma_-}{K_-} +\frac{\sigma_+}{K_+}\right).} \label{eq:4.23} \tag{4.14} \end{equation}

For an atomic target, v is piecewise C^1 in the reference coordinate, with finitely many jumps. Apply (4.14) to the Poisson–Abel regularization F_\lambda=f_0+2\sum\lambda^nf_n\cos(n\theta). For fixed \lambda<1, the identity first holds for finite cosine truncations and then for F_\lambda by absolute convergence. As \lambda\uparrow1, the exterior integrals converge by domination because d-x_\pm is bounded away from zero. The endpoint term also converges: F_\lambda(\pi)\to F(\pi) because the even periodic extension of \theta\mapsto F(c-r\cos\theta) is continuous at \pi. For the interior pairing, the zero principal value of the constant numerator gives, at every continuity point, g(c-r\cos\theta)=\frac1{\pi r}\operatorname {pv}\int_0^\pi \frac{F(c-r\cos\phi)-F(c-r\cos\theta)} {\cos\theta-\cos\phi}\,d\phi. We now give a direct uniform-integrability estimate; no commutation of the finite Hilbert transform with Poisson convolution is used. Write \widetilde F(\theta)=F(c-r\cos\theta). Since v(d)=d_i-d on each spatial target block, \widetilde F is piecewise real analytic with finitely many interior jumps. If those jumps are \Delta_\ell at \theta_\ell, decompose \widetilde F(\theta)=F_c(\theta)+ \sum_\ell\Delta_\ell\mathbf 1_{[\theta_\ell,\pi]}(\theta). The function F_c is continuous and piecewise real analytic. Its cosine coefficients c_n satisfy |c_n|\le Cn^{-2}: integrate once by parts, and use that the piecewise analytic derivative F_c' has bounded variation, whose Fourier coefficients are O(n^{-1}).

For a single jump term, its nth cosine coefficient is -\Delta_\ell\sin(n\theta_\ell)/(\pi n). Using U_{n-1}(\cos\theta)=\sin(n\theta)/\sin\theta and summing \sum_{n\ge1}\lambda^n\cos(nt)/n=-\log|1-\lambda e^{it}| gives its regularized transform exactly as \frac{\Delta_\ell}{\pi r\sin\theta} \left\{\log|1-\lambda e^{i(\theta+\theta_\ell)}| -\log|1-\lambda e^{i(\theta-\theta_\ell)}|\right\}. The first logarithm is uniformly bounded near \theta_\ell; the second has the only singularity. Uniformly in 0<\lambda<1, \int_{|t|<\delta}|\log|1-\lambda e^{it}||\,dt \le C\delta(1+|\log\delta|). To verify this estimate, use |1-\lambda e^{it}|\asymp\sqrt{(1-\lambda)^2+t^2} for |t|\le1 and integrate separately over |t|\le1-\lambda and 1-\lambda<|t|<\delta; the part where the logarithm is positive is uniformly bounded. At 0 and \pi the two logarithms in braces have the same endpoint value; the mean-value theorem gives a difference O(\sin\theta) uniformly in \lambda, so division by \sin\theta remains bounded.

For the continuous part, put \delta_\theta=\min(\theta,\pi-\theta). Since |\sin(n\theta)|\le\min(n\delta_\theta,1) and \sin\theta\ge2\delta_\theta/\pi, the coefficient bound gives, uniformly in \lambda, \left|\sum_{n\ge1}\lambda^nc_nU_{n-1}(\cos\theta)\right| \le C\bigl(1+|\log\delta_\theta|\bigr). The last two bounds are integrable, B(\theta)=O(\theta^2) at zero, and B is bounded at \pi. Away from the finitely many jumps the displayed series converge pointwise as \lambda\uparrow1; the logarithmic estimates give uniform integrability on the omitted neighborhoods. Hence the pairings converge in L^1(B(\theta)\,d\theta), which passes (4.14) to the target velocity. Since the top target quantile is d_N\le2, v(2)=d_N-2\le0,\qquad \dot M_k\ge\int_Ig\,d\xi. Combining this one-sided derivative estimate with (4.9) gives the corrected supporting inequality \begin{equation} M_k(T)\ge M_k(T_0)+\int_Ig\,d\xi. \label{eq:4.25} \tag{4.15} \end{equation} ◻

Proof. For u\in I_i, every off-block target difference and reference difference has the same sign. The tangent inequality \frac{y-x}{x}\ge\log y-\log x\qquad(x,y>0) applies to their absolute values and to the external-field coordinates. Indeed, writing g(u) as shorthand for g(T_0(u)), the unreduced first variation is g(u)=k\frac{T(u)-T_0(u)}{T_0(u)} +\operatorname {pv}\int_0^1 \frac{[T(u)-T(v)]-[T_0(u)-T_0(v)]} {T_0(u)-T_0(v)}\,dv. For v\in I_j, j\ne i, the tangent inequality gives \frac{(d_i-d_j)-[T_0(u)-T_0(v)]}{T_0(u)-T_0(v)} \ge\log|d_i-d_j|-\log|T_0(u)-T_0(v)|. For v\in I_i the target difference is zero, so the quotient equals -1 almost everywhere. Applying the same inequality also to d_i/T_0(u) and integrating the preceding bounds yields \begin{align*} g(u)\ge{}&k\log d_i-k\log T_0(u) +\sum_{j\ne i}q_j\log|d_i-d_j|\\ &-\int_{(0,1)\setminus I_i}\log|T_0(u)-T_0(v)|\,dv-q_i. \end{align*} Now (2.3) makes the first and third terms equal to -q_i\log R_i, while the platform identity (4.5) supplies exactly the remaining constant -C. Therefore g(u)\ge-q_i\log R_i-C-q_i +\int_{I_i}\log|T_0(u)-T_0(v)|\,dv. Integrating, summing, and using (4.15) yields \mathfrak J_k\ge M_0+\sum_i\left\{ r_i[-q_i\log R_i-C-q_i]+\mathcal E(I_i)+2R_i\right\}, where M_0=M_k(T_0). As a function of R_i>0, the expression in braces has derivative -q_ir_i/R_i+2 and positive second derivative q_ir_i/R_i^2. Its unique minimum is therefore \mathcal E(I_i)-q_ir_i\log\frac{q_ir_i}{2}-Cr_i, attained at R_i=q_ir_i/2. Here r_i>0, because \xi is positive on every nonempty subinterval away from the single endpoint a. Put R_0=\xi(I)=\widehat\xi((0,1)),\qquad C_{\rm eff}=C+\frac{L-M_0}{R_0}. R_0 is strictly positive. It follows that the interval inequalities \begin{equation} \boxed{\mathcal E(I')\ge qr\log\frac{qr}{2}+C_{\rm eff}r} \label{eq:4.32} \tag{4.17} \end{equation} for every nonempty quantile interval I'\subset(0,1) (where q=|I'| and r=\widehat\xi(I')) imply, after summing over the target blocks and using \sum_i r_i=R_0, \mathfrak J_k\ge M_0+(C_{\rm eff}-C)R_0=L. This is the conclusion of the proposition. ◻

Proof. The certificate uses only interval evaluations of the following analytic equations and their differentiated forms. Near q=0, set \varrho=(-\log q)^{-1},\qquad z_\pm=qu_\pm. Then \begin{equation} G_\pm(\varrho,z)=A\log\frac{z-q^2}{|1-z|}-\log z-d(q)=0, \label{eq:6.9} \tag{6.6} \end{equation} with 0<z_+<1<z_-. The length is \begin{equation} \Lambda=\frac{2(z_--z_+)}{(1+q)^2} +\frac{2q^2}{(1+q)^2}\left(\frac1{z_-}-\frac1{z_+}\right). \label{eq:6.10} \tag{6.7} \end{equation} At \varrho=0, (z_+,z_-)=(1/2,3/2). On the complete tail 0\le\varrho\le0.02, direct interval substitution into (6.6), the strict G_z signs, and implicit differentiation give -0.788<\frac{d\Lambda}{d\varrho}<-0.737. To verify the stated continuation at \varrho=0, note that q=e^{-1/\varrho}\to0, A=1-\varrho d(q)\to1, and (6.6) tends to -\log(2|1-z|)=0. Its two solutions are exactly z=1/2 and z=3/2.

At the soft edge, put x=\log u_+ and y=x^2. After division by the identically present root x=0, the plus equation is \begin{equation} D(q,y)=A-1+2A\sum_{n\ge1}q^n \frac{\sinh(n\sqrt y)}{n\sqrt y}=0. \label{eq:6.12} \tag{6.8} \end{equation} For an exact, nonsingular evaluation of this series, define \begin{aligned} C(y)&=\cosh\sqrt y=\sum_{m\ge0}\frac{y^m}{(2m)!},\\ S(y)&=\frac{\sinh\sqrt y}{\sqrt y} =\sum_{m\ge0}\frac{y^m}{(2m+1)!},\\ b(q,y)&=\frac{qS(y)}{1-qC(y)},\qquad w(q,y)=y\,b(q,y)^2,\\ \operatorname{atanhc}(w)&=\frac{\operatorname{atanh}\sqrt w}{\sqrt w} =\sum_{m\ge0}\frac{w^m}{2m+1}, \end{aligned} with the continuous values C(0)=S(0)=\operatorname{atanhc}(0)=1. Since \sqrt w=\frac{q\sinh x}{1-q\cosh x} and 2\operatorname{atanh}\frac{q\sinh x}{1-q\cosh x} =\log\frac{1-qe^{-x}}{1-qe^x}, the series in (6.8) is exactly b(q,y)\operatorname{atanhc}(w(q,y)). Thus the expression actually evaluated is \begin{equation} D(q,y)=A(q)-1+2A(q)b(q,y)\operatorname{atanhc}(w(q,y)). \label{eq:6.12b} \tag{6.9} \end{equation} In particular D(q,0)=A(q)\frac{1+q}{1-q}-1, which vanishes exactly when A(q)=s(q), that is, at q=q_s. The certificate proves D_y>0 on the soft chart and 0\le w<0.03. It evaluates the three positive series displayed above through m=32 and appends explicit one-sided remainder bounds, including bounds for their first derivatives.

In the bulk, proposed root boxes are first certified by opposite endpoint signs and a strict G_z sign. A parametric interval-Newton contraction then encloses z_\pm(\varrho), after which z'=-G_\varrho/G_z,\qquad z''=-\frac{G_{\varrho\varrho}+2G_{\varrho z}z'+G_{zz}(z')^2}{G_z} is substituted into (6.7). No floating-point root is accepted without these sign and Newton checks.

The next two terminating decimals are retained at full precision because they are exact rational centers of tests with radii 10^{-30} and 10^{-48}, respectively; shortening either center without changing the verifier would change the certificate. Put \begin{aligned} q_c&= 0.02571553686652745032257637166391965344,\\ \varrho_c&=-\frac1{\log q_c},\\ c_s&= 0.1236306846493834978974060904264788695442437883724. \end{aligned} The exhaustive cover is:

The bulk right endpoint \varrho=1/\log10 is exactly q=1/10, so the bulk and soft charts meet with no gap. Inside the strictly convex tube, the certificate proves opposite signs for \Lambda_\varrho at \varrho=\varrho_c-10^{-30} and \varrho=\varrho_c+10^{-30}. Continuity gives a stationary point between them, and strict convexity gives at most one stationary point in the entire tube. The signs on the adjacent charts therefore show that it is the unique stationary point on (0,q_s] and is the global minimum. Outward substitution into (6.7) gives tighter boxes for q_* and L; rounding their endpoints outward gives the more readable enclosures in Theorem 1.1.

Finally, (6.4) proves that the function in (6.3) is strictly decreasing. Opposite endpoint signs on [c_s-10^{-48},c_s+10^{-48}] therefore isolate its unique zero q_s and give the stated enclosure for q_s. The endpoint soft chart is evaluated on the slightly larger interval through c_s+10^{-48} and is then intersected with q\le q_s; thus the isolation box does not create a coverage gap. All these checks are implemented by the one-cut section of numerical_verifier.py. ◻