Introduction and main results
Free stability does not imply stability under classical convolution. This distinction leads to a natural question: which free stable laws are classically infinitely divisible, self-decomposable, or generalized gamma convolutions? We answer these three questions and study the related bell-shape problem.
Bercovici and Pata established the role of free stable laws in free limit theory, with their density geometry and duality developed in Biane’s appendix [2 ] . Explicit Mellin and characteristic transforms were obtained by Hasebe and Kuznetsov [5 Theorems 1–3] . Hasebe, Simon and Wang proved classical infinite divisibility for indices at most one, excluded symmetric laws above one, and proved the extended Thorin property for indices at most 3 / 4 3/4 3/4 [6 Theorems 1 and 2] . They conjectured the optimal Thorin bound 4 / 5 4/5 4/5
Normalization and terminology
Write G μ ( z ) = ∫ R ( z − x ) − 1 μ ( d x ) G_\mu(z)=\int_{\mathbb{R}}(z-x)^{-1}\,\mu(\mathrm{d}x) G μ ( z ) = ∫ R ( z − x ) − 1 μ ( d x ) F μ = 1 / G μ F_\mu=1/G_\mu F μ = 1/ G μ ϕ μ ( z ) = F μ − 1 ( z ) − z \phi_\mu(z)=F_\mu^{-1}(z)-z ϕ μ ( z ) = F μ − 1 ( z ) − z [7 Introduction] : X α , ρ X_{\alpha,\rho} X α , ρ
ϕ α , ρ ( z ) = − e i π α ρ z 1 − α , Im z > 0 , (1) \phi_{\alpha,\rho}(z)=-e^{i\pi\alpha\rho}z^{1-\alpha}, \qquad\operatorname{Im}z>0, \tag*{(1)} ϕ α , ρ ( z ) = − e iπ α ρ z 1 − α , Im z > 0 , (1) with admissible parameters
A = { 0 < α ≤ 1 , 0 ≤ ρ ≤ 1 } ∪ { 1 < α ≤ 2 , 1 − 1 / α ≤ ρ ≤ 1 / α } . (2) \mathcal{A}=\{0<\alpha\leq1,\ 0\leq\rho\leq1\}\cup\{1<\alpha\leq2,\ 1-1/\alpha\leq\rho\leq1/\alpha\}. \tag*{(2)} A = { 0 < α ≤ 1 , 0 ≤ ρ ≤ 1 } ∪ { 1 < α ≤ 2 , 1 − 1/ α ≤ ρ ≤ 1/ α } . (2) Reflection replaces ρ \rho ρ 1 − ρ 1-\rho 1 − ρ classical additive convolution , denoted by ∗ * ∗
Definition 1.1 (Classical convolution classes). A probability law μ \mu μ n ≥ 1 n\geq1 n ≥ 1 μ = μ n ∗ n \mu=\mu_n^{*n} μ = μ n ∗ n μ n \mu_n μ n c ∈ ( 0 , 1 ) c\in(0,1) c ∈ ( 0 , 1 ) μ c \mu_c μ c μ = D c μ ∗ μ c \mu=D_c\mu*\mu_c μ = D c μ ∗ μ c
where D c μ D_c\mu D c μ c X cX c X X X X μ \mu μ [12 Sections 7 and 15] . An ID law belongs to the extended Thorin class, abbreviated GGC, when its Lévy measure has the form
ν ( d x ) = k + ( x ) x 1 { x > 0 } d x + k − ( − x ) ∣ x ∣ 1 { x < 0 } d x (3) \nu(\mathrm{d}x)=\frac{k_{+}(x)}{x}\mathbf{1}_{\{x>0\}}\mathrm{d}x+\frac{k_{-}(-x)}{|x|}\mathbf{1}_{\{x<0\}}\mathrm{d}x \tag*{(3)} ν ( d x ) = x k + ( x ) 1 { x > 0 } d x + ∣ x ∣ k − ( − x ) 1 { x < 0 } d x (3) with completely monotone kernels k + , k − k_{+},k_{-} k + , k − ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) ( − 1 ) n f ( n ) ( x ) ≥ 0 (-1)^n f^{(n)}(x)\ge0 ( − 1 ) n f ( n ) ( x ) ≥ 0 n ≥ 0 n\ge0 n ≥ 0 x > 0 x>0 x > 0 [4 Chapter 7, (7.1.5)] .
For an ID law with the representation (3) , the SD criterion is that both kernels are nonnegative and nonincreasing [12 Theorem 15.10] . Thus GGC implies SD, and SD implies ID. The GGC notation here includes two-sided laws; point masses satisfy all three properties.
The three classification theorems
The ID and SD regions above one consist of intervals adjoining the two extremal skewnesses. Their inner boundaries are different, as are their maximal indices. Below one, the GGC region is instead central. The following theorems give exact implicit classifications; § 3 defines their boundaries through explicit analytic transforms.
Theorem 1.2 (Classical infinite divisibility). There is a unique maximal ID index α I D \alpha_{\mathrm{ID}} α ID
1.52407396102878875153 < α I D < 1.52407396102878875155. (4) 1.52407396102878875153 < \alpha_{\mathrm{ID}} < 1.52407396102878875155. \tag*{(4)} 1.52407396102878875153 < α ID < 1.52407396102878875155. (4) Every admissible X α , ρ X_{\alpha,\rho} X α , ρ α ≤ 1 \alpha\le1 α ≤ 1 1 < α ≤ α I D 1<\alpha\le\alpha_{\mathrm{ID}} 1 < α ≤ α ID
ρ ∈ [ 1 − 1 / α , 1 − r ( α ) ] ∪ [ r ( α ) , 1 / α ] . (5) \rho\in[1-1/\alpha,1-r(\alpha)]\cup[r(\alpha),1/\alpha]. \tag*{(5)} ρ ∈ [ 1 − 1/ α , 1 − r ( α )] ∪ [ r ( α ) , 1/ α ] . (5) Here r r r 1 / 2 < r ( α ) < 1 / α 1/2<r(\alpha)<1/\alpha 1/2 < r ( α ) < 1/ α r ( α I D ) = 1 / α I D r(\alpha_{\mathrm{ID}})=1/\alpha_{\mathrm{ID}} r ( α ID ) = 1/ α ID r ( α ) → 1 / 2 r(\alpha)\to1/2 r ( α ) → 1/2 α ↓ 1 \alpha\downarrow1 α ↓ 1 α > α I D \alpha>\alpha_{\mathrm{ID}} α > α ID r r r (19) ; it is computable to arbitrary positive rational accuracy at every computable α ∈ ( 1 , α I D ) \alpha\in(1,\alpha_{\mathrm{ID}}) α ∈ ( 1 , α ID )
Theorem 1.3 (Extended generalized gamma convolutions). Every admissible skewness is GGC for 0 < α ≤ 4 / 5 0<\alpha\le4/5 0 < α ≤ 4/5 α = 1 \alpha=1 α = 1 α > 1 \alpha>1 α > 1 4 / 5 < α < 1 4/5<\alpha<1 4/5 < α < 1
[ g ( α ) , 1 − g ( α ) ] . (6) [g(\alpha),1-g(\alpha)]. \tag*{(6)} [ g ( α ) , 1 − g ( α )] . (6) The function g g g ( 0 , 1 ] (0,1] ( 0 , 1 ] g = 0 g=0 g = 0 α ≤ 4 / 5 \alpha\le4/5 α ≤ 4/5 α = 1 \alpha=1 α = 1 ( 4 / 5 , 1 ) (4/5,1) ( 4/5 , 1 ) 0 < g ( α ) < 1 / 2 0<g(\alpha)<1/2 0 < g ( α ) < 1/2 (21) , and g ( α ) g(\alpha) g ( α ) α \alpha α 4 / 5 4/5 4/5
Theorem 1.4 (Classical self-decomposability). Every admissible law with 0 < α ≤ 1 0<\alpha\le1 0 < α ≤ 1 α S D \alpha_{\mathrm{SD}} α SD
1.4283571142 < α S D < 1.4283571143 , (7) 1.4283571142 < \alpha_{\mathrm{SD}} < 1.4283571143, \tag*{(7)} 1.4283571142 < α SD < 1.4283571143 , (7) such that the SD section for 1 < α ≤ α S D 1<\alpha\le\alpha_{\mathrm{SD}} 1 < α ≤ α SD
[ 1 − 1 / α , 1 − s ( α ) ] ∪ [ s ( α ) , 1 / α ] . (8) [1-1/\alpha,1-s(\alpha)]\cup[s(\alpha),1/\alpha]. \tag*{(8)} [ 1 − 1/ α , 1 − s ( α )] ∪ [ s ( α ) , 1/ α ] . (8) The function s s s
r ( α ) < s ( α ) < 1 / α ( 1 < α < α S D ) , s ( α S D ) = 1 / α S D . r(\alpha)<s(\alpha)<1/\alpha\qquad(1<\alpha<\alpha_{\mathrm{SD}}), \qquad s(\alpha_{\mathrm{SD}})=1/\alpha_{\mathrm{SD}}. r ( α ) < s ( α ) < 1/ α ( 1 < α < α SD ) , s ( α SD ) = 1/ α SD . It is the global first-contact curve of the derivative transform − z F α ′ ( z ) -zF'_{\alpha}(z) − z F α ′ ( z ) § 3 . No law with α > α S D \alpha>\alpha_{\mathrm{SD}} α > α SD
s ( 1 + ε ) = 1 2 + 3 3 4 π ε + o ( ε ) . (9) s(1+\varepsilon)=\frac{1}{2}+\sqrt{\frac{3\sqrt{3}}{4\pi}}\sqrt{\varepsilon}+o(\sqrt{\varepsilon}). \tag*{(9)} s ( 1 + ε ) = 2 1 + 4 π 3 3 ε + o ( ε ) . (9) Bell-shape
Definition 1.5 (Bell-shape and whale-shape). A smooth probability density on R \mathbb{R} R n n n n n n n ≥ 0 n \ge0 n ≥ 0 [11 Introduction] ; they do not require all derivative zeros to be simple. A density on a half-line is whale-shaped if it is smooth in the interior, vanishes at both ends, and each derivative of positive order has exactly one zero in the interior, as in [7 Introduction, definition preceding Theorem 4] .
Smooth weakly bell-shaped densities are strictly bell-shaped [11 Theorems 1.1 and 1.3] . Their transform characterization replaces the derivative conditions by restrictions on crossings of integer phase levels. It permits bell-shape even when the phase itself is not monotone, and hence beyond the Thorin region.
Theorem 1.6 (Selected bell-shape results). The following statements hold.
(1) No admissible law with α > 1 \alpha> 1 α > 1 α = 1 \alpha= 1 α = 1
(2) For 0 < α < 1 0 < \alpha< 1 0 < α < 1 0 < ρ < 1 0 < \rho< 1 0 < ρ < 1 (39) . The GGC region satisfies these conditions, but there are nonempty open sets of strictly bell-shaped laws outside the GGC region.
(3) For every fixed 0 < α < 1 0 < \alpha< 1 0 < α < 1 ( 0 , 1 ) (0,1) ( 0 , 1 ) α ↑ 1 \alpha\uparrow1 α ↑ 1
(4) Let C C C T T T − log z -\log z − log z A > 0 A > 0 A > 0 B ≠ 0 B \ne0 B = 0 A C + B T AC + BT A C + B T
A π ∣ B ∣ ∈ N . (10) \frac{A}{\pi|B|} \in\mathbb{N}. \tag*{(10)} π ∣ B ∣ A ∈ N . (10) For A = 0 A = 0 A = 0 B = 0 B = 0 B = 0
The full bell-shape parameter set is not determined by these statements.
The sums in the exceptional family can be realized with independent classical summands: convolution by a Cauchy law agrees with free convolution by that Cauchy law. The case B = 0 B = 0 B = 0 A > 0 A > 0 A > 0 A = B = 0 A = B = 0 A = B = 0 X 1 , ρ X_{1,\rho} X 1 , ρ
Method of proof and organization
Two analytic representations govern the classifications. Above one, a signed Lévy representation reduces ID to positivity of one kernel and SD to positivity of its negative logarithmic derivative. Below one, Cauchy subordination expresses the Lévy densities through a single analytic phase. GGC asks that this phase be nondecreasing; bell-shape only restricts its integer-level crossings. § 2 gives the common criteria, and § 3 –§ 5 apply them to the free stable family.
In each boundary problem, estimates at zero and infinity reduce the remaining sign questions to compact domains. Finite interval certificates then provide the required signs on whole parameter boxes. This distinction is essential: a local double-root equation does not identify a global boundary. Theorem F.7 exhibits competing global ID contacts. Our boundary algorithms give any prescribed positive error; they do not decide equality to an arbitrary computable real input.
§ 6 compares the boundary scales and records the computational dependencies. Appendices § 7 –H contain the longer estimates and the specifications needed to reproduce the finite checks. All theorem statements include their boundary cases. The remaining bell-shape questions are stated at the end of § 6 .
We use N = { 1 , 2 , … } \mathbb{N}=\{1,2,\ldots\} N = { 1 , 2 , … } a a a α \alpha α
Analytic criteria
The proofs use two reductions: harmonic interpolation for signed Lévy kernels, and a phase representation for Cauchy subordination. We state them separately to make clear which properties are used above and below index one.
A sector principle for signed Lévy kernels
Lemma 2.1 (Sector interpolation). Let F F F θ 0 ≤ arg z ≤ θ 1 \theta_0 \le\arg z \le\theta_1 θ 0 ≤ arg z ≤ θ 1 z ≠ 0 z \ne0 z = 0 w w w p < π / w p < \pi/w p < π / w
∣ F ( z ) ∣ ≤ C ( ∣ z ∣ p + ∣ z ∣ − p ) |F(z)| \le C\left(|z|^p+|z|^{-p}\right) ∣ F ( z ) ∣ ≤ C ( ∣ z ∣ p + ∣ z ∣ − p ) If Re F ≥ 0 \operatorname{Re} F \ge0 Re F ≥ 0 Re F ≥ 0 \operatorname{Re} F \ge0 Re F ≥ 0 Re F \operatorname{Re} F Re F F F F − z F ′ ( z ) -zF'(z) − z F ′ ( z )
Proof. Choose p < q < π / w p<q<\pi/w p < q < π / w θ m = ( θ 0 + θ 1 ) / 2 \theta_m=(\theta_0+\theta_1)/2 θ m = ( θ 0 + θ 1 ) /2
( r q + r − q ) cos ( q ( θ − θ m ) ) \left(r^q+r^{-q}\right)\cos\left(q(\theta-\theta_m)\right) ( r q + r − q ) cos ( q ( θ − θ m ) ) is positive on the closed sector. Add ϵ \epsilon ϵ Re F \operatorname{Re} F Re F ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0 ∣ z ∣ |z| ∣ z ∣ z F ′ ( z ) zF'(z) z F ′ ( z )
Suppose a family has a signed Lévy representation, with nonnegative Gaussian coefficient and ∫ R ( 1 ∧ x 2 ) ∣ ν ∣ ( d x ) < ∞ \int_{\mathbb{R}}(1\wedge x^2)|\nu|(\mathrm{d}x)<\infty ∫ R ( 1 ∧ x 2 ) ∣ ν ∣ ( d x ) < ∞ K θ ( x ) = Re F ( x e i θ ) K_\theta(x)=\operatorname{Re}F(xe^{i\theta}) K θ ( x ) = Re F ( x e i θ ) K θ ≥ 0 K_\theta\ge0 K θ ≥ 0 K θ ( x ) → 0 K_\theta(x)\to0 K θ ( x ) → 0
− x K θ ′ ( x ) = Re [ − z F ′ ( z ) ] ≥ 0 , z = x e i θ . -xK_\theta'(x)=\operatorname{Re}[-zF'(z)]\ge0,\qquad z=xe^{i\theta}. − x K θ ′ ( x ) = Re [ − z F ′ ( z )] ≥ 0 , z = x e i θ . Indeed nonincreasingness and the zero limit already force K θ ≥ 0 K_\theta\ge0 K θ ≥ 0
A common phase for Cauchy subordination
Definition 2.2 (Complete Bernstein and Stieltjes functions). A complete Bernstein function is a nonnegative function on ( 0 , ∞ ) (0,\infty) ( 0 , ∞ ) C ∖ ( − ∞ , 0 ] \mathbb{C}\setminus(-\infty,0] C ∖ ( − ∞ , 0 ] [14 Theorem 6.2] . A Stieltjes function has a representation
G ( z ) = c + d / z + ∫ ( 0 , ∞ ) ( z + t ) − 1 σ ( d t ) , c , d ≥ 0 , ∫ ( 0 , ∞ ) ( 1 + t ) − 1 σ ( d t ) < ∞ , G(z)=c+d/z+\int_{(0,\infty)}(z+t)^{-1}\sigma(\mathrm{d}t),\qquad c,d\ge0,\qquad\int_{(0,\infty)}(1+t)^{-1}\sigma(\mathrm{d}t)<\infty, G ( z ) = c + d / z + ∫ ( 0 , ∞ ) ( z + t ) − 1 σ ( d t ) , c , d ≥ 0 , ∫ ( 0 , ∞ ) ( 1 + t ) − 1 σ ( d t ) < ∞ , where σ \sigma σ [14 Definition 2.1] .
Let Y ≥ 0 Y \ge0 Y ≥ 0 L ( z ) = e − ψ ( z ) L(z)=e^{-\psi(z)} L ( z ) = e − ψ ( z ) ψ \psi ψ ψ ( 0 ) = 0 \psi(0)=0 ψ ( 0 ) = 0
ψ ( z ) = O ( ∣ z ∣ γ ) ( z → 0 ) (11) \psi(z)=O\left(|z|^\gamma\right)\quad(z\to0) \tag*{(11)} ψ ( z ) = O ( ∣ z ∣ γ ) ( z → 0 ) (11) for some γ > 0 \gamma>0 γ > 0 C ρ ( t ) C_\rho(t) C ρ ( t ) t sin ( π ρ ) t\sin(\pi\rho) t sin ( π ρ ) − t cos ( π ρ ) -t\cos(\pi\rho) − t cos ( π ρ ) Y Y Y 0 < ρ < 1 0<\rho<1 0 < ρ < 1 θ ∈ ( 0 , π ) \theta\in(0,\pi) θ ∈ ( 0 , π )
P θ ( u ) = π − 1 Im ψ ( u e i θ ) , u > 0. (12) P_\theta(u)=\pi^{-1}\operatorname{Im}\psi(ue^{i\theta}),\qquad u>0. \tag*{(12)} P θ ( u ) = π − 1 Im ψ ( u e i θ ) , u > 0. (12) Theorem 2.3 (Cauchy-subordination criteria). The law μ ρ = L ( C ρ ( Y ) ) \mu_\rho=\mathcal{L}(C_\rho(Y)) μ ρ = L ( C ρ ( Y ))
ν + ( x ) = ∫ 0 ∞ e − u x P π ρ ( u ) d u , ν − ( x ) = ∫ 0 ∞ e − u x P π ( 1 − ρ ) ( u ) d u , x > 0. (13) \nu_+(x)=\int_0^\infty e^{-ux}P_{\pi\rho}(u)\,du,\qquad
\nu_-(x)=\int_0^\infty e^{-ux}P_{\pi(1-\rho)}(u)\,du,\qquad x>0. \tag*{(13)} ν + ( x ) = ∫ 0 ∞ e − ux P π ρ ( u ) d u , ν − ( x ) = ∫ 0 ∞ e − ux P π ( 1 − ρ ) ( u ) d u , x > 0. (13) Consequently:
μ ρ \mu_\rho μ ρ P P P (13) are nondecreasing;
μ ρ \mu_\rho μ ρ
∫ 0 ∞ e − u x ( u x − 1 ) P ( u ) d u ≥ 0 for every x > 0 ; (14) \int_0^\infty e^{-ux}(ux-1)P(u)\,du\ge0\qquad\text{for every }x>0; \tag*{(14)} ∫ 0 ∞ e − ux ( ux − 1 ) P ( u ) d u ≥ 0 for every x > 0 ; (14) μ ρ \mu_\rho μ ρ P ( u ) − m P(u)-m P ( u ) − m m ∈ N m\in\mathbb{N} m ∈ N
In addition, if Y Y Y μ ρ \mu_\rho μ ρ [13 Theorem 3.1(i)] .
Proof. Use μ ^ ( ξ ) = ∫ R e − i ξ x μ ( d x ) \widehat{\mu}(\xi)=\int_{\mathbb{R}}e^{-i\xi x}\,\mu(dx) μ ( ξ ) = ∫ R e − i ξ x μ ( d x ) ξ > 0 \xi>0 ξ > 0
μ ^ ρ ( ξ ) = exp { − ψ ( e i π ( ρ − 1 / 2 ) ξ ) } . \widehat{\mu}_\rho(\xi)=\exp\left\{-\psi\left(e^{i\pi(\rho-1/2)}\xi\right)\right\}. μ ρ ( ξ ) = exp { − ψ ( e iπ ( ρ − 1/2 ) ξ ) } . The right-half-plane logarithm exists. Its boundary imaginary parts at ξ = i u \xi=iu ξ = i u ξ = − i u \xi=-iu ξ = − i u − π P π ρ ( u ) -\pi P_{\pi\rho}(u) − π P π ρ ( u ) π P π ( 1 − ρ ) ( u ) \pi P_{\pi(1-\rho)}(u) π P π ( 1 − ρ ) ( u )
A complete Bernstein function is O ( ∣ z ∣ ) O(|z|) O ( ∣ z ∣ ) (11) , this permits the twice-subtracted half-plane Cauchy formula. After the linear term is absorbed into the drift, that formula gives
log μ ^ ρ ( ξ ) = − i d ξ + ∫ R { 1 i ξ + u − 1 u + i ξ u 2 1 { ∣ u ∣ ≥ 1 } } φ ( u ) d u , (15) \log\widehat{\mu}_\rho(\xi)=-id\xi+\int_{\mathbb{R}}\left\{\frac{1}{i\xi+u}-\frac{1}{u}+\frac{i\xi}{u^2}\mathbf{1}_{\{|u|\ge1\}}\right\}\varphi(u)\,du, \tag*{(15)} log μ ρ ( ξ ) = − i d ξ + ∫ R { i ξ + u 1 − u 1 + u 2 i ξ 1 { ∣ u ∣ ≥ 1 } } φ ( u ) d u , (15) where the normalization at ξ = 0 \xi=0 ξ = 0
φ ( u ) = P π ρ ( u ) ( u > 0 ) , φ ( u ) = − P π ( 1 − ρ ) ( − u ) ( u < 0 ) . \varphi(u)=P_{\pi\rho}(u)\quad(u>0),\qquad
\varphi(u)=-P_{\pi(1-\rho)}(-u)\quad(u<0). φ ( u ) = P π ρ ( u ) ( u > 0 ) , φ ( u ) = − P π ( 1 − ρ ) ( − u ) ( u < 0 ) . One may equivalently use the regularization in [11 (1.1)] ; the difference is a drift and a real normalization constant. The origin integrals converge because P ( u ) = O ( u γ ) P(u)=O(u^\gamma) P ( u ) = O ( u γ ) ∫ 1 ∞ P ( u ) u − 3 d u < ∞ \int_1^\infty P(u)u^{-3}\,du<\infty ∫ 1 ∞ P ( u ) u − 3 d u < ∞ P ( u ) = O ( u ) P(u)=O(u) P ( u ) = O ( u ) ϵ \epsilon ϵ R R R (11) ; the twice-subtracted large arc tends to zero using the linear growth. Normalizing at zero yields (15) . The only remaining entire ambiguity is linear and purely imaginary on the Fourier axis, and hence is drift.
For u > 0 u>0 u > 0
∫ 0 ∞ ( e − i ξ x − 1 ) e − u x d x = 1 u + i ξ − 1 u \int_0^\infty(e^{-i\xi x}-1)e^{-ux}\,dx=\frac{1}{u+i\xi}-\frac{1}{u} ∫ 0 ∞ ( e − i ξ x − 1 ) e − ux d x = u + i ξ 1 − u 1 and its reflected version identify the Lévy densities as in (13) . Truncation at u = ϵ , R u=\epsilon,R u = ϵ , R P P P ψ \psi ψ
For a nonnegative continuous P P P P ( 0 ) = 0 P(0)=0 P ( 0 ) = 0 x ∫ 0 ∞ e − u x P ( u ) d u x\int_{0}^{\infty}e^{-ux}P(u)\,\mathrm{d}u x ∫ 0 ∞ e − ux P ( u ) d u P P P ∫ e − u x U ( d u ) \int e^{-ux}U(\mathrm{d}u) ∫ e − ux U ( d u ) x x x P ( u ) = U ( [ 0 , u ] ) P(u)=U([0,u]) P ( u ) = U ([ 0 , u ]) x ν ± ( x ) x\nu_{\pm}(x) x ν ± ( x ) (14) and the SD assertion by the Lévy-kernel criterion.
The representation (15) , with the integrability just checked, is the representation in [11 Theorems 1.1 and 1.3] . Its regularity condition holds because ∣ μ ^ ρ ∣ ≤ 1 |\widehat{\mu}_{\rho}|\leq1 ∣ μ ρ ∣ ≤ 1 ξ I m μ ^ ρ ( ξ ) → 0 \xi\mathop{\mathrm{Im}}\widehat{\mu}_{\rho}(\xi)\to0 ξ Im μ ρ ( ξ ) → 0 Y = d c Y ′ + Z c Y \mathrel{\stackrel{d}{=}} cY'+Z_c Y = d c Y ′ + Z c Z c ≥ 0 Z_c\geq0 Z c ≥ 0 Y ′ Y' Y ′ C ρ ( Y ) = d c C ρ ( Y ′ ) + C ρ ′ ( Z c ) C_{\rho}(Y) \mathrel{\stackrel{d}{=}} cC_{\rho}(Y')+C_{\rho}'(Z_c) C ρ ( Y ) = d c C ρ ( Y ′ ) + C ρ ′ ( Z c )
Remark 2.4. GGC asks for monotonicity of the entire phase, while bell-shape only forbids a downward passage through an integer. SD is the intermediate Laplace-average test (14) ; it is not a pointwise phase monotonicity condition. These distinctions explain why equality of the three parameter regions should not be expected.
Critical curves for ID and GGC
The Wright transform connects the abstract criteria with the free stable parameters. For a > 0 a>0 a > 0 [5 Theorem 3]
W a ( z ) = ∑ n = 0 ∞ z n n ! Γ ( 2 + ( a − 1 ) n ) . (16) W_a(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!\Gamma(2+(a-1)n)}. \tag*{(16)} W a ( z ) = n = 0 ∑ ∞ n ! Γ ( 2 + ( a − 1 ) n ) z n . (16) For reference, the Wright characteristic transform, in the plus-Fourier convention φ α , ρ ( t ) = E e i t X α , ρ \varphi_{\alpha,\rho}(t)=\mathbb{E}e^{itX_{\alpha,\rho}} φ α , ρ ( t ) = E e i t X α , ρ [5 Theorem 3, (21)]
φ α , ρ ( t ) = W α ( − e − i α ω t α ) , ω = π ( ρ − 1 2 ) , t > 0. (17) \varphi_{\alpha,\rho}(t)=W_{\alpha}\left(-e^{-i\alpha\omega}t^{\alpha}\right),\qquad\omega=\pi\left(\rho-\frac{1}{2}\right),\quad t>0. \tag*{(17)} φ α , ρ ( t ) = W α ( − e − i α ω t α ) , ω = π ( ρ − 2 1 ) , t > 0. (17) For 0 < α < 1 0<\alpha<1 0 < α < 1 X α = X α , 1 X_{\alpha}=X_{\alpha,1} X α = X α , 1 L α ( s ) = W α ( − s α ) L_{\alpha}(s)=W_{\alpha}(-s^{\alpha}) L α ( s ) = W α ( − s α ) b α = α ( 1 − α ) 1 / α − 1 b_{\alpha}=\alpha(1-\alpha)^{1/\alpha-1} b α = α ( 1 − α ) 1/ α − 1
In the appendices, α ∗ \alpha_{*} α ∗ α I D \alpha_{\mathrm{ID}} α ID [ z n ] f [z^n]f [ z n ] f ( u ) n = u ( u + 1 ) ⋯ ( u + n − 1 ) (u)_n=u(u+1)\cdots(u+n-1) ( u ) n = u ( u + 1 ) ⋯ ( u + n − 1 ) ( u ) 0 = 1 (u)_0=1 ( u ) 0 = 1 1 F 1 {}_1F_1 1 F 1 2 F 1 {}_2F_1 2 F 1 p F q ( a ; b ; z ) = ∑ n ≥ 0 ∏ i ( a i ) n z n ∏ j ( b j ) n n ! pF_q(a;b;z)=\sum_{n\geq0}\frac{\prod_i(a_i)_n z^n}{\prod_j(b_j)_n n!} p F q ( a ; b ; z ) = ∑ n ≥ 0 ∏ j ( b j ) n n ! ∏ i ( a i ) n z n [15 Section 16.2] . We use H m ( j ) = ∑ k = 1 m k − j H_m^{(j)}=\sum_{k=1}^{m}k^{-j} H m ( j ) = ∑ k = 1 m k − j H m = H m ( 1 ) H_m=H_m^{(1)} H m = H m ( 1 ) γ \gamma γ ζ \zeta ζ z / ( e z − 1 ) = ∑ n ≥ 0 B n z n / n ! z/(e^z-1)=\sum_{n\geq0}B_nz^n/n! z / ( e z − 1 ) = ∑ n ≥ 0 B n z n / n ! B 1 = − 1 / 2 B_1=-1/2 B 1 = − 1/2
For a > 1 a>1 a > 1 [1 Theorem 1] , applied with parameters a − 1 a-1 a − 1 2 2 2 1 / a < 1 1/a<1 1/ a < 1 W a W_a W a W a ( 0 ) = 1 W_a(0)=1 W a ( 0 ) = 1 − λ n -\lambda_n − λ n r n = λ n 1 / a r_n=\lambda_n^{1/a} r n = λ n 1/ a Proposition D.2 . Define
F a ( z ) = i z π ∑ n ∫ 0 ∞ e − z t { log ∣ 1 − ( t r n ) a ∣ + i π 1 { t > r n } } d t , Re z > 0. (18) \mathcal{F}_a(z)=\frac{iz}{\pi}\sum_n\int_0^{\infty}e^{-zt}\left\{\log\left|1-\left(\frac{t}{r_n}\right)^a\right|+i\pi\mathbf{1}_{\{t>r_n\}}\right\}\,\mathrm{d}t,\qquad\operatorname{Re}z>0. \tag*{(18)} F a ( z ) = π i z n ∑ ∫ 0 ∞ e − z t { log 1 − ( r n t ) a + iπ 1 { t > r n } } d t , Re z > 0. (18) The series of integrals converges normally on this half-plane. For upper-half skewness the only possibly negative Lévy kernel is K a , ρ ( x ) = Re F a ( x e − i π ( 1 − ρ ) ) K_{a,\rho}(x)=\operatorname{Re}\mathcal{F}_a(xe^{-i\pi(1-\rho)}) K a , ρ ( x ) = Re F a ( x e − iπ ( 1 − ρ ) )
ID: global positivity and the first zero ray
For 1 < a < α I D 1<a<\alpha_{\mathrm{ID}} 1 < a < α ID Theorem 1.2 is specified by
π ( 1 − r ( a ) ) = min { θ ∈ ( π ( 1 − 1 a ) , π 2 ) : Re F a ( x e − i θ ) = 0 for some x > 0 } . (19) \pi(1-r(a))=\min\left\{\theta\in\left(\pi\left(1-\frac{1}{a}\right),\frac{\pi}{2}\right):\operatorname{Re}F_a(xe^{-i\theta})=0\text{ for some }x>0\right\}. \tag*{(19)} π ( 1 − r ( a )) = min { θ ∈ ( π ( 1 − a 1 ) , 2 π ) : Re F a ( x e − i θ ) = 0 for some x > 0 } . (19) The cutoff is the unique zero minimum of the extremal kernel described in Theorem E.3 . In particular the number (4) is not defined by a numerical fit.
Proof of Theorem 1.2 . For a ≤ 1 a\leq1 a ≤ 1 [6 Theorem 1(a)] . For a > 1 a>1 a > 1 K ≥ 0 K\geq0 K ≥ 0 O ( ∣ z ∣ a + ∣ z ∣ − a ) O(|z|^a+|z|^{-a}) O ( ∣ z ∣ a + ∣ z ∣ − a ) π / 2 \pi/2 π /2 a < 2 a<2 a < 2 Lemma 2.1 proves that every nonempty upper-half section is an interval. This step does not assume that an extremal law is ID.
The complete extremal positivity theorem (Theorem E.6 ) proves that the endpoint kernels are strictly positive for 1 < a < α I D 1<a<\alpha_{\mathrm{ID}} 1 < a < α ID α I D \alpha_{\mathrm{ID}} α ID Theorems E.3 and E.4 establish the unique cutoff and exclude every other skewness above it, using separate skewness witnesses. Strict origin and infinite-tail bounds, together with compactness of the remaining radii, make strict extremal positivity stable under a small skewness perturbation. Hence the ID interval has positive width below the cutoff.
At its inner boundary the kernel must vanish at some finite nonzero radius; otherwise those same uniform bounds extend the interval. The zeros are finite in number, since the kernel is analytic in radius and strictly positive at both ends. At any boundary zero, K x = 0 K_x=0 K x = 0 K x x ≥ 0 K_{xx}\geq0 K xx ≥ 0 K ρ > 0 K_\rho>0 K ρ > 0 (19) . The full compactness and Hopf argument is given in Theorem E.8 .
Weak closedness of ID gives lower semicontinuity of the boundary. Strict positivity on every interior ray, with uniform end bounds, gives upper semicontinuity. The endpoint assertions follow by the same argument and the near-one ID rectangles. Finally, the finite-radius cutoffs and spectral remainder in Theorem G.2 allow simultaneous searches for strict positive covers and strict negative witnesses. Two rational trial rays eventually bracket the boundary to any requested accuracy. This proves computability without assuming unique contact or a decision algorithm at equality. □
GGC: the first loss of phase monotonicity
For 0 < a < 1 0<a<1 0 < a < 1
L a ( z ) = W a ( − z a ) , ψ a ( z ) = − log L a ( z ) , H a ( z ) = z ψ a ′ ( z ) , b a = a ( 1 − a ) 1 / a − 1 . (20) L_a(z)=W_a(-z^a),\qquad\psi_a(z)=-\log L_a(z),\qquad H_a(z)=z\psi_a'(z),\qquad b_a=a(1-a)^{1/a-1}. \tag*{(20)} L a ( z ) = W a ( − z a ) , ψ a ( z ) = − log L a ( z ) , H a ( z ) = z ψ a ′ ( z ) , b a = a ( 1 − a ) 1/ a − 1 . (20) The logarithm is continued from the positive axis. Its global slit-plane existence is justified in Lemma 4.1 below. For 4 / 5 < a < 1 4/5<a<1 4/5 < a < 1
π ( 1 − g ( a ) ) = min { arg z : π 2 < arg z < π , Im H a ( z ) = 0 } . (21) \pi(1-g(a))=\min\left\{\arg z:\frac{\pi}{2}<\arg z<\pi,\ \operatorname{Im}H_a(z)=0\right\}. \tag*{(21)} π ( 1 − g ( a )) = min { arg z : 2 π < arg z < π , Im H a ( z ) = 0 } . (21) Proof of Theorem 1.3 . The uniform sufficiency up to 4 / 5 4/5 4/5 Theorem C.4 : it combines the earlier GGC theorem with a complete Kanter phase certificate on the remaining compact parameter range and explicit infinite-tail estimates. Theorem A.3 proves failure of the one-sided GGC property for every 4 / 5 < a < 1 4/5<a<1 4/5 < a < 1
For each remaining index, Theorem B.14 proves the symmetric law GGC. The positive-real criterion and harmonic sector interpolation then give a central interval. Strict positivity on inner rays gives a nontrivial central neighborhood. At its outer boundary the first zero of Im H a \operatorname{Im}H_a Im H a
Im H a = Im ( z H a ′ ) = 0 , Re ( z H a ′ ) < 0. \operatorname{Im}H_a=\operatorname{Im}(zH_a')=0,\qquad\operatorname{Re}(zH_a')<0. Im H a = Im ( z H a ′ ) = 0 , Re ( z H a ′ ) < 0. Compactness of the contact set and the strict angular derivative yield continuity and local Lipschitz regularity, as detailed in Theorem B.15 . Theorem B.11 gives g ( a ) → 0 g(a)\to0 g ( a ) → 0 Theorem G.2 prove the computability assertion.
For a > 1 a>1 a > 1 Theorem A.2 ). At a = 1 a=1 a = 1
Self-decomposability
For a ≤ 1 a\leq1 a ≤ 1 a > 1 a>1 a > 1
The positive law and its analytic phase
Lemma 4.1 (Whale-shaped factorization). For 0 < a < 1 0<a<1 0 < a < 1 L a L_a L a ψ a = − log L a \psi_a=-\log L_a ψ a = − log L a
0 < Im { ψ a ( z ) − b a z } < 2 arg z , 0 < arg z < π . (22) 0<\operatorname{Im}\{\psi_a(z)-b_a z\}<2\arg z,\qquad0<\arg z<\pi. \tag*{(22)} 0 < Im { ψ a ( z ) − b a z } < 2 arg z , 0 < arg z < π . (22) Moreover X a , ρ = d C ρ ( X a , 1 ) X_{a,\rho}\mathrel{\overset{d}{=}}C_\rho(X_{a,1}) X a , ρ = d C ρ ( X a , 1 ) 0 < ρ < 1 0<\rho<1 0 < ρ < 1
Proof. The translated density of X a , 1 − b a X_{a,1}-b_a X a , 1 − b a [6 Theorem 4(a)] . By [11 Theorem 1.13] , it is the convolution of an exponential density and an integrable completely monotone function. On normalizing the second factor,
e b a z L a ( z ) = λ λ + z G ( z ) , G ( 0 ) = 1 , e^{b_a z}L_a(z)=\frac{\lambda}{\lambda+z}G(z),\qquad G(0)=1, e b a z L a ( z ) = λ + z λ G ( z ) , G ( 0 ) = 1 , where λ > 0 \lambda>0 λ > 0 G G G 0 < arg z = θ < π 0<\arg z=\theta<\pi 0 < arg z = θ < π − θ -\theta − θ G G G − arg G ∈ [ 0 , θ ] -\arg G\in[0,\theta] − arg G ∈ [ 0 , θ ] 0 < arg ( λ + z ) < θ 0<\arg(\lambda+z)<\theta 0 < arg ( λ + z ) < θ (22) , with the branch continued from z > 0 z>0 z > 0 [14 Theorem 6.2] applies. Finally, the Cauchy factorization is [7 (8)] ; it can also be checked directly by comparing the Wright characteristic transforms. □
Proposition 4.2 (Positive SD below one). For 4 / 5 ≤ a < 1 4/5\leq a<1 4/5 ≤ a < 1 X a , 1 X_{a,1} X a , 1 ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
Proof. Put β = 1 − a \beta=1-a β = 1 − a c = β 1 / a c=\beta^{1/a} c = β 1/ a Z = ( X a , 1 − b a ) / c Z=(X_{a,1}-b_a)/c Z = ( X a , 1 − b a ) / c j a ( y ) / y j_a(y)/y j a ( y ) / y
h a ( v ) = a a − 1 − e − v − e ( a − 1 ) v a − 1 = y . (23) h_a(v)=\frac{a}{a-1}-e^{-v}-\frac{e^{(a-1)v}}{a-1}=y. \tag*{(23)} h a ( v ) = a − 1 a − e − v − a − 1 e ( a − 1 ) v = y . (23) Its removable value at a = 1 a=1 a = 1 1 − e − v − v 1-e^{-v}-v 1 − e − v − v
p a ( y ) = 2 / a π y F a ( y ) , F a ( y ) = ∑ n ≥ 0 f n ( a ) y n , j a ( y ) = ∑ n ≥ 0 k n ( a ) y n . p_a(y)=\frac{\sqrt{2/a}}{\pi}\sqrt{y}F_a(y),\qquad F_a(y)=\sum_{n\geq0}f_n(a)y^n,\qquad j_a(y)=\sum_{n\geq0}k_n(a)y^n. p a ( y ) = π 2/ a y F a ( y ) , F a ( y ) = n ≥ 0 ∑ f n ( a ) y n , j a ( y ) = n ≥ 0 ∑ k n ( a ) y n . These are the rational parameter polynomials constructed in Lemma D.4, with f 0 = 1 f_0=1 f 0 = 1 k 0 = 3 / 2 k_0=3/2 k 0 = 3/2
We first justify an all-order remainder, including the endpoint a = 1 a=1 a = 1 ∣ Im v ∣ < 2 π / a |\operatorname{Im}v|<2\pi/a ∣ Im v ∣ < 2 π / a
C = e ∓ 2 π i / a B ( t ) , C = − a / β , B ( t ) = e − t − β − 1 e − β t . C=e^{\mp2\pi i/a}B(t),\qquad C=-a/\beta,\qquad B(t)=e^{-t}-\beta^{-1}e^{-\beta t}. C = e ∓ 2 π i / a B ( t ) , C = − a / β , B ( t ) = e − t − β − 1 e − β t . The range of B B B [ C , ∞ ) [C,\infty) [ C , ∞ )
a β sin ( 2 π β / a ) ≥ 4. \frac{a}{\beta}\sin(2\pi\beta/a)\geq4. β a sin ( 2 π β / a ) ≥ 4. The inequality follows from sin ( 2 π u ) ≥ 4 u \sin(2\pi u)\geq4u sin ( 2 π u ) ≥ 4 u 0 ≤ u ≤ 1 / 4 0\leq u\leq1/4 0 ≤ u ≤ 1/4 a = 1 a=1 a = 1 ∣ y ∣ < 4 |y| < 4 ∣ y ∣ < 4 F a F_a F a j a j_a j a
For n ≥ 1 n \ge1 n ≥ 1 R n = 4 n / ( n + 1 ) R_n = 4n/(n + 1) R n = 4 n / ( n + 1 ) M n = [ 3 ( n + 1 ) / 2 ] 5 M_n = [3(n + 1)/2]^5 M n = [ 3 ( n + 1 ) /2 ] 5 β = 1 / 5 \beta= 1/5 β = 1/5
1 − β − β ( R n + M n − 1 ) − M n − β = 2 15 ( n + 1 ) − 1 5 M n > 0. 1-\beta-\beta(R_n+M_n^{-1})-M_n^{-\beta}=\frac{2}{15(n+1)}-\frac{1}{5M_n}>0. 1 − β − β ( R n + M n − 1 ) − M n − β = 15 ( n + 1 ) 2 − 5 M n 1 > 0. The left side is concave in β \beta β 0 < β ≤ 1 / 5 0 < \beta\le1/5 0 < β ≤ 1/5 (23) excludes Re v ≥ log M n \operatorname{Re} v \ge\log M_n Re v ≥ log M n ∣ y ∣ = R n |y| = R_n ∣ y ∣ = R n ∣ F a ( y ) ∣ ≤ M n / 2 |F_a(y)| \le M_n/2 ∣ F a ( y ) ∣ ≤ M n /2
∣ f n ( a ) ∣ ≤ 12 ( n + 1 ) 5 4 − n , ∣ k n ( a ) ∣ ≤ 128 ( n + 1 ) 8 4 − n . (24) |f_n(a)| \le12(n+1)5^{4-n}, \qquad|k_n(a)| \le128(n+1)8^{4-n}. \tag*{(24)} ∣ f n ( a ) ∣ ≤ 12 ( n + 1 ) 5 4 − n , ∣ k n ( a ) ∣ ≤ 128 ( n + 1 ) 8 4 − n . (24) For completeness the second bound follows from the exact recurrence
k n = ( 3 / 2 ) n + 1 n ! f n − ∑ m = 1 n ( 3 / 2 ) m ( n − m ) ! n ! f m k n − m . (25) k_n=\frac{(3/2)^{n+1}}{n!}f_n-\sum_{m=1}^{n}\frac{(3/2)^m(n-m)!}{n!}f_mk_{n-m}. \tag*{(25)} k n = n ! ( 3/2 ) n + 1 f n − m = 1 ∑ n n ! ( 3/2 ) m ( n − m )! f m k n − m . (25) The first 128 cases are exact rational polynomial inequalities. For n ≥ 128 n \ge128 n ≥ 128 A m = 4 m sup a ∣ f m ( a ) ∣ A_m=4^m\sup_a|f_m(a)| A m = 4 m sup a ∣ f m ( a ) ∣ m ≤ 7 m \le7 m ≤ 7 (24) . Splitting the convolution at m = n / 2 m=n/2 m = n /2 ( n m ) ≥ ( n / m ) m \binom{n}{m}\ge(n/m)^m ( m n ) ≥ ( n / m ) m
S 1 = ∑ m = 1 7 A m ( m + 1 ) ( m / 128 ) m + 12 ∑ m = 8 64 ( m + 1 ) 6 ( m / 128 ) m + 12 ∑ m ≥ 65 ( m + 1 ) 6 2 − m , S_1=\sum_{m=1}^{7}A_m(m+1)(m/128)^m+12\sum_{m=8}^{64}(m+1)^6(m/128)^m+12\sum_{m\ge65}(m+1)6^{2-m}, S 1 = m = 1 ∑ 7 A m ( m + 1 ) ( m /128 ) m + 12 m = 8 ∑ 64 ( m + 1 ) 6 ( m /128 ) m + 12 m ≥ 65 ∑ ( m + 1 ) 6 2 − m , S 2 = 12 129 2 { 1 + ∑ k = 1 64 ( k + 1 ) 8 ( k / 128 ) k + ∑ k ≥ 65 ( k + 1 ) 8 2 − k } . S_2=\frac{12}{129^2}\left\{1+\sum_{k=1}^{64}(k+1)^8(k/128)^k+\sum_{k\ge65}(k+1)8^{2-k}\right\}. S 2 = 12 9 2 12 { 1 + k = 1 ∑ 64 ( k + 1 ) 8 ( k /128 ) k + k ≥ 65 ∑ ( k + 1 ) 8 2 − k } . The free term is at most 18 / ( 128 ⋅ 129 ) 18/(128\cdot129) 18/ ( 128 ⋅ 129 ) S 1 + S 2 + 18 / ( 128 ⋅ 129 ) < 0.083 < 1 S_1+S_2+18/(128\cdot129)<0.083<1 S 1 + S 2 + 18/ ( 128 ⋅ 129 ) < 0.083 < 1
On a ∈ [ 4 / 5 , 1 ] a\in[4/5,1] a ∈ [ 4/5 , 1 ] 0 ≤ y ≤ 5 / 2 0\le y\le5/2 0 ≤ y ≤ 5/2 r > 0 r>0 r > 0
128 r ∑ n > 128 ( n + 1 ) 9 ( r / 4 ) n . (26) \frac{128}{r}\sum_{n>128}(n+1)^9(r/4)^n. \tag*{(26)} r 128 n > 128 ∑ ( n + 1 ) 9 ( r /4 ) n . (26) The removable r = 0 r=0 r = 0 − j a ′ ( y ) > 0 -j_a'(y)>0 − j a ′ ( y ) > 0 k n = P n ( a ) / a n k_n=P_n(a)/a^n k n = P n ( a ) / a n P n , j + 1 = a P n , j ′ − ( n + j ) P n , j P_{n,j+1}=aP'_{n,j}-(n+j)P_{n,j} P n , j + 1 = a P n , j ′ − ( n + j ) P n , j (26) .
It remains to control every y ≥ 5 / 2 y\ge5/2 y ≥ 5/2
ϕ a ( t ) = − π − 1 Im log L a ( − t / c + i 0 ) . \phi_a(t)=-\pi^{-1}\operatorname{Im}\log L_a(-t/c+i0). ϕ a ( t ) = − π − 1 Im log L a ( − t / c + i 0 ) . The deterministic translation contributes no cut imaginary part, and Lemma 4.1 gives 0 ≤ ϕ a ≤ 2 0\le\phi_a\le2 0 ≤ ϕ a ≤ 2
j a ( y ) = y ∫ 0 ∞ e − y t ϕ a ( t ) d t . (27) j_a(y)=y\int_0^\infty e^{-yt}\phi_a(t)\,dt. \tag*{(27)} j a ( y ) = y ∫ 0 ∞ e − y t ϕ a ( t ) d t . (27) Set T = t a T=t^a T = t a z = T e − i π β z=Te^{-i\pi\beta} z = T e − iπ β G β ( z ) = e − z / β W a ( z / β ) G_\beta(z)=e^{-z/\beta}W_a(z/\beta) G β ( z ) = e − z / β W a ( z / β ) β = 0 \beta=0 β = 0 0 ≤ β ≤ 1 / 5 0\le\beta\le1/5 0 ≤ β ≤ 1/5 0 ≤ T ≤ 1 0\le T\le1 0 ≤ T ≤ 1
Re G β ( z ) > 0 , 2 3 < d d T ϕ a ( T 1 / a ) = sin ( π β ) π β − 1 π Im e − i π β G β ′ ( z ) G β ( z ) < 6 5 . (28) \operatorname{Re}G_\beta(z)>0,\qquad\frac{2}{3}<\frac{\mathrm{d}}{\mathrm{d}T}\phi_a(T^{1/a})=\frac{\sin(\pi\beta)}{\pi\beta}-\frac{1}{\pi}\operatorname{Im}\frac{e^{-i\pi\beta}G_\beta'(z)}{G_\beta(z)}<\frac{6}{5}. \tag*{(28)} Re G β ( z ) > 0 , 3 2 < d T d ϕ a ( T 1/ a ) = π β sin ( π β ) − π 1 Im G β ( z ) e − iπ β G β ′ ( z ) < 5 6 . (28) The real-half-plane check fixes the logarithm branch. Integrating from zero gives ( 2 / 3 ) t a ≤ ϕ a ( t ) ≤ ( 6 / 5 ) t a (2/3)t^a\le\phi_a(t)\le(6/5)t^a ( 2/3 ) t a ≤ ϕ a ( t ) ≤ ( 6/5 ) t a 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1 (27) , the factor y t − 1 yt-1 y t − 1 t < 1 / y t < 1/y t < 1/ y ( 1 / y , 1 ) (1/y,1) ( 1/ y , 1 ) t > 1 t > 1 t > 1
y a + 1 [ − j a ′ ( y ) ] ≥ 2 3 B ( a , y ) − 8 15 N ( a ) , y^{a+1}[-j_a'(y)] \ge\frac{2}{3}B(a,y)-\frac{8}{15}N(a), y a + 1 [ − j a ′ ( y )] ≥ 3 2 B ( a , y ) − 15 8 N ( a ) , where
B ( a , y ) = ∫ 0 y e − u ( u − 1 ) u a d u , N ( a ) = ∫ 0 1 e − u ( 1 − u ) u a d u . B(a,y)=\int_0^y e^{-u}(u-1)u^a\,\mathrm{d}u,\qquad N(a)=\int_0^1 e^{-u}(1-u)u^a\,\mathrm{d}u. B ( a , y ) = ∫ 0 y e − u ( u − 1 ) u a d u , N ( a ) = ∫ 0 1 e − u ( 1 − u ) u a d u . For y ≥ 5 / 2 y\ge5/2 y ≥ 5/2 B B B a a a y y y N N N a a a
2 3 B ( 4 / 5 , 5 / 2 ) − 8 15 N ( 4 / 5 ) > 0.02578 \frac{2}{3}B(4/5,5/2)-\frac{8}{15}N(4/5)>0.02578 3 2 B ( 4/5 , 5/2 ) − 15 8 N ( 4/5 ) > 0.02578 proves strict negativity of j a ′ j_a' j a ′ certify_sd_below_one.py verify. This completes the proof. □
Extremal kernels and the upper SD transition
We give the numerical obligations explicitly, since an extremal transition alone does not exclude a disjoint set of nonextremal SD laws. For a > 1 a>1 a > 1
c a = ( a − 1 ) 1 / a , q = ρ − 1 / 2 1 / a − 1 / 2 , D ( a , q , y ) = − ∂ y K a , ρ ( c a y ) . c_a=(a-1)^{1/a},\qquad q=\frac{\rho-1/2}{1/a-1/2},\qquad D(a,q,y)=-\partial_yK_{a,\rho}(c_ay). c a = ( a − 1 ) 1/ a , q = 1/ a − 1/2 ρ − 1/2 , D ( a , q , y ) = − ∂ y K a , ρ ( c a y ) . At q = 1 q=1 q = 1 j a ( y ) = K a , 1 / a ( c a y ) j_a(y)=K_{a,1/a}(c_ay) j a ( y ) = K a , 1/ a ( c a y ) Lemma D.4 .
Lemma 4.3 (Extremal SD and global exclusions). Put a L = 1.42834 a_L=1.42834 a L = 1.42834 a U = 1.42838 a_U=1.42838 a U = 1.42838 I = [ 4.99 , 5.04 ] I=[4.99,5.04] I = [ 4.99 , 5.04 ]
(1) D ( a , 1 , y ) > 0 D(a,1,y)>0 D ( a , 1 , y ) > 0 1 < a ≤ a L 1<a\le a_L 1 < a ≤ a L y > 0 y>0 y > 0
(2) For a L ≤ a ≤ a U a_L\le a\le a_U a L ≤ a ≤ a U D ( a , 1 , y ) > 0 D(a,1,y)>0 D ( a , 1 , y ) > 0 I I I I I I D y y > 1 / 50 D_{yy}>1/50 D y y > 1/50 D a < 0 D_a<0 D a < 0 D y D_y D y
(3) The minimum on I I I a L a_L a L a U a_U a U [ a L , a U ] × [ 0.98 , 1 ] × I [a_L,a_U]\times[0.98,1]\times I [ a L , a U ] × [ 0.98 , 1 ] × I D q > 0 D_q>0 D q > 0
(4) On [ a L , a U ] × [ 11 / 16 , 0.98 ] [a_L,a_U]\times[11/16,0.98] [ a L , a U ] × [ 11/16 , 0.98 ] D D D y = 5.0104 y=5.0104 y = 5.0104 [ a U , 1.43 ] × [ 11 / 16 , 1 ] [a_U,1.43]\times[11/16,1] [ a U , 1.43 ] × [ 11/16 , 1 ]
(5) For 1.43 ≤ a ≤ α I D 1.43\le a\le\alpha_{\mathrm{ID}} 1.43 ≤ a ≤ α ID 11 / 16 ≤ q ≤ 1 11/16\le q\le1 11/16 ≤ q ≤ 1 D ( a , q , 4.944 ) < 0 D(a,q,4.944)<0 D ( a , q , 4.944 ) < 0 q < 11 / 16 q<11/16 q < 11/16
Proof. We describe the analytic bounds, the exact covered domains, and the finite verification. On 1 ≤ a ≤ 1.3 1\le a\le1.3 1 ≤ a ≤ 1.3 ∣ k n ( a ) ∣ ≤ 128 ( n + 1 ) 2 ( 5 / 2 ) − n |k_n(a)|\le128(n+1)^2(5/2)^{-n} ∣ k n ( a ) ∣ ≤ 128 ( n + 1 ) 2 ( 5/2 ) − n [ 1.3 , 1.4 ] [1.3,1.4] [ 1.3 , 1.4 ] [ 1.4 , 1.53 ] [1.4,1.53] [ 1.4 , 1.53 ] 32 ( n + 1 ) 2 4 − n 32(n+1)^2 4^{-n} 32 ( n + 1 ) 2 4 − n 32 ( n + 1 ) 2 ( 51 / 10 ) − n 32(n+1)^2(51/10)^{-n} 32 ( n + 1 ) 2 ( 51/10 ) − n
For the infinite tail put e = a − 1 e=a-1 e = a − 1 h = sin π ( 1 / a − 1 / 2 ) h=\sin\pi(1/a-1/2) h = sin π ( 1/ a − 1/2 ) ξ = π ( 2 − a ) \xi=\pi(2-a) ξ = π ( 2 − a ) Q m = ∑ R n − m a Q_m=\sum R_n^{-ma} Q m = ∑ R n − ma R n = c a r n R_n=c_ar_n R n = c a r n a ≤ 3 / 2 a\le3/2 a ≤ 3/2 cos ξ ≤ 0 \cos\xi\le0 cos ξ ≤ 0
( 1 + 2 z cos ξ + z 2 ) − 1 ≥ 1 − 2 z cos ξ − z 2 (1+2z\cos\xi+z^2)^{-1}\ge1-2z\cos\xi-z^2 ( 1 + 2 z cos ξ + z 2 ) − 1 ≥ 1 − 2 z cos ξ − z 2 and the maximum of t 2 a + 1 e − h y t t^{2a+1}e^{-hyt} t 2 a + 1 e − h y t
y a + 1 D ( a , 1 , y ) ≥ a sin ( π e ) π e { 1 − e Γ ( 3 a + 1 ) Q 3 y − 2 a } − 2 ( 2 a + 1 e h ) 2 a + 1 Q 2 y − a . (29) y^{a+1}D(a,1,y)\ge a\frac{\sin(\pi e)}{\pi e}\left\{1-e\Gamma(3a+1)Q_3y^{-2a}\right\}
-2\left(\frac{2a+1}{eh}\right)^{2a+1}Q_2y^{-a}. \tag*{(29)} y a + 1 D ( a , 1 , y ) ≥ a π e sin ( π e ) { 1 − e Γ ( 3 a + 1 ) Q 3 y − 2 a } − 2 ( e h 2 a + 1 ) 2 a + 1 Q 2 y − a . (29) Here the removable moments at e = 0 e=0 e = 0 e e e y y y y = 2 y=2 y = 2 a = 1.1 a=1.1 a = 1.1 1.1 ≤ a ≤ 1.42 1.1\le a\le1.42 1.1 ≤ a ≤ 1.42 2 ≤ y ≤ 8 2\le y\le8 2 ≤ y ≤ 8 y ≥ 8 y\ge8 y ≥ 8 y a + 1 y^{a+1} y a + 1 h y R n > a + 1 hyR_n>a+1 h y R n > a + 1
Near the transition a sharper germ holds uniformly on [ 1.42 , 1.44 ] [1.42,1.44] [ 1.42 , 1.44 ]
∣ k n ( a ) ∣ ≤ 32 ( n + 1 ) 2 ( 20 / 107 ) n . (30) \lvert k_n(a)\rvert\le32(n+1)^2(20/107)^n. \tag*{(30)} ∣ k n ( a )∣ ≤ 32 ( n + 1 ) 2 ( 20/107 ) n . (30) The inverse-branch radius exceeds 5.3617 and the working radius is 107 / 20 = 5.35 107/20=5.35 107/20 = 5.35 (30) , validated spectral integrals and Rayleigh remainders to check items (1)–(4), except for D a < 0 D_a<0 D a < 0 D ( y 0 ) − D y ( y 0 ) 2 / ( 2 m ) D(y_0)-D_y(y_0)^2/(2m) D ( y 0 ) − D y ( y 0 ) 2 / ( 2 m ) m = 1 / 50 m=1/50 m = 1/50
The parameter derivative is checked independently. Differentiate the degree-768 spatial germ with respect to a a a a 0 = 1.42836 a_0=1.42836 a 0 = 1.42836 R = 0.0002 R=0.0002 R = 0.0002 M M M [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] ( w + w − 1 ) / 2 (w+w^{-1})/2 ( w + w − 1 ) /2 ∣ w ∣ = ϱ > 1 \lvert w\rvert=\varrho>1 ∣ w ∣ = ϱ > 1 r = 10 − 6 r=10^{-6} r = 1 0 − 6 τ = max ∣ a − a 0 ∣ / R \tau=\max\lvert a-a_0\rvert/R τ = max ∣ a − a 0 ∣ / R η = r / R \eta=r/R η = r / R
M R { η 18 ( 1 − η 18 ) ( 1 − τ ) 2 + 18 τ 17 1 − τ + τ 18 ( 1 − τ ) 2 } . (31) \frac{M}{R}\left\{\frac{\eta^{18}}{(1-\eta^{18})(1-\tau)^2}+\frac{18\tau^{17}}{1-\tau}+\frac{\tau^{18}}{(1-\tau)^2}\right\}. \tag*{(31)} R M { ( 1 − η 18 ) ( 1 − τ ) 2 η 18 + 1 − τ 18 τ 17 + ( 1 − τ ) 2 τ 18 } . (31) This follows by summing aliases of the Cauchy coefficients and the tail of their differentiated power series. Values are kept as full spatial polynomials until after this discrete transform, so interval dependence on y y y
The infinite spatial remainder is bounded separately. With Q n ( a ) = a n k n ( a ) Q_n(a)=a^n k_n(a) Q n ( a ) = a n k n ( a ) deg Q n ≤ 2 n \deg Q_n\le2n deg Q n ≤ 2 n [ L , U ] = [ 1.42 , 1.44 ] [L,U]=[1.42,1.44] [ L , U ] = [ 1.42 , 1.44 ] [9 Introduction, (1.1)] gives
∣ k n ′ ( a ) ∣ ≤ 32 n ( n + 1 ) 2 ( U a ( 107 / 20 ) ) n { 2 ( a − L ) ( U − a ) + 1 a } . \lvert k_n'(a)\rvert\le32n(n+1)^2\left(\frac{U}{a(107/20)}\right)^n\left\{\frac{2}{\sqrt{(a-L)(U-a)}}+\frac{1}{a}\right\}. ∣ k n ′ ( a )∣ ≤ 32 n ( n + 1 ) 2 ( a ( 107/20 ) U ) n { ( a − L ) ( U − a ) 2 + a 1 } . Multiplication by n y n − 1 ny^{n-1} n y n − 1 n > 768 n>768 n > 768 (31) , and all rounding errors gives
− 0.40472 < D a < − 0.25965 -0.40472<D_a<-0.25965 − 0.40472 < D a < − 0.25965 on the complete extremal core. All ten complex-parameter values are freshly recomputed in verification mode.
Lastly, the ID boundary locators on [ 1.43 , α I D ] [1.43,\alpha_{\mathrm{ID}}] [ 1.43 , α ID ] q = 11 / 16 q=11/16 q = 11/16 a L a_L a L § 6 . Thus all radii and all admissible skewnesses required by the lemma are accounted for.
Proof of the SD cutoff and its rational bracket. By Lemma 4.3 , every a ∈ [ a L , a U ] a\in[a_L,a_U] a ∈ [ a L , a U ] y ( a ) y(a) y ( a ) D y y > 0 D_{yy}>0 D y y > 0 m ( a ) = D ( a , 1 , y ( a ) ) m(a)=D(a,1,y(a)) m ( a ) = D ( a , 1 , y ( a )) m ′ ( a ) = D a ( a , 1 , y ( a ) ) < 0 m'(a)=D_a(a,1,y(a))<0 m ′ ( a ) = D a ( a , 1 , y ( a )) < 0 α S D \alpha_{\mathrm{SD}} α SD D q > 0 D_q>0 D q > 0 q q q
For the refined bracket, evaluate the degree-896 germ at y 0 = 5.01036765 y_0 = 5.01036765 y 0 = 5.01036765 (30) . Directed evaluations give
a certified quantity 1.4283571142 D ( a , 1 , y 0 ) − D y ( a , 1 , y 0 ) 2 / ( 2 / 50 ) > 3.94 ⋅ 10 − 12 1.4283571143 D ( a , 1 , y 0 ) < − 2.94 ⋅ 10 − 11 \begin{array}{c|c}
a & \text{certified quantity} \\
1.4283571142 & D(a,1,y_0)-D_y(a,1,y_0)^2/(2/50)>3.94\cdot10^{-12} \\
1.4283571143 & D(a,1,y_0)<-2.94\cdot10^{-11}
\end{array} a 1.4283571142 1.4283571143 certified quantity D ( a , 1 , y 0 ) − D y ( a , 1 , y 0 ) 2 / ( 2/50 ) > 3.94 ⋅ 1 0 − 12 D ( a , 1 , y 0 ) < − 2.94 ⋅ 1 0 − 11 Strong convexity turns the first into a lower bound on the entire core minimum. The unique-zero argument identifies the same global cutoff and proves (7) . □
The SD boundary, continuity and effective approximation
For 1 < a < α S D 1<a<\alpha_{\mathrm{SD}} 1 < a < α SD D a ( z ) = − z F a ′ ( z ) D_a(z)=-zF'_a(z) D a ( z ) = − z F a ′ ( z ) s s s
π ( 1 − s ( a ) ) = min { θ ∈ ( π ( 1 − 1 / a ) , π / 2 ) : Re D a ( x e − i θ ) = 0 for some x > 0 } . (32) \pi(1-s(a))=\min\left\{\theta\in\left(\pi(1-1/a),\pi/2\right):\operatorname{Re}D_a(xe^{-i\theta})=0\text{ for some }x>0\right\}. \tag*{(32)} π ( 1 − s ( a )) = min { θ ∈ ( π ( 1 − 1/ a ) , π /2 ) : Re D a ( x e − i θ ) = 0 for some x > 0 } . (32) Here the endpoint derivative is strictly positive, by Lemma 4.3 .
Proof of Theorem 1.4 , except for (9) . For a ≤ 4 / 5 a\leq4/5 a ≤ 4/5 Proposition 4.2 and Cauchy subordination prove all remaining a < 1 a<1 a < 1 Re D a ≥ 0 \operatorname{Re}D_a\geq0 Re D a ≥ 0 Lemma 2.1 proves interval convexity, with strict positivity on every interior ray. Weak closedness gives a closed interval.
At the ID boundary a nonnegative kernel has a finite positive zero and is strictly positive farther out. Such a kernel cannot be nonincreasing. Hence closedness of SD gives s ( a ) > r ( a ) s(a)>r(a) s ( a ) > r ( a ) s ( a ) < 1 / a s(a)<1/a s ( a ) < 1/ a δ = π ( 1 / a − ρ ) \delta=\pi(1/a-\rho) δ = π ( 1/ a − ρ ) B = a / ( a − 1 ) B=a/(a-1) B = a / ( a − 1 )
− ∂ y K a , ρ ( a y ) = − 2 Re { e − i δ j a ′ ( y e − i δ ) } + B sin δ π y 2 + P δ ′ ( y ) , -\partial_yK_{a,\rho}(ay)=-2\operatorname{Re}\{e^{-i\delta}j'_a(ye^{-i\delta})\}+\frac{B\sin\delta}{\pi y^2}+P'_\delta(y), − ∂ y K a , ρ ( a y ) = − 2 Re { e − i δ j a ′ ( y e − i δ )} + π y 2 B sin δ + P δ ′ ( y ) , P δ ′ ( y ) = sin δ π ∫ 0 ∞ u j a ′ ( u ) ( u − y cos δ ) 2 + y 2 sin 2 δ d u . (33) P'_\delta(y)=\frac{\sin\delta}{\pi}\int_0^\infty\frac{u j'_a(u)}{(u-y\cos\delta)^2+y^2\sin^2\delta}\,du. \tag*{(33)} P δ ′ ( y ) = π sin δ ∫ 0 ∞ ( u − y cos δ ) 2 + y 2 sin 2 δ u j a ′ ( u ) d u . (33) The endpoint has j a ( 0 ) = 3 / 2 j_a(0)=3/2 j a ( 0 ) = 3/2 j a ′ ( 0 ) = − c < 0 j'_a(0)=-c<0 j a ′ ( 0 ) = − c < 0 j a ′ < 0 j'_a<0 j a ′ < 0 j a ( ∞ ) = 0 j_a(\infty)=0 j a ( ∞ ) = 0 ∣ w ∣ ≤ d |w|\leq d ∣ w ∣ ≤ d ∣ j a ′ ( w ) + c ∣ ≤ c / 10 |j'_a(w)+c|\leq c/10 ∣ j a ′ ( w ) + c ∣ ≤ c /10 y ≤ d / 2 y\leq d/2 y ≤ d /2
11 c 10 { cos δ + sin δ π log d + y y } , \frac{11c}{10}\left\{\cos\delta+\frac{\sin\delta}{\pi}\log\frac{d+y}{y}\right\}, 10 11 c { cos δ + π sin δ log y d + y } , and its tail is at most 6 sin δ / ( π d ) 6\sin\delta/(\pi d) 6 sin δ / ( π d ) 3 / 2 3/2 3/2 cos δ ≥ 9 / 10 \cos\delta\geq9/10 cos δ ≥ 9/10 61 c / 100 61c/100 61 c /100 d d d
B y 2 − 11 c 10 log d + y y − 6 d ≥ 0 ( 0 < y ≤ d / 2 ) . \frac{B}{y^2}-\frac{11c}{10}\log\frac{d+y}{y}-\frac{6}{d}\geq0\qquad(0<y\leq d/2). y 2 B − 10 11 c log y d + y − d 6 ≥ 0 ( 0 < y ≤ d /2 ) . This proves uniform origin positivity for small δ \delta δ (32) . At the cutoff, the only SD law on the upper half is extremal, so closedness gives continuity there.
For effectivity, fix a trial 1 / 2 < ρ < 1 / a 1/2<\rho<1/a 1/2 < ρ < 1/ a 0 ≤ j a ≤ 3 / 2 0\leq j_a\leq3/2 0 ≤ j a ≤ 3/2
∣ P δ ′ ( y ) ∣ ≤ C δ y , C δ = 3 sin δ 4 ( 1 − cos δ ) 2 . |P'_\delta(y)|\leq\frac{C_\delta}{y},\qquad C_\delta=\frac{3\sin\delta}{4(1-\cos\delta)^2}. ∣ P δ ′ ( y ) ∣ ≤ y C δ , C δ = 4 ( 1 − cos δ ) 2 3 sin δ . If ∣ j a ′ ∣ ≤ M |j'_a|\leq M ∣ j a ′ ∣ ≤ M ∣ z ∣ ≤ d 0 |z|\leq d_0 ∣ z ∣ ≤ d 0 y 0 < d 0 y_0<d_0 y 0 < d 0 2 M y 0 2 + C δ y 0 < B sin δ / π 2My_0^2+C_{\delta}y_0<B\sin\delta/\pi 2 M y 0 2 + C δ y 0 < B sin δ / π (33) then controls all smaller radii. At infinity, put h = cos π ( 1 − ρ ) h=\cos\pi(1-\rho) h = cos π ( 1 − ρ ) ξ = π ( 1 − a ( 1 − ρ ) ) \xi=\pi(1-a(1-\rho)) ξ = π ( 1 − a ( 1 − ρ )) S 2 = ∑ r n − 2 a S_2=\sum r_n^{-2a} S 2 = ∑ r n − 2 a
− K a , ρ ′ ( x ) ≥ a sin ξ π x − a − 1 − C 2 x − 2 a − 1 , -K'_{a,\rho}(x)\ge\frac{a\sin\xi}{\pi}x^{-a-1}-C_2x^{-2a-1}, − K a , ρ ′ ( x ) ≥ π a sin ξ x − a − 1 − C 2 x − 2 a − 1 , C 2 = S 2 { 2 a sin ξ Γ ( 2 a + 1 ) π + 2 ( 2 a + 1 e h ) 2 a + 1 } . C_2=S_2\left\{\frac{2a\sin\xi\,\Gamma(2a+1)}{\pi}+2\left(\frac{2a+1}{eh}\right)^{2a+1}\right\}. C 2 = S 2 { π 2 a sin ξ Γ ( 2 a + 1 ) + 2 ( e h 2 a + 1 ) 2 a + 1 } . It follows by ( 1 + 2 z cos ξ + z 2 ) − 1 ≥ ( 1 + z ) − 2 ≥ 1 − 2 z (1+2z\cos\xi+z^2)^{-1}\ge(1+z)^{-2}\ge1-2z ( 1 + 2 z cos ξ + z 2 ) − 1 ≥ ( 1 + z ) − 2 ≥ 1 − 2 z S 2 S_2 S 2 X X X X a > 2 π C 2 / ( a sin ξ ) X^a>2\pi C_2/(a\sin\xi) X a > 2 π C 2 / ( a sin ξ ) x ≥ X x\ge X x ≥ X s ( a ) s(a) s ( a )
At the upper SD cutoff there is also a local analytic closing law. At ( α S D , 1 , y ( α S D ) ) (\alpha_{\mathrm{SD}},1,y(\alpha_{\mathrm{SD}})) ( α SD , 1 , y ( α SD )) ( D , D y ) (D,D_y) ( D , D y ) ( q , y ) (q,y) ( q , y ) D q D y y > 0 D_qD_{yy}>0 D q D y y > 0
1 a − s ( a ) = C S D ( α S D − a ) + O ( ( α S D − a ) 2 ) , C S D = ( 1 α S D − 1 2 ) − D a D q ∣ ( α S D , 1 , y ( α S D ) ) > 0. (34) \frac{1}{a}-s(a)=C_{\mathrm{SD}}(\alpha_{\mathrm{SD}}-a)+O((\alpha_{\mathrm{SD}}-a)^2),\qquad C_{\mathrm{SD}}=\left(\frac{1}{\alpha_{\mathrm{SD}}}-\frac{1}{2}\right)\left.\frac{-D_a}{D_q}\right|_{(\alpha_{\mathrm{SD}},1,y(\alpha_{\mathrm{SD}}))}>0. \tag*{(34)} a 1 − s ( a ) = C SD ( α SD − a ) + O (( α SD − a ) 2 ) , C SD = ( α SD 1 − 2 1 ) D q − D a ( α SD , 1 , y ( α SD )) > 0. (34) Uniform origin and tail bounds, and strict outside-core positivity, are what make this local contact the global boundary. No numerical neighborhood size or numerical value of C S D C_{\mathrm{SD}} C SD
The square-root SD boundary at index one
Lemma 4.4 (A sharp geometric-series bound). For all v > 0 v>0 v > 0 and t ∈ R t\in\mathbb{R} t ∈ R
2 v 2 Im ∑ n ≥ 2 n e − n v + i n t ≤ M , M = 3 3 4 . (35) 2v^2\operatorname{Im}\sum_{n\ge2}ne^{-nv+int}\le M,\qquad M=\frac{3\sqrt{3}}{4}. \tag*{(35)} 2 v 2 Im n ≥ 2 ∑ n e − n v + in t ≤ M , M = 4 3 3 . (35) The constant is optimal, as v ↓ 0 v\downarrow0 v ↓ 0 t = v / 3 t=v/\sqrt{3} t = v / 3
Proof. Reduce t t t 2 π 2\pi 2 π 0 ≤ t ≤ π 0\le t\le\pi 0 ≤ t ≤ π n = 1 n=1 n = 1
sinh v sin t 2 ( cosh v − cos t ) 2 = Im ∑ k ∈ Z ( v − i ( t − 2 π k ) ) − 2 . \frac{\sinh v\sin t}{2(\cosh v-\cos t)^2}=\operatorname{Im}\sum_{k\in\mathbb{Z}}(v-\mathrm{i}(t-2\pi k))^{-2}. 2 ( cosh v − cos t ) 2 sinh v sin t = Im k ∈ Z ∑ ( v − i ( t − 2 π k ) ) − 2 . If v ≤ 3 π v\le\sqrt{3}\pi v ≤ 3 π k k k − k -k − k k = 0 k=0 k = 0 u / ( v 2 + u 2 ) 2 u/(v^2+u^2)^2 u / ( v 2 + u 2 ) 2 u ≥ v / 3 u\ge v/\sqrt{3} u ≥ v / 3 2 π k − t ≥ π 2\pi k-t\ge\pi 2 π k − t ≥ π 2 v t / ( v 2 + t 2 ) 2 2vt/(v^2+t^2)^2 2 v t / ( v 2 + t 2 ) 2 n = 1 n=1 n = 1 2 v 2 2v^2 2 v 2 3 3 / 4 3\sqrt{3}/4 3 3 /4 − π ≤ t ≤ 0 -\pi\le t\le0 − π ≤ t ≤ 0 e − v e^{-v} e − v 2 v 2 e − v ≤ 8 e − 2 < M 2v^2e^{-v}\le8e^{-2}<M 2 v 2 e − v ≤ 8 e − 2 < M v > 3 π v>\sqrt{3}\pi v > 3 π 2 v 2 e − v ( 1 − e − v ) − 1 < M 2v^2e^{-v}(1-e^{-v})^{-1}<M 2 v 2 e − v ( 1 − e − v ) − 1 < M t = v / 3 t=v/\sqrt{3} t = v / 3
Proof of (9) . Write e = a − 1 e=a-1 e = a − 1 q = C 0 e q=C_0\sqrt{e} q = C 0 e C 0 > 0 C_0>0 C 0 > 0 h = sin { π ( 1 / a − 1 / 2 ) q } h=\sin\{\pi(1/a-1/2)q\} h = sin { π ( 1/ a − 1/2 ) q } c 0 = cos { π ( 1 / a − 1 / 2 ) q } c_0=\cos\{\pi(1/a-1/2)q\} c 0 = cos { π ( 1/ a − 1/2 ) q } p = h − i c 0 p=h-\mathrm{i}c_0 p = h − i c 0 v = h y v=hy v = h y h / e → π C 0 / 2 h/\sqrt{e}\to\pi C_0/2 h / e → π C 0 /2
R n , e = n / d e − A e + ϵ n , e , n = 2 , 3 , … , d e = 1 + O ( e 2 ) , A e = O ( e ) , ∣ ϵ n , e ∣ ≤ C e / n . (36) R_{n,e}=n/d_e-A_e+\epsilon_{n,e},\qquad n=2,3,\ldots,\qquad d_e=1+O(e^2),\qquad A_e=O(e),\qquad|\epsilon_{n,e}|\le Ce/n. \tag*{(36)} R n , e = n / d e − A e + ϵ n , e , n = 2 , 3 , … , d e = 1 + O ( e 2 ) , A e = O ( e ) , ∣ ϵ n , e ∣ ≤ C e / n . (36) The uniformity in n n n Lemma H.6 . A fixed-zero Taylor expansion would not suffice here.
Write the signed kernel in scaled radius as
K ( c a y ) = P e ( y ) − 2 Re ∑ n ≥ 2 e − p y R n , e . K(c_a y)=P_e(y)-2\operatorname{Re}\sum_{n\ge2}e^{-pyR_{n,e}}. K ( c a y ) = P e ( y ) − 2 Re n ≥ 2 ∑ e − p y R n , e . With ξ = π ( 1 − a / 2 ) ( 1 + q ) \xi=\pi(1-a/2)(1+q) ξ = π ( 1 − a /2 ) ( 1 + q )
− P e ′ ( y ) ≥ a sin ξ π e y − a − 1 − a sin ξ π Γ ( 3 a + 1 ) Q 3 ( e ) y − 3 a − 1 , (37) -P'_e(y)\ge\frac{a\sin\xi}{\pi e}y^{-a-1}-\frac{a\sin\xi}{\pi}\Gamma(3a+1)Q_3(e)y^{-3a-1}, \tag*{(37)} − P e ′ ( y ) ≥ π e a sin ξ y − a − 1 − π a sin ξ Γ ( 3 a + 1 ) Q 3 ( e ) y − 3 a − 1 , (37) where Q 3 Q_3 Q 3 cos ξ < 0 \cos\xi<0 cos ξ < 0 e e e π e y 2 [ − P e ′ ( y ) ] ≥ 1 + o ( 1 ) \pi e y^2[-P'_e(y)]\ge1+o(1) π e y 2 [ − P e ′ ( y )] ≥ 1 + o ( 1 ) 1 ≤ y ≤ V / h 1\le y\le V/h 1 ≤ y ≤ V / h V V V y − e y^{-e} y − e y y y 1 + o ( 1 ) 1+o(1) 1 + o ( 1 )
Compare (36) with R ˉ n = n / d e − A e \bar R_n=n/d_e-A_e R ˉ n = n / d e − A e 1 ≤ y ≤ V / h 1\le y\le V/h 1 ≤ y ≤ V / h
∣ ∑ R ˉ n e − p y R n − ∑ R ˉ n e − p y R ˉ n ∣ ≤ C e ( 1 + ∣ log v ∣ ) + C e y e c 1 v − 1 = O ( e + e ∣ log e ∣ ) . \left|\sum\bar R_n e^{-pyR_n}-\sum\bar R_n e^{-py\bar R_n}\right|\le Ce(1+|\log v|)+\frac{Cey}{e^{c_1v}-1}
=O(\sqrt{e}+e|\log e|). ∑ R ˉ n e − p y R n − ∑ R ˉ n e − p y R ˉ n ≤ C e ( 1 + ∣ log v ∣ ) + e c 1 v − 1 C ey = O ( e + e ∣ log e ∣ ) . Indeed v ≥ h v\ge h v ≥ h ∣ p y e n ∣ ≤ C e y / n = O ( e ) |py e_n|\le Cey/n=O(\sqrt{e}) ∣ p y e n ∣ ≤ C ey / n = O ( e ) c 1 v n c_1vn c 1 v n
e p y A e { d e − 1 ∑ n ≥ 2 n e − p y n / d e − A e ∑ n ≥ 2 e − p y n / d e } . e^{pyA_e}\left\{d_e^{-1}\sum_{n\ge2}ne^{-pyn/d_e}-A_e\sum_{n\ge2}e^{-pyn/d_e}\right\}. e p y A e { d e − 1 n ≥ 2 ∑ n e − p y n / d e − A e n ≥ 2 ∑ e − p y n / d e } . Since e p y A e = 1 + O ( e ) e^{pyA_e}=1+O(\sqrt{e}) e p y A e = 1 + O ( e ) p = − i + O ( e ) p=-i+O(\sqrt{e}) p = − i + O ( e )
y 2 h 2 2 Re ( p ∑ n ≥ 2 R n e − p y R n ) ≤ M + o ( 1 ) . (38) y^2h^2\,2\operatorname{Re}\left(p\sum_{n\ge2}R_ne^{-pyR_n}\right)\le M+o(1). \tag*{(38)} y 2 h 2 2 Re ( p n ≥ 2 ∑ R n e − p y R n ) ≤ M + o ( 1 ) . (38) The absolute geometric derivative sums multiplied by v 2 v^2 v 2
For y ≥ V / h y\ge V/h y ≥ V / h n / 2 ≤ R n ≤ 2 n n/2\le R_n\le2n n /2 ≤ R n ≤ 2 n C e − v ( 1 − e − v / 2 ) − 2 Ce^{-v}(1-e^{-v/2})^{-2} C e − v ( 1 − e − v /2 ) − 2 e y a + 1 e y^{a+1} e y a + 1 V V V e h − a − 1 eh^{-a-1} e h − a − 1 (37) holds throughout this infinite range and has positive limiting normalized constant.
The origin is controlled separately. In (33) , δ = π ( 1 / a − 1 / 2 ) ( 1 − q ) → π / 2 \delta=\pi(1/a-1/2)(1-q)\to\pi/2 δ = π ( 1/ a − 1/2 ) ( 1 − q ) → π /2 δ ≥ π / 3 \delta\ge\pi/3 δ ≥ π /3 j a ′ j'_a j a ′ ∣ z ∣ ≤ 1 |z|\le1 ∣ z ∣ ≤ 1
4 π ∫ 0 ∞ j a ( u ) u 2 + y 2 d u ≤ 3 y + C . \frac{4}{\pi}\int_0^\infty\frac{j_a(u)}{u^2+y^2}\,\mathrm{d}u\le\frac{3}{y}+C. π 4 ∫ 0 ∞ u 2 + y 2 j a ( u ) d u ≤ y 3 + C . Here j a ≤ 3 / 2 j_a\le3/2 j a ≤ 3/2 ∫ 2 ∞ j a ( u ) u − 1 d u \int_2^\infty j_a(u)u^{-1}\,\mathrm{d}u ∫ 2 ∞ j a ( u ) u − 1 d u
− ∂ y K ( c a y ) ≥ a sin δ π e y 2 − 3 y − C > 0 ( 0 < y ≤ 1 ) -\partial_yK(c_ay)\ge\frac{a\sin\delta}{\pi e y^2}-\frac{3}{y}-C>0\qquad(0<y\le1) − ∂ y K ( c a y ) ≥ π e y 2 a sin δ − y 3 − C > 0 ( 0 < y ≤ 1 ) for all sufficiently small e e e
If C 0 2 > 3 3 / π C_0^2>3\sqrt{3}/\pi C 0 2 > 3 3 / π h 2 / e → π 2 C 0 2 / 4 > π M h^2/e\to\pi^2C_0^2/4>\pi M h 2 / e → π 2 C 0 2 /4 > π M − ∂ y K > 0 -\partial_yK>0 − ∂ y K > 0 q q q C 0 2 < 3 3 / π C_0^2<3\sqrt{3}/\pi C 0 2 < 3 3 / π y y y Y = 2 π Y=2\pi Y = 2 π c 0 y / d e = 2 π + ( h y / d e ) / 3 c_0y/d_e=2\pi+(hy/d_e)/\sqrt{3} c 0 y / d e = 2 π + ( h y / d e ) / 3 (35) and the fixed-radius equality in (37) give
e y 2 ∂ y K ( c a y ) ⟶ − 1 π + 4 M π 2 C 0 2 > 0. e y^2\partial_yK(c_ay)\longrightarrow-\frac{1}{\pi}+\frac{4M}{\pi^2C_0^2}>0. e y 2 ∂ y K ( c a y ) ⟶ − π 1 + π 2 C 0 2 4 M > 0. Thus this q q q q S D / e q_{\mathrm{SD}}/\sqrt{e} q SD / e 3 3 / π \sqrt{3\sqrt{3}/\pi} 3 3 / π s ( a ) − 1 / 2 = ( 1 / a − 1 / 2 ) q S D s(a)-1/2=(1/a-1/2)q_{\mathrm{SD}} s ( a ) − 1/2 = ( 1/ a − 1/2 ) q SD (9) . No explicit smallness threshold is used. □
Bell-shape: exact tests and structural consequences
We first derive the two-ray integer-level test and a uniform tail bound. These yield the exclusions, the exceptional-family criterion, and open BS regions outside GGC. We then study the geometry of the skewness sections and the bands that accumulate near index one.
The phase test and an explicit tail
For 0 < a < 1 0<a<1 0 < a < 1 0 < θ < π 0<\theta<\pi 0 < θ < π
Φ a , θ ( r ) = π − 1 Im ψ a ( r e i θ ) . \Phi_{a,\theta}(r)=\pi^{-1}\operatorname{Im}\psi_a(re^{i\theta}). Φ a , θ ( r ) = π − 1 Im ψ a ( r e i θ ) . The continuous logarithm is fixed by Lemma 4.1 . Theorem 2.3 and the analyticity of the interior free stable densities give the exact equivalence
X a , ρ is strictly BS ⟺ there are no 0 < u < v and m ∈ N with X_{a,\rho}\text{ is strictly BS}\quad\Longleftrightarrow\quad\text{there are no }0<u<v\text{ and }m\in\mathbb{N}\text{ with} X a , ρ is strictly BS ⟺ there are no 0 < u < v and m ∈ N with Φ a , θ ( u ) > m > Φ a , θ ( v ) , θ ∈ { π ρ , π ( 1 − ρ ) } . (39) \Phi_{a,\theta}(u)>m>\Phi_{a,\theta}(v),\qquad\theta\in\{\pi\rho,\pi(1-\rho)\}. \tag*{(39)} Φ a , θ ( u ) > m > Φ a , θ ( v ) , θ ∈ { π ρ , π ( 1 − ρ )} . (39) Indeed each phase starts at zero and tends to infinity on a proper ray, by (22) and b a > 0 b_a>0 b a > 0
Lemma 5.1 (A whole-ray derivative bound). For every 0 < a < 1 0<a<1 0 < a < 1 r > 0 r>0 r > 0 0 < θ < π 0<\theta<\pi 0 < θ < π
Im H a ( r e i θ ) ≥ b a r sin θ − tan ( θ / 2 ) . (40) \operatorname{Im}H_a(re^{i\theta})\geq b_a r\sin\theta-\tan(\theta/2). \tag*{(40)} Im H a ( r e i θ ) ≥ b a r sin θ − tan ( θ /2 ) . (40) In particular Φ a , θ ′ ( r ) > 0 \Phi'_{a,\theta}(r)>0 Φ a , θ ′ ( r ) > 0 b a r ( 1 + cos θ ) > 1 b_a r(1+\cos\theta)>1 b a r ( 1 + cos θ ) > 1
Proof. The bounded imaginary part in (22) gives a bounded-density Herglotz representation
ψ a ( z ) − b a z = ∫ 0 ∞ z ϕ ( t ) t ( z + t ) d t , 0 ≤ ϕ ( t ) ≤ 2. \psi_a(z)-b_a z=\int_0^\infty\frac{z\phi(t)}{t(z+t)}\,\mathrm{d}t,\qquad0\leq\phi(t)\leq2. ψ a ( z ) − b a z = ∫ 0 ∞ t ( z + t ) z ϕ ( t ) d t , 0 ≤ ϕ ( t ) ≤ 2. There is no singular part: the imaginary part of the holomorphic function is bounded in the upper half-plane. The constant is zero by ψ a ( 0 ) = 0 \psi_a(0)=0 ψ a ( 0 ) = 0 H a ( z ) − b a z = ∫ ϕ ( t ) ( z + t ) − 2 d t H_a(z)-b_a z=\int\phi(t)(z+t)^{-2}\,\mathrm{d}t H a ( z ) − b a z = ∫ ϕ ( t ) ( z + t ) − 2 d t t = r u t=ru t = r u
Im e i θ ( u + e i θ ) 2 = sin θ ( u 2 − 1 ) ∣ u + e i θ ∣ 4 . \operatorname{Im}\frac{e^{i\theta}}{(u+e^{i\theta})^2}=\frac{\sin\theta(u^2-1)}{\lvert u+e^{i\theta}\rvert^4}. Im ( u + e i θ ) 2 e i θ = ∣ u + e i θ ∣ 4 sin θ ( u 2 − 1 ) . It is negative only on 0 < u < 1 0<u<1 0 < u < 1 ϕ ≤ 2 \phi\leq2 ϕ ≤ 2 ϕ ≥ 0 \phi\geq0 ϕ ≥ 0
2 Im ∫ 0 1 e i θ ( u + e i θ ) 2 d u = − tan ( θ / 2 ) . 2\operatorname{Im}\int_0^1\frac{e^{i\theta}}{(u+e^{i\theta})^2}\,\mathrm{d}u=-\tan(\theta/2). 2 Im ∫ 0 1 ( u + e i θ ) 2 e i θ d u = − tan ( θ /2 ) . This proves the assertion, including all radii beyond the stated cutoff. □
The phase test permits rigorous numerical witnesses without a numerical logarithm-unwrapping assumption. Put β = 1 − a \beta=1-a β = 1 − a T = β r a T=\beta r^a T = β r a η = π ( 1 − a ρ ) \eta=\pi(1-a\rho) η = π ( 1 − a ρ ) G β ( t ) = e − t / β W a ( t / β ) G_\beta(t)=e^{-t/\beta}W_a(t/\beta) G β ( t ) = e − t / β W a ( t / β ) t = T e − i η t=Te^{-i\eta} t = T e − i η
Φ a , π ρ ( r ) = T sin η π β − Arg G β ( t ) π + 2 k . (41) \Phi_{a,\pi\rho}(r)=\frac{T\sin\eta}{\pi\beta}-\frac{\operatorname{Arg}G_\beta(t)}{\pi}+2k. \tag*{(41)} Φ a , π ρ ( r ) = π β T sin η − π Arg G β ( t ) + 2 k . (41) There is exactly one integer lift satisfying
0 < Φ a , π ρ ( r ) − b a r sin ( π ρ ) π < 2 ρ . 0<\Phi_{a,\pi\rho}(r)-\frac{b_a r\sin(\pi\rho)}{\pi}<2\rho. 0 < Φ a , π ρ ( r ) − π b a r sin ( π ρ ) < 2 ρ . Its interval has length less than two, so strict interval checks select the correct branch. An inconclusive check is not accepted as a phase value. The retained-circle Hankel evaluator, including both banks and their tails, is described in the appendix.
Exclusion above one and the exceptional family
For a > 1 a>1 a > 1 (17) vanishes at t = λ n 1 / a e i π ( ρ − 1 / 2 ) t=\lambda_n^{1/a}e^{i\pi(\rho-1/2)} t = λ n 1/ a e iπ ( ρ − 1/2 ) [11 Remark 1.2(b) and Theorem 1.3] , a weakly bell-shaped probability law has a zero-free holomorphic characteristic transform there. The identity theorem proves the contradiction. This exclusion does not use either numerical cutoff. At one the interior strictly stable laws are Cauchy. Below one the endpoint densities are whale-shaped by Hasebe–Simon–Wang, and hence weakly BS by [11 Theorem 1.13] ; they are not strictly BS.
For the exceptional family scale and reflect to ∣ B ∣ = 1 |B|=1 ∣ B ∣ = 1 ℓ = A / ( π ∣ B ∣ ) \ell=A/(\pi|B|) ℓ = A / ( π ∣ B ∣ ) [7 Theorem 3 and Remark 7] has, besides its Cauchy part, the kernel
1 x − ∑ n ≥ 2 e − n x . \frac{1}{x}-\sum_{n\ge2}e^{-nx}. x 1 − n ≥ 2 ∑ e − n x . The Laplace densities of the two Lévy densities are therefore
ℓ u , ( ℓ + 1 ) u − N ( u ) , N ( u ) = # { n ≥ 2 : n ≤ u } , u > 0. (42) \ell u,\qquad(\ell+1)u-N(u),\qquad N(u)=\#\{n\ge2:n\le u\},\qquad u>0. \tag*{(42)} ℓ u , ( ℓ + 1 ) u − N ( u ) , N ( u ) = # { n ≥ 2 : n ≤ u } , u > 0. (42) At an integer n ≥ 2 n\ge2 n ≥ 2 ℓ n + 2 \ell n+2 ℓ n + 2 ℓ n + 1 \ell n+1 ℓ n + 1 ℓ n \ell n ℓ n [11 Theorems 1.1 and 1.3] . Values at the single jump point do not remove the strict inequalities on its two sides. Consequently 2 ℓ 2\ell 2 ℓ 3 ℓ 3\ell 3 ℓ ℓ \ell ℓ ℓ ≥ 0 \ell\ge0 ℓ ≥ 0 A > 0 A>0 A > 0 A = 0 A=0 A = 0 [6 Theorem 4(d)] .
All these exceptional laws are SD: after unit scaling, the derivative of the non-Cauchy kernel is
− x − 2 + e x ( e x − 1 ) 2 − e − x < 0 , -x^{-2}+\frac{e^x}{(e^x-1)^2}-e^{-x}<0, − x − 2 + ( e x − 1 ) 2 e x − e − x < 0 , since 2 sinh ( x / 2 ) > x 2\sinh(x/2)>x 2 sinh ( x /2 ) > x B ≠ 0 B\ne0 B = 0 (42) preclude GGC, equivalently by uniqueness of the Laplace spectral measure. This also recovers the known non-GGC conclusion in [7 Remark 7] .
Bell-shape beyond GGC
Proposition 5.2 (Noninteger contacts leave room for bell-shape ). Consider a parameter family satisfying Theorem 2.3 , with phases starting at zero and tending to infinity. Suppose every decreasing part of each phase is contained in finitely many disjoint compact intervals, the phase on each such interval is in a single open integer strip ( m , m + 1 ) (m,m+1) ( m , m + 1 )
Proof. No interval containing a decreasing part contains an integer level. Between these intervals each integer passage is upward. Since the phase starts below each positive integer and ends above it, there is exactly one change of sign for every level. Apply Theorem 2.3 . Strict margins on the finitely many compact pieces, and the stated controls at both ends, persist under small parameter perturbations. □ \square □
At a GGC boundary with finitely many contacts, if none of the contact phase values is an integer, take small disjoint cores around them. The phase stays in an open integer strip in each core. Strict outside-core GGC positivity and the end estimates then imply a BS neighborhood, even if the radial derivative becomes negative inside a core. This gives a general explanation for the separation between GGC and bell-shape.
The following examples are independently certified:
( a , ρ ) = ( 99 100 , 9869504901 10 10 ) , ( 19 20 , 99340401760 10 11 ) , ( 9 10 , 99993759565 10 11 ) . (43) (a,\rho)=\left(\frac{99}{100},\frac{9869504901}{10^{10}}\right),\quad\left(\frac{19}{20},\frac{99340401760}{10^{11}}\right),\quad\left(\frac{9}{10},\frac{99993759565}{10^{11}}\right). \tag*{(43)} ( a , ρ ) = ( 100 99 , 1 0 10 9869504901 ) , ( 20 19 , 1 0 11 99340401760 ) , ( 10 9 , 1 0 11 99993759565 ) . (43) For the upper ray the respective compact T T T [ 2.005 , 2.011 ] [2.005,2.011] [ 2.005 , 2.011 ] [ 2.06 , 2.10 ] [2.06,2.10] [ 2.06 , 2.10 ] [ 2.99 , 3.12 ] [2.99,3.12] [ 2.99 , 3.12 ] ( 4 , 5 ) (4,5) ( 4 , 5 ) ( 1 , 2 ) (1,2) ( 1 , 2 ) ( 1 , 2 ) (1,2) ( 1 , 2 ) certify_bell_beyond_ggc.py verify, with the correct branch fixed by (41) . The opposite ray has angle below π / 2 \pi/2 π /2 Proposition 5.2 proves strict BS and non-GGC. The strict witnesses and locally uniform end estimates give nonempty open neighborhoods with the same properties.
For comparison, at a = 999 / 1000 a=999/1000 a = 999/1000 ρ = 1999 / 2000 \rho=1999/2000 ρ = 1999/2000
Φ ( T = 29 / 10 ) > 3.30774 > 3 > 2.69471 > Φ ( T = 31 / 10 ) . \Phi(T=29/10)>3.30774>3>2.69471>\Phi(T=31/10). Φ ( T = 29/10 ) > 3.30774 > 3 > 2.69471 > Φ ( T = 31/10 ) . Here T = β r a T=\beta r^a T = β r a
Endpoint bands
Proposition 5.3 (Strict BS next to each one-sided law). For each 0 < a < 1 0<a<1 0 < a < 1 η ( a ) > 0 \eta(a)>0 η ( a ) > 0 X a , ρ X_{a,\rho} X a , ρ 0 < ρ < η ( a ) 0<\rho<\eta(a) 0 < ρ < η ( a ) 1 − η ( a ) < ρ < 1 1-\eta(a)<\rho<1 1 − η ( a ) < ρ < 1 a a a ( 0 , 1 ) (0,1) ( 0 , 1 )
Proof. The GGC theorem handles a ≤ 4 / 5 a\le4/5 a ≤ 4/5 β = 1 − a \beta=1-a β = 1 − a L a L_a L a t = r a t=r^a t = r a
Im L a ( − r + i 0 ) = ∑ n ≥ 1 Γ ( β n − 1 ) sin 2 ( π β n ) π n ! t n . (44) \operatorname{Im} L_a(-r+i0)=\sum_{n\ge1}\frac{\Gamma(\beta n-1)\sin^2(\pi\beta n)}{\pi n!}t^n. \tag*{(44)} Im L a ( − r + i 0 ) = n ≥ 1 ∑ π n ! Γ ( β n − 1 ) sin 2 ( π β n ) t n . (44) Removable Gamma factors are interpreted by continuity. The nonzero coefficients are negative before β n = 1 \beta n=1 β n = 1 m m m 1 / β 1/\beta 1/ β t m t^m t m (44) has exactly one positive zero t 0 t_0 t 0
There is no zero of L a L_a L a r 0 = t 0 1 / a r_0=t_0^{1/a} r 0 = t 0 1/ a 0 ≤ Φ a , π ≤ 2 0\le\Phi_{a,\pi}\le2 0 ≤ Φ a , π ≤ 2 (44) place the phase in ( 0 , 1 ) (0,1) ( 0 , 1 ) r 0 r_0 r 0 ( 1 , 2 ) (1,2) ( 1 , 2 ) − log L a -\log L_a − log L a π \pi π L a ( − r 0 ) < 0 L_a(-r_0)<0 L a ( − r 0 ) < 0 Φ a , π ( r 0 ) = 1 \Phi_{a,\pi}(r_0)=1 Φ a , π ( r 0 ) = 1 Φ a , π ′ ( r 0 ) > 0 \Phi'_{a,\pi}(r_0)>0 Φ a , π ′ ( r 0 ) > 0
We need a tail estimate uniform up to that boundary. Choose an integer N N N β N > 1 \beta N>1 β N > 1 E N E_N E N N N N
J N ( z ) = ∫ 0 ∞ e − u u − 2 E N ( z u β ) d u . J_N(z)=\int_0^\infty e^{-u}u^{-2}E_N(zu^\beta)\,du. J N ( z ) = ∫ 0 ∞ e − u u − 2 E N ( z u β ) d u . Both endpoints converge, and reflection gives exactly
W a ( z ) = P N − 1 ( z ) + i 2 π { J N ( z e i π β ) − J N ( z e − i π β ) } . (45) W_a(z)=P_{N-1}(z)+\frac{i}{2\pi}\{J_N(ze^{i\pi\beta})-J_N(ze^{-i\pi\beta})\}. \tag*{(45)} W a ( z ) = P N − 1 ( z ) + 2 π i { J N ( z e iπ β ) − J N ( z e − iπ β )} . (45) For s = r e i ( π − δ ) s=re^{i(\pi-\delta)} s = r e i ( π − δ ) r a e − i a δ r^ae^{-ia\delta} r a e − ia δ u = e − i δ v u=e^{-i\delta}v u = e − i δ v − v + r a v β -v+r^av^\beta − v + r a v β v 0 = β 1 / a r v_0=\beta^{1/a}r v 0 = β 1/ a r b a r b_ar b a r − a / v 0 -a/v_0 − a / v 0 δ \delta δ
J N ( r a e − i a δ ) = 2 π a v 0 − 3 / 2 e 3 i δ / 2 b a r e − i δ ( 1 + O ( r − 1 ) ) . J_N(r^a e^{-ia\delta})=\sqrt{\frac{2\pi}{a}}v_0^{-3/2}e^{3i\delta/2}b_a r e^{-i\delta}(1+O(r^{-1})). J N ( r a e − ia δ ) = a 2 π v 0 − 3/2 e 3 i δ /2 b a r e − i δ ( 1 + O ( r − 1 )) . Strict concavity gives an exponential gap off that neighborhood. The subtractions make the small arc vanish; their contributions are polynomial or exp ( O ( r a ) ) \exp(O(r^a)) exp ( O ( r a )) b a r max ( cos ( 2 π β + a δ ) , 0 ) 1 / a b_a r\max(\cos(2\pi\beta+a\delta),0)^{1/a} b a r max ( cos ( 2 π β + a δ ) , 0 ) 1/ a b a r cos δ b_a r\cos\delta b a r cos δ δ \delta δ C a = β − 3 / ( 2 a ) / 2 π a C_a=\beta^{-3/(2a)}/\sqrt{2\pi a} C a = β − 3/ ( 2 a ) / 2 π a
L a ( s ) = C a e − b a s s − 3 / 2 ( 1 + O ( r − 1 ) ) , ψ a ′ ( s ) = b a + 3 2 s + O ( r − 2 ) , (46) L_a(s)=C_a e^{-b_a s}s^{-3/2}(1+O(r^{-1})),\qquad\psi_a'(s)=b_a+\frac{3}{2s}+O(r^{-2}), \tag*{(46)} L a ( s ) = C a e − b a s s − 3/2 ( 1 + O ( r − 1 )) , ψ a ′ ( s ) = b a + 2 s 3 + O ( r − 2 ) , (46) uniformly for 0 ≤ δ ≤ δ 0 0\leq\delta\leq\delta_0 0 ≤ δ ≤ δ 0 (22) : at δ = 0 \delta=0 δ = 0 3 / 2 3/2 3/2
It follows that, for some C , R C,R C , R
∣ Φ a , π − δ ( r ) − b a r sin δ π − 3 2 ( 1 − δ / π ) ∣ ≤ C / r , \left|\Phi_{a,\pi-\delta}(r)-\frac{b_a r\sin\delta}{\pi}-\frac{3}{2}(1-\delta/\pi)\right|\leq C/r, Φ a , π − δ ( r ) − π b a r sin δ − 2 3 ( 1 − δ / π ) ≤ C / r , ∣ Φ a , π − δ ′ ( r ) − b a sin δ π ∣ ≤ C / r 2 . \left|\Phi'_{a,\pi-\delta}(r)-\frac{b_a\sin\delta}{\pi}\right|\leq C/r^2. Φ a , π − δ ′ ( r ) − π b a sin δ ≤ C / r 2 . Take δ 0 < π / 12 \delta_0<\pi/12 δ 0 < π /12 C / R < 1 / 16 C/R<1/16 C / R < 1/16 R > 4 C R>4C R > 4 C r ≥ R r\geq R r ≥ R D = b a r sin δ / π D=b_a r\sin\delta/\pi D = b a r sin δ / π D ≤ 1 / 4 D\leq1/4 D ≤ 1/4 D ≥ 1 / 4 D\geq1/4 D ≥ 1/4 1 / ( 4 r ) − C / r 2 > 0 1/(4r)-C/r^2>0 1/ ( 4 r ) − C / r 2 > 0 δ \delta δ D ( R ) < 1 / 4 D(R)<1/4 D ( R ) < 1/4
On the remaining compact radii, isolate the transverse boundary crossing and use continuity. Away from it the boundary phase has positive distance from every integer; the origin Wright series controls a uniform neighborhood of zero. Small δ \delta δ δ < π / 2 \delta<\pi/2 δ < π /2 (39) proves the result. All constants and the integer N N N a a a a = 4 / 5 a=4/5 a = 4/5
Finite components and boundary contacts
We use two standard notions from real analytic geometry. A semianalytic set is locally a finite union of sets described by finitely many real analytic equalities and inequalities. A subanalytic set is locally a projection of a relatively compact semianalytic set; see [3 Sections 2 and 3] . These notions will be applied only on compact parameter and radius charts.
Proposition 5.4 (Finite geometry on compact index ranges). For a fixed 0 < a < 1 0<a<1 0 < a < 1 ( 0 , 1 ) (0,1) ( 0 , 1 ) a a a m ≥ 1 m\geq1 m ≥ 1
Φ a , θ ( r ) = m , ∂ r Φ a , θ ( r ) = 0. (47) \Phi_{a,\theta}(r)=m,\qquad\partial_r\Phi_{a,\theta}(r)=0. \tag*{(47)} Φ a , θ ( r ) = m , ∂ r Φ a , θ ( r ) = 0. (47) Proof. Fix a compact index interval inside ( 0 , 1 ) (0,1) ( 0 , 1 ) ρ ∈ [ η , 1 − η ] \rho\in[\eta,1-\eta] ρ ∈ [ η , 1 − η ] η > 0 \eta>0 η > 0 r ∈ [ r − , r + ] r\in[r_-,r_+] r ∈ [ r − , r + ] m ≤ M m\leq M m ≤ M
The phase is jointly real analytic in a , ρ , r a,\rho,r a , ρ , r m ≤ M m\leq M m ≤ M
{ ( a , ρ , u , v ) : r − ≤ u < v ≤ r + , Φ ( u ) > m > Φ ( v ) } \{(a,\rho,u,v):r_-\leq u<v\leq r_+,\ \Phi(u)>m>\Phi(v)\} {( a , ρ , u , v ) : r − ≤ u < v ≤ r + , Φ ( u ) > m > Φ ( v )} is relatively compact and semianalytic. Its projection onto ( a , ρ ) (a,\rho) ( a , ρ ) [3 Definition 3.1, Theorems 3.10 and 3.14] . Strict witnesses persist, so the BS section is relatively closed. Adding the endpoint bands proves the stated form.
If no contact (47) existed at a boundary, all the finitely many compact-range integer crossings would be transverse and upward. The implicit function theorem and compactness would preserve them, as well as all intervening integer strips, in a parameter neighborhood. This contradicts boundary status. The contact condition is necessary; a local contact need not be a global boundary. □
For a fixed parameter, an equivalent finite test lists the maximal decreasing runs ( u j , v j ) (u_j,v_j) ( u j , v j )
( Φ ( v j ) , Φ ( u j ) ) ∩ N = ∅ . (\Phi(v_j),\Phi(u_j))\cap\mathbb{N}=\varnothing. ( Φ ( v j ) , Φ ( u j )) ∩ N = ∅ . Negative derivative intervals must be joined across an isolated zero if no positive derivative interval separates them; otherwise an odd higher-order downward crossing could be missed. Equality to an integer at an extremum is permitted: a tangency is not a sign change. This finite reduction does not locate all parameter boundaries.
Proposition 5.3, the symmetric GGC law, and the strict non-BS witness at a = 999 / 1000 a=999/1000 a = 999/1000 ρ = 1999 / 2000 \rho=1999/2000 ρ = 1999/2000 a = 999 / 1000 a=999/1000 a = 999/1000
A quantization obstruction near one
Suppose a j ↑ 1 a_j\uparrow1 a j ↑ 1 β j = 1 − a j \beta_j=1-a_j β j = 1 − a j min ( ρ j , 1 − ρ j ) / β j → ℓ < ∞ \min(\rho_j,1-\rho_j)/\beta_j\to\ell<\infty min ( ρ j , 1 − ρ j ) / β j → ℓ < ∞
ℓ ∈ { 0 , 1 , 2 , … } . (48) \ell\in\{0,1,2,\ldots\}. \tag*{(48)} ℓ ∈ { 0 , 1 , 2 , … } . (48) Indeed the Cauchy factorization and the affine endpoint limit give, after reflection if necessary,
β j β j − X a j , 1 − min ( ρ j , 1 − ρ j ) β j ⟹ T + π ℓ C . \frac{\beta_j^{\beta_j}-X_{a_j,1-\min(\rho_j,1-\rho_j)}}{\beta_j}\Longrightarrow T+\pi\ell C. β j β j β j − X a j , 1 − m i n ( ρ j , 1 − ρ j ) ⟹ T + π ℓ C . The positive-law endpoint limit is the exceptional-law limit in [7 Section 2.4, (15)] , also recovered from the compact Hankel limit in the appendix. Weak BS is affine invariant and closed under weak convergence [11 Lemma 3.1] . The exceptional classification therefore gives (48) . Uniformly on a compact set of noninteger ratios, this excludes BS for all sufficiently small β \beta β
Integer-ratio bands and an unbounded number of components
The quantization obstruction has a partial converse with a uniform width. Only a qualitative eventual statement is needed here.
Proposition 5.5. For every integer ℓ ≥ 1 \ell\ge1 ℓ ≥ 1 β ℓ , k ℓ > 0 \beta_\ell,k_\ell>0 β ℓ , k ℓ > 0 0 < β < β ℓ 0<\beta<\beta_\ell 0 < β < β ℓ a = 1 − β a=1-\beta a = 1 − β
∣ π ( 1 − ρ ) − δ ℓ ( β ) ∣ ≤ k ℓ β 2 , δ ℓ ( β ) = arctan ℓ sin c cos c 1 + ℓ sin 2 c , c = π β 1 − β , (49) |\pi(1-\rho)-\delta_\ell(\beta)|\le k_\ell\beta^2,\qquad
\delta_\ell(\beta)=\arctan\frac{\ell\sin c\cos c}{1+\ell\sin^2 c},\qquad
c=\frac{\pi\beta}{1-\beta}, \tag*{(49)} ∣ π ( 1 − ρ ) − δ ℓ ( β ) ∣ ≤ k ℓ β 2 , δ ℓ ( β ) = arctan 1 + ℓ sin 2 c ℓ sin c cos c , c = 1 − β π β , (49) the law X a , ρ X_{a,\rho} X a , ρ δ ℓ ( β ) / π = ℓ β + ℓ β 2 + O ( β 3 ) \delta_\ell(\beta)/\pi=\ell\beta+\ell\beta^2+O(\beta^3) δ ℓ ( β ) / π = ℓ β + ℓ β 2 + O ( β 3 ) a ↑ 1 a\uparrow1 a ↑ 1 β ℓ \beta_\ell β ℓ
The required estimate must keep the two saddles together. Separate fixed-index Wright expansions would not control their cancellation.
Lemma 5.6 (Uniform two-saddle estimate). Let 0 < β ≤ 1 / 10 0 < \beta\le1/10 0 < β ≤ 1/10 a = 1 − β a = 1-\beta a = 1 − β N = ⌈ 2 / β ⌉ N = \lceil2/\beta\rceil N = ⌈ 2/ β ⌉ E N ( z ) = e z − ∑ k < N z k / k ! E_N(z)=e^z-\sum_{k<N}z^k/k! E N ( z ) = e z − ∑ k < N z k / k !
J N ( z ) = ∫ 0 ∞ e − u u − 2 E N ( z u β ) d u . J_N(z)=\int_0^\infty e^{-u}u^{-2}E_N(zu^\beta)\,du. J N ( z ) = ∫ 0 ∞ e − u u − 2 E N ( z u β ) d u . There are constants C , V 0 , ϵ > 0 C,V_0,\epsilon>0 C , V 0 , ϵ > 0 β \beta β ∣ v ∣ ≥ V 0 |v|\ge V_0 ∣ v ∣ ≥ V 0 ∣ arg v ∣ ≤ ϵ |\arg v|\le\epsilon ∣ arg v ∣ ≤ ϵ
J N ( v a / β ) = 2 π / a v − 3 / 2 e a v / β A β ( v ) , (50) J_N(v^{a/\beta})=\sqrt{2\pi/a}\,v^{-3/2}e^{av/\beta}A_\beta(v), \tag*{(50)} J N ( v a / β ) = 2 π / a v − 3/2 e a v / β A β ( v ) , (50) ∣ A β − 1 ∣ ≤ C / ∣ v ∣ , ∣ ( log A β ) ( j ) ∣ ≤ C ∣ v ∣ − j − 1 ( j = 1 , 2 ) . |A_\beta-1|\le C/|v|,\qquad|(\log A_\beta)^{(j)}|\le C|v|^{-j-1}\quad(j=1,2). ∣ A β − 1∣ ≤ C /∣ v ∣ , ∣ ( log A β ) ( j ) ∣ ≤ C ∣ v ∣ − j − 1 ( j = 1 , 2 ) . The amplitude is holomorphic in v v v s = r e i ( π − δ ) s=re^{i(\pi-\delta)} s = r e i ( π − δ ) δ = O ( β ) \delta=O(\beta) δ = O ( β )
v = β 1 / a r , R = a v / β , x = R { sin ( δ + 2 c ) − sin δ } − 3 c , κ = tan ( δ + c ) , q = e − κ ( x + 3 c ) . v=\beta^{1/a}r,\qquad R=av/\beta,\qquad x=R\{\sin(\delta+2c)-\sin\delta\}-3c,\qquad\kappa=\tan(\delta+c),\qquad q=e^{-\kappa(x+3c)}. v = β 1/ a r , R = a v / β , x = R { sin ( δ + 2 c ) − sin δ } − 3 c , κ = tan ( δ + c ) , q = e − κ ( x + 3 c ) . Then, uniformly for real v ≥ V 0 v\ge V_0 v ≥ V 0
L a ( s ) = i 2 π J N ( r a e − i a δ ) { 1 − q e − i x + e β ( v ) } , ∣ e β ( v ) ∣ ≤ C β / v , ∣ e β ′ ( v ) ∣ ≤ C β / v 2 . (51) L_a(s)=\frac{i}{2\pi}J_N(r^ae^{-ia\delta})\{1-qe^{-ix}+e_\beta(v)\},\qquad|e_\beta(v)|\le C\beta/v,\qquad|e'_\beta(v)|\le C\beta/v^2. \tag*{(51)} L a ( s ) = 2 π i J N ( r a e − ia δ ) { 1 − q e − i x + e β ( v )} , ∣ e β ( v ) ∣ ≤ C β / v , ∣ e β ′ ( v ) ∣ ≤ C β / v 2 . (51) The constants in the last statement can depend on a fixed bound for δ / β \delta/\beta δ / β N N N
Proof. The regularization makes the integral entire, since β N ≥ 2 \beta N\ge2 β N ≥ 2 u = v w u=vw u = v w
f β ( w ) = w − 1 − w β − 1 β , f 0 ( w ) = w − 1 − log w . f_\beta(w)=w-1-\frac{w^\beta-1}{\beta},\qquad f_0(w)=w-1-\log w. f β ( w ) = w − 1 − β w β − 1 , f 0 ( w ) = w − 1 − log w . On [ 1 / 2 , 2 ] [1/2,2] [ 1/2 , 2 ] a a a (50) , provided the two ends and the polynomial subtractions are controlled uniformly.
Write R 0 = ∣ v ∣ R_0=|v| R 0 = ∣ v ∣ w 0 = ( 8 / R 0 ) 1 / β w_0=(8/R_0)^{1/\beta} w 0 = ( 8/ R 0 ) 1/ β p = β N ∈ [ 2 , 2 + β ] p=\beta N\in[2,2+\beta] p = β N ∈ [ 2 , 2 + β ] V 0 V_0 V 0 cos ϵ > 3 / 4 \cos\epsilon>3/4 cos ϵ > 3/4
∣ E N ( z ) ∣ ≤ ∣ z ∣ N e max ( Re z , 0 ) / N ! |E_N(z)|\le|z|^N e^{\max(\operatorname{Re}z,0)}/N! ∣ E N ( z ) ∣ ≤ ∣ z ∣ N e m a x ( Re z , 0 ) / N ! bounds the normalized integral on ( 0 , w 0 ) (0,w_0) ( 0 , w 0 )
exp { [ log R 0 + ( p − 1 ) log 8 + p ( 1 − log p ) + 8 − a Re v ] / β } . \exp\{[\log R_0+(p-1)\log8+p(1-\log p)+8-a\operatorname{Re}v]/\beta\}. exp {[ log R 0 + ( p − 1 ) log 8 + p ( 1 − log p ) + 8 − a Re v ] / β } . This is O ( e − c 0 R 0 / β ) O(e^{-c_0R_0/\beta}) O ( e − c 0 R 0 / β ) [ w 0 , 1 / 2 ] [w_0,1/2] [ w 0 , 1/2 ] w − 2 e − Re v f β ( w ) w^{-2}e^{-\operatorname{Re}v f_\beta(w)} w − 2 e − Re v f β ( w ) Re v ( w β − w ) − 2 \operatorname{Re}v(w^\beta-w)-2 Re v ( w β − w ) − 2 w β − w w^\beta-w w β − w R 0 ( w 0 β − w 0 ) ≥ 7 R_0(w_0^\beta-w_0)\ge7 R 0 ( w 0 β − w 0 ) ≥ 7 V 0 V_0 V 0 2 − β − 1 / 2 > 0.43 2^{-\beta}-1/2>0.43 2 − β − 1/2 > 0.43 O ( e − c 0 R 0 ) O(e^{-c_0R_0}) O ( e − c 0 R 0 ) [ 2 , ∞ ) [2,\infty) [ 2 , ∞ ) f β f_\beta f β
To estimate the subtractions on [ w 0 , 1 ] [w_0,1] [ w 0 , 1 ]
∫ w 0 1 w β k − 2 d w ≤ log ( 1 / w 0 ) max ( 1 , w 0 β k − 1 ) . \int_{w_0}^{1}w^{\beta k-2}\,dw\le\log(1/w_0)\max(1,w_0^{\beta k-1}). ∫ w 0 1 w β k − 2 d w ≤ log ( 1/ w 0 ) max ( 1 , w 0 β k − 1 ) . Their coefficient sums are bounded by N ( e R 0 / p ) p / β N(eR_0/p)^{p/\beta} N ( e R 0 / p ) p / β ( R 0 / 8 ) 1 / β e 8 / β (R_0/8)^{1/\beta}e^{8/\beta} ( R 0 /8 ) 1/ β e 8/ β [ 1 , ∞ ) [1,\infty) [ 1 , ∞ ) w β k − 2 ≤ 1 + w 1 / 10 w^{\beta k-2}\le1+w^{1/10} w β k − 2 ≤ 1 + w 1/10 e − Re v w e^{-\operatorname{Re}vw} e − Re v w e a Re v / β e^{a\operatorname{Re}v/\beta} e a Re v / β O ( e − c 0 R 0 / β ) O(e^{-c_0R_0/\beta}) O ( e − c 0 R 0 / β ) 1 / β 1/\beta 1/ β R 0 R_0 R 0 log R 0 \log R_0 log R 0 c 0 c_0 c 0
This proves the uniform expansion. Prove it first on a slightly larger sector; Cauchy estimates on relative disks give both logarithmic derivative bounds. In particular, for small real d 1 , d 2 d_1,d_2 d 1 , d 2
∣ Im log A β ( v e − i d 1 ) ∣ ≤ C ∣ d 1 ∣ / v , ∣ log A β ( v e − i d 2 ) − log A β ( v e − i d 1 ) ∣ ≤ C ∣ d 2 − d 1 ∣ / v , |\operatorname{Im}\log A_\beta(ve^{-id_1})|\le C|d_1|/v,\qquad|\log A_\beta(ve^{-id_2})-\log A_\beta(ve^{-id_1})|\le C|d_2-d_1|/v, ∣ Im log A β ( v e − i d 1 ) ∣ ≤ C ∣ d 1 ∣/ v , ∣ log A β ( v e − i d 2 ) − log A β ( v e − i d 1 ) ∣ ≤ C ∣ d 2 − d 1 ∣/ v , and their v v v 1 / v 1/v 1/ v
Apply the exact reflection identity (45) . The two saddle coordinates are v e − i δ ve^{-i\delta} v e − i δ v e − i ( δ + 2 c ) ve^{-i(\delta+2c)} v e − i ( δ + 2 c ) q e − i x qe^{-ix} q e − i x O ( β / v ) O(\beta/v) O ( β / v )
::: The reciprocal-Gamma coefficients in the polynomial P N − 1 P_{N-1} P N − 1 0 ≤ β k ≤ 2 + β 0 \le\beta k \le2+\beta 0 ≤ β k ≤ 2 + β e − c 0 v / β e^{-c_0v/\beta} e − c 0 v / β q q q − log q = O ( β v ) -\log q=O(\beta v) − log q = O ( β v ) e β e_\beta e β (51) . □
Proof of Proposition 5.5. We treat three radial ranges and then the perturbation of the center. All constants in this proof may depend on the fixed integer ℓ \ell ℓ T = β r a = v a T=\beta r^a=v^a T = β r a = v a
Bounded radii. The compact Hankel expansion gives G β ( t ) = e − t / β W a ( t / β ) → 1 / Γ ( 2 − t ) G_\beta(t)=e^{-t/\beta}W_a(t/\beta)\to1/\Gamma(2-t) G β ( t ) = e − t / β W a ( t / β ) → 1/Γ ( 2 − t ) n ≥ 2 n\ge2 n ≥ 2 G β ( t ) = ( t − ζ n ( β ) ) A n ( t , β ) G_\beta(t)=(t-\zeta_n(\beta))A_n(t,\beta) G β ( t ) = ( t − ζ n ( β )) A n ( t , β ) ζ n ( β ) = n + O ( β ) \zeta_n(\beta)=n+O(\beta) ζ n ( β ) = n + O ( β ) A n A_n A n δ = π ℓ β + O ( β 2 ) \delta=\pi\ell\beta+O(\beta^2) δ = π ℓ β + O ( β 2 ) ℓ n + 3 / 2 \ell n+3/2 ℓ n + 3/2 t − ζ n t-\zeta_n t − ζ n ℓ n + 1 , ℓ n + 2 \ell n+1,\ell n+2 ℓ n + 1 , ℓ n + 2
∣ T − ζ n ∣ + β ∣ T − ζ n ∣ + β , |T-\zeta_n|+\frac{\beta}{|T-\zeta_n|+\beta}, ∣ T − ζ n ∣ + ∣ T − ζ n ∣ + β β , on a fixed sufficiently small neighborhood, up to an O ( β ) O(\beta) O ( β ) β \sqrt{\beta} β ℓ + 1 > 0 \ell+1>0 ℓ + 1 > 0
The large-radius center. Equation (49) at equality in its center is equivalent to
2 sin δ sin ( δ + 2 c ) − sin δ = ℓ . \frac{2\sin\delta}{\sin(\delta+2c)-\sin\delta}=\ell. sin ( δ + 2 c ) − sin δ 2 sin δ = ℓ . The logarithm of (51) then yields the model phase
Φ = ℓ x 2 π + 3 2 − Arg ( 1 − q e − i x ) π + O ( β 3 ) + O ( β / v ) , (52) \Phi=\frac{\ell x}{2\pi}+\frac{3}{2}-\frac{\operatorname{Arg}(1-qe^{-ix})}{\pi}+O(\beta^3)+O(\beta/v), \tag*{(52)} Φ = 2 π ℓ x + 2 3 − π Arg ( 1 − q e − i x ) + O ( β 3 ) + O ( β / v ) , (52) together with a perturbation of size O ( β / v ) O(\beta/v) O ( β / v ) q e − i x qe^{-ix} q e − i x ℓ c − δ ℓ = O ( β 3 ) \ell c-\delta_\ell=O(\beta^3) ℓ c − δ ℓ = O ( β 3 ) O ( β 2 ) O(\beta^2) O ( β 2 )
Write x = 2 π n + u x=2\pi n+u x = 2 π n + u ∣ u ∣ ≤ π |u|\le\pi ∣ u ∣ ≤ π u 0 = π / ( ℓ + 1 ) u_0=\pi/(\ell+1) u 0 = π / ( ℓ + 1 ) ∣ u ∣ ≤ u 0 |u|\le u_0 ∣ u ∣ ≤ u 0 ( ℓ n + 1 , ℓ n + 2 ) (\ell n+1,\ell n+2) ( ℓ n + 1 , ℓ n + 2 )
c ℓ min ( 1 , ∣ u ∣ + 1 − q ∣ u ∣ + 1 − q ) . (53) c_\ell\min\left(1,|u|+\frac{1-q}{|u|+1-q}\right). \tag*{(53)} c ℓ min ( 1 , ∣ u ∣ + ∣ u ∣ + 1 − q 1 − q ) . (53) To see this for u ≥ 0 u\ge0 u ≥ 0 A = Arg ( 1 − q e − i u ) A=\operatorname{Arg}(1-qe^{-iu}) A = Arg ( 1 − q e − i u ) 0 ≤ A ≤ ( π − u ) / 2 0\le A\le(\pi-u)/2 0 ≤ A ≤ ( π − u ) /2 π / 2 − A = arctan ( ( 1 − q cos u ) / ( q sin u ) ) \pi/2-A=\arctan((1-q\cos u)/(q\sin u)) π /2 − A = arctan (( 1 − q cos u ) / ( q sin u )) u u u min ( 1 , ( 1 − q ) / u ) \min(1,(1-q)/u) min ( 1 , ( 1 − q ) / u ) 1 / [ 2 ( ℓ + 1 ) ] 1/[2(\ell+1)] 1/ [ 2 ( ℓ + 1 )] u u u
Here 1 − q ≍ min ( 1 , β v ) 1-q\asymp\min(1,\beta v) 1 − q ≍ min ( 1 , β v ) ∣ 1 − q e − i u ∣ ≥ c ( ∣ u ∣ + 1 − q ) |1-qe^{-iu}|\ge c(|u|+1-q) ∣1 − q e − i u ∣ ≥ c ( ∣ u ∣ + 1 − q ) w = β v ≤ 1 w=\beta v\le1 w = β v ≤ 1 (53) is O ( v − 2 ) O(v^{-2}) O ( v − 2 )
( ∣ u ∣ + w ) ( ∣ u ∣ + w ∣ u ∣ + w ) ≥ w . (|u|+w)\left(|u|+\frac{w}{|u|+w}\right)\ge w. ( ∣ u ∣ + w ) ( ∣ u ∣ + ∣ u ∣ + w w ) ≥ w . For w ≥ 1 w\ge1 w ≥ 1 O ( β / v ) O(\beta/v) O ( β / v ) (52) are smaller than c min ( 1 , β v ) c\min(1,\sqrt{\beta v}) c min ( 1 , β v ) v 0 v_0 v 0 β \beta β
Outside these cores,
d d x Arg ( 1 − q e − i x ) = q cos u − q 2 − κ q sin u 1 − 2 q cos u + q 2 . \frac{\mathrm{d}}{\mathrm{d}x}\operatorname{Arg}(1-qe^{-ix})=\frac{q\cos u-q^2-\kappa q\sin u}{1-2q\cos u+q^2}. d x d Arg ( 1 − q e − i x ) = 1 − 2 q cos u + q 2 q cos u − q 2 − κ q sin u . For cos u ≥ 0 \cos u \ge0 cos u ≥ 0 ( csc ∣ u ∣ − 1 ) / 2 (\csc|u|-1)/2 ( csc ∣ u ∣ − 1 ) /2 cos u ≤ 0 \cos u \le0 cos u ≤ 0 κ \kappa κ O ℓ ( β ) O_{\ell}(\beta) O ℓ ( β ) sin u 0 ≥ 2 / ( ℓ + 1 ) \sin u_0 \ge2/(\ell+1) sin u 0 ≥ 2/ ( ℓ + 1 ) ( ℓ + 1 ) / ( 4 π ) + o ( 1 ) > 0 (\ell+1)/(4\pi)+o(1)>0 ( ℓ + 1 ) / ( 4 π ) + o ( 1 ) > 0
There is no unproved logarithm lift in this argument. The exponent bound gives ∣ q e ε β ∣ ≤ e − c ℓ β v + C β / v < 1 |qe^{\varepsilon_\beta}|\le e^{-c_{\ell}\beta v+C\beta/v}<1 ∣ q e ε β ∣ ≤ e − c ℓ β v + C β / v < 1 V 0 V_0 V 0
∣ Arg ( 1 − q e − i x + ε β ) ∣ ≤ π / 2 − c ℓ min ( 1 , β v ) . \left|\operatorname{Arg}(1-qe^{-ix+\varepsilon_\beta})\right|\le\pi/2-c_{\ell}\min(1,\sqrt{\beta v}). Arg ( 1 − q e − i x + ε β ) ≤ π /2 − c ℓ min ( 1 , β v ) . Together with the amplitude argument bound, this puts the phase after removal of its drift strictly inside ( 0 , 2 ρ ) (0,2\rho) ( 0 , 2 ρ ) (22) , whose interval has length less than two, fixes the lift uniquely.
Width and the infinite tail. For ∣ δ − δ ℓ ∣ ≤ k ℓ β 2 |\delta-\delta_{\ell}|\le k_{\ell}\beta^2 ∣ δ − δ ℓ ∣ ≤ k ℓ β 2 ℓ + O ( k ℓ β ) \ell+O(k_{\ell}\beta) ℓ + O ( k ℓ β ) V 0 ≤ v ≤ M ℓ / β V_0\le v\le M_{\ell}/\beta V 0 ≤ v ≤ M ℓ / β C ℓ k ℓ β v C_{\ell}k_{\ell}\beta v C ℓ k ℓ β v C ℓ k ℓ max ( M ℓ , M ℓ ) C_{\ell}k_{\ell}\max(\sqrt{M_{\ell}},M_{\ell}) C ℓ k ℓ max ( M ℓ , M ℓ ) k ℓ k_{\ell} k ℓ v ≥ M ℓ / β v\ge M_{\ell}/\beta v ≥ M ℓ / β Lemma 5.1 gives positive derivative once
( a v / β ) ( 1 − cos δ ) > 1. (av/\beta)(1-\cos\delta)>1. ( a v / β ) ( 1 − cos δ ) > 1. A sufficiently large fixed M ℓ M_{\ell} M ℓ δ / ( π β ) → ℓ > 0 \delta/(\pi\beta)\to\ell>0 δ / ( π β ) → ℓ > 0 δ < π / 2 \delta<\pi/2 δ < π /2 T = 2 T=2 T = 2
Finally fix any K K K ℓ = 1 , … , K \ell=1,\ldots,K ℓ = 1 , … , K K + 1 K+1 K + 1 (48) . The endpoint band, the K K K 2 K + 3 2K+3 2 K + 3 β \beta β K K K
The arguments in this section prove Theorem 1.6 .
Consequences and computational dependencies
Within the strictly free stable family the results give a strict hierarchy, with weak bell-shape understood as a property of laws:
G G C ⊊ weakly BS ⊊ S D ⊊ I D . (54) \mathrm{GGC}\subsetneq\text{weakly BS}\subsetneq\mathrm{SD}\subsetneq\mathrm{ID}. \tag*{(54)} GGC ⊊ weakly BS ⊊ SD ⊊ ID . (54) The first inclusion is the general Thorin consequence of the phase criterion and is strict by (43) . Every weakly BS member has index at most one, hence is SD by Theorem 1.4 . The law X 6 / 5 , 5 / 6 X_{6/5,5/6} X 6/5 , 5/6 X 3 / 2 , 2 / 3 X_{3/2,2/3} X 3/2 , 2/3 (54) is a conclusion about this family, not a general implication for arbitrary probability laws.
Boundary scales near index one
Corollary 6.1 (Different boundary scales at index one). With ε , β ↓ 0 \varepsilon,\beta\downarrow0 ε , β ↓ 0
r ( 1 + ε ) = 1 2 + 2 ε + C I D ε 2 + o ( ε 2 ) , r(1+\varepsilon)=\frac{1}{2}+2\varepsilon+C_{\mathrm{ID}}\varepsilon^2+o(\varepsilon^2), r ( 1 + ε ) = 2 1 + 2 ε + C ID ε 2 + o ( ε 2 ) , s ( 1 + ε ) = 1 2 + 3 3 4 π ε + o ( ε ) , s(1+\varepsilon)=\frac{1}{2}+\frac{3\sqrt{3}}{4\pi}\sqrt{\varepsilon}+o(\sqrt{\varepsilon}), s ( 1 + ε ) = 2 1 + 4 π 3 3 ε + o ( ε ) , g ( 1 − β ) = β π 2 − β + 2 π 2 + 13 12 π 2 β 3 / 2 − β 2 + O ( β 5 / 2 ) , g(1-\beta)=\frac{\sqrt{\beta}}{\pi\sqrt{2}}-\beta+\frac{2\pi^2+13}{12\pi\sqrt{2}}\beta^{3/2}-\beta^2+O(\beta^{5/2}), g ( 1 − β ) = π 2 β − β + 12 π 2 2 π 2 + 13 β 3/2 − β 2 + O ( β 5/2 ) , where
C I D = 2 log ( 2 π ) − 12 π 2 + 2 ∫ 0 ∞ v 2 4 π 2 + v 2 e − 2 v 1 − e − v d v = − 114.740012808536 … C_{\mathrm{ID}}=2\log(2\pi)-12\pi^2+2\int_0^\infty\frac{v^2}{4\pi^2+v^2}\frac{e^{-2v}}{1-e^{-v}}\,\mathrm{d}v=-114.740012808536\ldots C ID = 2 log ( 2 π ) − 12 π 2 + 2 ∫ 0 ∞ 4 π 2 + v 2 v 2 1 − e − v e − 2 v d v = − 114.740012808536 … The ID and GGC contacts are each uniquely controlling for sufficiently small positive parameters in these expansions. No common numerical smallness threshold is claimed.
Proof. The ID expansion and its uniform spectral argument are Theorem H.7 ; the GGC expansion, including its whole-radius localization, is Theorem H.5 . The SD assertion is (9) .
Convolution powers and classical embeddings
The ID domain is exactly the domain admitting a classical Lévy process with one-time law X a , ρ X_{a,\rho} X a , ρ [12 Theorems 17.5 and 17.9] . For every 1 < a < α S D 1<a<\alpha_{\mathrm{SD}} 1 < a < α SD r ( a ) ≤ ρ < s ( a ) r(a)\leq\rho<s(a) r ( a ) ≤ ρ < s ( a ) a = α I D a=\alpha_{\mathrm{ID}} a = α ID
There is also an explicit construction at every SD parameter. For unit mean reversion, the background driving Lévy process Z Z Z d U t = − U t d t + d Z t \mathrm{d}U_t=-U_t\mathrm{d}t+\mathrm{d}Z_t d U t = − U t d t + d Z t
ν Z ( d x ) = − k + ′ ( x ) 1 { x > 0 } d x − k − ′ ( − x ) 1 { x < 0 } d x . (55) \nu_Z(\mathrm{d}x)=-k_+'(x)\mathbf{1}_{\{x>0\}}\mathrm{d}x-k_-'(-x)\mathbf{1}_{\{x<0\}}\mathrm{d}x. \tag*{(55)} ν Z ( d x ) = − k + ′ ( x ) 1 { x > 0 } d x − k − ′ ( − x ) 1 { x < 0 } d x . (55) Indeed the stationary stochastic integral ∫ 0 ∞ e − t d Z t \int_0^\infty e^{-t}\,\mathrm{d}Z_t ∫ 0 ∞ e − t d Z t n + n_+ n +
∫ 0 ∞ e t n + ( e t x ) d t = 1 x ∫ x ∞ n + ( y ) d y = k + ( x ) x , \int_0^\infty e^t n_+(e^t x)\,\mathrm{d}t=\frac{1}{x}\int_x^\infty n_+(y)\,\mathrm{d}y=\frac{k_+(x)}{x}, ∫ 0 ∞ e t n + ( e t x ) d t = x 1 ∫ x ∞ n + ( y ) d y = x k + ( x ) , and likewise on the negative side. The zero limit of the kernels at infinity justifies the last identity. The drift is chosen by the usual compensated Lévy formula. Thus the same derivative tests used in the classification give nonnegative driving densities; a failed SD test rules out this stationary representation.
The complete GGC classification also characterizes the laws whose every classical convolution power is weakly bell-shaped, by [11 Corollary 1.5] . In particular the strictly BS, non-GGC examples (43) have a classical integer convolution power which fails bell-shape. This is a concrete distinction between bell-shape of one distribution and preservation of bell-shape along its entire convolution semigroup. This implication is constructive in the phase framework. If P ( u ) > P ( v ) P(u)>P(v) P ( u ) > P ( v ) u < v u<v u < v n n n n ( P ( u ) − P ( v ) ) > 1 n(P(u)-P(v))>1 n ( P ( u ) − P ( v )) > 1 n P ( v ) nP(v) n P ( v ) n P ( u ) nP(u) n P ( u ) n n n n P nP n P P P P
The framework in § 2 is a synthesis of Lévy-kernel, Pick-function and integer-level criteria. The characterization of general bell-shaped functions belongs to Kwaśnicki–Simon; the contribution here is its compatible use with the ID, GGC and SD ray tests, and the global contact and tail arguments needed for the free stable family.
What the finite certificates establish
A certificate is a finite binary partition of an explicitly specified rational parameter box, or an exact list of root brackets and moment remainders. Verification reconstructs the partition from its roots, checks that both children exhaust each parent, and recomputes the required interval inequality on every leaf. It does not rely on a saved decimal value. Integration is carried out with directed complex ball arithmetic and analytic quadrature, with an explicit independent bound for every omitted contour piece; see [8 ] . Analyticity flags are forwarded to logarithms in the quadrature callbacks. Root completeness uses moments or argument principles, never a finite sign scan alone.
The finite computational inputs essential to the main classification statements can be organized as follows. Counts refer to accepted whole-box inequalities. All filenames below are relative to the research archive containing this manuscript directory. Run the following commands from that archive with Python and python-flint installed (prefix each line with python). The archived Kanter run used python-flint 0.9.0; the working precisions and truncations are specified in the individual verifiers. The principal reproduction commands are
Table 1.
certify_kanter_optimal.py verify
certify_symmetric_ggc.py verify
certify_global_id_cutoff.py verify
certify_sd_below_one.py verify
certify_near_one_extremal_sd.py verify
certify_extremal_sd_extension.py verify
certify_sd_transition_global.py verify
certify_sd_transition_alpha_derivative.py verify
certify_large_alpha_non_sd.py verify
certify_sd_cutoff_refinement.py verify
certify_bell_beyond_ggc.py verify
check_bell_phase_witnesses.pyThe mathematical specifications in the appendix and § 4 explain their formulas and remainder bounds. The full archive retains the exact rational partitions, coefficient polynomials, root certificates, programs and versioned verification records. Files under manuscript_audit_runs/ record a fresh run of all twelve commands on 4 October 2026; all returned success, and the original dependencies were preserved. Eight additional upstream recomputations, recorded under manuscript_dependency_runs/, also passed: these check the exact germ generation, the remaining endpoint ID covers, the SD transition germ, the competing ID contacts, and the three whole-ray GGC slices used by the non-GGC BS examples. Some commands also rerun their prerequisite covers. The archived provenance audit reports 1,069 recorded dependency hashes across 134 verification records. These are the execution records of 3–4 October 2026. References in the appendices to successful or fresh runs refer to those archived computations. Their saved outputs and dependency hashes are included in the accompanying archive. Hash consistency, numerical recomputation, and the analytic arguments are distinct parts of the evidence.
Scope of the assertions
The main classification results do not assume a unique global radial contact at every index. The explicit ID crossover in the appendix shows why such an assumption would be false. The convergence and strict-margin searches prove boundary computability, while remaining silent about finite-time equality decisions for arbitrary real inputs.
The bell-shape statements proved here leave a finite contact problem at each fixed index, but they do not solve all of those problems. In particular, no globally earliest bell-shape cusp, exact global component count, or complete bell-shape classification is asserted. Unfinished exploratory covers are not inputs to any theorem in this manuscript. These limitations concern the scope of the claims, rather than an extrapolation from numerical evidence.
Obstructions to the Thorin property
The first obstruction uses zeros of the characteristic transform above one. The second transfers a cut-phase oscillation to the positive free stable law below one. These arguments prove necessity independently of the positive certificates in the next two appendices.
A zero-free obstruction for every alpha > 1
Lemma A.1. An extended GGC characteristic function, restricted to t > 0 t > 0 t > 0 Re t > 0 \operatorname{Re} t > 0 Re t > 0 Ψ = − log φ \Psi= -\log\varphi Ψ = − log φ
Re Ψ ′ ( t ) ≥ 0 , Re t > 0. (56) \operatorname{Re}\Psi'(t) \ge0,\qquad\operatorname{Re} t > 0. \tag*{(56)} Re Ψ ′ ( t ) ≥ 0 , Re t > 0. (56) Proof. The canonical representation is
φ ( t ) = exp { i a t − σ 2 t 2 2 − ∫ R ∖ { 0 } [ log ( 1 − i t / u ) + i t u 1 + u 2 ] U ( d u ) } . \varphi(t)=\exp\left\{iat-\frac{\sigma^{2}t^{2}}{2}-\int_{\mathbb{R}\setminus\{0\}}\left[\log(1-it/u)+\frac{itu}{1+u^{2}}\right]U(\mathrm{d}u)\right\}. φ ( t ) = exp { ia t − 2 σ 2 t 2 − ∫ R ∖ { 0 } [ log ( 1 − i t / u ) + 1 + u 2 i t u ] U ( d u ) } . where U U U u − 2 u^{-2} u − 2 [10 Section 3, p. 4] . The logarithm has a holomorphic branch on the right half-plane, since its argument never vanishes there. The compensated integrand is O ( u − 2 ) O(u^{-2}) O ( u − 2 ) O ( 1 + ∣ log ∣ u ∣ ∣ ) O(1+|\log|u||) O ( 1 + ∣ log ∣ u ∣∣ ) t t t t = x + i y t=x+iy t = x + i y
Re Ψ ′ ( t ) = σ 2 x + ∫ x ( u + y ) 2 + x 2 U ( d u ) ≥ 0. \operatorname{Re}\Psi'(t)=\sigma^{2}x+\int\frac{x}{(u+y)^{2}+x^{2}}U(\mathrm{d}u)\ge0. Re Ψ ′ ( t ) = σ 2 x + ∫ ( u + y ) 2 + x 2 x U ( d u ) ≥ 0. Theorem A.2. Every admissible X α , ρ X_{\alpha,\rho} X α , ρ 1 < α ≤ 2 1 < \alpha\le2 1 < α ≤ 2
Proof. For α > 1 \alpha> 1 α > 1 W α W_{\alpha} W α − λ n -\lambda_n − λ n [1 Theorem 1] ; see also [7 Remark 3] . Set r n = λ n 1 / α r_n=\lambda_n^{1/\alpha} r n = λ n 1/ α (17) vanishes at
t n = r n e i ω , − e − i α ω t n α = − λ n . t_n=r_ne^{i\omega},\qquad-e^{-i\alpha\omega}t_n^\alpha=-\lambda_n. t n = r n e iω , − e − i α ω t n α = − λ n . Admissibility gives ∣ ω ∣ < π / 2 |\omega|<\pi/2 ∣ ω ∣ < π /2 Re t > 0 \operatorname{Re}t>0 Re t > 0 Lemma A.1 and the identity theorem. □ \square □
No determination of the ID region above one is used in this proof.
The one-sided obstruction at 4/5
Theorem A.3. For every 4 / 5 < α < 1 4/5 < \alpha< 1 4/5 < α < 1 X α , 1 X_{\alpha,1} X α , 1 X α , 0 X_{\alpha,0} X α , 0
This improves the non-explicit exclusion near one in Corollary 1 of [7 ] . The proof below includes the step needed to transfer the oscillation; non-GGC of the Kanter factor alone would not imply the result.
An exact differential relation
Fix α ∈ ( 4 / 5 , 1 ) \alpha\in(4/5,1) α ∈ ( 4/5 , 1 )
β = 1 − α , p = α / β , v 0 = β 1 / α , b = α β β / α , ϑ = 2 π β / α ∈ ( 0 , π / 2 ) . \beta= 1-\alpha,\qquad p=\alpha/\beta,\qquad v_0=\beta^{1/\alpha},\qquad b=\alpha\beta^{\beta/\alpha},\qquad\vartheta=2\pi\beta/\alpha\in(0,\pi/2). β = 1 − α , p = α / β , v 0 = β 1/ α , b = α β β / α , ϑ = 2 π β / α ∈ ( 0 , π /2 ) . Let U U U ( 0 , 1 ) (0,1) ( 0 , 1 )
K α = sin ( π α U ) sin ( 1 − α ) / α ( π ( 1 − α ) U ) sin 1 / α ( π U ) , 0 < α < 1. K_\alpha=\frac{\sin(\pi\alpha U)\sin^{(1-\alpha)/\alpha}(\pi(1-\alpha)U)}{\sin^{1/\alpha}(\pi U)},\qquad0<\alpha<1. K α = sin 1/ α ( π U ) sin ( π α U ) sin ( 1 − α ) / α ( π ( 1 − α ) U ) , 0 < α < 1. Its normalization and transform are those of [7 (6) and Section 2.3] :
F β ( s ) = E e − s K α = ∑ n ≥ 0 ( − s α ) n n ! Γ ( 1 − β n ) . F_\beta(s)=\mathbb{E}e^{-sK_\alpha}=\sum_{n\geq0}\frac{(-s^\alpha)^n}{n!\Gamma(1-\beta n)}. F β ( s ) = E e − s K α = n ≥ 0 ∑ n ! Γ ( 1 − β n ) ( − s α ) n . The coefficient identity for reciprocal Gamma functions gives
F β ( s ) = L α ( s ) − β α s L α ′ ( s ) . (57) F_\beta(s)=L_\alpha(s)-\frac{\beta}{\alpha}sL'_\alpha(s). \tag*{(57)} F β ( s ) = L α ( s ) − α β s L α ′ ( s ) . (57) Indeed 1 / Γ ( 1 − β n ) = ( 1 − β n ) / Γ ( 2 − β n ) 1/\Gamma(1-\beta n)=(1-\beta n)/\Gamma(2-\beta n) 1/Γ ( 1 − β n ) = ( 1 − β n ) /Γ ( 2 − β n )
The complex saddle
The lower-cut representation from [7 ] reads
F β ( − r − i 0 ) = 1 + 1 2 π i ∫ 0 ∞ e − r u ( e r u β e 2 π i β − e − r u β ) d u u . (58) F_\beta(-r-i0)=1+\frac{1}{2\pi i}\int_0^\infty e^{-ru}\left(e^{ru^\beta e^{2\pi i\beta}}-e^{-ru^\beta}\right)\frac{\mathrm{d}u}{u}. \tag*{(58)} F β ( − r − i 0 ) = 1 + 2 π i 1 ∫ 0 ∞ e − r u ( e r u β e 2 π i β − e − r u β ) u d u . (58) Let q ( v ) = v β − v q(v)=v^\beta-v q ( v ) = v β − v
J ( λ ) = ∫ 0 ∞ e λ q ( v ) − e − λ v v d v , Re λ > 0. J(\lambda)=\int_0^\infty\frac{e^{\lambda q(v)}-e^{-\lambda v}}{v}\,\mathrm{d}v,\qquad\operatorname{Re}\lambda>0. J ( λ ) = ∫ 0 ∞ v e λ q ( v ) − e − λ v d v , Re λ > 0. Near zero its integrand is O ( v β − 1 ) O(v^{\beta-1}) O ( v β − 1 ) e − r u e^{-ru} e − r u u = e i ϑ v u=e^{i\vartheta}v u = e i ϑ v β ϑ + 2 π β = ϑ \beta\vartheta+2\pi\beta=\vartheta β ϑ + 2 π β = ϑ
B ( r ) : = Re F β ( − r − i 0 ) = 1 + 1 2 π Im J ( r e i ϑ ) . (59) B(r):=\operatorname{Re}F_\beta(-r-i0)=1+\frac{1}{2\pi}\operatorname{Im}J(re^{i\vartheta}). \tag*{(59)} B ( r ) := Re F β ( − r − i 0 ) = 1 + 2 π 1 Im J ( r e i ϑ ) . (59) The subtracted integrand makes the small arc vanish. The large arc vanishes because ϑ < π / 2 \vartheta<\pi/2 ϑ < π /2 β < 1 \beta<1 β < 1 q q q v 0 v_0 v 0
q ( v 0 ) = b , q ′ ′ ( v 0 ) = − α / v 0 . q(v_0)=b,\qquad q''(v_0)=-\alpha/v_0. q ( v 0 ) = b , q ′′ ( v 0 ) = − α / v 0 . Laplace’s method in the fixed complex direction arg λ = ϑ \arg\lambda=\vartheta arg λ = ϑ
J ( r e i ϑ ) = 2 π α v 0 r − 1 / 2 e − i ϑ / 2 e b r e i ϑ + O ( r − 3 / 2 e b r cos ϑ ) . (60) J(re^{i\vartheta})=\sqrt{\frac{2\pi}{\alpha v_0}}\,r^{-1/2}e^{-i\vartheta/2}e^{bre^{i\vartheta}}+O(r^{-3/2}e^{br\cos\vartheta}). \tag*{(60)} J ( r e i ϑ ) = α v 0 2 π r − 1/2 e − i ϑ /2 e b r e i ϑ + O ( r − 3/2 e b r c o s ϑ ) . (60) Here constants may depend on the fixed α \alpha α v 0 v_0 v 0 q q q 1 / v 1/v 1/ v r v β − 1 rv^{\beta-1} r v β − 1 q ( v ) < 0 q(v)<0 q ( v ) < 0
Set
C = ( 2 π α v 0 ) − 1 / 2 , κ = b cos ϑ > 0 , ν = b sin ϑ > 0. C=(2\pi\alpha v_0)^{-1/2},\qquad\kappa=b\cos\vartheta>0,\qquad\nu=b\sin\vartheta>0. C = ( 2 π α v 0 ) − 1/2 , κ = b cos ϑ > 0 , ν = b sin ϑ > 0. It follows that
B ( r ) = C r − 1 / 2 e κ r sin ( ν r − ϑ / 2 ) + O ( r − 3 / 2 e κ r ) . (61) B(r)=Cr^{-1/2}e^{\kappa r}\sin(\nu r-\vartheta/2)+O(r^{-3/2}e^{\kappa r}). \tag*{(61)} B ( r ) = C r − 1/2 e κ r sin ( ν r − ϑ /2 ) + O ( r − 3/2 e κ r ) . (61) Transferring the oscillation
Let A ( r ) = Re L α ( − r − i 0 ) A(r)=\operatorname{Re}L_\alpha(-r-i0) A ( r ) = Re L α ( − r − i 0 )
B = A − r p A ′ , A ( r ) = r p { c − p ∫ r 0 r u − p − 1 B ( u ) d u } . B=A-\frac{r}{p}A',\qquad A(r)=r^p\left\{c-p\int_{r_0}^{r}u^{-p-1}B(u)\,\mathrm{d}u\right\}. B = A − p r A ′ , A ( r ) = r p { c − p ∫ r 0 r u − p − 1 B ( u ) d u } . For λ = b e i ϑ \lambda=be^{i\vartheta} λ = b e i ϑ
∫ r 0 r e λ u u − p − 3 / 2 d u = λ − 1 e λ r r − p − 3 / 2 ( 1 + O ( r − 1 ) ) + O ( 1 ) . \int_{r_0}^{r}e^{\lambda u}u^{-p-3/2}\,\mathrm{d}u=\lambda^{-1}e^{\lambda r}r^{-p-3/2}\left(1+O(r^{-1})\right)+O(1). ∫ r 0 r e λ u u − p − 3/2 d u = λ − 1 e λ r r − p − 3/2 ( 1 + O ( r − 1 ) ) + O ( 1 ) . The integrated error from (A.6) is O ( e κ r r − p − 5 / 2 ) + O ( 1 ) O(e^{\kappa r}r^{-p-5/2})+O(1) O ( e κ r r − p − 5/2 ) + O ( 1 ) κ > 0 \kappa>0 κ > 0
Re L α ( − r − i 0 ) = − p C b r − 3 / 2 e κ r { sin ( ν r − 3 ϑ / 2 ) + O ( r − 1 ) } . (62) \operatorname{Re}L_\alpha(-r-i0)=-\frac{pC}{b}r^{-3/2}e^{\kappa r}\left\{\sin(\nu r-3\vartheta/2)+O(r^{-1})\right\}. \tag*{(62)} Re L α ( − r − i 0 ) = − b pC r − 3/2 e κ r { sin ( ν r − 3 ϑ /2 ) + O ( r − 1 ) } . (62) Taking sequences where the sine is + 1 +1 + 1 − 1 -1 − 1
The Thorin contradiction
If X α X_\alpha X α b b b ( x − b ) 1 / 2 (x-b)^{1/2} ( x − b ) 1/2 [7 ] , and Watson’s lemma also for the first derivative, give
e b s L α ( s ) ∼ c α s − 3 / 2 , lim s → ∞ s ( − L α ′ ( s ) L α ( s ) − b ) = 3 2 . e^{bs}L_\alpha(s)\sim c_\alpha s^{-3/2},\qquad\lim_{s\to\infty}s\left(-\frac{L_\alpha'(s)}{L_\alpha(s)}-b\right)=\frac{3}{2}. e b s L α ( s ) ∼ c α s − 3/2 , s → ∞ lim s ( − L α ( s ) L α ′ ( s ) − b ) = 2 3 . For a GGC this limit is the total Thorin mass, so U ( ( 0 , ∞ ) ) = 3 / 2 U((0,\infty))=3/2 U (( 0 , ∞ )) = 3/2 r r r
L α ( − r − i 0 ) = exp { b r − ∫ 0 ∞ log ∣ 1 − r / u ∣ U ( d u ) } e i π U ( ( 0 , r ) ) . L_\alpha(-r-i0)=\exp\left\{br-\int_0^\infty\log|1-r/u|U(\mathrm{d}u)\right\}e^{i\pi U((0,r))}. L α ( − r − i 0 ) = exp { b r − ∫ 0 ∞ log ∣1 − r / u ∣ U ( d u ) } e iπ U (( 0 , r )) . For all sufficiently large r r r 1 / 2 < U ( ( 0 , r ) ) ≤ 3 / 2 1/2<U((0,r))\leq3/2 1/2 < U (( 0 , r )) ≤ 3/2 Theorem A.3 .
Remark A.4. At α = 4 / 5 \alpha=4/5 α = 4/5 ϑ = π / 2 \vartheta=\pi/2 ϑ = π /2 κ = 0 \kappa=0 κ = 0
Corollary A.5. If c < 1 c<1 c < 1 X α , ρ X_{\alpha,\rho} X α , ρ 0 < α ≤ c 0<\alpha\leq c 0 < α ≤ c ρ ∈ [ 0 , 1 ] \rho\in[0,1] ρ ∈ [ 0 , 1 ] c ≤ 4 / 5 c\leq4/5 c ≤ 4/5 Theorem C.4 below.
Corollary A.6. For each fixed 4 / 5 < α < 1 4/5<\alpha<1 4/5 < α < 1 ε α > 0 \varepsilon_\alpha>0 ε α > 0
ρ ∈ [ 0 , ε α ] ∪ [ 1 − ε α , 1 ] ⟹ X α , ρ is not extended GGC . \rho\in[0,\varepsilon_\alpha]\cup[1-\varepsilon_\alpha,1]\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is not extended GGC}. ρ ∈ [ 0 , ε α ] ∪ [ 1 − ε α , 1 ] ⟹ X α , ρ is not extended GGC . Proof. Extended GGC laws are closed under weak convergence. A sequence of GGC laws approaching either one-sided endpoint would contradict Theorem A.3 . □
No explicit ε α \varepsilon_\alpha ε α
A certified two-sided example
For 0 < α < 1 0<\alpha<1 0 < α < 1
H α ( s ) = − s L α ′ ( s ) L α ( s ) = − α z W α ′ ( z ) W α ( z ) , z = − s α , H_\alpha(s)=-s\frac{L_\alpha'(s)}{L_\alpha(s)}=-\alpha z\frac{W_\alpha'(z)}{W_\alpha(z)},\qquad z=-s^\alpha, H α ( s ) = − s L α ( s ) L α ′ ( s ) = − α z W α ( z ) W α ′ ( z ) , z = − s α , at regular points. For q = e i π ( 1 / 2 − ρ ) q=e^{i\pi(1/2-\rho)} q = e iπ ( 1/2 − ρ ) φ α , ρ ( t ) = L α ( q t ) \varphi_{\alpha,\rho}(t)=L_\alpha(qt) φ α , ρ ( t ) = L α ( q t ) t > 0 t>0 t > 0 t = − i r t=-ir t = − i r
Im H α ( r e i π ρ ) ≥ 0 ( r > 0 ) (63) \operatorname{Im}H_\alpha(re^{i\pi\rho})\geq0\qquad(r>0) \tag*{(63)} Im H α ( r e iπ ρ ) ≥ 0 ( r > 0 ) (63) where the boundary expression is regular. Reflection gives the corresponding ray with angle π ( 1 − ρ ) \pi(1-\rho) π ( 1 − ρ )
Proposition A.7. X 9 / 10 , 999999 / 1000000 X_{9/10,999999/1000000} X 9/10 , 999999/1000000
@ аҩ? ID follows from the original theorem. The script certify_two_sided_non_ggc.py gives the outward enclosure
− 0.000858987369327726 < Im H 9 / 10 ( 60 e i π 999999 / 1000000 ) < − 0.000858987369327724 , -0.000858987369327726 < \operatorname{Im} H_{9/10}\left(60e^{i\pi999999/1000000}\right) < -0.000858987369327724, − 0.000858987369327726 < Im H 9/10 ( 60 e iπ 999999/1000000 ) < − 0.000858987369327724 , which violates (A.8). This is an Arb interval computation, with explicit truncation error as follows.
For n ≥ N > 1 / β n \ge N > 1/\beta n ≥ N > 1/ β β = 1 − α \beta= 1-\alpha β = 1 − α
∣ z n n ! Γ ( 2 − β n ) ∣ ≤ A n = r α n Γ ( β n − 1 ) π n ! . \left|\frac{z^n}{n!\Gamma(2-\beta n)}\right| \le A_n = \frac{r^{\alpha n}\Gamma(\beta n-1)}{\pi n!}. n ! Γ ( 2 − β n ) z n ≤ A n = π n ! r α n Γ ( β n − 1 ) . Concavity under a Gamma law gives Γ ( x + β ) / Γ ( x ) ≤ x β \Gamma(x+\beta)/\Gamma(x) \le x^\beta Γ ( x + β ) /Γ ( x ) ≤ x β [15 (5.6.4)] . Therefore
A n + 1 A n ≤ q N = r α β β N − α < 1. \frac{A_{n+1}}{A_n} \le q_N = r^\alpha\beta^\beta N^{-\alpha} < 1. A n A n + 1 ≤ q N = r α β β N − α < 1. Truncation before n = N n=N n = N
A N 1 − q N , A N { N 1 − q N + q N ( 1 − q N ) 2 } \frac{A_N}{1-q_N}, \qquad A_N\left\{\frac{N}{1-q_N}+\frac{q_N}{(1-q_N)^2}\right\} 1 − q N A N , A N { 1 − q N N + ( 1 − q N ) 2 q N } for W W W z W ′ zW' z W ′ N = 400 N=400 N = 400 r = 60 r=60 r = 60 α = 9 / 10 \alpha=9/10 α = 9/10 4.14 ⋅ 10 − 185 4.14\cdot10^{-185} 4.14 ⋅ 1 0 − 185 1.657 ⋅ 10 − 182 1.657\cdot10^{-182} 1.657 ⋅ 1 0 − 182
Appendix B. The central GGC interval
We prove the ray criterion, establish a nonempty central interval, and characterize its first angular contact. The final moment construction is an independent obstruction and is not needed for the classification.
The GGC skewness set is a central interval
This section first gives the interval structure. Nonemptiness at every index is proved subsequently in Theorem B.14 .
Lemma B.1 (Uniform logarithmic derivative bounds). For every fixed 0 < α < 1 0<\alpha<1 0 < α < 1 ∣ arg s ∣ ≤ π − ε |\arg s|\le\pi-\varepsilon ∣ arg s ∣ ≤ π − ε
H α ( s ) = s α Γ ( α ) + O ( ∣ s ∣ 2 α ) ( s → 0 ) , H α ( s ) = b α s + 3 2 + o ( 1 ) ( ∣ s ∣ → ∞ ) . H_\alpha(s)=\frac{s^\alpha}{\Gamma(\alpha)}+O(|s|^{2\alpha})\qquad(s\to0),\qquad H_\alpha(s)=b_\alpha s+\frac{3}{2}+o(1)\qquad(|s|\to\infty). H α ( s ) = Γ ( α ) s α + O ( ∣ s ∣ 2 α ) ( s → 0 ) , H α ( s ) = b α s + 2 3 + o ( 1 ) ( ∣ s ∣ → ∞ ) . In particular L α L_\alpha L α
Proof. The first assertion is the Wright series. For the second, the proof of Theorem 1 in [7 ] shows that y − 1 f α ( b α + y ) y^{-1}f_\alpha(b_\alpha+y) y − 1 f α ( b α + y ) ν \nu ν
f α ( b α + y ) = y ∫ 0 ∞ e − y u ν ( d u ) , L α ( s ) = e − b α s g ( s ) , g ( s ) = ∫ 0 ∞ ( s + u ) − 2 ν ( d u ) . f_\alpha(b_\alpha+y)=y\int_0^\infty e^{-yu}\nu(\mathrm{d}u),\qquad L_\alpha(s)=e^{-b_\alpha s}g(s),\qquad g(s)=\int_0^\infty(s+u)^{-2}\nu(\mathrm{d}u). f α ( b α + y ) = y ∫ 0 ∞ e − y u ν ( d u ) , L α ( s ) = e − b α s g ( s ) , g ( s ) = ∫ 0 ∞ ( s + u ) − 2 ν ( d u ) . The identity for g ( s ) g(s) g ( s ) f α ( b α + y ) ∼ c y f_\alpha(b_\alpha+y)\sim c\sqrt{y} f α ( b α + y ) ∼ c y c > 0 c>0 c > 0 ν R ( A ) = R − 1 / 2 ν ( R A ) \nu_R(A)=R^{-1/2}\nu(RA) ν R ( A ) = R − 1/2 ν ( R A )
∫ 0 ∞ e − t u ν R ( d u ) = R − 1 / 2 f α ( b α + t / R ) t / R ⟶ c t − 1 / 2 , ν R ⟶ c π u − 1 / 2 d u \int_0^\infty e^{-tu}\nu_R(\mathrm{d}u)=R^{-1/2}\frac{f_\alpha(b_\alpha+t/R)}{t/R}\longrightarrow ct^{-1/2},\qquad\nu_R\longrightarrow\frac{c}{\sqrt{\pi}}u^{-1/2}\,\mathrm{d}u ∫ 0 ∞ e − t u ν R ( d u ) = R − 1/2 t / R f α ( b α + t / R ) ⟶ c t − 1/2 , ν R ⟶ π c u − 1/2 d u vaguely on [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) ν ( [ 0 , T ] ) ≤ e ∫ e − u / T ν ( d u ) = O ( T ) \nu([0,T])\le e\int e^{-u/T}\nu(\mathrm{d}u)=O(\sqrt{T}) ν ([ 0 , T ]) ≤ e ∫ e − u / T ν ( d u ) = O ( T ) T → ∞ T\to\infty T → ∞ ( z + u ) − 2 (z+u)^{-2} ( z + u ) − 2 ( z + u ) − 3 (z+u)^{-3} ( z + u ) − 3 ∣ z ∣ = 1 |z|=1 ∣ z ∣ = 1 ∣ arg z ∣ ≤ π − ε |\arg z|\le\pi-\varepsilon ∣ arg z ∣ ≤ π − ε
g ( s ) ∼ c π 2 s − 3 / 2 , g ′ ( s ) ∼ − 3 c π 4 s − 5 / 2 . g(s)\sim\frac{c\sqrt{\pi}}{2}s^{-3/2},\qquad g'(s)\sim-\frac{3c\sqrt{\pi}}{4}s^{-5/2}. g ( s ) ∼ 2 c π s − 3/2 , g ′ ( s ) ∼ − 4 3 c π s − 5/2 . Thus − s L α ′ ( s ) / L α ( s ) = b α s − s g ′ ( s ) / g ( s ) -sL'_{\alpha}(s)/L_{\alpha}(s)=b_{\alpha}s-sg'(s)/g(s) − s L α ′ ( s ) / L α ( s ) = b α s − s g ′ ( s ) / g ( s )
Lemma B.2 (Positive-real converse). Suppose a characteristic function restricted to t > 0 t>0 t > 0 exp ( − Ψ ( t ) ) \exp(-\Psi(t)) exp ( − Ψ ( t )) Ψ ( 0 + ) = 0 \Psi(0+)=0 Ψ ( 0 + ) = 0 D = Ψ ′ D=\Psi' D = Ψ ′ Re t > 0 \operatorname{Re}t>0 Re t > 0 Re D ≥ 0 \operatorname{Re}D\ge0 Re D ≥ 0 D ( t ) = O ( t δ − 1 ) D(t)=O(t^{\delta-1}) D ( t ) = O ( t δ − 1 ) t ↓ 0 t\downarrow0 t ↓ 0 δ > 0 \delta>0 δ > 0
Proof. The half-plane Herglotz representation has the form
D ( t ) = σ 2 t + i c + ∫ R ( 1 t + i u + i u 1 + u 2 ) U ( d u ) , U ≥ 0 , ∫ R U ( d u ) 1 + u 2 < ∞ . D(t)=\sigma^2t+ic+\int_{\mathbb{R}}\left(\frac{1}{t+iu}+\frac{iu}{1+u^2}\right)U(du),\quad U\ge0,\quad\int_{\mathbb{R}}\frac{U(du)}{1+u^2}<\infty. D ( t ) = σ 2 t + i c + ∫ R ( t + i u 1 + 1 + u 2 i u ) U ( d u ) , U ≥ 0 , ∫ R 1 + u 2 U ( d u ) < ∞. The bound at zero and positivity give
∫ 0 1 Re D ( x ) d x = σ 2 2 + 1 2 ∫ R log ( 1 + u − 2 ) U ( d u ) < ∞ . \int_0^1\operatorname{Re}D(x)\,dx=\frac{\sigma^2}{2}+\frac{1}{2}\int_{\mathbb{R}}\log(1+u^{-2})U(du)<\infty. ∫ 0 1 Re D ( x ) d x = 2 σ 2 + 2 1 ∫ R log ( 1 + u − 2 ) U ( d u ) < ∞. In particular U ( { 0 } ) = 0 U(\{0\})=0 U ({ 0 }) = 0 Ψ ( 0 + ) = 0 \Psi(0+)=0 Ψ ( 0 + ) = 0 Lemma A.1 . This also verifies the needed integrability, rather than treating positive-realness alone as sufficient without a normalization at zero. □
Proposition B.3 (An exact ray criterion). Fix 0 < α < 1 0<\alpha<1 0 < α < 1 0 < ρ < 1 0<\rho<1 0 < ρ < 1 m = max ( ρ , 1 − ρ ) m=\max(\rho,1-\rho) m = max ( ρ , 1 − ρ ) X α , ρ X_{\alpha,\rho} X α , ρ
L α ( s ) ≠ 0 L_{\alpha}(s)\ne0 L α ( s ) = 0 s ≠ 0 s\ne0 s = 0 ∣ arg s ∣ ≤ π m |\arg s|\le\pi m ∣ arg s ∣ ≤ π m
For all r > 0 r>0 r > 0
Im H α ( r e i π ρ ) ≥ 0 , Im H α ( r e i π ( 1 − ρ ) ) ≥ 0. \operatorname{Im}H_{\alpha}(re^{i\pi\rho})\ge0,\qquad\operatorname{Im}H_{\alpha}(re^{i\pi(1-\rho)})\ge0. Im H α ( r e iπ ρ ) ≥ 0 , Im H α ( r e iπ ( 1 − ρ ) ) ≥ 0. The zero condition is part of this criterion; checking ray signs on a grid is not sufficient.
Proof. Put q = e i π ( 1 / 2 − ρ ) q=e^{i\pi(1/2-\rho)} q = e iπ ( 1/2 − ρ ) D ρ ( t ) = H α ( q t ) / t D_{\rho}(t)=H_{\alpha}(qt)/t D ρ ( t ) = H α ( q t ) / t Lemma A.1 gives zero-freeness in − π ρ < arg s < π ( 1 − ρ ) -\pi\rho<\arg s<\pi(1-\rho) − π ρ < arg s < π ( 1 − ρ ) k k k t 0 ∈ i R ∖ { 0 } t_0\in i\mathbb{R}\setminus\{0\} t 0 ∈ i R ∖ { 0 } D ρ ( t 0 + x ) = − k / x + O ( 1 ) D_{\rho}(t_0+x)=-k/x+O(1) D ρ ( t 0 + x ) = − k / x + O ( 1 ) Re D ρ ≥ 0 \operatorname{Re}D_{\rho}\ge0 Re D ρ ≥ 0 0 < ρ < 1 0<\rho<1 0 < ρ < 1 t = i r t=ir t = i r t = − i r t=-ir t = − i r
Conversely the zero condition makes D ρ D_{\rho} D ρ Lemma B.1 gives D ρ ( t ) = O ( 1 ) D_{\rho}(t)=O(1) D ρ ( t ) = O ( 1 ) D ρ ( t ) = O ( ∣ t ∣ α − 1 ) D_{\rho}(t)=O(|t|^{\alpha-1}) D ρ ( t ) = O ( ∣ t ∣ α − 1 ) 1 − α < p < 1 1-\alpha<p<1 1 − α < p < 1 ϵ Re ( t p + t − p ) \epsilon\operatorname{Re}(t^p+t^{-p}) ϵ Re ( t p + t − p ) Re D ρ ≥ 0 \operatorname{Re}D_{\rho}\ge0 Re D ρ ≥ 0 ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0 cos ( p arg t ) ≥ cos ( p π / 2 ) > 0 \cos(p\arg t)\ge\cos(p\pi/2)>0 cos ( p arg t ) ≥ cos ( p π /2 ) > 0 Lemma B.2 finishes the proof. □
Theorem B.4 (Connectedness of the GGC skewness section). For each fixed 0 < α < 1 0<\alpha<1 0 < α < 1
G α = { ρ ∈ [ 0 , 1 ] : X α , ρ is extended GGC } \mathcal{G}_{\alpha}=\{\rho\in[0,1]:X_{\alpha,\rho}\text{ is extended GGC}\} G α = { ρ ∈ [ 0 , 1 ] : X α , ρ is extended GGC } is either empty or a closed interval [ g α , 1 − g α ] [g_{\alpha},1-g_{\alpha}] [ g α , 1 − g α ] 0 ≤ g α ≤ 1 / 2 0\le g_{\alpha}\le1/2 0 ≤ g α ≤ 1/2 4 / 5 < α < 1 4/5<\alpha<1 4/5 < α < 1 g α > 0 g_{\alpha}>0 g α > 0
Proof. Suppose first that ρ 0 ∈ G α ∩ ( 1 / 2 , 1 ) \rho_0\in\mathcal{G}_{\alpha}\cap(1/2,1) ρ 0 ∈ G α ∩ ( 1/2 , 1 ) Proposition B.3 makes u ( s ) = Im H α ( s ) u(s)=\operatorname{Im}H_{\alpha}(s) u ( s ) = Im H α ( s ) 0 < arg s < π ρ 0 0<\arg s<\pi\rho_0 0 < arg s < π ρ 0 Lemma B.1 are O ( ∣ s ∣ α ) O(|s|^{\alpha}) O ( ∣ s ∣ α ) O ( ∣ s ∣ ) O(|s|) O ( ∣ s ∣ ) 1 < p < 1 / ρ 0 1<p<1/\rho_0 1 < p < 1/ ρ 0
ϵ ( r p + r − p ) cos ( p ( θ − π ρ 0 2 ) ) , s = r e i θ . (64) \epsilon(r^p+r^{-p})\cos\left(p\left(\theta-\frac{\pi\rho_0}{2}\right)\right),\qquad s=re^{i\theta}. \tag*{(64)} ϵ ( r p + r − p ) cos ( p ( θ − 2 π ρ 0 ) ) , s = r e i θ . (64) Its cosine has a strictly positive lower bound on the closed sector. The minimum principle on truncated sectors, followed by removal of the barrier, gives u ( s ) ≥ 0 u(s) \ge0 u ( s ) ≥ 0 ρ ∈ [ 1 − ρ 0 , ρ 0 ] \rho\in[1-\rho_0,\rho_0] ρ ∈ [ 1 − ρ 0 , ρ 0 ] Proposition B.3 proves GGC on that whole interval.
If an endpoint 0 or 1 belongs to G α \mathcal{G}_\alpha G α H α ( s ) = b α s + ∫ s / ( s + t ) U ( d t ) H_\alpha(s)=b_\alpha s+\int s/(s+t)\,U(\mathrm{d}t) H α ( s ) = b α s + ∫ s / ( s + t ) U ( d t ) L α L_\alpha L α G α = [ 0 , 1 ] \mathcal{G}_\alpha=[0,1] G α = [ 0 , 1 ] G α \mathcal{G}_\alpha G α { 1 / 2 } \{1/2\} { 1/2 } α > 4 / 5 \alpha>4/5 α > 4/5 Theorem A.3 excludes both endpoints; closedness then gives g α > 0 g_\alpha>0 g α > 0
The interval theorem reduces the classification to nonemptiness and the location of g α g_\alpha g α Theorem B.14 supplies nonemptiness throughout the remaining range, and Theorem B.15 identifies the boundary.
Proposition B.5 (Stability away from the one-sided endpoints). Suppose 0 < α 0 < 1 0<\alpha_0<1 0 < α 0 < 1 X α 0 , 1 X_{\alpha_0,1} X α 0 , 1 η ∈ ( 0 , 1 / 2 ) \eta\in(0,1/2) η ∈ ( 0 , 1/2 ) ϵ > 0 \epsilon>0 ϵ > 0
∣ α − α 0 ∣ < ϵ , η ≤ ρ ≤ 1 − η ⟹ X α , ρ is extended GGC. |\alpha-\alpha_0|<\epsilon,\qquad\eta\le\rho\le1-\eta\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is extended GGC.} ∣ α − α 0 ∣ < ϵ , η ≤ ρ ≤ 1 − η ⟹ X α , ρ is extended GGC. In particular, for each such η \eta η [6 Theorem 2] extends to some rectangle ( 3 / 4 , 3 / 4 + ϵ η ) × [ η , 1 − η ] (3/4,3/4+\epsilon_\eta)\times[\eta,1-\eta] ( 3/4 , 3/4 + ϵ η ) × [ η , 1 − η ] ϵ η \epsilon_\eta ϵ η
Proof. The small- and large-s s s Lemma B.1 hold uniformly when α \alpha α α 0 \alpha_0 α 0 ( 0 , 1 ) (0,1) ( 0 , 1 ) s s s α \alpha α [7 ] , or from the analytic inverse at its nondegenerate quadratic critical point. Its leading coefficient is positive and continuous. The positive-measure scaling proof of Lemma B.1 then applies uniformly, using the uniform bound ν α ( [ 0 , T ] ) ≤ e f α ( b α + 1 / T ) T \nu_\alpha([0,T])\le e f_\alpha(b_\alpha+1/T)T ν α ([ 0 , T ]) ≤ e f α ( b α + 1/ T ) T ( α , s ) (\alpha,s) ( α , s )
For q = e i π ( 1 / 2 − ρ ) q=e^{i\pi(1/2-\rho)} q = e iπ ( 1/2 − ρ ) α = α 0 \alpha=\alpha_0 α = α 0 Re t ≥ 0 \operatorname{Re}t\ge0 Re t ≥ 0 t ≠ 0 t\ne0 t = 0
D α 0 , ρ ( t ) = b α 0 q + ∫ 1 t + u / q U ( d u ) , Re D α 0 , ρ ( t ) ≥ b α 0 sin ( π η ) > 0. D_{\alpha_0,\rho}(t)=b_{\alpha_0}q+\int\frac{1}{t+u/q}\,U(\mathrm{d}u),\qquad\operatorname{Re}D_{\alpha_0,\rho}(t)\ge b_{\alpha_0}\sin(\pi\eta)>0. D α 0 , ρ ( t ) = b α 0 q + ∫ t + u / q 1 U ( d u ) , Re D α 0 , ρ ( t ) ≥ b α 0 sin ( π η ) > 0. All the corresponding q t qt q t ∣ arg s ∣ ≤ π ( 1 − η ) |\arg s|\le\pi(1-\eta) ∣ arg s ∣ ≤ π ( 1 − η ) L α 0 L_{\alpha_0} L α 0 α \alpha α D α , ρ ( t ) = b α q + o ( 1 ) D_{\alpha,\rho}(t)=b_\alpha q+o(1) D α , ρ ( t ) = b α q + o ( 1 )
D α , ρ ( t ) = q α t α − 1 Γ ( α ) ( 1 + O ( ∣ t ∣ α ) ) , ∣ arg ( q α t α − 1 ) ∣ ≤ π 2 − α π η , D_{\alpha,\rho}(t)=\frac{q^\alpha t^{\alpha-1}}{\Gamma(\alpha)}\left(1+O(|t|^\alpha)\right),\qquad|\arg(q^\alpha t^{\alpha-1})|\le\frac{\pi}{2}-\alpha\pi\eta, D α , ρ ( t ) = Γ ( α ) q α t α − 1 ( 1 + O ( ∣ t ∣ α ) ) , ∣ arg ( q α t α − 1 ) ∣ ≤ 2 π − α π η , again giving uniform strict positivity. On the intervening compact half-annulus, joint continuity preserves the displayed positive margin at α 0 \alpha_0 α 0 ϵ \epsilon ϵ Lemma B.2 proves the claim. Taking α 0 = 3 / 4 \alpha_0=3/4 α 0 = 3/4
Corollary B.6 (Certified skewness intervals). For each row in the following table, all ρ ∈ [ 0 , ϵ ] ∪ [ 1 − ϵ , 1 ] \rho\in[0,\epsilon]\cup[1-\epsilon,1] ρ ∈ [ 0 , ϵ ] ∪ [ 1 − ϵ , 1 ]
Table 2.
Proof. The script certify_ggc_skewness_exclusions.py verifies these outward enclosures with 150-digit Arb arithmetic and the explicit Wright tail bounds given earlier. The reusable evaluator is wright_interval.py; the certificate records each rational input and strict enclosing decimal interval. Each negative value excludes ρ ∗ = 1 − ϵ \rho_* = 1-\epsilon ρ ∗ = 1 − ϵ ρ ∈ [ ρ ∗ , 1 ] \rho\in[\rho_*,1] ρ ∈ [ ρ ∗ , 1 ] Theorem B.4 would force ρ ∗ \rho_* ρ ∗
Theorem B.7 (Endpoint exclusion on every linear scale). For each finite M > 0 M>0 M > 0 α M < 1 \alpha_M<1 α M < 1
α M < α < 1 , min ( ρ , 1 − ρ ) ≤ M ( 1 − α ) ⟹ X α , ρ is not extended GGC. \alpha_M<\alpha<1,\qquad\min(\rho,1-\rho)\le M(1-\alpha)\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is not extended GGC.} α M < α < 1 , min ( ρ , 1 − ρ ) ≤ M ( 1 − α ) ⟹ X α , ρ is not extended GGC. Consequently, along any sequence of nonempty sections G α = [ g α , 1 − g α ] G_\alpha=[g_\alpha,1-g_\alpha] G α = [ g α , 1 − g α ] α ↑ 1 \alpha\uparrow1 α ↑ 1
g α 1 − α ⟶ + ∞ . \frac{g_\alpha}{1-\alpha}\longrightarrow+\infty. 1 − α g α ⟶ + ∞. The threshold α M \alpha_M α M
Proof. Let β = 1 − α \beta=1-\alpha β = 1 − α δ = 1 − ρ \delta=1-\rho δ = 1 − ρ C C C X α X_\alpha X α [7 ] gives
X α , 1 − δ = d X α { cos ( π δ ) + sin ( π δ ) C } . X_{\alpha,1-\delta}\mathrel{\stackrel{d}{=}}X_\alpha\{\cos(\pi\delta)+\sin(\pi\delta)C\}. X α , 1 − δ = d X α { cos ( π δ ) + sin ( π δ ) C } . The known exceptional-index limit is ( β β − X α ) / β ⇒ T (\beta^\beta-X_\alpha)/\beta\Rightarrow T ( β β − X α ) / β ⇒ T δ / β → ℓ < ∞ \delta/\beta\to\ell<\infty δ / β → ℓ < ∞ X α → 1 X_\alpha\to1 X α → 1 C C C
β β − X α , 1 − δ β ⇒ T − π ℓ C = d T + π ℓ C . \frac{\beta^\beta-X_{\alpha,1-\delta}}{\beta}\Rightarrow T-\pi\ell C\mathrel{\stackrel{d}{=}}T+\pi\ell C. β β β − X α , 1 − δ ⇒ T − π ℓ C = d T + π ℓ C . Indeed the additional location correction is X α ( 1 − cos ( π δ ) ) / β = o P ( 1 ) X_\alpha(1-\cos(\pi\delta))/\beta=o_P(1) X α ( 1 − cos ( π δ )) / β = o P ( 1 ) [7 ] , the negative-side Lévy kernel of this limit is
k − ( x ) = 1 + ℓ x − e − 2 x 1 − e − x = ∫ 0 ∞ e − u x ( ( 1 + ℓ ) d u − ∑ n ≥ 2 δ n ( d u ) ) . k_-(x)=\frac{1+\ell}{x}-\frac{e^{-2x}}{1-e^{-x}}=\int_0^\infty e^{-ux}\left((1+\ell)\,du-\sum_{n\ge2}\delta_n(du)\right). k − ( x ) = x 1 + ℓ − 1 − e − x e − 2 x = ∫ 0 ∞ e − ux ( ( 1 + ℓ ) d u − n ≥ 2 ∑ δ n ( d u ) ) . Uniqueness of the Laplace transform shows that it cannot be CM: the negative atoms cannot be canceled by a multiple of Lebesgue measure. Thus the limit is not extended GGC for any finite ℓ \ell ℓ M M M α ↑ 1 \alpha\uparrow1 α ↑ 1 δ / β ∈ [ 0 , M ] \delta/\beta\in[0,M] δ / β ∈ [ 0 , M ] δ / β → ℓ ∈ [ 0 , M ] \delta/\beta\to\ell\in[0,M] δ / β → ℓ ∈ [ 0 , M ] ρ = g α \rho=g_\alpha ρ = g α
This result does not decide GGC for a fixed interior ρ \rho ρ α ↑ 1 \alpha\uparrow1 α ↑ 1
A stronger square-root endpoint obstruction
The first zero of the reciprocal Gamma function gives a sharper scale than the preceding weak-limit argument.
Theorem B.8. For every 0 < c < 1 / ( π 2 ) 0<c<1/(\pi\sqrt{2}) 0 < c < 1/ ( π 2 ) α c < 1 \alpha_c<1 α c < 1
α c < α < 1 , min ( ρ , 1 − ρ ) ≤ c 1 − α ⟹ X α , ρ is not extended GGC. \alpha_c<\alpha<1,\qquad\min(\rho,1-\rho)\le c\sqrt{1-\alpha}\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is not extended GGC.} α c < α < 1 , min ( ρ , 1 − ρ ) ≤ c 1 − α ⟹ X α , ρ is not extended GGC. In particular, along any sequence of nonempty GGC sections approaching one,
lim inf α ↑ 1 g α 1 − α ≥ 1 π 2 . \liminf_{\alpha\uparrow1}\frac{g_\alpha}{\sqrt{1-\alpha}}\ge\frac{1}{\pi\sqrt{2}}. α ↑ 1 lim inf 1 − α g α ≥ π 2 1 . A matching upper bound and a sharper expansion are proved in Theorem H.5 .
Proof. Write β = 1 − α \beta= 1-\alpha β = 1 − α f ( x ) = 1 / Γ ( 2 − x ) f(x) = 1/\Gamma(2-x) f ( x ) = 1/Γ ( 2 − x )
G β ( t ) = e − t / β W 1 − β ( t / β ) . G_{\beta}(t)=e^{-t/\beta}W_{1-\beta}(t/\beta). G β ( t ) = e − t / β W 1 − β ( t / β ) . We first show, uniformly for t = 2 e − i η t=2e^{-i\eta} t = 2 e − i η ∣ η ∣ ≤ C β |\eta|\le C\sqrt{\beta} ∣ η ∣ ≤ C β C C C
G β ( t ) = − ( t − 2 ) + O ( β ) , G β ′ ( t ) = − 1 + O ( β ) . (65) G_{\beta}(t)=-(t-2)+O(\beta),\qquad G'_{\beta}(t)=-1+O(\sqrt{\beta}). \tag*{(65)} G β ( t ) = − ( t − 2 ) + O ( β ) , G β ′ ( t ) = − 1 + O ( β ) . (65) Let λ = 2 / β \lambda=2/\beta λ = 2/ β z = t / β z=t/\beta z = t / β N N N λ \lambda λ e λ − Re z e^{\lambda-\operatorname{Re}z} e λ − Re z λ − Re z = 2 ( 1 − cos η ) / β = O ( 1 ) \lambda-\operatorname{Re}z=2(1-\cos\eta)/\beta=O(1) λ − Re z = 2 ( 1 − cos η ) / β = O ( 1 ) 1 ≤ x ≤ 3 1\le x\le3 1 ≤ x ≤ 3 f ( 2 ) = 0 f(2)=0 f ( 2 ) = 0 f ′ ( 2 ) = − 1 f'(2)=-1 f ′ ( 2 ) = − 1
f ( x ) = − ( x − 2 ) + O ( ( x − 2 ) 2 ) . f(x)=-(x-2)+O((x-2)^2). f ( x ) = − ( x − 2 ) + O (( x − 2 ) 2 ) . The full complex Poisson sum of the linear term is exactly − ( t − 2 ) -(t-2) − ( t − 2 ) E ( β N − 2 ) 2 = 2 β \mathbb{E}(\beta N-2)^2=2\beta E ( β N − 2 ) 2 = 2 β β N < 1 \beta N<1 β N < 1 f f f 1 / β 1/\beta 1/ β n ≥ n 0 = ⌈ 3 / β ⌉ n\ge n_0=\lceil3/\beta\rceil n ≥ n 0 = ⌈ 3/ β ⌉ ∣ f ( β n ) ∣ ≤ Γ ( β n − 1 ) / π |f(\beta n)|\le\Gamma(\beta n-1)/\pi ∣ f ( β n ) ∣ ≤ Γ ( β n − 1 ) / π λ n Γ ( β n − 1 ) / n ! \lambda^n\Gamma(\beta n-1)/n! λ n Γ ( β n − 1 ) / n ! λ β β n − 1 \lambda\beta^{\beta n-1} λ β β n − 1 n 0 n_0 n 0 β \beta β n 0 n_0 n 0 O ( e − k / β ) O(e^{-k/\beta}) O ( e − k / β ) e − λ e^{-\lambda} e − λ 0 < k < 2 − 3 ( 1 + log ( 2 / 3 ) ) 0<k<2-3(1+\log(2/3)) 0 < k < 2 − 3 ( 1 + log ( 2/3 )) k k k
Termwise differentiation of the entire series gives
G β ′ ( t ) = e − z ∑ n ≥ 0 z n n ! f ( β ( n + 1 ) ) − f ( β n ) β . G'_{\beta}(t)=e^{-z}\sum_{n\ge0}\frac{z^n}{n!}\frac{f(\beta(n+1))-f(\beta n)}{\beta}. G β ′ ( t ) = e − z n ≥ 0 ∑ n ! z n β f ( β ( n + 1 )) − f ( β n ) . On the middle interval, its final factor equals − 1 + O ( ∣ β n − 2 ∣ + β ) -1+O(|\beta n-2|+\beta) − 1 + O ( ∣ β n − 2∣ + β ) E ∣ β N − 2 ∣ ≤ 2 β \mathbb{E}|\beta N-2|\le\sqrt{2\beta} E ∣ β N − 2∣ ≤ 2 β 1 / β 1/\beta 1/ β
Now choose δ = c β \delta=c\sqrt{\beta} δ = c β ρ = 1 − δ \rho=1-\delta ρ = 1 − δ
s β = ( 2 / β ) 1 / ( 1 − β ) e i π ρ , η = π ( β + ( 1 − β ) δ ) , z = − s β 1 − β = ( 2 / β ) e − i η , t = β z . s_{\beta}=(2/\beta)^{1/(1-\beta)}e^{i\pi\rho},\qquad\eta=\pi(\beta+(1-\beta)\delta),\qquad z=-s_{\beta}^{1-\beta}=(2/\beta)e^{-i\eta},\qquad t=\beta z. s β = ( 2/ β ) 1/ ( 1 − β ) e iπ ρ , η = π ( β + ( 1 − β ) δ ) , z = − s β 1 − β = ( 2/ β ) e − i η , t = β z . Since t − 2 = − 2 i π c β + O ( β ) t-2=-2i\pi c\sqrt{\beta}+O(\beta) t − 2 = − 2 iπ c β + O ( β ) G β ( t ) ≠ 0 G_{\beta}(t)\ne0 G β ( t ) = 0 β \beta β
t G β ′ ( t ) G β ( t ) = i π c β + O ( 1 ) . \frac{tG'_{\beta}(t)}{G_{\beta}(t)}=\frac{i}{\pi c\sqrt{\beta}}+O(1). G β ( t ) t G β ′ ( t ) = π c β i + O ( 1 ) . The exact logarithmic derivative identity is
H 1 − β ( s β ) = − ( 1 − β ) ( z + t G β ′ ( t ) G β ( t ) ) . H_{1-\beta}(s_{\beta})=-(1-\beta)\left(z+\frac{tG'_{\beta}(t)}{G_{\beta}(t)}\right). H 1 − β ( s β ) = − ( 1 − β ) ( z + G β ( t ) t G β ′ ( t ) ) . Consequently
lim β ↓ 0 β Im H 1 − β ( s β ) = 2 π c − 1 π c < 0. (66) \lim_{\beta\downarrow0}\sqrt{\beta}\,\operatorname{Im}H_{1-\beta}(s_{\beta})=2\pi c-\frac{1}{\pi c}<0. \tag*{(66)} β ↓ 0 lim β Im H 1 − β ( s β ) = 2 π c − π c 1 < 0. (66) The necessary ray condition excludes ρ = 1 − c β \rho=1-c\sqrt{\beta} ρ = 1 − c β c ↑ 1 / ( π 2 ) c\uparrow1/(\pi\sqrt{2}) c ↑ 1/ ( π 2 )
The file check_sqrt_endpoint_scale.py supplies strict point enclosures consistent with (B.2); these checks are not needed for the analytic proof and do not establish a converse. For example, at β = 10 − 4 \beta=10^{-4} β = 1 0 − 4 c = 1 / 5 c=1/5 c = 1/5 α = 9999 / 10000 \alpha=9999/10000 α = 9999/10000 ρ ∈ [ 0 , 1 / 500 ] ∪ [ 499 / 500 , 1 ] \rho\in[0,1/500]\cup[499/500,1] ρ ∈ [ 0 , 1/500 ] ∪ [ 499/500 , 1 ]
GGC returns on every fixed interior interval near index one
The endpoint obstruction does not persist at a fixed distance from ρ = 0 , 1 \rho= 0,1 ρ = 0 , 1
Lemma B.9 (Compact Hankel limit). Let G β ( t ) = e − t / β W 1 − β ( t / β ) G_\beta(t)=e^{-t/\beta}W_{1-\beta}(t/\beta) G β ( t ) = e − t / β W 1 − β ( t / β ) β ↓ 0 \beta\downarrow0 β ↓ 0
G β ( t ) = 1 Γ ( 2 − t ) + O ( β ) , G β ′ ( t ) = ( 1 Γ ( 2 − t ) ) ′ + O ( β ) , G_\beta(t)=\frac{1}{\Gamma(2-t)}+O(\beta),\qquad G_\beta'(t)=\left(\frac{1}{\Gamma(2-t)}\right)'+O(\beta), G β ( t ) = Γ ( 2 − t ) 1 + O ( β ) , G β ′ ( t ) = ( Γ ( 2 − t ) 1 ) ′ + O ( β ) , locally uniformly on the whole complex plane.
Proof. Take a Hankel contour H \mathcal{H} H 0 < β < 1 0<\beta<1 0 < β < 1
G β ( t ) = 1 2 π i ∫ H e u u − 2 exp ( t u β − 1 β ) d u . G_\beta(t)=\frac{1}{2\pi i}\int_{\mathcal{H}}e^u u^{-2}\exp\left(t\frac{u^\beta-1}{\beta}\right)\,du. G β ( t ) = 2 π i 1 ∫ H e u u − 2 exp ( t β u β − 1 ) d u . The powers use the corresponding boundary logarithms along the two straight parts of H \mathcal{H} H 0 < β ≤ 1 / 4 0<\beta\leq1/4 0 < β ≤ 1/4
∣ u β − 1 β − log u ∣ ≤ β 2 ∣ log u ∣ 2 e β ∣ log u ∣ . \left|\frac{u^\beta-1}{\beta}-\log u\right|\leq\frac{\beta}{2}|\log u|^2e^{\beta|\log u|}. β u β − 1 − log u ≤ 2 β ∣ log u ∣ 2 e β ∣ l o g u ∣ . On either straight part u = − r ± i 0 u=-r\pm i0 u = − r ± i 0 r ≥ 1 r\geq1 r ≥ 1 β ( log r + π ) 2 r 1 / 4 \beta(\log r+\pi)^2r^{1/4} β ( log r + π ) 2 r 1/4 ∣ t ∣ ≤ M |t|\leq M ∣ t ∣ ≤ M r r r ∣ e u ∣ = e − r |e^u|=e^{-r} ∣ e u ∣ = e − r t t t β \beta β β \beta β e − r / 2 e^{-r/2} e − r /2 r r r O ( β ) O(\beta) O ( β ) β = 0 \beta=0 β = 0 1 / Γ ( 2 − t ) 1/\Gamma(2-t) 1/Γ ( 2 − t )
Lemma B.10 (Uniform endpoint scaling). For 0 < β ≤ 1 / 5 0<\beta\leq1/5 0 < β ≤ 1/5 a = 1 − β a=1-\beta a = 1 − β c β = β 1 / a c_\beta=\beta^{1/a} c β = β 1/ a
Y β = X a − b a c β , P β ( v ) = E e − v Y β = e a v / β W a ( − v a / β ) . Y_\beta=\frac{X_a-b_a}{c_\beta},\qquad P_\beta(v)=\mathbb{E}e^{-vY_\beta}=e^{av/\beta}W_a(-v^a/\beta). Y β = c β X a − b a , P β ( v ) = E e − v Y β = e a v / β W a ( − v a / β ) . For every fixed η > 0 \eta>0 η > 0 0 < β ≤ 1 / 5 0<\beta\leq1/5 0 < β ≤ 1/5 ∣ arg v ∣ ≤ π − η |\arg v|\leq\pi-\eta ∣ arg v ∣ ≤ π − η
P β ( v ) ∼ v − 3 / 2 2 π a , P β ′ ( v ) ∼ − 3 v − 5 / 2 2 2 π a ( ∣ v ∣ → ∞ ) . P_\beta(v)\sim\frac{v^{-3/2}}{\sqrt{2\pi a}},\qquad P_\beta'(v)\sim-\frac{3v^{-5/2}}{2\sqrt{2\pi a}}\qquad(|v|\to\infty). P β ( v ) ∼ 2 π a v − 3/2 , P β ′ ( v ) ∼ − 2 2 π a 3 v − 5/2 ( ∣ v ∣ → ∞ ) . In particular, sufficiently far out in any such sector, P β ≠ 0 P_\beta\neq0 P β = 0 − v P β ′ / P β = 3 / 2 + o ( 1 ) -vP_\beta'/P_\beta=3/2+o(1) − v P β ′ / P β = 3/2 + o ( 1 )
Proof. Let p β p_\beta p β Y β Y_\beta Y β [7 ] , with u = − G Y β u=-G_{Y_\beta} u = − G Y β
y = h β ( u ) : = 1 − 1 u + u − β − 1 β . y=h_\beta(u):=1-\frac{1}{u}+\frac{u^{-\beta}-1}{\beta}. y = h β ( u ) := 1 − u 1 + β u − β − 1 . This is analytic jointly near u = 1 u=1 u = 1 β = 0 \beta=0 β = 0 h 0 ( u ) = 1 − 1 / u − log u h_0(u)=1-1/u-\log u h 0 ( u ) = 1 − 1/ u − log u h β ( 1 ) = h β ′ ( 1 ) = 0 h_\beta(1)=h_\beta'(1)=0 h β ( 1 ) = h β ′ ( 1 ) = 0 h β ′ ′ ( 1 ) = − a h_\beta''(1)=-a h β ′′ ( 1 ) = − a
p β ( y ) = 1 π 2 a y + O ( y 3 / 2 ) ( y ↓ 0 ) , p_\beta(y)=\frac{1}{\pi}\sqrt{\frac{2}{a}}\sqrt{y}+O(y^{3/2})\qquad(y\downarrow0), p β ( y ) = π 1 a 2 y + O ( y 3/2 ) ( y ↓ 0 ) , with a uniform remainder for 0 < β ≤ 1 / 5 0<\beta\leq1/5 0 < β ≤ 1/5 a ≥ 4 / 5 a\geq4/5 a ≥ 4/5 β > 0 \beta>0 β > 0 ℑ u > 0 \Im u>0 ℑ u > 0 y \sqrt{y} y [7 Proof of Theorem 1] , with shape two. The unit-rate gamma component has Laplace transform ( 1 + s ) − 2 (1+s)^{-2} ( 1 + s ) − 2
p β ( y ) y = ∫ 0 ∞ e − y u ν β ( d u ) , P β ( v ) = ∫ 0 ∞ ( v + u ) − 2 ν β ( d u ) . \frac{p_{\beta}(y)}{y}=\int_{0}^{\infty}e^{-yu}\nu_{\beta}(\mathrm{d}u),\qquad P_{\beta}(v)=\int_{0}^{\infty}(v+u)^{-2}\nu_{\beta}(\mathrm{d}u). y p β ( y ) = ∫ 0 ∞ e − y u ν β ( d u ) , P β ( v ) = ∫ 0 ∞ ( v + u ) − 2 ν β ( d u ) . The uniform density estimate implies ν β ( [ 0 , T ] ) = O ( T ) \nu_{\beta}([0,T])=O(\sqrt{T}) ν β ([ 0 , T ]) = O ( T ) T T T ν β ( [ 0 , T ] ) ≤ e T p β ( 1 / T ) \nu_{\beta}([0,T])\leq eT p_{\beta}(1/T) ν β ([ 0 , T ]) ≤ e T p β ( 1/ T ) T T T ν β ( [ 0 , T ] ) ≤ ( 1 + T ) 2 P β ( 1 ) ≤ ( 1 + T ) \nu_{\beta}([0,T])\leq(1+T)^2P_{\beta}(1)\leq(1+T) ν β ([ 0 , T ]) ≤ ( 1 + T ) 2 P β ( 1 ) ≤ ( 1 + T ) R → ∞ R\to\infty R → ∞ R − 1 / 2 ν β ( R d u ) R^{-1/2}\nu_{\beta}(R\,\mathrm{d}u) R − 1/2 ν β ( R d u ) Lemma B.1 . Taking a subsequence of β \beta β [ 0 , 1 / 5 ] [0,1/5] [ 0 , 1/5 ]
1 π π 2 1 − β ∗ u − 1 / 2 d u . \frac{1}{\pi\sqrt{\pi}}\sqrt{\frac{2}{1-\beta_*}}u^{-1/2}\,\mathrm{d}u. π π 1 1 − β ∗ 2 u − 1/2 d u . The uniform cumulative-mass bounds control both ends of the Stieltjes integrals and their derivatives on every closed slit sector. This proves the two uniform asymptotics and their ratio. No unproved uniform steepest-descent expansion is used here. □
Theorem B.11 (Fixed interior skewness near one ). For every 0 < η < 1 / 2 0<\eta<1/2 0 < η < 1/2 β η > 0 \beta_{\eta}>0 β η > 0
1 − β η < α < 1 , η ≤ ρ ≤ 1 − η ⟹ X α , ρ is extended GGC. 1-\beta_{\eta}<\alpha<1,\qquad\eta\leq\rho\leq1-\eta\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is extended GGC.} 1 − β η < α < 1 , η ≤ ρ ≤ 1 − η ⟹ X α , ρ is extended GGC. Thus the GGC sections are nonempty for all indices sufficiently close to one, and their critical endpoints satisfy
g α ⟶ 0 ( α ↑ 1 ) , lim inf α ↑ 1 g α 1 − α ≥ 1 π 2 . g_{\alpha}\longrightarrow0\quad(\alpha\uparrow1),\qquad\liminf_{\alpha\uparrow1}\frac{g_{\alpha}}{\sqrt{1-\alpha}}\geq\frac{1}{\pi\sqrt{2}}. g α ⟶ 0 ( α ↑ 1 ) , α ↑ 1 lim inf 1 − α g α ≥ π 2 1 . This argument gives no explicit β η \beta_{\eta} β η Theorem H.5 .
Proof. Write β = 1 − α \beta=1-\alpha β = 1 − α a = α a=\alpha a = α v = c β s v=c_{\beta}s v = c β s t = − v a = − β s a t=-v^{a}=-\beta s^{a} t = − v a = − β s a Lemma B.10 gives, uniformly on ∣ arg s ∣ ≤ π ( 1 − η ) |\arg s|\leq\pi(1-\eta) ∣ arg s ∣ ≤ π ( 1 − η )
H a ( s ) = a β v − v P β ′ ( v ) P β ( v ) = a β v + 3 2 + o ( 1 ) ( ∣ v ∣ → ∞ ) . H_a(s)=\frac{a}{\beta}v-\frac{vP_{\beta}'(v)}{P_{\beta}(v)}=\frac{a}{\beta}v+\frac{3}{2}+o(1)\qquad(|v|\to\infty). H a ( s ) = β a v − P β ( v ) v P β ′ ( v ) = β a v + 2 3 + o ( 1 ) ( ∣ v ∣ → ∞ ) . Choose R ≥ 1 R\geq1 R ≥ 1 ∣ v ∣ ≥ R |v|\geq R ∣ v ∣ ≥ R 0 < β ≤ 1 / 5 0<\beta\leq1/5 0 < β ≤ 1/5 P β ≠ 0 P_{\beta}\ne0 P β = 0 4 R sin ( π η ) > 1 4R\sin(\pi\eta)>1 4 R sin ( π η ) > 1 H a H_a H a π η \pi\eta π η π ( 1 − η ) \pi(1-\eta) π ( 1 − η )
For ∣ v ∣ ≤ R |v|\leq R ∣ v ∣ ≤ R t = − v a t=-v^{a} t = − v a
K = { 0 } ∪ { ∣ t ∣ ≤ R , ∣ arg t ∣ ≥ π η } . \mathcal{K}=\{0\}\cup\{|t|\leq R,\ |\arg t|\geq\pi\eta\}. K = { 0 } ∪ { ∣ t ∣ ≤ R , ∣ arg t ∣ ≥ π η } . The function f ( t ) = 1 / Γ ( 2 − t ) f(t)=1/\Gamma(2-t) f ( t ) = 1/Γ ( 2 − t ) 2 , 3 , … 2,3,\ldots 2 , 3 , … Lemma B.9 shows that for small enough β \beta β G β G_{\beta} G β G β ′ / G β G_{\beta}'/G_{\beta} G β ′ / G β
H a ( s ) = a s a ( 1 + β G β ′ ( t ) G β ( t ) ) H_a(s)=as^{a}\left(1+\beta\frac{G_{\beta}'(t)}{G_{\beta}(t)}\right) H a ( s ) = a s a ( 1 + β G β ( t ) G β ′ ( t ) ) therefore preserves strictly positive imaginary part on both rays: the arguments of s a s^{a} s a 4 π η / 5 4\pi\eta/5 4 π η /5 π ( 1 − η ) \pi(1-\eta) π ( 1 − η ) s s s L a L_a L a v v v Proposition B.3 hold. The limit of g α g_{\alpha} g α η \eta η Theorem B.8 provides the lower bound. □
A certificate covering every symmetric law below one
The two compact integrals below remove the singular dependence on β \beta β 4 / 5 4/5 4/5
Lemma B.12 (An explicit symmetric tail). For 0 < β ≤ 1 / 5 0 < \beta\le1/5 0 < β ≤ 1/5 a = 1 − β a = 1-\beta a = 1 − β r ≥ 200 r \ge200 r ≥ 200
Im H a ( i r / c β ) > 4 r − 450 > 0. \operatorname{Im} H_a(ir/c_\beta) > 4r - 450 > 0. Im H a ( i r / c β ) > 4 r − 450 > 0. Moreover, L a L_a L a Re s ≥ 0 \operatorname{Re} s \ge0 Re s ≥ 0
Proof. Use p β , P β , h β p_\beta,P_\beta,h_\beta p β , P β , h β Lemma B.10 , and set j β ( y ) = p β ( y ) / y j_\beta(y)=p_\beta(y)/y j β ( y ) = p β ( y ) / y
p β ( y ) < y , − j β ′ ( y ) > 1 10 y − 3 / 2 , 0 < y ≤ 1 / 200. (67) p_\beta(y)<\sqrt{y},\qquad-j_\beta'(y)>\frac{1}{10}y^{-3/2},\qquad0<y\le1/200. \tag*{(67)} p β ( y ) < y , − j β ′ ( y ) > 10 1 y − 3/2 , 0 < y ≤ 1/200. (67) Write u = 1 + w u=1+w u = 1 + w
h β ( 1 + w ) = ∑ n ≥ 2 h n w n , h n = ( − 1 ) n ( ( 1 + β ) n − 1 n ! − 1 ) . h_\beta(1+w)=\sum_{n\ge2}h_nw^n,\qquad h_n=(-1)^n\left(\frac{(1+\beta)^{n-1}}{n!}-1\right). h β ( 1 + w ) = n ≥ 2 ∑ h n w n , h n = ( − 1 ) n ( n ! ( 1 + β ) n − 1 − 1 ) . They satisfy h 2 = − a / 2 h_2=-a/2 h 2 = − a /2 ∣ h 3 ∣ / a ≤ 7 / 10 |h_3|/a\le7/10 ∣ h 3 ∣/ a ≤ 7/10 ∣ h 4 ∣ / a ≤ 81 / 100 |h_4|/a\le81/100 ∣ h 4 ∣/ a ≤ 81/100 ∣ h n ∣ / a ≤ 5 / 4 |h_n|/a\le5/4 ∣ h n ∣/ a ≤ 5/4 n ≥ 5 n\ge5 n ≥ 5 R = 3 / 20 R=3/20 R = 3/20
4 5 ( R 2 2 − 7 10 R 3 − 81 100 R 4 − 5 4 R 5 1 − R ) > 0.00669 > 1 / 200. \frac{4}{5}\left(\frac{R^2}{2}-\frac{7}{10}R^3-\frac{81}{100}R^4-\frac{5}{4}\frac{R^5}{1-R}\right)>0.00669>1/200. 5 4 ( 2 R 2 − 10 7 R 3 − 100 81 R 4 − 4 5 1 − R R 5 ) > 0.00669 > 1/200. Rouché’s theorem puts the two inverse roots in ∣ w ∣ < R |w|<R ∣ w ∣ < R y y y y = t 2 y=t^2 y = t 2 w = t V ( t ) w=tV(t) w = t V ( t )
h β ( 1 + w ) = − a 2 w 2 ( 1 + R ( w ) ) , h β ′ ( 1 + w ) = − a w ( 1 + S ( w ) ) . h_\beta(1+w)=-\frac{a}{2}w^2(1+\mathcal{R}(w)),\qquad h_\beta'(1+w)=-aw(1+\mathcal{S}(w)). h β ( 1 + w ) = − 2 a w 2 ( 1 + R ( w )) , h β ′ ( 1 + w ) = − a w ( 1 + S ( w )) . The same coefficient bounds give
∣ R ∣ < 0.26 , ∣ S ∣ < 0.414 , ∣ S − R ∣ < 0.158. |\mathcal{R}|<0.26,\qquad|\mathcal{S}|<0.414,\qquad|\mathcal{S}-\mathcal{R}|<0.158. ∣ R ∣ < 0.26 , ∣ S ∣ < 0.414 , ∣ S − R ∣ < 0.158. For clarity, the respective bounds before rounding are
2 ( 7 10 R + 81 100 R 2 + 5 4 R 3 1 − R ) , 2\left(\frac{7}{10}R+\frac{81}{100}R^2+\frac{5}{4}\frac{R^3}{1-R}\right), 2 ( 10 7 R + 100 81 R 2 + 4 5 1 − R R 3 ) , 21 10 R + 81 25 R 2 + 5 4 R 3 { 5 1 − R + R ( 1 − R ) 2 } , \frac{21}{10}R+\frac{81}{25}R^2+\frac{5}{4}R^3\left\{\frac{5}{1-R}+\frac{R}{(1-R)^2}\right\}, 10 21 R + 25 81 R 2 + 4 5 R 3 { 1 − R 5 + ( 1 − R ) 2 R } , 7 10 R + 81 50 R 2 + 5 4 R 3 { 3 1 − R + R ( 1 − R ) 2 } . \frac{7}{10}R+\frac{81}{50}R^2+\frac{5}{4}R^3\left\{\frac{3}{1-R}+\frac{R}{(1-R)^2}\right\}. 10 7 R + 50 81 R 2 + 4 5 R 3 { 1 − R 3 + ( 1 − R ) 2 R } . Thus
V = i 2 a ( 1 + R ) − 1 / 2 , ∣ V ∣ < 1.84 , Im V > 1.24 , ∣ t V ′ ∣ = ∣ V R − S 1 + S ∣ < 0.5. V=i\sqrt{\frac{2}{a}}(1+\mathcal{R})^{-1/2},\qquad|V|<1.84,\qquad\operatorname{Im}V>1.24,\qquad|tV'|=\left|V\frac{\mathcal{R}-\mathcal{S}}{1+\mathcal{S}}\right|<0.5. V = i a 2 ( 1 + R ) − 1/2 , ∣ V ∣ < 1.84 , Im V > 1.24 , ∣ t V ′ ∣ = V 1 + S R − S < 0.5. Since p β ( t 2 ) = t Im V / π p_\beta(t^2)=t\operatorname{Im}V/\pi p β ( t 2 ) = t Im V / π − j β ′ ( y ) y 3 / 2 = Im ( V − t V ′ ) / ( 2 π ) > 0.1 -j_\beta'(y)y^{3/2}=\operatorname{Im}(V-tV')/(2\pi)>0.1 − j β ′ ( y ) y 3/2 = Im ( V − t V ′ ) / ( 2 π ) > 0.1
Write j β ( y ) = ∫ e − y u ν β ( d u ) j_\beta(y)=\int e^{-yu}\nu_\beta(du) j β ( y ) = ∫ e − y u ν β ( d u ) T ≥ 200 T\ge200 T ≥ 200 ν β ( [ 0 , T ] ) ≤ e T \nu_\beta([0,T])\le e\sqrt{T} ν β ([ 0 , T ]) ≤ e T r , 2 r , 4 r , … r,2r,4r,\ldots r , 2 r , 4 r , …
∣ P β ′ ( i r ) ∣ ≤ 2 e ( 1 + 2 1 − 2 − 5 / 2 ) r − 5 / 2 . |P_\beta'(ir)|\le2e\left(1+\frac{\sqrt{2}}{1-2^{-5/2}}\right)r^{-5/2}. ∣ P β ′ ( i r ) ∣ ≤ 2 e ( 1 + 1 − 2 − 5/2 2 ) r − 5/2 . The elementary bound ( 1 + t 2 ) 2 ≤ 6 e t (1+t^2)^2\le6e^t ( 1 + t 2 ) 2 ≤ 6 e t
∣ P β ( i r ) ∣ ≥ − Im P β ( i r ) = 2 r 2 ∫ u / r ( 1 + ( u / r ) 2 ) 2 ν β ( d u ) ≥ − j β ′ ( 1 / r ) 3 r 3 > 1 30 r − 3 / 2 . |P_\beta(ir)|\ge-\operatorname{Im}P_\beta(ir)=\frac{2}{r^2}\int\frac{u/r}{(1+(u/r)^2)^2}\nu_\beta(du)\ge-\frac{j_\beta'(1/r)}{3r^3}>\frac{1}{30}r^{-3/2}. ∣ P β ( i r ) ∣ ≥ − Im P β ( i r ) = r 2 2 ∫ ( 1 + ( u / r ) 2 ) 2 u / r ν β ( d u ) ≥ − 3 r 3 j β ′ ( 1/ r ) > 30 1 r − 3/2 . Consequently
∣ r P β ′ ( i r ) P β ( i r ) ∣ < 60 e ( 1 + 2 1 − 2 − 5 / 2 ) < 450. \left|\frac{rP_\beta'(ir)}{P_\beta(ir)}\right|<60e\left(1+\frac{\sqrt{2}}{1-2^{-5/2}}\right)<450. P β ( i r ) r P β ′ ( i r ) < 60 e ( 1 + 1 − 2 − 5/2 2 ) < 450. [#eq:B.3]
The identity H a ( i r / c β ) = a i r / β − i r P β ′ ( i r ) / P β ( i r ) H_a(ir/c_\beta)=air/\beta-irP'_\beta(ir)/P_\beta(ir) H a ( i r / c β ) = ai r / β − i r P β ′ ( i r ) / P β ( i r ) a / β ≥ 4 a/\beta\ge4 a / β ≥ 4 P β ( s ) = ∫ ( s + u ) − 2 ν β ( d u ) P_\beta(s)=\int(s+u)^{-2\nu_\beta}(\mathrm{d}u) P β ( s ) = ∫ ( s + u ) − 2 ν β ( d u ) Re s > 0 \operatorname{Re}s>0 Re s > 0 L a ( 0 ) = 1 L_a(0)=1 L a ( 0 ) = 1
Lemma B.13 (Validated parabolic contour). For 0 ≤ β ≤ 1 / 5 0\le\beta\le1/5 0 ≤ β ≤ 1/5 E β ( z ) = ( e β z − 1 ) / β E_\beta(z)=(e^{\beta z}-1)/\beta E β ( z ) = ( e β z − 1 ) / β E 0 ( z ) = z E_0(z)=z E 0 ( z ) = z r > 0 r>0 r > 0
v = i r , q = 1 + x / r , w = 1 + i q , u = r w 2 / 2. v=ir,\qquad q=1+x/\sqrt{r},\qquad w=1+iq,\qquad u=rw^2/2. v = i r , q = 1 + x / r , w = 1 + i q , u = r w 2 /2. ℓ = 2 log ( 1 + ( 1 + i ) x 2 r ) , Φ = u − v − v E β ( ℓ ) . \ell=2\log\left(1+\frac{(1+i)x}{2\sqrt{r}}\right),\qquad\Phi=u-v-vE_\beta(\ell). ℓ = 2 log ( 1 + 2 r ( 1 + i ) x ) , Φ = u − v − v E β ( ℓ ) . The logarithm displayed here has its continuous principal value on the real x x x
F = ∫ − ∞ ∞ e Φ w − 3 d x , J = ∫ − ∞ ∞ E β ( ℓ ) e Φ w − 3 d x . F=\int_{-\infty}^{\infty}e^\Phi w^{-3}\,\mathrm{d}x,\qquad J=\int_{-\infty}^{\infty}E_\beta(\ell)e^\Phi w^{-3}\,\mathrm{d}x. F = ∫ − ∞ ∞ e Φ w − 3 d x , J = ∫ − ∞ ∞ E β ( ℓ ) e Φ w − 3 d x . For β > 0 \beta>0 β > 0
P β ( i r ) = 2 π r 3 / 2 F , P β ′ ( i r ) = − 2 a π r 3 / 2 J , β Im H a ( i r / c β ) a r = 1 + β Re ( J / F ) . P_\beta(ir)=\frac{2}{\pi r^{3/2}}F,\qquad P'_\beta(ir)=-\frac{2a}{\pi r^{3/2}}J,\qquad\frac{\beta\operatorname{Im}H_a(ir/c_\beta)}{ar}=1+\beta\operatorname{Re}(J/F). P β ( i r ) = π r 3/2 2 F , P β ′ ( i r ) = − π r 3/2 2 a J , a r β Im H a ( i r / c β ) = 1 + β Re ( J / F ) . The errors on truncating the two integrals to [ − T , T ] [-T,T] [ − T , T ]
B T = e − 3 T 2 / 10 3 T / 10 , 17 B T , B_T=\frac{e^{-3T^2/10}}{3T/10},\qquad17B_T, B T = 3 T /10 e − 3 T 2 /10 , 17 B T , respectively, uniformly in both parameters. These formulas and bounds also have a regular β = 0 \beta=0 β = 0
Proof. Deform the Hankel contour to u = ( i r + i x / 2 ) 2 u=(\sqrt{ir}+ix/\sqrt{2})^2 u = ( i r + i x / 2 ) 2 e u e^u e u β → 0 \beta\to0 β → 0 Φ \Phi Φ v v v − a E β ( ℓ ) -aE_\beta(\ell) − a E β ( ℓ )
For q ≥ 0 q\ge0 q ≥ 0 Re Φ / r \operatorname{Re}\Phi/r Re Φ/ r q = 1 q=1 q = 1
− 1 + ( 2 β − 1 ) 2 1 − β Re { e i a π / 2 ( 1 + i q ) − 2 a } ≤ − 1. -1+(2\beta-1)2^{1-\beta}\operatorname{Re}\left\{e^{ia\pi/2}(1+iq)^{-2a}\right\}\le-1. − 1 + ( 2 β − 1 ) 2 1 − β Re { e iaπ /2 ( 1 + i q ) − 2 a } ≤ − 1. For q ≤ 0 q\le0 q ≤ 0 Re ( e i a π / 2 ( 1 + i q ) 2 β ) \operatorname{Re}(e^{ia\pi/2}(1+iq)^{2\beta}) Re ( e iaπ /2 ( 1 + i q ) 2 β ) sin ( β π / 2 ) \sin(\beta\pi/2) sin ( β π /2 )
Re Φ r ≤ 1 − q 2 2 − 2 − β sin ( β π / 2 ) β < 4 5 − q 2 2 . \frac{\operatorname{Re}\Phi}{r}\le\frac{1-q^2}{2}-2^{-\beta}\frac{\sin(\beta\pi/2)}{\beta}<\frac45-\frac{q^2}{2}. r Re Φ ≤ 2 1 − q 2 − 2 − β β sin ( β π /2 ) < 5 4 − 2 q 2 . The ratio at zero is interpreted by continuity. Its lower bound 13 / 10 13/10 13/10 sin z / z ≥ 1 − z 2 / 6 \sin z/z\ge1-z^2/6 sin z / z ≥ 1 − z 2 /6 Re Φ ≤ − 3 x 2 / 10 \operatorname{Re}\Phi\le-3x^2/10 Re Φ ≤ − 3 x 2 /10
If z = 1 + q 2 ≥ 1 z=\sqrt{1+q^2}\ge1 z = 1 + q 2 ≥ 1
∣ E β ( ℓ ) w − 3 ∣ ≤ e d / 5 ( 2 log z + d ) z − 13 / 5 < 17 , d = log 2 + 3 π / 2. \lvert E_\beta(\ell)w^{-3}\rvert\le e^{d/5}(2\log z+d)z^{-13/5}<17,\qquad d=\log2+3\pi/2. ∣ E β ( ℓ ) w − 3 ∣ ≤ e d /5 ( 2 log z + d ) z − 13/5 < 17 , d = log 2 + 3 π /2. The right side is decreasing in z ≥ 1 z\ge1 z ≥ 1 17 17 17 ∣ w − 3 ∣ ≤ 1 \lvert w^{-3}\rvert\le1 ∣ w − 3 ∣ ≤ 1
Theorem B.14 (Every symmetric law below one is GGC). For every 0 < α ≤ 1 0<\alpha\le1 0 < α ≤ 1 X α , 1 / 2 X_{\alpha,1/2} X α , 1/2 4 / 5 ≤ α < 1 4/5\le\alpha<1 4/5 ≤ α < 1 Lemma B.12 . In particular, G α \mathcal{G}_\alpha G α 0 < α < 1 0<\alpha<1 0 < α < 1
Proof. Theorem C.4 covers α ≤ 4 / 5 \alpha\le4/5 α ≤ 4/5 α = 1 \alpha= 1 α = 1 β = 1 − α ∈ ( 0 , 1 / 5 ] \beta= 1-\alpha\in(0,1/5] β = 1 − α ∈ ( 0 , 1/5 ]
For small scaled radius, use G β G_\beta G β t = − T e i α π / 2 t=-Te^{i\alpha\pi/2} t = − T e i α π /2
Q 0 ( β , T ) = Im [ e i α π / 2 { 1 + β G β ′ ( t ) / G β ( t ) } ] > 0 , 0 ≤ β ≤ 1 / 5 , 0 ≤ T ≤ 1. Q_0(\beta,T)=\operatorname{Im}\left[e^{i\alpha\pi/2}\left\{1+\beta G_\beta'(t)/G_\beta(t)\right\}\right]>0,\quad0\le\beta\le1/5,\quad0\le T\le1. Q 0 ( β , T ) = Im [ e i α π /2 { 1 + β G β ′ ( t ) / G β ( t ) } ] > 0 , 0 ≤ β ≤ 1/5 , 0 ≤ T ≤ 1. This covers 0 < r ≤ 1 0<r\le1 0 < r ≤ 1 T = r α T=r^\alpha T = r α G G G G ′ G' G ′ u = − z ± i 0 u=-z\pm i0 u = − z ± i 0 z ≥ 80 z\ge80 z ≥ 80
∣ E β ( log u ) ∣ ≤ ( log z + π ) z 1 / 5 ≤ z / 2. \lvert E_\beta(\log u)\rvert\le(\log z+\pi)z^{1/5}\le z/2. ∣ E β ( log u )∣ ≤ ( log z + π ) z 1/5 ≤ z /2. The resulting two-sided tail errors for G , G ′ G,G' G , G ′ 2 e − 40 / ( π 80 2 ) 2e^{-40}/(\pi80^2) 2 e − 40 / ( π 8 0 2 ) e − 40 / ( π 80 ) e^{-40}/(\pi80) e − 40 / ( π 80 ) 1 ≤ r ≤ 200 1\le r\le200 1 ≤ r ≤ 200 T = 12 T=12 T = 12
Q 1 ( β , r ) = 1 + β Re ( J / F ) > 0. Q_1(\beta,r)=1+\beta\operatorname{Re}(J/F)>0. Q 1 ( β , r ) = 1 + β Re ( J / F ) > 0. For numerical stability only, the phase is evaluated in the exactly equivalent form
Φ = v ℓ 2 2 { 1 F 1 ( 1 ; 3 ; ℓ ) − β 1 F 1 ( 1 ; 3 ; β ℓ ) } , E β ( ℓ ) = ℓ 1 F 1 ( 1 ; 2 ; β ℓ ) . \Phi=\frac{v\ell^2}{2}\left\{{}_1F_1(1;3;\ell)-\beta{}_1F_1(1;3;\beta\ell)\right\},\qquad E_\beta(\ell)=\ell{}_1F_1(1;2;\beta\ell). Φ = 2 v ℓ 2 { 1 F 1 ( 1 ; 3 ; ℓ ) − β 1 F 1 ( 1 ; 3 ; β ℓ ) } , E β ( ℓ ) = ℓ 1 F 1 ( 1 ; 2 ; β ℓ ) . Thus neither division by a parameter interval containing zero nor subtraction of the large saddle terms is needed.
The file symmetric_ggc_certificate.json records a recursive binary partition of the two complete rectangles into 720 boxes. The functions in symmetric_hankel.py use directed complex ball arithmetic and validated analytic quadrature at 45 decimal digits, adding every displayed truncation error. All logarithms in quadrature forward its analyticity flag. The separate command certify_symmetric_ggc.py verify reconstructs the exact partition from its roots and reevaluates every box, accepting only strictly positive whole-box lower bounds. That verification passed; its hashes and constant bounds are saved in symmetric_ggc_verification.json. Finally r ≥ 200 r\ge200 r ≥ 200 ρ = 1 / 2 \rho=1/2 ρ = 1/2
The critical curve and its first angular contact
The structural statements can now be made without an empty-section qualification. They also specify which contact must be selected in any numerical investigation of the critical curve.
Theorem B.15 (Critical GGC curve). There is a continuous function g : ( 0 , 1 ] → [ 0 , 1 / 2 ) g:(0,1]\to[0,1/2) g : ( 0 , 1 ] → [ 0 , 1/2 )
X α , ρ is extended GGC ⟺ g ( α ) ≤ ρ ≤ 1 − g ( α ) , 0 < α ≤ 1. X_{\alpha,\rho}\ \text{is extended GGC}\ \Longleftrightarrow\ g(\alpha)\le\rho\le1-g(\alpha),\qquad0<\alpha\le1. X α , ρ is extended GGC ⟺ g ( α ) ≤ ρ ≤ 1 − g ( α ) , 0 < α ≤ 1. It satisfies
g ( α ) = 0 ( 0 < α ≤ 4 / 5 or α = 1 ) , 0 < g ( α ) < 1 / 2 ( 4 / 5 < α < 1 ) , g(\alpha)=0\ (0<\alpha\le4/5\ \text{or}\ \alpha=1),\qquad0<g(\alpha)<1/2\ (4/5<\alpha<1), g ( α ) = 0 ( 0 < α ≤ 4/5 or α = 1 ) , 0 < g ( α ) < 1/2 ( 4/5 < α < 1 ) , is locally Lipschitz on ( 4 / 5 , 1 ) (4/5,1) ( 4/5 , 1 )
lim inf α ↑ 1 g ( α ) 1 − α ≥ 1 π 2 . \liminf_{\alpha\uparrow1}\frac{g(\alpha)}{\sqrt{1-\alpha}}\ge\frac{1}{\pi\sqrt{2}}. α ↑ 1 lim inf 1 − α g ( α ) ≥ π 2 1 . For 4 / 5 < α < 1 4/5<\alpha<1 4/5 < α < 1
π ( 1 − g ( α ) ) = min { arg s : π / 2 < arg s < π , L α ( s ) ≠ 0 , Im H α ( s ) = 0 } . (68) \pi(1-g(\alpha))=\min\{\arg s:\ \pi/2<\arg s<\pi,\ L_\alpha(s)\ne0,\ \operatorname{Im}H_\alpha(s)=0\}. \tag*{(68)} π ( 1 − g ( α )) = min { arg s : π /2 < arg s < π , L α ( s ) = 0 , Im H α ( s ) = 0 } . (68) Every minimizer s ∗ s_* s ∗
Im H α ( s ∗ ) = 0 , Im { s ∗ H α ′ ( s ∗ ) } = 0 , Re { s ∗ H α ′ ( s ∗ ) } < 0. (69) \operatorname{Im}H_\alpha(s_*)=0,\qquad\operatorname{Im}\{s_*H_\alpha'(s_*)\}=0,\qquad\operatorname{Re}\{s_*H_\alpha'(s_*)\}<0. \tag*{(69)} Im H α ( s ∗ ) = 0 , Im { s ∗ H α ′ ( s ∗ )} = 0 , Re { s ∗ H α ′ ( s ∗ )} < 0. (69) Thus the relevant contact is the smallest angular one, not an arbitrarily chosen solution of the two equality equations. The theorem is an exact variational characterization. Theorem G.2 below proves arbitrary-accuracy computability, and Theorem H.5 gives the sharp endpoint expansion.
Proof. Existence of a closed nonempty central section follows from Theorems B.4 and B.14 . The positive and negative one-sided results give its stated zero and positive ranges. At every 4 / 5 < α < 1 4/5 < \alpha< 1 4/5 < α < 1 Im H α ( i r ) \operatorname{Im} H_\alpha(ir) Im H α ( i r ) r > 0 r > 0 r > 0 Lemma B.1 stay strictly positive on rays in a neighborhood of π / 2 \pi/2 π /2 g ( α ) < 1 / 2 g(\alpha) < 1/2 g ( α ) < 1/2
Write θ ∗ = π ( 1 − g ( α ) ) \theta_* = \pi(1-g(\alpha)) θ ∗ = π ( 1 − g ( α )) Theorem B.4 , U α ( s ) = Im H α ( s ) U_\alpha(s) = \operatorname{Im} H_\alpha(s) U α ( s ) = Im H α ( s ) 0 ≤ arg s ≤ θ ∗ 0 \le\arg s \le\theta_* 0 ≤ arg s ≤ θ ∗ Lemma B.1 , and continuity handles the compact annulus. Such an increase contradicts the definition of the central section. There is therefore a contact on the outer ray. No smaller angle can give a zero, proving (B.4). At a contact, the restriction to the outer ray has a local minimum, so ∂ log r U = Im ( s H ′ ) = 0 \partial_{\log r} U = \operatorname{Im}(sH') = 0 ∂ l o g r U = Im ( s H ′ ) = 0 ∂ θ U = Re ( s H ′ ) < 0 \partial_\theta U = \operatorname{Re}(sH') < 0 ∂ θ U = Re ( s H ′ ) < 0
To prove continuity, weak closedness of the extended GGC class and continuity of this free stable family imply g ( α 0 ) ≤ lim inf g ( α n ) g(\alpha_0) \le\liminf g(\alpha_n) g ( α 0 ) ≤ lim inf g ( α n ) α n → α 0 < 1 \alpha_n \to\alpha_0 < 1 α n → α 0 < 1 π / 2 \pi/2 π /2 θ ∗ ( α 0 ) \theta_*(\alpha_0) θ ∗ ( α 0 ) Lemma B.10 at infinity, and compactness in between. Taking such angles up to θ ∗ ( α 0 ) \theta_*(\alpha_0) θ ∗ ( α 0 ) α 0 = 4 / 5 \alpha_0 = 4/5 α 0 = 4/5 Theorem B.11 .
For completeness the Hopf inequality also gives local Lipschitz regularity on ( 4 / 5 , 1 ) (4/5, 1) ( 4/5 , 1 ) − ∂ θ U -\partial_\theta U − ∂ θ U θ = Θ ( α , r ) \theta= \Theta(\alpha,r) θ = Θ ( α , r ) ∣ ∂ α Θ ∣ |\partial_\alpha\Theta| ∣ ∂ α Θ∣ U U U Theorem B.8 .
Exact rational moment certificates
This is an independent check on a subset of Theorem A.3 and a reusable method, not a prerequisite for its analytic proof. Put Y α = X α − b α Y_\alpha= X_\alpha-b_\alpha Y α = X α − b α
c α = 2 α ( 1 − α ) 1 / α , h α ( v ) = 2 F 1 ( α + 1 , 1 ; 3 ; v ) . c_\alpha= \frac{2}{\alpha(1-\alpha)^{1/\alpha}}, \qquad h_\alpha(v) = {}_2F_1(\alpha+1,1;3;v). c α = α ( 1 − α ) 1/ α 2 , h α ( v ) = 2 F 1 ( α + 1 , 1 ; 3 ; v ) . Define
A n = 1 2 n + 1 [ v 2 n ] ( 1 − v ) − 2 h α ( v ) − n − 1 / 2 , d n = ( − 1 ) n ( 3 / 2 ) n A n . A_n = \frac{1}{2n+1}[v^{2n}](1-v)^{-2}h_\alpha(v)^{-n-1/2}, \qquad d_n = (-1)^n(3/2)_n A_n. A n = 2 n + 1 1 [ v 2 n ] ( 1 − v ) − 2 h α ( v ) − n − 1/2 , d n = ( − 1 ) n ( 3/2 ) n A n . The endpoint expansion in [7 ] gives the finite-order large-s s s
E e − s Y α ∼ a 0 Γ ( 3 / 2 ) s − 3 / 2 ∑ n ≥ 0 d n ( c α / s ) n . \mathbb{E}e^{-sY_\alpha} \sim a_0\Gamma(3/2)s^{-3/2}\sum_{n\ge0}d_n(c_\alpha/s)^n. E e − s Y α ∼ a 0 Γ ( 3/2 ) s − 3/2 n ≥ 0 ∑ d n ( c α / s ) n . No convergence of this infinite asymptotic series is assumed. Let
log ( ∑ n ≥ 0 d n t n ) = ∑ n ≥ 1 ℓ n t n , M 0 = 3 2 , M n = ( − 1 ) n n ℓ n ( n ≥ 1 ) . \log\left(\sum_{n\ge0}d_nt^n\right)=\sum_{n\ge1}\ell_nt^n,\qquad M_0=\frac{3}{2},\qquad M_n=(-1)^n n\ell_n\quad(n\ge1). log ( n ≥ 0 ∑ d n t n ) = n ≥ 1 ∑ ℓ n t n , M 0 = 2 3 , M n = ( − 1 ) n n ℓ n ( n ≥ 1 ) . If X α X_\alpha X α M n = c − n ∫ u n U ( d u ) M_n=c^{-n}\int u^nU(\mathrm{d}u) M n = c − n ∫ u n U ( d u )
[ M i + j ] i , j = 0 m ⪰ 0 , [ M i + j + 1 ] i , j = 0 m ⪰ 0. [M_{i+j}]_{i,j=0}^{m}\succeq0,\qquad[M_{i+j+1}]_{i,j=0}^{m}\succeq0. [ M i + j ] i , j = 0 m ⪰ 0 , [ M i + j + 1 ] i , j = 0 m ⪰ 0. Each M n M_n M n α \alpha α 2 n 2n 2 n
M 1 = ( 2 α 2 − 23 α + 47 ) / 48. M_1=(2\alpha^2-23\alpha+47)/48. M 1 = ( 2 α 2 − 23 α + 47 ) /48. The coefficient recurrences use exact rational arithmetic. A rational vector producing a negative quadratic form gives a finite obstruction. On a parameter interval, strictly negative coefficients of that polynomial in the Bernstein basis { ( n j ) t j ( 1 − t ) n − j : 0 ≤ j ≤ n } \{\binom{n}{j}t^j(1-t)^{n-j}:0\le j\le n\} { ( j n ) t j ( 1 − t ) n − j : 0 ≤ j ≤ n } 0 ≤ t ≤ 1 0\le t\le1 0 ≤ t ≤ 1
The saved certificate has 18 adjacent rational closed intervals covering 9 / 10 ≤ α ≤ 1 9/10\le\alpha\le1 9/10 ≤ α ≤ 1 α = 1 \alpha=1 α = 1 9 / 10 ≤ α < 1 9/10\le\alpha<1 9/10 ≤ α < 1
The verifier certify_thorin_exclusion.py checks thorin_exclusion_certificate.json. The exact moment construction is in thorin_moment_probe.py.
The uniform GGC bound
This appendix proves the remaining compact part of the uniform 4 / 5 4/5 4/5
A complete Kanter phase certificate up to the critical endpoint
This section gives a computer-assisted proof of the missing compact parameter range. The infinite tails are controlled by explicit analytic inequalities, including at the critical endpoint. The checker uses directed ball arithmetic and validated integration, rather than sampled function values. Its source and certificate are part of the paper’s reproducibility bundle.
Put β = 1 − α \beta=1-\alpha β = 1 − α a = 1 − β a=1-\beta a = 1 − β F β ( s ) = E e − s K 1 − β F_\beta(s)=\mathbb{E}e^{-sK_{1-\beta}} F β ( s ) = E e − s K 1 − β s > 0 s>0 s > 0
F β ( − r − i 0 ) = R ( r ) + i I ( r ) F_\beta(-r-i0)=R(r)+iI(r) F β ( − r − i 0 ) = R ( r ) + i I ( r ) for its analytic continuation to the lower cut. More explicitly,
F β ( − r − i 0 ) = ∑ n ≥ 0 r a n e i π β n n ! Γ ( 1 − β n ) , I ( r ) = ∑ n ≥ 1 Γ ( β n ) sin 2 ( π β n ) π n ! r a n > 0. (70) F_\beta(-r-i0)=\sum_{n\ge0}\frac{r^{an}e^{i\pi\beta n}}{n!\Gamma(1-\beta n)},\qquad
I(r)=\sum_{n\ge1}\frac{\Gamma(\beta n)\sin^2(\pi\beta n)}{\pi n!}r^{an}>0. \tag*{(70)} F β ( − r − i 0 ) = n ≥ 0 ∑ n ! Γ ( 1 − β n ) r an e iπ β n , I ( r ) = n ≥ 1 ∑ π n ! Γ ( β n ) sin 2 ( π β n ) r an > 0. (70) In particular I ′ / I ≥ a / r I'/I\ge a/r I ′ / I ≥ a / r
Q β ( r ) = R ( r ) I ′ ( r ) I ( r ) − R ′ ( r ) , Q_\beta(r)=R(r)\frac{I'(r)}{I(r)}-R'(r), Q β ( r ) = R ( r ) I ( r ) I ′ ( r ) − R ′ ( r ) , the phase derivative has the sign of Q β Q_\beta Q β [7 ] says precisely that K 1 − β K_{1-\beta} K 1 − β
Lemma C.1 (Uniform imaginary-part estimate). For 1 / 5 ≤ β ≤ 1 / 4 1/5\le\beta\le1/4 1/5 ≤ β ≤ 1/4 r ≥ 100 r\ge100 r ≥ 100
I ′ ( r ) I ( r ) ≥ b 2 > 47 200 , b = a β / a . \frac{I'(r)}{I(r)}\ge\frac{b}{2}>\frac{47}{200},\qquad b=a^{\beta/a}. I ( r ) I ′ ( r ) ≥ 2 b > 200 47 , b = a β / a . Proof. Set q ( u ) = u β − u q(u)=u^\beta-u q ( u ) = u β − u ζ = e 2 π i β \zeta=e^{2\pi i\beta} ζ = e 2 π i β
J = ∫ 0 ∞ ( e r q ( u ) − e − r u ) d u u , J c = ∫ 0 ∞ ( e r ( ζ u β − u ) − e − r u ) d u u . J=\int_0^\infty\left(e^{r q(u)}-e^{-ru}\right)\frac{\mathrm{d}u}{u},\qquad
J_c=\int_0^\infty\left(e^{r(\zeta u^\beta-u)}-e^{-ru}\right)\frac{\mathrm{d}u}{u}. J = ∫ 0 ∞ ( e r q ( u ) − e − r u ) u d u , J c = ∫ 0 ∞ ( e r ( ζ u β − u ) − e − r u ) u d u . Then 2 π I = J − Re J c 2\pi I=J-\operatorname{Re}J_c 2 π I = J − Re J c c = b / 2 c=b/2 c = b /2 u 0 = β 1 / a u_0=\beta^{1/a} u 0 = β 1/ a
47 100 < b < 27 50 , q ( u ) ≥ 9 10 b ( u 0 / 2 ≤ u ≤ 3 u 0 / 2 ) , 3 u 0 / 2 < 1. \frac{47}{100}<b<\frac{27}{50},\qquad q(u)\geq\frac{9}{10}b\quad(u_0/2\leq u\leq3u_0/2),\qquad3u_0/2<1. 100 47 < b < 50 27 , q ( u ) ≥ 10 9 b ( u 0 /2 ≤ u ≤ 3 u 0 /2 ) , 3 u 0 /2 < 1. The middle inequality follows from concavity and its two endpoint inequalities. All these elementary parameter inequalities are enclosed on the fifty rational subintervals of length 1 / 1000 1/1000 1/1000 J ′ − c J J'-cJ J ′ − c J 2 5 b log 3 e 9 b r / 10 > 1 5 e 9 b r / 10 \frac{2}{5}b\log3\,e^{9br/10}>\frac{1}{5}e^{9br/10} 5 2 b log 3 e 9 b r /10 > 5 1 e 9 b r /10 0 < u < 1 0<u<1 0 < u < 1 q < c q<c q < c
( q − c ) e r q + ( u + c ) e − r u = ( q − c ) ( e r q − e − r u ) + u β e − r u ≥ − c r u β e c r . (q-c)e^{rq}+(u+c)e^{-ru}=(q-c)(e^{rq}-e^{-ru})+u\beta e^{-ru}\geq-cru^\beta e^{cr}. ( q − c ) e r q + ( u + c ) e − r u = ( q − c ) ( e r q − e − r u ) + u β e − r u ≥ − cr u β e cr . For u ≥ 1 u\geq1 u ≥ 1 q ( u ) ≤ − a ( u − 1 ) q(u)\leq-a(u-1) q ( u ) ≤ − a ( u − 1 )
J ′ − c J ≥ 1 5 e 9 b r / 10 − 27 20 r e b r / 2 − 17 20 r . J'-cJ\geq\frac{1}{5}e^{9br/10}-\frac{27}{20}re^{br/2}-\frac{17}{20r}. J ′ − c J ≥ 5 1 e 9 b r /10 − 20 27 r e b r /2 − 20 r 17 . Also 0 ≤ cos ( 2 π β ) < 31 / 100 0\leq\cos(2\pi\beta)<31/100 0 ≤ cos ( 2 π β ) < 31/100 cos ( 2 π β ) u β − u / 2 \cos(2\pi\beta)u^\beta-u/2 cos ( 2 π β ) u β − u /2
cos ( 2 π β ) u β − u ≤ 3 b / 10 − u / 2 , \cos(2\pi\beta)u^\beta-u\leq3b/10-u/2, cos ( 2 π β ) u β − u ≤ 3 b /10 − u /2 , since ( 31 / 100 ) 5 / 4 2 1 / 3 < 3 / 10 (31/100)^{5/4}2^{1/3}<3/10 ( 31/100 ) 5/4 2 1/3 < 3/10 ∣ e r ζ u β − 1 ∣ ≤ r u β e r cos ( 2 π β ) u β |e^{r\zeta u^\beta}-1|\leq ru^\beta e^{r\cos(2\pi\beta)u^\beta} ∣ e r ζ u β − 1∣ ≤ r u β e r c o s ( 2 π β ) u β
∣ J c ′ − c J c ∣ ≤ Γ ( β ) ( 2 / r ) β ( 2 β + 1 + c r ) e 3 b r / 10 ≤ 5 ( 3 / 2 + 27 r / 100 ) e 3 b r / 10 . |J'_c-cJ_c|\leq\Gamma(\beta)(2/r)^\beta(2\beta+1+cr)e^{3br/10}\leq5(3/2+27r/100)e^{3br/10}. ∣ J c ′ − c J c ∣ ≤ Γ ( β ) ( 2/ r ) β ( 2 β + 1 + cr ) e 3 b r /10 ≤ 5 ( 3/2 + 27 r /100 ) e 3 b r /10 . Here Γ ( β ) ≤ 1 / β ≤ 5 \Gamma(\beta)\leq1/\beta\leq5 Γ ( β ) ≤ 1/ β ≤ 5 Γ \Gamma Γ e 9 b r / 10 e^{9br/10} e 9 b r /10
1 5 − 27 20 r e − 2 b r / 5 − 17 20 r e − 9 b r / 10 − 5 ( 3 / 2 + 27 r / 100 ) e − 3 b r / 5 . \frac{1}{5}-\frac{27}{20}re^{-2br/5}-\frac{17}{20r}e^{-9br/10}-5(3/2+27r/100)e^{-3br/5}. 5 1 − 20 27 r e − 2 b r /5 − 20 r 17 e − 9 b r /10 − 5 ( 3/2 + 27 r /100 ) e − 3 b r /5 . Replace b b b 47 / 100 47/100 47/100 r ≥ 100 r\geq100 r ≥ 100 I ′ − c I > 0 I'-cI>0 I ′ − c I > 0
Lemma C.2 (Uniform real-part estimates at infinity). For 1 / 5 ≤ β ≤ 1 / 4 1/5\leq\beta\leq1/4 1/5 ≤ β ≤ 1/4 r ≥ 500 r\geq500 r ≥ 500
R ( r ) > 1 , ∣ R ′ ( r ) ∣ < 1 5 + 9 r . R(r)>1,\qquad|R'(r)|<\frac{1}{5}+\frac{9}{r}. R ( r ) > 1 , ∣ R ′ ( r ) ∣ < 5 1 + r 9 . In particular Q β ( r ) > 17 / 1000 Q_\beta(r)>17/1000 Q β ( r ) > 17/1000
Proof. Set
ψ = 5 π β / 2 , d = − cot ψ , v = ( β sin ψ ) 1 / a , λ = r v , ϕ ( t ) = t β / β − t , E ( t ) = e − λ d t β / β . \psi=5\pi\beta/2,\qquad d=-\cot\psi,\qquad v=(\beta\sin\psi)^{1/a},\qquad\lambda=rv,\qquad\phi(t)=t^\beta/\beta-t,\qquad E(t)=e^{-\lambda d t^\beta/\beta}. ψ = 5 π β /2 , d = − cot ψ , v = ( β sin ψ ) 1/ a , λ = r v , ϕ ( t ) = t β / β − t , E ( t ) = e − λ d t β / β . Rotation to the positive imaginary axis in the preceding integral for J c J_c J c
R ( r ) = 5 4 + 1 2 π ∫ 0 ∞ E ( t ) sin ( λ ϕ ( t ) ) d t t . (71) R(r)=\frac{5}{4}+\frac{1}{2\pi}\int_0^\infty E(t)\sin(\lambda\phi(t))\frac{dt}{t}. \tag*{(71)} R ( r ) = 4 5 + 2 π 1 ∫ 0 ∞ E ( t ) sin ( λ ϕ ( t )) t d t . (71) At β = 1 / 5 \beta=1/5 β = 1/5 e − ϵ t e^{-\epsilon t} e − ϵ t ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0 T T T cos ( β ( 2 π + θ ) ) ≤ cos θ \cos(\beta(2\pi+\theta))\leq\cos\theta cos ( β ( 2 π + θ )) ≤ cos θ 0 ≤ θ ≤ π / 2 0\leq\theta\leq\pi/2 0 ≤ θ ≤ π /2 1 / 5 ≤ β ≤ 1 / 4 1/5\leq\beta\leq1/4 1/5 ≤ β ≤ 1/4 T T T − r T cos θ / 2 -rT\cos\theta/2 − r T cos θ /2 O ( ( r T ) − 1 ) O((rT)^{-1}) O (( r T ) − 1 ) e − i r x e^{-irx} e − i r x e − r x e^{-rx} e − r x i π / 2 i\pi/2 iπ /2 J c J_c J c 5 / 4 5/4 5/4
The following elementary inequalities, including their endpoint values, are checked on the same fifty parameter intervals:
3 25 < v < 29 200 , 0 ≤ d < 21 50 , J 0 : = 2 β − 2 − β β 2 < 7 , \frac{3}{25}<v<\frac{29}{200},\qquad0\leq d<\frac{21}{50},\qquad J_0:=\frac{2^\beta-2^{-\beta}}{\beta^2}<7, 25 3 < v < 200 29 , 0 ≤ d < 50 21 , J 0 := β 2 2 β − 2 − β < 7 , M L : = 1 2 − β − 1 / 2 < 3 , M R : = 1 2 − 2 β < 5 4 , 1 1 − 2 − a < 5 2 , M_L:=\frac{1}{2^{-\beta}-1/2}<3,\qquad M_R:=\frac{1}{2-2^\beta}<\frac{5}{4},\qquad\frac{1}{1-2^{-a}}<\frac{5}{2}, M L := 2 − β − 1/2 1 < 3 , M R := 2 − 2 β 1 < 4 5 , 1 − 2 − a 1 < 2 5 , 2 d β ( 2 a − 1 ) < 31 5 , ∣ 1 − a β ( 2 a − 1 ) ∣ < 5. \frac{2d}{\beta(2^a-1)}<\frac{31}{5},\qquad\left|1-\frac{a}{\beta(2^a-1)}\right|<5. β ( 2 a − 1 ) 2 d < 5 31 , 1 − β ( 2 a − 1 ) a < 5. For a monotone nonnegative g g g 0 ≤ E ≤ 1 0 \le E \le1 0 ≤ E ≤ 1 E g Eg E g 2 sup g 2\sup g 2 sup g d ( E g ) = E d g + g d E d(Eg)=E\,dg+g\,dE d ( E g ) = E d g + g d E
Let t L = β 1 / a t_L=\beta^{1/a} t L = β 1/ a ( 0 , t L ) (0,t_L) ( 0 , t L ) y = ϕ ( t ) y=\phi(t) y = ϕ ( t ) E ( t ) / ( t ϕ ′ ( t ) ) E(t)/(t\phi'(t)) E ( t ) / ( t ϕ ′ ( t )) t ϕ ′ ( t ) = t β − t t\phi'(t)=t^\beta-t t ϕ ′ ( t ) = t β − t 1 − cos ( λ y ) 1-\cos(\lambda y) 1 − cos ( λ y ) [ t L , 1 / 2 ] [t_L,1/2] [ t L , 1/2 ] 2 M L / λ 2M_L/\lambda 2 M L / λ [ 2 , ∞ ) [2,\infty) [ 2 , ∞ ) 2 M R / λ 2M_R/\lambda 2 M R / λ log 4 \log4 log 4
R ( r ) ≥ 5 4 − log 4 2 π − 17 / 2 2 π ( 3 / 25 ) r > 1 ( r ≥ 500 ) . R(r)\ge\frac{5}{4}-\frac{\log4}{2\pi}-\frac{17/2}{2\pi(3/25)r}>1 \qquad(r\ge500). R ( r ) ≥ 4 5 − 2 π log 4 − 2 π ( 3/25 ) r 17/2 > 1 ( r ≥ 500 ) . The final lower bound at 500 is greater than 1.006817.
For the derivative, first insert e − ε t e^{-\varepsilon t} e − εt
G ( t ) = − d t β / β + i ϕ ( t ) t ϕ ′ ( t ) = − d / β + i ( 1 / β − t a ) 1 − t a . G(t)=\frac{-dt^\beta/\beta+i\phi(t)}{t\phi'(t)}=\frac{-d/\beta+i(1/\beta-t^a)}{1-t^a}. G ( t ) = t ϕ ′ ( t ) − d t β / β + i ϕ ( t ) = 1 − t a − d / β + i ( 1/ β − t a ) . On [ 0 , 1 / 2 ] [0,1/2] [ 0 , 1/2 ] 21 / 4 21/4 21/4 11 11 11 65 / ( 2 λ ) 65/(2\lambda) 65/ ( 2 λ ) [ 2 , ∞ ) [2,\infty) [ 2 , ∞ ) 31 / ( 5 λ ) 31/(5\lambda) 31/ ( 5 λ ) 12 / λ 12/\lambda 12/ λ v / ( 2 π ) v/(2\pi) v / ( 2 π )
65 / 2 + 91 / 5 2 π r < 9 r . \frac{65/2+91/5}{2\pi r}<\frac{9}{r}. 2 π r 65/2 + 91/5 < r 9 . On [ 1 / 2 , 2 ] [1/2,2] [ 1/2 , 2 ] ϕ > 0 \phi>0 ϕ > 0
v 2 π [ ( 1 + d ) J 0 − 3 2 ] < 29 / 200 2 π [ ( 71 / 50 ) 7 − 3 2 ] < 1 5 . \frac{v}{2\pi}\left[(1+d)J_0-\frac{3}{2}\right]<\frac{29/200}{2\pi}\left[(71/50)7-\frac{3}{2}\right]<\frac{1}{5}. 2 π v [ ( 1 + d ) J 0 − 2 3 ] < 2 π 29/200 [ ( 71/50 ) 7 − 2 3 ] < 5 1 . All estimates are independent of ε \varepsilon ε 1 / t 1/t 1/ t ε ↓ 0 \varepsilon\downarrow0 ε ↓ 0 r r r R R R d = 0 d=0 d = 0 Q > 47 / 200 − 1 / 5 − 9 / 500 = 17 / 1000 Q>47/200-1/5-9/500=17/1000 Q > 47/200 − 1/5 − 9/500 = 17/1000
Proposition C.3 (Finite certificate specification). The finite certificate and exact recursive cover verified by certify_kanter_optimal.py establish
R ( r ) > 0 , Q β ( r ) > 0 ( 1 / 5 ≤ β ≤ 1 / 4 , 1 ≤ r ≤ 500 ) . R(r)>0,\qquad Q_\beta(r)>0\qquad(1/5\le\beta\le1/4,\ 1\le r\le500). R ( r ) > 0 , Q β ( r ) > 0 ( 1/5 ≤ β ≤ 1/4 , 1 ≤ r ≤ 500 ) . For 0 < r ≤ 1 0<r\le1 0 < r ≤ 1
Certificate algorithm and small-radius proof. Let A n = Γ ( β n ) r a n / ( π n ! ) A_n=\Gamma(\beta n)r^{an}/(\pi n!) A n = Γ ( β n ) r an / ( π n !) [15 (5.6.4)] gives
A n + 1 A n ≤ r a β β n − a . \frac{A_{n+1}}{A_n}\le r^a\beta^\beta n^{-a}. A n A n + 1 ≤ r a β β n − a . For any truncation index N N N q N = r a β β N − a < 1 q_N=r^a\beta^\beta N^{-a}<1 q N = r a β β N − a < 1
∑ n ≥ N A n ≤ A N 1 − q N , ∑ n ≥ N n A n ≤ A N [ N 1 − q N + q N ( 1 − q N ) 2 ] . \sum_{n\ge N}A_n\le\frac{A_N}{1-q_N},\qquad
\sum_{n\ge N}nA_n\le A_N\left[\frac{N}{1-q_N}+\frac{q_N}{(1-q_N)^2}\right]. n ≥ N ∑ A n ≤ 1 − q N A N , n ≥ N ∑ n A n ≤ A N [ 1 − q N N + ( 1 − q N ) 2 q N ] . For all β ∈ [ 1 / 5 , 1 / 4 ] \beta\in[1/5,1/4] β ∈ [ 1/5 , 1/4 ] N = 80 N=80 N = 80 r = 1 r=1 r = 1
S : = ∑ n ≥ 2 ( n − 1 ) Γ ( β n ) π n ! < 3 / 5. S:=\sum_{n\ge2}(n-1)\frac{\Gamma(\beta n)}{\pi n!}<3/5. S := n ≥ 2 ∑ ( n − 1 ) π n ! Γ ( β n ) < 3/5. The first real coefficient in (C.1) is positive. It follows that R ≥ 1 − S > 0 R \ge1-S > 0 R ≥ 1 − S > 0 R − ( r / a ) R ′ ≥ 1 − S > 0 R-(r/a)R' \ge1-S > 0 R − ( r / a ) R ′ ≥ 1 − S > 0 0 < r ≤ 1 0<r\le1 0 < r ≤ 1 I ′ / I ≥ a / r I'/I \ge a/r I ′ / I ≥ a / r
For the compact rectangle the program bisects rational intervals in β \beta β r r r [ β 0 , β 1 ] × [ r 0 , r 1 ] [\beta_0,\beta_1]\times[r_0,r_1] [ β 0 , β 1 ] × [ r 0 , r 1 ] R R R R ′ R' R ′ r 0 ≥ 100 r_0\ge100 r 0 ≥ 100 I ′ / I > 47 / 200 I'/I>47/200 I ′ / I > 47/200 I I I c n r a n c_n r^{an} c n r an m ( r ) m(r) m ( r ) m ′ ( r ) = a Var ( n ) / r m'(r)=a\operatorname{Var}(n)/r m ′ ( r ) = a Var ( n ) / r
I ′ ( r ) I ( r ) ≥ ( 1 − β 1 ) m ∗ ( r 0 ) r 1 . \frac{I'(r)}{I(r)}\ge(1-\beta_1)\frac{m_*(r_0)}{r_1}. I ( r ) I ′ ( r ) ≥ ( 1 − β 1 ) r 1 m ∗ ( r 0 ) . where m ∗ m_* m ∗ 1 + ∑ 1 ≤ n < N ( n − 1 ) ( c n r 0 a n ) l o w e r / I u p p e r 1+\sum_{1\le n<N}(n-1)(c_n r_0^{an})_{\mathrm{lower}}/I_{\mathrm{upper}} 1 + ∑ 1 ≤ n < N ( n − 1 ) ( c n r 0 an ) lower / I upper I u p p e r I_{\mathrm{upper}} I upper R R R R l o w e r ( I ′ / I ) l o w e r − ( R ′ ) u p p e r > 0 R_{\mathrm{lower}}(I'/I)_{\mathrm{lower}}-(R')_{\mathrm{upper}}>0 R lower ( I ′ / I ) lower − ( R ′ ) upper > 0
For completeness the stable integral used near β = 1 / 5 \beta=1/5 β = 1/5 θ = π / 2 − 1 / 20 \theta=\pi/2-1/20 θ = π /2 − 1/20 c = cos θ c=\cos\theta c = cos θ B = tan θ B=\tan\theta B = tan θ p = 1 / β p=1/\beta p = 1/ β A = r a c − β A=r^{a}c^{-\beta} A = r a c − β ζ = exp ( i β ( 2 π + θ ) ) \zeta=\exp(i\beta(2\pi+\theta)) ζ = exp ( i β ( 2 π + θ ))
R = 1 + θ 2 π + 1 2 π β 3 Im ∫ 0 ∞ e − ( 1 + i B ) u p + A ζ u − e − u p u d u , R=1+\frac{\theta}{2\pi}+\frac{1}{2\pi\beta^3}\operatorname{Im}\int_0^\infty\frac{e^{-(1+iB)u^p+A\zeta u}-e^{-u^p}}{u}\,du, R = 1 + 2 π θ + 2 π β 3 1 Im ∫ 0 ∞ u e − ( 1 + i B ) u p + A ζ u − e − u p d u , R ′ = a A 2 π β r Im ∫ 0 ∞ ζ e − ( 1 + i B ) u p + A ζ u d u . R'=\frac{aA}{2\pi\beta r}\operatorname{Im}\int_0^\infty\zeta e^{-(1+iB)u^p+A\zeta u}\,du. R ′ = 2 π β r a A Im ∫ 0 ∞ ζ e − ( 1 + i B ) u p + A ζ u d u . Away from the endpoint, use θ = π ( 1 / ( 2 β ) − 2 ) \theta=\pi(1/(2\beta)-2) θ = π ( 1/ ( 2 β ) − 2 ) ζ = i \zeta=i ζ = i e − u p sin ( A u − B u p ) / u e^{-u^p}\sin(Au-Bu^p)/u e − u p sin ( A u − B u p ) / u e − u p cos ( A u − B u p ) e^{-u^p}\cos(Au-Bu^p) e − u p cos ( A u − B u p ) [ 10 − 9 , 3 ] [10^{-9},3] [ 1 0 − 9 , 3 ] A Re ζ < 3 p min − 1 / 2 A\operatorname{Re}\zeta<3^{p_{\min}-1/2} A Re ζ < 3 p m i n − 1/2 − u p min / 2 -u^{p_{\min}}/2 − u p m i n /2
2 e − 3 p min / 2 p min 3 p min , 2 e − 3 p min / 2 p min 3 p min − 1 . \frac{2e^{-3^{p_{\min}}/2}}{p_{\min}3^{p_{\min}}},\qquad
\frac{2e^{-3^{p_{\min}}/2}}{p_{\min}3^{p_{\min}-1}}. p m i n 3 p m i n 2 e − 3 p m i n /2 , p m i n 3 p m i n − 1 2 e − 3 p m i n /2 . At the origin, bounds are ( ∣ A ∣ ε + ∣ B ∣ ε p min / p min ) e ∣ A ∣ ε (|A|\varepsilon+|B|\varepsilon^{p_{\min}}/p_{\min})e^{|A|\varepsilon} ( ∣ A ∣ ε + ∣ B ∣ ε p m i n / p m i n ) e ∣ A ∣ ε ε e ∣ A ∣ ε \varepsilon e^{|A|\varepsilon} ε e ∣ A ∣ ε
Theorem C.4 (Optimal uniform GGC bound). For every 0 < α ≤ 4 / 5 0<\alpha\le4/5 0 < α ≤ 4/5 0 ≤ ρ ≤ 1 0\le\rho\le1 0 ≤ ρ ≤ 1 X α , ρ X_{\alpha,\rho} X α , ρ 0 < α < 1 0<\alpha<1 0 < α < 1 K α K_\alpha K α α ≤ 4 / 5 \alpha\le4/5 α ≤ 4/5 4 / 5 4/5 4/5
Proof. Lemmas C.1–C.2 and Proposition C.3 give increasing cut phase for β ∈ [ 1 / 5 , 1 / 4 ] \beta\in[1/5,1/4] β ∈ [ 1/5 , 1/4 ] [7 Theorem 2] handles β ≥ 1 / 4 \beta\ge1/4 β ≥ 1/4 [7 Corollary 2] . The multiplicative factorization and the Cauchy subordination argument of [7 Section 2.3] transfer Kanter GGC to X α , ρ X_{\alpha,\rho} X α , ρ α = 1 \alpha=1 α = 1 4 / 5 < α < 1 4/5<\alpha<1 4/5 < α < 1
Remark C.5 (Execution record). The complete cover has 20,261 leaves. It was built and then verified again in a fresh process on 3 October 2026, using python-flint 0.9.0 with 55 decimal digits of working precision. Both executions passed every interval assertion. The file kanter_optimal_certificate.json contains the exact recursive cover; the separate verification record stores hashes of the checker, its integral module, and the certificate. The computation is an essential part of the proof on the compact rectangle. The quadrature algorithm and its analytic-domain requirements are documented in [8 ] .
Corollary C.6. For every fixed 0 < η < 1 / 2 0<\eta<1/2 0 < η < 1/2 ε η > 0 \varepsilon_\eta>0 ε η > 0 X α , ρ X_{\alpha,\rho} X α , ρ 4 / 5 < α < 4 / 5 + ε η 4/5<\alpha<4/5+\varepsilon_\eta 4/5 < α < 4/5 + ε η η ≤ ρ ≤ 1 − η \eta\le\rho\le1-\eta η ≤ ρ ≤ 1 − η
Proof. Apply Proposition B.5 at α 0 = 4 / 5 \alpha_0 = 4/5 α 0 = 4/5
Signed Lévy kernels and extremal germs
We derive the signed kernel representation and prove angular interpolation. Convergent extremal germs then supply explicit origin bounds and the complete index-3 / 2 3/2 3/2 § 11 .
An analytic ID anchor above one
The following estimate supplies a useful explicit ID region.
Proposition D.1. For 1 < α ≤ 53 / 50 1 < \alpha\le53/50 1 < α ≤ 53/50 X α , 3 / 4 X_{\alpha,3/4} X α , 3/4 X α , 1 / 4 X_{\alpha,1/4} X α , 1/4
Proof. Write W α ( z ) = ∏ n ( 1 + z / λ n ) W_\alpha(z) = \prod_n(1+z/\lambda_n) W α ( z ) = ∏ n ( 1 + z / λ n ) r n = λ n 1 / α r_n = \lambda_n^{1/\alpha} r n = λ n 1/ α log W α ( s α ) \log W_\alpha(s^\alpha) log W α ( s α ) X α , 1 / α X_{\alpha,1/\alpha} X α , 1/ α
∑ n 1 ( 1 + ( r n x ) α ) 2 ≤ α − 1 α ∑ n 1 1 + ( r n x ) α . (72) \sum_n \frac{1}{(1+(r_nx)^\alpha)^2} \le\frac{\alpha-1}{\alpha}\sum_n \frac{1}{1+(r_nx)^\alpha}. \tag*{(72)} n ∑ ( 1 + ( r n x ) α ) 2 1 ≤ α α − 1 n ∑ 1 + ( r n x ) α 1 . (72) For ρ > 1 / 2 \rho> 1/2 ρ > 1/2
ω = π ( ρ − 1 2 ) , ξ = π ( 1 − α ( 1 − ρ ) ) , B α , ξ ( v ) = α sin ξ π ∫ 0 ∞ e − v u u α − 1 1 + 2 cos ξ u α + u 2 α d u . \omega= \pi\left(\rho-\frac{1}{2}\right), \qquad\xi= \pi(1-\alpha(1-\rho)), \qquad B_{\alpha,\xi}(v) = \frac{\alpha\sin\xi}{\pi}\int_0^\infty\frac{e^{-vu}u^{\alpha-1}}{1+2\cos\xi\,u^\alpha+u^{2\alpha}}\,\mathrm{d}u. ω = π ( ρ − 2 1 ) , ξ = π ( 1 − α ( 1 − ρ )) , B α , ξ ( v ) = π α sin ξ ∫ 0 ∞ 1 + 2 cos ξ u α + u 2 α e − v u u α − 1 d u . Single-factor Fourier inversion gives the negative-side signed Lévy kernel
K α , ρ ( x ) = ∑ n B α , ξ ( r n x ) − 2 ∑ n e − r n x sin ω cos ( r n x cos ω ) . K_{\alpha,\rho}(x)=\sum_n B_{\alpha,\xi}(r_nx)-2\sum_n e^{-r_nx\sin\omega}\cos(r_nx\cos\omega). K α , ρ ( x ) = n ∑ B α , ξ ( r n x ) − 2 n ∑ e − r n x s i n ω cos ( r n x cos ω ) . The positive-side kernel is ∑ n B α , η ( r n x ) ≥ 0 \sum_n B_{\alpha,\eta}(r_nx)\ge0 ∑ n B α , η ( r n x ) ≥ 0 η = π ( 1 − α ρ ) \eta=\pi(1-\alpha\rho) η = π ( 1 − α ρ ) η = 0 \eta=0 η = 0
d d t log ( 1 − e − i α ω ( t / r ) α ) = α t α − 1 t α − r α e i α ω . \frac{\mathrm{d}}{\mathrm{d}t}\log\left(1-e^{-i\alpha\omega(t/r)^\alpha}\right)=\frac{\alpha t^{\alpha-1}}{t^\alpha-r^\alpha e^{i\alpha\omega}}. d t d log ( 1 − e − i α ω ( t / r ) α ) = t α − r α e i α ω α t α − 1 . The upward rotation crosses r e i ω re^{i\omega} r e iω ∫ ( x 2 ∧ ∣ x ∣ ) ∣ ν r ∣ ( d x ) = O ( r − α ) \int(x^2\wedge|x|)|\nu_r|(\mathrm{d}x)=O(r^{-\alpha}) ∫ ( x 2 ∧ ∣ x ∣ ) ∣ ν r ∣ ( d x ) = O ( r − α ) ∑ n r n − α = 1 / Γ ( α + 1 ) \sum_n r_n^{-\alpha}=1/\Gamma(\alpha+1) ∑ n r n − α = 1/Γ ( α + 1 ) K K K
For ρ = 3 / 4 \rho=3/4 ρ = 3/4 ω = π / 4 \omega=\pi/4 ω = π /4 cos ξ < 0 \cos\xi<0 cos ξ < 0
1 + 2 cos ξ u α + u 2 α ≤ ( 1 + u 2 ) α ≤ e α u . 1+2\cos\xi\,u^\alpha+u^{2\alpha}\le(1+u^2)^\alpha\le e^{\alpha u}. 1 + 2 cos ξ u α + u 2 α ≤ ( 1 + u 2 ) α ≤ e α u . Hölder’s inequality then gives
B α , ξ ( v ) ≥ Γ ( α + 1 ) sin ( π α / 4 ) π ( 1 + α α / ( α − 1 ) ) α − 1 1 1 + v α > 1 5 ( 1 + v α ) . B_{\alpha,\xi}(v)\ge\frac{\Gamma(\alpha+1)\sin(\pi\alpha/4)}{\pi(1+\alpha^{\alpha/(\alpha-1)})^{\alpha-1}}\frac{1}{1+v^\alpha}>\frac{1}{5(1+v^\alpha)}. B α , ξ ( v ) ≥ π ( 1 + α α / ( α − 1 ) ) α − 1 Γ ( α + 1 ) sin ( π α /4 ) 1 + v α 1 > 5 ( 1 + v α ) 1 . Here Γ ( α + 1 ) ≥ 1 \Gamma(\alpha+1)\ge1 Γ ( α + 1 ) ≥ 1 sin ( π α / 4 ) ≥ 1 / 2 \sin(\pi\alpha/4)\ge1/\sqrt{2} sin ( π α /4 ) ≥ 1/ 2 11 / 10 11/10 11/10
sup v ≥ 0 e − v / 2 ( cos ( v / 2 ) + ( 1 + v α ) 2 ) < 7 / 4. \sup_{v\ge0}e^{-v/\sqrt{2}}\left(\cos(v/\sqrt{2})+(1+v^\alpha)^2\right)<7/4. v ≥ 0 sup e − v / 2 ( cos ( v / 2 ) + ( 1 + v α ) 2 ) < 7/4. On the first positive cosine interval, bound this by e − 7 v / 10 − v 2 / 4 ( 1 + v ) 53 / 25 e^{-7v/10-v^2/4}(1+v)^{53/25} e − 7 v /10 − v 2 /4 ( 1 + v ) 53/25 v = ( 107 − 6 ) / 5 v=(\sqrt{107}-6)/5 v = ( 107 − 6 ) /5 7 / 4 7/4 7/4 v > 6 v>6 v > 6 e − 7 v / 10 ( 1 + v ) 53 / 25 < 1 e^{-7v/10}(1+v)^{53/25}<1 e − 7 v /10 ( 1 + v ) 53/25 < 1
K α , 3 / 4 ( x ) > ( 1 5 − 7 α − 1 2 α ) ∑ n 1 1 + ( r n x ) α ≥ 1 530 ∑ n 1 1 + ( r n x ) α > 0. K_{\alpha,3/4}(x)>\left(\frac{1}{5}-\frac{7\alpha-1}{2\alpha}\right)\sum_n\frac{1}{1+(r_nx)^\alpha}\ge\frac{1}{530}\sum_n\frac{1}{1+(r_nx)^\alpha}>0. K α , 3/4 ( x ) > ( 5 1 − 2 α 7 α − 1 ) n ∑ 1 + ( r n x ) α 1 ≥ 530 1 n ∑ 1 + ( r n x ) α 1 > 0. Reflection proves the other case. These examples contradict the conjectured exclusion of every index above one discussed after Theorem 1 of [6 ] . □
Sector interpolation for ID
The following argument proves interval convexity. Inclusion of the extremal skewness requires the separate endpoint positivity theorem.
Proposition D.2 (Each half of the ID section is an interval). For fixed 1 < α < 2 1<\alpha<2 1 < α < 2
I α = { ρ ∈ ( 1 / 2 , 1 / α ] : X α , ρ is ID } I_{\alpha}=\{\rho\in(1/2,1/\alpha]:X_{\alpha,\rho}\text{ is ID}\} I α = { ρ ∈ ( 1/2 , 1/ α ] : X α , ρ is ID } is either empty or a closed interval [ ρ − ( α ) , ρ + ( α ) ] [\rho_{-}(\alpha),\rho_{+}(\alpha)] [ ρ − ( α ) , ρ + ( α )] 1 / 2 < ρ − ≤ ρ + ≤ 1 / α 1/2<\rho_{-}\leq\rho_{+}\leq1/\alpha 1/2 < ρ − ≤ ρ + ≤ 1/ α x > 0 x>0 x > 0 ρ + = 1 / α \rho_{+}=1/\alpha ρ + = 1/ α
Proof. First, nonnegativity of the negative-side kernel is also necessary for ID. Its absolutely summable compensated representation above is a signed Lévy–Khintchine representation, and uniqueness of that representation forces its signed measure to agree with the positive Lévy measure of any ID law. Thus
X α , ρ is ID ⟺ K α , ρ ( x ) ≥ 0 ( x > 0 ) . X_{\alpha,\rho}\text{ is ID}\quad\Longleftrightarrow\quad K_{\alpha,\rho}(x)\geq0\quad(x>0). X α , ρ is ID ⟺ K α , ρ ( x ) ≥ 0 ( x > 0 ) . Let Q r ( t ) = log ∣ 1 − ( t / r ) α ∣ + i π 1 { t > r } Q_r(t)=\log\lvert1-(t/r)^\alpha\rvert+i\pi\mathbf{1}_{\{t>r\}} Q r ( t ) = log ∣ 1 − ( t / r ) α ∣ + iπ 1 { t > r } Re z > 0 \operatorname{Re}z>0 Re z > 0
F α ( z ) = i z π ∑ n ∫ 0 ∞ e − z t Q r n ( t ) d t . F_\alpha(z)=\frac{iz}{\pi}\sum_n\int_0^\infty e^{-zt}Q_{r_n}(t)\,\mathrm{d}t. F α ( z ) = π i z n ∑ ∫ 0 ∞ e − z t Q r n ( t ) d t . This uses a continuous logarithmic boundary value rather than resetting its imaginary part after each zero. Its integrable logarithmic singularities cause no difficulty. On each closed subsector of the right half-plane the sum converges normally and
∣ F α ( z ) ∣ = O ( ∣ z ∣ − α ) both at zero and at infinity . (73) \lvert F_\alpha(z)\rvert=O(\lvert z\rvert^{-\alpha})\quad\text{both at zero and at infinity}. \tag*{(73)} ∣ F α ( z )∣ = O (∣ z ∣ − α ) both at zero and at infinity . (73) Indeed for all s > 0 s>0 s > 0
s ∫ 0 ∞ e − s u ∣ Q 1 ( u ) ∣ d u ≤ C α s − α . s\int_0^\infty e^{-su}\lvert Q_1(u)\rvert\,\mathrm{d}u\leq C_\alpha s^{-\alpha}. s ∫ 0 ∞ e − s u ∣ Q 1 ( u )∣ d u ≤ C α s − α . For s ≥ 1 s\geq1 s ≥ 1 s ≤ 1 s\leq1 s ≤ 1 O ( 1 + ∣ log s ∣ ) O(1+\lvert\log s\rvert) O ( 1 + ∣ log s ∣) C s − α Cs^{-\alpha} C s − α r n r_n r n Re z \operatorname{Re}z Re z ∣ z ∣ \lvert z\rvert ∣ z ∣ ∑ r n − α < ∞ \sum r_n^{-\alpha}<\infty ∑ r n − α < ∞
Set θ = π ( 1 − ρ ) ∈ ( 0 , π / 2 ) \theta=\pi(1-\rho)\in(0,\pi/2) θ = π ( 1 − ρ ) ∈ ( 0 , π /2 )
K α , ρ ( x ) = Re F α ( x e − i θ ) . (74) K_{\alpha,\rho}(x)=\operatorname{Re}F_\alpha(xe^{-i\theta}). \tag*{(74)} K α , ρ ( x ) = Re F α ( x e − i θ ) . (74) For an explicit sign check, integration by parts replaces z Q r zQ_r z Q r
α t α − 1 ( t α − r α − i 0 ) . \frac{\alpha t^{\alpha-1}}{(t^\alpha-r^\alpha-i0)}. ( t α − r α − i 0 ) α t α − 1 . Rotating this path from just below the positive ray to angle θ \theta θ r r r i / π i/\pi i / π − 2 e − r x e − i θ -2e^{-rxe^{-i\theta}} − 2 e − r x e − i θ B α , π − α θ ( r x ) B_{\alpha,\pi-\alpha\theta}(rx) B α , π − α θ ( r x )
Suppose ρ 1 < ρ 2 \rho_1<\rho_2 ρ 1 < ρ 2 Re F α \operatorname{Re}F_\alpha Re F α π / 2 \pi/2 π /2 θ m \theta_m θ m
ϵ ( r 2 + r − 2 ) cos ( 2 ( θ − θ m ) ) . \epsilon(r^2+r^{-2})\cos(2(\theta-\theta_m)). ϵ ( r 2 + r − 2 ) cos ( 2 ( θ − θ m )) . Since α < 2 \alpha<2 α < 2 ϵ ↓ 0 \epsilon\downarrow0 ϵ ↓ 0 x x x ρ \rho ρ 1 / 2 1/2 1/2 □ \square □
Extremal ID at index three halves
We first give the complete extremal positivity argument at this index, including its finite certificate and both radial ends.
Theorem D.3 (Extremal index 3 / 2 3/2 3/2 X 3 / 2 , 2 / 3 X_{3/2,2/3} X 3/2 , 2/3 X 3 / 2 , 1 / 3 X_{3/2,1/3} X 3/2 , 1/3 k ( x ) = K 3 / 2 , 2 / 3 ( x ) k(x)=K_{3/2,2/3}(x) k ( x ) = K 3/2 , 2/3 ( x ) b = 3 2 2 1 / 3 b=\frac{3}{2}2^{1/3} b = 2 3 2 1/3
0 < k ( x ) ≤ 3 2 ( x > 0 ) , ∫ 0 ∞ k ( x ) d x = b , Re k ( z ) > 1 ( ∣ z ∣ ≤ 1 / 2 ) . 0<k(x)\le\frac{3}{2}\quad(x>0),\qquad\int_{0}^{\infty} k(x)\,\mathrm{d}x=b,\qquad\operatorname{Re}k(z)>1\quad(|z|\le1/2). 0 < k ( x ) ≤ 2 3 ( x > 0 ) , ∫ 0 ∞ k ( x ) d x = b , Re k ( z ) > 1 ( ∣ z ∣ ≤ 1/2 ) . The last statement uses the analytic germ at zero and includes k ( 0 ) = 3 / 2 k(0)=3/2 k ( 0 ) = 3/2
Proof. Set c = 2 − 2 / 3 c=2^{-2/3} c = 2 − 2/3 Z = ( b − X 3 / 2 , 2 / 3 ) / c > 0 Z=(b-X_{3/2,2/3})/c>0 Z = ( b − X 3/2 , 2/3 ) / c > 0 p ( y ) = 2 π 3 y A ( y ) p(y)=\frac{2}{\pi\sqrt{3}}\sqrt{y}A(y) p ( y ) = π 3 2 y A ( y ) y = 3 − u − 1 − 2 u y=3-u^{-1}-2\sqrt{u} y = 3 − u − 1 − 2 u v = u v=\sqrt{u} v = u
2 v 3 + ( y − 3 ) v 2 + 1 = 0. 2v^{3}+(y-3)v^{2}+1=0. 2 v 3 + ( y − 3 ) v 2 + 1 = 0. The two roots coalescing at v = 1 v=1 v = 1 y = 0 y=0 y = 0 − ( 1 + V ) / 2 -(1+V)/2 − ( 1 + V ) /2
y = V ( V + 3 ) 2 ( V + 1 ) 2 , V ( 0 ) = 0 , A ( y ) = 1 + V + V 2 / 3 ( 1 + V ) 3 ( 1 + V / 3 ) . y=\frac{V(V+3)^{2}}{(V+1)^{2}},\qquad V(0)=0,\qquad A(y)=\frac{\sqrt{1+V+V^{2}/3}}{(1+V)^{3}(1+V/3)}. y = ( V + 1 ) 2 V ( V + 3 ) 2 , V ( 0 ) = 0 , A ( y ) = ( 1 + V ) 3 ( 1 + V /3 ) 1 + V + V 2 /3 . The normalized density factor A A A ∣ y ∣ < 3 3 |y|<3\sqrt{3} ∣ y ∣ < 3 3 9 / 2 ± ( 3 3 / 2 ) i 9/2\pm(3\sqrt{3}/2)i 9/2 ± ( 3 3 /2 ) i y \sqrt{y} y
Put R = 51 / 10 R=51/10 R = 51/10 B = 41 / 10 B=41/10 B = 41/10 A ( y ) = ∑ f n y n A(y)=\sum f_n y^n A ( y ) = ∑ f n y n ∣ y ∣ = R |y|=R ∣ y ∣ = R B B B 2 B 3 > ( R + 3 ) B 2 + 1 2B^3>(R+3)B^2+1 2 B 3 > ( R + 3 ) B 2 + 1 A A A 2 i 4 / 3 y 2i\sqrt{4/3}\sqrt{y} 2 i 4/3 y
∣ A ( y ) ∣ ≤ 3 B 2 2 R < 7 , ∣ f n ∣ ≤ 7 R − n . |A(y)|\le\frac{\sqrt{3}B^2}{2\sqrt{R}}<7,\qquad|f_n|\le7R^{-n}. ∣ A ( y ) ∣ ≤ 2 R 3 B 2 < 7 , ∣ f n ∣ ≤ 7 R − n . These constants are verified exactly by the checker.
Let j ( y ) = k ( c y ) = ∑ n ≥ 0 k n y n j(y)=k(cy)=\sum_{n\ge0}k_ny^n j ( y ) = k ( cy ) = ∑ n ≥ 0 k n y n p 0 = 3 / 2 p_0=3/2 p 0 = 3/2 p ∗ j = y p p*j=yp p ∗ j = y p
k 0 = p 0 , k n = ( p 0 ) n + 1 n ! f n − ∑ m = 1 n ( p 0 ) m ( n − m ) ! n ! f m k n − m . (75) k_0=p_0,\qquad k_n=\frac{(p_0)^{n+1}}{n!}f_n-\sum_{m=1}^{n}\frac{(p_0)^m(n-m)!}{n!}f_mk_{n-m}. \tag*{(75)} k 0 = p 0 , k n = n ! ( p 0 ) n + 1 f n − m = 1 ∑ n n ! ( p 0 ) m ( n − m )! f m k n − m . (75) This is an exact local inverse argument, not an inference from an asymptotic expansion alone. More explicitly, the analytic logarithmic derivative of the extremal Laplace transform has a causal inverse Laplace distribution. Multiplication by the transform of p p p p ∗ j = y p p*j=yp p ∗ j = y p y > 0 y>0 y > 0 y p 0 − 1 y^{p_0-1} y p 0 − 1
Here is a transparent all-order bound for those coefficients:
∣ k n ∣ ≤ 16 ( n + 1 ) 2 R − n ( n ≥ 0 ) . (76) |k_n|\le16(n+1)^2R^{-n}\qquad(n\ge0). \tag*{(76)} ∣ k n ∣ ≤ 16 ( n + 1 ) 2 R − n ( n ≥ 0 ) . (76) The base 0 ≤ n < 100 0\le n<100 0 ≤ n < 100 n ≥ 100 n\ge100 n ≥ 100 ( 3 / 2 ) m ≤ ( m + 1 ) ! (3/2)^m\le(m+1)! ( 3/2 ) m ≤ ( m + 1 )!
1 ( n + 1 ) 2 ∑ m = 1 n ( m + 1 ) ( n − m + 1 ) 2 ( n m ) < 5 n . \frac{1}{(n+1)^2}\sum_{m=1}^{n}\frac{(m+1)(n-m+1)^2}{\binom{n}{m}}<\frac{5}{n}. ( n + 1 ) 2 1 m = 1 ∑ n ( m n ) ( m + 1 ) ( n − m + 1 ) 2 < n 5 . To verify the latter elementary estimate, separate m = 1 , 2 , n − 2 , n − 1 , n m=1,2,n-2,n-1,n m = 1 , 2 , n − 2 , n − 1 , n ( n m ) ≥ ( n 3 ) \binom{n}{m}\ge\binom{n}{3} ( m n ) ≥ ( 3 n ) ( m + 1 ) ( n − m + 1 ) 2 ≤ 4 ( n + 2 ) 3 / 27 (m+1)(n-m+1)^2\le4(n+2)^3/27 ( m + 1 ) ( n − m + 1 ) 2 ≤ 4 ( n + 2 ) 3 /27 n n n
3 + 6 99 + 4 100 + 18 10000 + 8 9 ( 102 / 100 ) 3 ( 99 / 100 ) ( 98 / 100 ) < 5. 3+\frac{6}{99}+\frac{4}{100}+\frac{18}{10000}+\frac{8}{9}\frac{(102/100)^3}{(99/100)(98/100)}<5. 3 + 99 6 + 100 4 + 10000 18 + 9 8 ( 99/100 ) ( 98/100 ) ( 102/100 ) 3 < 5. Consequently the induction step in (D.4) is bounded, after division by ( n + 1 ) 2 R − n (n+1)^2R^{-n} ( n + 1 ) 2 R − n 7 ( 102 / 101 ) + 560 / 100 < 16 7(102/101)+560/100<16 7 ( 102/101 ) + 560/100 < 16 ∣ y ∣ < R |y|<R ∣ y ∣ < R
The checker computes f n , k n f_n,k_n f n , k n N = 128 N=128 N = 128 q = 4 / R q=4/R q = 4/ R
E N = 16 q N + 1 { ( N + 2 ) 2 1 − q + 2 ( N + 2 ) q ( 1 − q ) 2 + q ( 1 + q ) ( 1 − q ) 3 } < 3.25 ⋅ 10 − 8 . E_N=16q^{N+1}\left\{\frac{(N+2)^2}{1-q}+\frac{2(N+2)q}{(1-q)^2}+\frac{q(1+q)}{(1-q)^3}\right\}<3.25\cdot10^{-8}. E N = 16 q N + 1 { 1 − q ( N + 2 ) 2 + ( 1 − q ) 2 2 ( N + 2 ) q + ( 1 − q ) 3 q ( 1 + q ) } < 3.25 ⋅ 1 0 − 8 . All Bernstein coefficients of the kernel polynomial on the single interval [ 0 , 4 ] [0,4] [ 0 , 4 ] 1 / 50 1/50 1/50 ( 3 / 2 − ∑ n = 0 N k n y n ) / y (3/2-\sum_{n=0}^{N}k_ny^n)/y ( 3/2 − ∑ n = 0 N k n y n ) / y E N / 4 E_N/4 E N /4 0 < j ( y ) ≤ 3 / 2 0<j(y)\le3/2 0 < j ( y ) ≤ 3/2 [ 0 , 4 ] [0,4] [ 0 , 4 ] 4 / 5 4/5 4/5 0.403 < 1 / 2 0.403<1/2 0.403 < 1/2 Re j ( y ) > 1 \operatorname{Re}j(y)>1 Re j ( y ) > 1 ∣ y ∣ ≤ 4 / 5 |y|\le4/5 ∣ y ∣ ≤ 4/5
For the entire remaining tail in the original variable, take x 0 = 5 / 2 x_0=5/2 x 0 = 5/2
( 2.7407302 , 2.7407303 ) , ( 4.7166213 , 4.7166214 ) , ( 6.6596278 , 6.6596279 ) . (2.7407302,2.7407303),\qquad(4.7166213,4.7166214),\qquad(6.6596278,6.6596279). ( 2.7407302 , 2.7407303 ) , ( 4.7166213 , 4.7166214 ) , ( 6.6596278 , 6.6596279 ) . Set S m = ( − 1 ) m + 1 m [ z m ] log W 3 / 2 ( z ) S_m=(-1)^{m+1}m[z^m]\log W_{3/2}(z) S m = ( − 1 ) m + 1 m [ z m ] log W 3/2 ( z ) T m = S m − ∑ j = 1 3 r j − 3 m / 2 T_m=S_m-\sum_{j=1}^{3}r_j^{-3m/2} T m = S m − ∑ j = 1 3 r j − 3 m /2 8 6 T 4 < 1 8^6T_4<1 8 6 T 4 < 1 ξ = π / 2 \xi=\pi/2 ξ = π /2 1 + u 3 1+u^3 1 + u 3
k ( x ) ≥ 1 π x 3 / 2 − ( 3 / 2 ) Γ ( 9 / 2 ) S 3 π x 9 / 2 − 2 ∑ j = 1 3 e − r j x / 2 − 2 ⋅ 8 3 e − 4 x T 2 . k(x)\ge\frac{1}{\pi x^{3/2}}-\frac{(3/2)\Gamma(9/2)S_3}{\pi x^{9/2}}-2\sum_{j=1}^{3}e^{-r_jx/2}-2\cdot8^3e^{-4x}T_2. k ( x ) ≥ π x 3/2 1 − π x 9/2 ( 3/2 ) Γ ( 9/2 ) S 3 − 2 j = 1 ∑ 3 e − r j x /2 − 2 ⋅ 8 3 e − 4 x T 2 . At x 0 x_0 x 0 0.008 0.008 0.008 x 3 / 2 x^{3/2} x 3/2 x ≥ x 0 x\ge x_0 x ≥ x 0
k ( x ) ≤ 1 π x 3 / 2 + 2 ∑ j = 1 3 e − r j x / 2 + 2 ⋅ 8 3 e − 4 x T 2 k(x)\le\frac{1}{\pi x^{3/2}}+2\sum_{j=1}^{3}e^{-r_jx/2}+2\cdot8^3e^{-4x}T_2 k ( x ) ≤ π x 3/2 1 + 2 j = 1 ∑ 3 e − r j x /2 + 2 ⋅ 8 3 e − 4 x T 2 decreases and is less than 1 / 5 1/5 1/5 x 0 x_0 x 0 4 c > 5 / 2 4c>5/2 4 c > 5/2 ( 1 / 2 ) / c < 4 / 5 (1/2)/c<4/5 ( 1/2 ) / c < 4/5
The file certify_extremal_three_halves.py verified every finite inequality just used, with exact rational polynomial calculations and 100-digit outward ball evaluations of the remaining constants and zero signs. The zero Taylor tails are explicitly bounded using the decreasing ratio Γ ( t ) / Γ ( t + 1 / 2 ) \Gamma(t)/\Gamma(t+1/2) Γ ( t ) /Γ ( t + 1/2 ) extremal_three_halves_certificate.json. Nonnegative k k k b − X b-X b − X b b b ∫ k = b \int k=b ∫ k = b
A convergent extremal kernel with variable index.
Lemma D.4 (The logarithmic-strip inverse). Let 1 < a ≤ 2 1<a\le2 1 < a ≤ 2 ε = a − 1 \varepsilon=a-1 ε = a − 1 c a = ε 1 / a c_a=\varepsilon^{1/a} c a = ε 1/ a b a = a ε 1 / a − 1 b_a=a\varepsilon^{1/a-1} b a = a ε 1/ a − 1 Z a = ( b a − X a , 1 / a ) / c a Z_a=(b_a-X_{a,1/a})/c_a Z a = ( b a − X a , 1/ a ) / c a
p a ( y ) = 2 / a π y F a ( y ) , F a ( 0 ) = 1 , p_a(y)=\frac{\sqrt{2/a}}{\pi}\sqrt{y}F_a(y),\qquad F_a(0)=1, p a ( y ) = π 2/ a y F a ( y ) , F a ( 0 ) = 1 , where the germ F a F_a F a
∣ y ∣ < R a , R a = 2 a a − 1 sin π a . |y|<R_a,\qquad R_a=\frac{2a}{a-1}\sin\frac{\pi}{a}. ∣ y ∣ < R a , R a = a − 1 2 a sin a π . The corresponding signed Lévy kernel j a ( y ) = K a , 1 / a ( c a y ) j_a(y)=K_{a,1/a}(c_ay) j a ( y ) = K a , 1/ a ( c a y ) F a ( y ) = ∑ f n ( a ) y n F_a(y)=\sum f_n(a)y^n F a ( y ) = ∑ f n ( a ) y n j a ( y ) = ∑ k n ( a ) y n j_a(y)=\sum k_n(a)y^n j a ( y ) = ∑ k n ( a ) y n a n f n ( a ) a^nf_n(a) a n f n ( a ) a n k n ( a ) a^nk_n(a) a n k n ( a ) a a a 3 / 2 ≤ a ≤ 153 / 100 3/2\le a\le153/100 3/2 ≤ a ≤ 153/100 n ≥ 0 n\ge0 n ≥ 0
∣ f n ( a ) ∣ ≤ 7 ( 10 / 51 ) n , ∣ k n ( a ) ∣ ≤ 16 ( n + 1 ) 2 ( 10 / 51 ) n . (77) |f_n(a)|\le7(10/51)^n,\qquad|k_n(a)|\le16(n+1)^2(10/51)^n. \tag*{(77)} ∣ f n ( a ) ∣ ≤ 7 ( 10/51 ) n , ∣ k n ( a ) ∣ ≤ 16 ( n + 1 ) 2 ( 10/51 ) n . (77) This is a convergence and remainder theorem, not a claim of ID throughout the index interval.
Proof. In logarithmic Cauchy-transform coordinates u = e v u=e^v u = e v
y = h a ( v ) = a ε − e − v − 1 ε e ε v . y=h_a(v)=\frac{a}{\varepsilon}-e^{-v}-\frac{1}{\varepsilon}e^{\varepsilon v}. y = h a ( v ) = ε a − e − v − ε 1 e ε v . Consider the horizontal strip ∣ Im v ∣ < 2 π / a |\operatorname{Im}v|<2\pi/a ∣ Im v ∣ < 2 π / a v = t ± 2 π i / a v=t\pm2\pi i/a v = t ± 2 π i / a
h a ( v ) = a ε − e ∓ 2 π i / a ( e − t + 1 ε e ε t ) . h_a(v)=\frac{a}{\varepsilon}-e^{\mp2\pi i/a}\left(e^{-t}+\frac{1}{\varepsilon}e^{\varepsilon t}\right). h a ( v ) = ε a − e ∓ 2 π i / a ( e − t + ε 1 e εt ) . The bracketed real quantity is at least a / ε a/\varepsilon a / ε t = 0 t=0 t = 0 ∣ h a ( v ) ∣ ≥ R a |h_a(v)|\ge R_a ∣ h a ( v ) ∣ ≥ R a Re v = ± L \operatorname{Re}v=\pm L Re v = ± L L → ∞ L\to\infty L → ∞ a a a ( 1 , 2 ] (1,2] ( 1 , 2 ]
At a = 2 a=2 a = 2 y = 0 y=0 y = 0 cosh v = 1 \cosh v=1 cosh v = 1 v = 0 v=0 v = 0 a a a a a a ∣ y ∣ < R a |y|<R_a ∣ y ∣ < R a h a ( v ) = y h_a(v)=y h a ( v ) = y h a ′ ( v ) = e − v ( 1 − e a v ) h_a'(v)=e^{-v}(1-e^{av}) h a ′ ( v ) = e − v ( 1 − e a v ) v + , v − v_+,v_- v + , v − y = 0 y=0 y = 0 v ± ∼ ± i 2 y / a v_\pm\sim\pm i\sqrt{2y/a} v ± ∼ ± i 2 y / a
F a ( y ) = e v + ( y ) − e v − ( y ) 2 i 2 / a y F_a(y)=\frac{e^{v_+(y)}-e^{v_-(y)}}{2i\sqrt{2/a}\sqrt{y}} F a ( y ) = 2 i 2/ a y e v + ( y ) − e v − ( y ) extends holomorphically across zero and gives the asserted density. This argument identifies the relevant inverse branches; it does not assume that all critical values on every logarithmic sheet control this germ.
For the uniform estimate put R = 51 / 10 R=51/10 R = 51/10 R a R_a R a ( 1 , 2 ] (1,2] ( 1 , 2 ] − sin θ − ( π − θ ) cos θ -\sin\theta-(\pi-\theta)\cos\theta − sin θ − ( π − θ ) cos θ θ = π / a \theta=\pi/a θ = π / a R 153 / 100 > 5.114 > R R_{153/100}>5.114>R R 153/100 > 5.114 > R h a ( v ) = y h_a(v)=y h a ( v ) = y ∣ y ∣ = R |y|=R ∣ y ∣ = R Re v ≥ log 17 \operatorname{Re}v\ge\log17 Re v ≥ log 17
R + a ε ≥ ∣ e − v + ε − 1 e ε v ∣ ≥ 17 ε ε − 1 17 > R + a ε , R+\frac{a}{\varepsilon}\ge\left|e^{-v}+\varepsilon^{-1}e^{\varepsilon v}\right|\ge\frac{17^\varepsilon}{\varepsilon}-\frac{1}{17}>R+\frac{a}{\varepsilon}, R + ε a ≥ e − v + ε − 1 e ε v ≥ ε 1 7 ε − 17 1 > R + ε a , a contradiction. The last inequality holds uniformly for $1/2\le\varepsilon\le53/100: after multiplication by ε \varepsilon ε 1 / 2 1/2 1/2 17 log 17 − 1 − R − 1 / 17 > 0 \sqrt{17}\log17-1-R-1/17>0 17 log 17 − 1 − R − 1/17 > 0 ∣ e v ± ∣ < 17 |e^{v_\pm}|<17 ∣ e v ± ∣ < 17
∣ F a ( y ) ∣ ≤ 17 a 2 R ≤ 17 3 20 < 7. |F_a(y)|\le17\sqrt{\frac{a}{2R}}\le17\sqrt{\frac{3}{20}}<7. ∣ F a ( y ) ∣ ≤ 17 2 R a ≤ 17 20 3 < 7. Cauchy’s estimate proves the first bound in (D.6).
The exact convolution equation p a ∗ j a = y p a p_a*j_a=yp_a p a ∗ j a = y p a p 0 = 3 / 2 p_0=3/2 p 0 = 3/2
D a ( y ) = ∑ n ≥ 0 ( p 0 ) n n ! f n ( a ) y n , q a = D a ′ , d a = j a − p 0 . D_a(y)=\sum_{n\ge0}\frac{(p_0)^n}{n!}f_n(a)y^n,\qquad q_a=D_a',\qquad d_a=j_a-p_0. D a ( y ) = n ≥ 0 ∑ n ! ( p 0 ) n f n ( a ) y n , q a = D a ′ , d a = j a − p 0 . The equivalent local Volterra equation is d a = y q a − q a ∗ d a d_a=yq_a-q_a*d_a d a = y q a − q a ∗ d a D a D_a D a q a q_a q a ∣ y ∣ < R a |y|<R_a ∣ y ∣ < R a
Finally let
A a ( w ) = − 2 h a ( log ( 1 + w ) ) a w 2 , A a ( 0 ) = 1. A_a(w)=-\frac{2h_a(\log(1+w))}{aw^2},\qquad A_a(0)=1. A a ( w ) = − a w 2 2 h a ( log ( 1 + w )) , A a ( 0 ) = 1. Its coefficient of w m w^m w m a a a m m m
a n f n ( a ) = ( − 2 ) n 2 n + 1 [ w 2 n ] A a ( w ) − n − 1 / 2 . a^n f_n(a)=\frac{(-2)^n}{2n+1}[w^{2n}]A_a(w)^{-n-1/2}. a n f n ( a ) = 2 n + 1 ( − 2 ) n [ w 2 n ] A a ( w ) − n − 1/2 . The recurrence (D.4) then proves the same polynomial property for a n k n ( a ) a^n k_n(a) a n k n ( a ) 0 ≤ n < 100 0 \le n < 100 0 ≤ n < 100 a = 303 / 200 a = 303/200 a = 303/200 3 / 200 3/200 3/200 ( 3 / 2 ) n (3/2)^n ( 3/2 ) n a = 3 / 2 a = 3/2 a = 3/2 general_extremal_polynomials.py, with the checks in certify_general_extremal_germ.py.
Proposition D.5 (A certified non-ID endpoint). The extremal law X 153 / 100 , 100 / 153 X_{153/100,100/153} X 153/100 , 100/153
Proof. Set a = 153 / 100 a = 153/100 a = 153/100 x = 53 / 20 x = 53/20 x = 53/20 ω = π ( 100 / 153 − 1 / 2 ) \omega= \pi(100/153 - 1/2) ω = π ( 100/153 − 1/2 ) h = sin ω h = \sin\omega h = sin ω c = cos ω c = \cos\omega c = cos ω ξ = 47 π / 100 \xi= 47\pi/100 ξ = 47 π /100
( 2.6247913 , 2.6247914 ) , ( 4.5379757 , 4.5379758 ) , ( 6.4203113 , 6.4203114 ) . (2.6247913, 2.6247914),\quad(4.5379757, 4.5379758),\quad(6.4203113, 6.4203114). ( 2.6247913 , 2.6247914 ) , ( 4.5379757 , 4.5379758 ) , ( 6.4203113 , 6.4203114 ) . Let S j = ( − 1 ) j + 1 j [ z j ] log W a ( z ) S_j = (-1)^{j+1}j[z^j]\log W_a(z) S j = ( − 1 ) j + 1 j [ z j ] log W a ( z ) T j = S j − ∑ n = 1 3 r n − a T_j = S_j - \sum_{n=1}^{3} r_n^{-a} T j = S j − ∑ n = 1 3 r n − a R = T 4 − 1 / ( 4 a ) R = T_4^{-1/(4a)} R = T 4 − 1/ ( 4 a ) R > 7.779 R > 7.779 R > 7.779 R > a / ( h x ) R > a/(hx) R > a / ( h x ) cos ξ > 0 \cos\xi> 0 cos ξ > 0
K a , 1 / a ( x ) ≤ sin ξ π x a − 2 ∑ n = 1 3 e − h x r n cos ( c x r n ) + 2 R a e − h x R T 1 < − 0.003. K_{a,1/a}(x) \le\frac{\sin\xi}{\pi x^a} - 2\sum_{n=1}^{3}e^{-h x r_n}\cos(c x r_n) + 2R^a e^{-h x R}T_1 < -0.003. K a , 1/ a ( x ) ≤ π x a sin ξ − 2 n = 1 ∑ 3 e − h x r n cos ( c x r n ) + 2 R a e − h x R T 1 < − 0.003. The certificate non_id_endpoint_153_certificate.json was reconstructed at 90 decimal digits, including every zero-bracket Taylor tail and every unlisted-root contribution. Its checker uses no numerical quadrature. The final displayed strict bound is an upper bound for the kernel, which suffices to prove non-ID.
The complete ID slice at index three halves.
Theorem D.6 (All skewnesses at α = 3 / 2 \alpha= 3/2 α = 3/2 There is a unique solution ( ρ 0 , x 0 ) (\rho_0,x_0) ( ρ 0 , x 0 )
K 3 / 2 , ρ ( x ) = 0 , ∂ x K 3 / 2 , ρ ( x ) = 0 K_{3/2,\rho}(x) = 0,\qquad\partial_x K_{3/2,\rho}(x) = 0 K 3/2 , ρ ( x ) = 0 , ∂ x K 3/2 , ρ ( x ) = 0 in 0.652317 < ρ < 0.652318 0.652317 < \rho< 0.652318 0.652317 < ρ < 0.652318 12 / 5 < x < 27 / 10 12/5 < x < 27/10 12/5 < x < 27/10
0.6523173573678771199 < ρ 0 < 0.6523173573678771201 , 0.6523173573678771199 < \rho_0 < 0.6523173573678771201, 0.6523173573678771199 < ρ 0 < 0.6523173573678771201 , 2.53849362351816328 < x 0 < 2.53849362351816331. 2.53849362351816328 < x_0 < 2.53849362351816331. 2.53849362351816328 < x 0 < 2.53849362351816331. The complete classification, including equality at the boundary, is
X 3 / 2 , ρ is ID ⟺ ρ ∈ [ 1 / 3 , 1 − ρ 0 ] ∪ [ ρ 0 , 2 / 3 ] . X_{3/2,\rho}\ \text{is ID}\quad\Longleftrightarrow\quad\rho\in[1/3,1-\rho_0]\cup[\rho_0,2/3]. X 3/2 , ρ is ID ⟺ ρ ∈ [ 1/3 , 1 − ρ 0 ] ∪ [ ρ 0 , 2/3 ] . In particular X 3 / 2 , 33 / 50 X_{3/2,33/50} X 3/2 , 33/50 X 3 / 2 , 13 / 20 X_{3/2,13/20} X 3/2 , 13/20
Proof. Write K ( ρ , x ) = K 3 / 2 , ρ ( x ) K(\rho,x) = K_{3/2,\rho}(x) K ( ρ , x ) = K 3/2 , ρ ( x ) J = [ 0.652317 , 0.652318 ] J = [0.652317,0.652318] J = [ 0.652317 , 0.652318 ] ρ ∈ J \rho\in J ρ ∈ J
0 < x ≤ 1 / 2 K > 1 / 2 [ 1 / 2 , 12 / 5 ] ∪ [ 27 / 10 , 4 ] K > 1 / 1250 [ 12 / 5 , 27 / 10 ] K x x > 1 / 4 , K ρ > 1 x ≥ 4 K > 2 / ( 25 x 3 / 2 ) \begin{array}{c|c}
0 < x \le1/2 & K > 1/2 \\
[1/2,12/5]\cup[27/10,4] & K > 1/1250 \\
[12/5,27/10] & K_{xx} > 1/4,\quad K_\rho> 1 \\
x \ge4 & K > 2/(25x^{3/2})
\end{array} 0 < x ≤ 1/2 [ 1/2 , 12/5 ] ∪ [ 27/10 , 4 ] [ 12/5 , 27/10 ] x ≥ 4 K > 1/2 K > 1/1250 K xx > 1/4 , K ρ > 1 K > 2/ ( 25 x 3/2 ) Here is a proof and a complete specification of the numerical errors. For the origin, use the kernel k k k Theorem D.3 . Put δ = π ( 2 / 3 − ρ ) \delta= \pi(2/3-\rho) δ = π ( 2/3 − ρ ) w = x e − i δ w = xe^{-i\delta} w = x e − i δ
K ( ρ , x ) = 2 Re k ( w ) + b sin δ π x − x sin δ π ∫ 0 ∞ k ( y ) ( y − x cos δ ) 2 + x 2 sin 2 δ d y (78) K(\rho,x)=2\operatorname{Re}k(w)+\frac{b\sin\delta}{\pi x}-\frac{x\sin\delta}{\pi}\int_0^\infty\frac{k(y)}{(y-x\cos\delta)^2+x^2\sin^2\delta}\,dy \tag*{(78)} K ( ρ , x ) = 2 Re k ( w ) + π x b sin δ − π x sin δ ∫ 0 ∞ ( y − x cos δ ) 2 + x 2 sin 2 δ k ( y ) d y (78) follows by Cauchy continuation. Indeed the extremal characteristic logarithm has derivative
χ e ′ ( t ) = i b − i ∫ 0 ∞ e − i t y k ( y ) d y , \chi'_e(t)=ib-i\int_0^\infty e^{-ity}k(y)\,dy, χ e ′ ( t ) = ib − i ∫ 0 ∞ e − i t y k ( y ) d y , so for ℑ w > 0 \Im w>0 ℑ w > 0
i π ∫ 0 ∞ e i t w χ e ′ ( t ) d t = − i b π w + i π ∫ 0 ∞ k ( y ) w − y d y . \frac{i}{\pi}\int_0^\infty e^{itw}\chi'_e(t)\,dt=-\frac{ib}{\pi w}+\frac{i}{\pi}\int_0^\infty\frac{k(y)}{w-y}\,dy. π i ∫ 0 ∞ e i tw χ e ′ ( t ) d t = − π w ib + π i ∫ 0 ∞ w − y k ( y ) d y . Continuation through the positive axis adds 2 k ( w ) 2k(w) 2 k ( w ) χ ρ ( t ) = χ e ( e i δ t ) \chi_\rho(t)=\chi_e(e^{i\delta}t) χ ρ ( t ) = χ e ( e i δ t ) k k k x ≤ 1 / 2 x\le1/2 x ≤ 1/2 3 / 2 3/2 3/2 K > 1 / 2 K>1/2 K > 1/2
For the tail put a = 3 / 2 a=3/2 a = 3/2 S j = ∑ r n − a j S_j=\sum r_n^{-aj} S j = ∑ r n − aj h = sin ( π ( ρ − 1 / 2 ) ) h=\sin(\pi(\rho-1/2)) h = sin ( π ( ρ − 1/2 )) ξ = π ( a ρ − 1 / 2 ) \xi=\pi(a\rho-1/2) ξ = π ( a ρ − 1/2 ) cos ξ ≥ 0 \cos\xi\ge0 cos ξ ≥ 0
K ( ρ , x ) ≥ sin ξ π [ x − a − 2 a Γ ( 2 a ) cos ξ S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( 3 e h x ) 3 S 2 . K(\rho,x)\ge\frac{\sin\xi}{\pi}\left[x^{-a}-2a\Gamma(2a)\cos\xi S_2x^{-2a}-a\Gamma(3a)S_3x^{-3a}\right]-2\left(\frac{3}{ehx}\right)^3S_2. K ( ρ , x ) ≥ π sin ξ [ x − a − 2 a Γ ( 2 a ) cos ξ S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( e h x 3 ) 3 S 2 . Use ( 1 + 2 q cos ξ + q 2 ) − 1 ≥ 1 − 2 q cos ξ − q 2 (1+2q\cos\xi+q^2)^{-1}\ge1-2q\cos\xi-q^2 ( 1 + 2 q cos ξ + q 2 ) − 1 ≥ 1 − 2 q cos ξ − q 2 e − v ≤ ( 3 / e ) 3 v − 3 e^{-v}\le(3/e)^3v^{-3} e − v ≤ ( 3/ e ) 3 v − 3 x 3 / 2 x^{3/2} x 3/2 0.0110878 > 1 / 100 0.0110878>1/100 0.0110878 > 1/100
For the finite part, twenty disjoint rational Wright-zero brackets have width at most 10 − 80 10^{-80} 1 0 − 80 0 < r < 45 0<r<45 0 < r < 45 T m = S m − ∑ n = 1 20 r n − 3 m / 2 T_m=S_m-\sum_{n=1}^{20}r_n^{-3m/2} T m = S m − ∑ n = 1 20 r n − 3 m /2 T 20 ( 204 / 5 ) 30 < 1 T_{20}(204/5)^{30}<1 T 20 ( 204/5 ) 30 < 1 204 / 5 204/5 204/5
The listed positive integrals use validated quadrature after t = u 2 t=u^2 t = u 2 [ 0 , 12 ] [0,12] [ 0 , 12 ] e i ξ / ( 1 + q e i ξ ) e^{i\xi}/(1+qe^{i\xi}) e i ξ / ( 1 + q e i ξ )
a π ∑ m = 1 19 ( − 1 ) m + 1 sin ( m ξ ) Γ ( a m ) x − a m T m . \frac{a}{\pi}\sum_{m=1}^{19}(-1)^{m+1}\sin(m\xi)\Gamma(am)x^{-am}T_m. π a m = 1 ∑ 19 ( − 1 ) m + 1 sin ( m ξ ) Γ ( am ) x − am T m . Its j j j x x x a Γ ( 30 + j ) x − 30 − j T 20 / π a\Gamma(30+j)x^{-30-j}T_{20}/\pi a Γ ( 30 + j ) x − 30 − j T 20 / π j = 0 , 1 , 2 j=0,1,2 j = 0 , 1 , 2 ρ \rho ρ 21 a 2 Γ ( 30 ) x − 30 T 20 21a^2\Gamma(30)x^{-30}T_{20} 21 a 2 Γ ( 30 ) x − 30 T 20 ∣ 1 + q e i ξ ∣ ≥ 1 |1+qe^{i\xi}|\ge1 ∣1 + q e i ξ ∣ ≥ 1 q / ∣ 1 + q e i ξ ∣ ≤ 1 q/|1+qe^{i\xi}|\le1 q /∣1 + q e i ξ ∣ ≤ 1 1 ≤ m ≤ 20 1\le m\le20 1 ≤ m ≤ 20 h x R > a m + j hxR>am+j h x R > am + j R = 204 / 5 R=204/5 R = 204/5 2 R a m + j e − h x R T m 2R^{am+j}e^{-hxR}T_m 2 R am + j e − h x R T m ρ \rho ρ j = 1 j=1 j = 1 π x \pi x π x
The exact partition has 19 boxes for positivity off the contact interval and 3 boxes for both derivative inequalities on it. Endpoint checks give K x ( ρ , 12 / 5 ) < 0 < K x ( ρ , 27 / 10 ) K_x(\rho,12/5)<0<K_x(\rho,27/10) K x ( ρ , 12/5 ) < 0 < K x ( ρ , 27/10 ) x ( ρ ) x(\rho) x ( ρ ) m ( ρ ) = K ( ρ , x ( ρ ) ) m(\rho)=K(\rho,x(\rho)) m ( ρ ) = K ( ρ , x ( ρ )) m ′ ( ρ ) = K ρ ( ρ , x ( ρ ) ) > 1 m'(\rho)=K_\rho(\rho,x(\rho))>1 m ′ ( ρ ) = K ρ ( ρ , x ( ρ )) > 1
Let ρ L , ρ U \rho_L,\rho_U ρ L , ρ U x t = 2.5384936235181632931 x_t=2.5384936235181632931 x t = 2.5384936235181632931
− 1.607 ⋅ 10 − 19 < K ( ρ L , x t ) < − 1.530 ⋅ 10 − 19 , -1.607\cdot10^{-19}<K(\rho_L,x_t)<-1.530\cdot10^{-19}, − 1.607 ⋅ 1 0 − 19 < K ( ρ L , x t ) < − 1.530 ⋅ 1 0 − 19 , 1.799 ⋅ 10 − 19 < K ( ρ U , x t ) < 1.875 ⋅ 10 − 19 , 1.799\cdot10^{-19}<K(\rho_U,x_t)<1.875\cdot10^{-19}, 1.799 ⋅ 1 0 − 19 < K ( ρ U , x t ) < 1.875 ⋅ 1 0 − 19 , ∣ K x ( ρ U , x t ) ∣ < 3.46 ⋅ 10 − 19 . |K_x(\rho_U,x_t)|<3.46\cdot10^{-19}. ∣ K x ( ρ U , x t ) ∣ < 3.46 ⋅ 1 0 − 19 . Convexity supplies the global bound m ( ρ U ) ≥ K ( ρ U , x t ) − K x ( ρ U , x t ) 2 / ( 2 ( 1 / 4 ) ) > 0 m(\rho_U)\ge K(\rho_U,x_t)-K_x(\rho_U,x_t)^2/(2(1/4))>0 m ( ρ U ) ≥ K ( ρ U , x t ) − K x ( ρ U , x t ) 2 / ( 2 ( 1/4 )) > 0 m ( ρ L ) < 0 m(\rho_L)<0 m ( ρ L ) < 0 ρ 0 \rho_0 ρ 0 m ( ρ 0 ) = 0 m(\rho_0)=0 m ( ρ 0 ) = 0 x 0 x_0 x 0 ρ \rho ρ
At ρ 0 \rho_0 ρ 0 Theorem D.3 and sector interpolation give ID on [ ρ 0 , 2 / 3 ] [\rho_0,2/3] [ ρ 0 , 2/3 ] ρ 0 \rho_0 ρ 0 ρ 0 \rho_0 ρ 0
The command certify_three_halves_classification.py verify reruns the extremal prerequisite, twenty-zero verification, all 22 boxes, contact checks and tail constants at 180 decimal digits. It passed. Full enclosures and hashes are in three_halves_classification_verification.json. This slice alone does not prove the global cutoff; the independent argument in Theorem E.4 supplies that conclusion. □
The same spectral inequality gives a general local statement above one. It does not require monotonicity of the full kernel in the skewness.
Proposition D.7. Fix 0 < η ≤ 1 / 4 0 < \eta\le1/4 0 < η ≤ 1/4 s = sin ( π η ) s = \sin(\pi\eta) s = sin ( π η )
ε η = min { 1 2 , η 1 − η , s 4 96 π } > 0. \varepsilon_\eta= \min\left\{\frac{1}{2}, \frac{\eta}{1-\eta}, \frac{s^4}{96\pi}\right\} > 0. ε η = min { 2 1 , 1 − η η , 96 π s 4 } > 0. If
1 < α ≤ 1 + ε η , ρ ∈ [ η , 1 / 2 − η ] ∪ [ 1 / 2 + η , 1 − η ] , 1 < \alpha\le1+\varepsilon_\eta,\qquad\rho\in[\eta,1/2-\eta]\cup[1/2+\eta,1-\eta], 1 < α ≤ 1 + ε η , ρ ∈ [ η , 1/2 − η ] ∪ [ 1/2 + η , 1 − η ] , then the parameters are admissible and X α , ρ X_{\alpha,\rho} X α , ρ ρ ∈ ( 0 , 1 ) ∖ { 1 / 2 } \rho\in(0,1)\setminus\{1/2\} ρ ∈ ( 0 , 1 ) ∖ { 1/2 } α > 1 \alpha>1 α > 1
Proof. By reflection take 1 / 2 + η ≤ ρ ≤ 1 − η 1/2+\eta\le\rho\le1-\eta 1/2 + η ≤ ρ ≤ 1 − η α ≤ 1 / ( 1 − η ) \alpha\le1/(1-\eta) α ≤ 1/ ( 1 − η )
η ≤ α ( 1 − ρ ) ≤ 1 / 2 − η 1 − η < 1 2 . \eta\le\alpha(1-\rho) \le\frac{1/2-\eta}{1-\eta} < \frac{1}{2}. η ≤ α ( 1 − ρ ) ≤ 1 − η 1/2 − η < 2 1 . Hence sin ξ ≥ s \sin\xi\ge s sin ξ ≥ s sin ω ≥ s \sin\omega\ge s sin ω ≥ s 1 ≤ α ≤ 3 / 2 1 \le\alpha\le3/2 1 ≤ α ≤ 3/2 1 + u α ≤ e u 1+u^\alpha\le e^u 1 + u α ≤ e u u ≥ 0 u \ge0 u ≥ 0 u α ≤ u u^\alpha\le u u α ≤ u u α ≤ u 2 ≤ e u − 1 u^\alpha\le u^2 \le e^u-1 u α ≤ u 2 ≤ e u − 1
B α , ξ ( v ) ≥ Γ ( α + 1 ) sin ξ π ( v + 2 ) α ≥ s 4 π ( 1 + v α ) , B_{\alpha,\xi}(v) \ge\frac{\Gamma(\alpha+1)\sin\xi}{\pi(v+2)^\alpha} \ge\frac{s}{4\pi(1+v^\alpha)}, B α , ξ ( v ) ≥ π ( v + 2 ) α Γ ( α + 1 ) sin ξ ≥ 4 π ( 1 + v α ) s , where ( v + 2 ) α ≤ 2 α − 1 ( v α + 2 α ) ≤ 4 ( 1 + v α ) (v+2)^\alpha\le2^{\alpha-1}(v^\alpha+2^\alpha) \le4(1+v^\alpha) ( v + 2 ) α ≤ 2 α − 1 ( v α + 2 α ) ≤ 4 ( 1 + v α ) Γ ( α + 1 ) ≥ 1 \Gamma(\alpha+1)\ge1 Γ ( α + 1 ) ≥ 1
sup v ≥ 0 e − s v ( 1 + v α ) 2 ≤ max { 4 , 4 sup v ≥ 1 v 3 e − s v } ≤ 108 e 3 s 3 < 6 s 3 . \sup_{v\ge0}e^{-sv}(1+v^\alpha)^2\le\max\left\{4,4\sup_{v\ge1}v^3e^{-sv}\right\}\le\frac{108}{e^3s^3}<\frac{6}{s^3}. v ≥ 0 sup e − s v ( 1 + v α ) 2 ≤ max { 4 , 4 v ≥ 1 sup v 3 e − s v } ≤ e 3 s 3 108 < s 3 6 . Writing S j = ∑ n ( 1 + ( r n x ) α ) − j S_j=\sum_n(1+(r_nx)^\alpha)^{-j} S j = ∑ n ( 1 + ( r n x ) α ) − j x > 0 x>0 x > 0
K α , ρ ( x ) ≥ s 4 π S 1 − 12 s 3 S 2 ≥ ( s 4 π − 12 ε η s 3 ) S 1 ≥ s 8 π S 1 > 0. K_{\alpha,\rho}(x)\ge\frac{s}{4\pi}S_1-\frac{12}{s^3}S_2\ge\left(\frac{s}{4\pi}-\frac{12\varepsilon_\eta}{s^3}\right)S_1\ge\frac{s}{8\pi}S_1>0. K α , ρ ( x ) ≥ 4 π s S 1 − s 3 12 S 2 ≥ ( 4 π s − s 3 12 ε η ) S 1 ≥ 8 π s S 1 > 0. The opposite-side Lévy kernel is already nonnegative. The absolutely summable compensated representation above therefore proves ID. Theorem A.2 excludes extended GGC. □
The global ID region
The proof has three tasks: exclude every skewness at large indices, locate the extremal transition, and prove positivity at all smaller extremal indices. Only after these tasks are complete does sector interpolation identify the global interval and its inner boundary. We write α ∗ = α I D \alpha_*=\alpha_{\mathrm{ID}} α ∗ = α ID
A uniform upper exclusion for the ID index
Lemma E.1 (An elementary characteristic-function obstruction). Every classically ID characteristic function satisfies
∣ φ ( 2 t ) ∣ 2 ≥ ∣ φ ( t ) ∣ 8 , t ∈ R . \lvert\varphi(2t)\rvert^{2} \ge\lvert\varphi(t)\rvert^{8}, \qquad t \in\mathbb{R}. ∣ φ ( 2 t ) ∣ 2 ≥ ∣ φ ( t ) ∣ 8 , t ∈ R . For every admissible free stable law with 9 / 5 ≤ α ≤ 2 9/5 \le\alpha\le2 9/5 ≤ α ≤ 2 t = 89 / 100 t = 89/100 t = 89/100
∣ φ ( 178 / 100 ) ∣ 2 − ∣ φ ( 89 / 100 ) ∣ 8 < − 10 − 5 . \lvert\varphi(178/100)\rvert^{2} - \lvert\varphi(89/100)\rvert^{8} < -10^{-5}. ∣ φ ( 178/100 ) ∣ 2 − ∣ φ ( 89/100 ) ∣ 8 < − 1 0 − 5 . Consequently none of these laws is classically ID.
Proof. Write A ( t ) = − log ∣ φ ( t ) ∣ A(t) = -\log\lvert\varphi(t)\rvert A ( t ) = − log ∣ φ ( t )∣ 1 − cos ( 2 u ) ≤ 4 ( 1 − cos u ) 1-\cos(2u) \le4(1-\cos u) 1 − cos ( 2 u ) ≤ 4 ( 1 − cos u ) A ( 2 t ) ≤ 4 A ( t ) A(2t) \le4A(t) A ( 2 t ) ≤ 4 A ( t )
For the strict reverse inequality, use reflection and set
α = ( 9 + p ) / 5 , ρ = 1 2 + q ( 1 / α − 1 2 ) , 0 ≤ p , q ≤ 1. \alpha= (9+p)/5,\qquad\rho= \frac{1}{2} + q\left(1/\alpha-\frac{1}{2}\right),\qquad0 \le p,q \le1. α = ( 9 + p ) /5 , ρ = 2 1 + q ( 1/ α − 2 1 ) , 0 ≤ p , q ≤ 1. The Wright argument in (17) is − t α exp [ − i π q ( 1 − α / 2 ) ] -t^{\alpha}\exp[-i\pi q(1-\alpha/2)] − t α exp [ − iπ q ( 1 − α /2 )] 0 ≤ n < 32 0 \le n < 32 0 ≤ n < 32 ∣ z ∣ < 4 \lvert z\rvert< 4 ∣ z ∣ < 4
E = 4 32 32 ! Γ ( 2 + 4 5 32 ) 1 1 − 4 / 33 . E = \frac{4^{32}}{32!\,\Gamma\left(2+\frac{4}{5}32\right)}\frac{1}{1-4/33}. E = 32 ! Γ ( 2 + 5 4 32 ) 4 32 1 − 4/33 1 . Indeed, every consecutive absolute term ratio from index 32 onward is at most 4 / ( n + 1 ) 4/(n+1) 4/ ( n + 1 ) cf_exclusion_certificate.json has 4,289 leaves. On each whole rectangle, outward ball evaluation of the two finite series, with E E E certify_cf_exclusion.py checks every leaf and reconstructs the complete cover; the fresh 50-digit verification passed.
Theorem E.2 (All skewnesses above 153/100). For every admissible ( α , ρ ) (\alpha,\rho) ( α , ρ ) 153 / 100 ≤ α ≤ 2 153/100 \le\alpha\le2 153/100 ≤ α ≤ 2 X α , ρ X_{\alpha,\rho} X α , ρ
Proof. It remains to treat 153 / 100 ≤ α ≤ 9 / 5 153/100 \le\alpha\le9/5 153/100 ≤ α ≤ 9/5 ρ = 1 / 2 + q ( 1 / α − 1 / 2 ) \rho= 1/2 + q(1/\alpha- 1/2) ρ = 1/2 + q ( 1/ α − 1/2 ) 0 ≤ q ≤ 1 0 \le q \le1 0 ≤ q ≤ 1 q > 0 q > 0 q > 0
ω = π q ( 1 / α − 1 / 2 ) , h = sin ω , c = cos ω , ξ = π ( 1 − α / 2 ) ( 1 + q ) . \omega= \pi q(1/\alpha- 1/2),\qquad h = \sin\omega,\qquad c = \cos\omega,\qquad\xi= \pi(1-\alpha/2)(1+q). ω = π q ( 1/ α − 1/2 ) , h = sin ω , c = cos ω , ξ = π ( 1 − α /2 ) ( 1 + q ) . Here cos ξ > 0 \cos\xi> 0 cos ξ > 0
K ( x ) = ∑ n B α , ξ ( r n x ) − 2 ∑ n e − h r n x cos ( c r n x ) . K(x) = \sum_{n} B_{\alpha,\xi}(r_nx) - 2\sum_{n}e^{-hr_nx}\cos(cr_nx). K ( x ) = n ∑ B α , ξ ( r n x ) − 2 n ∑ e − h r n x cos ( c r n x ) . Its positive part obeys
∑ n B α , ξ ( r n x ) ≤ sin ξ π x α . (79) \sum_{n} B_{\alpha,\xi}(r_nx) \le\frac{\sin\xi}{\pi x^{\alpha}}. \tag*{(79)} n ∑ B α , ξ ( r n x ) ≤ π x α sin ξ . (79) because its denominator is at least one and ∑ n r n − α = 1 / Γ ( 1 + α ) \sum_n r_n^{-\alpha} = 1/\Gamma(1+\alpha) ∑ n r n − α = 1/Γ ( 1 + α ) [ j / 100 , ( j + 1 ) / 100 ] [j/100,(j+1)/100] [ j /100 , ( j + 1 ) /100 ] 153 ≤ j < 180 153 \le j < 180 153 ≤ j < 180 r j ∗ ( α ) r_j^*(\alpha) r j ∗ ( α ) e j e_j e j W α ( − ( r j ∗ ( α ) ± e j ) α ) W_\alpha(-(r_j^*(\alpha)\pm e_j)^\alpha) W α ( − ( r j ∗ ( α ) ± e j ) α ) α \alpha α
Let S j = ∑ r j − j α S_j=\sum r_j^{-j\alpha} S j = ∑ r j − j α R > r 3 R>r_3 R > r 3
( S 4 − ∑ j = 1 3 r j − 4 α ) R 4 α < 1. \left(S_4-\sum_{j=1}^{3}r_j^{-4\alpha}\right)R^{4\alpha}<1. ( S 4 − j = 1 ∑ 3 r j − 4 α ) R 4 α < 1. The left side is bounded above using the upper root polynomials, with the same Taylor method. Positivity of every Rayleigh summand shows that every unlisted zero exceeds R R R T 1 = S 1 − ∑ j = 1 3 r j − α T_1=S_1-\sum_{j=1}^{3}r_j^{-\alpha} T 1 = S 1 − ∑ j = 1 3 r j − α h x R > α h x R>\alpha h x R > α
K ( x ) ≤ sin ξ π x α − 2 ∑ j = 1 3 e − h r j x cos ( c r j x ) + 2 R α e − h x R T 1 . (80) K(x)\leq\frac{\sin\xi}{\pi x^\alpha}-2\sum_{j=1}^{3}e^{-h r_j x}\cos(c r_j x)+2R^\alpha e^{-h x R}T_1. \tag*{(80)} K ( x ) ≤ π x α sin ξ − 2 j = 1 ∑ 3 e − h r j x cos ( c r j x ) + 2 R α e − h x R T 1 . (80) On 1 / 20 ≤ q ≤ 1 1/20\leq q\leq1 1/20 ≤ q ≤ 1 x = τ / ( c r 1 ) x=\tau/(c r_1) x = τ / ( c r 1 ) τ = 2 π m + d \tau=2\pi m+d τ = 2 π m + d m ≥ 1 m\geq1 m ≥ 1 d d d 2 e − τ tan ω cos τ 2e^{-\tau\tan\omega}\cos\tau 2 e − τ t a n ω cos τ − 10 − 4 -10^{-4} − 1 0 − 4
The small-skewness region is covered analytically. The root certificates also verify uniformly that
r 1 < 3 , r 2 / r 1 > 5 / 3 , r 1 α S 1 − 1 < 3. r_1<3,\qquad r_2/r_1>5/3,\qquad r_1^\alpha S_1-1<3. r 1 < 3 , r 2 / r 1 > 5/3 , r 1 α S 1 − 1 < 3. If 0 < q ≤ 1 / 20 0<q\leq1/20 0 < q ≤ 1/20 tan ω < 1 / 40 \tan\omega<1/40 tan ω < 1/40 m = ⌈ 4 / ( 2 π tan ω ) ⌉ m=\lceil4/(2\pi\tan\omega)\rceil m = ⌈ 4/ ( 2 π tan ω )⌉ v = 2 π m tan ω v=2\pi m\tan\omega v = 2 π m tan ω x = 2 π m / ( c r 1 ) x=2\pi m/(c r_1) x = 2 π m / ( c r 1 ) 4 ≤ v < 4 + π / 20 4\leq v<4+\pi/20 4 ≤ v < 4 + π /20 x > 160 / 3 x>160/3 x > 160/3 r n ≥ ( 5 / 3 ) r 1 r_n\geq(5/3)r_1 r n ≥ ( 5/3 ) r 1
K ( x ) ≤ 1 π ( 160 / 3 ) 153 / 100 − 2 e − 4 − π / 20 + 6 ( 5 / 3 ) 9 / 5 e − 20 / 3 < − 1 100 . K(x)\leq\frac{1}{\pi(160/3)^{153/100}}-2e^{-4-\pi/20}+6(5/3)^{9/5}e^{-20/3}<-\frac{1}{100}. K ( x ) ≤ π ( 160/3 ) 153/100 1 − 2 e − 4 − π /20 + 6 ( 5/3 ) 9/5 e − 20/3 < − 100 1 . At q = 0 q=0 q = 0
The files variable_alpha_spectrum.py and certify_large_alpha_non_id.py implement the checks just described. A fresh 65-digit verification recomputed every moving root sign, all 27 Rayleigh bounds, all 962 kernel leaves, and the 4,289 characteristic-function leaves of Lemma E.1. The result and source hashes are recorded in large_alpha_non_id_verification.json. □
This first confines every possible larger-index ID law to α < 153 / 100 \alpha<153/100 α < 153/100
The extremal transition and the maximal ID index
Theorem E.3 (The extremal transition above 3 / 2 3/2 3/2 There is a unique critical index α ∗ \alpha_* α ∗ in the interval
1.52407396102878875153 < α ∗ < 1.52407396102878875155 (81) 1.52407396102878875153<\alpha_*<1.52407396102878875155 \tag*{(81)} 1.52407396102878875153 < α ∗ < 1.52407396102878875155 (81) For 3 / 2 ≤ α ≤ 153 / 100 3/2\leq\alpha\leq153/100 3/2 ≤ α ≤ 153/100
X α , 1 / α is classically ID ⟺ α ≤ α ∗ . X_{\alpha,1/\alpha}\text{ is classically ID}\iff\alpha\leq\alpha_*. X α , 1/ α is classically ID ⟺ α ≤ α ∗ . The same holds for its reflection. An exact definition of the critical index is the unique solution of
j α ( y ) = ∂ y j α ( y ) = 0 j_\alpha(y)=\partial_yj_\alpha(y)=0 j α ( y ) = ∂ y j α ( y ) = 0 in 38 / 25 ≤ α ≤ a U 38/25\leq\alpha\leq a_U 38/25 ≤ α ≤ a U 4 ≤ y ≤ 203 / 50 4\leq y\leq203/50 4 ≤ y ≤ 203/50 where a U a_U a U is the upper bound in (E.3) and j α j_\alpha j α is defined in Lemma D.4. Its contact satisfies
4.028954474287407697 < y ∗ < 4.028954474287407701. 4.028954474287407697<y_*<4.028954474287407701. 4.028954474287407697 < y ∗ < 4.028954474287407701. At the critical index the kernel is nonnegative on the entire positive half-line, with equality at the contact.
Proof. Write a L , a U a_L,a_U a L , a U Y = [ 4 , 203 / 50 ] Y=[4,203/50] Y = [ 4 , 203/50 ] Q n ( a ) = a n k n ( a ) Q_n(a)=a^n k_n(a) Q n ( a ) = a n k n ( a ) deg Q n ≤ 2 n \deg Q_n\le2n deg Q n ≤ 2 n
∥ Q n ∥ [ L , U ] ≤ 16 ( n + 1 ) 2 ( U / R ) n , L = 3 / 2 , U = 153 / 100 , R = 51 / 10. \lVert Q_n\rVert_{[L,U]}\le16(n+1)^2(U/R)^n,\qquad L=3/2,\quad U=153/100,\quad R=51/10. ∥ Q n ∥ [ L , U ] ≤ 16 ( n + 1 ) 2 ( U / R ) n , L = 3/2 , U = 153/100 , R = 51/10. The classical Bernstein and Markov inequalities [9 Introduction, (1.1)–(1.2)] give
∣ Q n ′ ( a ) ∣ ≤ min { 2 n ( a − L ) ( U − a ) , 8 n 2 U − L } ∥ Q n ∥ [ L , U ] , \lvert Q_n'(a)\rvert\le\min\left\{\frac{2n}{\sqrt{(a-L)(U-a)}},\frac{8n^2}{U-L}\right\}\lVert Q_n\rVert_{[L,U]}, ∣ Q n ′ ( a )∣ ≤ min { ( a − L ) ( U − a ) 2 n , U − L 8 n 2 } ∥ Q n ∥ [ L , U ] , with the Markov alternative at endpoints. Differentiating k n = Q n / a n k_n=Q_n/a^n k n = Q n / a n ∂ a j a \partial_a j_a ∂ a j a
The exact continuous cover in extremal_transition_certificate.json verifies:
j a ( y ) > 0 ( 3 2 ≤ a ≤ 38 25 , 0 ≤ y ≤ 21 5 ) , j_a(y)>0\qquad\left(\frac{3}{2}\le a\le\frac{38}{25},\quad0\le y\le\frac{21}{5}\right), j a ( y ) > 0 ( 2 3 ≤ a ≤ 25 38 , 0 ≤ y ≤ 5 21 ) , j a ( y ) > 0 ( 38 25 ≤ a ≤ a U , y ∈ [ 0 , 4 ] ∪ [ 203 50 , 21 5 ] ) , j_a(y)>0\qquad\left(\frac{38}{25}\le a\le a_U,\quad y\in[0,4]\cup\left[\frac{203}{50},\frac{21}{5}\right]\right), j a ( y ) > 0 ( 25 38 ≤ a ≤ a U , y ∈ [ 0 , 4 ] ∪ [ 50 203 , 5 21 ] ) , ∂ y 2 j a ( y ) > 1 10 , ∂ a j a ( y ) < − 1 2 ( 38 25 ≤ a ≤ a U , y ∈ Y ) . \partial_y^2j_a(y)>\frac{1}{10},\qquad\partial_a j_a(y)<-\frac{1}{2}\qquad\left(\frac{38}{25}\le a\le a_U,\quad y\in Y\right). ∂ y 2 j a ( y ) > 10 1 , ∂ a j a ( y ) < − 2 1 ( 25 38 ≤ a ≤ a U , y ∈ Y ) . It also checks ∂ y j a ( 4 ) < 0 \partial_yj_a(4)<0 ∂ y j a ( 4 ) < 0 ∂ y j a ( 203 / 50 ) > 0 \partial_yj_a(203/50)>0 ∂ y j a ( 203/50 ) > 0 Y Y Y a a a
Here is the infinite-radius control, independent of that compact calculation. On three index strips covering [ 3 / 2 , a U ] [3/2,a_U] [ 3/2 , a U ] Theorem E.2 encloses three roots and a lower bound R 0 R_0 R 0 ξ = π ( 2 − a ) \xi=\pi(2-a) ξ = π ( 2 − a ) h = sin [ π ( 1 / a − 1 / 2 ) ] h=\sin[\pi(1/a-1/2)] h = sin [ π ( 1/ a − 1/2 )] c = cos [ π ( 1 / a − 1 / 2 ) ] c=\cos[\pi(1/a-1/2)] c = cos [ π ( 1/ a − 1/2 )]
P a ( x ) = sin ξ π { x − a − 2 a cos ξ Γ ( 2 a ) S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a } . P_a(x)=\frac{\sin\xi}{\pi}\left\{x^{-a}-2a\cos\xi\,\Gamma(2a)S_2x^{-2a}-a\Gamma(3a)S_3x^{-3a}\right\}. P a ( x ) = π sin ξ { x − a − 2 a cos ξ Γ ( 2 a ) S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a } . This follows from ( 1 + s ) − 1 ≥ 1 − s (1+s)^{-1}\ge1-s ( 1 + s ) − 1 ≥ 1 − s s ≥ 0 s\ge0 s ≥ 0
P a ( x ) − 2 ∑ n = 1 3 e − h r n x cos ( c r n x ) − 2 R 0 2 e − h x R 0 T 1 P_a(x)-2\sum_{n=1}^{3}e^{-hr_nx}\cos(cr_nx)-2R_0^2e^{-hxR_0}T_1 P a ( x ) − 2 n = 1 ∑ 3 e − h r n x cos ( c r n x ) − 2 R 0 2 e − h x R 0 T 1 is positive on x = c a y x=c_ay x = c a y 21 / 5 ≤ y ≤ 5 21/5\le y\le5 21/5 ≤ y ≤ 5 x = 3 x=3 x = 3 x a x^a x a x ≥ 3 x\ge3 x ≥ 3 5 c a > 3 5c_a>3 5 c a > 3 x x x
To locate the transition, set y t = 4.02895447428740769884 y_t=4.02895447428740769884 y t = 4.02895447428740769884 a L , a U a_L,a_U a L , a U
j a L ( y t ) − ( ∂ y j a L ( y t ) ) 2 2 / 10 > 9.5691 ⋅ 10 − 21 , j a U ( y t ) < − 9.0056 ⋅ 10 − 21 . j_{a_L}(y_t)-\frac{(\partial_yj_{a_L}(y_t))^2}{2/10}>9.5691\cdot10^{-21},\qquad j_{a_U}(y_t)<-9.0056\cdot10^{-21}. j a L ( y t ) − 2/10 ( ∂ y j a L ( y t ) ) 2 > 9.5691 ⋅ 1 0 − 21 , j a U ( y t ) < − 9.0056 ⋅ 1 0 − 21 . Strong convexity makes the first expression a lower bound for the entire minimum on Y Y Y α ∗ \alpha_* α ∗ ∣ ∂ a ∂ y j a ∣ < 5 \lvert\partial_a\partial_yj_a\rvert<5 ∣ ∂ a ∂ y j a ∣ < 5 y ∗ y_* y ∗
Finally the cover verifies ∂ a j a ( y ) < − 1 / 4 \partial_a j_a(y)<-1/4 ∂ a j a ( y ) < − 1/4 a L ≤ a ≤ 153 / 100 a_L\le a\le153/100 a L ≤ a ≤ 153/100 y y y j a ( y ∗ ) < 0 j_a(y_*)<0 j a ( y ∗ ) < 0 a > α ∗ a>\alpha_* a > α ∗ 153 / 100 153/100 153/100 certify_extremal_transition.py; the results are in extremal_transition_verification.json. ∎
Theorem E.4 (The maximal classical ID index). The largest admissible stability index for which any free stable law is classically ID is exactly α ∗ \alpha_\ast α ∗ Theorem E.3 . More precisely,
α > α ∗ ⟹ X α , ρ is not ID for every admissible ρ . \alpha>\alpha_\ast\quad\Longrightarrow\quad X_{\alpha,\rho}\text{ is not ID for every admissible }\rho. α > α ∗ ⟹ X α , ρ is not ID for every admissible ρ . At α = α ∗ \alpha=\alpha_\ast α = α ∗
ρ = 1 / α ∗ , ρ = 1 − 1 / α ∗ . \rho=1/\alpha_\ast,\qquad\rho=1-1/\alpha_\ast. ρ = 1/ α ∗ , ρ = 1 − 1/ α ∗ . These laws are not extended GGC. The theorem does not claim the full skewness classification at indices below α ∗ \alpha_\ast α ∗
Proof. Theorem E.2 excludes α ≥ 153 / 100 \alpha\ge153/100 α ≥ 153/100 a L ≤ α ≤ 153 / 100 a_L\le\alpha\le153/100 a L ≤ α ≤ 153/100 a L a_L a L q q q 1 / 20 ≤ q ≤ 49 / 50 1/20\le q\le49/50 1/20 ≤ q ≤ 49/50 0 < q ≤ 1 / 20 0<q\le1/20 0 < q ≤ 1/20 38 / 25 38/25 38/25
K ( x ) ≤ 1 π ( 160 / 3 ) 38 / 25 − 2 e − 4 − π / 20 + 6 ( 5 / 3 ) 9 / 5 e − 20 / 3 < − 1 100 . K(x)\le\frac{1}{\pi(160/3)^{38/25}}-2e^{-4-\pi/20}+6(5/3)^{9/5}e^{-20/3}<-\frac{1}{100}. K ( x ) ≤ π ( 160/3 ) 38/25 1 − 2 e − 4 − π /20 + 6 ( 5/3 ) 9/5 e − 20/3 < − 100 1 . As before, q = 0 q=0 q = 0
The essential extra step concerns 49 / 50 ≤ q ≤ 1 49/50\le q\le1 49/50 ≤ q ≤ 1 x = c α y ∗ x=c_\alpha y_\ast x = c α y ∗ x x x ρ \rho ρ α \alpha α cos ξ > 0 \cos\xi>0 cos ξ > 0
∂ ξ B α , ξ ( v ) = α π ∫ 0 ∞ e − v u u α − 1 { cos ξ D + 2 sin 2 ξ u α D 2 } d u > 0 , D = 1 + 2 cos ξ u α + u 2 α . \partial_\xi B_{\alpha,\xi}(v)=\frac{\alpha}{\pi}\int_0^\infty e^{-vu}u^{\alpha-1}\left\{\frac{\cos\xi}{D}+\frac{2\sin^2\xi\,u^\alpha}{D^2}\right\}\,du>0,\qquad D=1+2\cos\xi\,u^\alpha+u^{2\alpha}. ∂ ξ B α , ξ ( v ) = π α ∫ 0 ∞ e − v u u α − 1 { D cos ξ + D 2 2 sin 2 ξ u α } d u > 0 , D = 1 + 2 cos ξ u α + u 2 α . Its contribution to ∂ ρ K \partial_\rho K ∂ ρ K
∂ ρ K ( x ) ≥ 2 π x ∑ n = 1 3 r n e − h r x cos ( c r n x + ω ) − 2 π x R 0 α + 1 e − h x R 0 T 1 . \partial_\rho K(x)\ge2\pi x\sum_{n=1}^{3}r_ne^{-h_rx}\cos(c_rnx+\omega)-2\pi xR_0^{\alpha+1}e^{-h xR_0}T_1. ∂ ρ K ( x ) ≥ 2 π x n = 1 ∑ 3 r n e − h r x cos ( c r n x + ω ) − 2 π x R 0 α + 1 e − h x R 0 T 1 . The last term bounds every unlisted residue derivative, since h x R 0 > α + 1 hxR_0>\alpha+1 h x R 0 > α + 1 y ∗ y_\ast y ∗ ∂ ρ K ( x ) > 1 / 2 \partial_\rho K(x)>1/2 ∂ ρ K ( x ) > 1/2
For α > α ∗ \alpha>\alpha_\ast α > α ∗ Theorem E.3 proved K α , 1 / α ( c α y ∗ ) = j α ( y ∗ ) < 0 K_{\alpha,1/\alpha}(c_\alpha y_\ast)=j_\alpha(y_\ast)<0 K α , 1/ α ( c α y ∗ ) = j α ( y ∗ ) < 0 ρ \rho ρ α = α ∗ \alpha=\alpha_\ast α = α ∗ Theorem A.2 .
The new cover contains 36 leaves, including the derivative box. certify_global_id_cutoff.py freshly verified them and reran both complete prerequisite covers: the 4,988 extremal-transition leaves and all the larger-index exclusions. The record is global_id_cutoff_verification.json. Thus the passage from an extremal transition to a global index cutoff is independently certified, rather than assumed from skewness interval convexity. □
The complete upper-index ID region
Theorem E.5 (All skewnesses between three halves and the cutoff). For 3 / 2 ≤ a ≤ α ∗ 3/2\le a\le\alpha_\ast 3/2 ≤ a ≤ α ∗
u = a − 3 2 , Q ( a ) = 114238 125000 + 341 100 u + 7 u 2 , ρ ( a , q ) = 1 2 + ( a − 1 − 1 2 ) q . u=a-\frac{3}{2},\qquad Q(a)=\frac{114238}{125000}+\frac{341}{100}u+7u^2,\qquad\rho(a,q)=\frac{1}{2}+\left(a^{-1}-\frac{1}{2}\right)q. u = a − 2 3 , Q ( a ) = 125000 114238 + 100 341 u + 7 u 2 , ρ ( a , q ) = 2 1 + ( a − 1 − 2 1 ) q . There is exactly one pair ( q ∗ ( a ) , x ∗ ( a ) ) (q_\ast(a),x_\ast(a)) ( q ∗ ( a ) , x ∗ ( a ))
∣ q − Q ( a ) ∣ < 1 / 500 , 47 / 20 < x < 57 / 20 |q-Q(a)|<1/500,\qquad47/20<x<57/20 ∣ q − Q ( a ) ∣ < 1/500 , 47/20 < x < 57/20 such that
K a , ρ ( a , q ) ( x ) = 0 , ∂ x K a , ρ ( a , q ) ( x ) = 0. K_{a,\rho(a,q)}(x)=0,\qquad\partial_xK_{a,\rho(a,q)}(x)=0. K a , ρ ( a , q ) ( x ) = 0 , ∂ x K a , ρ ( a , q ) ( x ) = 0. Both components are continuous in a a a 0 < q ∗ ( a ) ≤ 1 0 < q_*(a) \le1 0 < q ∗ ( a ) ≤ 1 r ( a ) = ρ ( a , q ∗ ( a ) ) r(a) = \rho(a, q_*(a)) r ( a ) = ρ ( a , q ∗ ( a ))
X a , ρ is ID ⟺ ρ ∈ [ 1 − a − 1 , 1 − r ( a ) ] ∪ [ r ( a ) , a − 1 ] . X_{a,\rho}\text{ is ID}\quad\Longleftrightarrow\quad\rho\in[1-a^{-1},1-r(a)]\cup[r(a),a^{-1}]. X a , ρ is ID ⟺ ρ ∈ [ 1 − a − 1 , 1 − r ( a )] ∪ [ r ( a ) , a − 1 ] . The endpoints are included. At a = 3 / 2 a=3/2 a = 3/2 r ( a ) = ρ 0 r(a)=\rho_0 r ( a ) = ρ 0 Theorem D.6 ; at a = α ∗ a=\alpha_* a = α ∗ q ∗ ( a ) = 1 q_*(a)=1 q ∗ ( a ) = 1 Q Q Q
∣ r ( a ) − [ 1 2 + ( a − 1 − 1 2 ) Q ( a ) ] ∣ < 1 / 3000. \left|r(a)-\left[\frac{1}{2}+\left(a^{-1}-\frac{1}{2}\right)Q(a)\right]\right|<1/3000. r ( a ) − [ 2 1 + ( a − 1 − 2 1 ) Q ( a ) ] < 1/3000. Proof. We use the spectral kernel formula also as an auxiliary analytic expression on the entire band q = Q ( a ) + v / 500 q=Q(a)+v/500 q = Q ( a ) + v /500 − 1 ≤ v ≤ 1 -1\le v\le1 − 1 ≤ v ≤ 1 q > 1 q>1 q > 1 q ≤ 1 q\le1 q ≤ 1 0 < ξ = π ( 1 − a / 2 ) ( 1 + q ) < π / 2 0<\xi=\pi(1-a/2)(1+q)<\pi/2 0 < ξ = π ( 1 − a /2 ) ( 1 + q ) < π /2 v = 1 v=1 v = 1 u = 0 u=0 u = 0 ξ / π = 0.478976 \xi/\pi=0.478976 ξ / π = 0.478976 u u u v = 1 v=1 v = 1 − 0.105452 + 0.09 u − 10.5 u 2 < 0 -0.105452+0.09u-10.5u^2<0 − 0.105452 + 0.09 u − 10.5 u 2 < 0 0 ≤ u ≤ 1 / 40 0\le u\le1/40 0 ≤ u ≤ 1/40
The following strict inequalities hold uniformly on this band for 3 / 2 ≤ a ≤ a U 3/2\le a\le a_U 3/2 ≤ a ≤ a U a U a_U a U
x ∈ [ 1 / 2 , 47 / 20 ] ∪ [ 57 / 20 , 4 ] K > 0 x ∈ [ 47 / 20 , 57 / 20 ] K x x > 1 / 25 , K ρ > 1 / 2 x = 47 / 20 K x < 0 x = 57 / 20 K x > 0 x ≥ 4 K > 0 \begin{array}{c|c}
x\in[1/2,47/20]\cup[57/20,4] & K>0\\
x\in[47/20,57/20] & K_{xx}>1/25,\quad K_\rho>1/2\\
x=47/20 & K_x<0\\
x=57/20 & K_x>0\\
x\ge4 & K>0
\end{array} x ∈ [ 1/2 , 47/20 ] ∪ [ 57/20 , 4 ] x ∈ [ 47/20 , 57/20 ] x = 47/20 x = 57/20 x ≥ 4 K > 0 K xx > 1/25 , K ρ > 1/2 K x < 0 K x > 0 K > 0 At q − = Q ( a ) − 1 / 500 q_-=Q(a)-1/500 q − = Q ( a ) − 1/500 x f ( a ) = 2.5385 + 4.16 u − 3.35 u 2 x_f(a)=2.5385+4.16u-3.35u^2 x f ( a ) = 2.5385 + 4.16 u − 3.35 u 2 q + = Q ( a ) + 1 / 500 q_+=Q(a)+1/500 q + = Q ( a ) + 1/500
K ( a , q + , x f ) − K x ( a , q + , x f ) 2 2 / 25 > 0. K(a,q_+,x_f)-\frac{K_x(a,q_+,x_f)^2}{2/25}>0. K ( a , q + , x f ) − 2/25 K x ( a , q + , x f ) 2 > 0. Strong convexity therefore makes the entire core minimum positive at q + q_+ q + certify_upper_id_region.py: 552 exact rational boxes cover the stated domains. We give the error controls below.
First we control the omitted origin interval. The convergent germ of Lemma D.4 and an exact coefficient bound give
∣ j a ( z ) − 3 / 2 ∣ < 9 / 20 ( ∣ z ∣ ≤ 4 / 5 ) , k a ( x ) < 2 ( x > 0 ) , 3 / 2 ≤ a ≤ 153 / 100. |j_a(z)-3/2|<9/20\quad(|z|\le4/5),\qquad k_a(x)<2\quad(x>0),\qquad3/2\le a\le153/100. ∣ j a ( z ) − 3/2∣ < 9/20 ( ∣ z ∣ ≤ 4/5 ) , k a ( x ) < 2 ( x > 0 ) , 3/2 ≤ a ≤ 153/100. For the first inequality, the rational coefficient sum through degree 128 plus its all-order tail is less than 0.408144. For the second, an eight-box cover proves j a ( y ) < 2 j_a(y)<2 j a ( y ) < 2 [ 0 , 21 / 5 ] [0,21/5] [ 0 , 21/5 ] x ≥ 5 / 2 x\ge5/2 x ≥ 5/2
k a ( x ) ≤ 1 π x a + 2 Γ ( 1 + a ) ( a e h x ) a < 1 k_a(x)\le\frac{1}{\pi x^a}+\frac{2}{\Gamma(1+a)}\left(\frac{a}{e h x}\right)^a<1 k a ( x ) ≤ π x a 1 + Γ ( 1 + a ) 2 ( e h x a ) a < 1 decreases with x x x ( 21 / 5 ) 2 − 2 / 3 > 5 / 2 (21/5)2^{-2/3}>5/2 ( 21/5 ) 2 − 2/3 > 5/2 ( 4 / 5 ) 2 − 2 / 3 > 1 / 2 (4/5)2^{-2/3}>1/2 ( 4/5 ) 2 − 2/3 > 1/2 Re k a ( z ) > 21 / 20 \operatorname{Re} k_a(z)>21/20 Re k a ( z ) > 21/20 ∣ z ∣ ≤ 1 / 2 |z|\le1/2 ∣ z ∣ ≤ 1/2 a ≤ α ∗ a\le\alpha_* a ≤ α ∗ Theorem E.3 . Apply (D.7) with b = b a b=b_a b = b a k = k a k=k_a k = k a δ = π ( a − 1 − ρ ) \delta=\pi(a^{-1}-\rho) δ = π ( a − 1 − ρ ) K a , ρ ( x ) > 1 / 10 K_{a,\rho}(x)>1/10 K a , ρ ( x ) > 1/10 0 < x ≤ 1 / 2 0<x\le1/2 0 < x ≤ 1/2 ρ > 1 / 2 \rho>1/2 ρ > 1/2 δ = 0 \delta=0 δ = 0 k a k_a k a certify_id_origin_strip.py.
For the finite spectral computations, ten distinct moving Wright zeros are enclosed on each of the three parameter strips [ 1.5 , 1.51 ] [1.5,1.51] [ 1.5 , 1.51 ] [ 1.51 , 1.52 ] [1.51,1.52] [ 1.51 , 1.52 ] [ 1.52 , a U ] [1.52,a_U] [ 1.52 , a U ] a a a 1 / 10 1/10 1/10 M ( h / ( 1 / 10 ) ) 64 / ( 1 − h / ( 1 / 10 ) ) M(h/(1/10))^{64}/(1-h/(1/10)) M ( h / ( 1/10 ) ) 64 / ( 1 − h / ( 1/10 )) h h h T m = S m − ∑ l i s t e d r − a m T_m=S_m-\sum_{\mathrm{listed}}r^{-am} T m = S m − ∑ listed r − am 15 4 T 4 < 1 15^4T_4<1 1 5 4 T 4 < 1 certify_ten_zero_id_spectra.py freshly checks every root sign and moment remainder.
The listed positive integrals use validated quadrature after u = t 2 / x u=t^{2}/x u = t 2 / x 0 < t < 1 / 4096 0<t<1/4096 0 < t < 1/4096 t > 12 t>12 t > 12 ( 1 + 2 z cos ξ + z 2 ) − 1 (1+2z\cos\xi+z^{2})^{-1} ( 1 + 2 z cos ξ + z 2 ) − 1 1 − 2 z cos ξ − z 2 1-2z\cos\xi-z^{2} 1 − 2 z cos ξ − z 2 1 1 1 T 2 T_{2} T 2 T 3 ≤ 15 − a T 2 T_{3}\le15^{-a}T_{2} T 3 ≤ 1 5 − a T 2 j j j
2 15 a m + j e − 15 h x T m 2\,15^{am+j}e^{-15hx}T_{m} 2 1 5 am + j e − 15 h x T m whenever 15 h x > a m + j 15hx>am+j 15 h x > am + j m ∈ { 1 , 2 , 4 } m\in\{1,2,4\} m ∈ { 1 , 2 , 4 } K x x K_{xx} K xx K ρ K_{\rho} K ρ x ≥ 4 x\ge4 x ≥ 4
K ≥ sin ξ π [ x − a − 2 a cos ξ Γ ( 2 a ) S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( 2 a e h x ) 2 a S 2 . K\ge\frac{\sin\xi}{\pi}\left[x^{-a}-2a\cos\xi\Gamma(2a)S_{2}x^{-2a}-a\Gamma(3a)S_{3}x^{-3a}\right]-2\left(\frac{2a}{ehx}\right)^{2a}S_{2}. K ≥ π sin ξ [ x − a − 2 a cos ξ Γ ( 2 a ) S 2 x − 2 a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( e h x 2 a ) 2 a S 2 . After multiplication by x a x^{a} x a 4 4 4
For each fixed a a a q q q q q q K ρ > 1 / 2 K_{\rho}>1/2 K ρ > 1/2 q − q_{-} q − q + q_{+} q + q + > 1 q_{+}>1 q + > 1 q ∗ ≤ 1 q_{*}\le1 q ∗ ≤ 1 q − < 1 < q + q_{-}<1<q_{+} q − < 1 < q + q ∗ q_{*} q ∗ q ≥ q ∗ q\ge q_{*} q ≥ q ∗ q < q ∗ q<q_{*} q < q ∗ q − q_{-} q − q − q_{-} q −
For completeness, the contact is continuous in a a a S 1 ( a ) = 1 / Γ ( 1 + a ) S_{1}(a)=1/\Gamma(1+a) S 1 ( a ) = 1/Γ ( 1 + a ) ( q , x ) (q,x) ( q , x ) x x x ρ \rho ρ r − a r^{-a} r − a q ∗ q_{*} q ∗ x ∗ x_{*} x ∗ a a a
Endpoint ID down to one and the global shape of the ID set
Theorem E.6 (The complete extremal classification). For every 1 < a ≤ 2 1<a\le2 1 < a ≤ 2
X a , 1 / a and X a , 1 − 1 / a are ID ⟺ a ≤ α ∗ . X_{a,1/a}\ \text{and}\ X_{a,1-1/a}\ \text{are ID}\quad\Longleftrightarrow\quad a\le\alpha_{*}. X a , 1/ a and X a , 1 − 1/ a are ID ⟺ a ≤ α ∗ . Their signed kernels are strictly positive at every x > 0 x>0 x > 0 1 < a < α ∗ 1<a<\alpha_{*} 1 < a < α ∗
Proof. It remains to extend Theorem E.3 downward. We give explicit uniform bounds, including the singular-looking limit a ↓ 1 a\downarrow1 a ↓ 1 ε = a − 1 \varepsilon=a-1 ε = a − 1 j a ( y ) = k a ( c a y ) j_{a}(y)=k_{a}(c_{a}y) j a ( y ) = k a ( c a y ) c a = ε 1 / a c_{a}=\varepsilon^{1/a} c a = ε 1/ a
Table 1 (PDF p. 56).
At a = 1 a=1 a = 1 ∣ e ν ∣ < 36 |e^{\nu}|<36 ∣ e ν ∣ < 36 log 36 > 1 + 5 / 2 + 1 / 36 \log36>1+5/2+1/36 log 36 > 1 + 5/2 + 1/36 36 ε 36^{\varepsilon} 3 6 ε ∣ F a ∣ < 36 ( 7 / 5 ) / 5 < 20 |F_{a}|<36\sqrt{(7/5)/5}<20 ∣ F a ∣ < 36 ( 7/5 ) /5 < 20 ∣ e ν ∣ < 25 |e^{\nu}|<25 ∣ e ν ∣ < 25
25 3 / 10 − 1 − 3 10 − 3 10 ( 4 + 1 / 25 ) > 0 , 25 3 / 10 log 25 − 1 − 4 − 1 / 25 > 0. 25^{3/10}-1-\frac{3}{10}-\frac{3}{10}(4+1/25)>0,\qquad25^{3/10}\log25-1-4-1/25>0. 2 5 3/10 − 1 − 10 3 − 10 3 ( 4 + 1/25 ) > 0 , 2 5 3/10 log 25 − 1 − 4 − 1/25 > 0. The resulting bound is 25 7 / 40 < 11 25\sqrt{7/40}<11 25 7/40 < 11 R = 4 R=4 R = 4 ε = 3 / 10 \varepsilon=3/10 ε = 3/10 R = 51 / 10 R=51/10 R = 51/10 ε = 2 / 5 \varepsilon=2/5 ε = 2/5 25 3 / 20 < 10 25\sqrt{3/20}<10 25 3/20 < 10 6 / 5 , 27 / 20 , 293 / 200 6/5,27/20,293/200 6/5 , 27/20 , 293/200 1 / 5 , 1 / 20 , 13 / 200 1/5, 1/20, 13/200 1/5 , 1/20 , 13/200 n n n 0 ≤ n < 128 0 \le n < 128 0 ≤ n < 128 0 ≤ n < 100 0 \le n < 100 0 ≤ n < 100
D n + 2 n + 1 + 5 C D n < C , D^{\frac{n+2}{n+1}}+\frac{5CD}{n}<C, D n + 1 n + 2 + n 5 C D < C , with ( D , C ) = ( 20 , 128 ) , ( 11 , 32 ) , ( 10 , 32 ) (D,C)=(20,128),(11,32),(10,32) ( D , C ) = ( 20 , 128 ) , ( 11 , 32 ) , ( 10 , 32 ) n = 128 n=128 n = 128 n = 100 n=100 n = 100
The convergent germs prove strict kernel positivity on
[ 1 , 13 / 10 ] × [ 0 , 2 ] , [ 13 / 10 , 7 / 5 ] × [ 0 , 3 ] , [ 7 / 5 , 3 / 2 ] × [ 0 , 87 / 20 ] . [1,13/10]\times[0,2],\qquad[13/10,7/5]\times[0,3],\qquad[7/5,3/2]\times[0,87/20]. [ 1 , 13/10 ] × [ 0 , 2 ] , [ 13/10 , 7/5 ] × [ 0 , 3 ] , [ 7/5 , 3/2 ] × [ 0 , 87/20 ] . It remains to prove the entire tails. Since ξ = π ( 2 − a ) \xi=\pi(2-a) ξ = π ( 2 − a ) cos ξ ≤ 0 \cos\xi\le0 cos ξ ≤ 0 a ≤ 3 / 2 a\le3/2 a ≤ 3/2
k a ( x ) ≥ sin ξ π [ x − a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( 2 a e h x ) 2 a S 2 , h = sin π ( a − 1 − 1 / 2 ) . (82) k_a(x)\ge\frac{\sin\xi}{\pi}\left[x^{-a}-a\Gamma(3a)S_3x^{-3a}\right]-2\left(\frac{2a}{\mathrm{e}hx}\right)^{2a}S_2,\qquad h=\sin\pi(a^{-1}-1/2). \tag*{(82)} k a ( x ) ≥ π sin ξ [ x − a − a Γ ( 3 a ) S 3 x − 3 a ] − 2 ( e h x 2 a ) 2 a S 2 , h = sin π ( a − 1 − 1/2 ) . (82) The lower bound multiplied by x a x^a x a x x x a ∈ [ 7 / 5 , 3 / 2 ] a\in[7/5,3/2] a ∈ [ 7/5 , 3/2 ] x = c a ( 87 / 20 ) x=c_a(87/20) x = c a ( 87/20 )
To include a = 1 a=1 a = 1 R j ( ε ) = S j ( 1 + ε ) / ε j R_j(\varepsilon)=S_j(1+\varepsilon)/\varepsilon^j R j ( ε ) = S j ( 1 + ε ) / ε j j = 2 , 3 j=2,3 j = 2 , 3 G j ( ε ) = 1 / Γ ( 2 + j ε ) G_j(\varepsilon)=1/\Gamma(2+j\varepsilon) G j ( ε ) = 1/Γ ( 2 + j ε )
S 2 = G 1 2 − G 2 , S 3 = G 1 3 − 3 2 G 1 G 2 + 1 2 G 3 . S_2=G_1^2-G_2,\qquad S_3=G_1^3-\frac{3}{2}G_1G_2+\frac{1}{2}G_3. S 2 = G 1 2 − G 2 , S 3 = G 1 3 − 2 3 G 1 G 2 + 2 1 G 3 . Direct differentiation at zero proves vanishing to orders two and three, respectively. The analytic quotients satisfy R 2 ( 0 ) = ζ ( 2 ) − 1 R_2(0)=\zeta(2)-1 R 2 ( 0 ) = ζ ( 2 ) − 1 R 3 ( 0 ) = ζ ( 3 ) − 1 R_3(0)=\zeta(3)-1 R 3 ( 0 ) = ζ ( 3 ) − 1 x = c a y x=c_a y x = c a y
k a ( c a y ) ≥ sin ( π ε ) π ε [ y − a − a ε Γ ( 3 a ) R 3 ( ε ) y − 3 a ] − 2 ( 2 a e h y ) 2 a R 2 ( ε ) . k_a(c_a y)\ge\frac{\sin(\pi\varepsilon)}{\pi\varepsilon}\left[y^{-a}-a\varepsilon\Gamma(3a)R_3(\varepsilon)y^{-3a}\right]-2\left(\frac{2a}{\mathrm{e}hy}\right)^{2a}R_2(\varepsilon). k a ( c a y ) ≥ π ε sin ( π ε ) [ y − a − a ε Γ ( 3 a ) R 3 ( ε ) y − 3 a ] − 2 ( e h y 2 a ) 2 a R 2 ( ε ) . This is positive at y = 2 y=2 y = 2 1 ≤ a ≤ 13 / 10 1\le a\le13/10 1 ≤ a ≤ 13/10 y = 3 y=3 y = 3 13 / 10 ≤ a ≤ 7 / 5 13/10\le a\le7/5 13/10 ≤ a ≤ 7/5 S j S_j S j R j R_j R j ∣ ε ∣ = 1 / 2 |\varepsilon|=1/2 ∣ ε ∣ = 1/2 ∣ S j ∣ |S_j| ∣ S j ∣ M j M_j M j ∣ ε ∣ ≤ r < 1 / 2 |\varepsilon|\le r<1/2 ∣ ε ∣ ≤ r < 1/2
M j ( 1 / 2 ) j ( 2 r ) 96 − j 1 − 2 r . \frac{M_j}{(1/2)^j}\frac{(2r)^{96-j}}{1-2r}. ( 1/2 ) j M j 1 − 2 r ( 2 r ) 96 − j . Thus the computation does not divide an interval containing zero by ε j \varepsilon^j ε j
The checker certify_near_one_extremal.py freshly verified all germ constants and a 165-box cover for 1 < a ≤ 7 / 5 1<a\le7/5 1 < a ≤ 7/5 certify_lower_extremal_id.py freshly verified a 623-box cover for 7 / 5 ≤ a ≤ 3 / 2 7/5\le a\le3/2 7/5 ≤ a ≤ 3/2 certify_lower_extremal_germ.py. These bounds, Theorem E.3 , and the global exclusion of Theorem E.4 prove the assertion. □
Theorem E.7 (A unique contact down to index seven fifths). For 7 / 5 ≤ a ≤ 3 / 2 7/5\le a\le3/2 7/5 ≤ a ≤ 3/2 v = a − 29 / 20 v=a-29/20 v = a − 29/20
Q − ( a ) = 0.758029 + 2.851767 v + 4.788686 v 2 + 9.650678 v 3 + 16.085231 v 4 , Q_-(a)=0.758029+2.851767v+4.788686v^2+9.650678v^3+16.085231v^4, Q − ( a ) = 0.758029 + 2.851767 v + 4.788686 v 2 + 9.650678 v 3 + 16.085231 v 4 , X − ( a ) = 2.321849 + 4.501192 v − 3.348444 v 2 − 0.569083 v 3 + 4.346122 v 4 . X_-(a)=2.321849+4.501192v-3.348444v^2-0.569083v^3+4.346122v^4. X − ( a ) = 2.321849 + 4.501192 v − 3.348444 v 2 − 0.569083 v 3 + 4.346122 v 4 . All terminating decimals here are exact rationals. There is a unique contact K = K x = 0 K=K_x=0 K = K x = 0
∣ q − Q − ( a ) ∣ < 1 / 500 , ∣ x − X − ( a ) ∣ < 1 / 5. |q-Q_-(a)|<1/500,\qquad|x-X_-(a)|<1/5. ∣ q − Q − ( a ) ∣ < 1/500 , ∣ x − X − ( a ) ∣ < 1/5. With this q ∗ ( a ) q_*(a) q ∗ ( a ) r ( a ) = 1 / 2 + ( 1 / a − 1 / 2 ) q ∗ ( a ) r(a)=1/2+(1/a-1/2)q_*(a) r ( a ) = 1/2 + ( 1/ a − 1/2 ) q ∗ ( a ) [ 1 − 1 / a , 1 − r ( a ) ] ∪ [ r ( a ) , 1 / a ] [1-1/a,1-r(a)]\cup[r(a),1/a] [ 1 − 1/ a , 1 − r ( a )] ∪ [ r ( a ) , 1/ a ]
Proof. The argument of Theorem E.5 applies with new, independently verified global inequalities. On the whole polynomial band, 0 < ξ < π / 2 0 < \xi< \pi/2 0 < ξ < π /2 0 < q < 1 0 < q < 1 0 < q < 1 1 / 100 1/100 1/100
The certificate certify_lower_id_region.py has 2,681 exact boxes. It proves K > 0 K > 0 K > 0 [ 11 / 20 , X − ( a ) − 1 / 5 ] ∪ [ X − ( a ) + 1 / 5 , 4 ] [11/20, X_{-}(a) - 1/5] \cup[X_{-}(a) + 1/5,4] [ 11/20 , X − ( a ) − 1/5 ] ∪ [ X − ( a ) + 1/5 , 4 ] K x x > 1 / 25 K_{xx} > 1/25 K xx > 1/25 K ρ > 1 / 2 K_{\rho} > 1/2 K ρ > 1/2 q = Q − ( a ) − 1 / 500 q = Q_{-}(a) - 1/500 q = Q − ( a ) − 1/500 K ( a , q , X − ( a ) ) < 0 K(a,q,X_{-}(a)) < 0 K ( a , q , X − ( a )) < 0 K ( a , q , X − ( a ) ) − K x ( a , q , X − ( a ) ) 2 / ( 2 / 25 ) > 0 K(a,q,X_{-}(a))-K_x(a,q,X_{-}(a))^2/(2/25) > 0 K ( a , q , X − ( a )) − K x ( a , q , X − ( a ) ) 2 / ( 2/25 ) > 0 x = 4 x = 4 x = 4
For the origin, a separate 64-box certificate proves k a ( x ) ≤ 3 / 2 k_a(x) \le3/2 k a ( x ) ≤ 3/2 ( 3 / 2 − j a ( y ) ) / y > 0 (3/2-j_a(y))/y > 0 ( 3/2 − j a ( y )) / y > 0 x ≥ 2 x \ge2 x ≥ 2
k a ( x ) ≤ 1 π sin ξ x a + 2 Γ ( 1 + a ) ( a e h x ) a < 3 / 2. k_a(x) \le\frac{1}{\pi\sin\xi\,x^a}+\frac{2}{\Gamma(1+a)}\left(\frac{a}{e h x}\right)^a < 3/2. k a ( x ) ≤ π sin ξ x a 1 + Γ ( 1 + a ) 2 ( e h x a ) a < 3/2. The ranges overlap since ( 87 / 20 ) ( 2 / 5 ) 5 / 7 > 2 (87/20)(2/5)^{5/7} > 2 ( 87/20 ) ( 2/5 ) 5/7 > 2 ∣ j a ( z ) − 3 / 2 ∣ < 0.684078 < 7 / 10 |j_a(z)-3/2| < 0.684078 < 7/10 ∣ j a ( z ) − 3/2∣ < 0.684078 < 7/10 ∣ z ∣ ≤ 11 / 10 |z| \le11/10 ∣ z ∣ ≤ 11/10 ( 11 / 10 ) ( 2 / 5 ) 5 / 7 > 11 / 20 (11/10)(2/5)^{5/7} > 11/20 ( 11/10 ) ( 2/5 ) 5/7 > 11/20 K > 1 / 10 K > 1/10 K > 1/10 0 < x ≤ 11 / 20 0 < x \le11/20 0 < x ≤ 11/20 Theorem E.6 . The convex core, global positivity, and interval interpolation now give the stated unique contact and full section. The complete verifier reran the root and origin prerequisites before all 2,681 boxes. □
Theorem E.8 (The global ID interval and its first contact). There is a continuous function r : ( 1 , α ∗ ] → ( 1 / 2 , 1 ) r:(1,\alpha_*] \to(1/2,1) r : ( 1 , α ∗ ] → ( 1/2 , 1 ) 1 < a ≤ α ∗ 1 < a \le\alpha_* 1 < a ≤ α ∗
[ 1 − 1 / a , 1 − r ( a ) ] ∪ [ r ( a ) , 1 / a ] . [1-1/a,1-r(a)] \cup[r(a),1/a]. [ 1 − 1/ a , 1 − r ( a )] ∪ [ r ( a ) , 1/ a ] . For 1 < a < α ∗ 1 < a < \alpha_* 1 < a < α ∗ 1 / 2 < r ( a ) < 1 / a 1/2 < r(a) < 1/a 1/2 < r ( a ) < 1/ a
lim a ↓ 1 r ( a ) = 1 / 2 , r ( α ∗ ) = 1 / α ∗ . \lim_{a\downarrow1}r(a)=1/2,\qquad r(\alpha_*)=1/\alpha_*. a ↓ 1 lim r ( a ) = 1/2 , r ( α ∗ ) = 1/ α ∗ . For 1 < a < α ∗ 1 < a < \alpha_* 1 < a < α ∗
π ( 1 − r ( a ) ) = min { θ ∈ ( π ( 1 − 1 / a ) , π / 2 ) : Re F a ( x e − i θ ) = 0 for some x > 0 } (83) \pi(1-r(a))=\min\left\{\theta\in\left(\pi(1-1/a),\pi/2\right):\operatorname{Re}F_a(xe^{-i\theta})=0\text{ for some }x>0\right\} \tag*{(83)} π ( 1 − r ( a )) = min { θ ∈ ( π ( 1 − 1/ a ) , π /2 ) : Re F a ( x e − i θ ) = 0 for some x > 0 } (83) where F a F_a F a x x x
K x = 0 , K x x ≥ 0 , K ρ > 0. K_x=0,\qquad K_{xx}\ge0,\qquad K_{\rho}>0. K x = 0 , K xx ≥ 0 , K ρ > 0. There are finitely many contact radii for each fixed index. Uniqueness and an explicit locating band are additionally proved on [ 7 / 5 , α ∗ ] [7/5,\alpha_*] [ 7/5 , α ∗ ] 7 / 5 7/5 7/5
Proof. Theorems E.6 and E.4, together with Proposition D.2 , give a nonempty interval reaching 1 / a 1/a 1/ a 1 < a ≤ α ∗ 1 < a \le\alpha_* 1 < a ≤ α ∗
On a compact index interval contained in ( 1 , α ∗ ] (1,\alpha_*] ( 1 , α ∗ ] k a ( 0 ) = 3 / 2 k_a(0)=3/2 k a ( 0 ) = 3/2 k a ≥ 0 k_a \ge0 k a ≥ 0 ∫ 0 ∞ k a = b a \int_0^\infty k_a=b_a ∫ 0 ∞ k a = b a b a b_a b a d > 0 d>0 d > 0 ∣ k a ( z ) − 3 / 2 ∣ < η |k_a(z)-3/2| < \eta ∣ k a ( z ) − 3/2∣ < η ∣ z ∣ ≤ d |z| \le d ∣ z ∣ ≤ d x ≤ d / 2 x \le d/2 x ≤ d /2 [ 0 , d ] [0,d] [ 0 , d ] 3 / 2 + η 3/2+\eta 3/2 + η [ d , ∞ ) [d,\infty) [ d , ∞ ) 4 x sup b a / ( π d 2 ) 4x\sup b_a/(\pi d^2) 4 x sup b a / ( π d 2 ) x > 0 x>0 x > 0
For ρ \rho ρ 1 / 2 1/2 1/2
K a , ρ ( x ) ≥ sin ξ π x a − C ( x − 2 a + x − 3 a ) > 0 K_{a,\rho}(x)\ge\frac{\sin\xi}{\pi x^a}-C\left(x^{-2a}+x^{-3a}\right)>0 K a , ρ ( x ) ≥ π x a sin ξ − C ( x − 2 a + x − 3 a ) > 0 for all sufficiently large x x x sin ξ \sin\xi sin ξ Theorem E.5 .
At a < α ∗ a < \alpha_\ast a < α ∗ 1 / a 1/a 1/ a r ( a ) < 1 / a r(a) < 1/a r ( a ) < 1/ a r ( a ) > 1 / 2 r(a) > 1/2 r ( a ) > 1/2 ρ = r ( a ) \rho= r(a) ρ = r ( a ) K x = 0 K_x = 0 K x = 0 K x x ≥ 0 K_{xx} \ge0 K xx ≥ 0 r ( a ) r(a) r ( a ) 1 / a 1/a 1/ a Re F a \operatorname{Re} F_a Re F a K ρ > 0 K_\rho> 0 K ρ > 0
For continuity it is convenient to normalize q ∗ ( a ) = ( r ( a ) − 1 / 2 ) / ( 1 / a − 1 / 2 ) q_\ast(a) = (r(a) - 1/2)/(1/a - 1/2) q ∗ ( a ) = ( r ( a ) − 1/2 ) / ( 1/ a − 1/2 ) q ∗ q_\ast q ∗ a < α ∗ a < \alpha_\ast a < α ∗ q ∈ ( q ∗ ( a ) , 1 ) q \in(q_\ast(a),1) q ∈ ( q ∗ ( a ) , 1 ) q ∗ ≤ 1 = q ∗ ( α ∗ ) q_\ast\le1 = q_\ast(\alpha_\ast) q ∗ ≤ 1 = q ∗ ( α ∗ ) a − 1 > 0 a - 1 > 0 a − 1 > 0 r ( a ) → 1 / 2 r(a) \to1/2 r ( a ) → 1/2 a ↓ 1 a \downarrow1 a ↓ 1 a a a □ \square □
The scaled Hankel formula gives stable root enclosures as the index approaches one. We use it to distinguish local contacts from the global boundary and to exhibit a change between two controlling radial branches.
Lemma F.1 (A scaled Hankel transform above one). For ε > 0 \varepsilon> 0 ε > 0
V ε ( z ) = e z / ε W 1 + ε ( − z / ε ) . V_\varepsilon(z) = e^{z/\varepsilon}W_{1+\varepsilon}(-z/\varepsilon). V ε ( z ) = e z / ε W 1 + ε ( − z / ε ) . On a Hankel contour H \mathcal{H} H
V ε ( z ) = 1 2 π i ∫ H e t t 2 exp { z F ε ( log t ) } d t , F ε ( ℓ ) = 1 − e − ε ℓ ε . (84) V_\varepsilon(z) = \frac{1}{2\pi i}\int_{\mathcal{H}} \frac{e^t}{t^2}\exp\{zF_\varepsilon(\log t)\}\,\mathrm{d}t,\qquad F_\varepsilon(\ell) = \frac{1-e^{-\varepsilon\ell}}{\varepsilon}. \tag*{(84)} V ε ( z ) = 2 π i 1 ∫ H t 2 e t exp { z F ε ( log t )} d t , F ε ( ℓ ) = ε 1 − e − ε ℓ . (84) The apparent singularity at ε = 0 \varepsilon= 0 ε = 0 F 0 ( ℓ ) = ℓ F_0(\ell) = \ell F 0 ( ℓ ) = ℓ V 0 ( z ) = 1 / Γ ( 2 − z ) V_0(z) = 1/\Gamma(2-z) V 0 ( z ) = 1/Γ ( 2 − z ) ( ε , z ) (\varepsilon,z) ( ε , z ) ( 0 , z 0 ) (0,z_0) ( 0 , z 0 ) z z z z z z
V ε ( z ) = G ( z ) − ε z 2 G ′ ′ ( z ) + ε 2 ( z 6 G ′ ′ ′ ( z ) + z 2 8 G ′ ′ ′ ′ ( z ) ) + O ( ε 3 ) , G ( z ) = 1 Γ ( 2 − z ) . V_\varepsilon(z) = G(z) - \frac{\varepsilon z}{2}G''(z) + \varepsilon^2\left(\frac{z}{6}G'''(z) + \frac{z^2}{8}G''''(z)\right) + O(\varepsilon^3),\qquad G(z) = \frac{1}{\Gamma(2-z)}. V ε ( z ) = G ( z ) − 2 ε z G ′′ ( z ) + ε 2 ( 6 z G ′′′ ( z ) + 8 z 2 G ′′′′ ( z ) ) + O ( ε 3 ) , G ( z ) = Γ ( 2 − z ) 1 . Proof. Hankel’s reciprocal-Gamma formula and termwise integration give (F.1); on the two straight rays the factor e − r e^{-r} e − r
F ε ( ℓ ) = ℓ ∫ 0 1 e − s ε ℓ d s F_\varepsilon(\ell) = \ell\int_0^1 e^{-s\varepsilon\ell}\,\mathrm{d}s F ε ( ℓ ) = ℓ ∫ 0 1 e − s ε ℓ d s removes the singularity and gives a stable entire expression. For real ε ≥ 0 \varepsilon\ge0 ε ≥ 0 Re ℓ ≥ 0 \operatorname{Re}\ell\ge0 Re ℓ ≥ 0
∣ F ε ( ℓ ) ∣ ≤ ∣ ℓ ∣ , ∣ F ε ( ℓ ) − ℓ ∣ ≤ 1 2 ε ∣ ℓ ∣ 2 . \lvert F_\varepsilon(\ell)\rvert\le\lvert\ell\rvert,\qquad\lvert F_\varepsilon(\ell)-\ell\rvert\le\frac{1}{2}\varepsilon\lvert\ell\rvert^2. ∣ F ε ( ℓ )∣ ≤ ∣ ℓ ∣ , ∣ F ε ( ℓ ) − ℓ ∣ ≤ 2 1 ε ∣ ℓ ∣ 2 . For complex ∣ ε ∣ ≤ d < 1 \lvert\varepsilon\rvert\le d < 1 ∣ ε ∣ ≤ d < 1 ∣ F ε ( log r ± i π ) ∣ ≤ C ( log r + π ) d \lvert F_\varepsilon(\log r \pm i\pi)\rvert\le C(\log r+\pi)^d ∣ F ε ( log r ± iπ )∣ ≤ C ( log r + π ) d e r e^r e r z z z F ε ( ℓ ) = ℓ − ε ℓ 2 / 2 + ε 2 ℓ 3 / 6 + O ( ε 3 ) F_\varepsilon(\ell) = \ell- \varepsilon\ell^2/2 + \varepsilon^2\ell^3/6 + O(\varepsilon^3) F ε ( ℓ ) = ℓ − ε ℓ 2 /2 + ε 2 ℓ 3 /6 + O ( ε 3 ) □ \square □
Corollary F.2 (Fixed scaled zero asymptotics). For every fixed integer m ≥ 0 m \ge0 m ≥ 0 a simple positive zero z m ( ε ) z_m(\varepsilon) z m ( ε ) of V ε V_\varepsilon V ε tends to n = m + 2 n=m+2 n = m + 2 Put p m = H m − γ p_m=H_m-\gamma p m = H m − γ with H 0 = 0 H_0=0 H 0 = 0 Then
z m ( ε ) = n + n p m ε + { n 2 ( p m 2 + H m ( 2 ) + ζ ( 2 ) ) + n 2 ( ζ ( 3 ) − H m ( 3 ) ) } ε 2 + O m ( ε 3 ) . z_m(\varepsilon)=n+np_m\varepsilon
+\left\{\frac{n}{2}\left(p_m^2+H_m^{(2)}+\zeta(2)\right)+n^2\left(\zeta(3)-H_m^{(3)}\right)\right\}\varepsilon^2+O_m(\varepsilon^3). z m ( ε ) = n + n p m ε + { 2 n ( p m 2 + H m ( 2 ) + ζ ( 2 ) ) + n 2 ( ζ ( 3 ) − H m ( 3 ) ) } ε 2 + O m ( ε 3 ) . The corresponding Wright zero is − z m ( ε ) / ε -z_m(\varepsilon)/\varepsilon − z m ( ε ) / ε No uniformity as m → ∞ m\to\infty m → ∞ is asserted.
Proof. Apply the holomorphic implicit function theorem at the simple zero n n n G G G s = z − n s=z-n s = z − n
G ( n + s ) = ( − 1 ) m + 1 m ! s exp { p m s − H m ( 2 ) + ζ ( 2 ) 2 s 2 + H m ( 3 ) − ζ ( 3 ) 3 s 3 + O ( s 4 ) } . G(n+s)=(-1)^{m+1}m!s\exp\left\{p_ms-\frac{H_m^{(2)}+\zeta(2)}{2}s^2+\frac{H_m^{(3)}-\zeta(3)}{3}s^3+O(s^4)\right\}. G ( n + s ) = ( − 1 ) m + 1 m ! s exp { p m s − 2 H m ( 2 ) + ζ ( 2 ) s 2 + 3 H m ( 3 ) − ζ ( 3 ) s 3 + O ( s 4 ) } . Substitution into Lemma F.1 gives the coefficients. Reality and simplicity persist for sufficiently small positive ε \varepsilon ε n > 0 n>0 n > 0
For clarity, the rigorous numerical implementation of (F.1) uses real z ≥ 0 z\ge0 z ≥ 0 0 ≤ ε ≤ 1 / 2 0\le\varepsilon\le1/2 0 ≤ ε ≤ 1/2 F = F ε F=F_\varepsilon F = F ε
V ε ( z ) = 1 π Re ∫ 0 π e i θ − i θ + z F ( i θ ) d θ − 1 π Im ∫ 1 ∞ r − 2 e − r + z F ( log r + i π ) d r . V_\varepsilon(z)=\frac{1}{\pi}\operatorname{Re}\int_0^\pi e^{i\theta-i\theta+zF(i\theta)}\,d\theta-\frac{1}{\pi}\operatorname{Im}\int_1^\infty r^{-2}e^{-r+zF(\log r+i\pi)}\,dr. V ε ( z ) = π 1 Re ∫ 0 π e i θ − i θ + z F ( i θ ) d θ − π 1 Im ∫ 1 ∞ r − 2 e − r + z F ( l o g r + iπ ) d r . The sign and orientation of both contour pieces are included here. Along the straight ray, Re F ( log r + i π ) ≤ log r + ε π 2 / 2 \operatorname{Re}F(\log r+i\pi)\le\log r+\varepsilon\pi^2/2 Re F ( log r + iπ ) ≤ log r + ε π 2 /2 U ≥ max ( 8 , 2 z ) U\ge\max(8,2z) U ≥ max ( 8 , 2 z ) ∂ z j V \partial_z^jV ∂ z j V j = 0 , 1 j=0,1 j = 0 , 1
E j ≤ 2 π e − U + z ε π 2 / 2 U z − 2 + j . E_j\le\frac{2}{\pi}e^{-U+z\varepsilon\pi^2/2}U^{z-2+j}. E j ≤ π 2 e − U + z ε π 2 /2 U z − 2 + j . For j = 1 j=1 j = 1 ∣ F ′ ∣ ≤ log r + π ≤ r |F'|\le\log r+\pi\le r ∣ F ′ ∣ ≤ log r + π ≤ r scaled_wright_hankel.py evaluates the finite integrals with directed complex balls and these explicit tails. The identity F ε ( ℓ ) = ℓ 1 F 1 ( 1 ; 2 ; − ε ℓ ) F_\varepsilon(\ell)=\ell\,{}_1F_1(1;2;-\varepsilon\ell) F ε ( ℓ ) = ℓ 1 F 1 ( 1 ; 2 ; − ε ℓ )
Lemma F.3 (Parameter derivatives on the scaled Hankel contour). Let I ⋐ ( 1 , 3 / 2 ] I\Subset(1,3/2] I ⋐ ( 1 , 3/2 ] be a real interval and let z ( a ) > 0 z(a)>0 z ( a ) > 0 be a real quadratic polynomial on I I I Set f ( a ) = V a − 1 ( z ( a ) ) f(a)=V_{a-1}(z(a)) f ( a ) = V a − 1 ( z ( a )) All Taylor coefficients of f f f can be obtained by differentiating the finite circular and straight Hankel integrals above. The following bound explicitly controls the omitted straight-ray coefficient of order n n n
At a real parameter a ∈ I a\in I a ∈ I write z ( a + u ) = z 0 + z 1 u + z 2 u 2 z(a+u)=z_0+z_1u+z_2u^2 z ( a + u ) = z 0 + z 1 u + z 2 u 2 and choose uniform bounds z 0 ≤ Z z_0\le Z z 0 ≤ Z a − 1 ≤ E a-1\le E a − 1 ≤ E and ∣ z k ∣ ≤ b k |z_k|\le b_k ∣ z k ∣ ≤ b k Define
C j = ∑ k = 0 min ( 2 , j ) b k ( j − k + 1 ) ! , P 0 = 1 , P n = 1 n ∑ j = 1 n j C j P n − j . C_j=\sum_{k=0}^{\min(2,j)}\frac{b_k}{(j-k+1)!},\qquad P_0=1,\qquad P_n=\frac{1}{n}\sum_{j=1}^n jC_jP_{n-j}. C j = k = 0 ∑ m i n ( 2 , j ) ( j − k + 1 )! b k , P 0 = 1 , P n = n 1 j = 1 ∑ n j C j P n − j . If U ≥ 8 U\ge8 U ≥ 8 and U > 2 ( Z + 2 n ) U>2(Z+2n) U > 2 ( Z + 2 n ) the coefficient error is at most
2 P n π exp ( − U + Z E π 2 / 2 ) U Z + 2 n − 2 . (85) \frac{2P_n}{\pi}\exp(-U+ZE\pi^2/2)U^{Z+2n-2}. \tag*{(85)} π 2 P n exp ( − U + Z E π 2 /2 ) U Z + 2 n − 2 . (85) In particular a midpoint Taylor polynomial through degree N − 1 N-1 N − 1 together with an interval enclosure of f ( N ) / N ! f^{(N)}/N! f ( N ) / N ! on I I I gives a rigorous real Taylor enclosure on the entire interval.
Proof. For ℓ = log r + i π \ell=\log r+i\pi ℓ = log r + iπ r ≥ 1 r\ge1 r ≥ 1
F ε ( ℓ ) = ℓ ∫ 0 1 e − t ε ℓ d t , ∣ [ u j ] F ε + u ( ℓ ) ∣ ≤ ∣ ℓ ∣ j + 1 ( j + 1 ) ! . F_\varepsilon(\ell)=\ell\int_0^1e^{-t\varepsilon\ell}\,dt,\qquad\left|[u^j]F_{\varepsilon+u}(\ell)\right|\le\frac{|\ell|^{j+1}}{(j+1)!}. F ε ( ℓ ) = ℓ ∫ 0 1 e − tε ℓ d t , [ u j ] F ε + u ( ℓ ) ≤ ( j + 1 )! ∣ ℓ ∣ j + 1 . Here ε ≥ 0 \varepsilon\ge0 ε ≥ 0 r ≥ 8 r\ge8 r ≥ 8 ∣ ℓ ∣ ≤ log r + π ≤ r |\ell|\le\log r+\pi\le r ∣ ℓ ∣ ≤ log r + π ≤ r j ≥ 1 j\ge1 j ≥ 1 z ( a + u ) F a − 1 + u ( ℓ ) z(a+u)F_{a-1+u}(\ell) z ( a + u ) F a − 1 + u ( ℓ ) C j r j + 1 ≤ C j r 2 j C_jr^{j+1}\le C_jr^{2j} C j r j + 1 ≤ C j r 2 j P n r 2 n P_nr^{2n} P n r 2 n exp ( Z log r + Z E π 2 / 2 ) \exp(Z\log r+ZE\pi^2/2) exp ( Z log r + Z E π 2 /2 )
Beyond U U U e − r r Z + 2 n e^{-r}r^{Z+2n} e − r r Z + 2 n − 1 / 2 -1/2 − 1/2 r − 2 r^{-2} r − 2 U − 2 U^{-2} U − 2
The implementation scaled_hankel_spectrum.py uses N = 16 N=16 N = 16 z ( a ) ± η z(a)\pm\eta z ( a ) ± η certify_scaled_moving_spectra.py.
Proposition F.4 (The endpoint germ and the exceptional law). As a ↓ 1 a\downarrow1 a ↓ 1 Z a = ( b a − X a , 1 / a ) / c a Z_a=(b_a-X_{a,1/a})/c_a Z a = ( b a − X a , 1/ a ) / c a 1 − T 1-T 1 − T T T T 1 1 1 [7 ] . Moreover, locally uniformly on ∣ y ∣ < 2 π |y|<2\pi ∣ y ∣ < 2 π
j a ( y ) ⟶ j 1 ( y ) = e − y + 1 y − 1 e y − 1 = 1 y − e − 2 y 1 − e − y . (86) j_a(y)\longrightarrow j_1(y)=e^{-y}+\frac{1}{y}-\frac{1}{e^y-1}=\frac{1}{y}-\frac{e^{-2y}}{1-e^{-y}}. \tag*{(86)} j a ( y ) ⟶ j 1 ( y ) = e − y + y 1 − e y − 1 1 = y 1 − 1 − e − y e − 2 y . (86) The singularity at zero is removable, and
k n ( 1 ) = ( − 1 ) n n ! − B n + 1 ( n + 1 ) ! , n ≥ 0 , B 1 = − 1 2 . k_n(1)=\frac{(-1)^n}{n!}-\frac{B_{n+1}}{(n+1)!},\qquad n\geq0,\qquad B_1=-\frac{1}{2}. k n ( 1 ) = n ! ( − 1 ) n − ( n + 1 )! B n + 1 , n ≥ 0 , B 1 = − 2 1 . The limiting kernel is already contained in Theorem 3 of [7 ] ; the assertion here identifies it with the regular endpoint of the variable-index germ. Its nonzero poles are 2 π i Z ∖ { 0 } 2\pi i\mathbb{Z}\setminus\{0\} 2 π i Z ∖ { 0 } − 1 -1 − 1
Proof. The Voiculescu transform of Z a Z_a Z a ε = a − 1 \varepsilon=a-1 ε = a − 1
a ε − 1 ε e ε ( i π − log z ) ⟶ 1 + log z − i π , Im z > 0. \frac{a}{\varepsilon}-\frac{1}{\varepsilon}e^{\varepsilon(i\pi-\log z)}\longrightarrow1+\log z-i\pi,\qquad\operatorname{Im}z>0. ε a − ε 1 e ε ( iπ − l o g z ) ⟶ 1 + log z − iπ , Im z > 0. The limit is the transform of 1 − T 1-T 1 − T h 1 ( v ) = 1 − e − v − v h_1(v)=1-e^{-v}-v h 1 ( v ) = 1 − e − v − v ∣ Im v ∣ < 2 π |\operatorname{Im}v|<2\pi ∣ Im v ∣ < 2 π 2 π 2\pi 2 π R < 2 π R<2\pi R < 2 π a a a ∣ y ∣ ≤ R |y|\leq R ∣ y ∣ ≤ R t = Re v t=\operatorname{Re}v t = Re v R + 2 R+2 R + 2
e ε t ≤ 1 + ε ( 1 + R + e − t ) < 1 + ε t ≤ e ε t , e^{\varepsilon t}\leq1+\varepsilon(1+R+e^{-t})<1+\varepsilon t\leq e^{\varepsilon t}, e εt ≤ 1 + ε ( 1 + R + e − t ) < 1 + εt ≤ e εt , a contradiction. The same conclusion at ε = 0 \varepsilon=0 ε = 0 [7 ] identifies that kernel with (F.3). Expansion of y / ( e y − 1 ) y/(e^y-1) y / ( e y − 1 )
This compact analytic limit does not cover neighborhoods approaching the poles while a ↓ 1 a\downarrow1 a ↓ 1
Theorem F.5 (The complete index 11 / 10 11/10 11/10 q † q_{\dagger} q †
0.1220544 < q † < 0.1220545 0.1220544<q_{\dagger}<0.1220545 0.1220544 < q † < 0.1220545 with a global zero minimum of the signed kernel. With ρ † = 1 / 2 + ( 9 / 22 ) q † \rho_{\dagger}=1/2+(9/22)q_{\dagger} ρ † = 1/2 + ( 9/22 ) q †
1890389 3437500 < ρ † < 24196981 44000000 , 0.54993134 < ρ † < 0.54993139 , \frac{1890389}{3437500}<\rho_{\dagger}<\frac{24196981}{44000000},\qquad0.54993134<\rho_{\dagger}<0.54993139, 3437500 1890389 < ρ † < 44000000 24196981 , 0.54993134 < ρ † < 0.54993139 , and the exact classification is
X 11 / 10 , ρ is ID ⟺ ρ ∈ [ 1 / 11 , 1 − ρ † ] ∪ [ ρ † , 10 / 11 ] . X_{11/10,\rho}\text{ is ID}\quad\Longleftrightarrow\quad\rho\in[1/11,1-\rho_{\dagger}]\cup[\rho_{\dagger},10/11]. X 11/10 , ρ is ID ⟺ ρ ∈ [ 1/11 , 1 − ρ † ] ∪ [ ρ † , 10/11 ] . Both boundary points are included. If c = ( 1 / 10 ) 10 / 11 c=(1/10)^{10/11} c = ( 1/10 ) 10/11 x = c y x=cy x = cy 6.68969 < y < 6.68973 6.68969<y<6.68973 6.68969 < y < 6.68973
There is also a different unique local contact K = K x = 0 K=K_x=0 K = K x = 0 0.0574 < q < 0.0575 0.0574<q<0.0575 0.0574 < q < 0.0575 3.7 < y < 3.9 3.7<y<3.9 3.7 < y < 3.9 q q q K ( c 7 ) < 0 K(c7)<0 K ( c 7 ) < 0
Proof. At a = 11 / 10 a=11/10 a = 11/10 V 1 / 10 V_{1/10} V 1/10
446 12 a ( S 12 − ∑ l i s t e d r − 12 a ) < 1 , 446^{12a}\left(S_{12}-\sum_{\mathrm{listed}}r^{-12a}\right)<1, 44 6 12 a ( S 12 − listed ∑ r − 12 a ) < 1 , so every unlisted root exceeds 446. No completeness of the root scan is assumed. The checker certify_scaled_point_spectra.py ignores the proposal and Newton traces and recomputes all 120 endpoint signs and the moment remainder.
The global contact lies slightly beyond ξ = π / 2 \xi=\pi/2 ξ = π /2 K ρ K_\rho K ρ D = 1 + 2 u cos ξ + u 2 D=1+2u\cos\xi+u^2 D = 1 + 2 u cos ξ + u 2
D ≥ sin 2 ξ , 1 D = 1 − 2 u cos ξ + ( 4 cos 2 ξ − 1 ) u 2 + 2 u 3 cos ξ D . D\ge\sin^2\xi,\qquad\frac{1}{D}=1-2u\cos\xi+\frac{(4\cos^2\xi-1)u^2+2u^3\cos\xi}{D}. D ≥ sin 2 ξ , D 1 = 1 − 2 u cos ξ + D ( 4 cos 2 ξ − 1 ) u 2 + 2 u 3 cos ξ . The listed terms use validated Laplace quadrature; this identity bounds the unlisted positive terms with Rayleigh remainders T 3 , T 4 T_3,T_4 T 3 , T 4 2 R a m + j e − h x R T m 2R^{am+j}e^{-hxR}T_m 2 R am + j e − h x R T m h x R > a m + j hxR>am+j h x R > am + j
K ρ ≥ 2 π x ∑ l i s t e d r e − h r x cos ( x r cos ω + ω ) − π x E 1 + a min ( cos ξ , 0 ) sin 2 ξ x a , K_\rho\ge2\pi x\sum_{\mathrm{listed}}re^{-hrx}\cos(xr\cos\omega+\omega)-\pi xE_1+\frac{a\min(\cos\xi,0)}{\sin^2\xi\,x^a}, K ρ ≥ 2 π x listed ∑ r e − h r x cos ( x r cos ω + ω ) − π x E 1 + sin 2 ξ x a a min ( cos ξ , 0 ) , where E 1 E_1 E 1 sin ξ / D \sin\xi/D sin ξ / D
The origin is controlled by the extremal kernel. The convergent germ with R = 5 / 2 R=5/2 R = 5/2 C = 128 C=128 C = 128 k a ≤ 3 / 2 k_a\le3/2 k a ≤ 3/2 0 ≤ y ≤ 2 0\le y\le2 0 ≤ y ≤ 2 y = 2 y=2 y = 2 0.895797 < 3 / 2 0.895797<3/2 0.895797 < 3/2 y y y k a ≤ 3 / 2 k_a\le3/2 k a ≤ 3/2 q q q Re j a ( y e − i δ ) > 4 / 5 \operatorname{Re}j_a(ye^{-i\delta})>4/5 Re j a ( y e − i δ ) > 4/5 0 ≤ y ≤ 6 / 5 0\le y\le6/5 0 ≤ y ≤ 6/5 δ = π ( a − 1 − ρ ) \delta=\pi(a^{-1}-\rho) δ = π ( a − 1 − ρ ) K ( c a y ) > 1 / 10 K(cay)>1/10 K ( c a y ) > 1/10
The finite certificate proves positivity on [ 6 / 5 , 6.4 ] ∪ [ 7 , 20 ] [6/5,6.4]\cup[7,20] [ 6/5 , 6.4 ] ∪ [ 7 , 20 ] [ 6.4 , 7 ] [6.4,7] [ 6.4 , 7 ] ∂ y 2 K ( c a y ) > 1 / 10 \partial_y^2K(cay)>1/10 ∂ y 2 K ( c a y ) > 1/10 K ρ > 1 K_\rho>1 K ρ > 1 ∂ y K \partial_yK ∂ y K q = 0.1220544 q=0.1220544 q = 0.1220544 y = 6.689706 y=6.689706 y = 6.689706 q = 0.1220545 q=0.1220545 q = 0.1220545 ( ∂ y K ) 2 / ( 2 / 10 ) (\partial_yK)^2/(2/10) ( ∂ y K ) 2 / ( 2/10 ) y = 6.6896 , 6.68973 y=6.6896,6.68973 y = 6.6896 , 6.68973 y ≥ 20 y\ge20 y ≥ 20 h , ξ h,\xi h , ξ x a x^a x a
The smaller local contact is proved by the same strict convexity, skewness increase, and opposite minimum signs on its separate rectangle. The negative value at y = 7 y=7 y = 7 [ 0.0574 , 0.0575 ] [0.0574,0.0575] [ 0.0574 , 0.0575 ] certify_eleven_tenths_id.py: 416 rational boxes, including both contacts, plus the independent sixty-root prerequisite and explicit origin and infinite-tail controls. The record is eleven_tenths_id_verification.json. □
Theorem F.6 (The complete middle ID region). *For every 6 / 5 ≤ a ≤ 7 / 5 6/5\le a\le7/5 6/5 ≤ a ≤ 7/5
c a = ( a − 1 ) 1 / a c_a=(a-1)^{1/a} c a = ( a − 1 ) 1/ a q = ( ρ − 1 / 2 ) / ( 1 / a − 1 / 2 ) q=(\rho-1/2)/(1/a-1/2) q = ( ρ − 1/2 ) / ( 1/ a − 1/2 )
I j = [ j / 100 , ( j + 1 ) / 100 ] , j = 120 , … , 139. I_j=[j/100,(j+1)/100],\qquad j=120,\ldots,139. I j = [ j /100 , ( j + 1 ) /100 ] , j = 120 , … , 139.
On each I j I_j I j Q j , Y j Q_j,Y_j Q j , Y j middle_id_region_manifest.json define the band
∣ q − Q j ( a ) ∣ ≤ 1 / 500 , ∣ y − Y j ( a ) ∣ ≤ 3 / 10 , x = c a y . \lvert q-Q_j(a)\rvert\le1/500,\qquad\lvert y-Y_j(a)\rvert\le3/10,\qquad x=c_a y. ∣ q − Q j ( a )∣ ≤ 1/500 , ∣ y − Y j ( a )∣ ≤ 3/10 , x = c a y . There is exactly one pair ( q ∗ ( a ) , y ∗ ( a ) ) (q_*(a),y_*(a)) ( q ∗ ( a ) , y ∗ ( a )) K a , ρ ( c a y ) = ∂ y K a , ρ ( c a y ) = 0 K_{a,\rho}(c_a y)=\partial_y K_{a,\rho}(c_a y)=0 K a , ρ ( c a y ) = ∂ y K a , ρ ( c a y ) = 0 r ( a ) = 1 / 2 + ( 1 / a − 1 / 2 ) q ∗ ( a ) r(a)=1/2+(1/a-1/2)q_*(a) r ( a ) = 1/2 + ( 1/ a − 1/2 ) q ∗ ( a )
X a , ρ is ID ⟺ ρ ∈ [ 1 − 1 / a , 1 − r ( a ) ] ∪ [ r ( a ) , 1 / a ] . X_{a,\rho}\text{ is ID}\quad\Longleftrightarrow\quad\rho\in[1-1/a,1-r(a)]\cup[r(a),1/a]. X a , ρ is ID ⟺ ρ ∈ [ 1 − 1/ a , 1 − r ( a )] ∪ [ r ( a ) , 1/ a ] . Equality is included. Combined with the preceding upper-region certificates, a unique controlling contact is now certified on the entire interval [ 6 / 5 , α ∗ ] [6/5,\alpha_*] [ 6/5 , α ∗ ]
Proof. On each strip, Lemma F.3 encloses twenty distinct moving Wright zeros using exact rational graph brackets. All endpoint signs are strict on complete real parameter covers; no completeness of the preliminary scan is assumed. A fourth Rayleigh remainder places every unlisted physical root above a positive constant R j R_j R j certify_scaled_moving_spectra.py.
The kernel calculation uses scaled roots r ^ n = ( a − 1 ) 1 / a r n = z n 1 / a \widehat r_n=(a-1)^{1/a}r_n=z_n^{1/a} r n = ( a − 1 ) 1/ a r n = z n 1/ a S ^ m = S m / ( a − 1 ) m \widehat S_m=S_m/(a-1)^m S m = S m / ( a − 1 ) m T 3 , T 4 T_3,T_4 T 3 , T 4 2 R ^ a m + k e − h y R ^ T ^ m 2\widehat R^{am+k}e^{-hy\widehat R}\widehat T_m 2 R am + k e − h y R T m h y R ^ > a m + k hy\widehat R>am+k h y R > am + k
sin ξ π [ y − a a − 1 − 2 a cos ξ Γ ( 2 a ) S ^ 2 y − 2 a − a Γ ( 3 a ) S ^ 3 y − 3 a ] . \frac{\sin\xi}{\pi}\left[\frac{y^{-a}}{a-1}-2a\cos\xi\,\Gamma(2a)\widehat S_2y^{-2a}-a\Gamma(3a)\widehat S_3y^{-3a}\right]. π sin ξ [ a − 1 y − a − 2 a cos ξ Γ ( 2 a ) S 2 y − 2 a − a Γ ( 3 a ) S 3 y − 3 a ] . For origin control, a convergent germ proves j a ≤ 3 / 2 j_a\le3/2 j a ≤ 3/2 [ 0 , 2 ] [0,2] [ 0 , 2 ] y = 2 y=2 y = 2 Re j a ( y e − i δ ) > 4 / 5 \operatorname{Re}j_a(ye^{-i\delta})>4/5 Re j a ( y e − i δ ) > 4/5 0 ≤ y ≤ 6 / 5 0\le y\le6/5 0 ≤ y ≤ 6/5 K ( c a y ) > 1 / 10 K(c_a y)>1/10 K ( c a y ) > 1/10
The finite cover proves positivity on [ 6 / 5 , Y j ( a ) − 3 / 10 ] ∪ [ Y j ( a ) + 3 / 10 , 7 ] [6/5,Y_j(a)-3/10]\cup[Y_j(a)+3/10,7] [ 6/5 , Y j ( a ) − 3/10 ] ∪ [ Y j ( a ) + 3/10 , 7 ]
∂ y 2 K ( c a y ) > 1 / 100 , K ρ ( c a y ) > 1 / 2 , \partial_y^2K(c_a y)>1/100,\qquad K_\rho(c_a y)>1/2, ∂ y 2 K ( c a y ) > 1/100 , K ρ ( c a y ) > 1/2 , and opposite radial derivative signs at the two endpoints. At q = Q j ( a ) − 1 / 500 q=Q_j(a)-1/500 q = Q j ( a ) − 1/500 y = Y j ( a ) y=Y_j(a) y = Y j ( a ) q = Q j ( a ) + 1 / 500 q=Q_j(a)+1/500 q = Q j ( a ) + 1/500 ( ∂ y K ) 2 / ( 2 / 100 ) (\partial_yK)^2/(2/100) ( ∂ y K ) 2 / ( 2/100 )
For the remaining infinite tail, at y = 7 y=7 y = 7 cos ξ < 0 \cos\xi<0 cos ξ < 0 7 h r ^ 1 > a 7h\widehat r_1>a 7 h r 1 > a 7 h R ^ > a 7h\widehat R>a 7 h R > a y a y^a y a y ≥ 7 y\ge7 y ≥ 7
The checker certify_middle_id_region.py independently recomputes the germ prerequisites, all spectral signs, and all 6733 kernel boxes. The manifest lists the twenty verified strips, their exact locators, and hashes of the unchanged proof inputs. Consequently the zero minimum is global and unique. Endpoint ID and the interval theorem give the entire skewness section and reflection gives its lower half. At common strip endpoints, global uniqueness identifies the two constructions. ▫
Theorem F.7 (Competing global ID contacts). Put I = [ 29 / 25 , 117 / 100 ] I = [29/25, 117/100] I = [ 29/25 , 117/100 ] t = a − 233 / 200 t = a - 233/200 t = a − 233/200 and define the following exact rational locator polynomials:
Q 1 ( a ) = 0.1751650016654 + 1.70766807201 t − 0.6820005682617 t 2 , Q 2 ( a ) = 0.1728668598909 + 0.8093353588945 t + 0.7122035515794 t 2 , Y 1 ( a ) = 3.865102673082 + 1.069198034134 t − 0.8285962464605 t 2 , Y 2 ( a ) = 6.964612825586 + 3.897040936013 t − 6.114433090261 t 2 . \begin{aligned}
Q_1(a) &= 0.1751650016654 + 1.70766807201t - 0.6820005682617t^2,\\
Q_2(a) &= 0.1728668598909 + 0.8093353588945t + 0.7122035515794t^2,\\
Y_1(a) &= 3.865102673082 + 1.069198034134t - 0.8285962464605t^2,\\
Y_2(a) &= 6.964612825586 + 3.897040936013t - 6.114433090261t^2.
\end{aligned} Q 1 ( a ) Q 2 ( a ) Y 1 ( a ) Y 2 ( a ) = 0.1751650016654 + 1.70766807201 t − 0.6820005682617 t 2 , = 0.1728668598909 + 0.8093353588945 t + 0.7122035515794 t 2 , = 3.865102673082 + 1.069198034134 t − 0.8285962464605 t 2 , = 6.964612825586 + 3.897040936013 t − 6.114433090261 t 2 . For every a ∈ I a \in I a ∈ I each band
∣ q − Q i ( a ) ∣ ≤ 1 / 2000 , ∣ y − Y i ( a ) ∣ ≤ 3 / 10 |q - Q_i(a)| \le1/2000,\qquad|y - Y_i(a)| \le3/10 ∣ q − Q i ( a ) ∣ ≤ 1/2000 , ∣ y − Y i ( a ) ∣ ≤ 3/10 contains a unique local contact ( q i ( a ) , y i ( a ) ) (q_i(a), y_i(a)) ( q i ( a ) , y i ( a )) satisfying K ( c a y ) = ∂ y K ( c a y ) = 0 K(c_a y) = \partial_y K(c_a y) = 0 K ( c a y ) = ∂ y K ( c a y ) = 0 Both contact curves are continuous, and the two radial cores are disjoint. The complete upper-half classification is
X a , ρ is ID ⟺ q ≥ max { q 1 ( a ) , q 2 ( a ) } , 0 < q ≤ 1. (87) X_{a,\rho}\text{ is ID}\quad\Longleftrightarrow\quad q \ge\max\{q_1(a),q_2(a)\},\qquad0 < q \le1. \tag*{(87)} X a , ρ is ID ⟺ q ≥ max { q 1 ( a ) , q 2 ( a )} , 0 < q ≤ 1. (87) Reflection gives the lower half and equality is included.
There is at least one a † ∈ ( 1.1613 , 1.1636 ) a_\dagger\in(1.1613,1.1636) a † ∈ ( 1.1613 , 1.1636 ) q 1 ( a † ) = q 2 ( a † ) q_1(a_\dagger) = q_2(a_\dagger) q 1 ( a † ) = q 2 ( a † ) a † a_\dagger a †
Proof. Forty moving Wright roots on I I I certify_crossover_spectrum.py.
Write J i ( a ) = [ Y i ( a ) − 3 / 10 , Y i ( a ) + 3 / 10 ] J_i(a) = [Y_i(a)-3/10,Y_i(a)+3/10] J i ( a ) = [ Y i ( a ) − 3/10 , Y i ( a ) + 3/10 ] η = 1 / 2000 \eta= 1/2000 η = 1/2000
∂ y 2 K ( c a y ) > 1 / 100 , \partial_y^2 K(c_a y) > 1/100, ∂ y 2 K ( c a y ) > 1/100 , opposite radial derivative signs at the endpoints of J i ( a ) J_i(a) J i ( a ) q = Q i ( a ) − η q = Q_i(a)-\eta q = Q i ( a ) − η y = Y i ( a ) y = Y_i(a) y = Y i ( a ) K − ( ∂ y K ) 2 / ( 2 / 100 ) K - (\partial_y K)^2/(2/100) K − ( ∂ y K ) 2 / ( 2/100 ) q = Q i ( a ) + η q = Q_i(a)+\eta q = Q i ( a ) + η y = Y i ( a ) y = Y_i(a) y = Y i ( a )
throughout the entire skewness envelope [ min ( Q 1 , Q 2 ) − η , max ( Q 1 , Q 2 ) + η ] [\min(Q_1,Q_2)-\eta,\max(Q_1,Q_2)+\eta] [ min ( Q 1 , Q 2 ) − η , max ( Q 1 , Q 2 ) + η ] a a a
The exterior certificate uses only the narrower band
q ∈ [ max ( Q 1 , Q 2 ) − η , max ( Q 1 , Q 2 ) + η ] . q \in[\max(Q_1,Q_2)-\eta,\max(Q_1,Q_2)+\eta]. q ∈ [ max ( Q 1 , Q 2 ) − η , max ( Q 1 , Q 2 ) + η ] . It proves strict positivity off J 1 ( a ) ∪ J 2 ( a ) J_1(a)\cup J_2(a) J 1 ( a ) ∪ J 2 ( a ) j a ≤ 3 / 2 j_a \le3/2 j a ≤ 3/2 y = 6 / 5 y = 6/5 y = 6/5 y = 10 y = 10 y = 10 y a y^a y a
Set q ∗ = max ( q 1 , q 2 ) q_* = \max(q_1,q_2) q ∗ = max ( q 1 , q 2 ) q i q_i q i J i J_i J i q ∗ q_* q ∗ q ∗ q_* q ∗ q ∗ q_* q ∗
At the endpoints of I I I
Q 2 ( 1.16 ) − Q 1 ( 1.16 ) > 0.0022283 > 2 η , Q 1 ( 1.17 ) − Q 2 ( 1.17 ) > 0.0067549 > 2 η . Q_{2}(1.16)-Q_{1}(1.16)>0.0022283>2\eta,\qquad Q_{1}(1.17)-Q_{2}(1.17)>0.0067549>2\eta. Q 2 ( 1.16 ) − Q 1 ( 1.16 ) > 0.0022283 > 2 η , Q 1 ( 1.17 ) − Q 2 ( 1.17 ) > 0.0067549 > 2 η . Thus the order of the true contact values reverses, proving existence of a double contact by continuity. For localization, D = Q 1 − Q 2 D=Q_{1}-Q_{2} D = Q 1 − Q 2 I I I
D ( 1.1613 ) < − 0.0010447 < − 2 η , D ( 1.1636 ) > 0.0010377 > 2 η . D(1.1613)<-0.0010447<-2\eta,\qquad D(1.1636)>0.0010377>2\eta. D ( 1.1613 ) < − 0.0010447 < − 2 η , D ( 1.1636 ) > 0.0010377 > 2 η . Every equality q 1 = q 2 q_{1}=q_{2} q 1 = q 2
The full verifier certify_id_crossover.py independently recomputes all spectral prerequisites and all 1924 kernel boxes. Its record is id_crossover_verification.json; the exact localization checks are in id_crossover_localization.json. □
Corollary F.8 (Complete lower contact cover). The complete ID classification is effectively certified throughout 11 / 10 ≤ a ≤ α ∗ 11/10\leq a\leq\alpha_* 11/10 ≤ a ≤ α ∗ [ 11 / 10 , 29 / 25 ] [11/10,29/25] [ 11/10 , 29/25 ] Theorem F.7 . On [ 117 / 100 , α ∗ ] [117/100,\alpha_*] [ 117/100 , α ∗ ] [ 29 / 25 , 117 / 100 ] [29/25,117/100] [ 29/25 , 117/100 ]
Proof. Nine further adjacent strips use the single-core specialization of the proof of Theorem F.7 . The quadratic locators are recorded as exact decimal rationals in lower_contact_region_manifest.json. Each has skewness half-width 1 / 2000 1/2000 1/2000 3 / 10 3/10 3/10 certify_single_contact_region.py recomputes the root signs, moment remainder, germ, convexity, skewness derivative, endpoint signs, and the entire exterior. The latter includes the origin and all y ≥ 10 y\geq10 y ≥ 10
Table 1 (PDF p. 65).
There are 147740 kernel boxes and 970 root-sign boxes in total, including the two-core strip. The other nine strips use thirty moving roots each. Their graph separation is checked at each common parameter, as in the two-core proof. The manifest checks the exact adjacent cover and the hashes of the completed independent verifications; it does not substitute for those verifications. Combining this cover with the middle and upper classifications proves the assertion. □
Effective approximation of the boundaries
The strict sign criteria become terminating approximation procedures once the origin, tail and omitted spectrum have explicit bounds. The ID and GGC constructions below do not require contact uniqueness.
Effective boundary computation
Lemma G.1 (Pole cancellation at the ID origin). Let 1 < a ≤ α ∗ 1<a\leq\alpha_* 1 < a ≤ α ∗ c a = ( a − 1 ) 1 / a c_a=(a-1)^{1/a} c a = ( a − 1 ) 1/ a ∣ k a ( z ) − 3 / 2 ∣ ≤ η < 1 / 2 |k_a(z)-3/2|\leq\eta<1/2 ∣ k a ( z ) − 3/2∣ ≤ η < 1/2 ∣ z ∣ ≤ D |z|\leq D ∣ z ∣ ≤ D
K a , ρ ( x ) ≥ 3 / 2 − 3 η > 0 ( 0 < x ≤ D / 2 ) . K_{a,\rho}(x)\geq3/2-3\eta>0 \qquad(0<x\leq D/2). K a , ρ ( x ) ≥ 3/2 − 3 η > 0 ( 0 < x ≤ D /2 ) . One may take D = c a / 200 D = c_a/200 D = c a /200 η = 12 / 25 \eta= 12/25 η = 12/25 1 < a ≤ α ∗ 1 < a \le\alpha_* 1 < a ≤ α ∗ D = c a / 3 D = c_a/3 D = c a /3 η = 17 / 40 \eta= 17/40 η = 17/40 1 < a ≤ 13 / 10 1 < a \le13/10 1 < a ≤ 13/10
Proof. Write δ = π ( 1 / a − ρ ) \delta= \pi(1/a-\rho) δ = π ( 1/ a − ρ ) b a = a c a / ( a − 1 ) = ∫ k a b_a = ac_a/(a-1) = \int k_a b a = a c a / ( a − 1 ) = ∫ k a
K ( x ) = 2 Re k a ( x e − i δ ) + b a sin δ π x − x sin δ π ∫ 0 ∞ k a ( u ) d u ( u − x cos δ ) 2 + x 2 sin 2 δ . K(x) = 2\operatorname{Re} k_a(xe^{-i\delta}) + \frac{b_a\sin\delta}{\pi x} - \frac{x\sin\delta}{\pi}\int_0^\infty\frac{k_a(u)\,du}{(u-x\cos\delta)^2+x^2\sin^2\delta}. K ( x ) = 2 Re k a ( x e − i δ ) + π x b a sin δ − π x sin δ ∫ 0 ∞ ( u − x cos δ ) 2 + x 2 sin 2 δ k a ( u ) d u . The normalized integral over [ 0 , D ] [0,D] [ 0 , D ] 3 / 2 + η 3/2+\eta 3/2 + η x b a sin δ / [ π ( D − x ) 2 ] xb_a\sin\delta/[\pi(D-x)^2] x b a sin δ / [ π ( D − x ) 2 ] k a ≥ 0 k_a \ge0 k a ≥ 0 x ≤ D / 2 x \le D/2 x ≤ D /2 3 − 2 η 3-2\eta 3 − 2 η δ = 0 \delta=0 δ = 0
For completeness the uniform germ constants follow from the inverse equation with radius R = 1 R=1 R = 1 M = 9 M=9 M = 9 log 9 > 2 + 1 / 9 \log9 > 2+1/9 log 9 > 2 + 1/9 e u ≥ 1 + u e^u \ge1+u e u ≥ 1 + u a > 1 a>1 a > 1 a ≤ 2 a \le2 a ≤ 2 ∣ j a ( y ) − 3 / 2 ∣ ≤ e 77 ∣ y ∣ − 1 \lvert j_a(y)-3/2\rvert\le e^{77\lvert y\rvert}-1 ∣ j a ( y ) − 3/2 ∣ ≤ e 77 ∣ y ∣ − 1 ∣ y ∣ ≤ 1 / 2 \lvert y\rvert\le1/2 ∣ y ∣ ≤ 1/2 e 77 / 200 − 1 < 12 / 25 e^{77/200}-1 < 12/25 e 77/200 − 1 < 12/25 ∣ y ∣ ≤ 1 / 3 \lvert y\rvert\le1/3 ∣ y ∣ ≤ 1/3 1 ≤ a ≤ 13 / 10 1 \le a \le13/10 1 ≤ a ≤ 13/10 0.423666806557 < 17 / 40 0.423666806557 < 17/40 0.423666806557 < 17/40 certify_universal_id_origin.py and certify_near_one_id_origin.py. □
Theorem G.2 (Constructive critical boundaries). For each computable 1 < a < α ∗ 1<a<\alpha_* 1 < a < α ∗ r ( a ) r(a) r ( a ) g ( a ) g(a) g ( a ) 4 / 5 < a < 1 4/5<a<1 4/5 < a < 1
ID part. Fix 1 / 2 < ρ < 1 / a 1/2<\rho<1/a 1/2 < ρ < 1/ a h = sin ω > 0 h=\sin\omega>0 h = sin ω > 0 S j = ∑ r n − a j S_j=\sum r_n^{-aj} S j = ∑ r n − aj ( 1 + u a ) − 2 ≥ 1 − 2 u a (1+u^a)^{-2}\ge1-2u^a ( 1 + u a ) − 2 ≥ 1 − 2 u a e − v ≤ ( 2 a / e ) 2 a v − 2 a e^{-v}\le(2a/e)^{2av-2a} e − v ≤ ( 2 a / e ) 2 a v − 2 a
K ( x ) ≥ sin ξ π x a − C 2 x − 2 a , C 2 = S 2 [ 2 a Γ ( 2 a ) sin ξ π + 2 ( 2 a e h ) 2 a ] . K(x)\ge\frac{\sin\xi}{\pi x^a}-C_2x^{-2a},\qquad C_2=S_2\left[\frac{2a\Gamma(2a)\sin\xi}{\pi}+2\left(\frac{2a}{eh}\right)^{2a}\right]. K ( x ) ≥ π x a sin ξ − C 2 x − 2 a , C 2 = S 2 [ π 2 a Γ ( 2 a ) sin ξ + 2 ( e h 2 a ) 2 a ] . Thus X = ( 2 π C 2 / sin ξ ) 1 / a X=(2\pi C_2/\sin\xi)^{1/a} X = ( 2 π C 2 / sin ξ ) 1/ a Lemma G.1 supplies a computable origin cutoff x 0 > 0 x_0>0 x 0 > 0
All Wright zeros are negative real. Computable entire-series tails and argument-principle isolation on rational rectangles enumerate them with multiplicities; enumerate nearby rectangle boundaries to avoid zeros on a contour. For any finite spectral part put T 1 = S 1 − ∑ i n c l u d e d r n − a T_1=S_1-\sum_{\mathrm{included}}r_n^{-a} T 1 = S 1 − ∑ included r n − a [ x 0 , X ] [x_0,X] [ x 0 , X ]
x − a T 1 [ a Γ ( a ) π sin ξ + 2 ( a / e ) a h − a ] , x^{-a}T_1\left[\frac{a\Gamma(a)}{\pi\sin\xi}+2(a/e)^a h^{-a}\right], x − a T 1 [ π sin ξ a Γ ( a ) + 2 ( a / e ) a h − a ] , which tends uniformly to zero. Finite integral terms have explicit exponential tails. Increasing precision, spectral truncation and radial subdivision therefore gives convergent whole-interval enclosures.
For ρ > r ( a ) \rho>r(a) ρ > r ( a ) ρ < r ( a ) \rho<r(a) ρ < r ( a ) p < q p<q p < q p p p q q q q − p q-p q − p
Lemma G.3 (All-angle parabolic descent). For 0 < b ≤ 1 / 4 0<b\le1/4 0 < b ≤ 1/4 a = 1 − b a=1-b a = 1 − b c = cos ( 2 π b ) c=\cos(2\pi b) c = cos ( 2 π b ) k = a / b k=a/b k = a / b
κ b = ( 1 − c ) ( c + k ) 2 { 1 + 3 c + ( 1 − c ) k } > 0. \kappa_b=\frac{(1-c)(c+k)}{2\{1+3c+(1-c)k\}}>0. κ b = 2 { 1 + 3 c + ( 1 − c ) k } ( 1 − c ) ( c + k ) > 0. :::`
The phase in the parabolic representation satisfies Re Φ ≤ − κ b x 2 \operatorname{Re} \Phi\le-\kappa_b x^2 Re Φ ≤ − κ b x 2 0 ≤ b ≤ 1 / 4 0 \le b \le1/4 0 ≤ b ≤ 1/4 ∣ arg v ∣ ≤ π ρ 0 < π |\arg v| \le\pi\rho_0 < \pi ∣ arg v ∣ ≤ π ρ 0 < π
Re Φ ≤ − ( 1 − ρ 0 ) x 2 / 4. \operatorname{Re} \Phi\le-(1-\rho_0)x^2/4. Re Φ ≤ − ( 1 − ρ 0 ) x 2 /4. The logarithm is the continuous parabolic-contour logarithm.
Proof. Set h = x / 2 h=x/\sqrt{2} h = x / 2 λ = v \lambda=\sqrt{v} λ = v Re λ > 0 \operatorname{Re}\lambda>0 Re λ > 0
( λ + i h ) 2 − λ 2 − λ 2 − 2 b { ( λ + i h ) 2 b − λ 2 b } / b . (\lambda+ih)^2-\lambda^2-\lambda^{2-2b}\{(\lambda+ih)^{2b}-\lambda^{2b}\}/b. ( λ + ih ) 2 − λ 2 − λ 2 − 2 b {( λ + ih ) 2 b − λ 2 b } / b . At λ = i y \lambda=iy λ = i y X = ∣ y + h ∣ X=|y+h| X = ∣ y + h ∣ Y = ∣ y ∣ Y=|y| Y = ∣ y ∣ X 2 + κ Y 2 − d X 2 b Y 2 a / b X^2+\kappa Y^2-dX^{2b}Y^{2a/b} X 2 + κ Y 2 − d X 2 b Y 2 a / b d = 1 d=1 d = 1 d = c d=c d = c X 2 b Y 2 a ≤ 2 b X Y + ( 1 − 2 b ) Y 2 X^{2b}Y^{2a}\le2bXY+(1-2b)Y^2 X 2 b Y 2 a ≤ 2 b X Y + ( 1 − 2 b ) Y 2 h 2 h^2 h 2 X + Y = ∣ h ∣ X+Y=|h| X + Y = ∣ h ∣ X 2 + [ c + ( 1 − c ) κ ] Y 2 − 2 c X Y X^2+[c+(1-c)\kappa]Y^2-2cXY X 2 + [ c + ( 1 − c ) κ ] Y 2 − 2 c X Y 2 κ b h 2 2\kappa_b h^2 2 κ b h 2 Y = 0 Y=0 Y = 0 2 κ b ≤ 1 2\kappa_b\le1 2 κ b ≤ 1 − 2 a h 2 -2ah^2 − 2 a h 2
For h > 0 h>0 h > 0 h 2 ( 1 / 2 + arg λ / π ) h^2(1/2+\arg\lambda/\pi) h 2 ( 1/2 + arg λ / π ) h 2 h^2 h 2 h < 0 h<0 h < 0 arg λ \arg\lambda arg λ ∣ arg λ ∣ ≤ π ρ 0 / 2 |\arg\lambda|\le\pi\rho_0/2 ∣ arg λ ∣ ≤ π ρ 0 /2 b = 0 b=0 b = 0
GGC part of Theorem G.2. For a fixed trial 1 / 2 < ρ < 1 1/2<\rho<1 1/2 < ρ < 1 s = ∣ v ∣ − 1 / 2 s=|v|^{-1/2} s = ∣ v ∣ − 1/2 s = 0 s=0 s = 0
A positive certificate consists of a complete radial partition with positive normalized ℑ H \Im H ℑ H Z ( v ) = ( 1 + v ) 3 / 2 e a v / b W a ( − v a / b ) Z(v)=(1+v)^{3/2}e^{av/b}W_a(-v^{a/b}) Z ( v ) = ( 1 + v ) 3/2 e a v / b W a ( − v a / b )
Every strict interior trial has such a finite certificate by compactness and strict positivity. If a trial lies beyond the boundary, the Hopf inequality at the first contact gives an open negative set at a smaller angle. Enumeration of rational angles and radii must find a strict negative witness. Two distinct trisection trials avoid stalling when one equals the boundary. Increasing precision, integration cutoffs and subdivision budgets therefore encloses the endpoint to any requested width. The finite strict tests also handle computable-real indices by interval names. The rational-index implementation is compute_ggc_boundary.py; its independent verifier repeats the cover, winding and negative witness. □
Appendix H. Boundary asymptotics at index one
We first obtain the leading ID scale from uniform spectral estimates. We then localize the GGC contact and refine the ID expansion to second order. The last subsection records finite slices and the unit-origin estimate used in the ID proof. Qualitative eventual assertions are kept separate from certified numerical thresholds.
The exact ID boundary scale at index one
H.1. The exact ID boundary scale at index one. The compact germ limit alone does not control a growing number of Wright zeros. The next two lemmas supply the missing uniform information: a counting estimate from endpoint ID, and a phase estimate from a parameter-uniform saddle integral.
Lemma H.1 (Uniform spectral heat trace). Put e = a − 1 e=a-1 e = a − 1 and let r ^ n = ( e λ n ) 1 / a \hat r_n=(e\lambda_n)^{1/a} r ^ n = ( e λ n ) 1/ a be the scaled positive Wright roots. Define
N a ( R ) = # { n : r ^ n ≤ R } , L a ( v ) = ∑ n e − v r ^ n . N_a(R)=\#\{n:\hat r_n\le R\},\qquad\mathcal{L}_a(v)=\sum_n e^{-v\hat r_n}. N a ( R ) = # { n : r ^ n ≤ R } , L a ( v ) = n ∑ e − v r ^ n . For a > 1 a>1 a > 1 sufficiently close to one,
N a ( R ) ≤ 4 R ( R > 0 ) . (88) N_a(R)\le4R\qquad(R>0). \tag*{(88)} N a ( R ) ≤ 4 R ( R > 0 ) . (88) Moreover,
v L a ( v ) ⟶ 1 ( a ↓ 1 , v ↓ 0 ) , sup v > 0 v a L a ( v ) ⟶ 1 ( a ↓ 1 ) . (89) v\mathcal{L}_a(v)\longrightarrow1\qquad(a\downarrow1,v\downarrow0),\qquad\sup_{v>0}v^a\mathcal{L}_a(v)\longrightarrow1\qquad(a\downarrow1). \tag*{(89)} v L a ( v ) ⟶ 1 ( a ↓ 1 , v ↓ 0 ) , v > 0 sup v a L a ( v ) ⟶ 1 ( a ↓ 1 ) . (89) The first limit is joint, with no restriction on the relative rates of its two parameters.
Proof. The scaled extremal variable Z a Z_a Z a j a ( u ) / u j_a(u)/u j a ( u ) / u j a ≥ 0 j_a\ge0 j a ≥ 0
Ψ a ( s ) = − log E e − s Z a = a s e − log W a ( s a / e ) , J a ( s ) = Ψ a ′ ( s ) = ∫ 0 ∞ e − s u j a ( u ) d u . \Psi_a(s)=-\log\mathbb{E}e^{-sZ_a}=\frac{as}{e}-\log W_a(s^a/e),\qquad J_a(s)=\Psi'_a(s)=\int_0^\infty e^{-su}j_a(u)\,du. Ψ a ( s ) = − log E e − s Z a = e a s − log W a ( s a / e ) , J a ( s ) = Ψ a ′ ( s ) = ∫ 0 ∞ e − s u j a ( u ) d u . For T j , a ( s ) = ∑ n ( 1 + ( r ^ n / s ) a ) − j T_{j,a}(s)=\sum_n(1+(\hat r_n/s)^a)^{-j} T j , a ( s ) = ∑ n ( 1 + ( r ^ n / s ) a ) − j
T 1 , a ( s ) = s e − s a J a ( s ) , T 2 , a ( s ) = s a − e s a 2 J a ( s ) + s 2 a 2 J a ′ ( s ) . T_{1,a}(s)=\frac{s}{e}-\frac{s}{a}J_a(s),\qquad T_{2,a}(s)=\frac{s}{a}-\frac{es}{a^2}J_a(s)+\frac{s^2}{a^2}J'_a(s). T 1 , a ( s ) = e s − a s J a ( s ) , T 2 , a ( s ) = a s − a 2 es J a ( s ) + a 2 s 2 J a ′ ( s ) . Both J a ≥ 0 J_a\ge0 J a ≥ 0 J a ′ ≤ 0 J'_a\le0 J a ′ ≤ 0 T 2 , a ( s ) ≤ s / a T_{2,a}(s)\le s/a T 2 , a ( s ) ≤ s / a j a ( u ) j_a(u) j a ( u )
J a ( 1 ) = a V e ′ ( − 1 ) V e ( − 1 ) = O ( 1 ) , J_a(1)=a\frac{V'_e(-1)}{V_e(-1)}=O(1), J a ( 1 ) = a V e ( − 1 ) V e ′ ( − 1 ) = O ( 1 ) , by Lemma F.1 , since V 0 ( − 1 ) = 1 / 2 V_0(-1)=1/2 V 0 ( − 1 ) = 1/2 s ≥ 2 s\ge2 s ≥ 2 J a ( s ) = O ( s − 1 ) J_a(s)=O(s^{-1}) J a ( s ) = O ( s − 1 ) J a ′ ( s ) = O ( s − 2 ) J'_a(s)=O(s^{-2}) J a ′ ( s ) = O ( s − 2 ) e − u j a ( u ) d u e^{-u}j_a(u)\,du e − u j a ( u ) d u
T 2 , a ( s ) = s / a + O ( 1 ) , s ≥ 2 , (90) T_{2,a}(s)=s/a+O(1),\qquad s\ge2, \tag*{(90)} T 2 , a ( s ) = s / a + O ( 1 ) , s ≥ 2 , (90) with a uniform constant.
Consider any sequences a ↓ 1 a\downarrow1 a ↓ 1 v ↓ 0 v\downarrow0 v ↓ 0 μ a , v = v ∑ n δ v r ^ n \mu_{a,v}=v\sum_n\delta_{v\hat r_n} μ a , v = v ∑ n δ v r ^ n μ a , v ( [ 0 , R ] ) ≤ 4 R \mu_{a,v}([0,R])\le4R μ a , v ([ 0 , R ]) ≤ 4 R t > 0 t>0 t > 0
∫ 0 ∞ μ a , v ( d u ) ( 1 + ( u / t ) a ) 2 = v T 2 , a ( t / v ) ⟶ t . \int_0^\infty\frac{\mu_{a,v}(du)}{(1+(u/t)^a)^2}=vT_{2,a}(t/v)\longrightarrow t. ∫ 0 ∞ ( 1 + ( u / t ) a ) 2 μ a , v ( d u ) = v T 2 , a ( t / v ) ⟶ t . Every vague subsequential limit therefore satisfies ∫ ( u + t ) − 2 μ ( d u ) = 1 / t \int(u+t)^{-2}\mu(du)=1/t ∫ ( u + t ) − 2 μ ( d u ) = 1/ t μ \mu μ ( u + t ) − 2 = ∫ 0 ∞ w e − w ( u + t ) d w (u+t)^{-2}=\int_0^\infty we^{-w(u+t)}\,dw ( u + t ) − 2 = ∫ 0 ∞ w e − w ( u + t ) d w e − u e^{-u} e − u
On every fixed compact interval of v > 0 v>0 v > 0 V e → 1 / Γ ( 2 − z ) V_e\to1/\Gamma(2-z) V e → 1/Γ ( 2 − z )
L a ( v ) ⟶ e − 2 v 1 − e − v . \mathcal{L}_a(v)\longrightarrow\frac{e^{-2v}}{1-e^{-v}}. L a ( v ) ⟶ 1 − e − v e − 2 v . The limit satisfies v L 1 ( v ) < 1 v\mathcal{L}_1(v)<1 v L 1 ( v ) < 1 v a L a ( v ) v^a\mathcal{L}_a(v) v a L a ( v ) v v v v ≤ 1 v\le1 v ≤ 1 v a − 1 ≤ 1 v^{a-1}\le1 v a − 1 ≤ 1 v = e v=e v = e e e → 1 e^e\to1 e e → 1 □ \square □
Lemma H.2 (Uniform large-root phase). There are constants e 0 > 0 e_0 > 0 e 0 > 0 R 0 R_0 R 0 C 0 C_0 C 0 r = r ^ n ≥ R 0 r = \widehat{r}_n \ge R_0 r = r n ≥ R 0 0 < e ≤ e 0 0 < e \le e_0 0 < e ≤ e 0
dist ( r , Z ) ≤ C 0 ( e + e 2 r + r − 1 ) . (91) \operatorname{dist}(r,\mathbb{Z}) \le C_0(e + e^2r + r^{-1}). \tag*{(91)} dist ( r , Z ) ≤ C 0 ( e + e 2 r + r − 1 ) . (91) Proof. Take e 0 < 1 / 2 e_0 < 1/2 e 0 < 1/2 W 1 + e ( − r 1 + e / e ) W_{1+e}(-r^{1+e}/e) W 1 + e ( − r 1 + e / e ) cos ( π e ) > 0 \cos(\pi e) > 0 cos ( π e ) > 0
W 1 + e ( − r 1 + e / e ) = − 1 π Im I e ( r ) , I e ( r ) = ∫ 0 ∞ u − 2 exp { − u − r 1 + e e − i π e u − e / e } d u . W_{1+e}(-r^{1+e}/e) = -\frac{1}{\pi}\operatorname{Im} I_e(r), \qquad I_e(r) = \int_0^\infty u^{-2}\exp\{-u-r^{1+e}e^{-i\pi e}u^{-e}/e\}\,du. W 1 + e ( − r 1 + e / e ) = − π 1 Im I e ( r ) , I e ( r ) = ∫ 0 ∞ u − 2 exp { − u − r 1 + e e − iπ e u − e / e } d u . Put η = π e / ( 1 + e ) \eta= \pi e/(1 + e) η = π e / ( 1 + e ) k = e − i η k = e^{-i\eta} k = e − i η u = r k t u = rkt u = r k t t > 0 t > 0 t > 0 Re k > 0 \operatorname{Re} k > 0 Re k > 0
I e ( r ) = e − ( 1 + e ) r k / e r k ∫ 0 ∞ t − 2 e − r k f e ( t ) d t , f e ( t ) = t − 1 + t − e − 1 e . I_e(r) = \frac{e^{-(1+e)rk/e}}{rk}\int_0^\infty t^{-2}e^{-rkf_e(t)}\,dt,\qquad f_e(t)=t-1+\frac{t^{-e}-1}{e}. I e ( r ) = r k e − ( 1 + e ) r k / e ∫ 0 ∞ t − 2 e − r k f e ( t ) d t , f e ( t ) = t − 1 + e t − e − 1 . The extension at zero is f 0 ( t ) = t − 1 − log t f_0(t)=t-1-\log t f 0 ( t ) = t − 1 − log t t > 0 t>0 t > 0 f e ( t ) ≥ f 0 ( t ) ≥ 0 f_e(t)\ge f_0(t)\ge0 f e ( t ) ≥ f 0 ( t ) ≥ 0 t = 1 t=1 t = 1 f e ′ ′ ( 1 ) = 1 + e f_e''(1)=1+e f e ′′ ( 1 ) = 1 + e
Uniform Laplace expansion at this minimum yields
∫ 0 ∞ t − 2 e − r k f e ( t ) d t = 2 π ( 1 + e ) r k { 1 + O ( r − 1 ) } , \int_0^\infty t^{-2}e^{-rkf_e(t)}\,dt
=\sqrt{\frac{2\pi}{(1+e)rk}}\{1+O(r^{-1})\}, ∫ 0 ∞ t − 2 e − r k f e ( t ) d t = ( 1 + e ) r k 2 π { 1 + O ( r − 1 )} , uniformly for 0 ≤ e ≤ e 0 0\le e\le e_0 0 ≤ e ≤ e 0 t = 1 t=1 t = 1 O ( r − 1 ) O(r^{-1}) O ( r − 1 ) f e ≥ f 0 f_e\ge f_0 f e ≥ f 0 Re k ≥ cos ( π e 0 / ( 1 + e 0 ) ) > 0 \operatorname{Re}k\ge\cos(\pi e_0/(1+e_0))>0 Re k ≥ cos ( π e 0 / ( 1 + e 0 )) > 0 ∫ t − 2 e − R Re ( k ) f 0 ( t ) d t \int t^{-2}e^{-R\operatorname{Re}(k)f_0(t)}\,dt ∫ t − 2 e − R Re ( k ) f 0 ( t ) d t R R R
It follows that
I e ( r ) = 2 π 1 + e ( r k ) − 3 / 2 e − ( 1 + e ) r k / e { 1 + O ( r − 1 ) } . I_e(r)=\sqrt{\frac{2\pi}{1+e}}(rk)^{-3/2}e^{-(1+e)rk/e}\{1+O(r^{-1})\}. I e ( r ) = 1 + e 2 π ( r k ) − 3/2 e − ( 1 + e ) r k / e { 1 + O ( r − 1 )} . At a zero of W W W π \pi π
d e r + 3 e 2 ( 1 + e ) , d e = sin η η = 1 + O ( e 2 ) . d_e r+\frac{3e}{2(1+e)},\qquad d_e=\frac{\sin\eta}{\eta}=1+O(e^2). d e r + 2 ( 1 + e ) 3 e , d e = η sin η = 1 + O ( e 2 ) . The distance of this quantity to an integer is O ( r − 1 ) O(r^{-1}) O ( r − 1 )
Theorem H.3 (Sharp ID boundary asymptotic). For the continuous boundary r ( a ) r(a) r ( a )
lim a ↓ 1 r ( a ) − 1 / 2 a − 1 = 2. (92) \lim_{a\downarrow1}\frac{r(a)-1/2}{a-1}=2. \tag*{(92)} a ↓ 1 lim a − 1 r ( a ) − 1/2 = 2. (92) Equivalently, its normalized skewness q ∗ ( a ) ∼ 4 ( a − 1 ) q_*(a)\sim4(a-1) q ∗ ( a ) ∼ 4 ( a − 1 ) C > 4 C>4 C > 4 q = C ( a − 1 ) q=C(a-1) q = C ( a − 1 ) a − 1 > 0 a-1>0 a − 1 > 0 0 < C < 4 0<C<4 0 < C < 4
Proof. Fix C > 0 C>0 C > 0 q = C e q=Ce q = C e h = sin ω h=\sin\omega h = sin ω c = cos ω c=\cos\omega c = cos ω
h / e ⟶ π C / 2 , c = 1 + O ( e 2 ) , ξ ⟶ π / 2. h/e\longrightarrow\pi C/2,\qquad c=1+O(e^2),\qquad\xi\longrightarrow\pi/2. h / e ⟶ π C /2 , c = 1 + O ( e 2 ) , ξ ⟶ π /2. Let P a ( y ) = ∑ n B a , ξ ( r ^ n y ) P_a(y)=\sum_n B_{a,\xi}(\widehat r_n y) P a ( y ) = ∑ n B a , ξ ( r n y ) K a , ρ ( c a y ) K_{a,\rho}(c_a y) K a , ρ ( c a y ) ∑ r ^ n − a = 1 / ( e Γ ( a + 1 ) ) \sum\widehat r_n^{-a}=1/(e\Gamma(a+1)) ∑ r n − a = 1/ ( e Γ ( a + 1 )) e ↓ 0 e\downarrow0 e ↓ 0 t h e r e f o r e i m p l i e s , f o r e a c h f i x e d therefore implies, for each fixed t h er e f or e im pl i es , f or e a c h f i x e d
inf y ≥ Y e y a P a ( y ) ≥ 1 / π − o ( 1 ) . (93) \inf_{y\ge Y}e^{y^a}P_a(y)\ge1/\pi-o(1). \tag*{(93)} y ≥ Y inf e y a P a ( y ) ≥ 1/ π − o ( 1 ) . (93) Here all error terms are bounded by constants times e Y − a eY^{-a} e Y − a e Y − 2 a eY^{-2a} e Y − 2 a y > 0 y>0 y > 0 e y a P a ( y ) → 1 / π e^{ya}P_a(y)\to1/\pi e y a P a ( y ) → 1/ π sin 2 ξ \sin^2\xi sin 2 ξ
For C > 4 C>4 C > 4 Lemma H.1 :
sup y > 0 2 e y a L a ( h y ) = 2 e h − a sup v > 0 v a L a ( v ) ⟶ 4 π C < 1 π . \sup_{y>0}2e^{ya}L_a(hy)=2eh^{-a}\sup_{v>0}v^aL_a(v)\longrightarrow\frac{4}{\pi C}<\frac{1}{\pi}. y > 0 sup 2 e y a L a ( h y ) = 2 e h − a v > 0 sup v a L a ( v ) ⟶ π C 4 < π 1 . Together with (H.6), this proves K ( c a y ) > 0 K(c_a y)>0 K ( c a y ) > 0 y ≥ Y y\ge Y y ≥ Y
There is no missing origin range. Choose Y = 1 Y=1 Y = 1 δ = π ( 1 / a − 1 / 2 ) ( 1 − q ) \delta=\pi(1/a-1/2)(1-q) δ = π ( 1/ a − 1/2 ) ( 1 − q ) δ → π / 2 \delta\to\pi/2 δ → π /2
K ( c a y ) = 2 Re j a ( y e − i δ ) + a sin δ π e y − y sin δ π ∫ 0 ∞ j a ( u ) d u u 2 − 2 u y cos δ + y 2 . K(c_a y)=2\operatorname{Re}j_a(ye^{-i\delta})+\frac{a\sin\delta}{\pi e y}-\frac{y\sin\delta}{\pi}\int_0^\infty\frac{j_a(u)\,du}{u^2-2uy\cos\delta+y^2}. K ( c a y ) = 2 Re j a ( y e − i δ ) + π ey a sin δ − π y sin δ ∫ 0 ∞ u 2 − 2 u y cos δ + y 2 j a ( u ) d u . The germ is uniformly bounded on ∣ z ∣ ≤ 1 |z|\le1 ∣ z ∣ ≤ 1
Ψ a ( 1 ) = 1 − log V e ( − 1 ) = O ( 1 ) , ∫ 1 ∞ j a ( u ) u − 1 d u = O ( 1 ) , \Psi_a(1)=1-\log V_e(-1)=O(1),\qquad\int_1^\infty j_a(u)u^{-1}\,du=O(1), Ψ a ( 1 ) = 1 − log V e ( − 1 ) = O ( 1 ) , ∫ 1 ∞ j a ( u ) u − 1 d u = O ( 1 ) , the second bound following from the positive Lévy representation of Ψ a ( 1 ) \Psi_a(1) Ψ a ( 1 ) e e e cos δ ≤ 1 / 2 \cos\delta\le1/2 cos δ ≤ 1/2 ( u 2 + y 2 ) / 2 (u^2+y^2)/2 ( u 2 + y 2 ) /2 0 < u < 1 0<u<1 0 < u < 1 j a j_a j a ∫ 0 1 ( u 2 + y 2 ) − 1 d u ≤ π / ( 2 y ) \int_0^1(u^2+y^2)^{-1}\,du\le\pi/(2y) ∫ 0 1 ( u 2 + y 2 ) − 1 d u ≤ π / ( 2 y ) u ≥ 1 u\ge1 u ≥ 1 0 < y ≤ 1 0<y\le1 0 < y ≤ 1
For the converse, put y m = 2 π m y_m=2\pi m y m = 2 π m m ≥ 1 m\ge1 m ≥ 1
e ∑ n e − h y m r ^ n { cos ( c y m r ^ n ) − 1 } ⟶ 0. e\sum_n e^{-hy_m\widehat r_n}\{\cos(cy_m\widehat r_n)-1\}\longrightarrow0. e n ∑ e − h y m r n { cos ( c y m r n ) − 1 } ⟶ 0. Roots below a fixed R R R O ( e R ) O(e^R) O ( e R ) R ≤ r ^ n ≤ M / e R\le\widehat r_n\le M/e R ≤ r n ≤ M / e Lemma H.2 and c = 1 + O ( e 2 ) c=1+O(e^2) c = 1 + O ( e 2 ) 2 π Z 2\pi\mathbb{Z} 2 π Z O m ( R − 1 + e M ) O_m(R^{-1}+eM) O m ( R − 1 + e M ) e e e M / e M/e M / e O C ( ( M + 1 ) e − κ C M ) O_C((M+1)e^{-\kappa_C M}) O C (( M + 1 ) e − κ C M ) e → 0 e\to0 e → 0 R , M → ∞ R,M\to\infty R , M → ∞
lim e ↓ 0 e y m K a , ρ ( c a y m ) = 1 π ( 1 − 4 C ) . (94) \lim_{e\downarrow0}ey_mK_{a,\rho}(c_ay_m)=\frac{1}{\pi}\left(1-\frac{4}{C}\right). \tag*{(94)} e ↓ 0 lim e y m K a , ρ ( c a y m ) = π 1 ( 1 − C 4 ) . (94) For C < 4 C<4 C < 4 q ∗ ( a ) / e q_*(a)/e q ∗ ( a ) / e C < 4 C<4 C < 4 C > 4 C>4 C > 4 r ( a ) − 1 / 2 = ( 1 / a − 1 / 2 ) q ∗ ( a ) r(a)-1/2=(1/a-1/2)q_*(a) r ( a ) − 1/2 = ( 1/ a − 1/2 ) q ∗ ( a )
Proposition H.4 (An explicit linear ID wedge). For 1 < a ≤ 501 / 500 1<a\le501/500 1 < a ≤ 501/500 q ≥ 6 ( a − 1 ) q\ge6(a-1) q ≥ 6 ( a − 1 )
Proof. It suffices to prove ID on q = 6 e q=6e q = 6 e 1 < a ≤ 3 / 2 1<a\le3/2 1 < a ≤ 3/2 ( 1 + u 2 ) a ≤ e 2 u (1+u^2)^a\le e^{2u} ( 1 + u 2 ) a ≤ e 2 u
B a , ξ ( v ) ≥ Γ ( a + 1 ) sin ξ π ( v + 2 ) a ≥ Γ ( a + 1 ) sin ξ h a π { 1 + ( 2 h ) a / e } { 1 + ( h v ) a } . B_{a,\xi}(v)\ge\frac{\Gamma(a+1)\sin\xi}{\pi(v+2)^a}\ge\frac{\Gamma(a+1)\sin\xi\,h^a}{\pi\{1+(2h)^{a/e}\} \{1+(hv)^a\}}. B a , ξ ( v ) ≥ π ( v + 2 ) a Γ ( a + 1 ) sin ξ ≥ π { 1 + ( 2 h ) a / e } { 1 + ( h v ) a } Γ ( a + 1 ) sin ξ h a . Put M a = sup w ≥ 0 e − w ( 1 + w a ) 2 M_a=\sup_{w\ge0}e^{-w}(1+w^a)^2 M a = sup w ≥ 0 e − w ( 1 + w a ) 2 h x hx h x
K ( x ) ≥ [ Γ ( a + 1 ) sin ξ h a π { 1 + ( 2 h ) a / e } e − 2 e M a a ] ∑ n 1 1 + ( h r n x ) a . K(x)\ge\left[\frac{\Gamma(a+1)\sin\xi\,h^a}{\pi\{1+(2h)^{a/e}\}e}-\frac{2eM_a}{a}\right]\sum_n\frac{1}{1+(hr_nx)^a}. K ( x ) ≥ [ π { 1 + ( 2 h ) a / e } e Γ ( a + 1 ) sin ξ h a − a 2 e M a ] n ∑ 1 + ( h r n x ) a 1 . For 0 < e < 1 / 2 0<e<1/2 0 < e < 1/2
M a ≤ 4 exp ( 1 ) exp ( − 4 e ) ( 1 − 2 e ) 2 . M_a\le\frac{4}{\exp(1)}\frac{\exp(-4e)}{(1-2e)^2}. M a ≤ exp ( 1 ) 4 ( 1 − 2 e ) 2 exp ( − 4 e ) . Indeed, on w ≤ 1 w \le1 w ≤ 1 w a ≤ w w^a \le w w a ≤ w w ≥ 1 w \ge1 w ≥ 1 1 + w a ≤ w e ( 1 + w ) 1+w^a \le w^e(1+w) 1 + w a ≤ w e ( 1 + w ) log w ≤ w − 1 \log w \le w-1 log w ≤ w − 1
Set E = 1 / 500 E=1/500 E = 1/500 Ω = 3 π E \Omega=3\pi E Ω = 3 π E d = 3 π ( 1 − E ) ( 1 + E ) − 1 ( 1 − Ω 2 / 6 ) d=3\pi(1-E)(1+E)^{-1}(1-\Omega^2/6) d = 3 π ( 1 − E ) ( 1 + E ) − 1 ( 1 − Ω 2 /6 ) 0 < e ≤ E 0<e\le E 0 < e ≤ E
h / e ≥ d , h a / e ≥ d ( d E ) E , sin ξ ≥ cos ( 5 π E / 2 ) , { 1 + ( 2 h ) a / e } e ≤ 2 E . h/e \ge d,\quad h^a/e \ge d(dE)^E,\quad\sin\xi\ge\cos(5\pi E/2),\quad\{1+(2h)^a/e\}^e \le2^E. h / e ≥ d , h a / e ≥ d ( d E ) E , sin ξ ≥ cos ( 5 π E /2 ) , { 1 + ( 2 h ) a / e } e ≤ 2 E . For the second estimate use that e log ( d e ) e\log(de) e log ( d e ) ( 0 , E ] (0,E] ( 0 , E ] 2 h < 1 2h<1 2 h < 1 Γ ( a + 1 ) ≥ 1 \Gamma(a+1)\ge1 Γ ( a + 1 ) ≥ 1 e e e
d ( d E ) E cos ( 5 π E / 2 ) π 2 E − 8 exp ( 1 ) exp ( − 4 E ) ( 1 − 2 E ) 2 > 0.0166 > 0. \frac{d(dE)^E\cos(5\pi E/2)}{\pi2^E}-\frac{8}{\exp(1)}\frac{\exp(-4E)}{(1-2E)^2}>0.0166>0. π 2 E d ( d E ) E cos ( 5 π E /2 ) − exp ( 1 ) 8 ( 1 − 2 E ) 2 exp ( − 4 E ) > 0.0166 > 0. The strict constant inequality is independently checked by certify_linear_id_wedge.py. This proves positivity for every x > 0 x>0 x > 0 q q q
The sharp GGC endpoint expansion
Theorem H.5. As b ↓ 0 b\downarrow0 b ↓ 0
g ( 1 − b ) = b π 2 − b + 2 π 2 + 13 12 π 2 b 3 / 2 − b 2 + O ( b 5 / 2 ) . g(1-b)=\frac{\sqrt{b}}{\pi\sqrt{2}}-b+\frac{2\pi^2+13}{12\pi\sqrt{2}}b^{3/2}-b^2+O(b^{5/2}). g ( 1 − b ) = π 2 b − b + 12 π 2 2 π 2 + 13 b 3/2 − b 2 + O ( b 5/2 ) . For all sufficiently small positive b b b w w w V V V w ( 0 ) = 1 / 2 w(0)=1/2 w ( 0 ) = 1/2 w ′ ( 0 ) = π 2 / 6 w'(0)=\pi^2/6 w ′ ( 0 ) = π 2 /6
g ( 1 − b ) = arcsin b w ( b ) − π b π ( 1 − b ) . g(1-b)=\frac{\arcsin\sqrt{bw(b)}-\pi b}{\pi(1-b)}. g ( 1 − b ) = π ( 1 − b ) arcsin b w ( b ) − π b . If ζ ( b ) \zeta(b) ζ ( b )
T ∗ ( b ) = ζ ( b ) 1 − b w ( b ) + b 2 V ( b ) = 2 + ( 2 γ − 1 / 2 ) b + O ( b 2 ) . T_*(b)=\zeta(b)\sqrt{1-bw(b)}+b^2V(b)=2+(2\gamma-1/2)b+O(b^2). T ∗ ( b ) = ζ ( b ) 1 − b w ( b ) + b 2 V ( b ) = 2 + ( 2 γ − 1/2 ) b + O ( b 2 ) . No numerical smallness threshold is asserted.
Proof. Set a = 1 − b a=1-b a = 1 − b θ = π − δ \theta=\pi-\delta θ = π − δ v = r e i θ v=re^{i\theta} v = r e i θ η = π b + a δ \eta=\pi b+a\delta η = π b + a δ T = r a T=r^a T = r a
t = − v a = T e − i η , G b ( t ) = e − t / b W a ( t / b ) , G 0 ( t ) = Γ ( 2 − t ) − 1 . t=-v^a=Te^{-i\eta},\qquad G_b(t)=e^{-t/b}W_a(t/b),\qquad G_0(t)=\Gamma(2-t)^{-1}. t = − v a = T e − i η , G b ( t ) = e − t / b W a ( t / b ) , G 0 ( t ) = Γ ( 2 − t ) − 1 . The normalized sign is
h b ( T , η ) = ℑ [ − e − i η { 1 + b G b ′ ( t ) / G b ( t ) } ] . h_b(T,\eta)=\Im\left[-e^{-i\eta}\left\{1+bG_b'(t)/G_b(t)\right\}\right]. h b ( T , η ) = ℑ [ − e − i η { 1 + b G b ′ ( t ) / G b ( t ) } ] . Fix a compact positive range of C C C δ = C b \delta=C\sqrt{b} δ = C b
Middle range. There are constants R 0 , c 0 , M , b 0 > 0 R_0,c_0,M,b_0>0 R 0 , c 0 , M , b 0 > 0 b 0 < 1 / 2 b_0<1/2 b 0 < 1/2 t = x − i y t=x-iy t = x − i y x ≥ R 0 x\ge R_0 x ≥ R 0 y > 0 y>0 y > 0 y 2 / x ≤ 1 y^2/x\le1 y 2 / x ≤ 1 b x ( log x ) 2 ≤ c 0 bx(\log x)^2\le c_0 b x ( log x ) 2 ≤ c 0
∣ G b ( j ) ( t ) − G 0 ( j ) ( t ) ∣ ∣ G 0 ( t ) ∣ ≤ M b x ( log x ) j + 2 ( 1 + 1 / y ) , j = 0 , 1. (95) \frac{\left|G_b^{(j)}(t)-G_0^{(j)}(t)\right|}{|G_0(t)|}\le Mbx(\log x)^{j+2}(1+1/y),\qquad j=0,1. \tag*{(95)} ∣ G 0 ( t ) ∣ G b ( j ) ( t ) − G 0 ( j ) ( t ) ≤ M b x ( log x ) j + 2 ( 1 + 1/ y ) , j = 0 , 1. (95) Here the retained-circle Hankel integrand uses E b ( ℓ ) = ( e b ℓ − 1 ) / b E_b(\ell)=(e^{b\ell}-1)/b E b ( ℓ ) = ( e b ℓ − 1 ) / b u ≥ 1 u\ge1 u ≥ 1
∣ E b ( log u ± i π ) − ( log u ± i π ) ∣ ≤ b 2 ( log u + π ) 2 u b . |E_b(\log u\pm i\pi)-(\log u\pm i\pi)|\le\frac{b}{2}(\log u+\pi)^2u^b. ∣ E b ( log u ± iπ ) − ( log u ± iπ ) ∣ ≤ 2 b ( log u + π ) 2 u b . Truncate first at U = D x 2 U=Dx^2 U = D x 2 D D D [ 1 , U ] [1,U] [ 1 , U ] M b x ( log x ) j + 2 e π y Γ ( x − 1 ) Mbx(\log x)^{j+2}e^{\pi y}\Gamma(x-1) M b x ( log x ) j + 2 e π y Γ ( x − 1 ) [ U , ∞ ) [U,\infty) [ U , ∞ ) before estimating, retaining the factor b b b 2 x ( log u + π ) u b ≤ u / 4 2x(\log u+\pi)u^b\le u/4 2 x ( log u + π ) u b ≤ u /4 M b x e − U / 2 Mbxe^{-U/2} M b x e − U /2 G 0 ( t ) = − Γ ( t − 1 ) sin ( π t ) / π G_0(t)=-\Gamma(t-1)\sin(\pi t)/\pi G 0 ( t ) = − Γ ( t − 1 ) sin ( π t ) / π exp ( − M y 2 / x ) \exp(-My^2/x) exp ( − M y 2 / x ) e π y / ∣ sin π t ∣ ≤ M ( 1 + 1 / y ) e^{\pi y}/|\sin\pi t|\le M(1+1/y) e π y /∣ sin π t ∣ ≤ M ( 1 + 1/ y )
Take R R R R o u t = K δ − 1 log ( 1 / δ ) R_{\mathrm{out}}=K\delta^{-1}\log(1/\delta) R out = K δ − 1 log ( 1/ δ ) R ≤ r ≤ R o u t R\le r\le R_{\mathrm{out}} R ≤ r ≤ R out x ≍ r x\asymp r x ≍ r y ≍ r δ y\asymp r\delta y ≍ r δ b b b ∣ G 0 ′ / G 0 ∣ ≤ log x + M + 1 / y \lvert G'_0/G_0\rvert\le\log x+M+1/y ∣ G 0 ′ / G 0 ∣ ≤ log x + M + 1/ y
b ∣ G b ′ / G b − G 0 ′ / G 0 ∣ = o ( δ ) , h b ≥ sin η − b ( log x + M + 1 / y ) − o ( δ ) > δ / 2. b\lvert G'_b/G_b-G'_0/G_0\rvert=o(\delta),\qquad h_b\ge\sin\eta-b(\log x+M+1/y)-o(\delta)>\delta/2. b ∣ G b ′ / G b − G 0 ′ / G 0 ∣ = o ( δ ) , h b ≥ sin η − b ( log x + M + 1/ y ) − o ( δ ) > δ /2. The last step follows by choosing R R R C C C
Outer range. Use the parabolic integrals F b , J b F_b,J_b F b , J b ℜ Φ ≤ − δ x 2 / ( 4 π ) \Re\Phi\le-\delta x^2/(4\pi) ℜΦ ≤ − δ x 2 / ( 4 π ) ∣ z ∣ ≥ δ / π \lvert z\rvert\ge\delta/\pi ∣ z ∣ ≥ δ / π ∣ x ∣ ≤ ε r \lvert x\rvert\le\varepsilon\sqrt r ∣ x ∣ ≤ ε r
ℜ Φ ≤ − c x 2 , ∣ Φ + a x 2 ∣ ≤ M ∣ x ∣ 3 r − 1 / 2 , ∣ z − 3 − 1 ∣ + ∣ E b ( 2 log z ) ∣ ≤ M ∣ x ∣ r − 1 / 2 . \Re\Phi\le-cx^2,\qquad\lvert\Phi+ax^2\rvert\le M\lvert x\rvert^3r^{-1/2},\qquad\lvert z^{-3}-1\rvert+\lvert E_b(2\log z)\rvert\le M\lvert x\rvert r^{-1/2}. ℜΦ ≤ − c x 2 , ∣ Φ + a x 2 ∣ ≤ M ∣ x ∣ 3 r − 1/2 , ∣ z − 3 − 1 ∣ + ∣ E b ( 2 log z )∣ ≤ M ∣ x ∣ r − 1/2 . Its integral error is O ( r − 1 / 2 ) O(r^{-1/2}) O ( r − 1/2 )
M δ − 4 ( 1 + log ( 1 / δ ) ) r − 1 / 2 e − c δ r M\delta^{-4}(1+\log(1/\delta))r^{-1/2}e^{-c\delta r} M δ − 4 ( 1 + log ( 1/ δ )) r − 1/2 e − cδ r for both omitted integrals. Choose K K K r ≥ R o u t r\ge R_{\mathrm{out}} r ≥ R out
F b = π / a + O ( r − 1 / 2 + δ 2 ) , J b = O ( r − 1 / 2 + δ 2 ) . (96) F_b=\sqrt{\pi/a}+O(r^{-1/2}+\delta^2),\qquad J_b=O(r^{-1/2}+\delta^2). \tag*{(96)} F b = π / a + O ( r − 1/2 + δ 2 ) , J b = O ( r − 1/2 + δ 2 ) . (96) Thus the normalized sign sin δ + b ℑ ( e i θ J b / F b ) \sin\delta+b\Im(e^{i\theta}J_b/F_b) sin δ + b ℑ ( e i θ J b / F b )
Compact range. The retained-circle formula is jointly holomorphic in ( b , t ) (b,t) ( b , t ) b = 0 b=0 b = 0 t t t
G b ( t ) = f ( t ) + 1 2 b t f ′ ′ ( t ) + O ( b 2 ) , f ( t ) = 1 / Γ ( 2 − t ) . G_b(t)=f(t)+\tfrac12btf''(t)+O(b^2),\qquad f(t)=1/\Gamma(2-t). G b ( t ) = f ( t ) + 2 1 b t f ′′ ( t ) + O ( b 2 ) , f ( t ) = 1/Γ ( 2 − t ) . Near each of its finitely many simple zeros n = 2 , 3 , … n=2,3,\ldots n = 2 , 3 , … G b ′ / G b = ( t − ζ n ( b ) ) − 1 + A n ( t , b ) G'_b/G_b=(t-\zeta_n(b))^{-1}+A_n(t,b) G b ′ / G b = ( t − ζ n ( b ) ) − 1 + A n ( t , b ) A n A_n A n ζ n ( b ) = n + O ( b ) \zeta_n(b)=n+O(b) ζ n ( b ) = n + O ( b )
ℑ − e − i η T e − i η − ζ = − ζ sin η T 2 − 2 ζ T cos η + ζ 2 . \Im\frac{-e^{-i\eta}}{Te^{-i\eta}-\zeta}=-\frac{\zeta\sin\eta}{T^2-2\zeta T\cos\eta+\zeta^2}. ℑ T e − i η − ζ − e − i η = − T 2 − 2 ζ T cos η + ζ 2 ζ sin η . Hence h b ≥ sin η { 1 − b / ( ζ n sin 2 η ) } − M b h_b\ge\sin\eta\{1-b/(\zeta_n\sin^2\eta)\}-Mb h b ≥ sin η { 1 − b / ( ζ n sin 2 η )} − M b C > 1 / 2 C>1/\sqrt{2} C > 1/ 2
Sector zero count. Write
P b ( v ) = e a v / b W a ( − v a / b ) , P 0 ( v ) = e v ( log v − 1 ) / Γ ( 2 + v ) . P_b(v)=e^{av/b}W_a(-v^a/b),\qquad P_0(v)=e^{v(\log v-1)}/\Gamma(2+v). P b ( v ) = e a v / b W a ( − v a / b ) , P 0 ( v ) = e v ( l o g v − 1 ) /Γ ( 2 + v ) . The three estimates also prove uniformly on the outer ray that P b / P 0 → 1 P_b/P_0\to1 P b / P 0 → 1 O ( b ) O(b) O ( b ) δ \delta δ − v -v − v − v a -v^a − v a M b r ( log r ) 2 + M b log r / δ = o ( 1 ) Mb r(\log r)^2+Mb\log r/\delta=o(1) M b r ( log r ) 2 + M b log r / δ = o ( 1 ) P b / P 0 = F b / F 0 → 1 P_b/P_0=F_b/F_0\to1 P b / P 0 = F b / F 0 → 1 b b b 1 1 1 1 / a 1/\sqrt a 1/ a
Unique contact and expansion. Choose 1 / 3 < C 0 < 1 / 2 < C 1 1/\sqrt{3}<C_0<1/\sqrt{2}<C_1 1/ 3 < C 0 < 1/ 2 < C 1 b b b C 0 C_0 C 0 C 1 C_1 C 1 ζ = ζ 2 ( b ) \zeta=\zeta_2(b) ζ = ζ 2 ( b ) c = cos η c=\cos\eta c = cos η
B ( T , c , b ) = ℑ [ ( − c + i sin η ) A 2 ( T ( c − i sin η ) , b ) ] sin η B(T,c,b)=\frac{\Im[(-c+i\sin\eta)A_2(T(c-i\sin\eta),b)]}{\sin\eta} B ( T , c , b ) = sin η ℑ [( − c + i sin η ) A 2 ( T ( c − i sin η ) , b )] extends real-analytically to ( 2 , 1 , 0 ) (2,1,0) ( 2 , 1 , 0 ) A 2 ( T , b ) + T ∂ T A 2 ( T , b ) A_2(T,b)+T\partial_TA_2(T,b) A 2 ( T , b ) + T ∂ T A 2 ( T , b )
h b / sin η = 1 − b ζ / D + b B , D = ( T − ζ c ) 2 + ζ 2 sin 2 η . h_b/\sin\eta=1-b\zeta/D+bB,\qquad D=(T-\zeta c)^2+\zeta^2\sin^2\eta. h b / sin η = 1 − b ζ / D + b B , D = ( T − ζ c ) 2 + ζ 2 sin 2 η . A nonpositive stationary value has first T − ζ c = O ( b ) T-\zeta c=O(\sqrt{b}) T − ζ c = O ( b ) T − ζ c = O ( b 2 ) T-\zeta c=O(b^{2}) T − ζ c = O ( b 2 ) b − 1 b^{-1} b − 1 sin 2 η = b w \sin^{2}\eta=bw sin 2 η = b w T = ζ 1 − b w + b 2 V T=\zeta\sqrt{1-bw+b^{2}V} T = ζ 1 − b w + b 2 V
The two analytic contact equations are
1 − ζ ζ 2 w + b 3 V 2 + b B = 0 , 2 ζ V ( ζ 2 w + b 3 V 2 ) 2 + B T = 0. 1-\frac{\zeta}{\zeta^{2}w+b^{3}V^{2}}+bB=0,\qquad\frac{2\zeta V}{(\zeta^{2}w+b^{3}V^{2})^{2}}+B_{T}=0. 1 − ζ 2 w + b 3 V 2 ζ + b B = 0 , ( ζ 2 w + b 3 V 2 ) 2 2 ζ V + B T = 0. At zero their Jacobian in ( w , V ) (w,V) ( w , V ) ( 2 , 1 ) (2,1) ( 2 , 1 )
ζ ′ ( 0 ) = 2 γ , A 2 ( 2 , 0 ) = − γ , ( A 2 ) T ( 2 , 0 ) = − π 2 6 , \zeta'(0)=2\gamma,\qquad A_{2}(2,0)=-\gamma,\qquad(A_{2})_{T}(2,0)=-\frac{\pi^{2}}{6}, ζ ′ ( 0 ) = 2 γ , A 2 ( 2 , 0 ) = − γ , ( A 2 ) T ( 2 , 0 ) = − 6 π 2 , so B ( 2 , 1 , 0 ) = − γ − π 2 / 3 B(2,1,0)=-\gamma-\pi^{2}/3 B ( 2 , 1 , 0 ) = − γ − π 2 /3 w ′ ( 0 ) = π 2 / 6 w'(0)=\pi^{2}/6 w ′ ( 0 ) = π 2 /6 δ = ( arcsin b w ( b ) − π b ) / ( 1 − b ) \delta=(\arcsin\sqrt{bw(b)}-\pi b)/(1-b) δ = ( arcsin b w ( b ) − π b ) / ( 1 − b ) − b / ( 1 − b ) -b/(1-b) − b / ( 1 − b )
Second-order ID asymptotics and eventual unique contact
The following refinement retains the small parameter in the phase error of Lemma H.2 . It also supplies complete root enumeration near index one, rather than proximity of selected roots.
Lemma H.6 (Uniform enumerated spectrum). Put e = a − 1 e=a-1 e = a − 1 η = π e / a \eta=\pi e/a η = π e / a d e = sin η / η d_e=\sin\eta/\eta d e = sin η / η A e = 3 η / ( 2 π d e ) A_e=3\eta/(2\pi d_e) A e = 3 η / ( 2 π d e ) e > 0 e>0 e > 0 n = 2 , 3 , … n=2,3,\ldots n = 2 , 3 , …
R n , e = n / d e − A e + δ n , e , ∣ δ n , e ∣ ≤ C e / n , R_{n,e}=n/d_e-A_e+\delta_{n,e},\qquad|\delta_{n,e}|\le Ce/n, R n , e = n / d e − A e + δ n , e , ∣ δ n , e ∣ ≤ C e / n , with a constant independent of n , e n,e n , e
Proof. In the exact rotated Laplace integral, regard k k k
A ( e , k , r ) = a k r 2 π ∫ 0 ∞ t − 2 e − r k f e ( t ) d t . A(e,k,r)=\sqrt{\frac{akr}{2\pi}}\int_{0}^{\infty}t^{-2}e^{-rkf_e(t)}\,dt. A ( e , k , r ) = 2 π ak r ∫ 0 ∞ t − 2 e − r k f e ( t ) d t . The common analytic Morse coordinate near t = 1 t=1 t = 1 A = 1 + O ( r − 1 ) A=1+O(r^{-1}) A = 1 + O ( r − 1 ) k k k r r r e e e f e ≥ f 0 f_e\ge f_0 f e ≥ f 0 ℜ ( r k ) ≥ c ∣ r ∣ \Re(rk)\ge c|r| ℜ ( r k ) ≥ c ∣ r ∣ ∣ r ∣ |r| ∣ r ∣
∂ k log A = O ( r − 1 ) , ∂ r ∂ k log A = O ( r − 2 ) . \partial_k\log A=O(r^{-1}),\qquad\partial_r\partial_k\log A=O(r^{-2}). ∂ k log A = O ( r − 1 ) , ∂ r ∂ k log A = O ( r − 2 ) . For real r > 0 r>0 r > 0 k = 1 k=1 k = 1 A A A A > 0 A>0 A > 0 1 1 1 e − i η e^{-i\eta} e − i η e e e
ℑ log A ( e , e − i η , r ) = O ( e / r ) , ℑ ∂ r log A ( e , e − i η , r ) = O ( e / r 2 ) . \Im\log A(e,e^{-i\eta},r)=O(e/r),\qquad\Im\partial_r\log A(e,e^{-i\eta},r)=O(e/r^{2}). ℑ log A ( e , e − i η , r ) = O ( e / r ) , ℑ ∂ r log A ( e , e − i η , r ) = O ( e / r 2 ) . The exact real-root phase is β e ( r ) = π d e r + 3 η / 2 + ℑ log A \beta_e(r)=\pi d_e r+3\eta/2+\Im\log A β e ( r ) = π d e r + 3 η /2 + ℑ log A β e ′ ( r ) = π d e + O ( e / r 2 ) > 0 \beta'_e(r)=\pi d_e+O(e/r^{2})>0 β e ′ ( r ) = π d e + O ( e / r 2 ) > 0 r ≥ R r\ge R r ≥ R R R R V e ( r a ) V_e(r^a) V e ( r a ) 1 / Γ ( 2 − r ) 1/\Gamma(2-r) 1/Γ ( 2 − r ) 2 ≤ n < R 2\le n<R 2 ≤ n < R R R R β e ( R ) → π R \beta_e(R)\to\pi R β e ( R ) → π R n n n O ( e ) O(e) O ( e ) C C C
Theorem H.7 (Second-order endpoint and uniqueness). Define
J ( y ) = ∫ 0 ∞ v 2 y 2 + v 2 e − 2 v 1 − e − v d v , Y = 2 π , J(y)=\int_{0}^{\infty}\frac{v^{2}}{y^{2}+v^{2}}\frac{e^{-2v}}{1-e^{-v}}\,dv,\qquad Y=2\pi, J ( y ) = ∫ 0 ∞ y 2 + v 2 v 2 1 − e − v e − 2 v d v , Y = 2 π , and D ∗ = 4 log Y + 8 − 12 π Y + 4 J ( Y ) D_* = 4\log Y + 8 - 12\pi Y + 4J(Y) D ∗ = 4 log Y + 8 − 12 π Y + 4 J ( Y )
q ∗ ( 1 + e ) = 4 e + D ∗ e 2 + o ( e 2 ) , q_*(1+e) = 4e + D_*e^2 + o(e^2), q ∗ ( 1 + e ) = 4 e + D ∗ e 2 + o ( e 2 ) , r ( 1 + e ) = 1 2 + 2 e + { 2 log ( 2 π ) − 12 π 2 + 2 J ( 2 π ) } e 2 + o ( e 2 ) . r(1+e) = \frac{1}{2} + 2e + \{2\log(2\pi) - 12\pi^2 + 2J(2\pi)\}e^2 + o(e^2). r ( 1 + e ) = 2 1 + 2 e + { 2 log ( 2 π ) − 12 π 2 + 2 J ( 2 π )} e 2 + o ( e 2 ) . The second coefficient in the latter formula equals − 114.740012808536192366 … -114.740012808536192366\ldots − 114.740012808536192366 … e e e O ( e 3 / 2 ) O(e^{3/2}) O ( e 3/2 ) 2 π 2\pi 2 π
Proof. Take q = 4 e + D e 2 q=4e+De^2 q = 4 e + D e 2 D D D p = h − i c p=h-ic p = h − i c h = sin ω h=\sin\omega h = sin ω c = cos ω c=\cos\omega c = cos ω Lemma H.6 gives, for 0 < v ≤ 1 0<v\le1 0 < v ≤ 1
L e ( v ) = e A e v − 2 v / d e 1 − e − v / d e + O ( e v ( 1 + ∣ log v ∣ ) ) = 1 v − 3 2 + O ( v + e + e 2 / v ) . \mathcal{L}_e(v)=\frac{e^{Ae}v^{-2v/d_e}}{1-e^{-v/d_e}}+O(ev(1+|\log v|))=\frac{1}{v}-\frac{3}{2}+O(v+e+e^2/v). L e ( v ) = 1 − e − v / d e e A e v − 2 v / d e + O ( e v ( 1 + ∣ log v ∣ )) = v 1 − 2 3 + O ( v + e + e 2 / v ) . Indeed the error in the n n n C e v n − 1 e − c 0 n v Cevn^{-1}e^{-c_0nv} C e v n − 1 e − c 0 n v v ≥ 1 v\ge1 v ≥ 1 y y y
S e ( y ) : = ∑ n ≥ 2 e − p y R n , e = e p y A e − 2 p y / d e 1 − e − p y / d e + O ( e ( 1 + ∣ log e ∣ ) ) . (97) S_e(y):=\sum_{n\ge2}e^{-pyR_{n,e}}=\frac{e^{py}A_e^{-2py/d_e}}{1-e^{-py/d_e}}+O(e(1+|\log e|)). \tag*{(97)} S e ( y ) := n ≥ 2 ∑ e − p y R n , e = 1 − e − p y / d e e p y A e − 2 p y / d e + O ( e ( 1 + ∣ log e ∣ )) . (97) At any fixed resonance y = 2 π m y=2\pi m y = 2 π m c y R n , e − 2 π m n = O ( e + e 2 n ) cyR_{n,e}-2\pi mn=O(e+e^2n) cy R n , e − 2 π mn = O ( e + e 2 n )
0 ≤ L e ( h y ) − ℜ S e ( y ) ≤ C ∑ e − c 0 e n ( e 2 + e 4 n 2 ) = O ( e ) . 0\le\mathcal{L}_e(hy)-\Re S_e(y)\le C\sum e^{-c_0en}(e^2+e^4n^2)=O(e). 0 ≤ L e ( h y ) − ℜ S e ( y ) ≤ C ∑ e − c 0 e n ( e 2 + e 4 n 2 ) = O ( e ) . In particular 2 ℜ S e ( y ) = 2 / ( h y ) − 3 + o ( 1 ) 2\Re S_e(y)=2/(hy)-3+o(1) 2ℜ S e ( y ) = 2/ ( h y ) − 3 + o ( 1 )
Let P e ( y ) = ∑ B a , ξ ( R n , e y ) P_e(y)=\sum B_{a,\xi}(R_{n,e}y) P e ( y ) = ∑ B a , ξ ( R n , e y )
P e ( y ) = sin ξ π e y a − J ( y ) π y + o ( 1 ) , (98) P_e(y)=\frac{\sin\xi}{\pi eya}-\frac{J(y)}{\pi y}+o(1), \tag*{(98)} P e ( y ) = π ey a sin ξ − π y J ( y ) + o ( 1 ) , (98) uniformly on fixed positive compact intervals. To check convergence, subtract one from the reciprocal denominator: its absolute value is bounded by C ( ∣ cos ξ ∣ z + z 2 ) C(|\cos\xi|z+z^2) C ( ∣ cos ξ ∣ z + z 2 ) Lemma H.6 . At a = 1 a=1 a = 1 ξ = π / 2 \xi=\pi/2 ξ = π /2
∑ n ≥ 2 [ B 1 , π / 2 ( n y ) − 1 π n y ] = − 1 π y ∫ 0 ∞ v 2 y 2 + v 2 e − 2 v 1 − e − v d v . \sum_{n\ge2}\left[B_{1,\pi/2}(ny)-\frac{1}{\pi ny}\right]
=-\frac{1}{\pi y}\int_0^\infty\frac{v^2}{y^2+v^2}\frac{e^{-2v}}{1-e^{-v}}\,dv. n ≥ 2 ∑ [ B 1 , π /2 ( n y ) − π n y 1 ] = − π y 1 ∫ 0 ∞ y 2 + v 2 v 2 1 − e − v e − 2 v d v . The same integral bounds give P e ′ ′ ( y ) = O ( e − 1 ) P_e''(y)=O(e^{-1}) P e ′′ ( y ) = O ( e − 1 ) sin ξ = 1 + O ( e 2 ) \sin\xi=1+O(e^2) sin ξ = 1 + O ( e 2 ) h = 2 π e + π ( D / 2 − 4 ) e 2 + O ( e 3 ) h=2\pi e+\pi(D/2-4)e^2+O(e^3) h = 2 π e + π ( D /2 − 4 ) e 2 + O ( e 3 )
C D ( y ) = 3 + D / 4 − 2 − log y − J ( y ) π y . (99) C_D(y)=3+\frac{D/4-2-\log y-J(y)}{\pi y}. \tag*{(99)} C D ( y ) = 3 + π y D /4 − 2 − log y − J ( y ) . (99) In particular C D ( Y ) = 0 C_D(Y)=0 C D ( Y ) = 0 D = D ∗ D=D_* D = D ∗
We next control all y ≥ 3 π y\ge3\pi y ≥ 3 π cos ξ < 0 \cos\xi<0 cos ξ < 0 e e e
P e ( y ) ≥ sin ξ π e y a − 2 y − 3 a , y ≥ 3 π . P_e(y)\ge\frac{\sin\xi}{\pi eya}-2y^{-3a},\qquad y\ge3\pi. P e ( y ) ≥ π ey a sin ξ − 2 y − 3 a , y ≥ 3 π . Here the scaled third moment tends to ζ ( 3 ) − 1 \zeta(3)-1 ζ ( 3 ) − 1 v = h y ≤ δ v=hy\le\delta v = h y ≤ δ e − e log y ≥ 1 − e log y e^{-e\log y}\ge1-e\log y e − e l o g y ≥ 1 − e log y
K ( c a y ) ≥ 3 + D / 4 − 2 − log y π y − 2 y − 3 a − O ( δ + e ) . K(cay)\ge3+\frac{D/4-2-\log y}{\pi y}-2y^{-3a}-O(\delta+e). K ( c a y ) ≥ 3 + π y D /4 − 2 − log y − 2 y − 3 a − O ( δ + e ) . Uniformly for − 230 ≤ D ≤ − 210 -230 \le D \le-210 − 230 ≤ D ≤ − 210 0.9123 0.9123 0.9123 3 π 3\pi 3 π ( 2 − D / 4 + log y ) / y (2-D/4+\log y)/y ( 2 − D /4 + log y ) / y δ \delta δ e e e δ ≤ v ≤ V \delta\le v \le V δ ≤ v ≤ V 2 / v 2/v 2/ v e − 2 v / ( 1 − e − v ) < 1 / v e^{-2v}/(1-e^{-v}) < 1/v e − 2 v / ( 1 − e − v ) < 1/ v v ≥ V v \ge V v ≥ V h a / e h^a/e h a / e V V V
The inequality 0 < J ( Y ) < 2 ( ζ ( 3 ) − 1 ) / Y 2 0 < J(Y) < 2(\zeta(3)-1)/Y^2 0 < J ( Y ) < 2 ( ζ ( 3 ) − 1 ) / Y 2 − 230 < D ∗ < − 210 -230 < D_* < -210 − 230 < D ∗ < − 210 D < D ∗ D < D_* D < D ∗ Y Y Y D > D ∗ D > D_* D > D ∗ Y Y Y K ≥ P e − 2 L e ( h y ) K \ge P_e-2L\mathcal{e}(hy) K ≥ P e − 2 L e ( h y ) [ 1 , 3 π ] [1,3\pi] [ 1 , 3 π ] P e ∼ 1 / ( π e y ) P_e \sim1/(\pi e y) P e ∼ 1/ ( π ey ) 0 < y ≤ 1 0 < y \le1 0 < y ≤ 1 q ∗ q_* q ∗
It remains to prove uniqueness, rather than infer it from the leading profile. For bounded D D D D ∗ D_* D ∗ Y Y Y w = p y / d e + 2 π i = u − i v w=py/d_e+2\pi i=u-iv w = p y / d e + 2 π i = u − i v A = e p y A e A=e^{py}A_e A = e p y A e u ≍ e u\asymp e u ≍ e ℜ A = 1 + O ( e 2 ) \Re A=1+O(e^2) ℜ A = 1 + O ( e 2 ) ℑ A = O ( e ) \Im A=O(e) ℑ A = O ( e )
ℜ S e ( y ) = ℜ A u − i v − 3 2 + O ( ∣ w ∣ ) + o ( 1 ) , P e ( y ) = 2 / u + O ( 1 ) . \Re S_e(y)=\Re\frac{A}{u-iv}-\frac{3}{2}+O(|w|)+o(1),\qquad P_e(y)=2/u+O(1). ℜ S e ( y ) = ℜ u − i v A − 2 3 + O ( ∣ w ∣ ) + o ( 1 ) , P e ( y ) = 2/ u + O ( 1 ) . If K ≤ 0 K\le0 K ≤ 0 2 / u − 2 ℜ [ A / ( u − i v ) ] ≤ C 2/u-2\Re[A/(u-iv)]\le C 2/ u − 2ℜ [ A / ( u − i v )] ≤ C
2 v 2 u ( u 2 + v 2 ) + 2 ( 1 − ℜ A ) u + 2 ℑ A v u 2 + v 2 . \frac{2v^2}{u(u^2+v^2)}+\frac{2(1-\Re A)u+2\Im A\,v}{u^2+v^2}. u ( u 2 + v 2 ) 2 v 2 + u 2 + v 2 2 ( 1 − ℜ A ) u + 2ℑ A v . Young’s inequality bounds it below by v 2 / [ u ( u 2 + v 2 ) ] − C e v^2/[u(u^2+v^2)]-Ce v 2 / [ u ( u 2 + v 2 )] − C e v 2 ≤ C u ( u 2 + v 2 ) v^2\le Cu(u^2+v^2) v 2 ≤ C u ( u 2 + v 2 ) e e e v 2 ≤ 2 C u 3 = O ( e 3 ) v^2\le2Cu^3=O(e^3) v 2 ≤ 2 C u 3 = O ( e 3 )
Twice differentiating the root-sum comparison produces error O ( e − 1 ) O(e^{-1}) O ( e − 1 ) n n n C e ( 1 + n ) e − c 0 e n Ce(1+n)e^{-c_0en} C e ( 1 + n ) e − c 0 e n ∣ δ n , e ∣ ≤ C e / n |\delta_{n,e}|\le Ce/n ∣ δ n , e ∣ ≤ C e / n
− 4 ℜ A ( p / d e ) 2 ( u − i v ) 3 + O ( e − 1 ) . -4\Re\frac{A(p/d_e)^2}{(u-iv)^3}+O(e^{-1}). − 4ℜ ( u − i v ) 3 A ( p / d e ) 2 + O ( e − 1 ) . Since A ( p / d e ) 2 = − 1 + O ( e ) A(p/d_e)^2=-1+O(e) A ( p / d e ) 2 = − 1 + O ( e ) v / u = O ( e ) v/u=O(\sqrt e) v / u = O ( e ) 4 u − 3 ( 1 + O ( e ) ) + O ( e − 1 ) > c / e 3 4u^{-3}(1+O(\sqrt e))+O(e^{-1})>c/e^3 4 u − 3 ( 1 + O ( e )) + O ( e − 1 ) > c / e 3 P e ′ ′ = O ( e − 1 ) P_e''=O(e^{-1}) P e ′′ = O ( e − 1 )
Finally the directed integral check gives J ( 2 π ) = 0.00974293585871004644 … J(2\pi)=0.00974293585871004644\ldots J ( 2 π ) = 0.00974293585871004644 … e − 2 U / [ 2 ( 1 − e − U ) ] e^{-2U}/[2(1-e^{-U})] e − 2 U / [ 2 ( 1 − e − U )] U = 80 U=80 U = 80 second_order_id_coefficient_fast_checks.json records the nonvacuous enclosure and the exterior constant. These finite checks supplement the uniform analytic proof above. □
H.3.1. A unit-origin estimate and finite slices. For 1 < a ≤ 11 / 10 1<a\le11/10 1 < a ≤ 11/10 0 < q ≤ 3 / 20 0<q\le3/20 0 < q ≤ 3/20 0 < y ≤ 1 0<y\le1 0 < y ≤ 1
K a , ρ ( c a y ) > 3 / 2. (100) K_{a,\rho}(c_a y)>3/2. \tag*{(100)} K a , ρ ( c a y ) > 3/2. (100) Here is the mass estimate needed in addition to the convergent germ. If N N N 1 / e 1/e 1/ e V e ( − 1 ) = E [ Γ ( 2 + e N ) − 1 ] ≥ 1 / 12 V_e(-1)=\mathbb{E}[\Gamma(2+eN)^{-1}]\ge1/12 V e ( − 1 ) = E [ Γ ( 2 + e N ) − 1 ] ≥ 1/12 Pr ( e N ≤ 2 ) ≥ 1 / 2 \Pr(eN\le2)\ge1/2 Pr ( e N ≤ 2 ) ≥ 1/2 Γ ≤ 6 \Gamma\le6 Γ ≤ 6 [ 2 , 4 ] [2,4] [ 2 , 4 ] Ψ a ( 1 ) < 7 / 2 \Psi_a(1)<7/2 Ψ a ( 1 ) < 7/2 ∫ 0 2 j a ( u ) d u / u < 49 / 12 \int_0^2 j_a(u)\,du/u<49/12 ∫ 0 2 j a ( u ) d u / u < 49/12 j a ( u ) ≤ 3 / 2 j_a(u)\le3/2 j a ( u ) ≤ 3/2 [ 0 , 2 ] [0,2] [ 0 , 2 ] ℜ j a ( y e − i u ) > 4 / 5 \Re j_a(ye^{-iu})>4/5 ℜ j a ( y e − i u ) > 4/5 π / 3 ≤ δ ≤ π / 2 \pi/3\le\delta\le\pi/2 π /3 ≤ δ ≤ π /2 3 / 2 3/2 3/2 49 y / ( 12 π ) 49y/(12\pi) 49 y / ( 12 π ) 5 3 / ( π y ) 5\sqrt{3}/(\pi y) 5 3 / ( π y )
K ( c a y ) > 1 10 + 5 3 / y − 49 y / 12 π > 3 2 . K(c_a y)>\frac{1}{10}+\frac{5\sqrt{3}/y-49y/12}{\pi}>\frac{3}{2}. K ( c a y ) > 10 1 + π 5 3 / y − 49 y /12 > 2 3 . The 1280-leaf independent verifier recomputes both germ covers. The normalized exact rotated Laplace integral, including its two infinite tails, now also provides efficient large-root signs. At a = 1.001 a = 1.001 a = 1.001
Table H.1.
Their global contacts are unique. The common checker is certify_id_point_contact.py. A separate finite-spectrum search, compute_near_one_id_boundary.py, obtained a q q q 10 − 6 10^{-6} 1 0 − 6 a = 1.001 a = 1.001 a = 1.001
The analogous global GGC slices, with zero sector winding, are
Table H.2.
The .90 and .95 certificates additionally have 2057 and 8 local derivative leaves. A generic .97 run independently verified a width below .001, without a uniqueness assumption.
References 