Proof. By 9 , \[\lim_{\beta\downarrow0}\Phi_{d,l}'(\beta)=+\infty, \qquad \lim_{\beta\uparrow1/d}\Phi_{d,l}'(\beta)=-\infty.\] Together with the strict concavity in 10 , this implies that \(\Phi_{d,l}'\) has a unique zero in \((0,1/d)\). The equation \(\Phi_{d,l}'(\beta)=0\) is equivalent to \[(1-d\beta)^l=\beta(1-\beta)^{l-1},\] which is ?? . Hence this unique zero is \(\beta_*\), and it is the unique maximizer of \(\Phi_{d,l}\).

The continuous extension of \(\Phi_{d,l}\) to the endpoints satisfies \[\lim_{\beta\downarrow0}\Phi_{d,l}(\beta)=0\] and \[\lim_{\beta\uparrow1/d}\Phi_{d,l}(\beta) =\frac{(l-1)(d-1)}{d}\ln\left(\frac{d-1}{d}\right) -\frac{1}{d}\ln\left(\frac{1}{d}\right) =f_l\left(\frac{1}{d}\right).\] If \(f_l(1/d)\geq0\), strict concavity implies that \(\Phi_{d,l}\) lies strictly above the chord joining its endpoint values; hence \(\Phi_{d,l}(\beta)>0\) for all \(\beta\in(0,1/d)\).

If \(f_l(1/d)<0\), then \(\Phi_{d,l}(\beta_*)>0\), because \(\Phi_{d,l}(0+)=0\) and \(\Phi_{d,l}\) initially increases. Since \(\Phi_{d,l}\) is strictly decreasing on \((\beta_*,1/d)\), there is a unique zero \(\beta_0\in(\beta_*,1/d)\). The asserted signs follow from the monotonicity on \((0,\beta_*)\) and \((\beta_*,1/d)\). ◻

Proof. See Exercise 3.1.9 of [34]. ◻

Proof. By 7 and ?? , for \(1\leq h<m/d\), \[\frac{B(m,h)e^{m\Phi_{d,l}(h/m)}}{\sqrt{2m\pi(h/m)(1-dh/m)}} \leq \mathbb{E}Z_h\leq \frac{A(m,h)e^{m\Phi_{d,l}(h/m)}}{\sqrt{2m\pi(h/m)(1-dh/m)}}, \label{azh}\tag{11}\] where \[\begin{align} \log A(m,h) &=\frac{1}{12m}+\frac{1}{12l(m-h)}+\frac{d}{12lm} -\frac{1}{12h+1}-\frac{1}{12(m-h)+1}\\ & -\frac{1}{12lm+1}-\frac{d}{12l(m-dh)+d},\\ \log B(m,h) &=\frac{1}{12m+1}+\frac{1}{12l(m-h)+1}+\frac{d}{12lm+d} -\frac{1}{12h}-\frac{1}{12(m-h)}\\ & -\frac{1}{12lm}-\frac{d}{12l(m-dh)}. \end{align}\] If \(h/m\) stays in a compact subset of \((0,1/d)\), then all denominators in the exponents above are of order \(m\), uniformly in \(h\). Hence \(A(m,h)\to1\) and \(B(m,h)\to1\) uniformly on such compact subsets, and the lemma follows from 11 . ◻

Proof. The bounds in ?? are immediate from the definitions. It remains to prove ?? . There are \[ls+2\{l(m\beta-s)-t\}=lm(2\beta-\rho)-2\theta m\] half-edges of type \(A\), together with those half-edges of \(B\cup C\) that do not collide with a half-edge from the other matching. No two of these half-edges are incident to the same vertex. The vertices containing them therefore account for \((lm(2\beta-\rho)-2\theta m)d\) half-edges. The \(2t\) colliding half-edges from \(B\cup C\) form \(t\) vertices, which account for a further \(\theta md\) half-edges, and these vertices are disjoint from the previous ones. Since the hypergraph has \(lm\) half-edges in total, \[(lm(2\beta-\rho)-2\theta m)d+\theta md\leq lm,\] which is ?? . ◻

Proof. First choose the two edge sets. The number of ordered pairs \((M_1,M_2)\) with \(|M_1|=|M_2|=m\beta\) and \(|M_1\cap M_2|=s\) is \[\binom ms\binom{m-s}{m\beta-s}\binom{m-m\beta}{m\beta-s} =\frac{m!}{s!(m\beta-s)!^2(m-2m\beta+s)!}.\] After the edge sets have been chosen, choose the \(t\) half-edges of \(B\) and the \(t\) half-edges of \(C\) that collide, and pair them. This gives \[\binom{l(m\beta-s)}t^2t! =\frac{[l(m\beta-s)]!^2}{t![l(m\beta-s)-t]!^2}\] choices, and the factor \((d-1)^t\) chooses the remaining half-edges at the \(t\) vertices where one \(B\)-half-edge and one \(C\)-half-edge collide. The remaining constrained vertices containing half-edges of \(M_1\cup M_2\) are then completed using half-edges from hyperedges in \(D\), and the still-unpaired half-edges are partitioned into vertices. Dividing by the total number of configurations gives the final factor \[\frac{[l(m-2m\beta+s)]!}{(lm)!} \frac{d^{2lm\beta-ls-t}\left(\frac{lm}{d}\right)!}{\left(\frac{lm}{d}-2lm\beta+ls+t\right)!}.\] Multiplying the displayed factors gives ?? ; summing over the admissible \(s,t\) gives ?? . ◻

Proof. This follows by applying the uniform Stirling bounds in Lemma 2 to the factorial expression ?? . The compactness assumption keeps all arguments of the logarithms and all denominator factors bounded away from zero. ◻

Proof. Let \[R(\rho):=\beta-\rho-(d-1)a(\rho), \qquad S(\rho):=1-2\beta d+\rho d+d(d-1)a(\rho).\] After the substitution 22 , the point \((\rho,l(d-1)a(\rho))\) belongs to \(\mathcal{R}_\beta\) if and only if \[\rho\in\mathcal{I}_\beta:=\{\rho\in(0,\beta): R(\rho)>0,\; S(\rho)>0\}. \label{eq:Ibeta}\tag{24}\] On this set \(h\) is well-defined and \[h(\rho)=\log a(\rho)+\log S(\rho)-2\log R(\rho).\] Notice the useful identity \[S(\rho)=1-\beta d-dR(\rho), \label{eq:S-R-identity}\tag{25}\] so that \[\mathcal{I}_\beta=\left\{\rho\in(0,\beta): 0<R(\rho)<\frac{1-\beta d}{d}\right\}.\] Moreover, \(a(\beta^2)=\beta^2\), and therefore \[R(\beta^2)=\beta(1-d\beta)>0, \qquad S(\beta^2)=(1-d\beta)^2>0.\] Thus \(\beta^2\in\mathcal{I}_\beta\), and direct substitution gives \(h(\beta^2)=0\).

We next prove that \(h\) is strictly increasing on every connected component of \(\mathcal{I}_\beta\). Put \[A(\rho):=\frac{a'(\rho)}{a(\rho)} =\frac{1}{l}\left\{\frac{1}{\rho}- \frac{2(l-1)}{\beta-\rho}-\frac{l-1}{1-2\beta+\rho}\right\}.\] Under the stated assumption on \(\beta\) we have \(\beta-\rho<\sqrt2(1-2\beta+\rho)\) for every \(\rho\in(0,\beta)\); when \(d\ge3\) this follows from \(\beta<1/d\le1/3<(4-\sqrt2)/7\). Hence \[\frac{d}{d\rho}A(\rho) =-\frac{1}{l}\left\{\frac{1}{\rho^2}+\frac{2(l-1)}{(\beta-\rho)^2} -\frac{l-1}{(1-2\beta+\rho)^2}\right\}<0.\] Since \(A(\rho)\to+\infty\) as \(\rho\downarrow0\) and \(A(\rho)\to-\infty\) as \(\rho\uparrow\beta\), the point \(c_0\) is the unique zero of \(A\); in particular \(A(\rho)>0\) on \((0,c_0)\) and \(A(\rho)<0\) on \((c_0,\beta)\).

Differentiating \(h\) gives, on \(\mathcal{I}_\beta\), \[\begin{align} h'(\rho) &=\frac{d}{S(\rho)}+\frac{2}{R(\rho)} +A(\rho)\left(1+\frac{d(d-1)a(\rho)}{S(\rho)} +\frac{2(d-1)a(\rho)}{R(\rho)}\right). \end{align}\] All factors \(R,S,a\) are positive on \(\mathcal{I}_\beta\). Therefore \(h'(\rho)>0\) whenever \(\rho\le c_0\). If \(\rho>c_0\), set \(s=\beta-\rho\). Since \(t_\beta(s)=a(\rho)/(\beta-\rho)\), a direct simplification gives \[h'(\rho)= \frac{G_\beta(\beta-\rho)}{l\rho(1-2\beta+\rho)R(\rho)S(\rho)} . \label{eq:hprime-G}\tag{26}\] The denominator in 26 is positive on \(\mathcal{I}_\beta\), and ?? gives \(G_\beta(\beta-\rho)>0\) for \(\rho>c_0\). Thus \(h'>0\) on \(\mathcal{I}_\beta\).

It remains only to check that the admissible domain \(\mathcal{I}_\beta\) has a single connected component. Let \(J\) be a connected component of \(\mathcal{I}_\beta\). Since \(h\) is strictly increasing on \(J\), a finite left endpoint of \(J\) cannot be a zero of \(R\); indeed \(R\downarrow0\) would force \(h\to+\infty\) from within \(J\), which is incompatible with monotonicity. Similarly, a finite right endpoint of \(J\) cannot be a zero of \(S\), since \(S\downarrow0\) would force \(h\to-\infty\) from within \(J\).

Suppose that there were two distinct components \(J_1<J_2\). By the preceding paragraph, the right endpoint of \(J_1\) must be a zero of \(R\), while the left endpoint of \(J_2\) must be a zero of \(S\). Using 25 , the latter means \(R=(1-\beta d)/d>0\). Starting from a zero of \(R\) at the right endpoint of \(J_1\) and moving towards the left endpoint of \(J_2\), continuity would then produce a first point after which \(R>0\); near that point \(S>0\) as well, so a component of \(\mathcal{I}_\beta\) would have a left endpoint at which \(R=0\), contradicting the previous paragraph. Hence \(\mathcal{I}_\beta\) is connected.

Since \(\beta^2\in\mathcal{I}_\beta\), \(h(\beta^2)=0\), and \(h\) is strictly increasing on the connected set \(\mathcal{I}_\beta\), this is the unique admissible solution of \(h(\rho)=0\). Every critical point in \(\mathcal{R}_\beta\) must satisfy 19 , hence 22 , and then \(h(\rho)=0\); conversely, 19 and 18 are exactly the exponentiated critical point equations. The unique critical point is therefore \((\beta^2,l(d-1)\beta^2)\). ◻

Proof. Extend \(\Psi_{d,l}\) continuously to the closure of \(\mathcal{R}_\beta\) by the convention \(x\log x=0\) at \(x=0\). The closure is compact, so the extension has a maximizer. We claim that no maximizer lies on the boundary. On the boundary piece \(\rho=0\), the inward derivative \(\partial_\rho\Psi\) tends to \(+\infty\). On the boundary piece \(\theta=0\), the inward derivative \(\partial_\theta\Psi\) tends to \(+\infty\). On the upper boundary \(\beta-\rho-\theta/l=0\), moving slightly inward by decreasing \(\theta\) increases \(\Psi\), since \(\partial_\theta\Psi\to-\infty\). On the lower boundary \(1-2\beta d+\rho d+\theta d/l=0\), moving slightly inward by increasing \(\theta\) increases \(\Psi\), since \(\partial_\theta\Psi\to+\infty\). The same one-sided perturbations, or their obvious combinations, handle the boundary corners. Therefore any maximizer of the continuous extension lies in the interior \(\mathcal{R}_\beta\).

An interior maximizer is a critical point. Lemma 7 shows that the only critical point in \(\mathcal{R}_\beta\) is \((\beta^2,l(d-1)\beta^2)\), so this point is the unique global maximizer. The value at this point is 21 . ◻

Proof. Let \(S=1-2\beta d+\rho d+\theta d/l\). Since \((\rho,\theta)\in\mathcal{R}_\beta\), \(S>0\) and \(\theta<l(\beta-\rho)<l/d\). Therefore \[\begin{align} \frac{l-1}{1-2\beta+\rho}-\frac{ld}{S} &=\frac{-l(d-1)+\theta d(l-1)/l-(1-2\beta d+\rho d)}{(1-2\beta+\rho)S}\\ &\leq \frac{-l(d-2)-S}{(1-2\beta+\rho)S}<0. \end{align}\] Also \[\frac{2(l-1)}{\beta-\rho}-\frac{2l}{\beta-\rho-\theta/l}<0.\] Combining these two inequalities with 15 proves ?? . ◻

Proof. At 20 , \[\begin{align} \det H(\beta,\beta^2,l(d-1)\beta^2) =\frac{1-2\beta+\beta^2(d+l-dl)}{l\beta^4(1-d\beta)^2(1-\beta)^2(d-1)}. \label{dth} \end{align}\tag{27}\] The numerator can be written as \[(1-\beta)^2-(d-1)(l-1)\beta^2.\] Thus ?? is exactly the condition that 27 be positive. Together with ?? and 16 , positivity of the determinant implies that the Hessian is negative definite, proving part (2). If \(l\leq d\) and \(0<\beta<1/d\), then ?? holds; this gives part (1). ◻

Proof. By ?? and 16 , it suffices to prove \(\det H>0\). Put \[\begin{align} A&=-\frac{1}{\rho}+\frac{1}{1-2\beta+\rho}+\frac{2}{\beta-\rho},\\ B&=-\frac{2}{\beta-\rho-\theta/2}-\frac{d}{1-2\beta d+\rho d+\theta d/2},\\ C&=-\frac{1}{\theta}, \qquad U=-\frac{1}{\rho}+\frac{1}{1-2\beta+\rho}. \end{align}\] Then \(B,C,U<0\) and \[\det H=(A+2B)\left(C+\frac{B}{2}\right)-B^2 =AC+B\left(\frac{A}{2}+2C\right).\] Since \(AC=CU-2/[\theta(\beta-\rho)]\geq -2/[\theta(\beta-\rho)]\), and since \(\theta<2(\beta-\rho)\), \[\begin{align} B\left(\frac{A}{2}+2C\right) &=\frac{BU}{2}+B\left(\frac{1}{\beta-\rho}-\frac{2}{\theta}\right)\\ &\geq \frac{4}{(\beta-\rho)\theta}, \end{align}\] we get \(\det H>0\). Strict concavity on the convex set \(\mathcal{R}_\beta\), together with the critical point equations, gives the unique global maximizer. ◻

Proof. Substituting the second derivatives 15 –17 into \(\det H(\beta,\rho,l\eta)=0\) and solving for \(\eta\) gives ?? . Differentiating ?? gives ?? . The asserted formula for the unique positive root of \(F\) follows from the displayed quadratic in ?? . ◻

Proof. By ?? , \[\xi''(\rho)=\frac{F'(\rho)D(\rho)-2F(\rho)D'(\rho)}{D(\rho)^3} =\frac{4(1-\beta d)^2(l-1)K(\rho)}{D(\rho)^3}, \label{eq:xi-second}\tag{29}\] where \(K\) is the cubic polynomial defined by the second equality. A direct calculation gives \[K'(\rho)=3(dl-d-l)L(\rho), \qquad L(\rho):=-2l(1-\beta)^2\rho^2+(\rho+\beta(1-2\beta))^2. \label{dkr}\tag{30}\] The positive root of \(L\) is \(\rho_3\), and \(L(\rho)>0\) on \((\rho_5,\rho_3)\) while \(L(\rho)<0\) on \((\rho_3,\beta)\). Hence \(K\) is increasing on \((\rho_5,\rho_3)\) and decreasing on \((\rho_3,\beta)\). Since \(F(\rho_3)=0\) and \(F'(\rho_3)>0\), \[\xi''(\rho_3)=\frac{F'(\rho_3)}{D(\rho_3)^2}>0.\] On the other hand, \(D(\rho)<0\) for \(\rho\in[\rho_5,\beta]\) because \[D(0)=\beta(1-2\beta)>0, \qquad D(\beta)=-2\beta(1-\beta)(l-1)(1-\beta d)<0,\] and \[D(\rho_5)=-\frac{\beta l(1-\beta)(1-2\beta)^2(l-1)(1-\beta d)}{(l(1-\beta)-1)^2}<0.\] Thus 29 implies \(K(\rho_3)<0\). Since \(K(\rho_3)\) is the maximum of \(K\) on \((\rho_5,\beta)\), \(K(\rho)<0\) throughout this interval. Together with \(D(\rho)<0\), 29 gives \(\xi''(\rho)>0\) on \((\rho_5,\beta)\). ◻

Proof. Let \(\rho_1\) be the unique zero of \(D\) in \((0,\beta)\). The uniqueness follows from \(D(0)>0\), \(D(\rho_5)<0\), \(D(\beta)<0\), and the fact that \(D\) is an upward-opening quadratic because \(dl-l-d=(d-1)(l-1)-1>0\); hence \(\rho_1<\rho_5\) and \(D<0\) on \((\rho_1,\beta)\). The equation \(\xi(\rho)=b_+(\rho)\) is equivalent to \(J(\rho)=D(\rho)\), and \[J(\rho)-D(\rho)=2(\beta d-1) \{\beta(1-2\beta)-[l(1-\beta)-1]\rho\}. \label{eq:J-minus-D}\tag{31}\] Thus \(\xi=b_+\) has the unique solution \(\rho_5\) in \((0,\beta)\), and also the endpoint solution \(\rho=\beta\).

We next show that no other branch of \(\xi\) enters the feasible region. If \(\rho<\rho_1\), then \(D(\rho)>0\) and \(\rho<\rho_3\), so ?? gives \(\xi'(\rho)<-1\). If \(\beta\leq1/(2d)\), then \(b_-(\rho)=0\) for all \(\rho\) and \(\xi(0)=\beta(2\beta d-1)\leq0\), whence \(\xi(\rho)<b_-(\rho)\) on \((0,\rho_1)\). Assume instead that \(\beta>1/(2d)\), and put \(r_0=2\beta-1/d\). On \([0,\min\{r_0,\rho_1\})\) we have \(b_-(\rho)=2\beta-\rho-1/d\) and \[\frac{d}{d\rho}\{\xi(\rho)-b_-(\rho)\}=\frac{F(\rho)}{D(\rho)^2}<0, \qquad \xi(0)-b_-(0)=\frac{(\beta d-1)(2\beta d-1)}{d}<0.\] Thus \(\xi<b_-\) on this interval. If \(r_0<\rho_1\), then \(\xi(r_0)<0\) and the inequality \(\xi'<-1\) gives \(\xi(\rho)<0=b_-(\rho)\) for \(r_0<\rho<\rho_1\). Hence \(\xi(\rho)<b_-(\rho)\) for every \(0<\rho<\rho_1\). For \(\rho_1<\rho<\rho_5\), the right-hand side of 31 is negative while \(D<0\), so \(J/D>1\) and hence \(\xi(\rho)>b_+(\rho)\). Therefore the only part of the curve that lies in the closed feasible region is the branch \(\rho\in[\rho_5,\beta]\).

It remains to prove ?? . First, \(\beta\leq1/(dl-d-l+2)\leq1/(d+1)\) for \(d\geq2,l\geq3\), and a direct calculation gives \[\mathcal{L}(0)-\left(2\beta-\frac{1}{d}\right) =-\frac{(\beta d-1)(\beta d+\beta-1)}{d(\beta-1)}\geq0.\] Since both \(\mathcal{L}(\rho)\) and \(2\beta-\rho-1/d\) are affine in \(\rho\), and \(\mathcal{L}'(\rho)=-{(d-1)\beta}/{(1-\beta)}>-1\), we have \(b_-(\rho)\leq\mathcal{L}(\rho)\) for all \(0\leq\rho\leq\beta\).

Next, \[\xi'(\beta)=-\frac{1}{l-1}.\] By ?? , \[-\frac{1}{l-1}\leq -\frac{(d-1)\beta}{1-\beta}=\mathcal{L}'(\rho).\] Since \(\xi'\) is strictly increasing on \((\rho_5,\beta)\) by Lemma 13, \(\xi'(\rho)<\mathcal{L}'(\rho)\) for \(\rho_5<\rho<\beta\). As \(\xi(\beta)=\mathcal{L}(\beta)=0\), integrating from \(\rho\) to \(\beta\) yields \(\mathcal{L}(\rho)<\xi(\rho)\) for \(\rho_5<\rho<\beta\). Finally, 31 and \(D<0\) imply \(\xi(\rho)<b_+(\rho)\) for \(\rho_5<\rho<\beta\). This proves ?? .

The sign assertion follows because the saddle \((\beta^2,(d-1)\beta^2)\) lies on \(\mathcal{L}\) and hence below \(\Gamma_\beta\); at the saddle, 27 is positive under ?? . Since the zero set in the feasible region is exactly the graph \(\Gamma_\beta\), the region below this graph is the connected component on which \(\det H>0\). ◻

Proof. Work in the \((\rho,\eta)\)-plane. Let \(z_0=(\beta^2,(d-1)\beta^2)\), and let \(S_\beta\) be the closed subset of \(\overline{\widehat{\mathcal{R}}_\beta}\) on which \(\det H\geq0\) and which contains \(z_0\). By Lemma 14, \(S_\beta\) is the feasible region below the graph \(\Gamma_\beta\).

We first prove that \(S_\beta\) is star-shaped with respect to \(z_0\). The feasible region \(\overline{\widehat{\mathcal{R}}_\beta}\) is convex, so only the determinant boundary \(\Gamma_\beta\) has to be checked. Fix \(u\in[\rho_5,\beta]\). Because \(\beta\leq1/(dl-d-l+2)\leq1/l\), we have \(\beta^2\leq\rho_5\leq u\). By Lemma 13, \(\xi\) is convex on \((\rho_5,\beta)\); hence its tangent at \(u\) lies below its graph. Furthermore, the point \(z_0\) lies below that tangent. Indeed, since \(\xi'(u)\leq\mathcal{L}'\) and \(\beta^2-u\leq0\), while \(\mathcal{L}(u)\leq\xi(u)\), \[(d-1)\beta^2=\mathcal{L}(\beta^2) =\mathcal{L}(u)+\mathcal{L}'(\beta^2-u) \leq \xi(u)+\xi'(u)(\beta^2-u).\] Therefore the segment from \(z_0\) to \((u,\xi(u))\) lies below the tangent at \(u\), and hence below the graph of \(\xi\). This proves the star-shapedness.

For any \(z\in S_\beta\setminus\{z_0\}\), let \(\gamma\) be the line segment from \(z_0\) to \(z\). The interior of \(\gamma\) lies in the region where \(\det H>0\). Since \(\partial_{\rho\rho}^2\widehat\Psi_\beta<0\) and \(\det H>0\) there, the Hessian of \(\widehat\Psi_\beta\) is negative definite along the interior of \(\gamma\). The first derivative of \(\widehat\Psi_\beta\) in the direction of \(\gamma\) is zero at \(z_0\) and then strictly decreases along the segment. Hence \[\widehat\Psi_\beta(z)<\widehat\Psi_\beta(z_0)=2\Phi_{d,l}(\beta)\] for every \(z\ne z_0\) in \(S_\beta\). ◻

Proof. By Lemma 15, the supremum over \(\mathcal{R}_\beta\cap\{\det H\geq0\}\) is \(2\Phi_{d,l}(\beta)\) and is attained only at the saddle. It remains to consider \(\det H<0\). An interior maximum in this region is impossible, because at such a point the Hessian would have to be negative semidefinite, whereas \(\det H<0\) makes it indefinite. The boundary \(\det H=0\) is already controlled by Lemma 15. By Lemma 14, the remaining boundary of the region \(\det H<0\) inside \(\widehat{\mathcal{R}}_\beta\) is the upper boundary \(\eta=\beta-\rho\) with \(\rho\in[\rho_5,\beta]\).

Along this boundary put \[T_{d,l,\beta}(\rho):=\Psi_{d,l}(\beta,\rho,l(\beta-\rho)). \label{dtr}\tag{32}\] Then \[\begin{align} T_{d,l,\beta}(\rho) &=-\rho\log\rho+(l-1)(1-2\beta+\rho)\log(1-2\beta+\rho) +l(\beta-\rho)\log(d-1)\\ &\quad +(l-2)(\beta-\rho)\log(\beta-\rho) -\frac{l}{d}(1-\beta d)\log(1-\beta d), \end{align}\] and \[T_{d,l,\beta}''(\rho) =-\frac{1}{\rho}+\frac{l-1}{1-2\beta+\rho}+\frac{l-2}{\beta-\rho} =\frac{[l(1-\beta)-1]\rho-\beta(1-2\beta)}{\rho(\beta-\rho)(1-2\beta+\rho)}. \label{dtr2}\tag{33}\] Thus \(T_{d,l,\beta}\) is convex on \([\rho_5,\beta)\). Its supremum on this interval is therefore the larger of its endpoint limits. The endpoint at \(\rho_5\) lies on \(\det H=0\) and is controlled by Lemma 15. At \(\rho=\beta\), \[\lim_{\rho\uparrow\beta}T_{d,l,\beta}(\rho)=\Phi_{d,l}(\beta)<2\Phi_{d,l}(\beta)\] by ?? . Hence the supremum over \(\det H<0\) is at most \(2\Phi_{d,l}(\beta)\), and equality can occur only at the saddle. ◻

Proof. Let \(z_\beta=(\beta^2,l(d-1)\beta^2)\) and write \(\Psi_\beta(\rho,\theta)=\Psi_{d,l}(\beta,\rho,\theta)\). Under either hypothesis, \(z_\beta\) is the unique point in \(\mathcal{R}_\beta\) at which the supremum of \(\Psi_\beta\) is equal to \(2\Phi_{d,l}(\beta)\), and the Hessian \(H_\beta:=H(\beta,z_\beta)\) is negative definite. In the case \(l=2\) this follows from Lemma 11; in the case \(l\ge3\) it follows from Lemma 16. Moreover 21 gives \(\Psi_\beta(z_\beta)=2\Phi_{d,l}(\beta)\).

We first make the contribution away from the saddle explicit. Extend \(\Psi_\beta\) continuously to the compact polygon \[\overline{\mathcal{R}_\beta}= \left\{0\le\rho\le\beta, \max\left\{0,2l\beta-l\rho-\frac{l}{d}\right\}\le \theta\le l\beta-l\rho \right\}\] by the convention \(u\log u=0\) at \(u=0\). The boundary analysis in Lemmas 11 and 16 shows that \(z_\beta\) is the unique maximizer of this continuous extension with value \(2\Phi_{d,l}(\beta)\). Therefore, for every fixed \(\delta>0\) small enough, there exists \(\eta_\delta>0\) such that \[\label{eq:l08-away-gap} \sup\{\Psi_\beta(z):z\in\overline{\mathcal{R}_\beta}, \|z-z_\beta\|\ge\delta\} \le 2\Phi_{d,l}(\beta)-\eta_\delta .\tag{34}\] The crude form of Stirling’s formula applied to the exact expression ?? gives a polynomially prefactored upper bound, uniformly over all admissible lattice points, \[\label{eq:l08-crude-bound} F_\beta(s,t) \le m^C\exp\left\{m\Psi_\beta(s/m,t/m)+o(m)\right\},\tag{35}\] with \(C\) depending only on \(d,l\) and \(\beta\). Combining 34 and 35 , and using that the number of admissible pairs \((s,t)\) is \(O(m^2)\), shows that the contribution from \(\| (s/m,t/m)-z_\beta\|\ge\delta\) is \(O(m^{C+2}e^{m(2\Phi_{d,l}(\beta)-\eta_\delta+o(1))})\), hence is exponentially smaller than the saddle contribution.

It remains to evaluate a small neighborhood of \(z_\beta\). Choose \(\delta>0\) so that the closed ball \(B_\delta(z_\beta)\) is contained in \(\mathcal{R}_\beta\) and the Hessian of \(\Psi_\beta\) is negative definite throughout this ball. Taylor’s formula and negative definiteness give, uniformly for \(\|z-z_\beta\|\leq\delta\), \[\Psi_\beta(z)=2\Phi_{d,l}(\beta)+\frac{1}{2}(z-z_\beta)H_\beta(z-z_\beta)^T +O(\|z-z_\beta\|^3),\] and also \[\Psi_\beta(z)\leq 2\Phi_{d,l}(\beta)-c\|z-z_\beta\|^2\] for some \(c>0\). Hence the part of this neighborhood outside an \(L/\sqrt m\)-neighborhood of \(z_\beta\) is bounded by a Gaussian tail, uniformly in \(m\), and vanishes as \(L\to\infty\).

On the \(L/\sqrt m\)-neighborhood, set \[x_1=\sqrt m\left(\frac{s}{m}-\beta^2\right), \qquad x_2=\sqrt m\left(\frac{t}{m}-l(d-1)\beta^2\right).\] Using Lemma 6 and then the Riemann-sum limit gives \[\begin{align} \mathbb{E} Z_{m\beta}^2 &=\frac{e^{2m\Phi_{d,l}(\beta)}}{m} \frac{1}{4\pi^2\beta^3(1-d\beta)^2\sqrt{l(d-1)}} \int_{\mathbb{R}^2}\exp\left\{\frac{1}{2} xH_\beta x^T\right\}\,dx\,(1+o(1))\\ &=\frac{e^{2m\Phi_{d,l}(\beta)}}{m} \frac{1}{2\pi\beta^3(1-d\beta)^2\sqrt{l(d-1)}\sqrt{\det H_\beta}} (1+o(1)). \end{align}\] By Lemma 3, \[(\mathbb{E} Z_{m\beta})^2= \frac{e^{2m\Phi_{d,l}(\beta)}}{m} \frac{1}{2\pi\beta(1-d\beta)}(1+o(1)).\] Finally, 27 gives \[\det H_\beta= \frac{\beta^2(d+l-dl)-2\beta+1}{l\beta^4(1-d\beta)^2(1-\beta)^2(d-1)}.\] Substitution yields ?? . ◻

Proof. All ingredients in the proof of Lemma 17 are compact-uniform in \(\beta\). The saddle remains a positive distance from the boundary on compact subsets, the Hessian determinant at the saddle is bounded away from zero by Lemma 10, and the gap estimate 34 is uniform by compactness and the uniqueness of the maximizer. The Stirling estimates in Lemmas 2, 3, and 6 are uniform on compact sets. The same discrete Laplace argument therefore gives the stated uniform convergence. ◻

Proof. From 37 , for \(t\ge1\) and fixed \(t,s,k\), \[\begin{align} \mathcal{E}^{(m)}_{\beta,k}(t,s) &=(1+o(1))\, \frac{(d-1)^k(l-1)^k(1-d\beta)^k}{2(1-\beta)^k} \binom{k-t-1}{t-1}\binom{k-2t}{s}\frac{1}{t} \\ &\quad\times \frac{\beta^{t+s}(1-\beta)^t(d-2)^s}{(1-d\beta)^{2t+s}}. \end{align}\] Equivalently, since \(t^{-1}\binom{k-t-1}{t-1}=(k-t)^{-1}\binom{k-t}{t}\), \[\begin{align} \sum_{s=0}^{k-2t}\mathcal{E}^{(m)}_{\beta,k}(t,s) &=(1+o(1))\, \frac{(d-1)^k(l-1)^k}{2(1-\beta)^k} \frac{1}{k-t}\binom{k-t}{t} \\ &\quad\times \beta^t(1-\beta)^t(1-2\beta)^{k-2t}. \end{align}\] Similarly, 38 gives \[\sum_{s=0}^{k}\mathcal{E}^{(m)}_{\beta,k}(0,s) =(1+o(1))\, \frac{(d-1)^k(l-1)^k}{2k(1-\beta)^k}(1-2\beta)^k .\] Thus, with \[\alpha=\frac{\beta(1-\beta)}{(1-2\beta)^2},\] we obtain \[\begin{align} \mathbb{E}(C_k\mid A_{M_0}) &=(1+o(1))\, \frac{(d-1)^k(l-1)^k(1-2\beta)^k}{2(1-\beta)^k} \sum_{t=0}^{\lfloor k/2\rfloor} \frac{1}{k-t}\binom{k-t}{t}\alpha^t . \end{align}\] The elementary coefficient identity \[\sum_{t=0}^{\lfloor k/2\rfloor}\frac{1}{k-t}\binom{k-t}{t}\alpha^t =\frac{(1-\beta)^k+(-1)^k\beta^k}{k(1-2\beta)^k}\] follows, for example, by extracting the coefficient of \(x^k\) in \(-\log(1-x(1+x)/\alpha)\). Substituting this identity proves ?? . ◻

Proof. Expose the selected cycles one at a time, recording for each cycle the pair \((t,s)\) used above. Since the total number of exposed hyperedges and vertices is bounded independently of \(m\), after any finite number of disjoint cycles has been exposed all populations appearing in 37 and 38 are changed by only \(O(1)\). Therefore the conditional contribution of the next disjoint \(k\)-cycle of type \((t,s)\) is \(\mathcal{E}^{(m)}_{\beta,k}(t,s)(1+O(m^{-1}))\), uniformly over the finitely many possible types. Multiplying these factors and summing over the finitely many type choices gives \[\frac{N^{\mathrm{dis}}_{(r_1,k_1),\ldots,(r_g,k_g)}}{N_{M_0}} =\prod_{i=1}^g \left(\sum_{t=0}^{\lfloor k_i/2\rfloor}\sum_{s=0}^{k_i-2t} \mathcal{E}^{(m)}_{\beta,k_i}(t,s)\right)^{r_i}(1+o(1)).\] The result follows from Lemma 18. ◻

Proof. The total length of the selected cycles is fixed. If the selected cycles are not disjoint in the incidence multigraph, then their union has at least one identification among the hyperedge- or vertex-nodes which would be distinct in the disjoint case. Thus the number of choices for the labelled hyperedges and vertices of the union is smaller by a factor \(O(m^{-1})\) than in the disjoint case. The probability that the required finite incidence pattern is realised, conditioned on \(A_{M_0}\), is of the same order as for a disjoint pattern with the same number of incidences, because only \(O(1)\) half-edges are exposed and all remaining pools have size \(\Theta(m)\). The constants in these \(O(\cdot)\) bounds depend only on the fixed cycle lengths and multiplicities; if \(\beta\) is restricted to a compact subset of \((0,1/d)\), all exposed matching and non-matching half-edge pools have size \(\Theta(m)\) uniformly in \(\beta\). Thus each finite overlap pattern has probability at most a constant multiple of the corresponding disjoint-pattern probability after accounting for the lost free vertex or hyperedge choice. Summing over the finitely many overlap patterns gives \(N^{\mathrm{ov}}_{(r_1,k_1),\ldots,(r_g,k_g)}=O(m^{-1}) N^{\mathrm{dis}}_{(r_1,k_1),\ldots,(r_g,k_g)}\), which proves the lemma. ◻

Proof. We prove the conditional statement; the unconditional one is the same argument with no distinguished matching, equivalently with \(\beta=0\), in which case \(\mu_{0,k}=\lambda_k\). For arbitrary fixed distinct \(k_1,\ldots,k_g\le b\) and \(r_1,\ldots,r_g\ge0\), \[\mathbb{E}\left[\prod_{i=1}^g (C_{k_i})_{r_i}\,\middle|\,A_{M_0}\right] =\frac{N^{\mathrm{dis}}_{(r_1,k_1),\ldots,(r_g,k_g)} +N^{\mathrm{ov}}_{(r_1,k_1),\ldots,(r_g,k_g)}}{N_{M_0}},\] where \((x)_r=x(x-1)\cdots(x-r+1)\). Lemmas 19 and 20 show that these mixed factorial moments converge to \(\prod_i\mu_{\beta,k_i}^{r_i}\), which are the mixed factorial moments of independent Poisson variables. Since Poisson laws are determined by their factorial moments, the conditional convergence follows. ◻

Proof. This is the standard Chernoff bound for Poisson random variables; see, for example, Theorem A.15 of [35]. ◻

Proof. Put \(z=y/(a\sqrt\lambda)\). The left side is \(e^{\lambda\phi(z)}\). If \(0<z\le1/2\), then \(\log(1+z)\ge z-z^2/2\), and hence \(\phi(z)\le -z^2/2+z^3/2\le -z^2/4\). This gives the first term in ?? . If \(1/2<z\le W\), then \(\phi\) is decreasing on \((0,\infty)\), so \(\phi(z)\le\phi(1/2)<0\); since \(\lambda=y^2/(a^2z^2)\ge y^2/(a^2W^2)\), this gives the second term. If \(z>W\), then \(\phi(z)\le -z(\log(1+z)-1)\le -z(\log(1+W)-1)\), giving the third term. ◻

Proof. By 42 and Lemma 21, for fixed \(b\) and \(y\), \[\begin{align} \lim_{m\to\infty} \frac{\sum_{\mathbf{c}\in S_b(y)}p_{\mathbf{c}}E_{\beta,\mathbf{c}}^2}{(\mathbb{E} Z_{m\beta})^2} &=\prod_{k=1}^b e^{(\mu_{\beta,k}-\lambda_k)^2/\lambda_k} \mathbb{P}\left\{|X_k-\lambda_k|\le y\sqrt{\lambda_k}\right\}, \end{align}\] where the \(X_k\) are independent Poisson variables with means \[\nu_k:=\frac{\mu_{\beta,k}^2}{\lambda_k} =\lambda_k\left(1+(-1)^kq_\beta^k\right)^2 .\] Let \(a_k:=1+(-1)^kq_\beta^k\). Then \(\gamma_\beta\le a_k\le A_\beta\). For sufficiently large \(y\), the lower-tail parameter comparing \(\lambda_k-y\sqrt{\lambda_k}\) with \(\nu_k\) is at least \(y/(2A_\beta^2\sqrt{\lambda_k})\) whenever the lower endpoint is non-negative; if the endpoint is negative, the lower tail is empty. Lemma 22 gives \[\mathbb{P}\{X_k<\lambda_k-y\sqrt{\lambda_k}\} \le \exp\left\{-\frac{\gamma_\beta^2y^2}{8A_\beta^4}\right\}.\] For the upper tail, the corresponding upper-tail parameter is at least \(y/(2A_\beta^2\sqrt{\lambda_k})\). Applying ?? with \(\lambda\) replaced by \(\gamma_\beta^2\lambda_k\) and \(y\) replaced by \(\gamma_\beta y/(2A_\beta^2)\) gives \[\begin{align} \mathbb{P}\{X_k>\lambda_k+y\sqrt{\lambda_k}\} &\le \exp\left\{-\frac{\gamma_\beta^2y^2}{16A_\beta^4}\right\} + \exp\left\{-\frac{|\phi(1/2)|\gamma_\beta^2y^2}{4A_\beta^4W^2}\right\} \\ &\quad+ \exp\left\{-(\log(1+W)-1) \frac{\gamma_\beta^2y\sqrt{\lambda_k}}{2A_\beta^2}\right\}. \end{align}\] By the union bound over \(1\le k\le b\), \[\prod_{k=1}^b \mathbb{P}\left\{|X_k-\lambda_k|\le y\sqrt{\lambda_k}\right\} \ge \Theta_{\beta,b,W}(y).\] Finally, \[\prod_{k=1}^b e^{(\mu_{\beta,k}-\lambda_k)^2/\lambda_k} =\exp\left\{\sum_{k=1}^b\frac{r_\beta^k}{2k}\right\} =\frac{1}{\sqrt{1-r_\beta}} \exp\left\{-\sum_{k>b}\frac{r_\beta^k}{2k}\right\}.\] Since \(\sum_{k>b} r_\beta^k/(2k)\le -\log(1-r_\beta^b)\) for \(0<r_\beta<1\), the last exponential is at least \(1-r_\beta^b\). This proves the lemma. ◻

Proof. By Lemma 21(1), \((C_1,\ldots,C_b)\) converges to independent Poisson variables with means \(\lambda_1,\ldots,\lambda_b\). The lower-tail bound in Lemma 22 gives \(e^{-y^2/2}\) for each \(k\). The upper-tail bound in Lemma 22, followed by ?? with \(a=1\), gives the remaining three terms. A union bound over \(k=1,\ldots,b\) completes the proof. ◻

Proof. Let \(\delta_k=(-1)^kq_\beta^k\). By 42 , uniformly over the finite set \(S_b(y)\), \[\frac{E_{\beta,\mathbf{c}}}{\mathbb{E} Z_{m\beta}} =(1+o(1))\prod_{k=1}^b e^{-\lambda_k\delta_k}(1+\delta_k)^{c_k}.\] Write \(c_k=\lambda_k+x_k\sqrt{\lambda_k}\) with \(|x_k|\le y\). Then \[\begin{align} \log\prod_{k=1}^b e^{-\lambda_k\delta_k}(1+\delta_k)^{c_k} &=\sum_{k=1}^b \lambda_k\{\log(1+\delta_k)-\delta_k\} +\sum_{k=1}^b x_k\sqrt{\lambda_k}\log(1+\delta_k). \end{align}\] Since \(|\delta_k|\le q_\beta^k\) and \(q_\beta<1\), \[\log(1+u)-u\ge -\frac{u^2}{1-q_\beta},\qquad |u|\le q_\beta,\] and \[|\log(1+u)|\le \frac{|u|}{1-q_\beta},\qquad |u|\le q_\beta.\] Therefore \[\sum_{k=1}^b \lambda_k\{\log(1+\delta_k)-\delta_k\} \ge -\frac{1}{1-q_\beta}\sum_{k=1}^{\infty}\lambda_k q_\beta^{2k}>-\infty,\] and \[\sum_{k=1}^b x_k\sqrt{\lambda_k}\log(1+\delta_k) \ge -\frac{y}{1-q_\beta} \sum_{k=1}^{\infty}\sqrt{\lambda_k}\,q_\beta^k> -\infty\cdot y.\] The last series converges exactly when \(r_\beta<1\). Absorbing the uniform \(o(1)\) into the constant \(A\) proves the result. ◻

Proof. The proofs expose only finitely many hyperedges, vertices and half-edges. For \(\beta\in K\), every pool appearing in the exposure process has cardinality \(c m+O(1)\) with \(c\) bounded above and below away from zero uniformly in \(\beta\). Hence every factorial ratio used in 37 –38 admits a uniform expansion of the form \((cm)^r(1+O_K(m^{-1}))\). This gives the uniform convergence of the conditional means, the disjoint factorial moments, the overlapping estimates and therefore the Poisson limits. Formula 42 is obtained from the same factorial-moment convergence by Bayes’ rule, so it is also uniform for fixed \(b\) and fixed \(\mathbf{c}\).

If \(\sup_{\beta\in K}r_\beta<1\), then also \(\sup_{\beta\in K}q_\beta<1\) and the series \(\sum_k\lambda_k q_\beta^{2k}\) and \(\sum_k\sqrt{\lambda_k}q_\beta^k\) are uniformly convergent on \(K\). The Poisson tail estimates in Lemmas 24 and 26 are then uniform after replacing \(\gamma_\beta=1-q_\beta\) by its positive infimum on \(K\) and \(A_\beta=1+q_\beta\) by its finite supremum on \(K\). ◻

Proof. We first check that \(\Phi_{d,l}(\beta)>0\). If \(l=2\), this follows from Lemma 1(2), since \(f_2(1/d)\ge0\). If \(l\ge3\), then \(L_1=1/(dl-d-l+2)\). Put \[B:=dl-d-l+2=(d-1)(l-1)+1.\] For \(0<u<1\) we have \((1-u)\log(1-u)\ge -u\) and \(-(1-u)\log(1-u)\ge u(1-u)\). Hence, for every \(0<\beta'\le 1/B\), \[\begin{align} \Phi_{d,l}(\beta') &\ge \beta'\log\frac{1}{\beta'}-(l-1)\beta' +l\beta'(1-d\beta') \\ &=\beta'\left(\log\frac{1}{\beta'}+1-ld\beta'\right) \ge \beta'\left(\log B+1-\frac{ld}{B}\right)>0. \end{align}\] The last inequality uses \(B\ge3\) and \(ld/B\le2\), which is equivalent to \((d-2)(l-2)\ge0\). Since \(\beta_m\to\beta\) and \(\beta_m\le L_1\) eventually in case \(l\ge3\), this proves the desired positivity.

By the compact-uniform first moment estimate in Lemma 3, \[\label{eq:section5-first-moment-log} \frac{1}{m}\log \mathbb{E} Z_{h_m}\longrightarrow \Phi_{d,l}(\beta).\tag{47}\] It remains to prove \[\label{lmb} \frac{1}{m}\log\frac{Z_{h_m}}{\mathbb{E} Z_{h_m}}\longrightarrow0 \quad\text{in probability}.\tag{48}\] The assumptions also imply \(r_\beta<1\). If \(l=2\), then for \(d=2\) we have \(r_\beta=q_\beta^2<1\), while for \(d\ge3\) the inequality \(\beta<1/d\) gives \(q_\beta<1/(d-1)\) and hence \(r_\beta=(d-1)q_\beta^2<1\). If \(l\ge3\), then \((d-1)(l-1)>1\) and the bound \(\beta_m\le L_1\) implies \(q_\beta\le 1/((d-1)(l-1))\), whence \(r_\beta\le 1/((d-1)(l-1))<1\).

Write \[X_m:=Z_{h_m},\qquad \mathcal{F}_{m,b}:=\sigma(C_1,\ldots,C_b),\qquad X_{m,b}:=\mathbb{E}(X_m\mid\mathcal{F}_{m,b}).\] All cycle estimates below are applied with \(\beta=\beta_m\); the compact-uniform form needed for this varying-density use is supplied by Lemma 27.

Fix \(b\ge1\). For every fixed \(\mathbf{c}=(c_1,\ldots,c_b)\in\mathbb{N}^b\), the asymptotic formula 42 , applied with \(\beta_m\) in place of \(\beta\), gives \[\label{eq:section5-Ebc-limit} \frac{E_{\beta_m,\mathbf{c}}}{\mathbb{E} Z_{h_m}} \longrightarrow \prod_{k=1}^b e^{-\lambda_k\delta_k}(1+\delta_k)^{c_k}, \qquad \delta_k:=(-1)^kq_\beta^k .\tag{49}\] The convergence is uniform over finite sets of \(\mathbf{c}\)’s by Lemma 27. Lemma 21(1) gives \[(C_1,\ldots,C_b)\Rightarrow (P_1,\ldots,P_b),\] where \(P_1,\ldots,P_b\) are independent Poisson random variables with means \(\lambda_1,\ldots,\lambda_b\). Combining this convergence with 49 and tightness of \((C_1,\ldots,C_b)\) yields \[\label{eq:section5-Xmb-limit} \frac{X_{m,b}}{\mathbb{E} X_m}\Rightarrow Y_b:=\prod_{k=1}^b e^{-\lambda_k\delta_k}(1+\delta_k)^{P_k}.\tag{50}\] Since \(q_\beta<1\), all factors in \(Y_b\) are positive.

The sequence \(Y_b\) converges almost surely to a finite strictly positive random variable. Indeed, \[\begin{align} \log Y_b &=\sum_{k=1}^b (P_k-\lambda_k)\log(1+\delta_k) +\sum_{k=1}^b \lambda_k\{\log(1+\delta_k)-\delta_k\}. \end{align}\] Because \[\label{eq:section5-l2-summable} \sum_{k\ge1}\lambda_k\delta_k^2 =\sum_{k\ge1}\frac{r_\beta^k}{2k}<\infty,\tag{51}\] the deterministic series is absolutely convergent, and the centred independent series is convergent in \(L^2\) and almost surely. We denote the limit by \(Y_\infty\); then \(Y_\infty\in(0,\infty)\) almost surely.

We next show that the residual \(X_m-X_{m,b}\) is negligible on the exponential scale. Since \(X_{m,b}=\mathbb{E}(X_m\mid\mathcal{F}_{m,b})\), \[\label{eq:section5-conditional-var-identity} \mathbb{E}(X_m-X_{m,b})^2 =\mathbb{E} X_m^2-\mathbb{E} X_{m,b}^2.\tag{52}\] The hypotheses of the lemma imply the second moment asymptotic ?? ; for \(\beta_m\to\beta\) the convergence is uniform by Corollary 1. Hence \[\label{eq:section5-second-ratio} \frac{\mathbb{E} X_m^2}{(\mathbb{E} X_m)^2} \longrightarrow R_\beta:=\frac{1}{\sqrt{1-r_\beta}}.\tag{53}\] On the other hand, fix any \(W_0>2\). Lemma 24, applied with \(W=W_0\) and with \(\beta_m\) in place of \(\beta\), and then using the compact-uniform form in Lemma 27 together with \(\beta_m\to\beta\), gives, for every fixed \(b\) and after letting its truncation parameter \(y\to\infty\), \[\label{eq:section5-truncated-second-lower} \liminf_{m\to\infty}\frac{\mathbb{E} X_{m,b}^2}{(\mathbb{E} X_m)^2} \ge (1-r_\beta^b)R_\beta .\tag{54}\] Combining 52 , 53 , and 54 , we obtain \[\label{eq:section5-residual-variance} \limsup_{m\to\infty} \frac{\mathbb{E}(X_m-X_{m,b})^2}{(\mathbb{E} X_m)^2} \le R_\beta r_\beta^b .\tag{55}\]

We now prove the two tails in 48 . The upper tail follows immediately from Markov’s inequality: \[\label{eq:section5-upper-tail} \mathbb{P}\left\{X_m>e^{m\varepsilon}\mathbb{E} X_m\right\} \le e^{-m\varepsilon}.\tag{56}\] For the lower tail, let \(\alpha>0\) be arbitrary. Choose \(a>0\) such that \[\mathbb{P}\{Y_\infty\le 4a\}<\alpha/3.\] Then choose \(b\) so large that \[\label{eq:section5-choose-b} \mathbb{P}\{|Y_b-Y_\infty|>a\}<\alpha/3, \qquad \frac{4R_\beta r_\beta^b}{a^2}<\alpha/3.\tag{57}\] For this fixed \(b\), the Portmanteau theorem, 50 , and 57 give \[\begin{align} \label{eq:section5-Xmb-lower-tight} \limsup_{m\to\infty} \mathbb{P}\left\{X_{m,b}<2a\,\mathbb{E} X_m\right\} &\le \mathbb{P}\{Y_b\le 2a\} \notag\\ &\le \mathbb{P}\{Y_\infty\le 4a\}+\mathbb{P}\{|Y_b-Y_\infty|>a\}\le 2\alpha/3. \end{align}\tag{58}\] If \(m\) is large enough that \(e^{-m\varepsilon}<a\), then \[\begin{align} \mathbb{P}\left\{X_m<e^{-m\varepsilon}\mathbb{E} X_m\right\} &\le \mathbb{P}\left\{X_{m,b}<2a\,\mathbb{E} X_m\right\} \\ &\quad+ \mathbb{P}\left\{|X_m-X_{m,b}|>a\,\mathbb{E} X_m\right\}. \end{align}\] By Chebyshev’s inequality and 55 , the limsup of the second probability is at most \(R_\beta r_\beta^b/a^2<\alpha/12\). Together with 58 , this gives \[\limsup_{m\to\infty} \mathbb{P}\left\{X_m<e^{-m\varepsilon}\mathbb{E} X_m\right\} \le \alpha .\] Since \(\alpha>0\) is arbitrary, the lower tail tends to zero. Combining this with 56 proves 48 . Finally, 47 and 48 imply ?? . ◻

Proof. By Lemma 1, \(\Phi'_{d,l}\) is strictly decreasing on \((0,1/d)\) and \(\beta_*\) is its unique zero. Hence \(\Phi'_{d,l}(L_1)\le0\) implies \(\beta_*\le L_1\). ◻

Proof. Differentiating 60 , \[\partial_\beta \Delta_{d,l}(\beta,x) =-\log\beta-(l-1)\log(1-\beta)+l\log(1-d\beta)+\log x,\] and \[\partial_{\beta\beta}^2\Delta_{d,l}(\beta,x) =-\frac{1}{\beta}- \frac{ld-d\beta-(l-1)}{(1-\beta)(1-d\beta)}<0, \qquad 0<\beta<\frac{1}{d} .\] Moreover, \[\lim_{\beta\downarrow0}\partial_\beta\Delta_{d,l}(\beta,x)=+\infty, \qquad \lim_{\beta\uparrow1/d}\partial_\beta\Delta_{d,l}(\beta,x)=-\infty .\] Hence \(\partial_\beta\Delta_{d,l}(\beta,x)\) has a unique zero in \((0,1/d)\), and strict concavity makes this zero the unique maximizer. The equation for this zero is ?? , which is equivalent to ?? . Since the continuous extension satisfies \(\Delta_{d,l}(0,x)=0\) and the right derivative at \(0\) is \(+\infty\), the maximum value is strictly positive. Finally, if \(l\ge3\) and ?? holds, then the strictly decreasing function \(\partial_\beta\Delta_{d,l}(\beta,x)\) is already non-positive at \(L_1\). Since its unique zero is \(\beta_*(x)\), we obtain ?? . ◻

Proof. Let \(\overline{\Phi}_{d,l}\) denote the continuous extension of \(\Phi_{d,l}\) to \([0,1/d]\). We shall use the following endpoint-uniform estimate. It follows from 7 and the Stirling bounds in Lemma 2, exactly as in the proof of Theorem 1(1): \[\label{eq:weighted-crude-first-moment} \sup_{0\le h\le H_m} \left| \frac{1}{m}\log \mathbb{E} Z_h- \overline{\Phi}_{d,l}\left(\frac{h}{m}\right) \right|=o(1),\tag{61}\] where \(Z_0=1\) and hence the term \(h=0\) is interpreted with \(\overline{\Phi}_{d,l}(0)=0\). Indeed, one may use the uniform form \(n!=\exp\{n\log n-n+O(\log(n+1))\}\), with \(0!=1\), in each factorial in 7 ; the total error is \(O(\log m)\), including the endpoint regimes \(h=O(1)\) and \(H_m-h=O(1)\). Therefore \[\begin{align} \mathbb{E} Z(x) &=\sum_{h=0}^{H_m}x^h\mathbb{E} Z_h \\ &\le (H_m+1) \exp\left\{m\sup_{0\le\beta\le1/d} \Delta_{d,l}(\beta,x)+o(m)\right\} \\ &=(H_m+1)\exp\left\{m\Delta_{d,l}(\beta_*(x),x)+o(m)\right\} . \end{align}\] This gives the upper bound for the logarithmic limit.

For the lower bound, choose any integer sequence \(h_m\) with \(h_m/m\to\beta_*(x)\), for instance \(h_m=\lfloor m\beta_*(x)\rfloor\). Since \(\beta_*(x)\in(0,1/d)\), the compact-uniform first-moment estimate proved in Section 2 gives \[\frac{1}{m}\log\mathbb{E} Z_{h_m}\longrightarrow \Phi_{d,l}(\beta_*(x)).\] Consequently, \[\frac{1}{m}\log\left(x^{h_m}\mathbb{E} Z_{h_m}\right) \longrightarrow \Phi_{d,l}(\beta_*(x))+\beta_*(x)\log x =\Delta_{d,l}(\beta_*(x),x).\] Since \(\mathbb{E} Z(x)\ge x^{h_m}\mathbb{E} Z_{h_m}\), the lower bound follows. ◻

Proof. A direct comparison of the first and third entries in 66 gives \(A_{d,l}<C_{d,l}\) for \(d\ge2,l\ge3\). Thus \(L_2<C_{d,l}\). Hence \(\beta<C_{d,l}\), and the negative definiteness of the Hessian follows from Lemma 10.

Next, \(A_{d,l}\le1/(2l)\). If \(d\ge2l\) this is automatic from the definition of \(A_{d,l}\). If \(d<2l\), it is equivalent, after squaring two positive quantities, to \[4l^2(d-1)\ge (2l-d)^2\bigl(d^{l/(l-1)}-1\bigr).\] This last inequality follows from the mean-value theorem applied to \(u\mapsto u^{l/(l-1)}\) on \([1,d]\) and the elementary bound \(\frac{l}{l-1}d^{1/(l-1)}(2l-d)^2\le4l^2\). Thus \(\beta\le1/(2l)\). On \((0,\beta^2]\), \[A'(\rho)=-\frac{1}{l}\left\{\frac{1}{\rho^2}+ \frac{2(l-1)}{(\beta-\rho)^2}-\frac{l-1}{(1-2\beta+\rho)^2}\right\}<0,\] because \(\beta\le1/(2l)\) implies \(\beta-\rho\le\beta<\sqrt2(1-2\beta+\rho)\). Therefore \(A\) is decreasing on this interval, and \[\beta^2(1-\beta)^2 l A(\beta^2)=1-2l\beta+l\beta^2>0.\] Hence \(A(\rho)>0\) for \(0<\rho\le\beta^2\).

For ?? , direct substitution into \(\partial_\rho\widehat\Psi_\beta\) gives the displayed expression. Exponentiating the inequality in ?? shows that it is equivalent to \[1-2\beta+d\beta^2\le d^{1/(l-1)}(1-d\beta)^2 .\] Solving this quadratic inequality in \(\beta\) gives precisely \(0<\beta\le A_{d,l}\).

Finally, another direct calculation gives \(\Phi'_{d,l}(A_{d,l})>0\). Since \(\Phi'_{d,l}\) is strictly decreasing, \(0<\beta\le L_2\le A_{d,l}\) implies \(\Phi'_{d,l}(u)>0\) for every \(u\in(0,\beta]\). With \(\Phi_{d,l}(0+)=0\), this yields ?? . ◻

Proof. Put \[g(\rho):=\partial_\rho\widehat\Psi_\beta(\rho,d\beta^2-\rho), \qquad \beta^2\le\rho\le d\beta^2.\] At the saddle, \(g(\beta^2)=0\). By ?? , \(g(d\beta^2)\le0\). Moreover \[\begin{align} g'(\rho) &=\partial_{\rho\rho}^2\widehat\Psi_\beta(\rho,d\beta^2-\rho) -\partial_{\rho\eta}^2\widehat\Psi_\beta(\rho,d\beta^2-\rho)\\ &=-\frac{1}{\rho}+\frac{l-1}{1-2\beta+\rho} +\frac{2(l-1)}{\beta-\rho} =\frac{V(\rho)}{\rho(\beta-\rho)(1-2\beta+\rho)}, \end{align}\] where \[V(\rho)=\rho^2+(l-2\beta-l\beta)\rho+2\beta^2-\beta.\] The denominator is positive on \([\beta^2,d\beta^2]\). The polynomial \(V\) is an upward-opening quadratic and \(V(0)=-\beta(1-2\beta)<0\); hence \(V\) has exactly one positive zero. Thus \(g\) decreases and then increases on the interval. A function with this monotonicity pattern has its maximum on a compact interval at one of the two endpoints. Since both endpoint values are at most zero, ?? follows. ◻

Proof. For fixed \(\rho\), the derivative in the \(\eta\) direction satisfies \[\partial_{\eta\eta}^2\widehat\Psi_\beta(\rho,\eta)<0.\] Moreover, \(\partial_\eta\widehat\Psi_\beta\to+\infty\) at the lower endpoint of the vertical section and \(\partial_\eta\widehat\Psi_\beta\to-\infty\) at the upper endpoint \(\eta=\beta-\rho\). Hence the vertical maximizer \(u_\beta(\rho)\) exists and is unique. It is characterized by \[\label{eq:vertical-critical} (d-1)(\beta-\rho-u_\beta(\rho))^2 =u_\beta(\rho)\{1-2\beta d+\rho d+d u_\beta(\rho)\}.\tag{67}\] The implicit function theorem applies, and the envelope theorem gives \[\label{eq:Mprime-envelope} M_\beta'(\rho)=\partial_\rho\widehat\Psi_\beta(\rho,u_\beta(\rho)).\tag{68}\]

We first treat \(0<\rho<\beta^2\). Let \(a(\rho)\) be as in 22 , and set \[R(\rho):=\beta-\rho-(d-1)a(\rho), \qquad S(\rho):=1-2\beta d+\rho d+d(d-1)a(\rho).\] Since \(A(\rho)=a'(\rho)/a(\rho)>0\) on \((0,\beta^2]\), the function \(R\) is decreasing there, and so \[R(\rho)>R(\beta^2)=\beta(1-d\beta)>0.\] If \(S(\rho)\le0\), then \[(d-1)R(\rho)^2-(d-1)a(\rho)S(\rho)>0.\] If \(S(\rho)>0\), then \(S(\rho')>0\) for every \(\rho'\in[\rho,\beta^2]\), because \(S'(\rho')=d+d(d-1)a'(\rho')>0\) on \((0,\beta^2]\). Hence the logarithm \[h(\rho):=\log a(\rho)+\log S(\rho)-2\log R(\rho)\] is well-defined on \([\rho,\beta^2]\). On this interval, \[h'(\rho') =\frac{d}{S(\rho')}+\frac{2}{R(\rho')} +A(\rho')\left(1+\frac{d(d-1)a(\rho')}{S(\rho')} +\frac{2(d-1)a(\rho')}{R(\rho')}\right)>0.\] Since \(h(\beta^2)=0\), we have \(h(\rho)<0\) for \(\rho<\beta^2\) in this case. Thus in all cases \[\label{eq:F-at-a-positive} (d-1)R(\rho)^2-(d-1)a(\rho)S(\rho)>0, \qquad 0<\rho<\beta^2.\tag{69}\] For fixed \(\rho\), the left-hand side of the vertical equation, \[F_\rho(\eta):=(d-1)(\beta-\rho-\eta)^2 -\eta(1-2\beta d+\rho d+d\eta),\] is strictly decreasing in \(\eta\) on the feasible vertical section. Since 69 says that \(F_\rho((d-1)a(\rho))>0\), the zero \(u_\beta(\rho)\) of \(F_\rho\) satisfies \[\label{eq:u-greater-a} u_\beta(\rho)>(d-1)a(\rho), \qquad 0<\rho<\beta^2.\tag{70}\] Using 67 to simplify \(\partial_\rho\widehat\Psi_\beta(\rho,u_\beta(\rho))\), we get \[\label{eq:Mprime-a-formula} M_\beta'(\rho)=l\log\frac{u_\beta(\rho)}{(d-1)a(\rho)}.\tag{71}\] Together with 70 , this proves \(M_\beta'(\rho)>0\) on \((0,\beta^2)\).

We next treat \(\rho>\beta^2\). Let \[z_0:=(\beta^2,(d-1)\beta^2).\] If \(u_\beta(\rho)\ge(d-1)\beta^2\), then \[\partial_\rho\widehat\Psi_\beta(\beta^2,u_\beta(\rho))<0,\] because \(\partial_\rho\widehat\Psi_\beta(z_0)=0\) and \(\partial_{\rho\eta}^2\widehat\Psi_\beta<0\). Since \(\partial_{\rho\rho}^2\widehat\Psi_\beta<0\) by Lemma 9, it follows that \(M_\beta'(\rho)<0\).

It remains to consider the case \(u_\beta(\rho)<(d-1)\beta^2\). We first note that \(u_\beta(\rho)\) lies above the line \(\eta=d\beta^2-\rho\). If \(d\beta^2-\rho\le0\) this is immediate. Otherwise, evaluating \(F_\rho\) at \(d\beta^2-\rho\) gives \[F_\rho(d\beta^2-\rho)=(1-d\beta)^2(\rho-\beta^2)>0.\] Since \(F_\rho\) is strictly decreasing and \(F_\rho(u_\beta(\rho))=0\), we obtain \(u_\beta(\rho)>d\beta^2-\rho\). Therefore \[\rho_0:=d\beta^2-u_\beta(\rho)\] belongs to \([\beta^2,d\beta^2]\), and \((\rho_0,u_\beta(\rho))\) lies on the diagonal segment considered in Lemma 33. That lemma gives \[\partial_\rho\widehat\Psi_\beta(\rho_0,u_\beta(\rho)) =\partial_\rho\widehat\Psi_\beta(\rho_0,d\beta^2-\rho_0)\le0.\] Using again \(\partial_{\rho\rho}^2\widehat\Psi_\beta<0\) and \(\rho\ge\rho_0\), we obtain \[M_\beta'(\rho)=\partial_\rho\widehat\Psi_\beta(\rho,u_\beta(\rho))<0.\] Thus \(M_\beta\) is strictly decreasing on \((\beta^2,\beta)\).

At \(\rho=\beta^2\), the vertical critical equation has the solution \(u_\beta(\beta^2)=(d-1)\beta^2\), and hence \(M_\beta(\beta^2)=\widehat\Psi_\beta(\beta^2,(d-1)\beta^2)=2\Phi_{d,l}(\beta)\). The asserted uniqueness follows from the strict monotonicity on the two sides of \(\beta^2\). ◻

Proof. The identity ?? and uniqueness follow immediately from Lemma 34, since every point in \(\widehat{\mathcal{R}}_\beta\) lies in one of the vertical sections. The negative definiteness of the Hessian and ?? follow from \(\beta\le L_2<C_{d,l}\) and Lemma 10. ◻

Proof. This is exactly the discrete two-dimensional Laplace estimate used in the proof of Lemma 17. The proof of that lemma uses the hypotheses there only to guarantee three facts: the saddle value is \(2\Phi_{d,l}(\beta)\), the saddle is the unique global maximizer, and the Hessian at the saddle is negative definite. These are precisely the assumptions above. The boundary and off-saddle terms are exponentially negligible because the maximum on \(\overline{\mathcal{R}_\beta}\) is unique and because \(\Phi_{d,l}(\beta)>0\) rules out endpoint contributions with the same exponential order. The local Gaussian computation at the saddle is unchanged. It gives the constant in ?? . ◻

Proof. The functions \(\Phi_{d,l}\), \(r_\beta\), \(z_\beta\), and the entries of \(H(\beta,z_\beta)\) are continuous in \(\beta\). Thus, after restricting to a small open interval \(I_0\subset(0,1/d)\) around \(\beta_0\), the inequalities \(\Phi_{d,l}(\beta)>0\), \(r_\beta<1\), and the negative definiteness of \(H(\beta,z_\beta)\) hold throughout \(I_0\). Also, \(z_\beta\) remains an interior critical point of \(\Psi_{d,l}(\beta,\cdot,\cdot)\) by the critical point equations 18 –19 .

We next prove the uniform exponent gap away from the saddle. Fix \(\delta>0\) small enough that \(z_\beta\in\mathcal{R}_\beta\) and \(B_{2\delta}(z_\beta)\subset\mathcal{R}_\beta\) for all \(\beta\) in a possibly smaller neighbourhood of \(\beta_0\). Suppose that no such gap holds outside \(B_\delta(z_\beta)\). Then there exist \(\beta_n\to\beta_0\) and \(w_n\in\overline{\mathcal{R}_{\beta_n}}\) such that \[\operatorname{dist}(w_n,z_{\beta_n})\ge\delta, \qquad \Psi_{d,l}(\beta_n,w_n) \ge 2\Phi_{d,l}(\beta_n)-o(1).\] The sets \(\overline{\mathcal{R}_{\beta_n}}\) are contained in the fixed compact rectangle \([0,1/d]\times[0,l/d]\). Passing to a subsequence, \(w_n\to w_0\). The defining inequalities of the closed feasible regions are continuous in \(\beta\), so \(w_0\in\overline{\mathcal{R}_{\beta_0}}\). The convention \(x\log x=0\) gives a jointly continuous extension of \(\Psi_{d,l}\) to these compact closures, and hence \[\Psi_{d,l}(\beta_0,w_0)=2\Phi_{d,l}(\beta_0).\] The assumed uniqueness of the maximizer at \(\beta_0\) gives \(w_0=z_{\beta_0}\). This contradicts the facts that \(w_n\to w_0\), \(z_{\beta_n}\to z_{\beta_0}\), and \(\operatorname{dist}(w_n,z_{\beta_n})\ge\delta\). Therefore, after shrinking \(I_0\) to an interval \(I\), for every such \(\delta\) there exists \(\eta_\delta>0\) such that \[\sup\{\Psi_{d,l}(\beta,w): w\in\overline{\mathcal{R}_\beta}, \operatorname{dist}(w,z_\beta)\ge\delta\} \le 2\Phi_{d,l}(\beta)-\eta_\delta, \qquad \beta\in I .\]

It remains only to identify the maximizer inside \(B_\delta(z_\beta)\). By continuity of the Hessian, we may choose \(\delta\) and then shrink \(I\) so that the Hessian of \(\Psi_{d,l}(\beta,\cdot,\cdot)\) is negative definite on \(B_\delta(z_\beta)\), uniformly for \(\beta\in I\). Hence \(\Psi_{d,l}(\beta,\cdot,\cdot)\) is strictly concave on this ball. Since \(z_\beta\) is an interior critical point, it is the unique maximizer in the ball. Together with the uniform exponent gap outside the ball, this proves that \(z_\beta\) is the unique global maximizer on \(\overline{\mathcal{R}_\beta}\) for all \(\beta\in I\).

The uniform version of ?? follows from the same compact-uniform discrete Laplace estimate used in Lemma 17: the exponent gap above controls the complement of a fixed neighbourhood of the saddle, while the local Hessian and all Stirling prefactors vary continuously and remain bounded away from zero and infinity on compact subsets of the neighbourhood. ◻

Proof. For all large \(m\), \(\beta_m:=h_m/m\) lies in the interval \(I\) supplied by Lemma 37. That lemma gives the second-moment ratio uniformly for \(\beta_m\), and also gives \(r_{\beta_m}<1\). The compact-uniform cycle estimates of Lemma 27 apply to the same sequence \(\beta_m\). Repeating the subgraph-conditioning proof of Lemma 28, with these uniform inputs and with \(h_m\) in place of \(m\beta\), gives \[\frac{1}{m}\log Z_{h_m}-\frac{1}{m}\log\mathbb{E} Z_{h_m}\longrightarrow0\] in probability. Lemma 3 gives \((1/m)\log\mathbb{E} Z_{h_m}\to\Phi_{d,l}(\beta_0)\). This proves the claim. ◻

Proof. By ?? and Lemma 35, the hypotheses of Lemma 36 hold uniformly on compact subsets of \(\{0<\beta\le L_2\}\). Hence the second moment ratio ?? holds with \(\beta_m\) in place of \(\beta\). In addition, ?? gives \(r_\beta<1\). The proof of Lemma 28 applies verbatim with this second-moment input: the alternatives in Lemma 28 are used only to invoke the second moment ratio and the inequality \(r_\beta<1\). Therefore ?? follows. ◻

Proof. Set \(\beta_m=k_m/m\). Then \[\beta_m=\beta_0+\frac{\log m}{2m\Phi'_{d,l}(\beta_0)}+\frac{r_m}{m},\] so \(\beta_m\to\beta_0\). Lemma 3, equivalently 11 , gives \[\mathbb{E} Z_{k_m} =\frac{e^{m\Phi_{d,l}(\beta_m)}}{\sqrt{2\pi m\beta_m(1-d\beta_m)}}(1+o(1)).\] Taylor expansion at \(\beta_0\), using \(\Phi_{d,l}(\beta_0)=0\), gives \[m\Phi_{d,l}(\beta_m) =\Phi'_{d,l}(\beta_0) \left(\frac{\log m}{2\Phi'_{d,l}(\beta_0)}+r_m\right) +O\left(\frac{(\log m)^2}{m}\right) =\frac{\log m}{2}+\Phi'_{d,l}(\beta_0)r_m+o(1).\] The factor \(e^{(\log m)/2}\) cancels the prefactor \(m^{-1/2}\), and the result follows. ◻

Proof. Choose \(\varepsilon_0>0\) so small that \([\beta_0/2,\beta_0+\varepsilon_0]\subset(0,1/d)\). By Lemma 3, there is a constant \(A<\infty\) such that, uniformly for \(h/m\in[\beta_0/2,\beta_0+\varepsilon_0]\), \[\mathbb{E} Z_h \le A m^{-1/2}\exp\left\{m\Phi_{d,l}(h/m)\right\} . \label{eq:tail-local-first-moment}\tag{76}\] Since \(\Phi_{d,l}\) is concave and \(\Phi_{d,l}(\beta_0)=0\), for every \(u\ge\beta_0\), \[\Phi_{d,l}(u)\le \Phi'_{d,l}(\beta_0)(u-\beta_0)=-a_0(u-\beta_0). \label{eq:tail-concavity}\tag{77}\] Let \[M_m:=\left\lceil \frac{5\log m}{2a_0}\right\rceil .\] For \(C\le \delta<M_m\), put \(h=k_m+\delta\). Then \(h-K_m=\delta+O(1)\), and the same Taylor expansion used in Proposition 3 gives, uniformly in this range, \[\mathbb{E} Z_{k_m+\delta} \le A_1 e^{-a_0\delta}\] for a constant \(A_1<\infty\). Hence \[\sum_{\delta=C}^{M_m-1}\mathbb{E} Z_{k_m+\delta} \le A_2 e^{-a_0 C}. \label{eq:near-tail-bound}\tag{78}\] For \(h\ge k_m+M_m\) and \(h/m\le\beta_0+\varepsilon_0\), we have \(h-m\beta_0\ge 2a_0^{-1}\log m+O(1)\). Combining 76 and 77 , \[\mathbb{E} Z_h\le A_3 m^{-5/2}\] for all sufficiently large \(m\), uniformly over such \(h\). Thus the contribution of these indices is \(O(m^{-3/2})\). Finally, on the remaining interval \(h/m>\beta_0+\varepsilon_0\), strict concavity and the fact that \(\Phi_{d,l}<0\) on \((\beta_0,1/d)\) imply that \(\Phi_{d,l}\le-\eta\) for some \(\eta>0\); the crude Stirling bounds used in Section 2 then show that the total contribution from this range is exponentially small. Together with 78 , this proves the first assertion. The probability bound follows from Markov’s inequality, \[\mathbb{P}\{\nu_m\ge k_m+C\}\le \mathbb{E} Z_{\ge k_m+C}.\] ◻

Proof. Suppose that \((d_n,l_n)\) is a sequence with \(d_n,l_n\ge2\) and \(\max\{d_n,l_n\}\to\infty\). We shall show that, after passing to a subsequence, the condition 79 fails for all sufficiently large \(n\). This rules out infinitely many admissible pairs.

Write \(\beta_n=\beta_*(d_n,l_n)\), and set \[z_n:=\frac{1-d_n\beta_n}{1-\beta_n}.\] The equation defining \(\beta_*\) is equivalent to \[(d_n-1)z_n^{l_n}+z_n-1=0, \qquad \beta_n=\frac{1-z_n}{d_n-z_n}. \label{eq:z-beta-relation}\tag{80}\] We repeatedly use the explicit expression for \(G_\beta\) from Section 3. All error terms below are along the chosen subsequence.

First assume that \[\frac{\log d_n}{l_n}\longrightarrow \alpha\in(0,\infty).\] Then 80 gives \[z_n=e^{-\alpha+o(1)}, \qquad d_n\beta_n=1-e^{-\alpha}+o(1), \qquad \frac{c_0(\beta_n)}{\beta_n}=o(1).\] Fix any \(\zeta\in(0,1)\), and put \(s_n=\zeta\beta_n\). Then \(0<s_n<\beta_n-c_0(\beta_n)\) for all large \(n\), and \[t_{\beta_n}(s_n)=\frac{\zeta e^\alpha(1-e^{-\alpha})}{d_n}(1+o(1)).\] Substitution in the formula for \(G_\beta\) gives the following expansion. The terms not containing \(t_{\beta_n}(s_n)\) contribute \[\frac{2e^{-\alpha}(1-e^{-\alpha})}{d_n}+l_n s_n(1-e^{-\alpha})+O(l_ns_n^2),\] whereas the terms containing \(t_{\beta_n}(s_n)\) contribute \[-l_n t_{\beta_n}(s_n) \left\{d_ns_n^2+s_ne^{-\alpha}+\frac{2e^{-\alpha}(1-e^{-\alpha})}{d_n} \right\} +o(l_n/d_n).\] Using \(s_n=\zeta(1-e^{-\alpha})d_n^{-1}(1+o(1))\) and the displayed expression for \(t_{\beta_n}(s_n)\), these terms combine to \[G_{\beta_n}(s_n) =-\frac{l_n}{d_n}\, \zeta^2(1-\zeta)(1-e^{-\alpha})^2(e^\alpha-1)(1+o(1))<0 \label{eq:G-alpha-finite}\tag{81}\] for all sufficiently large \(n\). Thus 79 fails in this case.

Next assume that \[\frac{\log d_n}{l_n}\longrightarrow0.\] Then \(l_n\to\infty\), \(z_n\to1\), \(d_n\beta_n\to0\), \(c_0(\beta_n)/\beta_n=o(1)\), and \[u_n:=l_n(d_n-1)\beta_n\longrightarrow\infty . \label{eq:u-to-infty}\tag{82}\] Indeed, if \(w_n=1-z_n\), then 80 gives \(w_n=(d_n-1)(1-w_n)^{l_n}\); this implies \(w_n\to0\) and \(l_nw_n\to\infty\), and 82 follows from \(\beta_n=w_n/(d_n-1+w_n)\). Fix \(\zeta\in(0,1)\) and put \(s_n=\zeta\beta_n\). Then \(s_n\) is admissible for all large \(n\), \[t_{\beta_n}(s_n)=\zeta\beta_n(1+o(1)),\] and the leading terms in \(G_{\beta_n}(s_n)\) are those of order \(l_n(d_n-1)\beta_n^2\). More explicitly, the first line of \(G_\beta\) is \(O(d_nl_ns_n^2)=O(l_nd_n\beta_n^2)\), the second line contributes \(2\beta_n\{1-(d_n-1)(l_n-1)t_{\beta_n}(s_n)\}+o(l_nd_n\beta_n^2)\), and the third line contributes \(s_n\{2l_n(d_n-1)t_{\beta_n}(s_n)+l_nd_n\beta_n\}+o(l_nd_n\beta_n^2)\). After inserting \(s_n=\zeta\beta_n\) and \(t_{\beta_n}(s_n)=\zeta\beta_n(1+o(1))\), the terms of smaller order cancel or are absorbed into the error, and \[G_{\beta_n}(s_n) =l_n(d_n-1)\beta_n^2\,\zeta(\zeta-1)(1+o(1))<0. \label{eq:G-small-ratio}\tag{83}\] This again contradicts 79 .

It remains to consider the case \[\frac{\log d_n}{l_n}\longrightarrow\infty.\] Put \(a_n=d_n^{-1/l_n}\). Then \(a_n\to0\), and 80 gives \[z_n=a_n(1+o(1)), \qquad d_n\beta_n=1-a_n+o(a_n). \label{eq:superexp-beta}\tag{84}\] If \(l_n\ge3\) eventually, take \(\zeta=1/3\) and set \(s_n=\zeta\beta_n\). The formula for \(c_0\) gives \(c_0(\beta_n)/\beta_n=(2l_n-1)^{-1}+o(1)\), so \(s_n\) is admissible. Moreover \[t_{\beta_n}(s_n) =(1-\zeta)^{1/l_n}\zeta^{1-2/l_n}\beta_n^{1-1/l_n}(1+o(1)).\] Substitution into \(G_\beta\), using 84 , yields the leading term \[s_n t_{\beta_n}(s_n)(d_n-1) \{-\beta_n l_n+3\beta_nd_n+2l_n-3\beta_nd_nl_n+o(l_n)\},\] while all terms not multiplied by \(t_{\beta_n}(s_n)\) are smaller by a factor \(a_n^{1/l_n}+o(1)\). Since \(d_n\beta_n=1-a_n+o(a_n)\), this is equivalent to \[G_{\beta_n}(s_n) =\zeta\,t_{\beta_n}(s_n) \left((l_n-1)\zeta-(l_n-2)+o(l_n)\right)<0, \label{eq:G-superexp-lge3}\tag{85}\] for all sufficiently large \(n\), because \(\zeta=1/3\) makes the coefficient negative for all \(l_n\ge3\).

Finally, if the subsequence has \(l_n=2\), choose \(\zeta_n=a_n^{1/2}=d_n^{-1/4}\) and set \(s_n=\zeta_n\beta_n\). Then \(s_n/\beta_n\to0\), so \(s_n<\beta_n-c_0(\beta_n)\) for all large \(n\). A direct substitution in \(G_\beta\), using 84 with \(l_n=2\), gives \[G_{\beta_n}(s_n)=-a_n^2(1+o(1))<0. \label{eq:G-superexp-l2}\tag{86}\] Thus the condition 79 fails in every possible divergent sequence of parameter pairs. Therefore it can hold for only finitely many pairs \((d,l)\). ◻