Proof. First suppose all variance increments being differentiated are strictly positive. Moving \(q_i\) lengthens the transition ending at level \(i\) and shortens the transition leaving that level. By (13), the incoming derivative, expressed at level \(i\) before propagation through the preceding kernels, is \(\{U_i''+\zeta_i(U_i')^2\}/2\). The outgoing derivative is \(-\{U_i''+\zeta_{i+1}(U_i')^2\}/2\). Their sum is \(-w_i(U_i')^2/2\). Formula (12) propagates this sum through the kernels ending at levels \(i,i-1,\ldots,0\), giving \(\partial_{q_i}V_f=-w_i\mathbb E[U_i'(Z_i)^2]/2\).

The same cancellation applies to \(h_i\) for \(i<k-1\). At the last field level the incoming derivative and the derivative of the subtraction \(-h_{k-1}/2\) combine to give \[\frac12\{(\log\cosh)''+ \zeta_{k-1}((\log\cosh)')^2-1\} =-\frac{1-\zeta_{k-1}}2\tanh^2.\] This is the same formula with \(w_{k-1}=1-\zeta_{k-1}\). Linearity yields both displayed variations.

For completeness, the differentiations remain valid at degenerate increments. A Gaussian transform of either terminal function is smooth in the spatial variable and has bounded derivatives of every positive order. This follows inductively by differentiating the tilted expectation: the resulting terms are finite sums of tilted moments of products of bounded derivatives of the next-level function. The functions themselves are bounded in the pattern case and have at most linear growth in the field case. On every compact variance-parameter set, the Gaussian exponential moments therefore dominate all terms in these formulas. Coupling the increments as \(\sqrt{s_i}G_i\) shows continuity as any \(s_i\) decreases to zero, both for recursion values and for the tilted moments that appear above. The heat-equation identity also has this continuous extension because its spatial derivatives do. Approximate the endpoints of a feasible segment by interior points on a common partition and integrate the already proved derivative formula along the approximating segments. Dominated convergence gives the same integral formula on the original segment and its one-sided endpoint derivatives.

Finally, the derivative martingale and conditional Jensen inequality show that its second moments increase with the level. The terminal derivative bounds are \(\|f'\|_\infty\) and \(1\), respectively; the same bounds hold at every earlier level by (12). An inserted zero-variance transition repeats the same derivative and chain value, so it does not change \(\rho_h\). ◻

Proof. Refine the two partitions and integrate Lemma 4 along the segment joining the paths. The terminal derivative bounds give (17). Step paths are dense in the indicated classes in \(L^1\), which gives the extensions.

For the order assertion on a common partition, set \(a_i=w_i(p_i-q_i)\) and \(A_j=\sum_{i=j}^{k-1}a_i\). The hypothesis gives \(A_j\ge0\). For every nonnegative nondecreasing sequence \(c_i\), \[ \sum_{i=0}^{k-1}a_ic_i =c_0 A_0+\sum_{j=1}^{k-1}(c_j-c_{j-1})A_j\ge0. \tag{20}\] At each point of the segment from \(q\) to \(p\), take \(c_i=\mathbb E[U_i'(Z_i)^2]\) and apply (14). The derivative along the segment is nonpositive, proving the claim for steps.

For general paths, replace each by its average on each interval of a common sequence of successively finer uniform partitions. These averages are nondecreasing and converge in \(L^1\) to the original paths. Their tail-integral differences at partition boundaries equal the original differences. Between boundaries the differences are affine and hence nonnegative as well. Apply the step result and then (17). ◻

Proof. The trial \(h=0\) gives \(S_{\mathrm I}(q)\ge0\). Each expression in its defining supremum is nondecreasing in \(q\) because \(h\ge0\). Also \(\ell(0)=0\) and (15), integrated along \(a\mapsto ah\), gives \(\ell(h)\le0\), whence \(S_{\mathrm I}(0)=0\). Every transform preserves the range of a bounded terminal function, so \(|V_f(q)|\le\|f\|_\infty\). In fact, the zero path has \(V_f(0)=\log\mathbb E e^{f(G)}\). These observations bound the infimum below by \(-\alpha\|f\|_\infty\) and provide a finite upper bound using \(q=0\).

For any quantile \(q\), let \(\underline q_m\) on each interval of the uniform \(m\)-partition be \((1-1/m)\) times the essential infimum of \(q\) on that interval. Then \(0\le\underline q_m\le q\) almost everywhere, \(\sup\underline q_m<1\), and \(\|\underline q_m-q\|_{L^1}\to0\). The last assertion follows, for example, from convergence at every continuity point of a monotone representative, followed by bounded convergence. Consequently \(S_{\mathrm I}(\underline q_m)\le S_{\mathrm I}(q)\), while \(V_f(\underline q_m)\to V_f(q)\). Taking a limit superior for each finite-entropy trial \(q\) proves the nontrivial inequality in (21); the other inequality follows from inclusion of the trial classes. No continuity assertion for the extended-valued entropy is needed. ◻

Proof. Evenness follows by reflection, and \(s=0\) is immediate. For \(s>0\), the law of the next absolute value \(y\ge0\) when the starting point is \(x\ge0\) has density proportional to \[e^{-y^2/(2s)}\cosh(xy/s)e^{dU(y)}.\] For \(x_2\ge x_1\ge0\), the ratio of these normalized densities is a positive constant times \(\cosh(x_2y/s)/\cosh(x_1y/s)\). Its logarithmic derivative in \(y\) is \[\frac{x_2}{s}\tanh(x_2y/s) -\frac{x_1}{s}\tanh(x_1y/s)\ge0,\] because \(x\mapsto x\tanh(xy/s)\) is nondecreasing on \([0,\infty)\). Thus the density ratio \(L(y)\) is increasing. If \(Y,Y'\) are independent with the law corresponding to \(x_1\), then \[\operatorname{Cov}(A(Y),L(Y)) =\tfrac12\mathbb E[(A(Y)-A(Y'))(L(Y)-L(Y'))]\ge0.\] Since \(\mathbb E L(Y)=1\), this proves \(K A(x_2)\ge K A(x_1)\). Negating \(A\) proves the reversed statement. The normalizations are included in \(L\) and do not alter its monotonicity. ◻

Proof. Fix a partition. We first work in its open cone \(0<h_0<h_1<\cdots<h_{k-1}\), where all parameter derivatives below exist. Write \(m_i=U_i'\) and \(\rho_i=\mathbb E m_i(Z_i)^2\) as in (16). By the first variation, the matrix \[A_{ij}=w_i\partial_{h_j}\rho_i =-2\partial_{h_j}\partial_{h_i}\ell\] is symmetric. We will prove that its off-diagonal entries are nonpositive and its row sums are nonnegative. These two properties make \(A\) positive semidefinite, as the decomposition (27) below shows. They correspond to increasing one later field level and to increasing all field levels together, respectively.

Every \(U_i\) is even and convex. Evenness is preserved by the Gaussian transforms. To check convexity when \(d>0\), use convexity of \(U\) and Hölder’s inequality to obtain, for \(0\le a\le1\), \[\mathbb E e^{dU(ax+(1-a)y+\sqrt sG)} \le (\mathbb E e^{dU(x+\sqrt sG)})^a (\mathbb E e^{dU(y+\sqrt sG)})^{1-a}.\] Taking logarithms proves convexity of the transform; expectation preserves it when \(d=0\). The terminal subtraction is constant in \(x\) and does not affect the argument. In particular, \(m_i\) is odd, \(m_i(x)\ge0\) for \(x\ge0\), and \(m_i(x)^2\) is even and nondecreasing in \(|x|\). All \(U_i\) have at most linear growth, so the normalizers in Lemma 7 are finite.

We first establish the cross-derivative sign \[ \partial_{h_j}\rho_i\le0\qquad(j>i). \tag{23}\] Repeat the switch calculation from Lemma 4, but stop its backwards propagation at the function \(U_i(x)\). This gives \[ \partial_{h_j}U_i(x) =-\frac{w_j}{2} (K_{i+1}\cdots K_j m_j^2)(x),\qquad j>i. \tag{24}\] Lemma 7 says that this function is even and nonincreasing in \(|x|\). Its derivative is therefore nonpositive for \(x>0\) and nonnegative for \(x<0\). Since \(m_i\) has the opposite signs, the direct derivative of the terminal test in the expression \(\rho_i=K_0\cdots K_i(m_i^2)(0)\) is pointwise nonpositive: \[\partial_{h_j}(m_i^2) =2m_i(\partial_{h_j}U_i)'\le0.\]

It remains to account for all derivatives of the probability kernels in this expression. If \(l\le i<j\), the variance of \(K_l\) is independent of \(h_j\). Put \(A_l=\partial_{h_j}U_l\). Differentiating the normalized kernel at a fixed starting point gives the score \[ \partial_{h_j}\log K_l(x,dy) =\zeta_l\{A_l(y)-(K_lA_l)(x)\}. \tag{25}\] Indeed the derivative of \(U_{l-1}\) in its normalizing factor equals \(K_l A_l\), by (12). In the product rule for \(K_0\cdots K_i(m_i^2)\), the test at level \(l\) is \(H_l=K_{l+1}\cdots K_i(m_i^2)\), with \(H_i=m_i^2\). It is even and nondecreasing in absolute value by Lemma 7. On the other hand, \(A_l\) is even and nonincreasing in absolute value by (24). Thus differentiating this kernel contributes, at its starting point, \[\zeta_l\operatorname{Cov}_{K_l(x,\cdot)}(H_l,A_l)\le0.\] The sign follows from the same independent-copy covariance identity as in Lemma 7, now for two functions of \(|y|\) with opposite monotonicities. Propagation through preceding probability kernels preserves the sign. This includes \(l=0\), whose score is zero because \(\zeta_0=0\). Together with the direct derivative, these are all terms in the product rule, proving (23).

Next increase every field value by the same number \(a\ge0\). All increments after level zero remain unchanged. Each \(U_i\), \(i\ge0\), changes only by the additive constant \(-a/2\), so its spatial derivative and all kernels \(K_l\) for \(l\ge1\) remain unchanged. The sole change in the chain is that its first, untilted Gaussian variance increases from \(h_0\) to \(h_0+a\). The function \(K_1\cdots K_i(m_i^2)\) is even and nondecreasing in absolute value. Coupling the initial values as \(\sqrt{h_0+a}G\) therefore proves \[ \sum_{j=0}^{k-1}\partial_{h_j}\rho_i\ge0. \tag{26}\]

We can now apply the matrix criterion. By (23) and symmetry every off-diagonal entry of \(A\) is nonpositive, and (26) gives nonnegative row sums. For every \(v\in\mathbb R^k\), \[ v^{\mathsf T}Av =\sum_i\left(\sum_jA_{ij}\right)v_i^2 +\sum_{i<j}(-A_{ij})(v_i-v_j)^2\ge0. \tag{27}\] Thus the Hessian of \(\ell\) is negative semidefinite on the open cone. Restricting \(\ell\) to a line segment in that cone proves concavity and, using its derivative at the starting point, the inequality (22) there.

For arbitrary field steps \(h,h'\) on this partition, replace \(h_i,h_i'\) by \(h_i+\varepsilon(i+1)\) and \(h_i'+\varepsilon(i+1)\). Both perturbed paths lie in the open cone and their difference remains \(h'-h\). Let \(\varepsilon\downarrow0\). The value and coefficient continuity in Lemma 4 proves both concavity and (22) on the closed cone. Common refinement handles distinct partitions. Finally, approximate any two bounded field paths in \(L^1\) by monotone steps on common partitions and use Proposition 5 to pass the concavity inequality to their limits. ◻

Proof. First suppose \(b=\mathbb P(B_{12}<-\varepsilon)>0\) for some \(\varepsilon>0\). On the event that all distinct pairs among the first \(n\) indices have overlap below \(-\varepsilon\), each of the \(n\) links to index \(n+1\) fails this condition with conditional probability \((1-b)/n\), by (28). A union bound therefore gives conditional probability at least \(b\) that all the new links satisfy it. Starting with \(n=2\), such blocks have positive probability for every size. Their quadratic form on the all-ones vector is at most \(nD-n(n-1)\varepsilon\), contradicting positive semidefiniteness for large \(n\). Taking a countable sequence of \(\varepsilon\) proves positivity.

We next prove a support duplication assertion. If \((a_{\ell m})_{\ell,m\leq n}\) belongs to the support of a finite block, \(n\geq2\), there is a supported extension to \(n+1\) such that \[ a_{i,n+1}=a_{in}\quad(i<n),\qquad a_{n,n+1}\leq\max_{i<n}a_{in}. \tag{31}\] Fix \(b_*>\max_{i<n}a_{in}\) and a positive-probability neighborhood event \(A\) for the first \(n\) indices, chosen so that all specified links to \(n\) are below \(b_*\). Let \(A_0\) impose just the restrictions among the first \(n-1\) indices. An index \(j\geq n\) matches if its links to those \(n-1\) indices belong to the neighborhoods prescribed for \(n\). Thus \(A=A_0\cap\{n\text{ matches}\}\).

Suppose that, on \(A\), a fresh index cannot both match and have its link to \(n\) below \(b_*\). Let \(W_B\) be the limiting empirical frequency of matching indices, and \(W\) the frequency of indices \(j>n\) with \(B_{nj}\geq b_*\). These frequencies exist in \(L^2\). Indeed, for an exchangeable sequence of indicators the second moments of empirical averages and their mixed second moments have the same limit, so the averages are \(L^2\)-Cauchy. The limits are unchanged by finite permutations of the relevant unobserved indices. The supposition and exchangeability imply \(W_B\leq W\) on \(A\).

Put \(b=\mathbb P(B_{12}\geq b_*)\). We have \(b<1\), since \(A\) has positive probability and specifies links below \(b_*\). Conditional on \(A\), the first high link to \(n\) has probability \(b/n\); after \(j\) consecutive high links the next has probability \((b+j)/(n+j)\). Applying (28) successively, and then expressing empirical frequency moments by products at distinct fresh indices, gives \[ \mathbb E[\mathbf1_A W^d] =\mathbb P(A)\prod_{j=0}^{d-1}\frac{b+j}{n+j}, \qquad d\geq1. \tag{32}\] The binomial series therefore yields \[ \mathbb E[\mathbf1_A(1+zW)^{-n}] =\mathbb P(A)(1+z)^{-b},\qquad z>0. \tag{33}\] Initially the series proves this for \(0<z<1\). Both sides are real analytic on \((0,\infty)\), the left side by boundedness of \(W\) and local power-series domination, so the identity holds throughout that interval. It also includes \(b=0\).

To compare the matching frequency with the high-link frequency, write \(I_j=\mathbf1_{\{j\text{ matches}\}}\). The event \(A_0\) uses only the first \(n-1\) indices, and \(W_B\) is unchanged by every finite permutation of the remaining indices, including index \(n\). Hence, for every bounded continuous \(\varphi\), exchangeability gives \[\mathbb E[\mathbf1_{A_0}I_n\varphi(W_B)] =\mathbb E\left[\mathbf1_{A_0} \frac1M\sum_{j=n}^{n+M-1}I_j\varphi(W_B)\right] \longrightarrow \mathbb E[\mathbf1_{A_0}W_B\varphi(W_B)].\] The limit uses the \(L^2\) convergence defining \(W_B\); adding index \(n\) does not change that frequency. Taking \(\varphi(x)=(1+zx)^{-n}\) and using \(W_B\le W\) on \(A\) now gives \[\begin{align*} \mathbb E[\mathbf1_A(1+zW)^{-n}] &\leq\mathbb E[\mathbf1_A(1+zW_B)^{-n}]\\ &=\mathbb E[\mathbf1_{A_0}W_B(1+zW_B)^{-n}] \leq z^{-1}. \tag{34}\end{align*}\] The last bound uses \(x/(1+zx)^n\leq1/z\). Equations (33) and (34) contradict \(b<1\) as \(z\to\infty\). Consequently the desired extension event has positive probability. Shrinking the neighborhoods and decreasing \(b_*\) to its target value, compactness of the bounded finite-block support proves (31). Permutation symmetry permits duplication of any chosen index.

If ultrametricity failed, a supported triple could be chosen with \(a_{12}<a_{13}\leq a_{23}\). Repeated duplication makes three groups of \(m\) indices, preserving the three cross-group entries \(a_{12},a_{13},a_{23}\). Within each group all off-diagonal entries stay at most its largest cross-group entry: this follows inductively from (31). Represent the resulting finite Gram matrix by vectors and let \(u_1,u_2,u_3\) be their group averages. Then \[\|u_2\|^2,\|u_3\|^2 \leq \frac Dm+\left(1-\frac1m\right)a_{23}, \qquad u_2\cdot u_3=a_{23},\qquad \|u_1\|\leq\sqrt D.\] Hence \(\|u_2-u_3\|^2\leq2(D-a_{23})/m\to0\), whereas \(u_1\cdot(u_3-u_2)=a_{13}-a_{12}>0\). Cauchy–Schwarz gives a contradiction. Applying exchangeability to all triples completes the proof. ◻

Proof. We give the Poisson and sampling calculations, since the distribution of a descendant tree under sampling requires a power bias. If a Poisson process has intensity \(C x^{-1-b}\,dx\), \(0<b<1\), and its points are multiplied by independent positive marks \(X\) with \(0<\mathbb E X^b<\infty\), the image process has intensity \(C\mathbb E X^b\,x^{-1-b}\,dx\). More precisely, after this change the attached mark law is biased by \(X^b\), independently of the image point. This follows by substituting \(y=xX\) in the marked intensity. Independent marking and mapping follow directly from the Poisson count law on disjoint sets, first for simple marks and then by approximation.

The sum \(S\) of points of intensity \(C x^{-1-b}\,dx\) has Laplace transform \[ \mathbb E e^{-tS}=e^{-ct^b}, \qquad c=C\Gamma(1-b)/b>0. \tag{37}\] Indeed the Poisson exponential formula gives the exponent \(-C\int_0^\infty(1-e^{-tx})x^{-1-b}\,dx=-ct^b\). The integral is finite; the formula proves both finiteness and positivity of the sum. For \(0<r<b\), integrating \(1-\mathbb E e^{-tS}\) against \(t^{-1-r}\,dt\) gives \(\mathbb E S^r<\infty\). For every \(r>0\), the identity \[\mathbb E S^{-r} =\frac1{\Gamma(r)}\int_0^\infty t^{r-1}e^{-ct^b}\,dt\] gives a finite negative moment. These moments imply a finite second moment for \(\log S\). Summing the tree from its last level upward and using the marking calculation proves the assertions about \(T\): the descendant exponent is strictly larger than the exponent of the edge above it, so every required positive fractional moment exists.

At a parent, let \(b\) be the exponent on its outgoing edges. Multiply each edge point by its descendant total. The resulting child masses are Poisson points of intensity \(C x^{-1-b}\,dx\), and their attached normalized descendant trees are independent of the points and have law biased by the descendant total to power \(b\). If the parent tree itself is biased by its total to a power \(\lambda<b\), only the point process is further biased by the \(\lambda\)th power of its sum; the conditional descendant description does not change. At the root \(\lambda=0\), and thereafter \(\lambda\) is the incoming exponent.

Here is the resulting partition law. Denote unbiased child masses by \((X_j)\) and their sum by \(S\). Under the bias \(S^\lambda\), the probability of a specified partition of \(m\geq1\) visits into \(d\) nonempty child classes of sizes \(m_1,\ldots,m_d\) is \[\begin{align*} p_\lambda(m_1,\ldots,m_d) &=\frac{\mathbb E\left[S^{-m+\lambda} \sum_{j_1,\ldots,j_d\ \mathrm{distinct}} \prod_{r=1}^dX_{j_r}^{m_r}\right]} {\mathbb E S^\lambda}\\ &=\frac{\prod_{r=1}^d C\Gamma(m_r-b)} {\Gamma(m-\lambda)\mathbb E S^\lambda} \int_0^\infty t^{db-\lambda-1}e^{-ct^b}\,dt. \tag{38}\end{align*}\] To derive this, insert the Laplace integral for \(S^{-m+\lambda}\) and use the Poisson exponential formula with \(d\) distinct point insertions. Each insertion contributes \(C\int_0^\infty x^{m_r-b-1}e^{-tx}\,dx =C\Gamma(m_r-b)t^{b-m_r}\). For completeness, the insertion rule follows by first restricting to a finite-intensity process, summing its Poisson count series, and then letting the truncation grow; Tonelli’s theorem applies to the nonnegative integrands.

The ratio of (38) after and before a visit to an existing class \(r\) is \[ \frac{m_r-b}{m-\lambda}. \tag{39}\] The complementary probability of creating a new child is \((db-\lambda)/(m-\lambda)\). Observed deeper shapes do not alter (39), because the attached descendant trees are independent of the child masses. Equivalently the probabilities of complete shapes factor into the successive partition probabilities. Along an existing class at level \(i\), these ratios telescope: after \(n\) replicas, the probability that the next replica enters that class is \[ \frac{m_*-\zeta_i}{n}, \tag{40}\] where \(m_*\) is its size. At level zero the class is the whole tree and the probability is one. With one previous replica, (40) gives \(\mathbb P(\gamma^1\wedge\gamma^2\geq i)=1-\zeta_i\), proving (36). With \(n\) previous replicas, applying (40) to the class of replica \(1\) gives exactly (28), first for upper-threshold tests and hence for every test on the finite set of levels.

Conversely, the upper-threshold relations of a finite ultrametric array are nested equivalence relations. Conditional on its first \(n\) indices, the Ghirlanda–Guerra property gives (40) for each existing class, using a representative of that class. These probabilities specify the distribution of the deepest existing class joined by the new index, and therefore of all its new pair entries. They coincide with the cascade probabilities. Induction on \(n\), starting from the prescribed two-replica law, proves uniqueness. ◻

Proof. Let the distinct quantized values be \(c_0<\cdots<c_s\). For a finite replica restriction, let \(A_j\) be the matrix indicating that two indices belong to the same threshold class at level \(c_j\), with diagonal entries one. Every \(A_j\) is positive semidefinite: its quadratic form is the sum over classes of the squared sum of the coordinates in that class. The quantized matrix is \[ c_0\mathbf1\mathbf1^{\mathsf T} +\sum_{j=1}^s(c_j-c_{j-1})A_j+(D-c_s)I. \tag{41}\] All coefficients are nonnegative. The Ghirlanda–Guerra property passes through the quantization map by the tower rule, so Lemma 11 identifies its law.

For the joint assertion, bin \(r\) and each \(a_j(r)\) on meshes of size \(\varepsilon\) and take their common refinement. Since every coordinate is nondecreasing, the nonempty joint bins are totally ordered. Choosing one actual vector \((r,a_1(r),\ldots,a_d(r))\) in each bin gives ordered representatives and error at most \(\varepsilon\) in each coordinate. Use (41) separately for each coordinate, with its own diagonal \(D_j\). The bins depend only on \(r\), including at atoms. If the functions are specified only on a full-measure ordered domain, the construction on that domain suffices for every sampled array entry. ◻

Proof. For uniqueness, round two candidate arrays on increasingly fine finite meshes. Their rounded marginals coincide, so Lemma 11 gives identical rounded array laws. Uniformly vanishing rounding error then identifies their original finite-block laws. For existence, approximate the desired quantile by ordered step functions, use finite cascades, and extract a limit of the bounded arrays. Their Gram property follows from (41); ultrametricity and the Ghirlanda–Guerra property pass to the limit. The limit has the desired marginal.

If \(q_m\to q\) in \(L^1\), their marginals converge weakly (in fact, coupling them by the same uniform variable bounds their Wasserstein distance by \(\|q_m-q\|_1\)). Any subsequential limit of the array laws has the Ghirlanda–Guerra property, is nonnegative and ultrametric, and has this limiting marginal. Uniqueness identifies it, proving convergence of the whole sequence. Weak convergence on the compact space of finite blocks is equivalent to convergence in bounded-Lipschitz distance. This is a continuity assertion; no rate independent of the finite-block size is needed. ◻

Proof. At every edge from level \(i-1\) to \(i\), multiply its Poisson point by \(e^{X_i-X_{i-1}}\) and reorder children. By (42), the conditional \(\zeta_i\)th moment of this multiplier is one. The marked Poisson mapping calculation in the proof of Lemma 11 therefore leaves the point intensity unchanged and changes the attached mark kernel exactly by (44).

The independence assertion requires doing this from the last level upward. Inductively, conditional on the path to a child, its transformed unmarked subtree has the ordinary unmarked law, independently of that path; its marks follow the transformed descendant kernels. Attach the entire transformed subtree as part of the child’s mark in the mapping at its parent. The multiplier depends only on the path through that child, so the mapping tilts the child kernel while keeping this conditional subtree law. This proves at the root that the transformed unmarked tree has its original law independently of root data, with marks generated by the stated conditional kernels.

Products of the edge multipliers telescope. If \(T^*\) is the transformed raw total, then \[ Z=e^{X_0}\frac{T^*}{T}. \tag{45}\] Both \(T\) and \(T^*\) have the same conditional law given root data, with finite squared logarithms by Lemma 11. Taking the conditional expectation of the logarithm proves the first part of (43); the elementary square bound proves the second. No independence between \(T\) and \(T^*\) is asserted or required.

The reweighted normalized masses are precisely the transformed masses, up to a relabeling that preserves ancestry. Sampling them consequently gives the ordinary cascade shape law. Conditional on the transformed unmarked tree, its mark kernels have the description just proved; hence sampling or conditioning on a shape does not further tilt those kernels. This independence concerns the transformed tree, not the original weights. Finally, factoring the expectations in (42) for an additive terminal log factor shows that the transforms add and the tilted component kernels factor. ◻

Proof. On any fixed number of base replicas, the conditional joint Gaussian law is a continuous function of its covariance matrix, also on the boundary of the positive semidefinite cone. For example, couple it as \(C^{1/2}g\); the nonnegative matrix square root is continuous and the Gaussian vector \(g\) has fixed dimension. Bounded continuous tests therefore pass to the joint array limit.

Write the positive reweighting factor as \(A\), and let \(Z=P(A)\) for the base probability \(P\). If \(0<c\leq A\leq C\), then \(c\leq Z\leq C\). An \(r\)-replica reweighted test has expectation \[\mathbb E\left[ Z^{-r}P^{\otimes r}\left(F\prod_{\ell=1}^r A_\ell\right) \right].\] Uniformly approximate \(z^{-r}\) on \([c,C]\) by polynomials. Each monomial \(Z^j\) is represented by \(j\) additional base replicas, evaluated in the same fresh fields. The resulting terms are finite-replica tests covered by the first part, and the uniform approximation controls the error. Approximating \(\log z\) on the same interval proves the logarithmic assertion. Finite-spin variables are simply additional independent base variables in this argument.

Finally quantize all the ordered covariance coordinates jointly. Their finite-replica matrices converge by Lemma 12, and the other observables and tests converge by the additional approximation hypothesis. Gaussian continuity and the same polynomial argument therefore apply. The shared Gaussian construction with independent residuals has exactly those matrices, including on repeated leaf visits. Lemma 11 identifies the quantized shape law, and Lemma 14 evaluates its bounded reweightings. Unbounded factors require an additional uniform-integrability argument; the cavity calculation below supplies the particular truncation estimate it needs. ◻

Proof. Theorem 10 gives positivity and ultrametricity. Lemma 15 passes the bounded positive factor through the array limit. Round the off-diagonal overlap into levels \(p_0\leq\cdots\leq p_{k-1}\) with probabilities \(w_i\), leaving its diagonal one. On the corresponding cascade, first integrate the independent variance \(1-p_{k-1}\) at each visit. The terminal shared-field log factor is consequently \[U_{k-1}(x)=\log\mathbb E e^{f(x+\sqrt{1-p_{k-1}}G)}.\] The remaining branch transforms have coefficients \(\zeta_{k-1},\ldots,\zeta_1\), by Lemma 14, and the common root Gaussian is averaged with coefficient zero. These are precisely the defining transforms of \(V_f\) on the rounded step quantile: its final coefficient is one, and its initial coefficient is zero. Sending the rounding error to zero proves (47), by Lemma 15 and continuity of the recursion in its quantile. For bounded continuous \(f\) this also supplies a direct description of that extension: Lemma 13 and the bounded reweighting argument show that the cascade values have a unique continuous limit. ◻

Proof. For finitely many states, differentiate the coupling \(H_t\) and apply ordinary Gaussian integration by parts. This contracts one half of the energy Hessian with \(\Delta\). A derivative of a normalized \(r\)-replica integral inserts the sum over its \(r\) sampled states minus \(r\) times the insertion at one extra replica. A second derivative and contraction give (49); the sum of its absolute coefficients, including the factor \(1/2\), is \(2r^2+r\). The first derivative of the log partition gives (48), and a single integration by parts gives (50).

To pass to countable labels, truncate to increasing finite label sets chosen independently of the added fields and renormalize their reference weights. Uniform integrability follows from elementary Gaussian moment bounds. For example, if a Gaussian energy has pointwise variance at most \(K\), then, conditionally on its independent probability reference, \[ \mathbb E Z^p\leq e^{p^2K/2},\qquad \mathbb E Z^{-p}\leq e^{p^2K/2},\qquad Z=P(e^H),\quad p\geq1. \tag{51}\] The first inequality is Jensen’s inequality for \(x^p\); the second uses \([P(e^H)]^{-p}\leq P(e^{-pH})\). Cauchy–Schwarz and the same single-state Gaussian moments control polynomial Gaussian insertions, and exponentials of sums of absolute values of finitely many such fields. Thus Gibbs derivatives, their Gaussian insertions, and the logarithms have uniformly integrable bounds on compact coefficient sets. The truncated integrals converge almost surely and in the necessary moments. Bounded pattern log factors can be absorbed into the reference; if estimates retain them explicitly, their exponential costs are integrable for either a fixed count or an independent Poisson count.

Integrate the finite-state covariance derivative first and then pass to the limit using its uniform bound. Gaussian coupling and the same moment bounds give continuity at segment endpoints. Ordinary energy parameter derivatives and (50) pass to the limit by the polynomial-insertion bounds. This proves the stated extensions. ◻

Proof. For a smooth function of Gaussian data, interpolate two standard Gaussian copies from independent to identical, with cross covariance \(tI\). Gaussian integration by parts differentiates their product expectation into the expectation of their gradient inner product. Integrating from zero to one and using the gradient bound proves the first claim. Truncation and smoothing extend it to Lipschitz functions. For the second, reveal the independent inputs successively and sum the variances of the martingale differences, each bounded by \(c_i^2\). For the third, let \(K'\) be an independent copy and use \[\operatorname{Var}(a(K)) =\tfrac12\mathbb E[(a(K)-a(K'))^2] \leq\tfrac{c^2}{2}\mathbb E[(K-K')^2]=c^2\lambda.\] ◻

Proof. Write \(L_N=NP_N\). Put the patterns, their count, and all common root Gaussian variables into the root data. Apply Lemma 14 to the leaf spin log integral. If \(X_0\) is the resulting recursion value at the root, that lemma gives \[\mathbb E[L_N\mid\text{root data}]=X_0, \qquad \mathbb E[(L_N-X_0)^2\mid\text{root data}]\le C_{\zeta}.\] The second assertion can also be seen directly: \(L_N-X_0\) is the difference between the logarithms of the transformed and original unnormalized cascade totals. Each total has the original conditional law, whose logarithm has finite second moment depending only on the fixed cascade. The two totals need not be independent for this bound.

Conditional on \(K\), the norm of the coefficient vector of the common Gaussian variables in the energy at any fixed state is at most \(C_H\sqrt N\). Indeed its squared norm is the sum of the common variances, bounded by \(NH+CNe_N^2\). The spin log integral has the same Lipschitz constant in these variables: the difference of two log integrals is bounded by the supremum of the difference of their energies. Every subsequent log-exponential transform preserves such a uniform change bound. Thus \(X_0\) has Gaussian variance at most \(C_HN\) by Lemma 18. Replacing one of the \(K\) patterns changes \(X_0\) by at most \(2\|f\|_\infty\), because this bound holds for every leaf energy and is preserved by each recursion. The independent-input variance bound therefore adds at most \(C K\). Combining these independent blocks gives \[\operatorname{Var}(X_0\mid K)\le C_HN+C K.\] The conditional mean \(\mathbb E[X_0\mid K=m]\) changes by at most \(\|f\|_\infty\) between \(m\) and \(m+1\), by adding one pattern under a coupling. If \(K'\) is an independent copy, the independent-copy variance formula bounds the variance of this conditional mean by \(\|f\|_\infty^2\mathbb E(K-K')^2/2 =\|f\|_\infty^2Nt\). The total variance of \(L_N\) is consequently \(O_H(N)\). Division by \(N^2\) proves (57). ◻

Proof. Increasing \(h\) by \(a\lambda\) adds an independent Gaussian energy with covariance \[aN R(x,x')\lambda_{\gamma\wedge\eta}.\] It also changes the deterministic subtraction by \(-aN\lambda_{k-1}/2\). The diagonal covariance term in Lemma 17 cancels this subtraction, proving (58) after reversing the pressure sign. Poisson differentiation gives \[\frac{d}{dt}\mathbb E L_N(K) =N\mathbb E[L_N(K+1)-L_N(K)],\] whose increment is the logarithm in (59). Its absolute value is bounded by \(\|f\|_\infty\), so differentiating the Poisson series and passing to its limits are justified. Continuity in \(t\) follows also by coupling Poisson counts. Continuity in \(h\) follows by coupling the vertex increments and the Gaussian interpolation estimates, including at zero increments; continuity in \(u\) follows from ordinary Gaussian-coupling differentiation. The countable label set is handled by the field-independent finite truncations and moment bounds in Lemma 17. ◻

Proof. Start from \(\log\cosh x\le|x|\). For \(d>0\), \[\mathcal T_{s,d}(|\cdot|+c)(x) \le |x|+c+\frac{ds}{2}+\frac{\log2}{d},\] because \(e^{d|x+\sqrt sG|}\le e^{d(x+\sqrt sG)}+e^{-d(x+\sqrt sG)} \le e^{d|x|}(e^{d\sqrt sG}+e^{-d\sqrt sG})\) after taking expectations. Propagate this bound through levels \(k-1,\ldots,1\) and then average \(|\sqrt{h_0}G|\) at the root. The variance-dependent constant is \[-\frac{h_{k-1}}2 +\frac12\sum_{i=1}^{k-1}\zeta_i(h_i-h_{i-1}) =-\frac12\sum_{i=0}^{k-1}w_ih_i,\] which proves (61), with \(C_\zeta=\sum_{i=1}^{k-1}(\log2)/\zeta_i\); the sum is zero if \(k=1\).

Adding the patterns changes pressure by at most \(\alpha\|f\|_\infty\) in expectation, and the perturbation cost is \(O(e_N^2)\). Thus \[F_N(t,h,u)\ge-\ell(h)-\alpha\|f\|_\infty-O(e_N^2).\] Since \(S_{\mathrm I}(q)\ge0\), \(|V_f(q)|\le\|f\|_\infty\), and the quadratic penalty is nonnegative, on \(h_{k-1}=H\) this gives the uniform lower bound \[ J_N(t,h,u)\ge \frac{w_{k-1}(1-q_{k-1})}{2}H -(\mathbb E|G|)\sqrt H-C_\zeta -2\alpha\|f\|_\infty-O(e_N^2). \tag{62}\] The linear coefficient is strictly positive, so choose \(H\) large enough that this lower bound is positive for all large \(N\). At \(t=0\), the entropy definition and (55) instead give \(J_N(0,h,u)\ge-O(e_N^2)\). Finally, testing \(t=\alpha\), \(h=0\), and \(u_j=3/2\) gives \(J_N\le-\eta/2\) for all sufficiently large dimensions in the violating sequence, by the Poisson coupling, the choice of \(\delta\), and the perturbation bound. Hence a minimizing triple lies on neither of the excluded faces. ◻

Proof. Fix \(j\). All derivatives in this proof hold \((t_N,h_N)\) and the other perturbation coordinates fixed. Abbreviate \(a_{N,j}=\sqrt N e_N2^{-j}\) and \(g(v)=\mathbb E P_N(t_N,h_N,u_1,\ldots,v,\ldots,u_N)\). Ordinary Gibbs differentiation gives \[\begin{align*} \partial_{u_j}P_N&=\frac{e_N2^{-j}}{\sqrt N} \langle Y_j\rangle,\tag{64}\\ g''(u_{N,j})&=e_N^24^{-j} \mathbb E\langle(Y_j-\langle Y_j\rangle)^2\rangle. \tag{65}\end{align*}\] The random function \(P_N\) is convex in \(u_j\), since it is a log integral of an energy affine in that coordinate. We first control the Gibbs fluctuation of \(Y_j\) around \(\langle Y_j\rangle\) by the curvature at the minimum. We then control the disorder fluctuation of \(\langle Y_j\rangle\) by finite differences. Both estimates hold at the chosen deterministic coordinate \(u_{N,j}\), without averaging it.

Interior minimality of \(J_N\) implies \(g'(u_{N,j})=2\cdot2^{-j}(u_{N,j}-3/2)\) and \(g''(u_{N,j})\le2\cdot2^{-j}\). Consequently \[ \mathbb E\langle|Y_j-\langle Y_j\rangle|\rangle \le C_j e_N^{-1}. \tag{66}\] For the disorder fluctuation, use minimality over an interval around \(u_{N,j}\). Cancellation of the linear term gives \[ 0\le g(u_{N,j}\pm s)-g(u_{N,j}) \mp s g'(u_{N,j})\le2^{-j}s^2 \tag{67}\] whenever both displaced coordinates lie in \([1,2]\). The first inequality uses convexity of \(g\).

For any differentiable convex random function, its derivative at a point lies between the backward and forward difference quotients. Apply this fact to \(P_N\) with step \(s\), and subtract \(g'(u_{N,j})\). The mean quotient errors are at most \(2^{-j}s\) by (67). Replacing each of the three random function values by its expectation costs at most \(C_H/(s\sqrt N)\) in \(L^1\), by Lemma 20. Taking the positive and negative parts therefore yields \[ \mathbb E|\partial_{u_j}P_N-g'(u_{N,j})| \le C_js+\frac{C_H}{s\sqrt N}. \tag{68}\] By (63), \(s=N^{-1/4}\) is eventually allowed. Then the right side is \(O_j(N^{-1/4})=o(e_N^2)\). Multiplying by \(\sqrt N/(e_N2^{-j})\) in (64), and adding (66), gives \[ \mathbb E\langle|Y_j-\mathbb E\langle Y_j\rangle|\rangle =o(\sqrt N e_N). \tag{69}\] Explicitly, the two contributions are \(O_j(N^{1/16})\) and \(O_j(N^{5/16})\), whereas \(\sqrt N e_N=N^{7/16}\). All assertions concern one fixed \(j\) at a time; no bound uniform in \(j\le N\) is needed.

Let \(D\) be a bounded continuous function of the first \(n\times n\) joint overlap block, \(n\ge2\). Gaussian integration by parts in the field \(Y_j\) gives \[ \mathbb E\langle Y_j(1)D\rangle =a_{N,j}u_{N,j}\, \mathbb E\left\langle D\left( \sum_{l=1}^n\kappa_j(R_{1l},T_{1l}) -n\kappa_j(R_{1,n+1},T_{1,n+1})\right)\right\rangle. \tag{70}\] The overlap test has no direct dependence on the Gaussian coordinates of \(Y_j\), so Lemma 17 applies. By (69), the left side equals \(\mathbb E\langle Y_j\rangle\mathbb E\langle D\rangle +o(\sqrt N e_N)\). The single-replica version of (70) reads \[\mathbb E\langle Y_j\rangle =a_{N,j}u_{N,j} \{\kappa_j(1,(k-1)/k) -\mathbb E\langle\kappa_j(R_{12},T_{12})\rangle\}.\] Substitute this expression and cancel the constant diagonal. After division by \(n a_{N,j}u_{N,j}\), one obtains \[\begin{align*} \mathbb E\langle D\kappa_j(R_{1,n+1},T_{1,n+1})\rangle ={}&\frac1n\mathbb E\langle D\rangle \mathbb E\langle\kappa_j(R_{12},T_{12})\rangle\\ &+\frac1n\sum_{l=2}^n \mathbb E\langle D\kappa_j(R_{1l},T_{1l})\rangle+o(1). \end{align*}\] Pass to a joint array subsequence, which exists because the entries range over a compact set. The identity holds for every fixed monomial and hence, by linearity and uniform polynomial approximation, for every continuous function of a pair \((R,T)\). Continuous tests of the finite block determine its measures, so the same identities hold with every bounded measurable block test. This is the full conditional-law formulation in Definition 9. ◻

Proof. The arrays \(R\) and \(R+T\) are Gram arrays of constant diagonal, and their scalar identities follow from the joint identities. Theorem 10 therefore makes both arrays ultrametric and makes the off-diagonal entries of \(R\) nonnegative. They are at most one by its Gram diagonal. The tree array \(T\) is already ultrametric.

Suppose the two-replica support contained two strictly crossed points. Choose small neighborhoods \(A,B\) of these points so that every \((r,t)\in A\) and \((r',t')\in B\) satisfy \(r<r'\) and \(t>t'\). Both neighborhoods have positive two-replica probability. The joint conditional law with \(n=2\) then gives \[\mathbb P\big((R_{12},T_{12})\in A, (R_{13},T_{13})\in B\big) \ge\tfrac12\mathbb P((R_{12},T_{12})\in A) \mathbb P((R_{12},T_{12})\in B)>0.\] On this event, ultrametricity of \(R\) forces \(R_{23}=R_{12}\): the inequalities for edges \(12\) and \(23\) give the two opposite inequalities once \(R_{12}<R_{13}\) is used. Likewise, ultrametricity of \(T\) forces \(T_{23}=T_{13}\). It follows that \(R_{23}+T_{23}\) is strictly smaller than both \(R_{12}+T_{12}\) and \(R_{13}+T_{13}\), contradicting ultrametricity of \(R+T\). Hence such support points cannot exist.

By (56), the successive values \(i/k\) of \(T_{12}\) have masses \(w_i\). The conditional spin-overlap supports at these levels are ordered: every value at an earlier level is at most every value at a later level. Thus sorting their mixture simply concatenates these conditional laws in blocks of lengths \(w_i\). This is exactly the claimed quantile description, including any atoms shared by adjacent conditional supports. ◻

Proof of Proposition 19. Continue along the violating sequence and its minimizing triples. Extract a joint array limit and let \(p\) be the spin-overlap quantile given by Lemma 24. For any nonnegative nondecreasing step direction \(\lambda\) on the fixed partition, sufficiently small positive displacements of \(h_N\) are feasible, since \(h_{N,k-1}<H\). Minimality and (58) yield \[\mathbb E\langle R_{12} \lambda_{\gamma^1\wedge\gamma^2}\rangle \ge\sum_{i=0}^{k-1}w_iq_i\lambda_i.\] Pass to the joint array limit and use its ordered conditional laws to obtain \[ \int_0^1p(u)\lambda(u)\,du \ge\int_0^1q(u)\lambda(u)\,du. \tag{71}\] Choose successively \(\lambda_i=\mathbf 1_{\{i\ge j\}}\), \(0\le j\le k-1\). This gives \[\int_{\zeta_j}^1p(u)\,du \ge\int_{\zeta_j}^1q(u)\,du \qquad(0\le j\le k),\] where \(j=k\) is equality. The function \(s\mapsto\int_s^1p(u)\,du\) is concave, because its almost-everywhere derivative is the nonincreasing function \(-p(s)\). On each partition interval the corresponding tail integral of \(q\) is affine. Concavity places the first tail above the chord joining its endpoint values, which in turn lies above the second tail. Therefore the tail inequalities hold for every \(s\in[0,1]\). By Proposition 5, \[ V_f(p)\le V_f(q). \tag{72}\]

At every minimizing triple \(t_N>0\), a small displacement to the left in time is feasible, even if \(t_N=\alpha\) or \(t_N\downarrow0\). Thus the derivative of \(J_N\) with respect to \(t\) at that triple is nonpositive. On the other hand, the fresh pattern in (59) has covariance \(R\) and diagonal one. Lemma 16, applied to the marginal limiting spin-overlap array, gives \[\partial_t J_N(t_N,h_N,u_N) \longrightarrow -V_f(p)+V_f(q)+\delta\ge\delta>0.\] This contradicts the required sign at the minimizers and proves (52) for every step trial with \(q_{k-1}<1\) and finite entropy. Infinite-entropy trials give no restriction. Taking the infimum over finite step trials and using Corollary 6 proves the claimed bound for the full variational infimum, and hence for every admissible \(q\). ◻

Proof. Let \(P_N^\circ\) be the omitted system’s random normalized log partition function. Uniformly when each parameter lies in \([1/2,5/2]\), we have \[ \operatorname{Var}(P_N^\circ)\le \frac C N. \tag{78}\] Indeed, replacing one pattern changes its unnormalized log partition function by at most \(2\|f\|_\infty\). Conditional on the patterns, its Lipschitz constant in the independent Gaussian tensors defining the perturbation is at most \(s_N(\sum_j4^{-j}v_j^2)^{1/2}\le Cs_N\). The conditional Gaussian variance bound and the bounded-difference bound for the conditional mean therefore give \(\operatorname{Var}(\log Z_N^\circ)\le C(N+s_N^2)\le CN\). Both bounds are instances of Lemma 18.

Fix \(j\), with \(N\ge j\), and put \(t_j=s_N2^{-j}\). Finite-dimensional Gaussian integration by parts and differentiation of the spin sum give \[\begin{align*} \partial_{v_j}P_N^\circ &=\frac{t_j}{N}\langle Y_j\rangle_\circ, \tag{79}\\ \partial_{v_j}\mathbb E P_N^\circ &=\frac{t_j^2v_j}{N} \bigl(1-\mathbb E\langle R_{12}^j\rangle_\circ\bigr), \tag{80}\\ \partial_{v_j}^2\mathbb E P_N^\circ &=\frac{t_j^2}{N}\mathbb E \big\langle(Y_j-\langle Y_j\rangle_\circ)^2\big\rangle_\circ. \tag{81}\end{align*}\] These operations are legitimate since the state space is finite and Gaussian exponential moments control all derivatives on a compact parameter interval. Formula (80) is bounded in absolute value by \(Ct_j^2/N\) on the enlarged interval. Integrating (81) from \(1\) to \(2\) consequently yields \[ \int_1^2\mathbb E \big\langle(Y_j-\langle Y_j\rangle_\circ)^2\big\rangle_\circ\,dv_j \le C. \tag{82}\] The other parameters remain arbitrary throughout this estimate.

For completeness, the concentration estimate needed for the quenched mean of \(Y_j\) follows from integrated convexity, without selecting a parameter in advance. Let \(F=\mathbb E P_N^\circ\) and \(0<\eta<1/4\). The forward and backward difference quotients bracket \(\partial_{v_j}P_N^\circ\). Replacing each random function value in these quotients by its expectation costs at most \(C/(\eta\sqrt N)\) in expected absolute value, by (78). For the mean function, the forward error is bounded by \(F'(v_j+\eta)-F'(v_j)\), and the backward error by \(F'(v_j)-F'(v_j-\eta)\). Integrating these differences over \([1,2]\) leaves intervals of total length at most \(4\eta\) at the ends. Since \(F'\) is bounded by \(Ct_j^2/N\), this proves \[ \int_1^2\mathbb E \big|\partial_{v_j}P_N^\circ-\partial_{v_j}\mathbb E P_N^\circ\big| \,dv_j \le C_j\eta\frac{s_N^2}{N}+\frac{C}{\eta\sqrt N}. \tag{83}\] Choose \(\eta=N^{-1/8}\). Both terms are \(O_j(N^{-3/8})\), whereas \(s_N^2/N=N^{-1/4}\). Combining (79), (82), and Cauchy–Schwarz gives \[\begin{align*} &\int\mathbb E\big\langle |Y_j-\mathbb E\langle Y_j\rangle_\circ| \big\rangle_\circ\,\pi(dv) \\ &\qquad\le C+\frac{N}{t_j}\,O_j(N^{-3/8}) =O_j(N^{1/4})=o(s_N). \tag{84}\end{align*}\] Here and below \(j\) is fixed before \(N\to\infty\).

Set \[a_{j,N}(v)=s_N^{-1}\mathbb E\big\langle |Y_j-\mathbb E\langle Y_j\rangle_\circ| \big\rangle_\circ.\] By (84), choose integers \(J_N\le N\) increasing to infinity sufficiently slowly that \[e_N:=\sum_{j\le J_N}\int a_{j,N}\,d\pi\longrightarrow0.\] With \(\delta_N=\sqrt{e_N}+N^{-1}\), define \[\mathcal G_N=\left\{v:\sum_{j\le J_N}a_{j,N}(v)\le\delta_N\right\}.\] Markov’s inequality gives \(\pi(\mathcal G_N^c)\le e_N/\delta_N\to0\), and each fixed \(a_{j,N}\) tends to zero uniformly on \(\mathcal G_N\).

To obtain the identities, let \(D\) be a bounded continuous test of the first \(n\times n\) overlap block, \(n\ge2\). Integration by parts gives \[\mathbb E\langle Y_j(x^1)D\rangle_\circ =t_jv_j\mathbb E\left\langle D\left( \sum_{l=1}^nR_{1l}^j-nR_{1,n+1}^j\right)\right\rangle_\circ.\] The left side differs from \(\mathbb E\langle Y_j\rangle_\circ\,\mathbb E\langle D\rangle_\circ\) by at most \(\|D\|_\infty s_Na_{j,N}(v)\). Using the corresponding one-replica integration-by-parts identity and \(R_{11}=1\) therefore gives \[\begin{align*} &\mathbb E\langle D R_{1,n+1}^j\rangle_\circ -\frac1n\mathbb E\langle D\rangle_\circ \mathbb E\langle R_{12}^j\rangle_\circ -\frac1n\sum_{l=2}^n\mathbb E\langle D R_{1l}^j\rangle_\circ =o(1), \end{align*}\] uniformly on \(\mathcal G_N\) for each fixed \(j,n,D\). Polynomial approximation on \([-1,1]\), followed by determination of finite measures by continuous tests, gives the full identities in every array limit. This proves the stated quantifier for arbitrary deterministic choices from the good sets. ◻

Proof. The required subsequences exist by boundedness and diagonal extraction. Lemma 25 and Theorem 10 show that the omitted limiting overlap array is nonnegative and ultrametric.

Proof. Compact support makes all scale derivatives needed below bounded. Uniformly in \(s\in\mathbb R\), Taylor expansion in the scale parameter therefore gives \[\begin{align*} f(\lambda_Ns)&=f(s)-\frac L{2N}s f'(s)+O_L(N^{-2}),\\ f^{(j)}(\lambda_Ns)&=f^{(j)}(s)+O_L(N^{-1}),\qquad j=1,2. \end{align*}\] For example, derivatives of \(f(\lambda s)\) with respect to \(\lambda\) are \(s^r f^{(r)}(\lambda s)\), uniformly bounded when \(\lambda\) stays near one. Next expand the translation by \((y_a\cdot z)/\sqrt{N+L}\) through second order, and replace its denominators by \(\sqrt N\) and \(N\). Its third-order remainder is bounded by \(C_LN^{-3/2}|y_a|^3\), uniformly over \(x,z\). The scale and denominator replacements add terms bounded by \(C_L(N^{-2}+N^{-3/2}|y_a|+N^{-2}|y_a|^2)\). Summing over the \(M_N=O(N)\) old rows thus gives an expected uniform error \(O_L(N^{-1/2})\).

The deterministic part of the quadratic term is \(L\sum_a f''(b_a(x))/(2N)\); together with the dilation term it is \(LC_N(x)/2\). The remaining quadratic term is precisely \(Q\), and the linear term is \(z\cdot\mathcal Y(x)\). Each new row can be replaced by \(f(b_a(x))\): its expected uniform cost is at most \(C_L(N^{-1}+N^{-1/2}\mathbb E|y_a|)\). Since there are only \(d=O_L(1)\) new rows, this is \(o(1)\) in total. We have proved that, after the perturbation comparison, the energy equals \[ \sum_{a=1}^{M_N}f(b_a(x))+H_N^v(x) +\log W(x,z)+Q(x,z)+\operatorname{rem}_N(x,z), \qquad \mathbb E\sup_{x,z}|\operatorname{rem}_N(x,z)|=o(1). \tag{94}\] The elementary log-partition bound by the uniform energy difference removes this remainder. Differentiating a normalized \(r\)-replica expectation along the energy segment bounds its change by \(2r\|D\|_\infty\sup|\operatorname{rem}_N|\), so the same removal applies to the asserted tests.

Put \(A(x)=LC_N(x)/2+\sum_{a>M_N}f(b_a(x))\). Uniformly, \(|A(x)|\le C_L\). Since the cavity prior is a product of symmetric signs, \[G_0(W)=G_N^v\left(e^{A(x)} \prod_{i=1}^L\cosh(\mathcal Y_i(x))\right) \ge e^{-C_L}=:\kappa_L.\] This bound concerns the untilted probability (92); it will control all subsequent normalizations.

It remains to remove \(Q\), which need not be uniformly small over all spin states. For each fixed \(x,z\) and fixed bulk data, the variables \(y_a\cdot z\) are independent \(N(0,L)\) random variables. Thus \[ \mathbb E_y Q(x,z)^2\le\frac{C_L}{N},\qquad \mathbb E_y e^{t|Q(x,z)|}\le C_{L,t}\quad(t>0), \tag{95}\] uniformly for large \(N\). To check the second assertion directly, put \(c_a=f''(b_a(x))/(2N)\). For fixed \(t\) and sufficiently large \(N\), \[\log\mathbb E_y e^{t c_a((y_a\cdot z)^2-L)} =-tLc_a-\tfrac12\log(1-2tLc_a)=O_{L,t}(c_a^2).\] The linear terms cancel. Summing over \(a\) is \(O_{L,t}(N^{-1})\); using \(e^{t|Q|}\le e^{tQ}+e^{-tQ}\) proves the claim. Also \[ \mathbb E_y e^{t z\cdot\mathcal Y(x)}\le C_{L,t}, \tag{96}\] because its conditional variance is \(LB_N(x,x)\le C_L\).

Write \(P_W=W\,dG_0/G_0(W)\) for the normalized approximation. The bounds just proved imply \[ \mathbb E P_W(|Q|)+\mathbb E P_W(|e^Q-1|) \le C_LN^{-1/2}. \tag{97}\] Indeed \(G_0(W)\ge\kappa_L\) and the non-linear-field part of \(W\) is bounded. For the second term, use \(|e^Q-1|\le|Q|e^{|Q|}\) and Cauchy–Schwarz to bound its fixed-state \(y\) expectation by \[(\mathbb E_yQ^2)^{1/2} \bigl(\mathbb E_y e^{2z\cdot\mathcal Y(x)+2|Q|}\bigr)^{1/2} \le C_LN^{-1/2}.\] A further Cauchy–Schwarz inequality controls the last exponential by (95) and (96). This uses no independence between \(Q\) and \(\mathcal Y\).

Jensen’s inequality and \(\log t\le t-1\) now give \[-P_W(|Q|)\le\log P_W(e^Q)\le P_W(|e^Q-1|),\] so removing \(Q\) changes the expected log partition by \(o(1)\). For normalized tests, the following elementary bound avoids any additional lower bound for \(P_W(e^Q)\). For every probability \(P\) and positive integrable \(Y\), writing \(m=P(Y)>0\), \[ P\left|\frac Ym-1\right| \le P\left|\frac Ym-Y\right|+P|Y-1| =|1-m|+P|Y-1|\le2P|Y-1|. \tag{98}\] Apply this with \(P=P_W,Y=e^Q\) and then use the product-measure bound for a fixed number of replicas. Together with (88) and (94), this proves (93) and all the claimed sample comparisons. Uniform continuity on \([-1,1]\) handles the replacement of \(R^+\) by \(R\). The deterministic denominator lower bound proves the final uniform lower bound for the increments. ◻

Proof. We identify the field-recursion derivative using one distinguished cavity coordinate. For a fixed rounding \((\bar p_m,\bar h_m)\), let \(\rho_{\bar h_m}\) be the bounded nondecreasing derivative path of Section 3; on level \(i\) it equals \(\mathbb E[U_i'(Z_i)^2]\) for the field recursion.

Under the cavity tilt, the conditional mean of \(z_1^1z_1^2\) given pair level \(i\) is precisely \((\rho_{\bar h_m})_i\). To verify this statement including the residual noise, fix the shared leaf field \(t\) for coordinate one. If \(\eta\) is the independent residual with variance \(a_{\mathrm d}-\bar h_{m,\mathrm{top}}\), then \[\mathbb E_\eta e^{z_1(t+\eta)} =e^{z_1t}e^{(a_{\mathrm d}-\bar h_{m,\mathrm{top}})/2}.\] The second factor is independent of \(z_1\), so after residual averaging the spin magnetization is \(\tanh t\). Two visits, even to the same leaf, use separate residuals and separate spin draws. Below their shared level their mark paths are conditionally independent. Propagating the leaf magnetization backwards along the tilted kernels therefore gives the conditional mean \(U_i'\) at level \(i\), by the spatial derivative martingale for the field recursion. The conditional product mean is its averaged square. All other coordinates and all new pattern marks have separate kernels by the additive-component property. The shape law itself is unchanged by the tilt.

At finite size the disorder law of the original \((N+L)\)-spin model, including its perturbation at fixed \(v\), is invariant under all coordinate permutations. Thus for every bounded continuous \(D_0:[-1,1]\to\mathbb R\), \[ \mathbb E\langle D_0(R^+_{12})z_1^1z_1^2\rangle_{N+L} =\mathbb E\langle D_0(R^+_{12})R^+_{12}\rangle_{N+L}. \tag{103}\] Indeed the left side is unchanged if the distinguished coordinate is replaced by any of the \(N+L\) coordinates, and their average product is \(R^+_{12}\). This symmetry is used before the cavity approximations. Lemma 27, the truncation passage above, and the finite cascade calculation then imply \[ \lim_{m\to\infty}\int_0^1 D_0(\bar p_m(u))\rho_{\bar h_m}(u)\,du =\int_0^1 D_0(p(u))p(u)\,du. \tag{104}\] For the right side, the unclipped cascade tilt preserves the rounded overlap law by Lemma 14. The truncation estimates above transfer this evaluation to the finite-size limit, whose overlap law has quantile \(p\).

The paths \(\rho_{\bar h_m}\) take values in \([0,1]\) and are nondecreasing. Bounded monotone selection and dominated convergence give a subsequence converging in \(L^1(0,1)\) to a function \(\rho\). Every \(\rho_{\bar h_m}\) is measurable with respect to the same completed sigma-algebra \(\sigma(p)\): the bins were formed from \((p,a(p))\) and hence are functions of \(p\), with no splitting of its flat intervals. The subspace of \(\sigma(p)\)-measurable functions is closed in \(L^1(0,1)\). Explicitly, conditional expectation on the Lebesgue probability space \((0,1)\) onto \(\sigma(p)\) is an \(L^1\) contraction, so \[\|\rho-\mathbb E[\rho\mid\sigma(p)]\|_1 \le2\|\rho-\rho_{\bar h_m}\|_1\longrightarrow0.\] Thus \(\rho\) itself is \(\sigma(p)\)-measurable.

Pass to this subsequence in (104). Uniform continuity of \(D_0\) and the \(L^1\) convergence of \(\bar p_m\) justify replacing \(D_0(\bar p_m)\) by \(D_0(p)\). We obtain \[\int_0^1D_0(p(u))(\rho(u)-p(u))\,du=0 \quad\hbox{for every continuous }D_0.\] Continuous tests determine finite signed measures on \([0,1]\). Therefore the conditional expectation of \(\rho-p\) given \(p\) is zero. Since \(\rho-p\) is already \(\sigma(p)\)-measurable, it follows that \(\rho=p\) almost everywhere. In particular, along this subsequence, \[ \rho_{\bar h_m}\longrightarrow p\quad\hbox{in }L^1(0,1). \tag{105}\]

Apply the supporting inequality (22) with base \(\bar h_m\) and an arbitrary fixed nonnegative nondecreasing field step \(h'\), refining the partitions if necessary: \[\ell(h')\le\ell(\bar h_m) -\frac12\int_0^1(h'-\bar h_m)\rho_{\bar h_m}.\] The fields \(\bar h_m\) are uniformly bounded and converge to \(h\) in \(L^1\). Using (105) and continuity of \(\ell\), we may pass to the limit to get \[\ell(h')+\frac12\int_0^1h'p \le\ell(h)+\frac12\int_0^1hp.\] Taking the supremum over all allowed field steps gives the upper bound in (102). Conversely, approximate the bounded nondecreasing \(h\) in \(L^1\) by nonnegative nondecreasing step functions. Continuity of \(\ell\) and boundedness of \(p\) show that their dual values converge to the right side of (102). The definition of \(S_{\mathrm I}\) gives the reverse inequality. ◻

Proof. Write \(\mathcal P=\inf_q\{S_{\mathrm I}(q)+\alpha V_f(q)\}\), a finite number by the elementary bounds for the variational functional. The limit (101) and Proposition 28 give, on every subsequence constructed above, \[\lim_N\Delta_{N,L}(v(N)) =L\{S_{\mathrm I}(p)+\alpha V_f(p)\} +(d-\alpha L)V_f(p) \ge L\mathcal P-\|f\|_\infty.\] We used \(|V_f(p)|\le\|f\|_\infty\) and \(|d-\alpha L|\le1\). Crucially, the starting good parameter sequence was arbitrary. Consequently \[ \liminf_{N\to\infty}\inf_{v\in\mathcal G_N}\Delta_{N,L}(v) \ge L\mathcal P-\|f\|_\infty. \tag{106}\] For otherwise a fixed positive violation could be selected at successive dimensions with deterministic \(v(N)\in\mathcal G_N\); compactness would supply the array and constant-\(d\) subsequence above, contradicting its evaluated limit.

The uniform all-parameter lower bound of Lemma 27 controls the complement of \(\mathcal G_N\). Specifically, for every \(\varepsilon>0\) and all large \(N\), the parameter integral is at least \[\pi(\mathcal G_N) (L\mathcal P-\|f\|_\infty-\varepsilon) -C_L\pi(\mathcal G_N^c).\] The use of the same product probability in both dimensions gives the exact identity \[\int\Delta_{N,L}(v)\,\pi(dv)=\mathcal F_{N+L}-\mathcal F_N.\] No good-parameter condition in dimension \(N+L\) is imposed or needed. We conclude that \[\liminf_N(\mathcal F_{N+L}-\mathcal F_N) \ge L\mathcal P-\|f\|_\infty.\] For fixed \(L\) and \(\varepsilon>0\), telescope the eventual bound on these increments separately in each of the \(L\) residue classes modulo \(L\). The finitely many initial terms in each class vanish upon division by \(N\), so \[\liminf_N\frac{\mathcal F_N}{N} \ge\mathcal P-\frac{\|f\|_\infty+\varepsilon}{L}.\] First let \(\varepsilon\downarrow0\) and then let \(L\to\infty\). Finally (75) transfers the result to the unperturbed pressure. Together with the upper bound this proves the variational formula for smooth compactly supported \(f\). ◻

Proof. Interpolate linearly between \(f\) and \(g\). The derivative of the expected pressure is a sum of Gibbs averages of \((g-f)(b_a(x))\), divided by \(N\). For each summand, remove pattern \(a\). Restoring it changes a nonnegative Gibbs average by a factor at most \(e^{2K}\), because the removed log factor lies in \([-K,K]\). Conditional on every other pattern and on a sampled spin \(x\), the fresh field \(g^a\cdot x/\sqrt N\) has law \(\gamma\). Hence the absolute derivative is at most \((M_N/N)e^{2K}\|g-f\|_{L^1(\gamma)}\). Integration proves (107).

For a fixed step path, terminal interpolation in the recursion propagates \(g-f\) by its tilted Gaussian kernels. This differentiation requires only bounded terminal functions: it follows directly by differentiating the finite expectations. In the notation of the pattern recursion, the density of the entire tilted path relative to its independent Gaussian increments has logarithm \[ \sum_{i=0}^{k}\zeta_i\big(U_i(Z_i)-U_{i-1}(Z_{i-1})\big) =f_t(Z_k)-\sum_{i=0}^{k-1}w_iU_i(Z_i), \qquad f_t=(1-t)f+tg. \tag{109}\] All the functions on the right take values in \([-K,K]\), so the density is at most \(e^{2K}\). Without the tilt, \(Z_k\) is standard normal, since the total variance is \(q_k-q_{-1}=1\). This proves (108), uniformly over partitions and path values.

Choose \(f_j\in C_c^\infty(\mathbb R)\) with \(|f_j|\le K\) and \(\|f_j-f\|_{L^1(\gamma)}\to0\). Such approximations follow by spatial truncation, approximation on compact intervals, and mollification with a nonnegative smooth kernel. Each \(V_{f_j}\) is continuous in \(L^1\) by Proposition 5. Estimate (108) makes them uniformly Cauchy on the dense set of step paths and hence on \(\mathcal Q\). Their uniform limit is continuous and agrees with the original finite recursion on every step path. Density also proves uniqueness. The elementary bound \(|V_f|\le\|f\|_\infty\) persists. ◻

Proof. Use the approximants from Lemma 30. Since the entropy functional is the same for every activation, \[|\mathcal P_f(\alpha)-\mathcal P_{f_j}(\alpha)| \le\alpha\sup_{q\in\mathcal Q}|V_f(q)-V_{f_j}(q)|\longrightarrow0.\] Together with the uniform pressure estimate, the already proved smooth formula gives convergence of \(\mathbb E p_N(f)\) to \(\mathcal P_f(\alpha)\).

Changing a single whole pattern vector changes \(\log Z_N(f)\) by at most \(2\|f\|_\infty\). The independent-input variance bound of Lemma 18 therefore gives \[ \operatorname{Var}(p_N(f))\le \frac{4M_N\|f\|_\infty^2}{N^2}=O(N^{-1}). \tag{110}\] Chebyshev’s inequality proves convergence in probability. ◻

Proof. We establish estimates uniform in both the dimension and the step path. Let \(P\) be the base probability for one fresh pattern, and let \(t\) be the value of \(f\) on its field. In the spin model, \(P\) is the Gibbs measure with that whole pattern omitted, independent of the fresh row. For a step path, let \((v_\lambda)\) be the cascade weights and let \((y_\lambda)\) be the Gaussian field built from shared tree increments, independent of those weights and with variance \(q_{k-1}\) at each leaf. Use the product probability and scalar function \[P=\sum_\lambda v_\lambda\delta_\lambda\otimes\gamma, \qquad t(\lambda,\eta)=f\bigl(y_\lambda+\sqrt{1-q_{k-1}}\,\eta\bigr).\] Thus the residual \(\eta\) is an integration coordinate of \(P\), as in Lemma 16. For a bounded Borel terminal function, integrating \(\eta\) gives the coefficient-one transform, and Lemma 14, followed by the root Gaussian average, evaluates the expected log integral as its pattern recursion. This identity uses measurability and boundedness, not the continuity needed for an array limit. Zero residual variance and repeated covariance levels are allowed. In the spin case, the fresh field at each fixed state is standard normal. In the cascade case the same marginal results after both averaging \(y_\lambda\) and integrating \(\eta\). Independence of the weights and the shared field, followed by Tonelli’s theorem, therefore gives, for every nonnegative Borel \(F\), \[ \mathbb E P[F(t)]=\mathbb E F(f(G)). \tag{113}\] Here the expectation includes the independent pattern and any base data. The log integral \(\log P(e^t)\) is finite almost surely and integrable: Jensen bounds it below by \(P(t)\), while its positive part is bounded by \(P(e^t)\). These are integrable by (111) and (113). Adding the rows successively also proves integrability of the finite-volume pressure.

Put \(t_J=\max(t,-J)\) and \(r=(-J-t)_+\). The nonnegative lower-clipping error satisfies \[\begin{align*} 0\le \log P(e^{t_J})-\log P(e^t) &=-\log\frac{P(e^{t_J}e^{-r})}{P(e^{t_J})} \le\frac{P(e^{t_J}r)}{P(e^{t_J})} \le P(r). \tag{114}\end{align*}\] The bound by the tilted mean of \(r\) is Jensen’s inequality under the \(e^{t_J}\) tilt. For the last inequality, \(e^{\max(t,-J)}\) is increasing in \(t\), while \((-J-t)_+\) is decreasing. Their covariance under \(P\) is nonpositive. This follows also by expanding the expectation of \((a(t)-a(t'))(b(t)-b(t'))\le0\) for two independent \(P\) samples.

After the lower clip, the upper-clipped exponential is at least \(e^{-J}\). The inequality \(\log(1+x)\le x\) gives \[ 0\le \log P(e^{t_J})-\log P(e^{\min(t_J,K)}) \le e^J P[(e^t-e^K)_+]. \tag{115}\] Define \[\varepsilon_{J,K}= \mathbb E(-J-f(G))_+ +e^J\mathbb E(e^{f(G)}-e^K)_+.\] Replacing one pattern at a time in the finite spin system, allowing the other factors to be arbitrary existing clipped or unclipped factors, yields \[ \mathbb E|p_N(f)-p_N(f_{J,K})|\le\alpha\varepsilon_{J,K}. \tag{116}\] The omitted factor is independent of that base probability at each replacement, so (113) continues to apply. The same argument for the single cascade log integral yields \[ \sup_{q\text{ step}}|V_f^{\mathrm{cas}}(q)-V_{f_{J,K}}(q)| \le\varepsilon_{J,K}, \tag{117}\] where \(V_f^{\mathrm{cas}}(q)\) denotes its integrable expected log value.

At fixed \(J\), the upper-clipping bound compares each bounded recursion uniformly on step paths with the cascade value for \(\max(f,-J)\); its error tends to zero as \(K\to\infty\). Hence these bounded continuous functionals are uniformly Cauchy on the dense set of step paths and therefore on \(\mathcal Q\). Their uniform limit is continuous. For \(J'\ge J\), apply the lower-clipping bound with \(f\) replaced by \(\max(f,-J')\). The difference between the corresponding limits is at most \(\mathbb E(-J-f(G))_+\), which tends to zero as \(J\to\infty\). This proves the ordered uniform limit (112). It agrees with \(V_f^{\mathrm{cas}}\) on step paths, and (117), with \(V_f\) on the left, passes to all paths by density.

Two applications of Jensen’s inequality to the same cascade integral give \[\mathbb E f(G)\le V_f(q)\le\log\mathbb E e^{f(G)}.\] The bounds hold first for step paths and then for every \(q\in\mathcal Q\). At \(q=0\) the upper bound is an equality. Since \(S_{\mathrm I}\ge0\) and \(S_{\mathrm I}(0)=0\), the variational value is finite, with \[\alpha\mathbb E f(G)\le\mathcal P_f(\alpha) \le\alpha\log\mathbb E e^{f(G)}.\] Estimate (117) also bounds the difference between the clipped and unclipped variational infima by \(\alpha\varepsilon_{J,K}\).

Apply Proposition 31 to \(f_{J,K}\) and then (116). This proves convergence in expectation. For convergence in probability, Markov’s inequality controls \(p_N(f)-p_N(f_{J,K})\) uniformly in \(N\); convergence for the bounded model handles the middle term. Take \(N\to\infty\), then \(K\to\infty\) at fixed \(J\), and finally \(J\to\infty\). No variance estimate for the unbounded model is required. ◻

Proof. First let \(f\in C_c^\infty(\mathbb R)\). We use the coordinate-replacement form of the Lindeberg principle; see Chatterjee (Chatterjee 2006, Theorem 1.1). Regard normalized pressure as a function of a single pattern entry. Three differentiations of the finite spin sum give a uniform bound \(C_fN^{-5/2}\) for its third derivative. Indeed, each derivative of the pattern argument supplies \(x_i/\sqrt N\), and the normalized log partition supplies the factor \(1/N\); the remaining terms are bounded Gibbs moments and cumulants of \(f',f'',f'''\). Taylor expansion through degree two and matching of the first two Gaussian and sign moments therefore bound each entry replacement by \(C_fN^{-5/2}\). There are \(M_NN=O(N^2)\) entries, so replacing them one at a time gives an expected-pressure error \(O_f(N^{-1/2})\).

For a general \(f\) as in the statement, take smooth approximants \(f_j\) with \(|f_j|,|f|\le K\) and \(\|f_j-f\|_{L^1(\gamma)}\to0\). The omitted-pattern argument from Lemma 30 still applies. Conditional on any hypercube state, however, the fresh sign field now has the common law \[X_N=N^{-1/2}\sum_{i=1}^N\epsilon_i, \qquad \mathbb P(\epsilon_i=1)=\mathbb P(\epsilon_i=-1)=1/2.\] Consequently, \[ \big|\mathbb E p_N^{\mathrm{sign}}(f) -\mathbb E p_N^{\mathrm{sign}}(f_j)\big| \le\alpha e^{2K}\mathbb E|f(X_N)-f_j(X_N)|. \tag{118}\] The same third-order replacement calculation on smooth compactly supported tests, together with \(\mathbb EX_N^2=1\), proves \(X_N\Rightarrow G\). The bounded function \(|f-f_j|\) is continuous outside a Gaussian-null set. The usual bounded almost-everywhere-continuous test criterion for weak convergence therefore gives \[\mathbb E|f(X_N)-f_j(X_N)| \longrightarrow\mathbb E|f(G)-f_j(G)|.\] Take \(N\to\infty\) in (118) for fixed \(j\), use the smooth replacement result and the Gaussian formula, and then take \(j\to\infty\). This proves expected-pressure convergence. The bounded-pattern variance bound (110) holds for independent sign rows as well, which proves convergence in probability. ◻