A spectral gap throughout the high-temperature Sherrington–Kirkpatrick phase

Proof. We first bound the expected quadratic process that defines the smallest eigenvalue of S_A. Concentration will give a margin at each fixed z, and a finite mesh will extend it along the path.

We use two standard Gaussian facts. A real-valued L-Lipschitz function of a standard Gaussian vector satisfies \Prob\{|F-\E F|>u\}\le2\exp(-u^2/(2L^2)). Also, if two centered Gaussian processes have ordered squared increments, the process with larger squared increments has the larger expected supremum (Sudakov–Fernique). The latter comparison remains valid after the same deterministic function is added to both processes; see the deterministic-shift comparison in Vitale [22]. Both facts apply by finite nets and continuity to the compact index sets below. The linear map from the independent standard Gaussian coordinates to W, in Frobenius norm, has norm \sqrt{2j/n}.

For the comparison, write w=A^{1/2}p for \norm p=1 and consider

where g is standard Gaussian in \R^n. If w'=A^{1/2}p', direct calculation gives

Consequently, for each z\in[0,1],

Set T_g=\sqrt{j/n}\norm{A^{1/2}g}. Since g^Tw=p^TA^{1/2}g\le\norm{A^{1/2}g}, the last supremum is at most

The remaining supremum depends on the Gaussian vector only through T_g. The variable T_g^2 has mean j\chi_A and variance 2j^2\Tr(A^2)/n^2\le C/n. The inequality |\sqrt x-\sqrt y|\le\sqrt{|x-y|} therefore shows, uniformly in A, that \E|T_g-\sqrt{j\chi_A}|=o(1). On replacing T_g by \sqrt{j\chi_A} in the preceding display, the expression becomes 2d-d^2, with 0\le d=z\sqrt{j\chi_A}v\le\sqrt j\,a_+<1. Thus, with \delta=(1-\sqrt j\,a_+)^2>0,

uniformly in A,z. The supremum in (3.9) is a_+-Lipschitz in W for Frobenius norm. Gaussian concentration gives an exponentially small probability that it exceeds 1-\delta/2, uniformly for each fixed A,z.

To pass from a fixed z to the entire interval, we also need a bound on \norm W. Taking A=I in the comparison without deterministic offsets gives

The same holds for -W. Concentration of these two eigenvalues proves the stated norm tail. On \norm W\le R, the function of z inside the supremum in (3.9) is uniformly Lipschitz, with constant at most a_+R+2ja_+^2. A sufficiently fine constant mesh of [0,1], followed by a union bound, therefore gives (3.8), for example with c_0=\delta/4.□

Proof. The calculation has two parts. We first justify integration by parts for the truncated words and compute the inverse factor alone. We then obtain a recursion in the number of ordinary W factors; the same recursion determines the formal prediction exactly and the expectation up to O(n^{-1}).

All constants in this proof depend only on the fixed length and parameter bounds. Lemma 3.2 shows that, for every fixed deterministic parameter tuple, the truncations agree with the actual matrices for all z\in[0,1], except on an event of probability Ce^{-c_2n}. Gaussian integration by parts can be applied to the globally Lipschitz truncated factors, using weak derivatives. More explicitly, for a scalar Lipschitz function H,

When a=b, the elementary direction on the right is doubled, as required by the diagonal variance 2j/n.

For (3.15), remove the projection from the distinguished ordinary W factor. The resulting expectation error is exponentially small, since the other factors are bounded and Gaussian norm tails control the removed factor. On the event of agreement, the differential of the inverse is

We may use the expressions on the right, with truncated factors, on the whole probability space. The error is exponentially small even after the index sums: all factors and their true truncated differentials are bounded, there are at most a fixed polynomial number of indices, and the disagreement event has exponentially small probability. More precisely, the elementary direction E_{ab}+E_{ba} has Frobenius norm at most 2. Both the true weak differential and its formal replacement are therefore bounded by C in each entry; their difference vanishes on the agreement event. Each such differential error has expected absolute value at most Ce^{-c_2n}. Expanding the diagonal entry of a length-L word and applying the product rule uses at most (L-1)n^{L-1} scalar contributions before any index identifications; the factor j/n in integration by parts only improves this bound. Thus their total error is at most a fixed polynomial in n times e^{-c_2n}, hence at most C'e^{-c'n}. Subsequent words have bounded lengths depending only on the original length. We therefore suppress hats during the loop calculation, with this convention understood.

We now compute the base case, in which the only nondiagonal factor is K=K_A(z;W). Write a_i=A_{ii}, k_i(z)=\E K_{ii}, and r(z)=n^{-1}\sum_i a_i k_i(z). Differentiating K_{bi} in (3.15) gives

For the second sum, the absolute value of the sum over b is at most \norm{KAe_i}\norm{Ke_i}\le C. The factorization in the first term follows from (3.13); its covariance is at most C n^{-3/2}, which is more than enough. The inverse identity K=I+zAWK-z^2B_AAK now yields

The loop equations must select the solution near k_i=1. We establish this by continuation from z=0, where the inverse is the identity. Put h(z)=r(z)-\chi_A. The truncated k_i are uniformly bounded and continuous in z. If |h(z)|\le\delta_1, the second equation in (3.17) gives \max_i|k_i(z)-1|\le C\delta_1+C/n. Choose \delta_1>0 small, using (3.1), so that throughout this region and for large n,

for some fixed \delta_2>0. Averaging the second equation in (3.17) with weights a_i gives

and hence |h(z)|\le C/n while |h(z)|\le\delta_1. Since h(0)=0, continuity rules out a first exit from that region for large n. It follows uniformly in z that

To complete the base case, we verify the formal identity \mathsf P(K_A(z;W))=I independently of the expectation calculation. The cancellation takes place at each coefficient of the series. In the expansion (3.6), a block -z^2B_AA may be regarded as the negative contraction of two adjacent blocks zAW,zAW, since their contraction is z^2j\chi_AA=z^2B_AA. Fix a completely noncrossing-paired chain (AW)^{2m} with m>0. Let \mathcal A be the nonempty set of its pairs which join two originally adjacent occurrences of W. It is nonempty because every nonempty noncrossing pairing of a linear order has an adjacent pair. All pairs in \mathcal A are disjoint. The terms of (3.6) which give this paired chain are in bijection with subsets T\subseteq\mathcal A: contract the pairs in T into the negative deterministic blocks, and pair the remaining occurrences normally. The magnitude of the final diagonal contribution is unchanged, and its sign is (-1)^{|T|}. Thus the total multiplier is \sum_{T\subseteq\mathcal A}(-1)^{|T|}=0. This argument is finite at each power of z and proves the asserted formal identity. Multiplication by surrounding diagonal matrices proves both claims of the lemma for words with no ordinary W letters.

For the induction step, select an ordinary occurrence of W in a general word and apply integration by parts at that occurrence. We describe the resulting finite recursion directly, without requiring convergence of the formal series. In each product-rule term, mark the partner of the selected occurrence: for a derivative falling on an ordinary factor, this is the W being differentiated. If the derivative falls on K_A, replace that factor by zK_AA\,\boxed W\,K_A and take the boxed occurrence as the partner. In either case, list the two marked positions in their linear order, so the marked product reads

The two orientations in (3.15) contribute respectively

There is an additional factor z when the derivative fell on K_A. The second expression in (3.19) is O(n^{-1}) by operator-norm bounds, including when it contains two copies of K_A. The first expression is

again by (3.13). The index sums explain the two terms. The opposite orientation identifies the two interior indices and gives \Tr V; the equal orientation reverses the interior path and gives V^T.

Consequently \E\widehat Q_{ii} is the finite sum of the leading products (3.20) over all ordinary partners, plus z times the corresponding leading product for the one K_A factor if present, with an O(n^{-1}) remainder. Each word V and UQ in this recursion has fewer ordinary W letters than the original word. Each contains at most one K_A: if the derivative fell on K_A, the splitting places one of its two copies in each word. Extra diagonal factors A cause no problem, since at most one is added per decrease of the induction index.

The decrease in the number of ordinary letters holds whether the selected occurrence lies before or after the inverse. The two possible orders can be checked explicitly. If the original word is P K_A R W T, the contraction with K_A contributes

If it is P W R K_A T, that contribution is

Here P,R,T contain no inverse factor. The displayed outer and inner words each have one inverse factor, even though the original distinguished occurrence may lie anywhere among the ordinary letters.

It remains to compare this expectation recursion with the formal prediction. Sort the noncrossing pairings by the partner of the selected ordinary occurrence. The inside and outside may then be paired independently; their predictions are respectively \mathsf p(V) and \mathsf P(UQ). If the partner occurs inside the series for K_A(z;W), marking that occurrence in the formal series is exactly the formal differentiation identity \dd K_A[E]=zK_AAEK_A. Thus the entire sum over positions in that series is the single z-weighted product just described.

We can now induct on the number r of ordinary W letters. An ordinary partner leaves a total of r-2 such letters in the two smaller words; a partner in K_A leaves r-1. The base prediction is constant. The recursion therefore proves that the formal prediction is a polynomial of degree at most r. It also proves (3.14): use the induction hypothesis for the two expectations in each leading product, and average the diagonal induction hypothesis to handle its normalized trace. There are only finitely many products and all factors are uniformly bounded. The same induction, with no inverse factor, proves the statement for ordinary words.□

Proof of Lemma 3.1. We first make the centered estimates uniform over the diagonal parameters, then add the expectation estimate from Lemma 3.3. Fix a letter pattern and let \theta range over its parameter box, which has at most C_Ln real coordinates. Work initially with the truncated words at z=1. Conjugation by any diagonal sign matrix preserves the law of W and conjugates \widehat Q_\theta by that same sign matrix. The diagonal parameters commute with the conjugation, and both the projection and the functional calculus respect it. Thus \E(\widehat Q_\theta)_{ab}=0 whenever a\ne b.

Define the two matrix seminorms

Both are 1-Lipschitz with respect to Frobenius norm. The entry moment bounds following (3.13) give

Indeed, sum the n second moments for k=2, and the at most n^2 fourth moments for k=4, then apply Jensen’s inequality. For two parameter tuples at distance at most \Delta, the same argument, using (3.12), gives

The random norms in this last display concentrate around their means at Gaussian scale C\sqrt{\Delta/n}, again by (3.12). The centered expectations are deterministic and do not change Lipschitz constants.

To control every parameter tuple, approximate it successively on finer meshes. For each \ell\ge0, choose a sup-norm mesh \mathcal N_\ell of spacing 2^{-\ell} in the parameter box. It may be chosen with

For a tuple \theta, choose nearest mesh points \theta_\ell\in\mathcal N_\ell. Then \norm{\theta_\ell-\theta_{\ell-1}}_\infty\le C2^{-\ell}. For all pairs of mesh points at this distance, concentration and a union bound show that their centered increments have both N_2 and N_4 norm at most

except with probability Ce^{-c(\ell+1)n}, provided C' is chosen sufficiently large. Specifically, the squared ratio of this threshold to the Gaussian scale is a constant multiple of (\ell+1)n, and its constant can be chosen to dominate the logarithm of the number of mesh pairs. The initial mesh is bounded in the same way using (3.21). Summing the failure probabilities over \ell, and summing the convergent increment bounds along each chain, proves, with probability 1-Ce^{-c_1n},

At each fixed n, continuity of the truncated word and its expectation in \theta justifies taking the limit of the mesh chain. There are only finitely many letter patterns of length at most L, so their events may be intersected.

The centered words are now controlled on one event for every parameter tuple. Lemma 3.3 supplies the remaining bias estimate, N_2(\E\widehat Q_\theta-\mathsf P(Q_\theta))\le C/\sqrt n, uniformly in \theta. Sign symmetry makes the off-diagonal of \E\widehat Q_\theta identically zero. Combining these facts with (3.22) proves the desired diagonal and off-diagonal assertions for the truncated words. Their operator norms are already bounded by construction. On \norm W\le R and at parameters with S_A(1;W)\succeq cI, the truncations equal the exact words, proving the GOE version.

The GOE estimate transfers to the zero-diagonal model through J=W-D_{\diag W}. This last step uses a Frobenius bound on the removed diagonal. Since

it is bounded by a fixed constant except with exponentially small probability. The norm tail for W, and a sufficiently large choice of R, ensure also that both W and J lie in the operator-norm ball except with exponentially small probability. For every parameter tuple, (3.11) then bounds the Frobenius norm of the difference of the truncated word evaluated at W and at J by a fixed constant. Both N_2 and N_4 of this difference are bounded by its Frobenius norm. The prediction is unchanged by this substitution. At parameters with S_A(1;J)\succeq cI, the truncated factors evaluated at J equal the exact factors. This proves all of (3.7). The polynomial assertion was proved in Lemma 3.3.□

Proof. For each of the specified bilinear forms, we calculate its expectation and then apply concentration. Use the truncations above with a cutoff agreeing with the inverse on the path-stability event. Sign conjugation makes the off-diagonal expectations of each relevant matrix vanish. Equation (3.18) at z=1 gives

The identities, valid on the event of agreement,

therefore give the three diagonal expectation predictions A, B_AA, and B_AI+B_A^2A, with entrywise errors O(n^{-1}). Their use with truncated factors changes expectations only exponentially little, by the same argument as in Lemma 3.3. In particular, their errors in any of the specified bilinear forms are O(n^{-1}): for a diagonal error matrix E, one has |v_r^TEv_s|\le\norm E\norm{v_r}\norm{v_s}.

The expectation calculation has identified the three required predictions. To control fluctuations, each truncated bilinear form is C M_1^2-Lipschitz in W for Frobenius norm, by (3.11). Gaussian concentration gives failure probability Ce^{-c_3n} at any fixed positive tolerance. Intersect these events for the 3m^2 forms and with Lemma 3.2. For large n the expectation errors are less than half the specified tolerance, proving the result.□

Proof. We first compute the mean log determinant with E=0 and a fixed deterministic V. Regularization will then provide a global Lipschitz bound, allowing a finite mesh in V and finally the finite-rank perturbation. Set

Since V\le I, we use the proof of Lemma 3.2 with the diagonal upper bound equal to 1; its stability and inverse-norm constants here depend only on j. On the path-stability event from Lemma 3.2 with A=V, the determinant of L_z equals the positive determinant of S_V(z;W) by \det(I-UV)=\det(I-VU), and both L_z^{-1} and K(z) have uniformly bounded operator norm. Indeed L_z=K(z)^{-T}, and (3.3) applies. Differentiating the normalized log determinant on this event gives

Here VL_z^{-1}=K(z)V, which follows by multiplying (I-zVW+z^2BV)V=V L_z, and cyclicity of trace was used.

The derivative is expressed in the two matrices whose diagonal expectations were computed in (3.18). Let \mathcal E_V be the path-stability event intersected with \norm W\le R, for a fixed R>2\sqrt j. This one event applies to the entire interval of z. Equations (3.18), and replacement by the truncated factors off \mathcal E_V, show uniformly in z,V that the expectation of the right-hand side of (3.27), multiplied by \one_{\mathcal E_V}, is

All differentiated quantities are bounded on \mathcal E_V, so integration and expectation may be interchanged. Since L_0=I, it follows that

On \mathcal E_V, the singular values of L_1 are bounded below by a positive constant. Therefore

Off \mathcal E_V, the absolute value of the regularized normalized log determinant is bounded by

for 0<\gamma\le1. The norm tails of W and (3.8) make its expected contribution exponentially small, for each fixed \gamma. Consequently, uniformly in deterministic V,

The mean bound is uniform in deterministic V. To obtain a single event for all V, we use the concentration supplied by regularization. Define on all real matrices

Its Frobenius gradient is 2n^{-1}M(M^TM+\gamma I)^{-1}. Each singular value of this gradient is at most 1/(n\sqrt\gamma), and hence

For fixed V, the map W\mapsto L_1 has Frobenius Lipschitz constant at most one. Gaussian concentration therefore gives

Choose \gamma_0\le1 small enough that C\gamma_0 in (3.29) is at most \varepsilon/8. For fixed \gamma\le\gamma_0 and sufficiently large n, the expectation bound and (3.31) bound all points of any fixed sup-norm mesh of the cube [0,1]^n by jb^2+\varepsilon/2, with failure \exp(-c_\gamma n^2). Indeed a mesh of spacing \Delta>0 has at most (C/\Delta)^n points, whose logarithm is only of order n. On \norm W\le R, if \norm{V-V'}\le\Delta, then

By (3.30), the corresponding F_\gamma values differ by at most (2j+R)\Delta/\sqrt\gamma, whereas |jb^2-j(b')^2|\le2j\Delta. Choose \Delta, depending only on j,R,\gamma,\varepsilon, so that these errors sum to at most \varepsilon/4. Including the exponentially likely norm event proves, simultaneously for all V,

We have obtained the simultaneous bound for E=0 with room for the perturbation. The same Lipschitz estimate (3.30) gives, for every allowed E,

This is at most \varepsilon/4 for large n, uniformly even when E is chosen from the observed matrix. It proves (3.26) and the lemma.□

Proof. The second-moment assumption makes the right-hand side finite, so it suffices to consider the case D:=D(P\Vert N(d,s))<\infty. In this case P is absolutely continuous, and hence q>0: otherwise P would be the point mass at zero. Also q<1, because \tanh^2 y<1 for every finite y. We use the following dimensionless parameters:

In these parameters, multiplying the desired inequality by j gives the target

We will combine two entropy bounds. A monotone change of variables handles one range of the discrepancy C-B; comparison with a Gaussian of a different variance handles the remaining range. We retain strictness in both arguments to identify all equality cases.

A change-of-variables bound. For the change of variables, fix u<1. The map

is an increasing C^1 diffeomorphism of \R onto itself, with derivative T_u'(y)=1-u(1-m(y)^2)>0. Change of variables in relative entropy yields

The entropy identity is well defined: for fixed u<1, \log T_u' is bounded, and the second-moment assumption makes the difference of the Gaussian log densities integrable. To estimate the logarithmic term, apply concavity on the segment joining 1 and 1-u. For 0\le v\le1, this gives

Consequently, with \ell(u)=-\log(1-u)-u,

Completing the square isolates the amount by which this bound can exceed the target in (4.3):

We first locate a range in which the extra term h(u) is strictly positive:

For the negative range, integrate v/(1+v)\le v over 0\le v\le -u to obtain \ell(u)\le u^2/2. Since B<1, this proves strict positivity for u<0. For the positive range, use \ell(u)=\sum_{r\ge2}u^r/r: it shows that \ell(u)/u^2 is increasing on 0<u<1. Moreover, using B=1-a_0,

It follows that h(u)/u^2\ge h(a_0)/a_0^2>0 for 0<u\le a_0, proving (4.6).

We now choose the transport parameter. If u_*=(C-B)/(1+k)\le a_0 and C\ne B, set u=u_* in (4.5). The square vanishes and h(u_*)>0, so (4.3) is strict. If u_*>a_0, put

In this case C>C_0. The endpoint choice u=a_0 in (4.5) still gives

This proves strictness throughout C_0<C<C_1 and leaves only the range beginning at C_1.

The remaining range. For C\ge C_1, we use the second moment to obtain the remaining entropy bound. Put v=\E_P[(y-d)^2]/s>0. Comparison with the Gaussian N(d,sv) gives the exact identity

and hence 2D\ge v-1-\log v. Cauchy–Schwarz gives

where the last inequality follows from C\ge C_1>C_0>1 and jk<1. The function v\mapsto v-1-\log v is increasing for v\ge1, so

The second inequality removes j from the lower bound. Indeed, with C,k held fixed, the derivative of C^2/k-\lambda-\lambda\log(C^2/(\lambda k)) is -\log(C^2/(\lambda k))\le0 for j\le\lambda\le1. Thus the remaining comparison with the target reduces to

We estimate f at C_0 and control its increase from C_0 to C_1. Direct simplification at C_0 gives

Indeed 1/k-1+\log k\ge0 for 0<k\le1, and

The last bound follows by integrating (1+x)^{-1}\le1-x+x^2 from zero to a_0. Differentiation also gives

For completeness, the first inequality follows from

The derivative bounds compensate for the possible negative value at C_0. Write \Delta=C_1-C_0\ge a_0^{3/2}/\sqrt3 and integrate them to obtain

The strict inequality uses 0<a_0<1 and 1/\sqrt3+1/6>2/3. Since f' is positive at C_0 and increasing thereafter, f(C)>0 for every C\ge C_1.

The two ranges together prove strictness in (4.3) whenever C\ne B. To finish the equality case, suppose C=B. The target then has right-hand side zero, so equality is equivalent to D=0, or P=N(d,s). Conversely, Gaussian integration by parts gives a=s\E_P\sech^2 y=sb for this law, which verifies equality.□

Proof. We first show that a consistent Gaussian must have equal mean and variance. We then solve the resulting one-variable fixed-point equation and prove the bounds in the statement.

Two Gaussian symmetry identities provide the first step. For s>0, the density \rho_s of N(s,s) satisfies

Pairing the integrals over y and -y therefore gives

Second, if d\ge0 and s>0, pairing the positive and negative half-lines shows that

Here the integrand is odd and nonnegative for y\ge0, while the density at y is at least its value at -y when d\ge0.

Now suppose a Gaussian law N(d,s) with s>0 satisfies (4.7) with \sigma=0. We begin by locating its mean. For fixed s, the map

is a contraction with Lipschitz constant at most j, because its derivative is j\E_{N(d,s)}\sech^2 y\in[0,j]. Iteration from zero stays nonnegative, so its unique fixed point d is nonnegative.

To compare this nonnegative mean with the variance, keep s fixed and define

For d\ge0, Gaussian differentiation and (4.11) give

Indeed the integrand is bounded below by -1, and its expectation is at most \E(1-m^2)\le1 after subtracting 2\E[m(1-m^2)]\ge0. Thus H_s is 1-Lipschitz on [0,\infty). By (4.10), H_s(s)=0. On the other hand, consistency says d-s=j(M-q)=jH_s(d). Consequently

forcing H_s(d)=0 and d=s. The consistent Gaussian law must therefore satisfy (4.8).

It remains to solve (4.8). Set

It is continuous at zero by bounded convergence. For s>0, differentiating both the Gaussian mean and variance gives

The lower bound is pointwise; the upper bound follows from (4.11). Hence F is nonnegative and 1-Lipschitz on [0,\infty), with F(s)\le s. Consequently, s\mapsto t+jF(s) is a contraction of [0,\infty) into itself and has exactly one fixed point s_*(t). The identity (4.10) then verifies both consistency equations for N(s_*(t),s_*(t)). If a consistent law has s=0, then 0=t+jq forces t=q=0, and hence the law is the point mass at zero and d=0; this agrees with the solution just constructed.

Existence and uniqueness are now established, including the degenerate case. To obtain the dependence on time, compare the fixed-point equations at t,t':

which proves the asserted Lipschitz bound. The size estimates follow from 0\le F(s)\le s, which gives t\le s_*(t)\le t/(1-j), and (4.10) gives q_*(t)=F(s_*(t))=(s_*(t)-t)/j\le t/(1-j). For t>0, the variance s_*(t) is positive, so its Gaussian law is nondegenerate and q_*(t)>0. The equality assertion follows from Lemma 4.1.□

Proof. We express failure of stability through a witness set that is fixed before observing the path. Let \mathcal W_J consist of pairs (y,A) satisfying 0\le A\le(1+\eta)I and \norm{A-V_y}_{\Frob}\le\epsilon_A\sqrt n for which \mathsf H_J(y,A)\not\succ cI. Define the least normalized residual among these witnesses by

If \mathcal W_J is nonempty, d_J is finite and n^{-1/2}-Lipschitz, because F_{J,h}(y)=F_{J,0}(y)-h. Let

with value +\infty when the set is empty. Continuity makes the first hitting time an observation stopping time. Before the hit, every violating witness has residual greater than \rho_0\sqrt n. At a hit, the definition of the infimum supplies a violating witness with residual strictly below 2\rho_0\sqrt n; attainment of the infimum is unnecessary. Proposition 5.1 therefore bounds the hitting probability.□

Proof. We first evaluate the integral without its indicator. For fixed m, the GOE covariance is

Convolution with the heat kernel of variance \sigma^2 and with \sqrt t\,g therefore produces the Gaussian vector Wm+\sqrt{t+\sigma^2}\,g', of covariance sI+jmm^{\mathsf T}/n. Its determinant is s^nS/s, and its inverse is

Moreover, F_{\wt J,h_0}(y)=z-Wm-\sqrt t\,g. The Gaussian density at z, multiplied by e^{njb^2/2}, differs from \prod_i\varphi_{d,s}(y_i) by the prefactor \sqrt{s/S} and the exponential whose logarithm divided by n is

This calculation gives the identity for the unrestricted integral. To retain the indicator, condition the joint Gaussian pair formed by W and the convolved observation on the value z of the latter. The resulting conditional probability is exactly the factor in (5.8). All integrands are nonnegative, so Tonelli’s theorem justifies the interchange of integrations.□

Proof. We first obtain a Gaussian tail envelope, then prove a compact-set bound by a finite cover. Strict negativity away from the equality laws will give the second and third assertions; an exponential tilt will give the fourth.

Throughout, |d|\le T+j and s\le T+1+j. Also a(P)^2\le q\int(y-d)^2\,\dd P and jq/S\le1/2. The elementary inequality (r-u)^2\le\frac32r^2+3u^2 consequently gives

On s\ge s_{\min}, it follows that

where c>0 and C<\infty depend only on j,T,s_{\min}. On \sum_i y_i^2>nR^2, use half of the remaining Gaussian square for integration and the other half for the required exponential decay. This proves the first assertion of (i).

The estimates on bounded second moments follow from one compactness argument, which we now give in full. Put

Tightness and preservation of the second-moment bound under weak limits make this set weakly compact. Integration of a continuous function of at most linear growth is continuous on it: the second-moment bound makes the contribution of |y|>K uniformly O(K^{-1}), while bounded truncations are weakly continuous. In particular, d,s,a(P),\psi are continuous on \mathcal P_R\times[0,T]\times[0,1] where s\ge s_{\min}.

For every compact subset \mathcal C of that region,

where P_y=n^{-1}\sum_i\delta_{y_i}; empty integrals are zero. To establish this bound, fix a strict finite upper bound r for its right-hand side. At each \xi=(P,t,\sigma)\in\mathcal C, choose a bounded continuous test function \varphi_\xi from the variational formula for relative entropy so that

Such a test also exists when the entropy is infinite. Retain positive slack in the inequality and then choose a neighborhood of \xi on which continuity controls \psi, the empirical average of \varphi_\xi, and d,s. Moreover, on \sum_i y_i^2\le nR^2, the logarithmic density ratio between N(d',s')^{\otimes n} and the fixed law N(d,s)^{\otimes n} is at most n times a quantity tending uniformly to zero as (d',s')\to(d,s); this follows directly by expanding their Gaussian log densities. For y in that neighborhood, insert e^{\sum_i\varphi_\xi(y_i)} and bound its expectation under the fixed product Gaussian. The integral on the neighborhood is then at most e^{nr}, after using less than the reserved slack for the three continuity errors. Cover \mathcal C by finitely many such neighborhoods. An arbitrarily small enlargement of r absorbs their number and proves (5.11). Combining this bound with (5.9) on the ball and (5.10) outside it proves the second assertion of (i).

To obtain a strictly negative rate, observe that the same variational formula expresses relative entropy as a supremum of continuous functions. Thus it is jointly lower semicontinuous in P,d,s when s is bounded away from zero, and \psi-D is upper semicontinuous. At \sigma=0, its zero set is precisely the equality curve identified above. On a compact set disjoint from any open neighborhood of that curve its supremum is strictly negative. The same remains true for all sufficiently small \sigma: otherwise a convergent sequence with \sigma\downarrow0 and rates approaching zero would give an equality point outside that neighborhood. Apply (5.11), with s_{\min}=t_0, and then choose the second-moment tail rate in (i) strictly negative. This proves (ii).

Near time zero, the restriction on q supplies the positive variance needed for the same argument. For (iii), choose t_0 so small that t_0/(1-j)<q_0/2. The set q\ge q_0 contains no equality point at \sigma=0 and 0\le t\le t_0. On this set s\ge jq_0>0, so the preceding compactness argument and the tail envelope apply without change, including at t=0.

It remains to control the contribution of large coordinates in (iv). Fix t_0,\nu>0, and take 0<\lambda<1/[8(T+1+j)]. Let h_{R_1}(y)=y^2\one_{\{|y|>R_1\}}. On a fixed second-moment ball, repeat the proof of (5.11), this time inserting e^{\lambda\sum_i h_{R_1}(y_i)}. Begin with any small positive rate bound \varepsilon in that proof and its finite collection of tests \varphi_\xi. Uniformly for |d|\le T+j and t_0\le s\le T+1+j,

by Gaussian tails; the choice of \lambda leaves a uniformly negative quadratic exponent. The same holds with the integrand multiplied by any of the finitely many bounded e^{\varphi_\xi}. The additional logarithmic moment can therefore be made arbitrarily small in every member of the finite cover. For sufficiently large R_1, the tilted integral on the ball is at most e^{2\varepsilon n}. On \av{h_{R_1}(y)}>\nu, remove the tilt at cost e^{-\lambda\nu n}. Choosing 2\varepsilon<\lambda\nu/2 gives a strictly negative rate there. The integral off the ball is estimated without the tilt, using (i) with a larger negative rate. This proves (iv).□

Proof. We first compute the Gaussian regression and the finite-rank correction at an equality law. We then show that the empirical-law and diagonal hypotheses preserve its positive margin, uniformly over all admissible diagonals.

Gaussian regression and whitening. In a sufficiently small neighborhood of the equality curve, s and q have positive lower bounds depending on j,t_0,T. Indeed s_*(t)\ge t_0, and q(P_t)=\E_{P_t}\tanh^2 y is positive and continuous there. Set

The norms of \one/\sqrt n,u,Z_1 are bounded uniformly by the assumed second-moment bound. Write

The three blocks in this decomposition are independent before conditioning, with \Cov(w)=(j/n)P_u and \Var(\alpha)=2j/n. The projection of the observation in (5.8) perpendicular to u observes \sqrt{nq}\,w with independent noise of covariance (t+\sigma^2)P_u; projecting along u observes \sqrt{nq}\,\alpha with noise variance t+\sigma^2. The elementary Gaussian conditional-mean and conditional-variance formulas therefore give the following representation, using a fresh centered GOE W_0:

For example, the perpendicular conditional mean is j\sqrt{q/s}\,P_uZ_1, which equals the displayed \bar w because u^{\mathsf T}Z_1=a_y/\sqrt{qs}. The parallel conditional mean follows from u^{\mathsf T}z=\sqrt{n/q}(a_y+Bq). The parallel residual variance is (2j/n)(t+\sigma^2)/S. These conditional means and residual variances verify all three blocks of (5.12).

For the moment, fix a deterministic admissible A. We compare the conditional Hessian with the positive matrix built from the fresh GOE by setting

Lemma 3.2 gives S_A\succeq c_0I except with exponentially small probability, where c_0>0 is uniform in A. Let G_A=A^{1/2}S_A^{-1}A^{1/2}. Lemma 3.4, applied to the three deterministic vectors \one/\sqrt n,u,Z_1, says that their Gram entries after \mathcal Y are as close as desired to their A-weighted Gram entries, and that the extra vector \mathcal Y(W_0u) has Gram entries as close as desired to those of

where N is a unit vector orthogonal to the three deterministic images in this comparison. Specifically, its mixed entries are B_Ap^{\mathsf T}Au and its squared norm is B_A+B_A^2u^{\mathsf T}Au. Here N describes only Gram entries; this notation does not assert a coupling with an additional random or deterministic vector. The comparison concerns finite Gram matrices; the associated symmetric finite-rank eigenvalues depend continuously on their entries, as we verify when returning to the empirical parameters. Each fixed Gram tolerance has failure probability at most Ce^{-a n} uniformly over the bounded deterministic vectors and over A. Also u^{\mathsf T}W_0u is smaller than any fixed tolerance with exponentially high probability, since its variance is 2j/n.

The correction at an equality law. We now compute the finite-rank perturbation in the ideal Gram comparison. Set \sigma=0, take s=s_*(t), and replace each weighted average n^{-1}\sum_i A_{ii}f(y_i) by \E_{N(s,s)}[\sech^2 y\,f(y)]. Evaluate all other empirical averages under the same law. Thus a_y=sb, B_A=B=jb, and \kappa^2=1-jq/s. Write

using the Gram interpretation just described. Subtracting W_0 from (5.12) gives exactly

At equality, \bar w=j\sqrt{q/s}\,Z_1-Bu and \alpha=2B after discarding the vanishing Gaussian scalar. Whiten by S_A, and use (5.13) with B_A=B. The perturbation to subtract from the identity consists of the planted term jX_1X_1^{\mathsf T}, the term 2jqUU^{\mathsf T}, and the whitened version of (5.14). The terms -2BUU^{\mathsf T} and +2BUU^{\mathsf T} cancel. Eliminating the orthogonal direction N by Schur complement adds (1-\kappa)^2BUU^{\mathsf T} to the perturbation. Its total UU^{\mathsf T} coefficient becomes

Consequently the Schur complement is the identity minus

To prove positivity of this Schur complement, it suffices to examine its three-vector Gram matrix. Put e=\E_{N(s,s)}[m^2(1-m^2)]. The Gram matrix of (X_1,M_1,G_1) equals

We derive the entries of this matrix explicitly. The density of N(s,s) satisfies p(y)/p(-y)=e^{2y}. Pairing y and -y shows that every integrable odd function f satisfies \E f(y)=\E[f(y)\tanh y]. This gives \E[m(1-m^2)]=e. Gaussian integration by parts gives

Applying it twice, and using (1-m^2)''=4m^2(1-m^2)-2(1-m^2)^2, gives the final diagonal entry b/s-2b+6e. The remaining entries are immediate, proving (5.16).

The coefficient matrix in (5.15) is

Set d_0=1-jb and g_1=je. The first and third columns of I-HC are (d_0,-g_1,2g_1)^{\mathsf T} and (-g_1,-g_1,d_0+3g_1)^{\mathsf T}. Adding B/s times the third column to the second makes the second column (0,d_0+g_1,0)^{\mathsf T}. Hence

The same determinant is obtained from the identity minus (5.15) on any ambient space containing its three vectors. To determine its sign as a quadratic form, first subtract only jX_1X_1^{\mathsf T}; the result is positive definite because j\norm{X_1}^2=jb<1. The remaining two-vector perturbation has at most one positive eigenvalue, since its coefficient block j\left(\begin{smallmatrix}2-B/s&1\\1&0\end{smallmatrix}\right) has one positive and one negative eigenvalue. The resulting matrix can consequently have at most one nonpositive direction. The positive determinant in (5.17) excludes both a zero eigenvalue and a single negative eigenvalue. Hence the matrix is positive definite. The Schur complement then also gives positivity before N was eliminated.

Perturbing the scalar law and the diagonal. The equality calculation supplies the positive margin we need. We first explain why the hypotheses put the actual weighted Gram entries near these equality values.

Each deterministic vector has coordinates n^{-1/2} times one of 1,m/\sqrt q,(y-d)/\sqrt s. Products of two such coordinate functions are bounded by C(1+y^2). For any such product f, the tail and Frobenius bounds imply

On |y_i|\le R_1, use Cauchy–Schwarz and boundedness of f; outside, use |A_{ii}-(V_y)_{ii}|\le2+\eta and the tail hypothesis, with R_1\ge1. The corresponding V_y-weighted functions are bounded and continuous, even when multiplied by y^2, because \sech^2 y decays exponentially. Their averages are therefore close to the equality-law averages in a sufficiently small weak neighborhood. The parameters d,s,q,a_y are continuous on the second-moment ball, so their denominators and coefficients are close as well. Finally |B_A-B|\le j\norm{A-V_y}_{\Frob}/\sqrt n.

We retain the positivity uniformly for t\in[t_0,T] and under small Gram errors. If Q has finitely many columns, the nonzero eigenvalues of QCQ^{\mathsf T} agree with those of (Q^{\mathsf T}Q)^{1/2}C(Q^{\mathsf T}Q)^{1/2}, including multiplicity. The largest eigenvalue, with zero adjoined, is therefore a continuous function of the Gram matrix and the coefficients. All equality parameters vary continuously on the compact interval [t_0,T], and s,q stay positive there. The exact calculation thus gives a uniform positive whitened margin. Choose a fixed tolerance \xi>0 on coefficients and Gram entries small enough to preserve, say, half of it. Lemmas 3.2 and 3.4, the Gaussian scalar estimate, and (5.18) show that this tolerance is achieved for each fixed admissible A with failure probability at most Ce^{-a n}, after choosing the scalar neighborhood, tail tolerance, \sigma_*, and Frobenius distance sufficiently small. Returning from whitening multiplies the margin by at least c_0. The omitted mass (B-B_A)A has norm at most j(1+\eta)\norm{A-V_y}_{\Frob}/\sqrt n and can be absorbed by decreasing that distance again.

At this stage the margin is proved for each fixed A. To make it simultaneous, we use that its failure exponent is uniform over the bounded deterministic vector norms. Those bounds were fixed before choosing the Frobenius distance. Also (5.12) implies \norm W\le K_1 with failure Ce^{-a_1n} for some uniform K_1: the norms of the deterministic means are bounded, and the remaining terms are controlled by \norm{W_0}. On this event, A\mapsto\mathsf H_{\wt J}(y,A) is uniformly 1/2-Hölder in diagonal sup norm. Indeed \norm{A^{1/2}-\widehat A^{1/2}}\le \norm{A-\widehat A}^{1/2}, while the middle factor in (5.2) has bounded norm.

Choose a fixed mesh size \zeta>0 small enough that the Hölder error is below one quarter of the margin. Given A with normalized Frobenius distance at most \epsilon from V_y, let \mathcal S_A=\{i:|A_{ii}-(V_y)_{ii}|>\zeta\}. It has at most \alpha n elements, where \alpha=\epsilon^2/\zeta^2. Outside \mathcal S_A replace A_{ii} by (V_y)_{ii}; on \mathcal S_A, round it within \zeta to a grid of [0,1+\eta], including the endpoints. The resulting \widehat A obeys

For fixed y, these approximants are deterministic and number at most

Its logarithm divided by n tends to zero as \alpha\downarrow0; for instance it is at most -\alpha\log\alpha-(1-\alpha)\log(1-\alpha) +\alpha\log N_\zeta+o(1) when \alpha<1/2. Choose \epsilon_* so small that 2\epsilon_* meets all moment tolerances and this counting rate is less than half the fixed-A failure exponent. A union bound, followed by the Hölder comparison, proves the desired uniform margin.

For later use, record the order that preserves the concentration exponents. First fix R,t_0,T,\eta, the limiting margin, and the Gram tolerance \xi. This fixes the vector norm bounds, conditional concentration exponents, K_1, and the mesh \zeta. Next take the equality neighborhood and tail tolerance \nu small. Choose any sufficiently large finite R_1, then take \epsilon_* small enough for (5.18) and the union bound. Finally decrease \sigma_* so all scalar coefficients lie within the required tolerance. No concentration exponent is lost when \epsilon_* is decreased. This also proves the last assertion of the lemma.□

Proof. We integrate over a Gaussian ellipsoid centered at the approximate zero. Its covariance is chosen to control the linearized residual. Let g'' be standard Gaussian and put \Delta=\sigma Q^{-1/2}g'', and restrict to \mathcal B=\{\norm{g''}\le2\sqrt n\}. For 2\sigma/\sqrt\gamma\le r, all resulting points y+\Delta belong to E. Changing variables in the integral gives the lower bound

Here and for the remainder of this proof F=F_{\wt J,h_0}. The conditioning has probability tending to one, and all estimates below remain valid under it with a uniform factor, since \Prob(\mathcal B)\ge1/2 for large n.

We must control the nonlinear residual on this ellipsoid uniformly in its center. Coordinatewise Taylor expansion gives

Writing b(y+\Delta)=b+\ell_b+r_b, its linear part is \ell_b=-2\av{m(1-m^2)\Delta}, and |r_b|\le C\norm\Delta^2/n. Thus, when the product jb,m in F is expanded, its remainder is bounded in norm by

For the product of first differences use |b(y+\Delta)-b(y)|\le C\norm\Delta/\sqrt n and \norm{m(y+\Delta)-m(y)}\le\norm\Delta. The remaining term -\wt Jm has remainder bounded by K\norm{r_m}. Since the covariance of \Delta is bounded above by \sigma^2I/\gamma, Gaussian fourth moments give

Combining these Taylor estimates with the Gaussian moments, and writing R_F=F(y+\Delta)-F(y)-L\Delta, yields

All constants depend only on j,K.

The choice of ellipsoid now controls the linear term: the singular values of LQ^{-1/2} are at most one. Hence \norm{L\Delta}\le\sigma\norm{g''} pointwise, including on \mathcal B. Expanding F(y+\Delta)=F(y)+L\Delta+R_F and applying Cauchy–Schwarz to its cross terms yields

The last mixed term may also be absorbed into the two squares. Apply Jensen’s inequality to the conditional expectation in (5.21). Its logarithm divided by n is at least

The determinant hypothesis cancels jb(y)^2/2, with a loss of at most \varepsilon_1/2. Choose the constants in the order in the statement; the other errors then have total size smaller than \varepsilon. This proves the lower bound.□

Proof of Proposition 5.1. We assemble the scalar, matrix, and local-volume estimates under the planted law (5.4). After excluding unstable approximate zeros at a fixed time, we remove the auxiliary diagonal, control the whole path by a fixed mesh, and undo the planting.

Let J be \wt J with its diagonal zeroed. Choose \eta,\theta,q_0>0 sufficiently small that j(1+\eta)^2<1 and, whenever q\le2q_0,

for some c_{\mathrm{sm}}>0. These choices are possible because at \eta=\theta=q=0 the lower scalar bound is 1+j-2\sqrt j=(1-\sqrt j)^2>0. On

we have \norm{\wt J}\le2\sqrt j+2\theta. Since mm^{\mathsf T}/n\preceq qI, (5.23) implies \mathsf H_{\wt J}(y,A)\succeq c_{\mathrm{sm}}I for all q\le2q_0. Choose t_0\in(0,T] small enough for Lemma 5.4(iii).

For the remaining times t\in[t_0,T], the conditional Hessian lemma provides the margin near the equality curve. Choose a second-moment ball whose complement has a fixed negative rate, using Lemma 5.4(i) with s_{\min}=t_0. Apply Lemma 5.5 on this ball. Take its equality neighborhood and tail tolerance, choose its tail cutoff R_1 sufficiently large for Lemma 5.4(iv), and then choose its diagonal-distance radius. Decrease \sigma_*>0 so that Lemma 5.4(ii) excludes the complement of that equality neighborhood, and so that the small-time scalar bound also holds. The scalar regions excluded by these choices all have fixed strictly negative rates, uniformly for 0<\sigma\le\sigma_*. On the region left over, the conditional matrix margin fails with probability at most e^{-a_*n}.

Let \epsilon_*>0 denote the full diagonal-distance radius just chosen. Choose c_E>0 smaller than both c_*/2 and c_{\mathrm{sm}}/2. For t\ge t_0, define

For 0\le t<t_0, instead put E_t(W)=\{y:q(y)\ge q_0\}. There is a_0>0 such that

for all sufficiently large n, uniformly in t,\sigma. For small times, this follows from the scalar bound and Lemma 5.3. For larger times, split the conditional integral into the excluded scalar regions and their complement. The excluded regions have negative rates. On the complement multiply the conditional probability e^{-a_*n} by the unrestricted scalar integral, whose rate is at most any prescribed positive number, by Lemma 5.4(i). Taking that number below a_*/2 proves (5.25). Finitely many pieces can be absorbed by decreasing a_0.

We now use local volume to exclude every unstable approximate zero. An individual witness will force a ball of the excluded set to contribute too much to the integral. Suppose

for some 0\le A\le(1+\eta)I, and assume (5.24). There is a constant r>0, independent of this witness and of n,t, such that \norm{y'-y}\le r\sqrt n implies y'\in E_t(W). For t\ge t_0, the same A witnesses instability throughout a sufficiently small ball. Indeed,

and

preserve the full diagonal-distance constraint and the bound \lambda_{\min}\le c_E if r is small enough. For t<t_0, (5.23) and the chosen c_E imply q(y)>2q_0; the same Lipschitz bound for q then gives q(y')\ge q_0 for a smaller r.

Set L=F'_{\wt J,h_0}(y). It is a bounded-rank, bounded-norm perturbation of

Here n^{-1}\Tr V_y=b(y), so the determinant rate in Lemma 3.5 matches the weight in (5.5). That lemma says that, for any chosen \varepsilon_1>0, one may choose a fixed \gamma>0 sufficiently small so that, except on an event of probability Ce^{-a_1n}, simultaneously at every y,

The lemma’s rank-perturbation allowance applies here with two rank-one terms of norms at most j and 2j. Choose \varepsilon<a_0/2, choose \varepsilon_1 for Lemma 5.6, then choose \gamma by the log-determinant lemma. Next fix \sigma\in(0,\sigma_*] sufficiently small for the local-volume lemma, and finally fix \rho>0 sufficiently small for it. A witness (5.26) on the determinant event forces I_{t,\sigma}(E_t)\ge e^{-\varepsilon n}. Markov’s inequality and (5.25) show that the probability of such a witness, under (5.24), is at most Ce^{-a_2n}, uniformly in t\in[0,T]. Here the uniformity of the negative rates permits the later choice of \sigma required by the local-volume estimate.

The integral argument used a full GOE matrix. We now pass to its zero-diagonal version, retaining slack in both residual and Hessian bounds. Choose a fixed \delta_D>0 so small that

The diagonal entries of \wt J are independent N(j/n,2j/n), so \Prob(\max_i|\wt J_{ii}|>\delta_D)\le Ce^{-a_Dn} for large n. On its complement,

Thus a zero-diagonal witness with residual at most \rho\sqrt n/2, radius \epsilon_*\sqrt n/2, and Hessian minimum at most c_E/4 would give a witness (5.26). We have proved an exponentially small fixed-time failure probability for the zero-diagonal matrix, restricted only to \norm J\le2\sqrt j+\theta.

To pass from a fixed time to the whole interval, we control observation increments on a deterministic mesh. Take a mesh of size \Delta>0 that includes both endpoints. For a scalar Brownian motion, the reflection bound implies that \sup_{0\le u\le\Delta}|B(u)|^2/\Delta has an exponential moment at a fixed positive argument. The coordinate Brownian motions are independent. Exponential Markov applied to the sum of their squared suprema consequently gives, for each desired \kappa_0>0, a mesh \Delta small enough that

The deterministic drift contributes at most \Delta\sqrt n to each increment. Choose the mesh so the full normalized increment is at most \rho/4. Set \rho_0=\rho/8, \epsilon_A=\epsilon_*/2, and c=c_E/4. A pathwise violation of (5.3) then gives a fixed-time violation at a mesh point: the Hessian and its witness do not use the field, and the residual increases by at most \rho\sqrt n/4. A union bound over the fixed finite mesh proves an annealed path failure bound Ce^{-a_4n} under the planted law, restricted to \norm J\le2\sqrt j+\theta.

We have obtained a path estimate under planting. To recover the original disorder law, introduce the partition-function density

Size-bias the original joint law of J,X, with X\sim\mu_0, by L(J). Its density in J,X is proportional to e^{x^{\mathsf T}Jx/2} times the original Gaussian density. Gaussian tilting makes X uniform and, conditionally on X, makes the off-diagonal entries of J independent with means jX_iX_j/n and variance j/n. Add independent diagonal entries with law N(j/n,2j/n)—a centered GOE diagonal plus jI/n—and conjugate by D_X; the resulting matrix and observation have exactly the law (5.4). All witness conditions are equivariant under this conjugation, because m(D_Xy)=D_Xm(y), V_{D_Xy}=V_y, and diagonal A commutes with D_X. Conditional observation paths given J are unchanged by size bias.

Undoing this change of law requires a lower bound for L(J) on typical original matrices. For independent uniform signs \varepsilon_i, Gaussian averaging gives

To obtain the inequality, introduce an independent standard Gaussian g and use \E_\varepsilon e^{\sqrt{j/n}g\sum_i\varepsilon_i} =\cosh(\sqrt{j/n}g)^n\le e^{jg^2/2}. Paley–Zygmund then yields a constant p_j>0 with \Prob(Z\ge\E Z/2)\ge p_j. As a function of the independent standard off-diagonal Gaussians, \log Z has gradient squared at most (j/n)\binom n2\le jn/2. Gaussian concentration about a median therefore has scale O(\sqrt n). The preceding positive-probability lower bound implies that this median is at least \log\E Z-O_j(\sqrt n): otherwise its upper concentration tail would be smaller than p_j. It follows that, for every fixed \delta>0,

We can now turn the annealed planted estimate into the required conditional observation bound. Let p_n(J) denote the conditional observation-path failure probability for (5.3). The planted estimate is

Choose \delta<a_4/2. On L(J)\ge e^{-\delta n}, this gives an original expectation bound Ce^{-(a_4-\delta)n} for the restricted p_n(J). Markov’s inequality makes p_n(J)\le e^{-an} outside an event of probability tending to zero, for a fixed sufficiently small a>0. Equation (5.27) and the usual GOE norm bound, including diagonal removal, show that the restrictions on L(J) and \norm J hold with probability tending to one. Intersect these events to obtain \mathcal J_n and finish the proof.

We finish by checking measurability of the witness events used in the argument. At fixed y, the matrix-witness sets are projections over the compact diagonal range. For existence of y,A,t along a continuous path, additionally restrict \norm y, \norm J, and \sup_{t\le T}\norm{Y(t)} by integers. The witness inequalities are closed and the witness parameter space is compact, so these restricted projections are closed. Taking a countable union removes the bounds. This also justifies the conditional probabilities and the restricted integrals above.□

Proof. We first bound the derivative inverse on the region in the hypothesis. We then prove existence and uniqueness of a zero, and integrate the inverse bound along a path to that zero.

Put L_1=BI-J-2jmm^{\mathsf T}/n. Since F'(y)=I+L_1V_y, the elementary inverse identity gives

Under (5.28), this inverse exists and has norm at most 1+(K+3j)/c, since 0<V_y\le I and \norm{L_1}\le K+3j.

For existence and uniqueness, pass to the coordinates m\in(-1,1)^n and define the potential

Its critical-point equation is F_{J,h_0}(\atanh m)=0. At any critical point,

so the critical point has positive definite Hessian and is isolated. Choose a\in(0,1) close enough to one that

Every critical point lies in the interior of [-a,a]^n, and on each face of this box the negative gradient -\nabla\Phi points strictly inward. The minimum of \Phi on the box is therefore attained at an interior critical point. This proves existence.

To prove uniqueness, consider the negative gradient flow from any point of the box. The inward-pointing condition keeps the trajectory in the box for all positive times. Its potential decreases and is bounded below, so \int_0^\infty\norm{\nabla\Phi(m(t))}^2\,\dd t<\infty. The squared gradient is uniformly continuous in time, because the gradient and Hessian are bounded on the box. It follows that the gradient tends to zero: a nonzero limsup would, by uniform continuity, force disjoint intervals of fixed positive integral. Every limit point of a trajectory is therefore critical. The critical set is finite, since it is a compact set of isolated zeros. The limit set of a trajectory is connected: it is the intersection of the nested compact connected closures of its tails. Each trajectory therefore converges to one critical point.

It remains to show that all trajectories converge to the same point. The positive Hessian makes every critical point a strict local attractor. For clarity, in a sufficiently small ball about it one has (m-m_*)^{\mathsf T}\nabla\Phi(m)\ge c_*\norm{m-m_*}^2, which gives exponential contraction of the distance along the flow and invariance of the ball. Continuous dependence of the flow on its starting point, up to the time it enters that ball, shows that each attraction basin is relatively open in [-a,a]^n. These disjoint nonempty basins cover the connected box. Hence there is only one basin and one critical point, proving uniqueness of r.

We now derive the distance estimate. Suppose \norm{F(y)}<r_0\sqrt n. Starting at y(0)=y, solve on the open region \norm{F}<r_0\sqrt n the ordinary differential equation

The chain rule gives F(y(s))=(1-s)F(y) while the solution exists. Its velocity has norm at most C\norm{F(y)}. If its maximal existence interval ended at or before time one, this velocity bound would give a finite limiting point, whose residual is still strictly less than r_0\sqrt n and whose derivative is invertible. Local existence would extend the solution, a contradiction. The solution consequently reaches time one, at which its value is the unique zero r. Integrating the velocity bound along this path gives (5.29).□

Proof. We first bound the formal matrices in (6.3), then compare them with the actual discrete derivatives. Expanding H_{l+1} produces the chain

and the terms obtained by contracting disjoint pairs of originally adjacent J’s. Contracting J D_{1-(m^s)^2}J contributes -j b_s I. This description follows inductively from the two terms defining H_{l+1}: either keep its first link, or contract that link with the next one. Multiplicities are retained when different contractions happen to give the same matrix.

The contraction terms cancel the diagonal predictions. To see this, fix a complete noncrossing pairing of the original chain. Reinserting each contracted adjacent pair identifies its contribution with the prediction of this full pairing, multiplied by the signs of the contractions. The originally adjacent pairs belonging to a pairing are mutually disjoint, and every nonempty complete noncrossing pairing contains at least one such pair. Summing over their subsets gives (1-1)^a=0, where a\geq1 is their number. An odd chain has no complete pairing. Lemma 3.1 therefore bounds the diagonal \ell^2 norm of H_l, and bounds its operator and off-diagonal \ell^4 norms term by term. Its full \ell^4 norm is bounded as well, since \norm{\diag H_l}_4\leq\norm{\diag H_l}_2.

We now compare these formal bounds with the actual derivatives. For a scalar function \psi with bounded second derivative, Taylor’s theorem gives

The Frobenius norm of the remainder is at most C\norm{\nabla h^l}_{\ell^4}^2. This proves the magnetization part of (6.4) once the field part is known. The same argument for \psi(z)=\sech^2z gives an operator bound for its site derivative matrix Q_l. Averaging its rows yields

In differentiating b_lm^{l-1}, the term from db_l is a rank-one matrix of Frobenius norm at most \norm{m^{l-1}}\norm{db_l}\leq C. The discrete product error is \nabla m^{l-1} times a diagonal whose diagonal has norm at most 2\norm{db_l}. Its Frobenius norm is bounded, since \norm{\nabla m^{l-1}}\leq C. Starting at \nabla h^1=J, these estimates close the induction. The resulting bounded Frobenius perturbation preserves the operator, diagonal \ell^2, and entrywise \ell^4 bounds of the formal matrices.

For the final assertion, apply the same comparison to a smooth function of several arguments. Expand along the segment between the arguments at x and x^{(j)}. The preceding bounds control the primary linear terms. A linear term involving db_s is a rank-one matrix with a bounded-entry left vector, hence has bounded Frobenius norm. The squared-average remainder is controlled by

Mixed remainders are bounded by sums of pure squares, and primary quadratic remainders are controlled by the entrywise \ell^4 bounds. This proves the assertion.□

Proof. We first control the coefficients and solve the implicit linear system. Only then do we differentiate it, using the larger region to cover spin flips from the smaller one. The uniform kernel and quotient bounds proved above will control the coefficients even when some A_i are small.

Put \ell_i=(\partial_z\Phi/\Phi)(h_i^k;r_i,a). At z=r_0 this ratio equals -\tanh r_0, for every a. Comparison with z=r_0,a=0, followed by (6.11), gives

In particular, the comparisons put A among the diagonals allowed by the stable-field implication, once \rho is decreased.

Let u_0=w_0/A coordinatewise and p_0=A\partial_k u_0. The quotient bounds just proved give \norm{u_0}+\norm{p_0}\leq C; their local derivatives have coordinatewise bounds proportional to |e_i| in the first case and to n^{-1/2} in the second. Direct differentiation with the specified partial convention gives

We can now eliminate y^{\rm imp} and c_1 and solve for the source alone. The resulting linear equation is

Write the matrix on the left as I+AL_2. Its symmetrically normalized version I+A^{1/2}L_2A^{1/2} differs in operator norm by O(\rho) from

For the rank term, use (6.11) and (6.17) in the elementary inequality \norm{uv^{\mathsf T}-u'v'^{\mathsf T}} \leq\norm{u-u'}\norm v+\norm{u'}\norm{v-v'}. The displayed symmetric matrix has the fixed positive margin supplied by (5.3). Decrease \rho until the perturbation is less than half that margin. The normalized matrix then has bounded inverse, even though it need not itself be symmetric. The identity

gives the same bound for the unnormalized inverse, since \norm{L_2} is bounded. The right side of (6.18) is bounded in norm, because |\av{p_0}|\leq\norm{p_0}/\sqrt n. This establishes the size bounds. To obtain the last assertion of (6.15), remove the negative rank term from the stable-field matrix and change b_k to \chi. The constants used to solve the system and obtain these bounds depend only on the fixed stability and operator margins.

It remains to bound the discrete derivatives. Compare the equations at x and x^{(i)}. On the stated inner region both configurations belong to \Omega_{2\rho} for large n, by Lemma 6.1. In every coefficient-difference term, retain the solution vector at the common point x. In particular,

Use the same product arrangement for A and \chi in

For clarity, put v=m^{k-1}+t_* and \mathcal B=u_0+(J-j\chi I)w-jc_1v, and use primes for evaluation at x^{(i)} in the next display. With \zeta_i=n^{-1}\{(d_i\ell)^{\mathsf T}w+\one^{\mathsf T}d_ip_0\}, the exact equation is

Thus the left matrix is precisely the matrix in (6.18) evaluated at x^{(i)}. The bracket \mathcal B on the right is consequently one common-point vector for all columns. This allows the following estimates to control the Frobenius norm of the entire right side.

We bound its terms in turn. A site coefficient difference times a common vector s of bounded norm has bounded Frobenius norm, since

This applies to A,\ell,u_0,p_0, using their coefficient bounds and Lemma 6.1; the seed factors are retained for u_0,p_0. The bracket multiplying d_iA is bounded in norm by the already established size estimates, so no bound on 1/A_i is needed. Also \norm{d\chi}\leq C/\sqrt n, because A_i is a primary site function with an average parameter of that derivative scale. If \zeta_i=\av{(d_i\ell)w+d_i p_0}, then

Its multipliers m^{k-1}(x^{(i)})+t_* have norm at most 2\sqrt n, which still gives a bounded Frobenius norm after summing the squared column norms. Finally |c_1|\norm{\nabla m^{k-1}}_{\Frob}\leq C, by its operator bound and |c_1|\leq C/\sqrt n. Applying the uniformly bounded flipped inverses column by column proves \norm{\nabla w}_{\Frob}\leq C_k. Returning to the two displayed difference identities gives \norm{dc_1}\leq C_k/\sqrt n and the required bound for \nabla y^{\rm imp}. The operator in (6.18), the regions \Omega_\rho, and the required buffer inclusions do not depend on the directional seed e. All bounds for its right side and its derivatives use only \norm e\leq1; indeed u_0=e and p_0=0 for that source. The inverse margins, the threshold on n, and all these estimates are therefore uniform over the full Euclidean unit ball of seeds.□

Proof. The proof has two parts. We bound the errors introduced by discrete differentiation, then cancel the trace predictions of the retained formal terms. Write \norm E_{\rm nuc} for the sum of the singular values of E. The error estimates repeatedly use

Every bounded-length diagnostic word P has bounded operator norm and bounded off-diagonal entrywise \ell^4 norm. For the errors arising before we solve the differentiated implicit equations, P may contain one inverse K from (3.2). Its required margin is supplied by the last assertion of (6.15).

Source errors. Consider one source term \phi_i s_i. Retain the derivative

and put parameter derivatives into the error rather than into the formal expression. Set

The finite sums have the bounds

Taylor expansion of d_j\phi_i has linear parameter terms and a remainder bounded entrywise by

After multiplication by s_i, the linear parameter terms have bounded nuclear norm: each is a rank-one matrix with a bounded-norm left vector and a bounded-norm right vector. For the Q^2 part, the matrix before multiplication by D_s has bounded Frobenius norm. Its product with D_s has bounded nuclear norm by (6.22) and \norm{D_s}_{\Frob}=\norm s\leq C. For the mixed part the sum of column norms is at most

For the last part it is at most C\norm s\sum_j\Theta_j^2\leq C. These separate estimates also apply to the combined remainder in (6.24): entry by entry, assign each summand in the bound its proportional share of the actual remainder.

The product rule also has a remainder with entries bounded by

Assume \norm{\nabla s}_{\Frob}\leq C, which is zero for a seed, holds for the initial implicit field by Lemma 6.3, and is inductive for later fields. The \Theta part has bounded sum of column norms by Cauchy–Schwarz. For the Q part, test against P and split the trace into diagonal and off-diagonal terms. Hölder’s inequality gives

The first line uses exponents 4,2,4. In particular, the argument does not assume the diagonal \ell^1 bound for \nabla s that we are still proving. Bounded entries of Q also give a bounded Frobenius norm for this error. We have therefore controlled each source error both in Frobenius norm and in trace against the indicated words.

Average corrections and size bounds. Next we control the averaged coefficients in the correction terms. Differentiating a coefficient function in a primary argument gives another source with the same inputs and bounds. Apply the size and Frobenius parts of the preceding estimates to that source, then average:

which proves the first line of (6.21). The implicit coefficient c_1 already has these bounds from its direct construction. For an auxiliary partial the scale is different: it is a bounded coefficient vector, rather than a small vector. Averaging its difference estimate |d_j\phi_i|\leq C(Q_{ij}+\Theta_j) and using the row bounds above gives |\av\phi|+\norm{d\av\phi}\leq C. This proves the second line, including the self-coefficient \chi when it occurs.

For a primary correction \alpha_lm^{l-1} retain \alpha_l\nabla m^{l-1}. The dropped rank-one matrix has nuclear norm at most \norm{m^{l-1}}\norm{d\alpha_l}\leq C. The dropped product error is \nabla m^{l-1} times a diagonal of signed entries 2d_i\alpha_l. Its nuclear norm is bounded by

This also holds for m^0=x. For an auxiliary correction \gamma_bw^b, the analogous rank-one and product errors have bounded nuclear norm using \norm{w^b}+\norm{\nabla w^b}_{\Frob}\leq C and |\gamma_b|+\norm{d\gamma_b}\leq C. When primary derivatives in retained terms are replaced by H_l,M_l, their Frobenius errors from Lemma 6.1 are multiplied either by a small-vector diagonal or by an O(n^{-1/2}) scalar. In the former case (6.22) applies; in the latter use \norm E_{\rm nuc}\leq\sqrt n\norm E_{\Frob}. Both give bounded nuclear norm.

We can now close the induction for the size and Frobenius parts of (6.20). In (6.7), the terms are a bounded-norm leading term, primary corrections bounded by C\sqrt n\,n^{-1/2}, and bounded-norm auxiliary corrections. The retained source derivative (6.23) is Frobenius-bounded: use the preceding auxiliary derivative bounds and \norm{D_{s\partial_l\phi}\nabla h^l}_{\Frob} \leq\norm{s\partial_l\phi}\norm{\nabla h^l}. The retained primary correction is bounded in Frobenius norm by |\alpha_l|\norm{\nabla m^{l-1}}_{\Frob}\leq C. Thus the size and Frobenius induction closes before we use or prove the diagonal \ell^1 assertion.

Solving the differentiated self-dependence. The initial implicit source needs one additional step before the trace cancellation. Retain its site partials, freeze the parameters and averages, and replace the primary derivatives by their formal matrices. This gives

The errors have bounded Frobenius norm and bounded trace against words containing at most one K. The parameter c_1 causes no exception: write the corresponding source term as -j(\sqrt n c_1)A(t_*/\sqrt n), so that it is exactly of the source form above. Its difference bound follows from (6.16). Solving (6.25) yields

Testing a propagated error against an ordinary word now puts at most one K into the test word. The preceding error estimates therefore give both bounded trace and bounded Frobenius norm. Substitution gives the same conclusion for \nabla y^{\rm imp}. Subsequent fields are ordinary: propagating these errors multiplies them only by ordinary words. Induction therefore bounds the trace of every final error against an ordinary word. Since the recipe is fixed and finite, the required increases in word length are finite as well.

Cancellation of the formal traces. All differentiated errors are now controlled. To finish, we must bound the diagonal \ell^1 norm of the retained formal derivative. A single ordinary field already illustrates the cancellation. From w_i=\phi(h_i^1)s_i we obtain y=Jw-j\av{\phi'(h^1)s}x. If P=D_{\phi'(h^1)s}, its retained derivative is JPJ-j\av{\phi'(h^1)s}I, whose two diagonal predictions cancel. We now show that the correction at each node gives this same cancellation along every dependency path, including paths through the implicit field.

At the point x, freeze the local partial vectors and all their averages. Index each primary or auxiliary field by a node a, and attach its source to the same node: for example, h^l has source m^{l-1}, and y^a has source w^a. Let p_{ad} be the partial of the source at node a with respect to the field at node d. Introduce a formal variable z, and denote the formal source and field derivatives by \mathfrak S_a and \mathfrak H_a, respectively. They obey

At the spin leaf, the source derivative for h^1 is I and its field derivative is zJ. A terminal source uses just the first equation. At z=1 these are precisely the retained formal derivatives. In the implicit case the only cyclic node is y^{\rm imp} and its source w; closing it gives

Every other dependency is acyclic and linear. Solving the single cyclic node therefore leaves at most one inverse K(z) in each product of the closed formulas.

The averaged term in (6.28) is essential even for k=2. Write D=D_{1-(m^1)^2} and P=D_{\partial_2w}, so that b_1=n^{-1}\Tr D and c_1=n^{-1}\Tr P. Then

and the implicit derivative is K(z)\{z^2P(JDJ-jb_1I)-z^3jc_1ADJ\}. At degree four, the two noncrossing pairings of AJAJPJDJ have predictions

The first is canceled by the terms from -j\chi A in the quadratic coefficient of K(z) and -jb_1I in \mathfrak H_{h^2}. The second is canceled by the degree-four term -jc_1AJADJ from the averaged correction. This calculation exhibits the first cancellation involving both the self-dependent node and its primary input. We now account for every degree by following dependency paths.

We work coefficient by coefficient. An uncontracted dependency path, followed backwards from a field to the spin leaf, has successive links JD_{p_{ad}} and ends with the J link of h^1. A terminal source contributes an initial partial diagonal and then such a path. The choices of input nodes index these paths separately, preserving all multiplicities. Expanding (6.27) is inclusion–exclusion over disjoint originally adjacent link pairs: the first term keeps the first link, while the second contracts it with the next link, with factor -j\av{p_{ad}}, and continues from the input node’s source. That continuation skips exactly the link just contracted. Every contraction has degree two in z, equal to the degree of the replaced links. Consequently a coefficient of degree L involves only uncontracted paths of length L, of which there are finitely many even with the self-dependence.

To compute the trace prediction, test the formal derivative by a bounded diagonal D. Fix a nonempty path and a full noncrossing pairing of its links. Reinserting contracted adjacent pairs restores that same full-pairing prediction. Each contraction contributes its minus sign. Conversely, all subsets of the originally adjacent pairs of that full pairing occur, with precisely these signs. Such a full pairing has at least one originally adjacent pair, so its total coefficient is zero. Paths of odd length have no full pairing. There is no empty path contributing a derivative of a small seed, because all seeds are fixed with respect to x; parameter derivatives were already put into the error. Thus the trace prediction vanishes as a formal power series. The polynomial-prediction assertion for single-inverse words in Lemma 3.1 then makes it zero at z=1 in the closed formulas.

To pass from this zero prediction to a bounded trace, we use the small-vector structure. Every product in a closed formal derivative of a small vector contains either a small-vector diagonal or a scalar of size O(n^{-1/2}): a path giving a nonzero derivative must enter the primary nodes before reaching the spin leaf. Entry from a small source uses either its primary partial, a small vector, or the average of that partial in a correction term. The latter has the indicated scalar size. For the implicit starting node these are exactly \partial_k w and c_1 in (6.28).

For a term containing a small diagonal D_s, rotate the tested trace to put that factor first. The word estimate for the remaining factors and Cauchy–Schwarz give

The normalized-trace prediction is cyclically invariant by Lemma 3.1, so the rotation does not change the prediction being cancelled. For a term with a scalar of size O(n^{-1/2}), its unscaled trace error is at most C\sqrt n by the diagonal \ell^2 word estimate, and multiplication gives the same O(1) bound. The closed formula has finitely many terms. Combining their errors with the zero prediction and the already controlled differentiated errors gives

Taking D_{ii} to be the sign of d_iU_i proves the remaining assertion \norm{\diag\nabla U}_1\leq C. All arguments were pointwise inside the prescribed buffers, which proves the local version as well.□

Proof. We induct strongly on p over all finite admissible recipes, proving the two assertions together. Each step may enlarge the original recipe, but it adds only finitely many auxiliary operations and invokes only smaller residual indices. Consequently every particular instance of the induction uses finitely many operations and a finite word-length bound.

At the base level p=1, condition on all spins except the one being differentiated. Its conditional mean is m_i^1 and its conditional variance is v_i^1. A function of this single spin is affine, with linear coefficient d_i of the function. Taking its conditional covariance with the spin gives the exact identity

By (1.2), its right side is

The diagonal derivative bound makes the first term at most C\E|G|. For the second term, the coefficient vector has pointwise norm at most \norm U+2\norm{\diag\nabla U}_2\leq C. Weighted Cauchy–Schwarz and 0<v_i^1\leq1 therefore bound that term by C\D_{h_0}(G)^{1/2}. This proves (6.29). For a supported local test, its derivatives vanish outside the one-flip neighborhood of its support; all bounds used here are available there.

For (6.30), absorb the bounded multiplier by replacing U in the base identity with \wt U=H_0U. This vector still has bounded norm and bounded diagonal derivative \ell^1 norm: the additional term obeys \sum_i|U_i d_iH_0|\leq\norm U\norm{dH_0}, and the product-error term is bounded by the same argument using \norm{\diag\nabla U}_2. Apply (6.31) with G=f^2 and use

The diagonal-derivative term is bounded by C\E f^2; the remaining terms are bounded by C\sqrt{\E f^2\,\D_{h_0}(f)}+C\D_{h_0}(f). The base case is now proved in both forms. Before proceeding to higher residuals, we record how bounded scalar multipliers affect the tests. If |\vartheta|+\norm{d\vartheta}\leq C, then

Indeed arrange the product rule as d_i(\vartheta G)=\vartheta(x^{(i)})d_iG+G(x)d_i\vartheta, square, and sum with weights v_i^1\leq1. In the square-density case such a multiplier can be absorbed into H_0; the product retains its bounded size and difference norm. For local tests only the buffer bounds are needed, since multiplication does not enlarge their support.

For the induction step, suppose both assertions hold through level p and take U admissible from level p+1. The scalar integral identity suggests the following new source. Set

This is a source admissible from level p. Its coefficient and all needed derivatives are bounded because |a_p|\leq j and the bounds for \Phi proved with (6.9) hold uniformly in its two real field arguments. The level condition on U makes its site partial at level p vanish: \partial_pU differentiates only explicit primary arguments. Parameters and auxiliary arguments remain fixed, including when their values depend on earlier primary fields. Applying (6.9) coordinatewise therefore yields

The primary recursion gives

where the last term is absent for p=1. Add a last auxiliary field y' by (6.7) with source U', and write \alpha_l=\av{\partial_lU'} and \gamma_b=\av{\partial_{y^b}U'}. Symmetry of J now gives

The first line now consists of earlier residuals with admissible sources and bounded multipliers. After testing, (6.32), (6.21), and the level constraints allow us to apply the induction hypothesis to each term. In particular the enlarged recipe is admissible from level p, which is also sufficient for its R_{p-1} term.

On the second line, the terms with l>p use the terminal source m^{l-1}/\sqrt n, admissible from level p, and multiplier \sqrt n\alpha_l. The multiplier bounds follow from Lemma 6.4. The remaining term is handled by the identity

This follows from (m^{p-1}-m^p)^{\mathsf T}(m^{p-1}+m^p) =\norm{m^{p-1}}^2-\norm{m^p}^2. The first term on the right of (6.34) cancels a_pn\alpha_p in (6.33); the remaining term is handled with the admissible source m^p/\sqrt n and multiplier \sqrt n\alpha_p. This completes the induction for both (6.29) and (6.30). In the local version it stops at level k; each added source respects the support and buffer requirements of the initial implicit construction.□

Proof. The vector in braces is JR_k-jb_{k-1}(R_{k-1}+R_k). To express its pairing with the implicit source in terms of earlier residuals, use symmetry of J, the equation Jw=y^{\rm imp}+j\chi w+jc_1m^{k-1}, and (6.34). The bracketed expression becomes

Apply Lemma 6.5 to the first three terms, absorbing their bounded parameters into the test. For the last term, use source m^k/\sqrt n and multiplier \sqrt n c_1. The bounds needed for this multiplier are precisely (6.15) and (6.16).□

Proof. We use the scalar residual pairing S_l=R_l^{\mathsf T}m^l/\sqrt n to control the total squared increment. First, the primary derivative bounds give \norm{dS_l}\le C_l pointwise. In the product rule, the linear terms are controlled by the operator norms of \nabla R_l and \nabla m^l, multiplied by \norm{m^l}/\sqrt n and \norm{R_l}/\sqrt n. The cross term for coordinate i has size at most 2\norm{d_i R_l}\norm{d_i m^l}/\sqrt n; summing its square uses the bounded column norms and gives the same scale.

Apply Lemma 6.5 with target m^l/\sqrt n and test G=S_l. It gives

For the square-density version use H_0=S_l/\sqrt n, which is bounded with bounded difference norm. Dividing the resulting estimate by N\sqrt n gives

We now use the squared-increment identity (6.2). Choose a fixed d with \lfloor d/2\rfloor\rho^2/4>2, followed by \eps>0 with 2d\eps<1. Whenever \max_{l\le d}|S_l|/\sqrt n\le\eps, telescoping gives \sum_{l=1}^d\norm{R_l}^2/n<2. If every consecutive pair contained a residual larger than \rho\sqrt n/2, the disjoint pairs (1,2),(3,4),\ldots would force the sum to exceed this bound. At least one pair therefore meets the required threshold.

To implement the selection smoothly, apply a scalar cutoff equal to one on [0,1/4] and zero on [1,\infty) to each of \norm{R_k}^2/(\rho^2n) and \norm{R_{k-1}}^2/(\rho^2n), and multiply the results to define C_k. The primary operator and column bounds, with the product rule, give \norm{d(\norm{R_l}^2/n)}\le C_l/\sqrt n, hence the claimed derivative bound. Finally D=\prod_{k=2}^d(1-C_k): positive deficit implies \max_{l\le d}|S_l|/\sqrt n>\eps. The preceding two moment bounds and a finite union bound give the ordinary and square-reweighted estimates.□

Proof. Fix an index k\ge2 and work on the small-residual region (6.10), with the buffered implicit construction (6.14). Abbreviate m=m^k, p=m-t_*, and put

Here \chi=\av{A_i} and A_i=\Phi(h_i^k;r_i,a) are defined globally by (6.12), even though the implicit vector is only needed locally. Thus |L_*|\le C\sqrt n and \norm{dL_*}\le C_k pointwise, by the primary derivative bounds and \norm{db_{k-1}}=O(n^{-1/2}).

For either of the two allowed sources w_0, the following pointwise identity holds on the buffered region:

Here the zeroth-order term in the first sum acts by multiplication on w_{0,i}. For the verification, subtract the root equation to write the vector on the left as h^k-r-am-Jp+j(1-q_*)p, then substitute Jw=y^{\rm imp}+j\chi w+jc_1m^{k-1}. Apply (6.9) to the contribution A(y^{\rm imp}-jc_1t_*), holding its auxiliary arguments and parameters fixed, and use nc_1=\sum_i\partial_k w_i. The remaining scalar terms cancel by

which follows by expanding p=m-t_* and R_k=m^{k-1}-m.

Lemma 6.6 controls the tested left-hand side of (). The last term on the right also has tested expectation at most C\norm{G}_*: apply Lemma 6.5 to target p/\sqrt n and multiplier \sqrt n c_1. The latter is bounded with bounded difference norm on the buffer. Meanwhile the size estimates for the implicit solution and Lemma 5.7 give

The source size and stability margins determine C_0 independently of k and the selection depth. This lets us choose \rho first so that C_0\rho<1/2 and all stability and flip-buffer conditions hold. The selection depth in Lemma 7.1 is chosen afterward. This order will permit absorption of the scalar error without a self-dependent choice of depth.

First control the averaged coefficient by taking w_{0,i}=\partial_{r_0}\Phi(h_i^k;r_i,a)/\sqrt n. The second identity in (6.13) makes the first sum in () equal to L_*. Test with G=C_k^2L_*. The cutoff and difference bounds give

Indeed its difference vector is bounded pointwise: |L_*|=O(\sqrt n) multiplies \norm{d(C_k^2)}=O(n^{-1/2}), and \norm{dL_*}=O(1). Absorbing (7.4) now yields

Now test a direction by taking w_0=Ae, where e is any deterministic vector with \norm e\le1. By the first identity in (6.13), the first sum in () is p^{\mathsf T}e. Test with \widehat C_kG. The residual estimates control the left side and the last term by C\norm G_*; the middle term is bounded by (7.5) and Cauchy–Schwarz, since 0\le\widehat C_k\le C_k. Thus

Summing Lemma 6.5 with the constant target e over l=1,\ldots,k gives the same bound for \E[\widehat C_kG(x-m^k)^{\mathsf T}e]. Sum over k. The deficit has contribution at most

Thus |\E[G(x-t_*)^{\mathsf T}e]|\le C\norm G_* uniformly over \norm e\le1. Centering G and taking the supremum over this seed class gives (7.2). The uncentered estimate also records the ordinary-mean consequence: take G=1 to obtain \norm{\E_{\mu_{h_0}}x-t_*}\le C. This norm is the unnormalized Euclidean norm. It compares the mean itself with t_*; no approximation of the derivative of the mean map is used here. The local construction required n above a fixed threshold. In the remaining dimensions, the elementary bounds \norm{\Cov(G,x)}^2\le n\Var(G) and \norm{\E_{\mu_{h_0}}x-t_*}\le2\sqrt n extend both conclusions after enlarging C.□

Proof. Select the cutoffs with a fixed threshold \rho\le\delta small enough for Lemma 5.7. On each cutoff support, the residual-to-root comparison gives

Here the constant comes from the recurrence and the inverse bound in Lemma 5.7, and is independent of k. For \norm e\le1, sum the square-density version of Lemma 6.5 with target e through k, using multiplier \widehat C_k. After division by N, this bounds

The cutoff weights sum to at most one, so all root-comparison terms together cost at most C\delta\sqrt n. The leftover mass contributes at most 2\sqrt n\Prob_\nu(D>0), which is controlled by the second part of Lemma 7.1. Add the terms, divide by \sqrt n, and take the supremum over e to obtain (7.6). The leading C comes from root stability; only C_\delta and the finite diagnostic length depend on the selected tolerance. Any dimensions below a fixed threshold required by this construction are covered by the elementary bound \norm{\E_\nu x-t_*}/\sqrt n\le2 after increasing C_\delta; this leaves the leading constant unchanged. The argument has used no implicit test vector.□

Proof. The curvature bound is obtained by summing an exact two-spin calculation. Fix i\ne j, condition on the other spins, and call the remaining spins x,y. Their conditional density is proportional to \exp(\alpha x+\gamma y+J_{ij}xy). Introduce the scalar parameters

The letters a,b,A,B have these scalar meanings only during the pair calculation. Use \E_0 for the product law of means a,b, and \E_{ij} for the interacting conditional law. The identity e^{J_{ij}xy}=\cosh(J_{ij})(1+uxy) gives their exact density ratio:

Expand f in the four-dimensional space of functions of x,y. Its two gradients then have the unique representation

with real p,s,r. The shared coefficient r is the mixed-spin term of this expansion. The conditional means and variances satisfy

For a function F of one spin, write F(z)=\overline F+z\,dF, and set

The identity (1-uxy)^2(1+uxy)=(1-u^2)(1-uxy) and independence under \E_0 give the following exact expression for the cross term:

Both terms follow from the one-spin identities \E_0[(x-a)F_2(x)]=A\,dF_2 and \E_0[(x-a)xF_2(x)]=A\overline F_2, together with their y counterparts. Thus the cross term has been reduced to four scalar coefficients.

The four coefficients appearing here are

Consequently the right-hand side of (8.5) is

where

The mixed coefficients rp and rs begin at order u^2: their first-order terms have canceled. Keeping this cancellation is what will leave only a quartic row cost after Young’s inequality.

For |u|\le1/4 and |a|,|b|\le1, all denominators above are bounded away from zero. There is thus a universal constant C_1 such that

Consequently the ps coefficient in (8.6) is -u+O(u^2), both mixed coefficients are O(u^2), and the r^2 coefficient is 1+O(|u|). These constants are uniform in the conditional means a,b: all the displayed rational functions and their required derivatives are bounded on the compact parameter range.

Return temporarily to the notation a_i=v_i^1g_i and a_j=v_j^1g_j for the two gradient weights. Equations (8.3) and (8.4) imply

Uniformly in a,b,x,y, one has

for universal positive constants c_1,C_2. Expanding g_i(y)g_j(x) and using that the centered factors have zero product-law mean gives

To obtain the four error bounds, expand the product into its ps, rp, rs, and r^2 terms. Each centered spin is bounded by two, and the uniform estimate for w_u-1 controls every resulting coefficient.

Adding u times (8.8) to (8.6), and decreasing the universal u_0>0 if necessary, yields

The last inequality is Young’s inequality, absorbing the mixed terms into r^2/2.

We must now express the two scalar costs in the heat-bath weights that can be summed over pairs. The lower bound for w_u in (8.7), followed by product independence, gives

Similarly, the density in (8.3) and the ratios v_i^1/A, v_j^1/B are bounded above and below by universal positive constants for |u|\le u_0. Hence

The comparison constants do not deteriorate when a conditional variance A or B is small. Substituting these bounds into the preceding scalar inequality proves the pair estimate

To assemble the curvature form, first condition on all but spin i in the diagonal term of \E(Hf)^2. Their sum is \sum_i\E[v_i^1g_i^2]=\D_{h_0}(f). Next average (8.9) over the outside spins and sum over ordered pairs i\ne j. The signed and absolute pair costs become the two quadratic forms in (8.2). The quartic cost is bounded by

Increasing C proves the lemma.□

Proof. For any deterministic unit vector z, the centered Gaussian variable z^{\mathsf T}Jz has variance

Thus

To pass from fixed quadratic forms to norms, choose a 1/4-net of the unit sphere in \R^k with at most 9^k points. Every symmetric k\times k matrix B then satisfies

to see this, approximate a unit vector maximizing the absolute quadratic form and bound the change by 2\|B\|/4. Taking k=n in (8.12) proves \|J\|\le M_0 with probability at least 1-e^{-c n} for a sufficiently large constant M_0=M_0(j).

For a fixed row i and a fixed set S of at most k indices, Gaussian independence within that row gives

Use exponential Markov with parameter n/(4j) to obtain this estimate: each nonzero Gaussian square contributes the factor \sqrt2 to the moment generating function. Choosing the full index set and taking a union bound over rows bounds every row square sum by a fixed M_1=M_1(j), with failure at most e^{-cn}. Separately, the Gaussian entry tail and a union bound give

Since |\tanh z|\le|z|, these bounds imply \|Q\|\le M_1 and r_4(U)\le M_1n^{-1/2}, using the maximal absolute row-sum bound for the operator norm of a symmetric matrix. Moreover,

The scalar inequality follows by integrating |1-\sech^2 z|=\tanh^2 z\le z^2. This proves (8.10), with a fixed M.

The global bounds have fixed M; we now select the small-set fraction \alpha. Put k=\lfloor\alpha n\rfloor. Bounding sets of size exactly k suffices, because a smaller set can be enlarged and restriction cannot increase a principal-submatrix norm. The \binom nk possible sets have logarithmic count at most n\mathfrak h(k/n), where

and \mathfrak h(t)\to0 as t\downarrow0. Applying the sphere-net argument on each support and using (8.12) gives

Choose \alpha>0 sufficiently small that this tends to zero exponentially in n. Then \|U-J\|=o(1) proves the assertion for U.

Finally, (8.13), with t=\varepsilon, and a union bound over all rows and all k-element sets give

Decreasing \alpha further makes this probability exponentially small. On its complement, the maximal row sum of every Q_{SS} is at most \varepsilon, which proves the remaining assertion.□

Proof. Choose the tolerances in the order needed to absorb the curvature costs. Let C be the constant in Lemma 8.1, and choose \varepsilon>0 with (1+C)\varepsilon\le1/8. Lemma 8.2 supplies \alpha and the fixed global bound M. Finally choose the fixed observation time T large enough that

Let \mathcal E_n^{\mathrm{term}} be the events supplied by that lemma. For all sufficiently large n, their bounds also ensure \max|U_{ij}|\le u_0 and Cr_4(U)\le1/8.

First count coordinates whose observation field remains small. Given X, the indicators \one_{\{|Y_i|<T/2\}} are independent; symmetry of the Gaussian noise gives the same success probability p_T for either sign of X_i. The Gaussian tail bound yields

For B_Y=\#\{i:|Y_i|<T/2\}, exponential Markov inequality yields

The same bound holds after averaging over X.

Suppose now that B_Y\le\alpha n/2. For every spin configuration x, \|Jx\|^2\le M^2n, and consequently

It follows, simultaneously for all x, that

At this fixed field Y, set q_f(x)=\sum_i v_i^1(x)g_i(x)^2 and \eta=\sech^2(T/4). Since 0<v_i^1\le1, one has pointwise

We may now use S(x) even though it depends on the configuration, because (8.11) holds for every small set on one matrix event. Split the quadratic form between S(x) and its complement. The small-set bound handles its first block, and the global norm handles the cross and complementary terms, giving

The identical argument, with |a| and Q, gives the same upper bound for |a|^{\mathsf T}Q|a|. Lemma 8.1 and (8.15) therefore imply

It remains to translate curvature into a gap. The kernel of the self-adjoint operator H contains exactly the constants: vanishing \E[fHf] forces every one-spin conditional variance to vanish, and the strict positivity of the Gibbs law propagates equality across every cube edge. For an eigenfunction with positive eigenvalue \lambda, the curvature estimate and (8.1) imply \lambda^2\ge\lambda/2, hence \lambda\ge1/2. Expand in an orthonormal eigenbasis to obtain (8.14). The observation failure bound was e^{-\alpha n/4}, uniformly in the initial spin law, so we may set c=\alpha/4.□