Proof. We expose one column at a time. The first diagonal entry, the rest of the first column, and the trailing square block are independent; this independence will persist after each rotation. The column vector has \(n-1\) independent \(N(0,1/n)\) entries. Choose an orthogonal transformation of the trailing coordinates that sends this vector to its nonnegative length times the first coordinate vector. Conjugation by this transformation produces the desired first off-diagonal entry.

To justify the next exposure, observe that the trailing block has density proportional to \(\exp(-n\mathop{\mathrm{Tr}}B^2/4)\). Both this density and Euclidean volume are invariant under orthogonal conjugation. Conditional on the column vector, the chosen transformation is fixed, so the conjugated trailing block still has the same GOE density and its conditional law does not depend on the column. It is therefore independent of all entries already exposed. Repeat the construction inside that block, then inside each successive trailing block. At step \(i\) the fresh column has \(n-i\) independent \(N(0,1/n)\) entries. This proves both the stated distributions and their joint independence. The conjugations preserve eigenvalues. ◻

Proof. We prove the lemma in three steps: the deterministic energy controls where a vector can concentrate, one noise event controls block prefixes at every relevant scale, and summation by parts combines these controls.

Energy controls spatial mass and gradients. For a unit vector \(u\), write \(e=u^TEu\). Expanding gives \[ e=\sum_{i<n}c_i(u_i-u_{i+1})^2 +\sum_{i=1}^n(2-c_i-c_{i-1})u_i^2. \tag{12}\] The row sums of \(D\) are at most two, so \(0\le E\le4I\). The coefficients \(2-c_i-c_{i-1}\) are at least \(i/(2n)\): for interior indices this follows from \(1-\sqrt{1-t}\ge t/2\), and the boundary indices satisfy the same inequality directly. Also \(c_i\ge1/\sqrt2\) when \(i\le n/2\). Thus (12) implies \[ \sum_i\frac{i}{n}u_i^2\le2e,\qquad \sum_{i>n/2}u_i^2+ \sum_{i<n}(u_i-u_{i+1})^2\le Ce. \tag{13}\] For the second inequality, control gradients in the first half by the weighted gradient term, and use \((u_i-u_{i+1})^2\le2u_i^2+2u_{i+1}^2\) for the remaining edges.

One event for every energy. The vector will be selected after the matrix has been realized, so the probability estimate must be simultaneous. We first prepare blocks at every dyadic energy scale. Set \[\sigma_j=2^js_*,\qquad 0\le j\le J_n:=\left\lceil\log_2(4/s_*)\right\rceil, \qquad d_j=\max\{1,\lfloor\sigma_j^{-1/2}\rfloor\}.\] For each of the two independent-entry sequences (diagonal entries and centered off-diagonal entries), partition its index set into consecutive blocks of lengths between \(d_j\) and \(2d_j\). Such a partition is obtained by merging a last short block with its predecessor. Assign label \(r=0\) when the first index of a block is at most \(2n\sigma_j\); otherwise assign \(r\ge1\) when that index is in \((2^rn\sigma_j,2^{r+1}n\sigma_j]\).

If \(\xi_i\) is either centered entry sequence and a block has length \(m\le2d_j\), its partial sums \(S_k\) obey \[\mathbb P\left\{\max_{k\le m}|S_k|\ge a\right\} \le2\exp\left(-\frac{na^2}{C d_j}\right).\] Indeed, \(\exp(tS_k-Ckt^2/n)\) is a nonnegative supermartingale by (9). Stopping it at its first crossing of \(a\) bounds the one-sided probability by \(\exp(-ta+Cmt^2/n)\); optimizing \(t>0\) and repeating for \(-S_k\) proves the displayed estimate.

At a fixed pair \((j,r)\), the number of blocks is at most \[C\left(1+\frac{2^rn\sigma_j}{d_j}\right) \le C2^rn\sigma_j^{3/2}=C2^{r+3j/2}.\] For \(\sigma_j\le1\), use \(d_j\ge\sigma_j^{-1/2}/2\); for \(\sigma_j>1\), use \(d_j=1\). The additive constant is absorbed because \(n\sigma_j^{3/2}\ge1\). Consequently, for a sufficiently large fixed constant \(K\), there is an event \(\mathcal F_n\) on which every block of both sequences has all its centered partial sums bounded by \[ K\sqrt{d_j/n}\,(h_n+\sqrt{j+r}), \tag{14}\] and \[\begin{aligned} \mathbb P(\mathcal F_n^c) &\le C\sum_{j,r\ge0}2^{r+3j/2} e^{-cK^2(h_n+\sqrt{j+r})^2}\\ &\le C e^{-c'h_n^2}=C e^{-c'L_n^{1/4}}. \end{aligned}\] Here \(K\) is chosen so that the factors in \(j\) and \(r\) form summable geometric series. This bound does not acquire a factor from the number of scales.

From block prefixes to quadratic forms. We may now fix the noise event and choose an arbitrary unit vector. Select the first \(j\) for which \(e\le\sigma_j\). Since \(e\le4\), this index exists, and \[\sigma_j\le2(e+s_*).\] For the weights \(w_i=u_i^2\) or \(w_i=u_iu_{i+1}\), the sum of their absolute values over all blocks with label \(r\) is at most \(C2^{-r}\). For \(r\ge1\), every coordinate used is beyond \(2^rn\sigma_j\), and (13) bounds its total squared mass by \(2e/(2^r\sigma_j)\le2^{1-r}\). The off-diagonal weights obey the same bound by \(|u_iu_{i+1}|\le(u_i^2+u_{i+1}^2)/2\). For \(r=0\), use the unit norm.

The summed variation of these weights within those blocks is at most \(C2^{-r/2}\sqrt e\). To verify this assertion, use \[u_{i+1}^2-u_i^2=(u_{i+1}-u_i)(u_{i+1}+u_i)\] for diagonal weights and \[u_{i+1}u_{i+2}-u_iu_{i+1} =u_{i+1}\bigl[(u_{i+2}-u_{i+1})+(u_{i+1}-u_i)\bigr]\] for off-diagonal weights. Cauchy–Schwarz combines the preceding tail mass bound with the total gradient bound in (13); each coordinate or edge is counted at most a fixed number of times.

For clarity, the summation-by-parts step can be made blockwise. Suppose all centered entry prefixes in a block are bounded by \(B\), and let \(\bar\xi\) be the entry average on that block. Then \(|\bar\xi|\le B/d_j\). The prefixes of \(\xi_i-\bar\xi\) are at most \(2B\) and vanish at the block end. Therefore \[\left|\sum_{i\text{ in block}}\xi_iw_i\right| \le\frac B{d_j}\sum_{i\text{ in block}}|w_i| +2B\sum_{\substack{i,i+1\text{ in block}}}|w_{i+1}-w_i|.\] Using (14) and summing by labels proves that either centered weighted sum is bounded by \[\begin{align*} &C n^{-1/2}\sum_{r\ge0}(h_n+\sqrt{j+r}) \left(d_j^{-1/2}2^{-r}+d_j^{1/2}2^{-r/2}\sqrt e\right)\\ &\hspace{35mm}\le C(h_n+\sqrt j)n^{-1/2}\sigma_j^{1/4}. \end{align*}\] For \(\sigma_j\le1\), the last step uses \(d_j\asymp\sigma_j^{-1/2}\) and \(e\le\sigma_j\). For \(1<\sigma_j<8\), it follows from \(d_j=1\) with a changed universal constant. The series in \(r\) converges geometrically, including its \(\sqrt r\) factor.

The centered entries have now been controlled uniformly in \(u\). The only remaining part of \(T-D\) is the deterministic error in the off-diagonal means. By (9), their difference from \(c_i\) is at most \(C/(\sqrt n\sqrt{n-i})\). It is at most \(C/n\) for \(i\le n/2\) and at most \(C/\sqrt n\) thereafter. Thus their quadratic-form contribution is at most \[C/n+Ce/\sqrt n,\] using again the tail mass bound. Finally, \[j\le C\log(2+e/s_*) \le C\bigl[\log(2+L_n)+\log(1+e/s_0)\bigr].\] Together with \(\sigma_j\le2(e+s_*)\) and \(s_*\le s_0\), this converts the centered estimate into (10). The mean errors are absorbed into the same bound because \(0\le e\le4\) and \(L_n\ge1\) for large \(n\). Both sequences, and the factor two from off-diagonal entries, are absorbed by choosing \(C_0\) once and for all.

Finally, the scalar comparison must be stable under changing the energy. Put \(q(x)=1+\log(1+x/s_0)\). Direct differentiation gives \[\Phi_n'(x)=C_0n^{-1/2}A_n(x+s_0)^{-3/4} \left(\frac{\sqrt{q(x)}}4+\frac1{2\sqrt{q(x)}}\right).\] Since \(y^{-3/4}\sqrt{1+\log y}\) is bounded for \(y\ge1\), its supremum is at most \(Cn^{-1/2}A_ns_0^{-3/4}=C\eta_n\). The equality \(\Phi_n(0)/s_0=C_0\eta_n\) proves the other assertion in (11). ◻

Proof. We first derive the differential equation and locate its relevant quadratic root, then absorb its finite-dimensional error at \(s\ge s_0\). Reverse the coordinates of \(D\). Its off-diagonal entries become \(\sqrt{k/n}\), \(1\le k<n\), so the characteristic polynomials of its successive principal minors satisfy \[P_0(z)=1,\quad P_1(z)=z,\quad P_{k+1}(z)=zP_k(z)-(k/n)P_{k-1}(z).\] Induction gives \(P_k'=kP_{k-1}\): differentiation of the recurrence and the recurrence one step earlier show that \(P_{k+1}'=P_k+kP_k\). Consequently \[P_n''-nzP_n'+n^2P_n=0.\] Since \(P_n\) is the characteristic polynomial of \(D\), \(R_D(s)=P_n'(2+s)/(nP_n(2+s))\). The differential equation yields \[ R_D(s)^2-(2+s)R_D(s)+1=\frac{T_D(s)}n>0. \tag{17}\] For every \(s>0\), \(R_D(s)\) is finite because \(E\ge0\). The two roots of the quadratic on the left are \(R(s)\) and \[R_+(s)=\frac{2+s+\sqrt{s(4+s)}}2.\] At large \(s\), \(R_D(s)\to0\) whereas \(R_+(s)\to\infty\). Positivity in (17) thus places \(R_D\) below the smaller root there, and continuity prevents it from crossing either root anywhere on \((0,\infty)\). If \(\delta(s)=R(s)-R_D(s)\), factoring the quadratic gives \[ 0\le\delta(s)\le \frac{T_D(s)}{n\sqrt{s(4+s)}}. \tag{18}\]

Termwise comparison of resolvents shows that \[T_D(s)\le\frac2s\bigl[R_D(s/2)-R_D(s)\bigr] \le\frac2s\bigl[R(s/2)-R(s)+\delta(s)\bigr].\] Differentiating the explicit expression for \(R\) gives \[0<-R'(s)=\frac{2}{\sqrt{s(4+s)}\, (2+s+\sqrt{s(4+s)})}\le\frac1{2\sqrt s}.\] Integrating from \(s/2\) to \(s\) therefore gives \(R(s/2)-R(s)\le\sqrt s-\sqrt{s/2}\le\sqrt s\). Hence \[T_D(s)\le\frac C{\sqrt s} +\frac{2T_D(s)}{ns\sqrt{s(4+s)}}.\] For \(s\ge s_0\), the coefficient of the last term is at most \((ns_0^{3/2})^{-1}=L_n^{-3/2}\to0\). Absorbing it proves the bound on \(T_D\), and (18) then proves the bound on \(\delta\). Notice that \(R_D(s/2)\le R(s/2)\) was established for every positive \(s/2\); it does not require \(s/2\ge s_0\).

The second-trace estimate also gives the counting estimate needed for the later dyadic transfer. For \(s_0\le t\le1\), each \(e_i\le t\) contributes at least \((2t)^{-2}\) to \(nT_D(t)\). Thus \(N(t)\le4nt^2T_D(t)\le Cnt^{3/2}\). For \(t\ge1\), the same bound follows from \(N(t)\le n\). ◻

Proof of Proposition 5. The previous lemmas supply the noise control and the reference traces. We now transfer them to an event defined entirely by the eigenvalues. There are three tasks: compare ordered eigenvalues, locate the top three, and compare the two traces.

Ordered eigenvalues. Work first in the tridiagonal realization. On \(\mathcal F_n\), (10) bounds the Rayleigh quotient of \(E-H=2I-T\) between \[f_-(e):=e-\Phi_n(e)\quad\hbox{and}\quad f_+(e):=e+\Phi_n(e).\] By (11), both functions are increasing for large \(n\). The span of the first \(i\) eigenvectors of \(E\) has every unit-vector energy at most \(e_i\), so the min–max principle gives \(2-\lambda_i\le f_+(e_i)\). On the span of eigenvectors \(i\) through \(n\), every energy is at least \(e_i\); the complementary max–min principle gives \(2-\lambda_i\ge f_-(e_i)\). Therefore \[ |2-\lambda_i-e_i|\le\Phi_n(e_i),\qquad 1\le i\le n. \tag{19}\] This argument uses monotonicity of scalar functions of Rayleigh quotients, and does not require an operator-monotonicity assertion.

Define \(\mathcal G_n(L_n)\) to be precisely the event (19), with the deterministic \(e_i\) and \(\Phi_n\) constructed above. It depends only on the ordered eigenvalues. The tridiagonal event \(\mathcal F_n\) implies it; hence Lemma 6 transfers its probability bound to \(W\). All remaining conclusions are deterministic on \(\mathcal G_n(L_n)\).

The first three directions. The conditioning argument will need three large coordinate variances. To locate the corresponding eigenvalues, take \(m=\lfloor n^{1/3}\rfloor\) and the span of the first three discrete Dirichlet sine modes on \(\{1,\ldots,m\}\), extended by zero. The Dirichlet Laplacian on this segment has eigenvalues \(4\sin^2(k\pi/[2(m+1)])\), so its quadratic form on this span is at most \(C/m^2\). On vectors supported on the segment, the form difference between \(E\) and that Laplacian has absolute value at most \[2\max_{1\le i<m}|1-c_i|\sum_{i<m}|u_iu_{i+1}| \le Cm/n.\] The min–max principle therefore gives \(0\le e_1,e_2,e_3\le C(m^{-2}+m/n)\le Cn^{-2/3}\). Equations (11) and (19) imply \[\max_{j\le3}\frac{|\lambda_j-2|}{s_0} \le C(L_n^{-1}+\eta_n).\] Also every \(e_i\in[0,4]\), so \(\max_i|\lambda_i|\le2+\Phi_n(4)\). The assumption \(L_n\le\log n\) gives \(\Phi_n(4)=o(1)\), proving the stated bound by three.

Resolvent traces. The eigenvalue comparison also controls every denominator: for every \(s\ge s_0\) and every \(e\in[0,4]\), \[ \frac{\Phi_n(e)}{s+e} \le\frac{\Phi_n(0)+C\eta_ne}{s+e} \le C\eta_n. \tag{20}\] Thus (19) makes \(2+s-\lambda_i\) positive and equal to \((s+e_i)(1+O(\eta_n))\), uniformly in both \(i\) and \(s\ge s_0\). The normalized second trace is consequently at most a constant times \(T_D(s)\), proving its bound in (8).

For the first trace, subtract term by term and use \((e_i+s_0)^{1/4}\le(s+e_i)^{1/4}\) to obtain \[\begin{align*} &\left|\frac1n\sum_i\frac1{2+s-\lambda_i}-R_D(s)\right|\\ &\qquad\le Cn^{-1/2}A_n\frac1n\sum_i (s+e_i)^{-7/4}\sqrt{1+\log(1+e_i/s_0)}. \tag{21}\end{align*}\] We spell out the dyadic sum because its uniformity near \(s_0\) is useful. The range \(e_i\le s\) contributes at most \[Cs^{-7/4}\frac{N(s)}n\sqrt{1+\log(1+s/s_0)} \le Cs^{-1/4}\sqrt{1+\log(1+s/s_0)}.\] For \(k\ge0\), the shell \(2^ks<e_i\le2^{k+1}s\) contributes at most \[Cs^{-1/4}2^{-k/4} \sqrt{1+\log(1+s/s_0)+(k+1)\log2},\] by (16). This counting estimate remains valid above one, so all shells are covered. The convergent geometric series, with its additional square-root factor, bounds the sum in (21) by \[C_Ms^{-1/4}\sqrt{1+\log(1+s/s_0)}.\] Dividing the resulting trace error by \(\sqrt s\) and writing \(r=s/s_0\) gives \[C_M\eta_n r^{-3/4}\sqrt{1+\log(1+r)}\le C_M\eta_n \qquad(r\ge1).\] Finally, Lemma 8 gives \[\frac{|R_D(s)-R(s)|}{\sqrt s} \le\frac{C_M}{ns^{3/2}}\le C_ML_n^{-3/2}.\] This proves the quantitative first-trace error and completes all of (8). The event was defined before \(M\) was chosen, and each displayed error is deterministic on that event. Consequently every deterministic sequence of spectra in \(\mathcal G_n(L_n)\) satisfies the estimates for every fixed finite \(M\), without any condition on eigenvector orientation. ◻

Proof. First choose the Gaussian variances so that the squared radius has mean \(n\). Then compare its radial density with the standard Gaussian density at that same radius. To implement the first step, denote the objective in (22) by \(F_a(z)\). On \(z>a\lambda_1\), \[F_a'(z)=\frac12-\frac1{2n}\sum_i(z-a\lambda_i)^{-1}, \qquad F_a''(z)=\frac1{2n}\sum_i(z-a\lambda_i)^{-2}>0.\] The derivative tends to \(-\infty\) at the left endpoint and to \(1/2\) at infinity. The function itself tends to \(+\infty\) at both ends. This proves existence, uniqueness, and (23).

The average of the reciprocals of \(z_a-a\lambda_i\) is one. Their minimum is therefore at most one, and their maximum is at least one, giving the stated bounds on \(z_a\). Consequently \[0<z_a-a\lambda_i \le1+a(\lambda_1-\lambda_n)\le13.\] The lower bound for \(k_i(a)\) follows; the upper bound follows from their positivity and sum \(n\). At \(a=0\) these formulas give \(z_0=1\), \(k_i(0)=1\), and \(B(0)=0\).

The minimizer is now fixed. The radial comparison itself is exact at any admissible \(z\): put \(Q=zI-a\Lambda\) and let \(Y^{z,a}\sim N(0,Q^{-1})\). Its density relative to a standard Gaussian vector at \(y\in\mathbb R^n\) is \[(\det Q)^{1/2} \exp\left(-\frac{z-1}{2}\|y\|^2+\frac a2y^T\Lambda y\right).\] The standard Gaussian has uniform conditional angular law. Integrating this ratio over the sphere of squared radius \(n\) gives \[ \frac{p_{z,a}(n)}{p_{\chi_n^2}(n)} =(\det Q)^{1/2}e^{-n(z-1)/2}S(a), \tag{25}\] where \(p_{z,a}\) is the density of \(\|Y^{z,a}\|^2\). This identity is a polar-coordinate identity between continuous radial densities at a positive radius. At \(z=z_a\), its rearrangement is (24). The same density ratio, with the radius fixed, proves the assertion about the conditional law. ◻

Proof. After normalization, it is enough to exclude a sequence of densities vanishing at zero. There are two ways such a sequence of weights can behave: all weights vanish individually, giving a Gaussian limit, or a positive largest weight remains and supplies a density bounded below near zero. The proof treats these cases separately.

Set \(\tau=(\sum_i k_i^2)^{1/2}\) and \(d_i=k_i/\tau\). The centered random variable \[T_d=\sum_i d_i(G_i^2-1) =\frac{\sum_i k_iG_i^2-n}{\tau}\] has density \(f_d\) with \(f_d(0)=\tau p(n)\), and \(\sum_i d_i^2=1\). Thus it suffices to bound \(f_d(0)\) below uniformly over all such finite arrays. Each array has at least three positive weights. Its characteristic function is integrable, since its absolute value decays as \(O(|t|^{-n/2})\) at infinity; we always take the resulting continuous version of its density.

Suppose, to the contrary, that arrays \(d^{(r)}\) satisfy \(f_{d^{(r)}}(0)\to0\). Sort each array in nonincreasing order and extend it by zeros. A diagonal subsequence has \(d_i^{(r)}\to c_i\) for every \(i\), with \(\sum_i c_i^2\le1\). We consider the two possible values of \(c_1\).

Case \(c_1=0\). For \(d\ge0\), define \[\psi_d(t)=\mathbb Ee^{itd(G^2-1)} =e^{-itd}(1-2itd)^{-1/2}.\] For logarithms of characteristic functions, use the continuous branch equal to zero at \(t=0\). For fixed \(t\) and \(|td|\) small, expansion gives \[\log\psi_d(t)=-t^2d^2+O(|t|^3d^3).\] It follows that the characteristic function \(\phi_r(t)=\prod_i\psi_{d_i^{(r)}}(t)\) converges pointwise to \(e^{-t^2}\), because \(\sum_i(d_i^{(r)})^3\le d_1^{(r)}\sum_i(d_i^{(r)})^2=d_1^{(r)}\).

Weak convergence alone does not control a density at one point. To pass through Fourier inversion we need an integrable bound uniform in \(r\). Concavity of \(\log(1+x)\) and \(\sum_i(d_i^{(r)})^2=1\) give \[|\phi_r(t)| =\prod_i\bigl(1+4(d_i^{(r)})^2t^2\bigr)^{-1/4} \le \bigl(1+4(d_1^{(r)})^2t^2\bigr)^{-1/(4(d_1^{(r)})^2)}.\] The function \(x\mapsto \log(1+Ax)/x\) is decreasing for \(A\ge0\). Once \((d_1^{(r)})^2\le1/8\), the last expression is at most \((1+t^2/2)^{-2}\). Fourier inversion and dominated convergence imply \[f_{d^{(r)}}(0) =\frac1{2\pi}\int_{\mathbb R}\phi_r(t)\,dt \longrightarrow \frac1{2\pi}\int_{\mathbb R}e^{-t^2}\,dt =\frac1{2\sqrt\pi},\] a contradiction.

Case \(c_1>0\). Remove the largest coordinate and write \[R_r=\sum_{i\ge2}d_i^{(r)}(G_i^2-1).\] We claim that \(R_r\) converges in distribution to \[ R=\sum_{i\ge2}c_i(G_i^2-1)+Z, \qquad Z\sim N\left(0,\,2\left(1-\sum_{i\ge1}c_i^2\right)\right), \tag{28}\] where \(Z\) and the displayed Gaussians are independent. The series converges in \(L^2\), since the tail beyond \(m\) has variance \(2\sum_{i>m}c_i^2\to0\). To identify the limit, note that the sorted weights satisfy \(d_{m+1}^{(r)}\le(m+1)^{-1/2}\). For fixed \(t\) and sufficiently large \(m\), the same logarithmic expansion as above gives, uniformly in \(r\), \[\log\mathbb Ee^{itR_r} = \sum_{i=2}^m\log\psi_{d_i^{(r)}}(t) -t^2\left(1-\sum_{i=1}^m(d_i^{(r)})^2\right) +O\left(\frac{|t|^3}{\sqrt{m+1}}\right).\] First let \(r\to\infty\) and then \(m\to\infty\). The resulting characteristic function is exactly that of (28), which proves the claim.

The residual limit has positive mass arbitrarily close to zero. This is the fact that allows the surviving largest coordinate to control the convolution density. For any \(\delta>0\), choose a finite truncation of its series whose remaining tail variance is less than \(\delta^2/16\). Each nonzero summand in the finite part has a density positive near zero, so the finite part together with \(Z\) has positive probability in \((-\delta/2,\delta/2)\). Zero summands or a degenerate \(Z\) cause no difficulty. The independent tail has positive probability in the same interval by Chebyshev’s inequality. Intersecting these events proves the assertion. This argument uses square summability of \((c_i)\), not summability.

In particular, by convergence in distribution and the open-set inequality, \[\liminf_{r\to\infty}\mathbb P\{|R_r|<c_1/3\} \ge\mathbb P\{|R|<c_1/3\}>0.\] The density of \(d(G^2-1)\), for \(d>0\), is \[g_d(x)= \frac{e^{-(x+d)/(2d)}}{\sqrt{2\pi d(x+d)}} \,\mathbf1_{\{x>-d\}}.\] For \(d\) in a sufficiently small neighborhood of \(c_1\), this density is bounded below by a positive constant depending on \(c_1\) whenever \(|x|\le c_1/3\). This interval stays strictly inside its support.

Independence now yields the pointwise convolution identity \[f_{d^{(r)}}(0)=\mathbb E\bigl[g_{d_1^{(r)}}(-R_r)\bigr].\] To justify it at this particular point, rather than only almost everywhere, the residual \(R_r\) has a bounded density by (26), using two of its positive coordinates. Convolution of that bounded density with the \(L^1\) function \(g_{d_1^{(r)}}\) is continuous: its change under translation by \(h\) is bounded by the residual density supremum times \(\|g_{d_1^{(r)}}(\,\cdot+h)-g_{d_1^{(r)}}\|_{L^1}\), which tends to zero. It therefore equals the continuous density \(f_{d^{(r)}}\) everywhere. Consequently \[\liminf_{r\to\infty}f_{d^{(r)}}(0) \ge c(c_1)\liminf_{r\to\infty}\mathbb P\{|R_r|<c_1/3\}>0,\] again a contradiction. The constant may depend on the subsequential limit: any positive liminf contradicts the chosen sequence tending to zero. The two cases prove a universal lower bound and hence (27). ◻

Proof. For the full sphere, the radial identity and the density-at-the-mean lemma give both bounds. For a hyperplane, only an upper density bound is needed; interlacing allows us to use the full-space saddle without asserting it is a saddle for the compression.

The chi-square density is \[p_{\chi_m^2}(r) =\frac{r^{m/2-1}e^{-r/2}}{2^{m/2}\Gamma(m/2)}, \qquad r>0.\] Stirling’s formula gives \[c n^{-1/2}\le p_{\chi_m^2}(n)\le C n^{-1/2} \quad\text{for }m=n\text{ or }m=n-1,\quad n\ge3;\] the constants absorb the finitely many small dimensions. By Lemma 10 and \(\bigl(\sum_i k_i(a)^2\bigr)^{1/2}\le\sum_i k_i(a)=n\), we have \(p_a(n)\ge c/n\). By (26) and \(k_i(a)\ge1/13\), we also have \(\sup_r p_a(r)\le C\). Lemma 9 proves the two full-space bounds.

We now pass to a hyperplane. Let \(\Lambda_H\) denote the compression of \(\Lambda\) to \(H\), and write its eigenvalues as \(\theta_1\ge\cdots\ge\theta_{n-1}\). The min–max principle gives \(\lambda_i\ge\theta_i\ge\lambda_{i+1}\). At the full-space minimizer \(z=z_a\), the matrix \(Q_H=z_a I_H-a\Lambda_H\) is positive definite. Since \(a\ge0\), interlacing gives \[ \begin{split} (\det Q_H)^{-1/2} &\le\prod_{i=1}^{n-1}(z_a-a\lambda_i)^{-1/2}\\ &=\left(\prod_{i=1}^n(z_a-a\lambda_i)^{-1/2}\right) \sqrt{z_a-a\lambda_n}. \end{split} \tag{29}\] The last square root is at most \(\sqrt{13}\).

Let \(p_{H,a}\) be the squared-norm density of a Gaussian on \(H\) with covariance \(Q_H^{-1}\). The same polar-coordinate calculation as in (25), now in dimension \(n-1\) but still at squared radius \(n\), yields \[S_H(a) =e^{n(z_a-1)/2}(\det Q_H)^{-1/2} \frac{p_{H,a}(n)}{p_{\chi_{n-1}^2}(n)}.\] Every eigenvalue of \(Q_H\) is at most \(z_a-a\lambda_n\le13\), so all its Gaussian variances are at least \(1/13\). There are at least two of them, and (26) gives \(p_{H,a}(n)\le C\). Combining this bound with (29) proves the assertion. Only an upper density bound is used: the compressed Gaussian need not have mean squared radius \(n\), and \(z_a\) need not minimize its own variational problem. ◻

Proof. We compare the exact saddle with the explicit point \(2+s_0\). The first trace locates the saddle and controls the total increase in variances when moving to it; the second trace bounds the variances at the explicit point. Set \(h_i=(2+s_0-\lambda_i)^{-1}\), whose denominators are positive by the edge estimates. Since \(R(s)=1-\sqrt{s}+O(s)\) as \(s\downarrow0\), Proposition 5 gives, uniformly over the good spectra, \[\sum_i h_i =n\bigl(1-\sqrt{s_0}+o(\sqrt{s_0})\bigr)<n.\] The function \(z\mapsto\sum_i(z-\lambda_i)^{-1}\) is strictly decreasing from infinity on \((\lambda_1,\infty)\). Its value at \(z_1\) is \(n\); hence \(\lambda_1<z_1<2+s_0\) and \(k_i\ge h_i\). Because \(\sum_i k_i=n\), the total increase is exactly \[\sum_i(k_i-h_i) =n-\sum_i h_i =n\sqrt{s_0}\bigl(1+o(1)\bigr).\] A nonnegative vector has Euclidean norm at most the sum of its coordinates. The triangle inequality and the second resolvent bound therefore give \[\begin{split} \left(\sum_i k_i^2\right)^{1/2} &\le \left(\sum_i h_i^2\right)^{1/2} +\sum_i(k_i-h_i)\\ &\le C\sqrt n\,s_0^{-1/4}+Cn\sqrt{s_0} \le Cn\sqrt{s_0}. \end{split}\] For the last step, \(n^{-1/2}s_0^{-3/4}=L_n^{-3/4}\to0\). For \(j\le3\), the edge estimates also give \[0<z_1-\lambda_j\le2+s_0-\lambda_j\le Cs_0,\] which proves the lower bounds on the first three variances. Lemma 10 now gives the asserted lower bound on \(p(n)\).

The three leading variances are needed for a uniform deletion estimate: after deleting any one index \(i\), at least two of the indices \(1,2,3\) remain. Applying (26) to their squares, and then convolving with the other squares, yields \(\sup_r p_{-i}(r)\le Cs_0\). ◻

Proof. Conditioning a single coordinate makes the role of the radial estimates transparent. Its unconditioned Gaussian mass is small, and the ratio of the residual radial density to the full radial density quantifies the cost of conditioning. Let \(\varphi_k\) be the density of \(N(0,k)\). Lemma 9 gives, for \(T>0\), \[\begin{split} \sigma_\Lambda\{|u_1|\le T\} &=\frac1{p(n)} \int_{|y|\le T}\varphi_{k_1}(y)p_{-1}(n-y^2)\,dy\\ &\le \frac{Cs_0}{p(n)}\,\mathbb P\{|Y_1|\le T\}. \end{split}\] The densities of the nonnegative residual sums are taken to be zero at negative arguments. By Lemma 12 and the Gaussian density bound, \[\frac{Cs_0}{p(n)}\le Cn s_0^{3/2}, \qquad \mathbb P\{|Y_1|\le b n^{1/3}\} \le \frac{Cb n^{1/3}}{\sqrt{k_1}} \le Cb n^{1/3}\sqrt{s_0} =Cb\sqrt{L_n}.\] Their product is \(CbL_n^2\), proving (31). For the final assertion, for example, \(L_n=\min\{\log n,b_n^{-1/4}\}\) for all sufficiently large \(n\) has the required properties. ◻

Proof. We reuse the same one-coordinate conditioning, now with the weight \(y^2\). First, the defining density of \(\sigma_\Lambda\) is unchanged by changing the sign of any individual coordinate. This proves the assertions about its mean and off-diagonal covariances. Conditioning as in the preceding proof, now with the nonnegative integrand \(y^2\), gives \[\int u_i^2\,\sigma_\Lambda(du) =\frac1{p(n)} \int_{\mathbb R}y^2\varphi_{k_i}(y)p_{-i}(n-y^2)\,dy \le\frac{Cs_0}{p(n)}\,\mathbb EY_i^2 =\frac{Cs_0 k_i}{p(n)}.\] Write \(m_i=\int u_i^2\,\sigma_\Lambda(du)\). Independence of \(U,V\) and diagonality give \[\mathbb E\left(\frac{U^TV}{n}\right)^2 =\frac1{n^2}\sum_i m_i^2 \le\frac{Cs_0^2}{n^2p(n)^2}\sum_i k_i^2.\] Lemma 10 bounds \(p(n)^{-2}\) by \(C\sum_i k_i^2\), and Lemma 12 bounds the latter sum by \(Cn^2s_0\). The resulting upper bound is \(Cn^2s_0^4=Cn^{-2/3}L_n^4\), as claimed. ◻

Proof. We estimate the derivative of the spherical variational value on both sides of the critical parameter \(a=1\). Close to that parameter the resolvent estimates give a quantitative margin; away from it a bounded-convergence argument suffices. The saddle equation in Lemma 9 and implicit differentiation show that \(z_a\) is continuously differentiable in \(a\). Differentiating the objective at its minimizer gives \[ B'(a)=\frac1{2n}\sum_{i=1}^n \frac{\lambda_i}{z_a-a\lambda_i}, \qquad B'(a)=\frac{z_a-1}{2a}\quad(a>0). \tag{40}\] The first identity also holds at \(a=0\), with a one-sided derivative at the endpoint. Since the positive weights \((z_a-a\lambda_i)^{-1}\) sum to \(n\), \[|B'(a)|\le\frac12\max_i|\lambda_i|\le\frac32 \qquad(0\le a\le2).\] For \(a>0\), put \(r_a=z_a/a\) and \[m_n(s)=\frac1n\sum_i(2+s-\lambda_i)^{-1}.\] The saddle equation is equivalently \[\frac1n\sum_i(r_a-\lambda_i)^{-1}=a.\] At \(s=s_0\), the spectral estimates give \(m_n(s_0)=1-\sqrt{s_0}+o(\sqrt{s_0})<1\). Thus, for \(a=1+u\) with \(0\le u\le1\), monotonicity of this resolvent gives \(r_a<2+s_0\) and hence \[ B'(1+u) \le \frac{1+u}{2}-\frac{u^2}{2(1+u)}+\frac{s_0}{2}. \tag{41}\]

The upper derivative estimate must be paired with a lower estimate below \(a=1\). The small-overlap range requires the following quantitative comparison. If \(4\sqrt{s_0}\le u\le1/4\), set \(t=u^2/(1-u)\). Then \(t\le1/12\) and \(0.9t\ge14.4s_0\). The function \(R\) from Proposition 5 satisfies \[R(t)=1-u,\qquad -R'(s)=\frac{2+s}{2\sqrt{s(4+s)}}-\frac12 \ge \frac{c_1}{\sqrt{s}} \quad(0<s\le1/12)\] for a universal \(c_1>0\). Integrating the last bound from \(0.9t\) to \(t\) therefore gives \[R(0.9t)-(1-u)\ge c_2\sqrt{t}\ge c_2u.\] The error in \(m_n(0.9t)=R(0.9t)+o(\sqrt{t})\) is uniform over the indicated range. It is smaller than this fixed positive margin for all sufficiently large \(n\). Consequently \(m_n(0.9t)>1-u\), so monotonicity implies \(r_{1-u}\ge2+0.9t\). Substitution in (40) yields \[ B'(1-u)\ge\frac{1-u}{2}-\frac{u^2}{20(1-u)} \qquad(4\sqrt{s_0}\le u\le1/4). \tag{42}\] For \(0\le u\le4\sqrt{s_0}\), use instead \(r_{1-u}>\lambda_1=2+o(s_0)\) to obtain \[B'(1-u) \ge\frac{1-u}{2}-\frac{u^2}{2(1-u)}-o(s_0) \ge\frac{1-u}{2}-Cs_0.\]

Write \(F(q)=B(1+q)+B(1-q)-2B(1)\). On \(0\le u\le1/4\), \[\frac1{2(1+u)}-\frac1{20(1-u)}\ge\frac13.\] Combining the preceding bounds, and absorbing \(u^2\le16s_0\) on the short interval \(u\le4\sqrt{s_0}\) into the error, gives \[F'(u)\le u-\frac{u^2}{3}+Cs_0 \qquad(0\le u\le1/4).\] Since \(F(0)=0\), \[ F(q)\le\frac{q^2}{2}-\frac{q^3}{9}+Cs_0q \qquad(0\le q\le1/4). \tag{43}\] For \(q\ge q_0\), the last error divided by \(q^3\) is at most \[C\frac{s_0}{q_0^2}=CL_n n^{-1/12}\longrightarrow0.\] Thus (39) holds on \([q_0,1/4]\), even with \(c=1/18\).

It remains to cover overlaps bounded away from zero. Here a uniform additive error tending to zero is enough. Fix \(a\in(0,1)\). The identity \(R(a+a^{-1}-2)=a\), the strict monotonicity of \(R\), and the spectral resolvent estimate imply \[r_a\longrightarrow a+a^{-1}, \qquad B'(a)\longrightarrow a/2.\] Indeed, evaluating the empirical resolvent at \(a+a^{-1}\pm\delta\), with \(\delta>0\) small enough that the lower point is above \(2\), traps \(r_a\) between these two points for all large \(n\). The upper resolvent parameter needed here is finite for each fixed \(a\). No estimate uniform as \(a\downarrow0\) is used.

The bound \(|B'|\le3/2\) now gives, by dominated convergence, \[\rho_n:=\int_0^1|B'(a)-a/2|\,da\longrightarrow0.\] Integrating (41) and bounding the subcritical error by \(\rho_n\) gives, uniformly for \(0\le q\le1\), \[\begin{align*} F(q) &\le \frac{q^2}{2} -\int_0^q\frac{u^2}{2(1+u)}\,du+\rho_n+\frac{s_0}{2} \\ &\le \frac{q^2}{2}-\frac{q^3}{12}+\rho_n+\frac{s_0}{2}. \tag{44}\end{align*}\] When \(q\ge1/4\), the cubic term is bounded away from zero, so the last two errors are absorbed with \(c=1/24\). This also covers \(q=1\).

The conclusion must hold uniformly over the entire deterministic spectral event. The estimates in the shrinking range already have that uniformity. The same is true of \(\rho_n\to0\): otherwise one could select spectra from \(\mathcal G_n\) along a violating subsequence. Their deterministic spectral bounds would still imply the pointwise convergence and bounded domination just proved, a contradiction. Thus the two ranges give the claimed uniform statement. ◻

Proof. The first-moment identity was established above. We now bound the exact second moment by treating the two overlap regions separately.

Large overlaps. Combining (38), the Chernoff bound \(p_I(q)\le e^{-nq^2/2}\), and Lemma 15 gives \[p_I(q)\frac{M(q)}{S(1)^2} \le Cn^2 e^{-cn|q|^3} \qquad(q\in\mathcal Q_n,\ |q|\ge q_0).\] There are at most \(n+1\) lattice points. Since \(nq_0^3=n^{1/8}\), \[ \sum_{\substack{q\in\mathcal Q_n\\ |q|\ge q_0}} p_I(q)\frac{M(q)}{S(1)^2} \le Cn^3e^{-c n^{1/8}}=: \tau_n. \tag{45}\] In particular, \(\tau_n\) is smaller than every fixed negative power of \(n\).

Central overlaps. The remaining obstacle is that \(M(q)\) may vary too rapidly for a derivative-based Riemann-sum estimate. We compare the unweighted overlap laws locally, then move each lattice mass outward, where \(M\) is larger. Let \(\Delta=2/n\). Stirling’s formula, uniformly for \(|q|<q_0\), gives \[\begin{gather*} p_I(q)= \sqrt{\frac{2}{\pi n(1-q^2)}}\, e^{-nJ(q)}\bigl(1+O(n^{-1})\bigr),\\ J(q)=\frac{(1+q)\log(1+q)+(1-q)\log(1-q)}2. \end{gather*}\] Here \(J(q)=q^2/2+O(q^4)\). Likewise \(c_n=\sqrt{n/(2\pi)}(1+O(n^{-1}))\) and \(\log(1-q^2)=-q^2+O(q^4)\). Dividing the two expressions shows that \[ \frac{p_I(q)}{\Delta f_S(q)} =\exp\!\bigl(O(n^{-1}+q^2+nq^4)\bigr)=1+o(1), \qquad |q|<q_0, \tag{46}\] with a uniform relative error, since \(nq_0^4\to0\).

Assign to each retained lattice point the outward bin \[ \mathcal B_q= \begin{cases} [q,q+\Delta],&q>0,\\ [q-\Delta,q],&q<0,\\ [-\Delta/2,\Delta/2],&q=0. \end{cases} \tag{47}\] For even \(n\), the lattice near zero is \(\ldots,-2\Delta,-\Delta,0,\Delta,2\Delta,\ldots\). The centered zero bin leaves gaps between \(\Delta/2\) and \(\Delta\), and between \(-\Delta\) and \(-\Delta/2\). For odd \(n\), the nearest points are \(-\Delta/2,\Delta/2\), whose outward bins leave the gap \((-\Delta/2,\Delta/2)\). In both cases all bin interiors are disjoint. Every \(r\in\mathcal B_q\) satisfies \(|r|\ge|q|\), and all these bins lie in \([-q_0-\Delta,q_0+\Delta]\subset(-1,1)\) for large \(n\).

On each such bin, \[\left|\log\frac{f_S(r)}{f_S(q)}\right| \le Cn(q_0+\Delta)\Delta\le C(q_0+\Delta),\] because \((\log f_S)'(r)=-(n-3)r/(1-r^2)\). It follows from (46) that there is a deterministic \(\delta_n\to0\) such that, for every nonnegative even function \(h\) nondecreasing in \(|q|\), \[ p_I(q)h(q) \le (1+\delta_n)\int_{\mathcal B_q}h(r)f_S(r)\,dr \qquad(q\in\mathcal Q_n,\ |q|<q_0). \tag{48}\] Indeed, the integral is at least \(\Delta h(q)f_S(q)e^{-C(q_0+\Delta)}\), whereas the lattice mass is at most \((1+e_n)\Delta f_S(q)\) for a deterministic \(e_n\to0\). One can take \(1+\delta_n=(1+e_n)e^{C(q_0+\Delta)}\). This estimate does not depend on how rapidly \(h\) varies. The outermost bin may extend beyond \(q_0\), which causes no difficulty because the comparison integral is over the full interval \([-1,1]\).

Combining the two regions. Apply (48) with \(h=M\). Because the bin interiors are disjoint, their sum is bounded by the full spherical integral. Together with (45) and (37), it yields \[\frac{\mathbb E_O I^2}{S(1)^2}\le1+\delta_n+\tau_n.\] The first-moment identity therefore gives \[ \mathbb E_O\!\left[\left(\frac I{S(1)}-1\right)^2\right] \le\delta_n+\tau_n\longrightarrow0. \tag{49}\] The bin error is independent of the spectrum, and the tail estimate is uniform on \(\mathcal G_n\) by Lemma 15. Thus the relative variance bound is uniform there. The unconditional assertion follows by Chebyshev’s inequality and \(\mathbb P(\mathcal G_n^c)\to0\). ◻

Proof. The restricted partition sum has a small first moment, while the unrestricted sum is close to its spherical mean. We choose the auxiliary spectral event after the prescribed threshold \(b_n\): take \(L_n\to\infty\) with \(L_n\le\log n\) and \(L_n^2b_n\to0\). For example, for all sufficiently large \(n\) one may take \(L_n=\min\{\log n,b_n^{-1/4}\}\). In the representation \(W=O\Lambda O^T\), take \(v=Oe_1\), and define the restricted normalized sum \[I_b=2^{-n}\sum_x \mathbf 1_{\{|v^Tx|\le b_n n^{1/3}\}}e^{x^TWx/2}.\] For every fixed \(x\), \(O^Tx\) is uniform on the sphere and its first coordinate is \(v^Tx\). Consequently Proposition 13 gives, conditional on any spectrum in \(\mathcal G_n\), \[ \mathbb E_O\frac{I_b}{S(1)} =\sigma_\Lambda\{|U_1|\le b_n n^{1/3}\} \le CL_n^2b_n. \tag{51}\] For every fixed \(\varepsilon>0\), Markov’s inequality and Proposition 16 imply \[\mathbb P_O\!\left(\frac{I_b}{I}>\varepsilon\right) \le \mathbb P_O\!\left(\frac{I}{S(1)}<\frac12\right) +\frac{2}{\varepsilon}\mathbb E_O\frac{I_b}{S(1)} \longrightarrow0\] uniformly over the good spectra. The ratio \(I_b/I\) is the desired Gibbs mass, so adding \(\mathbb P(\mathcal G_n^c)\) proves the claim. GOE has a simple spectrum almost surely: the set of matrices with repeated eigenvalues is the zero set of the nonzero discriminant polynomial and has Lebesgue measure zero, whereas GOE has a density. Thus any choice of the unit top eigenvector differs from \(Oe_1\) only by sign, which leaves the asserted event unchanged. ◻

Proof. To prove tightness rather than a bound with a fixed logarithmic loss, keep \(L_n\) free until the tail threshold has been specified. The same second-moment comparison works with the nonnegative overlap weight \(q^2\). Expanding two Gibbs weights gives the exact identity \[ \mathbb E_O\!\left[\left(\frac I{S(1)}\right)^2Q_n\right] =\frac1{S(1)^2}\sum_{q\in\mathcal Q_n}p_I(q)q^2M(q). \tag{53}\] The function \(q^2M(q)\) is nonnegative, even, and nondecreasing in \(|q|\). The central terms therefore satisfy the same bin comparison (48); the tails are at most \(\tau_n\) because \(q^2\le1\). Moreover, \[\frac1{S(1)^2}\int_{-1}^1q^2M(q)f_S(q)\,dq =\mathbb E_{\sigma_\Lambda^{\otimes2}} \left(\frac{U^TV}{n}\right)^2.\] Here \(U,V\) are independent under the tilted spherical law. Lemma 14 bounds this expectation by \(Cn^{-2/3}L_n^4\). Thus (53) gives \[ \mathbb E_O\!\left[\left(\frac I{S(1)}\right)^2 n^{2/3}Q_n\right] \le CL_n^4+n^{2/3}\tau_n \le C'L_n^4 \tag{54}\] for all large \(n\), uniformly on \(\mathcal G_n\).

The weighted estimate is now available for every admissible \(L_n\). Given an arbitrary deterministic \(T_n\to\infty\), choose \(L_n=\min\{\log n,T_n^{1/8}\}\) for all sufficiently large \(n\). Then \(L_n\to\infty\) and \(L_n^4/T_n\le T_n^{-1/2}\to0\). Writing \(\widehat I=I/S(1)\), conditional Markov’s inequality gives \[\mathbb P_O(n^{2/3}Q_n>T_n) \le \mathbb P_O(\widehat I<1/2) +\frac4{T_n}\mathbb E_O[\widehat I^{\,2} n^{2/3}Q_n] \le o(1)+\frac{CL_n^4}{T_n}.\] The first error is uniform on \(\mathcal G_n\) by Proposition 16. Adding \(\mathbb P(\mathcal G_n^c)\to0\) proves (52).

The random variable \(n^{2/3}Q_n\) has one fixed law for each \(n\) throughout this argument: it is a function of the off-diagonal disorder alone. Changing \(L_n\) changes only the auxiliary good event and estimates used to prove the assertion, not this law. To verify explicitly that (52) implies boundedness in probability, suppose tightness failed. Then there would be an \(\varepsilon>0\) and increasing indices \(n_j\) with \[\mathbb P(n_j^{2/3}Q_{n_j}>j)>\varepsilon.\] Define a deterministic staircase by \(T_n=j\) for \(n_j\le n<n_{j+1}\), with an arbitrary positive initial segment. This sequence tends to infinity and contradicts (52) at \(n=n_j\). Conversely, tightness directly implies (52) by comparing a divergent threshold with any fixed threshold. Finally, because the Gibbs law is invariant under adding a diagonal to \(W\), integrating out the auxiliary diagonal preserves the asserted probabilities. ◻

Proof. The upper and lower bounds use different outputs of the transfer: squared overlap bounds the covariance norm from above, while the top projection bounds it from below. Spin-flip symmetry gives \(\mu[x]=0\), so \(\Sigma=\mu[xx^T]\) and \[ \mathop{\mathrm{Tr}}\Sigma^2 =\sum_{i,j}\mu[x_ix_j]^2 =n^2\mathbb E_{\mu^{\otimes2}}q(x,y)^2 =n^2Q_n. \tag{55}\] Since \(\Sigma\) is positive semidefinite, \(\|\Sigma\|^2\le\mathop{\mathrm{Tr}}\Sigma^2\). The upper bound follows from Proposition 18 with threshold \(T_n=M_n^2\).

For the lower bound use Theorem 17 with \(b_n=M_n^{-1/4}\) and put \(B_n=\mu\{|v^Tx|\le b_n n^{1/3}\}\). Then \(B_n\to0\) in probability and \[\|\Sigma\|\ge v^T\Sigma v =\mu[(v^Tx)^2] \ge b_n^2 n^{2/3}(1-B_n).\] On \(B_n\le1/2\), for all sufficiently large \(n\) this is at least \(n^{2/3}/(2\sqrt{M_n})\ge n^{2/3}/M_n\). Both bounds hold simultaneously with probability tending to one by a union bound. The covariance is unchanged by the auxiliary diagonal, so these probabilities are also those of the original SK disorder. ◻

Proof. For either clock, reversibility diagonalizes the mean-square increment in the same basis as the Dirichlet form. In the inner product \(\langle f,h\rangle_\mu=\mu[fh]\), stationarity gives \[\mathbb E_\mu\bigl[(f(X_t)-f(X_0))^2\bigr] =2\langle f,(\mathrm{Id}-e^{t\mathcal L})f\rangle_\mu.\] The operator \(-\mathcal L\) is self-adjoint and nonnegative. Expanding in an orthonormal eigenbasis and using \(1-e^{-t\lambda}\le t\lambda\) gives (56), because \(\langle f,-\mathcal Lf\rangle_\mu=\mathcal D_\mu(f)\) by Lemma 4.

Each conditional expectation \(P_i\) is an orthogonal projection in \(L^2(\mu)\). Thus \(P=n^{-1}\sum_iP_i\) is self-adjoint with spectrum in \([0,1]\). The discrete stationary increment is \(2\langle f,(\mathrm{Id}-P^k)f\rangle_\mu\). For \(0\le\theta\le1\), \(1-\theta^k\le k(1-\theta)\), while \(\langle f,(\mathrm{Id}-P)f\rangle_\mu=\mathcal D_\mu(f)/n\). This proves (57). Finally, Lemma 4 gives \[\mathcal D_\mu(v^Tx) =\sum_{i=1}^n v_i^2\mu[\mathop{\mathrm{Var}}(x_i\mid x_{-i})] \le\sum_{i=1}^n v_i^2=1.\] ◻

Proof of Theorem 1. We prove the continuous-time statement in three steps. Choose a projection threshold above its root-mean-square increment scale, use its sign as a test for each fixed starting state, and average only the resulting intrinsic bad-state indicator. This last step will remove the auxiliary diagonal without averaging away the distance being tested.

Let \(t_n=o(n^{2/3})\) be deterministic and nonnegative. Choose a positive deterministic \(b_n\to0\) such that \[\frac{t_n}{b_n^2n^{2/3}}\longrightarrow0;\] for example, \(b_n=(t_n/n^{2/3})^{1/4}+n^{-1}\) works. Put \(a_n=b_nn^{1/3}\).

A sign that persists. Separate the auxiliary randomness by writing \(W=A+D\), where \(A\) has zero diagonal and \(D\) is the independent Gaussian diagonal. The Gibbs law \(\mu_A=\mu_W\) and the heat-bath transition laws depend only on \(A\). The unit top eigenvector \(v=v(A+D)\) is allowed to depend on both matrices. Set \(g(x)=v^Tx\). Conditional on \(A,D\), start the trajectory from \(\mu_A\). If \(|g(X_0)|>a_n\), failure of the two endpoint projections to have the same nonzero sign requires \(|g(X_{t_n})-g(X_0)|\ge a_n\). Hence Lemma 20 and Markov’s inequality give \[ \mathbb P_{\mu_A}\bigl(g(X_0),g(X_{t_n}) \text{ do not have the same nonzero sign}\mid A,D\bigr) \le \mu_A\{|g|\le a_n\}+\frac{2t_n}{a_n^2}. \tag{58}\] By Theorem 17, the first term tends to zero in probability over \(A,D\). It also tends to zero in expectation, since it lies in \([0,1]\). Therefore the deterministic quantity \[ \varepsilon_n :=\mathbb E_{A,D}\mu_A\{|g|\le a_n\}+\frac{2t_n}{a_n^2} \longrightarrow0 \tag{59}\] bounds the average sign-failure probability.

Testing each realized state. We now fix the initial configuration before defining its test event. For \(x\) with \(g(x)\ne0\), let \[H_{v,x}=\{y:g(y)g(x)>0\},\qquad r_n(A,D,x)=1-(e^{t_n\mathcal L}\mathbf1_{H_{v,x}})(x).\] For \(g(x)=0\), set \(r_n(A,D,x)=1\). This definition includes every zero projection in the failure event, and (58) implies \[\mathbb E_{A,D}\mu_A[r_n(A,D,\cdot)]\le\varepsilon_n.\] Spin-flip symmetry gives \(\mu_A(H_{v,x})\le1/2\). If \(g(x)\ne0\), testing total variation against this half-space yields \[d_{t_n}^{\mathrm{ct}}(A,x)\ge\frac12-r_n(A,D,x).\] When \(g(x)=0\) the following consequence remains valid because \(r_n=1\): \[ \mathbf1_{\{d_{t_n}^{\mathrm{ct}}(A,x)\le1/4\}} \le4r_n(A,D,x). \tag{60}\] Removing the auxiliary diagonal. The test half-space depends on \(D\), but the bad-state indicator on the left does not. We can therefore integrate (60) to obtain \[\mathbb E_A\mu_A\{x:d_{t_n}^{\mathrm{ct}}(A,x)\le1/4\} =\mathbb E_{A,D}\mu_A\{x:d_{t_n}^{\mathrm{ct}}(A,x)\le1/4\} \le4\varepsilon_n.\] In particular, for every fixed \(\delta>0\), \[ \mathbb P_A\!\left( \mu_A\{x:d_{t_n}^{\mathrm{ct}}(A,x)>1/4\}<1-\delta \right) \le\frac{4\varepsilon_n}{\delta}\longrightarrow0. \tag{61}\] This is the asserted convergence in probability over the original off-diagonal disorder. The argument integrates an intrinsic distance; it does not require choosing one auxiliary diagonal that works for a prescribed collection of initial states.

For deterministic integers \(k_n=o(n^{5/3})\), use \(b_n=(k_n/n^{5/3})^{1/4}+n^{-1}\) and the discrete increment bound. The same proof replaces \(2t_n/a_n^2\) by \(2k_n/(na_n^2)\) and proves (61) with \(d_{k_n}^{\mathrm{dt}}\). These are distances conditional on the realized initial configuration. Averaging the transition law itself over a Gibbs initial configuration would instead leave it equal to \(\mu_A\) by stationarity. ◻

Proof. The denominator in the inverse Rayleigh quotient of a linear function is diagonal in its coefficients. We first rewrite the quotient as a weighted covariance norm, then show that the sites with large weights form a sufficiently small exceptional set.

Weighted covariance. For each site define its local field and average conditional variance by \[g_i(x)=\sum_{j\ne i}W_{ij}x_j, \qquad \alpha_i=\mu[\mathop{\mathrm{sech}}^2g_i].\] Thus \(0<\alpha_i\le1\), and Lemma 4 gives \[\mathcal D_\mu(a^Tx)=\sum_i\alpha_i a_i^2, \qquad \mathop{\mathrm{Var}}_\mu(a^Tx)=a^T\Sigma a.\] It follows by the change of variables \(b_i=\sqrt{\alpha_i}a_i\) that \[ \mathcal R_{\mathrm{lin}}(\mu)=\|\Gamma\|, \qquad \Gamma=\mathop{\mathrm{diag}}(\alpha_i^{-1/2})\Sigma\mathop{\mathrm{diag}}(\alpha_i^{-1/2}). \tag{63}\] For every nonzero \(a\), the inequalities \(\alpha_i\le1\) and the nonnegativity of the variance give \[\frac{a^T\Sigma a}{\sum_i\alpha_i a_i^2} \ge \frac{a^T\Sigma a}{\|a\|^2}.\] Taking the supremum proves the lower bound in (62) deterministically.

Deleting one spin. To control a site’s weight, let \(\nu_i\) be the Gibbs law on the other \(n-1\) spins after deleting site \(i\) and all its incident couplings, without rescaling the remaining couplings. Write \(\bar\mu_i\) for the corresponding marginal of \(\mu\) and set \[Z_i=\nu_i[\cosh g_i],\qquad R_i=\mathbb E_{\nu_i\otimes\nu_i} \left[\left(\frac{x_{-i}^Ty_{-i}}{n}\right)^2\right].\] Summing over \(x_i\) gives the exact identities \[\frac{d\bar\mu_i}{d\nu_i}=\frac{\cosh g_i}{Z_i}, \qquad \alpha_i=\frac{\nu_i[(\cosh g_i)^{-1}]}{Z_i} \ge Z_i^{-2},\] where the last step is Jensen’s inequality for the reciprocal function. Since \(\cosh g_i\ge1\), the product marginal comparison and \((q-x_iy_i/n)^2\le2q^2+2n^{-2}\) imply \[ R_i\le Z_i^2\mathbb E_{\bar\mu_i\otimes\bar\mu_i} \left[\left(\frac{x_{-i}^Ty_{-i}}n\right)^2\right] \le2Z_i^2(Q_n+n^{-2}). \tag{64}\]

Let \(\mathcal F_i\) be the sigma-field generated by the matrix with row and column \(i\) deleted. Conditional on \(\mathcal F_i\), the law \(\nu_i\) is fixed and the incident couplings are independent \(N(0,1/n)\) variables. For a fixed spin configuration, \(g_i\) is consequently a centered Gaussian with variance \(\sigma_{\mathrm{loc}}^2=(n-1)/n\). For every fixed integer \(m\ge1\), Jensen’s inequality yields \[\mathbb E[Z_i^m\mid\mathcal F_i] \le \mathbb E_{G\sim N(0,\sigma_{\mathrm{loc}}^2)}[\cosh^mG] \le2e^{m^2/2}.\] Taking, for example, \(m=200\) and using a union bound gives \[ \mathbb P\left(\max_iZ_i>n^{1/100}\right)\longrightarrow0. \tag{65}\] Proposition 18 says that \(n^{2/3}Q_n\) is bounded in probability. Combining it with (64) and (65), we obtain \[ \mathbb P(E_n)\longrightarrow1, \qquad E_n=\left\{\max_iR_i\le r_n\right\}, \qquad r_n=n^{-7/12}. \tag{66}\] For clarity, the exponent slack here is \(1/12-1/50=19/300>0\): on \(\max_iZ_i\le n^{1/100}\) and \(Q_n\le\tfrac14 n^{-2/3+19/300}\), the first term on the right of (64) is at most \(r_n/2\), and its second term is at most \(r_n/2\) for all sufficiently large \(n\).

Counting exceptional sites. The uniform polynomial bound on \(Z_i\) is not by itself enough. We also need most \(Z_i\) to stay below a fixed constant. Conditional on \(\mathcal F_i\), for two configurations of the remaining spins put \(c=x_{-i}^Ty_{-i}/n\). Their two local fields have variances \(\sigma_{\mathrm{loc}}^2\) and covariance \(c\), so \[\mathbb E[\cosh g_i(x)\cosh g_i(y)\mid\mathcal F_i] =e^{\sigma_{\mathrm{loc}}^2}\cosh c.\] It follows that \[\begin{align*} \mathbb E[Z_i\mid\mathcal F_i]&=e^{\sigma_{\mathrm{loc}}^2/2},\tag{67}\\ \mathop{\mathrm{Var}}(Z_i\mid\mathcal F_i) &=e^{\sigma_{\mathrm{loc}}^2}\mathbb E_{\nu_i\otimes\nu_i}[\cosh c-1] \le C R_i, \tag{68}\end{align*}\] because \(|c|\le1\) and \(\cosh c-1\le Cc^2\) on this interval.

The event \(E_n\) depends on many incident rows. To use (68) with its stated conditioning, introduce the clipped count \[N_n=\sum_{i=1}^n \mathbf1_{\{R_i\le r_n\}}\mathbf1_{\{Z_i>4\}}.\] The variable \(R_i\) is \(\mathcal F_i\)-measurable, and \(4-e^{\sigma_{\mathrm{loc}}^2/2}\ge4-e^{1/2}>0\). Thus conditional Chebyshev gives \[\mathbb EN_n =\sum_i\mathbb E\!\left[ \mathbf1_{\{R_i\le r_n\}} \mathbb P(Z_i>4\mid\mathcal F_i)\right] \le Cnr_n=Cn^{5/12}.\] On \(E_n\), the clipped count equals \(|B|\), where \(B=\{i:Z_i>4\}\). Consequently, \[ \mathbb P(|B|>n^{1/2}) \le\mathbb P(E_n^c)+\mathbb P(N_n>n^{1/2}) \le\mathbb P(E_n^c)+Cn^{-1/12}\longrightarrow0. \tag{69}\] No independence between sites, or conditioning on \(E_n\) in the Gaussian calculation, is used.

Combining the covariance blocks. The probabilistic work is complete. Work on the event \(\{\max_iZ_i\le n^{1/100},\ |B|\le n^{1/2}\}\), whose probability tends to one by (65) and (69). Let \(G=B^c\). On \(G\), \(\alpha_i^{-1/2}\le Z_i\le4\), so the principal blocks of the positive semidefinite matrix \(\Gamma\) satisfy \[\|\Gamma_{GG}\|\le16\|\Sigma\|, \qquad \|\Gamma_{BB}\|\le\mathop{\mathrm{Tr}}\Gamma_{BB} =\sum_{i\in B}\alpha_i^{-1} \le |B|\max_i Z_i^2\le n^{13/25}.\] Here \(\Sigma_{ii}=1\) by spin-flip symmetry. Positivity also controls the cross block: for \(u\in\mathbb R^G\) and \(w\in\mathbb R^B\), \[|u^T\Gamma_{GB}w|^2 \le(u^T\Gamma_{GG}u)(w^T\Gamma_{BB}w).\] This follows by applying nonnegativity to the quadratic form at \((u,sw)\) for all real \(s\). If \(a=\|\Gamma_{GG}\|\) and \(b=\|\Gamma_{BB}\|\), it follows that \[(u,w)^T\Gamma(u,w) \le(\sqrt a\,\|u\|+\sqrt b\,\|w\|)^2 \le(a+b)(\|u\|^2+\|w\|^2).\] The same bound holds when either block is empty. Therefore \(\|\Gamma\|\le16\|\Sigma\|+n^{13/25}\), proving (62). Corollary 19 and \(13/25<2/3\) give the final assertion. ◻

Proof of Theorem 2. The factor in the linear comparison is fixed, so it can be absorbed while preserving every divergent threshold. Let \(M_n\to\infty\) be positive and deterministic. Apply Corollary 19 with the divergent threshold \(M_n/32\), and intersect its event with that of Proposition 21. With probability tending to one, \[\frac{32n^{2/3}}{M_n}\le\|\Sigma\| \le\frac{M_n n^{2/3}}{32}, \qquad \|\Sigma\|\le\mathcal R_{\mathrm{lin}}(\mu) \le\frac{M_n n^{2/3}}2+n^{13/25}.\] Since \(13/25<2/3\), the final term is at most \(M_nn^{2/3}/2\) for all sufficiently large \(n\). These inequalities give the claimed common interval for both quantities. ◻

Proof. The comparison works directly with least-squares characterizations; one need not compare the conditional kernels pointwise. For any full-support probability measure \(\eta\), \[ \mathop{\mathrm{Var}}_\eta(f)=\inf_{c\in\mathbb R}\eta[(f-c)^2], \qquad \mathcal D_\eta(f) =\sum_i\inf_{h_i}\eta[(f-h_i(x_{-i}))^2], \tag{71}\] where every \(h_i\) ranges independently over all functions of the other coordinates. For the second identity, conditional expectation gives the orthogonal decomposition \[\eta[(f-h_i)^2] =\eta[(f-\eta[f\mid x_{-i}])^2] +\eta[(\eta[f\mid x_{-i}]-h_i)^2].\] The first identity follows by the same decomposition with constants. The density bounds and (71) imply \[\mathop{\mathrm{Var}}_\rho(f)\le M\mathop{\mathrm{Var}}_\nu(f), \qquad \mathcal D_\rho(f)\ge m\mathcal D_\nu(f),\] which prove (70). For the exponential densities, \(\rho(x)/\nu(x)\) is a constant times \(e^{(1-\beta)H(x)}\); the common normalization cancels in the ratio of its maximum and minimum. ◻

Proof. We keep the order of parameters explicit: first the requested exponential tolerance, then a fixed subcritical temperature, then the dimension. Fix \(\epsilon>0\) and choose \(\beta\in(0,1)\) such that \(3(1-\beta)<\epsilon\). Let \(\mu_\beta(x)\propto e^{\beta H_W(x)}\). The zero-diagonal part of \(\beta W\) satisfies exactly the hypotheses of Theorem 22; omitting its diagonal changes the Hamiltonian by a constant. With probability tending to one, \[\mathop{\mathrm{Var}}_{\mu_\beta}(f)\le C_\beta\mathcal D_{\mu_\beta}(f) \qquad\text{for every }f.\] Intersect this event with \(\|W\|\le3\), whose probability tends to one by Proposition 5. On their intersection, \[\mathop{\mathrm{osc}}H_W\le3n, \qquad \mu_{\min}:=\min_x\mu(x) \ge2^{-n}e^{-\mathop{\mathrm{osc}}H_W}\ge e^{-(3+\log2)n}.\] Lemma 23 now gives \[ \mathop{\mathrm{Var}}_\mu(f)\le C_n\mathcal D_\mu(f), \qquad C_n=C_\beta e^{3(1-\beta)n}. \tag{72}\] We may increase \(C_\beta\) to be at least one.

The comparison has produced a full Poincaré inequality. To obtain worst-start total variation for both clocks, combine its spectral decay with the lower bound on the smallest Gibbs mass. We give the conversion explicitly so that the factor \(n\) in attempted updates remains visible. By Lemma 4, (72) gives a continuous-time gap at least \(1/C_n\) and a discrete-time gap at least \(1/(nC_n)\). Since \(P\) is an average of orthogonal projections, its spectrum is nonnegative. Spectral decomposition on the mean-zero subspace therefore yields \[\|e^{t\mathcal L}h\|_{L^2(\mu)} \le e^{-t/C_n}\|h\|_{L^2(\mu)}, \qquad \|P^kh\|_{L^2(\mu)} \le e^{-k/(nC_n)}\|h\|_{L^2(\mu)}.\] For a starting configuration \(x\), take \(h_x(y)=\mathbf1_{\{y=x\}}/\mu(x)-1\), whose squared norm is \(\mu(x)^{-1}-1\). Reversibility identifies the transition density relative to \(\mu\), minus one, with \(e^{t\mathcal L}h_x\) or \(P^kh_x\). The \(L^1\) expression for total variation and Cauchy–Schwarz give \[\begin{align*} \max_x d_t^{\mathrm{ct}}(W,x) &\le\tfrac12\mu_{\min}^{-1/2}e^{-t/C_n},\tag{73}\\ \max_x d_k^{\mathrm{dt}}(W,x) &\le\tfrac12\mu_{\min}^{-1/2}e^{-k/(nC_n)}. \tag{74}\end{align*}\] With \(B_n=\log2+\tfrac12(3+\log2)n\), these bounds imply \[t_{\mathrm{mix}}\le C_nB_n, \qquad k_{\mathrm{mix}}\le\lceil nC_nB_n\rceil.\] The two right-hand sides grow at most as a fixed constant depending on \(\beta\) times \(n^2e^{3(1-\beta)n}\), plus the rounding term. Since \(\beta\) is fixed and \(3(1-\beta)<\epsilon\), both are at most \(e^{\epsilon n}\) for all sufficiently large \(n\). The two events used above have an intersection of probability tending to one without any independence assumption. Finally, the mixing times depend only on the off-diagonal disorder, so averaging out the auxiliary diagonal preserves the conclusion. ◻

Proof of Theorem 3. The independent upper comparison is now proved. It remains to extract worst-start lower bounds from the stronger realized-start statement. Fix \(\epsilon>0\) and apply Theorem 1 to \[t_n=n^{2/3-\epsilon},\qquad k_n=\lfloor n^{5/3-\epsilon}\rfloor.\] For each clock, with probability tending to one there is a starting configuration whose distance at the displayed time exceeds \(1/4\). Worst-start distance is nonincreasing in time by contraction of total variation under a Markov kernel. Hence \(t_{\mathrm{mix}}\ge t_n\) and \(k_{\mathrm{mix}}>k_n\) on those events. Since \(k_{\mathrm{mix}}\) is an integer, the latter implies \(k_{\mathrm{mix}}\ge n^{5/3-\epsilon}\), including when \(k_n=0\). Intersect these two lower-bound events with the event in Proposition 24 to obtain all four asserted bounds simultaneously. ◻