64c3814
==== TABLE OF CONTENTS ====
Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/fixed-depth-crossing.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/polynomial-coverage.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//figures/narrow-increments.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//figures/protected-block.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//main.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//math-glyph-unicode.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//preamble.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//references.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/01-introduction.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/02-gaussian.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/03-scalar.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/04-locking.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/05-energy.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/07-coverage.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/08-narrow-locking.tex
Stretched-exponential-barriers-for-typical-SK-initial-states//sections/09-dynamics.tex
===========================
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/fixed-depth-crossing.tex ====
\section{A fixed number of crossings on polynomial time scales}
\label{fixed-sec:crossings}
The polynomial coverage estimate in
Proposition~\ref{poly-prop:coverage} has a direct dynamical application.
Together with the absolute-overlap locking estimate of
Proposition~\ref{static-prop:locking}, it rules out mixing on each fixed
polynomial time scale by using a fixed number of nested crossings.
We give the argument in full. Its number of levels and threshold
intervals remain fixed as $n$ grows.
\begin{theorem}[Polynomial-time obstruction from fixed-depth crossings]
\label{fixed-thm:polynomial}
For every fixed $\beta>1$ and every fixed $a>0$,
\begin{align}
\PP_{J,\,x_0\sim\pi^J}
\left\{d^J(x_0,n^a)>\frac14\right\}&\longrightarrow1,
\label{fixed-eq:continuous-conclusion}\\
\PP_{J,\,x_0\sim\pi^J}
\left\{d_{\mathrm{disc}}^J(x_0,\lfloor n^a\rfloor)
>\frac14\right\}
&\longrightarrow1.
\label{fixed-eq:discrete-conclusion}
\end{align}
Consequently both
$\PP\{T^J(x_0)>n^a\}$ and
$\PP\{T_{\mathrm{disc}}^J(x_0)>n^a\}$ tend to one under the same joint
law. Each assertion is equivalently a statement that the Gibbs mass of
the indicated initial states tends to one in probability over $J$.
\end{theorem}
As in the main theorem, the transition law in each distance starts from
the realized state $x_0$. The proof uses stationarity to sample possible
targets along a path, and then uses total variation from that fixed
initial state to force a sign passage. Independent random thresholds
will make such a passage unlikely.
\subsection{A stationary path and fixed overlap scales}
\label{fixed-subsec:path-scales}
Fix $\beta>1$ and $a>0$, and abbreviate $\pi^J$ to $\pi$. Put
\begin{equation}\label{fixed-eq:time-parameters}
t_n=n^a,\qquad m=\lceil2n^{a+1}\rceil,\qquad
e_*=\frac1{100},\qquad \delta_n=n^{-e_*}.
\end{equation}
Conditional on $J$, start $x_0$ with law $\pi$ and generate an infinite
discrete heat-bath chain $(x_k)_{k\ge0}$. At each step choose a uniform
site and use fresh resampling randomness, including possible
self-transitions. Let $N(t)$ be an independent rate-$n$ Poisson process.
The process $x_{N(t)}$ is the continuous-time chain with rate one at each
site. Indeed, if $K_i^J$ denotes resampling at site $i$, then
$K^J=n^{-1}\sum_iK_i^J$ and its generator is
$\sum_i(K_i^J-I)=n(K^J-I)$. Therefore
\[
\EE[(K^J)^{N(t)}]=\exp\{nt(K^J-I)\}=P_t^J.
\]
Stationarity gives, for every deterministic $k$,
\begin{equation}\label{fixed-eq:stationary-marginal}
\mathcal L(x_k\mid J)=\pi.
\end{equation}
Moreover, Chernoff's inequality at parameter $\log2$, with
$\EE N(t_n)=n^{a+1}$ and $m\ge2n^{a+1}$, gives
\begin{equation}\label{fixed-eq:poisson-tail}
\eta_n:=\PP\{N(t_n)>m\}
\le\exp\{-(2\log2-1)n^{a+1}\}=o(1).
\end{equation}
Use the common threshold $q_*=q_*(\beta)>0$ of
Proposition~\ref{poly-prop:coverage}. With the same energy cutoff $e_1$
as in the main coverage argument, its choice satisfies
\[
g=\frac{\beta^2}{4}-\frac{\beta e_1}{2}>0,\qquad
0<q_*<\min\{1,g/(8\beta^2)\}.
\]
Choose an integer $p\ge2$ such that
\begin{equation}\label{fixed-eq:depth-count}
pe_*>a+2.
\end{equation}
This integer is fixed throughout the limit. First choose $t_p>0$ with
$2t_p<q_*/2$. For $j=p,p-1,\ldots,2$, take a cutoff
$b_j\in(0,q_*/2)$ allowed by Proposition~\ref{static-prop:locking}
for large-overlap threshold $h=t_j/2$, and then choose $t_{j-1}>0$ with
\begin{equation}\label{fixed-eq:separated-scales}
t_{j-1}<t_j/10,\qquad 4t_{j-1}<b_j.
\end{equation}
Thus $0<t_1<\cdots<t_p$, and all scale gaps are fixed positive
constants. Independently of the disorder and of all other randomness
introduced below, draw
\begin{equation}\label{fixed-eq:thresholds}
r_j\sim\operatorname{Unif}[t_j,2t_j],\qquad 1\le j\le p,
\end{equation}
independently for different $j$.
\subsection{Polynomial reference banks cover every possible target}
\label{fixed-subsec:banks}
We next sample finite lists of Gibbs configurations. These lists provide
references for the later crossing construction. Their sizes are chosen
so that coverage holds for every earlier reference choice and every
deterministic index of the path segment $x_0,\ldots,x_m$.
Conditional on $J$, all entries in all banks are independent with law
$\pi$, and all banks are independent of the entire chain. Choose their
sizes recursively. Having chosen $D_{3,i}$ for $i<j$, set
\[
B_j=\sum_{i<j}(D_{3,i}+1),\qquad B_1=0,
\]
choose $D_{1,j}>B_j+a+3$, and use
Corollary~\ref{poly-cor:coverage-cells} with exponent $D_{1,j}$ to obtain
$D_{3,j}>0$. Bank $j$ consists of
\begin{equation}\label{fixed-eq:bank-sizes}
M_j=\lceil n^{D_{3,j}+1}\rceil
\quad\hbox{entries}\quad V_{j,1},\ldots,V_{j,M_j}.
\end{equation}
The constants are chosen before any bank entries or thresholds are
sampled. In particular, the number of tuples with one entry from each
of the first $j-1$ banks satisfies
\begin{equation}\label{fixed-eq:tuple-count}
Q_j:=\prod_{i<j}M_i\le2^{j-1}n^{B_j}.
\end{equation}
For the empty product, $Q_1=1$.
For such a tuple $\boldsymbol v=(v_1,\ldots,v_{j-1})$, define its cell by
\begin{equation}\label{fixed-eq:cell}
S_j(\boldsymbol v)
=\{z\in\Sig: |R(v_i,z)|\le3t_i\text{ for every }i<j\}.
\end{equation}
In particular $S_1=\Sig$. The cell restricts the small overlaps with
the earlier references, while coverage will supply a large overlap
with the current target inside that cell.
\begin{lemma}[Simultaneous bank coverage]
\label{fixed-lem:bank-coverage}
With probability $1-o(1)$, for every $1\le j\le p$, every tuple
$\boldsymbol v$ from banks $1,\ldots,j-1$, and every $0\le k\le m$,
\[
x_k\in S_j(\boldsymbol v)
\quad\Longrightarrow\quad
\text{there is }\ell\le M_j\text{ such that }
V_{j,\ell}\in S_j(\boldsymbol v),\quad
|R(V_{j,\ell},x_k)|\ge q_*.
\]
This event depends only on $J$, the chain and the bank lists, and is
independent of the thresholds.
\end{lemma}
\begin{proof}
Intersect the $p$ disorder events from
Corollary~\ref{poly-cor:coverage-cells}. Since $p$ is fixed, the
intersection has probability $1-o(1)$. On it, for every $S\subseteq\Sig$
the set
\[
\mathcal B_j(S)=
\{x\in S: \pi\{z\in S:|R(z,x)|\ge q_*\}<n^{-D_{3,j}}\}
\]
has $\pi$-mass less than $n^{-D_{1,j}}$.
Fix a disorder in this intersection and condition on the complete first
$j-1$ banks. Each cell in \eqref{fixed-eq:cell} is then fixed. The path
remains independent of these banks conditional on $J$, so
\eqref{fixed-eq:stationary-marginal} applies to every deterministic
$k$. A union bound over all tuples and all indices gives
\begin{equation}\label{fixed-eq:bad-target-bound}
\PP\!\left\{
x_k\in\mathcal B_j(S_j(\boldsymbol v))
\text{ for some }\boldsymbol v,k
\,\middle|\, J,\text{ banks }1,\ldots,j-1\right\}
\le Q_j(m+1)n^{-D_{1,j}}.
\end{equation}
This bound uses marginal stationarity, without requiring independence
between different times.
Now condition also on the entire chain. For a target $x_k$ in its cell
but outside its bad set, acceptable references have $\pi$-mass at least
$n^{-D_{3,j}}$. The entries of bank $j$ are still independent Gibbs
draws under this conditioning. The chance that this bank misses that
set is at most
\[
(1-n^{-D_{3,j}})^{M_j}
\le\exp(-M_jn^{-D_{3,j}})\le e^{-n}.
\]
Another union bound gives $Q_j(m+1)e^{-n}$. Combining the two bounds
and summing over the fixed number of banks, the total failure
probability is at most
\begin{equation}\label{fixed-eq:coverage-failure}
o(1)+\sum_{j=1}^pQ_j(m+1)
(n^{-D_{1,j}}+e^{-n})=o(1).
\end{equation}
Indeed $Q_j(m+1)=O(n^{B_j+a+1})$ and
$D_{1,j}>B_j+a+3$. Neither part of the event uses any threshold.
\end{proof}
\begin{lemma}[Simultaneous bank locking]
\label{fixed-lem:bank-locking}
With probability $1-o(1)$, for every $i<j\le p$, every entry $v$ of
bank $i$, every entry $w$ of bank $j$, and every $0\le k\le m$,
\begin{equation}\label{fixed-eq:all-bank-locking}
\begin{gathered}
|R(x_k,w)|\ge t_j/2,\qquad
|R(v,x_k)|,\ |R(v,w)|\le b_j
\\
\Longrightarrow\qquad
\bigl||R(v,x_k)|-|R(v,w)|\bigr|\le2\delta_n.
\end{gathered}
\end{equation}
\end{lemma}
\begin{proof}
For fixed bank indices and deterministic $k$, the triple $(v,x_k,w)$
has law $\pi^{\otimes3}$ conditional on $J$. This follows from the
independence of the two distinct banks and their independence from the
stationary chain. Proposition~\ref{static-prop:locking}, with
$h=t_j/2$ and cutoff $b_j$, bounds the averaged failure probability by
$e^{-n^{9/10}}$. There are only finitely many choices of $j$ for the
large-$n$ threshold in that proposition. A union bound therefore gives
the failure estimate
\begin{equation}\label{fixed-eq:locking-failure}
\sum_{i<j}(m+1)M_iM_j e^{-n^{9/10}}=o(1).
\end{equation}
Every bank exponent and $p$ are fixed, so the prefactor is polynomial.
\end{proof}
Let $\mathcal G_n$ be the intersection of the events in the two lemmas.
Then
\begin{equation}\label{fixed-eq:good-event}
\PP(\mathcal G_n^c)=o(1).
\end{equation}
It is determined by $J,x_0,\ldots,x_m$ and the bank lists. We have now
obtained coverage and locking simultaneously for every choice the
crossing construction can make; the thresholds remain independent of
this event.
\subsection{A measurable construction of nested crossings}
\label{fixed-subsec:construction}
For a reference $v\in\Sig$, write $D_v(k)=|R(v,x_k)|$. A single
update changes at most one spin, so
\begin{equation}\label{fixed-eq:one-step}
|D_v(k+1)-D_v(k)|\le2/n
\qquad\text{for every }v,k.
\end{equation}
We select the first acceptable entry in each bank, in its given order.
To keep every selected reference defined even on exceptional events,
we specify both successful selections and defaults.
At level $1$, choose $v_1$ as the first entry of bank $1$ satisfying
$D_{v_1}(0)\ge q_*$. If there is no such entry, set $v_1=V_{1,1}$,
$\tau_1=0$, and declare the construction inactive. If selection
succeeds, set
\begin{equation}\label{fixed-eq:first-crossing}
\tau_1=\min\{0\le k\le m:D_{v_1}(k)\le r_1\}
\end{equation}
when the set is nonempty, and declare the construction active. If that
set is empty, set $\tau_1=0$ and declare it inactive.
Suppose $j\ge2$ and the earlier levels are defined. If the construction
is active and $x_{\tau_{j-1}}\in S_j(v_1,\ldots,v_{j-1})$, select
$v_j$ as the first entry of bank $j$ in that cell with
\[
|R(v_j,x_{\tau_{j-1}})|\ge q_*.
\]
After a successful selection, search backwards from $\tau_{j-1}$:
\begin{equation}\label{fixed-eq:backward-crossing}
\tau_j=\max\{0\le k\le\tau_{j-1}:D_{v_j}(k)\le r_j\}
\end{equation}
if this set is nonempty, keeping the construction active. If it is
empty, retain the selected $v_j$, set $\tau_j=0$, and declare the
construction inactive. If the construction was already inactive, the
target was outside its cell, or the reference selection failed, set
$v_j=V_{j,1}$ and $\tau_j=0$, and declare it inactive. An inactive
construction stays inactive at all later levels. At every successful
selection, the depth at the target is at least $q_*>2t_j\ge r_j$;
hence any successful backwards crossing has $\tau_j<\tau_{j-1}$.
The information used in these choices is central to the proof. Define
\[
\mathcal W=\sigma(J,x_0,\ldots,x_m,\text{ all bank lists}),\qquad
\mathcal F_0=\mathcal W,\qquad
\mathcal F_j=\mathcal W\vee\sigma(r_1,\ldots,r_j).
\]
Induction through the explicit rules above gives
\begin{equation}\label{fixed-eq:filtration}
v_j\text{ is }\mathcal F_{j-1}\text{-measurable},\qquad
\tau_j\text{ and the active flag after level }j
\text{ are }\mathcal F_j\text{-measurable}.
\end{equation}
In particular, neither the selection of $v_j$ nor any default for it
uses $r_j$ or a later threshold.
Let $C_1$ denote successful selection of the first reference. On $C_1$
define $s_1=\operatorname{sgn}R(v_1,x_0)\in\{-1,1\}$ and let
\begin{equation}\label{fixed-eq:sign-passage}
\mathcal S_n=C_1\cap
\{\text{there exists }0\le k\le m: s_1R(v_1,x_k)\le0\}.
\end{equation}
Take this event to be false outside $C_1$. Starting from signed
overlap at least $q_*$, a sign passage must visit absolute overlap at
most $2/n$, because signed overlaps also change by at most $2/n$.
For sufficiently large $n$, it therefore forces the first crossing
in \eqref{fixed-eq:first-crossing}.
\begin{lemma}[Deterministic nesting]
\label{fixed-lem:nested-crossings}
For all sufficiently large $n$, on $\mathcal G_n\cap\mathcal S_n$
every reference selection and every crossing succeeds. For every
$1\le j\le p$,
\begin{equation}\label{fixed-eq:nested-shells}
|D_{v_j}(\tau_j)-r_j|\le2/n,\qquad
|D_{v_i}(\tau_j)-r_i|\le6j\delta_n
\quad(1\le i\le j).
\end{equation}
\end{lemma}
\begin{proof}
At level $1$, simultaneous coverage ensures successful selection and
$\mathcal S_n$ ensures a crossing. The initial depth exceeds $r_1$.
Minimality of $\tau_1$ and \eqref{fixed-eq:one-step} therefore put its
depth in $[r_1-2/n,r_1]$, proving the assertions at that level.
Assume the assertions through level $j-1$, where $j\ge2$. For every
$i<j$,
\[
D_{v_i}(\tau_{j-1})\le2t_i+6(j-1)\delta_n<3t_i
\]
for large $n$. Thus the target is in its cell, and simultaneous coverage
supplies an entry $v_j$ in that cell with new depth at least $q_*$.
Both $D_{v_i}(\tau_{j-1})$ and $|R(v_i,v_j)|$ are at most $3t_i$,
which is less than $b_j$ because $3t_i\le3t_{j-1}<b_j$. The new
depth at the target is at least $q_*>t_j/2$. Simultaneous locking gives
\begin{equation}\label{fixed-eq:reference-depth}
\bigl||R(v_i,v_j)|-r_i\bigr|
\le\bigl|D_{v_i}(\tau_{j-1})-r_i\bigr|+2\delta_n
\le(6(j-1)+2)\delta_n.
\end{equation}
We next verify the locking hypotheses throughout the backwards search.
At any visited index $k$ where the new depth is at least $t_j/2$ and
the old depths are at most $b_j$, locking implies
\[
D_{v_i}(k)\le |R(v_i,v_j)|+2\delta_n
\le3t_i+2\delta_n.
\]
The gap $b_j-3t_{j-1}$ is a fixed positive number. Hence
\eqref{fixed-eq:one-step} ensures that every old depth remains below
$b_j$ at the next backwards step, for all sufficiently large $n$.
Before a crossing, the new depth exceeds $r_j\ge t_j$; at the first
crossing step it is at least $r_j-2/n\ge t_j/2$. Starting from the
target, these observations prove by induction over the backwards
steps that locking continues to apply up to and including the first
crossing. If no crossing exists, the induction reaches index $0$.
Throughout that portion of the search, locking and
\eqref{fixed-eq:reference-depth} give, for $i<j$,
\begin{equation}\label{fixed-eq:old-depths}
|D_{v_i}(k)-r_i|\le(6(j-1)+4)\delta_n.
\end{equation}
A crossing must occur: otherwise this inequality at $k=0$ and $i=1$
would imply
\[
q_*\le D_{v_1}(0)
\le2t_1+(6(j-1)+4)\delta_n<q_*.
\]
At the backwards crossing, the next index towards the target has depth
strictly greater than $r_j$. Consequently
$D_{v_j}(\tau_j)\in[r_j-2/n,r_j]$ by the one-step bound.
Combining this with \eqref{fixed-eq:old-depths} proves
\eqref{fixed-eq:nested-shells}. Since $p$, $q_*$, and all $t_i,b_j$ are fixed,
one sufficiently large value of $n$ makes every buffer used in the
induction valid simultaneously.
\end{proof}
\subsection{Independent thresholds make the final shells unlikely}
\label{fixed-subsec:threshold-estimate}
The deterministic construction has reduced a sign passage to a single
time at which all selected reference depths lie in narrow intervals
about their thresholds. Define this event, using the references from
the construction including its defaults, by
\begin{equation}\label{fixed-eq:shell-event}
\mathcal A_n=
\{\text{there exists }0\le k\le m:
|D_{v_j}(k)-r_j|\le6p\delta_n
\text{ for every }1\le j\le p\}.
\end{equation}
Lemma~\ref{fixed-lem:nested-crossings}, applied at $\tau_p$, shows that
\begin{equation}\label{fixed-eq:sign-implies-shells}
\mathcal G_n\cap\mathcal S_n\subseteq\mathcal A_n.
\end{equation}
Fix a deterministic $k$ and set
$E_{j,k}=\{|D_{v_j}(k)-r_j|\le6p\delta_n\}$.
By \eqref{fixed-eq:filtration}, its center $D_{v_j}(k)$ is
$\mathcal F_{j-1}$-measurable, while $r_j$ is independent of
$\mathcal F_{j-1}$ and uniform on an interval of length $t_j$.
Thus
\begin{equation}\label{fixed-eq:one-threshold-bound}
\PP(E_{j,k}\mid\mathcal F_{j-1})
\le\min\{1,12p\delta_n/t_j\}.
\end{equation}
The earlier events $E_{i,k}$, $i<j$, are
$\mathcal F_{j-1}$-measurable. Applying conditional expectation
successively for $j=p,p-1,\ldots,1$ therefore gives
\[
\PP\!\left(\bigcap_{j=1}^pE_{j,k}\,\middle|\,\mathcal W\right)
\le\prod_{j=1}^p\min\{1,12p\delta_n/t_j\}.
\]
The centers may depend on earlier thresholds; the displayed filtration
is exactly what allows this product estimate. A union bound over the
deterministic indices $k=0,\ldots,m$ now yields
\begin{equation}\label{fixed-eq:simultaneous-shell-bound}
\PP(\mathcal A_n)
\le(m+1)\prod_{j=1}^p\frac{12p\delta_n}{t_j}
\le C_{\beta,a}n^{a+1-pe_*}=o(1).
\end{equation}
The exponent is less than $-1$ by \eqref{fixed-eq:depth-count}.
\subsection{A fast realized initial state forces a sign passage}
\label{fixed-subsec:tv-conclusion}
\begin{proof}[Proof of Theorem~\ref{fixed-thm:polynomial}]
First consider continuous time and write
\[
q_n=\PP_{J,\,x_0\sim\pi^J}
\{d^J(x_0,t_n)\le1/4\}.
\]
The first reference is chosen using only $J,x_0$ and bank $1$.
Conditional on these variables, the future chain retains its transition
law from the realized $x_0$. On $C_1$, define
\[
A=A(J,x_0,v_1)=\{z\in\Sig:s_1R(v_1,z)\le0\}.
\]
Spin reversal preserves $\pi$ and interchanges positive and negative
overlaps, so $\pi(A)\ge1/2$. For any realized $J,x_0$ with
$d^J(x_0,t_n)\le1/4$, and any bank $1$ for which $C_1$ holds, the
definition of total variation gives
\[
\PP\{x_{N(t_n)}\in A\mid J,x_0,\text{bank }1\}\ge1/4.
\]
If $N(t_n)\le m$, this endpoint event implies $\mathcal S_n$.
The Poisson process is independent of all the conditioning variables,
so the error in \eqref{fixed-eq:poisson-tail} is uniform under this
conditioning. Consequently
\begin{align}
\PP(\mathcal S_n)
&\ge(1/4-\eta_n)
\PP\{d^J(x_0,t_n)\le1/4,\ C_1\}\notag\\
&\ge q_n/4-\eta_n-\tfrac14\PP(C_1^c).
\label{fixed-eq:fast-start-sign-bound}
\end{align}
The first-bank case of Lemma~\ref{fixed-lem:bank-coverage} gives
$\PP(C_1^c)=o(1)$. Adding the independent thresholds does not change
any probability in this comparison.
Combining \eqref{fixed-eq:good-event},
\eqref{fixed-eq:sign-implies-shells},
\eqref{fixed-eq:simultaneous-shell-bound}, and
\eqref{fixed-eq:fast-start-sign-bound} gives the explicit estimate
\begin{equation}\label{fixed-eq:fast-start-final-bound}
\frac{q_n}{4}
\le\eta_n+\tfrac14\PP(C_1^c)+\PP(\mathcal G_n^c)
+C_{\beta,a}n^{a+1-pe_*}=o(1).
\end{equation}
This proves \eqref{fixed-eq:continuous-conclusion}.
For discrete time set $k_n=\lfloor n^a\rfloor\le m$ and
\[
q_n^{\mathrm{disc}}
=\PP_{J,\,x_0\sim\pi^J}
\{d_{\mathrm{disc}}^J(x_0,k_n)\le1/4\}.
\]
Use the same stationary discrete path, banks, scales and thresholds.
Whenever the realized initial state is in the event defining
$q_n^{\mathrm{disc}}$ and $C_1$ holds, total variation now gives
\[
\PP\{x_{k_n}\in A\mid J,x_0,\text{bank }1\}\ge1/4.
\]
This endpoint is among $x_0,\ldots,x_m$, so it forces $\mathcal S_n$
with no clock-tail error. The same argument therefore gives
\begin{equation}\label{fixed-eq:discrete-final-bound}
\frac{q_n^{\mathrm{disc}}}{4}
\le\tfrac14\PP(C_1^c)+\PP(\mathcal G_n^c)
+C_{\beta,a}n^{a+1-pe_*}=o(1),
\end{equation}
proving \eqref{fixed-eq:discrete-conclusion} directly.
For each realized finite chain, total variation is nonincreasing in
time, and in continuous time it is also continuous. Hence the first
conclusion implies $T^J(x_0)>n^a$. In discrete time it implies
$T_{\mathrm{disc}}^J(x_0)>\lfloor n^a\rfloor$; this integer inequality
gives $T_{\mathrm{disc}}^J(x_0)>n^a$. Finally, the Gibbs mass of each
exceptional set lies in $[0,1]$ and has expectation tending to zero.
Markov's inequality makes that mass tend to zero in probability over
$J$; conversely, convergence in probability of a bounded mass implies
convergence of its expectation. This proves the asserted equivalent
formulations.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/fixed-depth-crossing.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/polynomial-coverage.tex ====
\section{Coverage inside sets of polynomial Gibbs mass}
\label{poly-sec:coverage}
The fixed-depth crossing argument in Appendix~\ref{fixed-sec:crossings}
uses polynomially many reference samples. The neighborhood bound
$e^{-D_3n^\kap}$ in Proposition~\ref{prop:coverage} is smaller than every
fixed inverse power of $n$, so it does not supply those references.
We therefore apply the same energy and rotation mechanism with a
perturbation variance of order $\log n/n$. This yields polynomial
overlap mass inside sets of polynomial Gibbs mass, uniformly over every
subset of the cube. We give the argument from the energy and rotation
estimates in Section~\ref{sec:energy}, so that the fixed-depth route can
be read independently of the growing-depth construction.
Throughout this appendix, fix $\beta>1$ and use the off-diagonal
Hamiltonian of Equation~\eqref{eq:model}. Write
$H^J=H_n^J$, $Z^J=Z_{n,\beta}^J$, and $\pi^J=\pi_{n,\beta}^J$.
For a nonempty $S\subset\Sig$, define its conditional Gibbs measure by
\[
\pi_S^J(x)=\frac{\pi^J(x)\mathbf1_{\{x\in S\}}}{\pi^J(S)}.
\]
Every nonempty set has positive Gibbs mass. All Gaussian disorder vectors
below have independent standard normal coordinates indexed by $i<j$.
\begin{proposition}[Polynomial overlap coverage]
\label{poly-prop:coverage}
There exists $q_*=q_*(\beta)\in(0,1)$ such that, for every fixed $D_1>0$,
there exists a fixed $D_2>0$ for which
\begin{equation}\label{poly-eq:coverage-uniform}
\PP_J\!\left(
\text{for every }S\subset\Sig\text{ with }\pi^J(S)\ge n^{-D_1},\quad
(\pi_S^J)^{\otimes2}\{|R|\ge q_*\}\ge n^{-D_2}
\right)\longrightarrow1.
\end{equation}
The threshold $q_*$ does not depend on $D_1$.
\end{proposition}
The proof compares two estimates for the change in $\log Z$ under a small
Gaussian rotation. If a set has too few overlapping pairs, a fresh Gaussian
perturbation increases its restricted partition sum. The energy upper tail
limits the loss caused by decreasing the coefficient of the original
disorder. Together these give an increase that the rotation estimate makes
unlikely. We first prove the required gain for an arbitrary deterministic
probability measure, so that it can later be applied after conditioning on
$J$.
\subsection{Gain under independent Gaussian disorder}
\begin{lemma}[Gaussian gain from few overlapping pairs]
\label{poly-lem:coverage-gain}
Fix $M>0$, $q\in(0,1)$, and $D\ge\beta^2M/2+2$.
For all sufficiently large $n$, set $t=M\log n/n$.
If $\nu$ is any deterministic probability measure on $\Sig$ satisfying
\[
\nu^{\otimes2}\{|R|\ge q\}\le4n^{-D},
\]
and $G$ is standard Gaussian, define
$W=\nu(e^{\beta\sqrt t H^G})$.
There is a constant $C_\beta$, independent of $M,q,D,n,\nu$, such that
\begin{equation}\label{poly-eq:coverage-gain}
\PP_G\!\left(
\log W\ge\frac{\beta^2t(n-1)}4
-\beta^2ntq-C_\beta\sqrt{nt}-\log2
\right)\ge\frac12.
\end{equation}
\end{lemma}
\begin{proof}
For $x\in\Sig$ let $h_x=(n^{-1/2}x_ix_j)_{i<j}$, so that
$H^G(x)=h_x\cdot G$. In particular,
\[
|h_x|^2=\frac{n-1}{2},\qquad
\EE_G[H^G(x)H^G(y)]=\frac{nR(x,y)^2-1}{2}.
\]
Set
\[
m_0=\frac{\beta^2t(n-1)}4,\qquad
A=\frac{\beta^2ntq^2}{2}.
\]
The Gaussian exponential moment formula gives $\EE_GW=e^{m_0}$.
For the second moment, split pairs according to whether $|R|<q$:
\begin{align}
\frac{\EE_GW^2}{(\EE_GW)^2}
&=\nu^{\otimes2}\!\left(e^{\beta^2t(nR^2-1)/2}\right)\notag\\
&\le e^A+4n^{-D}e^{\beta^2nt/2}
\le e^A+4n^{-2}
\le2e^A.
\label{poly-eq:coverage-moments}
\end{align}
The exponent assumption on $D$ gives the second inequality, and the last
one holds for $n\ge2$. Paley--Zygmund therefore implies
\begin{equation}\label{poly-eq:coverage-pz}
\PP_G(\log W\ge m_0-\log2)\ge\frac18e^{-A}.
\end{equation}
This probability may be small. Gaussian concentration will turn it into
a lower bound of probability at least one half, while controlling the loss
in the value of $\log W$.
Write $Y(G)=\log W$. Differentiation expresses its gradient as
$\beta\sqrt t$ times a convex combination of the vectors $h_x$.
Consequently $Y$ is Lipschitz with constant at most
\[
L_W=\beta\sqrt{\frac{t(n-1)}2}.
\]
The Gaussian moment bound in Equation~\eqref{static-eq:gaussian-mgf}
of Lemma~\ref{static-lem:gaussian-tools}, followed by Chernoff's inequality,
gives the following one-sided tails for this deterministic function of $G$:
\[
\PP_G(Y-\EE_GY\ge s)\le e^{-s^2/(2L_W^2)},\qquad
\PP_G(Y-\EE_GY\le-s)\le e^{-s^2/(2L_W^2)}
\quad(s>0).
\]
If $m_0-\log2>\EE_GY$, comparison of the upper-tail bound with
Equation~\eqref{poly-eq:coverage-pz} gives
\[
\EE_GY\ge m_0-\log2-L_W\sqrt{2(A+\log8)}.
\]
If $m_0-\log2\le\EE_GY$, the same inequality holds immediately.
Using $\sqrt{a+b}\le\sqrt a+\sqrt b$, we thus obtain
\begin{equation}\label{poly-eq:coverage-mean-gain}
\EE_GY\ge m_0-\log2
-\frac{\beta^2ntq}{\sqrt2}
-\beta\sqrt{nt\log8}.
\end{equation}
The lower-tail bound also gives
$Y\ge\EE_GY-L_W\sqrt{2\log2}$ with probability at least $1/2$.
Combining this with Equation~\eqref{poly-eq:coverage-mean-gain}, and
bounding $\beta^2ntq/\sqrt2$ by $\beta^2ntq$, proves the stated estimate.
One admissible choice is
\[
C_\beta=\beta\bigl(\sqrt{\log8}+\sqrt{\log2}\bigr).
\]
This constant is independent of the measure $\nu$ and of the parameters
$M,q,D$.
\end{proof}
\subsection{Uniform coverage over all subsets}
\begin{proof}[Proof of Proposition~\ref{poly-prop:coverage}]
Let $e_1\in(0,\beta/2)$ and $c_E>0$ be the constants from
Lemma~\ref{static-lem:coverage-energy}. Thus, outside a disorder event
of probability at most $e^{-c_En}$,
\begin{equation}\label{poly-eq:energy-event}
\pi^J\{H^J>ne_1\}\le e^{-c_En}.
\end{equation}
Define
\[
g=\frac{\beta^2}{4}-\frac{\beta e_1}{2}>0.
\]
Choose the overlap threshold before choosing any mass exponent:
\begin{equation}\label{poly-eq:coverage-constants}
0<q_*<\min\left\{1,\frac{g}{8\beta^2}\right\}.
\end{equation}
It depends only on $\beta$. We choose this same threshold in
Proposition~\ref{prop:coverage}: both coverage arguments use the same
$e_1$ and permit the strict bound in
Equation~\eqref{poly-eq:coverage-constants}. Now fix $D_1>0$ and choose
\begin{equation}\label{poly-eq:coverage-parameters}
M\ge\frac{8(D_1+1)}g,\qquad
D_2=\frac{\beta^2M}{2}+2,\qquad
t=\frac{M\log n}{n},\qquad \theta=\arcsin\sqrt t.
\end{equation}
For all sufficiently large $n$, $0<t<1/2$.
All constants have been fixed independently of $n$ and of the disorder.
Let $\mathcal B_n$ be the event that
Equation~\eqref{poly-eq:energy-event} holds and that some $S\subset\Sig$
satisfies
\begin{equation}\label{poly-eq:violating-set}
\pi^J(S)\ge n^{-D_1},\qquad
(\pi_S^J)^{\otimes2}\{|R|\ge q_*\}<n^{-D_2}.
\end{equation}
We will bound the probability of this existence event by selecting one
violating set. Fix a deterministic ordering of the subsets of the finite
cube and take its first member satisfying
Equation~\eqref{poly-eq:violating-set}. For each fixed nonempty set, both
Gibbs quantities in that equation are continuous functions of $J$.
The resulting selector $S=S(J)$ is therefore measurable on
$\mathcal B_n$. In particular, it is fixed when we condition on $J$.
For such a fixed $J$, remove the part of $S$ above the energy threshold:
\[
T=S\cap\{x:H^J(x)\le ne_1\},\qquad \nu=\pi_T^J.
\]
For all sufficiently large $n$, uniformly over $J\in\mathcal B_n$,
\begin{equation}\label{poly-eq:trimmed-mass}
\pi^J(T)\ge\pi^J(S)-e^{-c_En}
\ge\frac12\pi^J(S)\ge\frac12n^{-D_1}.
\end{equation}
In particular, $T$ is nonempty. Since $T\subset S$, the two conditioning
denominators give
\begin{equation}\label{poly-eq:trimmed-pairs}
\nu^{\otimes2}\{|R|\ge q_*\}
\le\left(\frac{\pi^J(S)}{\pi^J(T)}\right)^2
(\pi_S^J)^{\otimes2}\{|R|\ge q_*\}
<4n^{-D_2}.
\end{equation}
The trimmed measure is measurable in $J$ as well. Draw $G$ independently
of $J$. Conditional on $J$, the measure $\nu$ is deterministic, so
Lemma~\ref{poly-lem:coverage-gain} applies with $q=q_*$ and $D=D_2$.
Write
\[
W=\nu(e^{\beta\sqrt t H^G}),\qquad
J_\theta=\sqrt{1-t}\,J+\sqrt t\,G.
\]
No part of the choice of $S,T$, or $\nu$ uses this fresh disorder.
We now compare this conditional gain with the full partition sum.
Linearity in the disorder, restriction to $T$, and
$\sqrt{1-t}-1<0$ imply
\begin{align*}
\log\frac{Z^{J_\theta}}{Z^J}
&=\log\pi^J\!\left(
e^{\beta(\sqrt{1-t}-1)H^J+\beta\sqrt t H^G}\right)\\
&\ge-D_1\log n-\log2
-\beta(1-\sqrt{1-t})ne_1+\log W.
\end{align*}
Thus Lemma~\ref{poly-lem:coverage-gain} gives, with conditional
probability at least $1/2$,
\begin{align}
\log\frac{Z^{J_\theta}}{Z^J}
&\ge-D_1\log n-2\log2
-\beta(1-\sqrt{1-t})ne_1
+\frac{\beta^2t(n-1)}4
-\beta^2ntq_*-C_\beta\sqrt{nt}\notag\\
&\ge nt\left(g-\beta^2q_*-\frac{D_1}{M}
-\frac{\beta e_1t}{2}\right)
-C_\beta\sqrt{nt}-2\log2-\frac{\beta^2t}{4}.
\label{poly-eq:partition-gain}
\end{align}
The last line uses
$1-\sqrt{1-t}\le t/2+t^2/2$ for $0\le t<1$.
Our choices ensure $\beta^2q_*<g/8$ and $D_1/M<g/8$.
Since $t\to0$ and $nt=M\log n\to\infty$, the remaining errors in
Equation~\eqref{poly-eq:partition-gain} are $o(nt)$. Consequently,
for all sufficiently large $n$ and every $J\in\mathcal B_n$,
\begin{equation}\label{poly-eq:bad-increment}
\PP_G\!\left(
\log Z^{J_\theta}-\log Z^J\ge\frac g2nt\,\middle|\,J
\right)\ge\frac12.
\end{equation}
The lower bound on $n$ depends only on the fixed parameters and is
uniform in $J$ and its selected set.
It remains to bound this increment under the joint law of $(J,G)$.
For $t\le1/2$, $\theta\le\sqrt{2t}$. The unconditional rotation
estimate in Lemma~\ref{static-lem:coverage-rotation} therefore yields
\begin{align*}
\PP_{J,G}\!\left(
\log Z^{J_\theta}-\log Z^J\ge\frac g2nt\right)
&\le\exp\!\left(-\frac{g^2nt}{8\beta^2}\right)\\
&=n^{-g^2M/(8\beta^2)}.
\end{align*}
Integrating Equation~\eqref{poly-eq:bad-increment} over
$\mathcal B_n$ bounds $\PP_J(\mathcal B_n)$ by twice this probability.
Adding the exceptional event from
Lemma~\ref{static-lem:coverage-energy}, we conclude that the failure
probability in Equation~\eqref{poly-eq:coverage-uniform} is at most
\begin{equation}\label{poly-eq:failure-bound}
e^{-c_En}+2n^{-g^2M/(8\beta^2)}\longrightarrow0.
\end{equation}
The selector controls the existence of any violating subset, without a
union bound over subsets of the cube. The choice of $q_*$ in
Equation~\eqref{poly-eq:coverage-constants} preceded the choice of
$D_1$, as required.
\end{proof}
\subsection{Target coverage within arbitrary subsets}
For the crossing argument, we need a bound on the total Gibbs mass of
targets lacking polynomial Gibbs mass of overlapping references in the
same subset. Both masses below use the original Gibbs measure. This is
the form needed when targets have Gibbs marginals and reference banks
contain unconditioned Gibbs samples. Uniformity of the pair statement
gives the target bound by applying it to the set of points that would
violate it.
\begin{corollary}[Target coverage within arbitrary subsets]
\label{poly-cor:coverage-cells}
Let $q_*$ be as in Proposition~\ref{poly-prop:coverage}.
For every fixed $D_1>0$, there exists a fixed $D_3>0$ such that, with
probability tending to one over $J$, every $S\subset\Sig$ satisfies
\begin{equation}\label{poly-eq:coverage-cells}
\pi^J\!\left(
\left\{x\in S:
\pi^J\{b\in S:|R(x,b)|\ge q_*\}<n^{-D_3}
\right\}\right)<n^{-D_1}.
\end{equation}
\end{corollary}
\begin{proof}
Fix $D_1>0$, take the corresponding $D_2$ from
Proposition~\ref{poly-prop:coverage}, and choose any fixed
$D_3>D_1+D_2$. Work on the simultaneous event in that proposition.
Given $S\subset\Sig$, denote the set inside the outer measure in
Equation~\eqref{poly-eq:coverage-cells} by $B$.
If $m=\pi^J(B)\ge n^{-D_1}$, then $B$ is nonempty, and $B\subset S$
implies
\begin{align*}
(\pi_B^J)^{\otimes2}\{|R|\ge q_*\}
&=\frac1{m^2}\sum_{x\in B}\pi^J(x)
\pi^J\{b\in B:|R(x,b)|\ge q_*\}\\
&<\frac{n^{-D_3}}m
\le n^{-D_3+D_1}<n^{-D_2}.
\end{align*}
This contradicts Proposition~\ref{poly-prop:coverage} applied to $B$.
Hence $\pi^J(B)<n^{-D_1}$. This implication is deterministic on the
same disorder event for every $S$, including cells selected after
observing $J$. For $S=\varnothing$ the assertion is immediate.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//appendices/polynomial-coverage.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//figures/narrow-increments.tex ====
\begin{figure}[t]
\centering
\begin{tikzpicture}[x=0.95cm,y=0.83cm,
every node/.style={font=\small},
active/.style={line width=1.3pt,blue!65!black},
other/.style={line width=1.3pt,red!65!black}]
\node[anchor=west] at (0,5.00) {Increments used in $Y=aY_0+\theta Y_1$};
\draw[gray] (0,2.9)--(3,2.9);
\draw[active] (1,2.9)--(3,2.9);
\draw[active] (3,2.9)--(3.7,3.6)--(5,3.6);
\draw[gray] (5,3.6)--(7,3.6);
\draw[other] (3,2.9)--(3.7,2.2)--(5,2.2);
\draw[gray] (5,2.2)--(7,2.2);
\fill (3,2.9) circle[radius=1.8pt];
\node[above] at (1.9,2.94) {$a/h$};
\node[above] at (4.35,3.65) {$-a/h$};
\node[below] at (4.35,2.15) {$\theta/h$};
\node[anchor=west] at (7.15,3.6) {$x,y$ branch};
\node[anchor=west] at (7.15,2.2) {$v$ branch};
\node[anchor=east,align=right] at (0.85,3.50) {common\\branch};
\foreach \x/\t in {1/{r-h},3/r,5/{r+h},7/q} {
\draw[gray,densely dotted] (\x,1.5)--(\x,3.85);
\node[below] at (\x,1.5) {$\t$};
}
\node[anchor=west,text=gray] at (8.2,1.12) {local scalar time};
\node[anchor=west] at (0,0.30) {Tent $T$ on the $x$ path};
\draw[->,gray] (0,-1.65)--(7.25,-1.65);
\draw[active] (0,-1.65)--(1,-1.65)--(3,-0.65)
--(5,-1.65)--(7,-1.65);
\node[above] at (3,-0.65) {$1$};
\foreach \x/\t in {1/{r-h},3/r,5/{r+h},7/q} {
\draw[gray] (\x,-1.60)--(\x,-1.70);
\node[below] at (\x,-1.72) {$\t$};
}
\node[anchor=west] at (8.05,0.30) {Ramp $S$ on the $v$ path};
\draw[->,gray] (8.1,-1.65)--(12.65,-1.65);
\draw[other] (8.5,-1.65)--(11.9,-0.65);
\node[above] at (11.9,-0.65) {$1$};
\foreach \x/\t in {8.5/r,11.9/{r+h}} {
\draw[gray] (\x,-1.60)--(\x,-1.70);
\node[below] at (\x,-1.72) {$\t$};
}
\end{tikzpicture}
\caption{The three increments in the perturbation. The upper labels are
coefficients of the raw field increments in $Y$; $a=1-\theta$.
Horizontal position records local scalar time, not the mass-sorted order
of the hierarchy. Below, the raw covariance of $Y_0$ with $X_s^x$,
divided by $\beta^2$, is the tent $T(s)$, which vanishes before the
endpoint $q$. Conditional on the common field $X_r$, the raw covariance
of $Y_1$ with $X_s^v$ on $[r,r+h]$, divided by $\beta^2$, is the ramp
$S(s)$. The tent isolates the small change in Parisi mass near $r$;
the other branch supplies the mixed susceptibility term. Lengths are
schematic.}
\label{fig:narrow-increments}
\end{figure}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//figures/narrow-increments.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//figures/protected-block.tex ====
\begin{figure}[t]
\centering
\begin{tikzpicture}[x=0.93cm,y=1cm,
every node/.style={font=\small},
kept/.style={circle,draw=blue!65!black,fill=blue!65!black,
inner sep=0pt,minimum size=4.5pt},
conflict/.style={circle,draw=red!70!black,fill=red!70!black,
inner sep=0pt,minimum size=4.5pt}]
\node[align=center] at (3.35,3.45) {Older blocks\\with earlier deletions};
\node[align=center] at (9.4,3.45)
{Immediately preceding block\\all $B$ references still retained};
\node[align=center] at (14.2,3.45) {New\\level $j$};
\draw[gray] (6.6,2.96)--(6.6,3.08)--(12.5,3.08)--(12.5,2.96);
\draw[gray,dashed] (13.0,-0.50)--(13.0,3.10);
\node[anchor=east,align=right] at (0.7,2.30) {Before\\reset};
\node[anchor=east,align=right] at (0.7,0.15) {After\\reset};
\foreach \y in {2.30,0.15} {
\foreach \x in {1.3,2.1,3.7,5.3}
\node[kept] at (\x,\y) {};
\foreach \x in {2.9,4.5} {
\draw[gray] (\x,\y) circle[radius=0.09];
\draw[gray] (\x-0.08,\y-0.08)--(\x+0.08,\y+0.08);
}
\node at (5.95,\y) {$\cdots$};
}
\node[above] at (1.3,2.40) {$v_1$};
\node[below] at (1.3,0.02) {$v_1$};
\fill[red!7] (9.80,1.89) rectangle (12.55,2.73);
\foreach \x in {6.9,7.8}
\node[kept] at (\x,2.30) {};
\node at (8.7,2.30) {$\cdots$};
\node[kept] at (9.35,2.30) {};
\foreach \x in {10.2,11.2,12.2}
\node[conflict] at (\x,2.30) {};
\node[above] at (6.9,2.40) {$v_{j-B}$};
\node[above] at (12.2,2.40) {$v_{j-1}$};
\draw[red!70!black] (9.80,1.83)--(9.80,1.68)--(12.55,1.68)--(12.55,1.83);
\node[text=red!70!black,align=center] at (11.18,1.28)
{Conflict suffix\\$d\le H_0<B$};
\foreach \x in {6.9,7.8,9.35}
\node[kept] at (\x,0.15) {};
\node at (8.7,0.15) {$\cdots$};
\foreach \x in {10.2,11.2,12.2} {
\draw[red!70!black] (\x,0.15) circle[radius=0.09];
\draw[red!70!black] (\x-0.08,0.07)--(\x+0.08,0.23);
}
\node[kept] at (14.2,0.15) {};
\node[below] at (14.2,0.02) {$v_j$};
\draw[->,gray] (14.2,2.50)--(14.2,0.50);
\node[align=center,fill=white,inner sep=2pt] at (14.2,1.52)
{Select\\and append};
\draw[->,gray] (1.15,-0.80)--(14.55,-0.80);
\node[fill=white,inner sep=2pt] at (7.8,-0.80) {Original level order};
\end{tikzpicture}
\caption{Why a reset preserves the first reference. The dots are schematic
slots in the original level order; omitted entries are indicated by
ellipsis, and crossed gray slots represent deletions at earlier resets.
Immediately before a reset at level $j$, the preceding block
$\{j-B,\ldots,j-1\}$ is complete. Conflicts form a suffix of the retained
list of length $d\le H_0<B$, illustrated in red. The suffix therefore
misses both the first reference of that complete block and all older
retained references. Thus $v_1$ survives at the first reset, when it
belongs to the preceding block, and at every later reset, as drawn.
The new reference $v_j$ is then appended.
Older blocks may contain holes and need not be complete. The diagram
illustrates a nonempty conflict suffix; the proof also permits $d=0$.}
\label{fig:protected-block}
\end{figure}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//figures/protected-block.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//main.tex ====
\documentclass[11pt]{article}
\input{preamble}
\input{math-glyph-unicode}
\title{Stretched-exponential barriers for typical SK initial states}
\author{OpenAI}
\date{September 24, 2026}
\begin{document}
\maketitle
\begin{abstract}
At every fixed inverse temperature $\beta>1$, we prove a stretched-exponential
obstruction to mixing for the zero-field Sherrington--Kirkpatrick model from
typical equilibrium configurations held fixed as initial states. The Gibbs
mass of states whose total-variation distance from equilibrium at time
$\exp(n^{1/10000})$ exceeds $1/4$ tends to one in probability over the
disorder. This holds for rate-one-per-site heat-bath dynamics and after the
same stated number of discrete update attempts.
\end{abstract}
\tableofcontents
\input{sections/01-introduction}
\input{sections/02-gaussian}
\input{sections/03-scalar}
\input{sections/04-locking}
\input{sections/05-energy}
\input{sections/07-coverage}
\input{sections/08-narrow-locking}
\input{sections/09-dynamics}
\appendix
\input{appendices/polynomial-coverage}
\input{appendices/fixed-depth-crossing}
\input{references}
\end{document}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//main.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//math-glyph-unicode.tex ====
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==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//preamble.tex ====
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==== END Stretched-exponential-barriers-for-typical-SK-initial-states//preamble.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//references.tex ====
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A.~S.~Bandeira, A.~El Alaoui and A.~R\"odder,
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\newblock preprint (2026), version~1, \href{https://arxiv.org/abs/2607.06813v1}{arXiv:2607.06813v1}.
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J.~Barbier and N.~Macris,
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G.~Ben Arous and A.~Jagannath,
\newblock Spectral gap estimates in mean field spin glasses,
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\newblock \href{https://doi.org/10.1007/s00220-018-3152-6}{doi:10.1007/s00220-018-3152-6}.
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A.~Billoire,
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\newblock \emph{J. Stat. Mech.} (2010), P11034.
\newblock \href{https://doi.org/10.1088/1742-5468/2010/11/P11034}{doi:10.1088/1742-5468/2010/11/P11034}.
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H.~J.~Brascamp and E.~H.~Lieb,
\newblock On extensions of the Brunn--Minkowski and Pr\'ekopa--Leindler theorems,
including inequalities for log concave functions, and with an application to the diffusion equation,
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W.-K.~Chen,
\newblock Variational representations for the Parisi functional and
the two-dimensional Guerra--Talagrand bound,
\newblock \emph{Annals of Probability} \textbf{45} (2017),
no.~6A, 3929--3966.
\newblock \href{https://doi.org/10.1214/16-AOP1154}{doi:10.1214/16-AOP1154};
\href{https://arxiv.org/abs/1501.06635v2}{arXiv:1501.06635v2}.
\bibitem{AMS22}
A.~El Alaoui, A.~Montanari and M.~Sellke,
\newblock Sampling from the Sherrington--Kirkpatrick Gibbs measure via
algorithmic stochastic localization,
\newblock \emph{Proc. 63rd IEEE FOCS} (2022), 323--334.
\newblock Extended version: \href{https://arxiv.org/abs/2203.05093v2}{arXiv:2203.05093v2}.
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F.~Guerra,
\newblock Broken replica symmetry bounds in the mean field spin glass model,
\newblock \emph{Comm. Math. Phys.} \textbf{233} (2003), 1--12.
\newblock \href{https://doi.org/10.1007/s00220-002-0773-5}{doi:10.1007/s00220-002-0773-5}.
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A.~Jagannath and I.~Tobasco,
\newblock Some properties of the phase diagram for mixed $p$-spin glasses,
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\href{https://arxiv.org/abs/1504.02731v2}{arXiv:1504.02731v2}.
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\newblock Eigenvalue method to compute the largest relaxation time of disordered systems,
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\bibitem{Mourrat2020}
J.-C.~Mourrat,
\newblock Free energy upper bound for mean-field vector spin glasses,
\newblock preprint (2020),
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\newblock The cited equations refer to the
\href{https://perso.ens-lyon.fr/jean-christophe.mourrat/genbound.pdf}{author-hosted manuscript}.
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\newblock Free energy in the mixed $p$-spin models with vector spins,
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\newblock A sequence of approximated solutions to the S-K model for spin glasses,
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\bibitem{Sellke25}
M.~Sellke,
\newblock Exponentially slow mixing of the low temperature SK model,
\newblock preprint (2025), version~2, \href{https://arxiv.org/abs/2511.22621v2}{arXiv:2511.22621v2}.
\bibitem{SK75}
D.~Sherrington and S.~Kirkpatrick,
\newblock Solvable model of a spin-glass,
\newblock \emph{Phys. Rev. Lett.} \textbf{35} (1975), no.~26, 1792--1796.
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M.~Talagrand,
\newblock The Parisi formula,
\newblock \emph{Ann. of Math. (2)} \textbf{163} (2006), no.~1, 221--263.
\newblock \href{https://doi.org/10.4007/annals.2006.163.221}{doi:10.4007/annals.2006.163.221}.
\bibitem{Talagrand07}
M.~Talagrand,
\newblock Mean field models for spin glasses: some obnoxious problems,
\newblock in E.~Bolthausen and A.~Bovier (eds.), \emph{Spin Glasses},
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63--80.
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\end{thebibliography}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//references.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/01-introduction.tex ====
\section{Introduction}\label{sec:introduction}
Local spin updates can preserve information about a realized equilibrium
configuration for a long time, even though the unconditioned chain is
stationary. We quantify this persistence throughout the low-temperature
phase of the zero-field Sherrington--Kirkpatrick model. The central issue
is the mass of slowly mixing initial states: a worst-state lower bound
alone gives no information about whether a Gibbs sample encounters the
obstruction.
For an integer $n\ge2$, let $\Sigma_n=\Sig$, and let
$(J_{ij})_{1\le i<j\le n}$ be independent standard Gaussian variables.
At inverse temperature $\beta>0$, the zero-field
Sherrington--Kirkpatrick model has Hamiltonian and Gibbs measure
\begin{equation}\label{eq:model}
H^J(x)=H_n^J(x)=\frac1{\sqrt n}\sum_{i<j}J_{ij}x_ix_j,
\qquad
\pi^J(x)=\pi_{n,\beta}^J(x)
=\frac{e^{\beta H^J(x)}}{Z_{n,\beta}^J},
\qquad x\in\Sigma_n,
\end{equation}
where $Z^J=Z_{n,\beta}^J=\sum_{x\in\Sigma_n}e^{\beta H^J(x)}$.
For $x,y\in\Sigma_n$, their overlap is
\[
R(x,y)=\frac1n\sum_{i=1}^n x_i y_i.
\]
The measure $\pi^J$ is invariant under the global spin reversal
$x\mapsto-x$. We fix $\beta>1$ before sending $n$ to infinity.
In continuous-time heat-bath dynamics, every site has an independent
rate-one clock. At a ring of site $i$, its spin is resampled according to
\[
\PP(x_i\gets s\mid x_{[n]\setminus\{i\}})
=\frac{\exp\{\beta s h_i^J(x)\}}
{2\cosh(\beta h_i^J(x))},
\qquad
h_i^J(x)=\frac1{\sqrt n}\sum_{j\ne i}J_{ij}x_j,
\quad s\in\{-1,1\},
\]
with $J_{ji}=J_{ij}$ and $J_{ii}=0$. Write $P_t^J$ for its transition
kernel. The total update-attempt rate is $n$, and an attempt may leave
the configuration unchanged. We also consider the discrete kernel $K^J$
that chooses one site uniformly and performs one such conditional
resampling. Both dynamics have stationary measure $\pi^J$.
For probability measures $\nu,\eta$ on $\Sigma_n$, set
$\norm{\nu-\eta}_{\TV}=\frac12\sum_x|\nu(x)-\eta(x)|$.
For a fixed realized state $x$, define
\begin{align}
d^J(x,t)&=\norm{P_t^J(x,\cdot)-\pi^J}_{\TV},
&T^J(x)&=\inf\{t\ge0:d^J(x,t)\le1/4\},
\label{eq:point-mixing-time}\\
d_{\mathrm{disc}}^J(x,k)&=
\norm{(K^J)^k(x,\cdot)-\pi^J}_{\TV},
&T_{\mathrm{disc}}^J(x)&=
\min\{k\ge0:d_{\mathrm{disc}}^J(x,k)\le1/4\}.
\label{eq:point-discrete-mixing-time}
\end{align}
Here $k$ is an integer, and neither mixing time takes a supremum over
the initial state.
\begin{theorem}\label{thm:main}
Set $\kap=1/10000$. For every fixed $\beta>1$,
\begin{align*}
\PP_{J,\,x_0\sim\pi^J}
\left\{d^J\bigl(x_0,\exp(n^\kap)\bigr)>\frac14\right\}
&\longrightarrow1,\\
\PP_{J,\,x_0\sim\pi^J}
\left\{d_{\mathrm{disc}}^J
\bigl(x_0,\lfloor\exp(n^\kap)\rfloor\bigr)>\frac14\right\}
&\longrightarrow1.
\end{align*}
\end{theorem}
The order of conditioning in this statement is essential. We first draw
$x_0$ from $\pi^J$, then regard it as a fixed argument of the transition
kernel. Integrating that kernel over $x_0$ would instead recover
$\pi^J$ at every time. The joint formulation above is equivalent to
convergence of the corresponding Gibbs mass to one in disorder
probability. Indeed, if $F_n(J)\in[0,1]$ is the mass of the complementary
set, the joint assertion says $\EE F_n\to0$. Markov's inequality gives
convergence in probability, and boundedness proves the converse.
Total variation contracts under a Markov kernel. In particular, for
$s,t\ge0$,
\[
d^J(x,t+s)
=\norm{\bigl(P_t^J(x,\cdot)-\pi^J\bigr)P_s^J}_{\TV}
\le d^J(x,t),
\]
and the discrete distances are nonincreasing in $k$ as well.
Since the state space is finite, $t\mapsto d^J(x,t)$ is continuous.
Consequently $d^J(x,t)>1/4$ implies $T^J(x)>t$; in discrete time,
$d_{\mathrm{disc}}^J(x,k)>1/4$ implies $T_{\mathrm{disc}}^J(x)>k$.
It follows that the theorem implies a lower bound at every fixed
polynomial time scale. Appendix~\ref{fixed-sec:crossings} will also give
an independent proof using a fixed number of crossing levels.
\begin{corollary}\label{cor:superpolynomial}
For every fixed $\beta>1$ and every fixed $a>0$,
\[
\PP_{J,\,x_0\sim\pi^J}\{T^J(x_0)>n^a\}\longrightarrow1,
\qquad
\PP_{J,\,x_0\sim\pi^J}
\{T_{\mathrm{disc}}^J(x_0)>n^a\}\longrightarrow1.
\]
Equivalently, each corresponding Gibbs mass tends to one in probability
over the disorder.
\end{corollary}
\begin{proof}
Fix $a>0$. Eventually $n^a$ is smaller than both
$\exp(n^\kap)$ and $\lfloor\exp(n^\kap)\rfloor$. Apply
Theorem~\ref{thm:main} at those larger times and use contraction of
total variation to pass to $n^a$.
\end{proof}
\subsection{Background and relation to earlier work}
Sherrington and Kirkpatrick introduced the model as a mean-field
description of a spin glass~\cite{SK75}. Parisi's variational
description of its equilibrium free energy~\cite{Parisi80} was
established rigorously through Guerra's interpolation bound and
Talagrand's matching lower bound for the limiting
pressure~\cite{Guerra03,Talagrand06}.
These equilibrium results provide the scalar variational structure
behind our argument. Passing from free energy to relaxation requires
additional information about which overlap configurations a local
trajectory can visit.
Ben Arous and Jagannath developed criteria that turn overlap
free-energy barriers into exponentially small spectral gaps for
mean-field spin glasses~\cite{BAJ18}. Their use of constrained replica
pressures is an important antecedent of the overlap comparisons used
here. Their barrier criteria require additional landscape hypotheses;
the explicit Ising sufficient conditions in that work exclude the
pure quadratic covariance of SK. They therefore do not directly give
the present result throughout $\beta>1$.
For SK itself, Sellke proves exponentially slow worst-case Glauber
mixing at sufficiently large inverse
temperature~\cite[Theorem~1.1]{Sellke25}.
That conclusion concerns the slowest initial states. It does not
determine the Gibbs mass of the states that remain far from equilibrium
after their initial configuration has been fixed.
The role of the external field is also substantial. Bandeira,
El Alaoui, and R\"odder prove polynomial worst-case mixing for lazy
discrete heat-bath dynamics at every fixed inverse temperature when
a sufficiently large constant uniform field is present, with high
probability over the disorder~\cite[Corollary~1.2]{BER26}.
Our result concerns zero field. Its distinctive combination is the
entire range of fixed $\beta>1$, Gibbs mass tending to one for
realized initial states, and a stretched-exponential time scale.
The exponent $1/10000$ is not asserted to be optimal, and the constants
in the proof may depend on $\beta$; the argument does not give
uniformity as $\beta$ decreases to one.
El Alaoui, Montanari and Sellke prove a complementary obstruction to
sampling at $\beta>1$ for algorithms stable under disorder perturbations,
in normalized $2$-Wasserstein distance~\cite[Theorem~2.6]{AMS22}. Their stability
hypothesis and approximation metric differ from the conditional Glauber
transition law considered here.
Numerical studies have examined much sharper predictions for
low-temperature relaxation. Billoire studies equilibrium
autocorrelation times for binary couplings and sequential Metropolis
updates~\cite{Billoire10}; Monthus and Garel compute inverse spectral
gaps for small SK systems with Metropolis
dynamics~\cite{MonthusGarel09}. Both report evidence associated with
an $n^{1/3}$ scale for logarithmic relaxation times. Their observables
and dynamics differ from the state-conditioned total variation
distance studied here, and these numerical predictions do not fix
the exponent in Theorem~\ref{thm:main}.
The static comparisons build on Guerra's Gaussian interpolation
framework~\cite{Guerra03} and Talagrand's extensions to systems with
prescribed replica overlaps~\cite[Propositions~3.4 and~4.4]{Talagrand07}.
Panchenko developed a general vector-spin formulation and proved the
asymptotic sharpness of these constrained-overlap
bounds~\cite{Panchenko18}. Talagrand's two- and three-copy bounds already
retain an explicit cost when the prescribed overlaps differ from the
covariance endpoint. In Section~\ref{static-sec:gaussian}, we derive the
finite-hierarchy matrix form needed here for the quadratic SK covariance,
allowing arbitrary finite replica dimension and nonunit endpoint
diagonals. The same section obtains a
quantitative scalar free-energy comparison by choosing an interpolation
path through a differential equation driven by overlap expectations.
This choice has a methodological parallel in the adaptive
interpolation method of Barbier and Macris~\cite{BarbierMacris19};
their theorems are not inputs to our finite-size SK estimate.
Gaussian concentration and the Brascamp--Lieb variance
inequality~\cite{BL76} supply the probabilistic estimates used in
these comparisons.
\subsection{Proof strategy}
Set $L=n^\kap$ and $\rho=L^{-1}$. Our task is to exclude an overlap
sign passage during the first $e^L$ units of time. To connect that task
to mixing, choose a reference $v_1$ with $R(v_1,x_0)\ge q_*$ for a
fixed $q_*>0$. The half-space $\{z:R(v_1,z)\le0\}$ has Gibbs mass at
least $1/2$ by spin reversal. A transition law within $1/4$ of
equilibrium must therefore enter it with probability at least $1/4$.
The reference will be chosen from independent equilibrium samples,
using the initial state but no future updates.
The first part of the paper establishes the static estimates that
make this path argument possible. Section~\ref{static-sec:gaussian}
compares the expected normalized log partition function
$n^{-1}\EE\log Z^J$ with the minimum of the scalar Parisi functional,
with error $O_\beta(n^{-1/24})$, and proves the matrix comparison
for constrained replica overlaps.
Section~\ref{static-sec:scalar} develops the scalar evolution and
the variational identities of a minimizing measure.
Section~\ref{static-sec:locking} uses these identities to compare
three replicas: when one pair has a substantial positive overlap, their
small signed overlaps with the third replica are forced to remain close,
apart from a set of very small averaged Gibbs probability.
Section~\ref{sec:energy} proves an equilibrium energy bound and a
quantitative estimate for small Gaussian rotations of the disorder.
All of these specialized inputs are proved within the paper.
There are two further static steps. First, the energy and rotation
bounds imply a coverage statement in Section~\ref{sec:coverage}:
simultaneously for every subset $S$, outside a set of targets of
absolute Gibbs mass less than $e^{-D_1L}$, each $x\in S$ has
Gibbs mass at least $e^{-D_3L}$ of $z\in S$ with $|R(x,z)|\ge q_*$.
Here $D_3$ is fixed after $D_1$. Sampling $\exp(O(L))$ independent
Gibbs configurations can therefore supply a reference inside $S$.
Second, Section~\ref{sec:narrow} treats triples whose two smaller
overlaps lie near a positive level $r$, while the larger overlap exceeds
$r$ by only a shrinking amount. It supplies locking at the scales
required by the backwards searches. The estimate uses a scalar
stationarity identity on a fixed interval; if that identity fails, a
pair-overlap pressure gap supplies a forbidden interval instead. This
alternative avoids assuming a complete description of the minimizing
measure's support.
Section~\ref{sec:dynamics} applies these estimates to a stationary
discrete path $(x_k)$, and obtains the continuous-time path by an
independent rate-$n$ Poisson clock. At a crossing of a small positive
overlap level, coverage supplies a new reference with large overlap
with the current configuration. Searching backwards for the new
reference's crossing, the locking estimates keep the overlaps with
previously retained references almost unchanged. Iterating creates
many simultaneous overlap constraints at a single path index.
The reference samples are arranged in blocks of a fixed length $B$.
Within a block, each new reference is chosen in the set of
configurations having small absolute overlap with all the earlier
references in that block. At the start of the next block this
restriction is reset. A conflict with older retained references can
then delete only a suffix of length at most a fixed $H_0<B$.
Since $H_0<B$, deletion cannot reach the first reference of the
preceding block or any older retained reference. In particular,
$v_1$ survives; Figure~\ref{fig:protected-block} shows this invariant.
This construction permits $p$ levels of order $L$
with only $\exp(O(L))$ total reference samples, while retaining a
fixed positive fraction of the levels.
The final probability estimate uses the order in which randomness is
revealed. At level $i$, the reference is already determined when the
independent threshold $r_i$ is drawn uniformly from an interval of
length $\rho$. Its overlap at any fixed path index is unlikely to fall
in the much narrower permitted window. Removing constraints by
conditional expectation in decreasing order of level bounds their
joint probability, even though later reference choices depend on
earlier thresholds. We sum over deterministic path indices and
deterministic subsets of retained levels. This avoids conditioning on
the adaptively chosen retained list and completes both clock versions.
Two auxiliary conclusions have a separate role. Appendix~\ref{poly-sec:coverage}
proves that every disorder-selected set of Gibbs mass at least
$n^{-D_1}$ has polynomially large conditional mass of pairs with
absolute overlap at least a fixed $q_*(\beta)$; this threshold is
chosen before $D_1$, and the resulting exponent $D_2$ is fixed before
$n$ grows. Its Gaussian-gain lemma and cell corollary are proved in
full. Appendix~\ref{fixed-sec:crossings} combines that result with the
fixed-scale locking estimate to obtain a second proof of the
polynomial-time obstruction. There the number of levels is fixed and
the reference banks have polynomial sizes. These appendices reuse the
energy, rotation and static locking proofs already established in the
body, while retaining their own complete coverage and crossing arguments.
\paragraph{Notation.}
We suppress $J$, $n$, and the fixed $\beta$ from $H_n^J$, $\pi_{n,\beta}^J$,
and $Z_{n,\beta}^J$ when harmless. All constants may depend on $\beta$
and on other parameters fixed before $n$ grows. Gibbs replicas and
reference samples are independent conditional on the disorder unless
stated otherwise. For matrices, $U:V=\sum_{a,b}U_{ab}V_{ab}$ and
$|U|^2=U:U$ denote the Frobenius inner product and squared norm.
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/01-introduction.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/02-gaussian.tex ====
\section{Gaussian hierarchies and quantitative free energy comparison}
\label{static-sec:gaussian}
The overlap estimates in the later sections require two pressure comparisons:
an upper bound with prescribed replica overlaps, and a lower bound for the
unconstrained pressure with a controlled finite-size error. We derive both
from finite Gaussian hierarchies. The upper bound allows the hierarchy's
covariance endpoint to differ from the prescribed overlap matrix; the cost
of that difference remains explicit. For the lower bound, we choose the
hierarchy along an overlap-dependent differential equation and estimate its
error before making that choice.
The constrained comparison belongs to Guerra's Gaussian interpolation
framework~\cite{Guerra03}, including Talagrand's bounds for coupled
replicas~\cite[Propositions~3.4 and~4.4]{Talagrand07} and Panchenko's
vector-spin formulation~\cite{Panchenko18}. Talagrand's bounds already
retain a cost when prescribed overlaps differ from the covariance
endpoint. We derive the finite-hierarchy matrix form needed here
directly, allowing nonunit endpoint diagonals as well. The quantitative
lower comparison is proved separately below.
We fix $\beta>0$ throughout. A constant $C_\beta$ may vary between
occurrences, but never depends on $n$.
We first remove a constant term from the covariance calculation. Add
independent standard Gaussian variables $J_{11},\ldots,J_{nn}$ to the
disorder and, only in free-energy calculations, use the Hamiltonian
\[
\widehat H_n(\sigma)
=H_n(\sigma)+\frac1{\sqrt{2n}}\sum_{i=1}^nJ_{ii}.
\]
The extra term has mean zero and is the same for every $\sigma$.
The Gibbs measure and the expected log partition function are therefore
unchanged. The covariance now takes the exact form
\begin{equation}\label{static-eq:hamiltonian-covariance}
\E\widehat H_n(\sigma)\widehat H_n(\tau)
=\frac n2R(\sigma,\tau)^2.
\end{equation}
\subsection{Two Gaussian tools and a hierarchy identity}
We will need concentration for Gaussian disorder and a variance estimate
after conditioning on a spin configuration. The following standard
Gaussian and Brascamp--Lieb inequalities supply these two estimates.
\begin{lemma}[Gaussian tools]\label{static-lem:gaussian-tools}
Let $G$ be a standard Gaussian vector in a finite-dimensional Euclidean space.
If $f$ is $L$-Lipschitz, then for $a>0$ and $\theta\in\R$,
\begin{align}
\Prob\bigl[|f(G)-\E f(G)|\ge a\bigr]
&\le 2\exp\left(-\frac{a^2}{2L^2}\right),
\label{static-eq:gaussian-concentration}\\
\log\E\exp\{\theta(f(G)-\E f(G))\}
&\le\frac{\theta^2L^2}{2}.
\label{static-eq:gaussian-mgf}
\end{align}
If a probability density on $\R^m$ is proportional to $e^{-V}$, where $V$ is
twice differentiable and $\nabla^2V\succeq I$, then
\begin{equation}\label{static-eq:brascamp-lieb}
\Var(f)\le\E\|\nabla f\|^2.
\end{equation}
In particular, the variance of a linear functional $a\cdot G$ under this
density is at most $\|a\|^2$.
\end{lemma}
The variance bound is the strongly log-concave case of the
Brascamp--Lieb inequality~\cite{BL76}. For clarity, we prove the
concentration estimate and the specialization of the variance bound to
linear functions, which is the only specialization used below.
\begin{proof}[Proof of the needed Gaussian tools]
For the concentration estimate, write $H_s$ for convolution with a
centered Gaussian of covariance $sI$, and let $B_t$ be standard Brownian
motion. The heat equation and It\^o's formula represent a smooth
$L$-Lipschitz function as
\[
f(B_1)-\E f(B_1)
=\int_0^1\nabla H_{1-t}f(B_t)\cdot\dd B_t,
\qquad |\nabla H_s f|\le L.
\]
The integrand is bounded, so its exponential martingale has expectation
one. Its quadratic variation is at most $L^2$; this proves
Equation~\eqref{static-eq:gaussian-mgf} for every real $\theta$.
Apply exponential Markov's inequality and optimize separately for the
two tails to obtain Equation~\eqref{static-eq:gaussian-concentration}.
Smooth convolution and localization, followed by Gaussian
integrability, extend the argument to Lipschitz functions of at most
linear growth. Every application here lies in that class.
For the linear variance estimate, we may work with the potentials that
occur in the proof. They are smooth and satisfy
$I\preceq\nabla^2V\preceq CI$, where the finite constant $C$ may depend
on the fixed finite system. To see the upper bound, note that the
nonquadratic terms are finite logarithmic partition functions or
hierarchy prefix values: their Hessians are bounded sums of
covariances. A linear tilt leaves the Hessian unchanged. Consider the
diffusion
\[
\dd Y_t=\sqrt2\,\dd B_t-\nabla V(Y_t)\dd t
\]
and denote its semigroup by $Q_t$. The drift is globally Lipschitz,
which gives a unique solution. Under synchronous coupling, strong
convexity contracts distances by $e^{-t}$. Thus an $L$-Lipschitz
function becomes $e^{-t}L$-Lipschitz under $Q_t$. Integration by parts
shows that $\nu(\dd y)\propto e^{-V(y)}\dd y$ is invariant and that
\[
\frac{\dd}{\dd t}\Var_\nu(Q_tf)
=-2\int|\nabla Q_tf|^2\dd\nu.
\]
The Gaussian tails supplied by strong convexity justify the same
identity for $f(y)=a\cdot y$ by a cutoff argument. For independent
$Y,Y'$ with law $\nu$, contraction also gives
\[
\Var_\nu(Q_tf)
\le\tfrac12 e^{-2t}|a|^2\E|Y-Y'|^2\longrightarrow0.
\]
We can therefore integrate the dissipation identity to infinity.
The gradient bound $|\nabla Q_tf|\le e^{-t}|a|$ yields
$\Var_\nu(a\cdot Y)\le2|a|^2\int_0^\infty e^{-2t}\dd t=|a|^2$.
This proves the linear estimate needed for the conditional disorder
laws. The finite upper Hessian bound justifies the diffusion argument;
at each application we will check its lower Hessian bound explicitly.
\end{proof}
We next define the hierarchy and its probability law. Besides any
quenched root randomness, take independent Gaussian fields
$g_0,\ldots,g_k$. The field $g_0$ belongs to the quenched root. Choose
masses
\[
0=x_{-1}\le x_0\le\cdots\le x_{k-1}\le x_k=1,
\qquad w_l=x_l-x_{l-1},
\]
and a terminal log partition function $F_k$ on a finite configuration
space. Integrate the fields backwards by setting, for $1\le j\le k$,
\begin{equation}\label{static-eq:hierarchy-recursion}
F_{j-1}=\frac1{x_{j-1}}\log\E_j e^{x_{j-1}F_j}.
\end{equation}
Here $\E_j$ integrates the field $g_j$. At a zero mass the right-hand
side is interpreted as $\E_jF_j$. After these recursive integrations,
the hierarchy pressure is $\E F_0$, with expectation also over all
quenched randomness.
The recursion also specifies a probability law on fields and terminal
configurations. Conditional on the prefix through level $j-1$, multiply
the original Gaussian law of $g_j$ by the density
\begin{equation}\label{static-eq:hierarchy-tilt}
W_j=\exp\{x_{j-1}(F_j-F_{j-1})\}.
\end{equation}
Equation~\eqref{static-eq:hierarchy-recursion} normalizes this density.
After sampling level $k$, draw a configuration from the terminal Gibbs
law. For a pair of paths indexed by a level $l$, sample their root and
fields through $l$ once, and then sample the two continuations
independently conditional on this prefix. We denote expectation under
that pair law by $\E_l^{(2)}$. At $l=k$ the only independent choices
are the terminal thermal draws. More generally, a finite tree is
sampled by drawing each shared prefix once and making its descendants
conditionally independent. In particular, every branch has the
single-path marginal law. This convention fixes the shared randomness
in all subsequent integration-by-parts calculations.
\begin{lemma}[Hierarchy differentiation]\label{static-lem:hierarchy-derivative}
Suppose an independent standard Gaussian vector $g$ at level $j$ enters the
terminal exponent as $\sqrt v\,g\cdot e(S)$, where $e$ is a fixed vector-valued
function of the terminal configuration. With all other parameters fixed,
\begin{equation}\label{static-eq:hierarchy-derivative}
\frac{\partial}{\partial v}\E F_0
=\frac12\left(
\E\|e(S)\|^2-
\sum_{l=j}^k w_l\E_l^{(2)}e(S^1)\cdot e(S^2)
\right).
\end{equation}
A quenched Gaussian is treated as level $0$. The identity holds for
$v>0$ and extends to interpolation endpoints by integration and continuity.
\end{lemma}
\begin{proof}
To differentiate the pressure, first add a deterministic vector in the
same direction as $\sqrt v\,g$. Differentiating the terminal log
partition function and then the hierarchy recursion shows that, for
$l\ge j$, the gradient of $F_l$ is the conditional mean $m_l^e$ of
$e(S)$ given the prefix through $l$. A second differentiation gives
\begin{equation}\label{static-eq:hessian-recursion-general}
\nabla^2F_l
=\E[\nabla^2F_{l+1}\mid l]
+x_l\Cov(m_{l+1}^e\mid l).
\end{equation}
All conditional expectations in this calculation use the path law.
At the terminal level, the Hessian is the thermal covariance of $e(S)$.
For $j\ge1$, differentiate the $j$th logarithmic expectation with
respect to the variance and integrate by parts in its Gaussian field. The result
is one half of the tilted expectation of
\[
\tr\nabla^2F_j+x_{j-1}\|m_j^e\|^2.
\]
At level $j=0$, ordinary Gaussian differentiation omits the second
term, exactly as prescribed by $x_{-1}=0$. Earlier levels propagate
the derivative by conditional expectation. Iteration of
Equation~\eqref{static-eq:hessian-recursion-general} now expresses twice
the pressure derivative as
\[
\E\|e(S)\|^2-\E\|m_k^e\|^2
+\sum_{l=j}^{k-1}x_l
\bigl(\E\|m_{l+1}^e\|^2-\E\|m_l^e\|^2\bigr)
+x_{j-1}\E\|m_j^e\|^2.
\]
The sum telescopes to
$\E\|e(S)\|^2-\sum_{l=j}^kw_l\E\|m_l^e\|^2$.
Two descendants are independent conditional on their common prefix,
so $\E\|m_l^e\|^2=\E_l^{(2)}e(S^1)\cdot e(S^2)$. This is the
claimed identity.
Gaussian integrability justifies each differentiation: the configuration
space and the number of levels are finite, the coupled gradients are
bounded, and the recursive values grow at most linearly in the Gaussian
variables. A zero mass can be handled directly by the expectation
recursion, or obtained as a limit. Integration and continuity give the
stated endpoint interpretation.
\end{proof}
\subsection{A matrix interpolation bound}
We now apply the differentiation identity to a prescribed overlap
constraint. For $d$ configurations $\sigma^1,\ldots,\sigma^d$, write
$D_{ab}=R(\sigma^a,\sigma^b)$ for their Gram matrix. If $D$ is feasible,
meaning that at least one such tuple realizes it, define
\begin{equation}\label{static-eq:constrained-pressure}
F_{d,n}(D)=\frac1n\E\log
\sum_{\substack{\sigma^1,\ldots,\sigma^d\in\Sig\\
R(\sigma^a,\sigma^b)=D_{ab}\ \forall a,b}}
\exp\left\{\beta\sum_{a=1}^dH_n(\sigma^a)\right\}.
\end{equation}
The sum is unnormalized. We use the Frobenius notation
$U:V=\sum_{a,b}U_{ab}V_{ab}$ and $|U|^2=U:U$.
To compare this pressure with a single-site recursion, choose a symmetric
matrix path
\[
Q_{-1}=0,\qquad Q_j-Q_{j-1}\succeq0\quad(0\le j\le k),
\qquad Q_k=Q_{\rm end}.
\]
Within these conditions, the endpoint is free: it may differ from $D$
and may have nonunit diagonal. Increments of zero covariance are
allowed. If the first nonzero increment should carry positive mass,
insert a quenched level $Q_0=0$; the first genuine increment is then
integrated at mass $x_0$. At a single site, take independent
$d$-dimensional Gaussian fields of covariance
$\beta^2(Q_j-Q_{j-1})$. For a symmetric multiplier $\lambda$, define
$\Psi(Q,x,\lambda)$ as the expected hierarchy value with terminal
function
\[
\log\sum_{s\in\{-1,1\}^d}
\exp\left\{\left(\sum_{j=0}^kg_j\right)\cdot s
+s^{\mathsf T}\lambda s\right\}.
\]
For the deterministic correction to the pressure, set
\begin{equation}\label{static-eq:matrix-path-integral}
\int x\,\dd|Q|^2
=\sum_{j=0}^kx_{j-1}\bigl(|Q_j|^2-|Q_{j-1}|^2\bigr).
\end{equation}
The initial quenched increment therefore carries mass zero in this
integral. The next bound keeps the cost of the free endpoint explicit.
\begin{proposition}[Matrix interpolation]\label{static-prop:matrix-bound}
For every feasible $D$ and every path and multiplier as above,
\begin{equation}\label{static-eq:matrix-bound}
F_{d,n}(D)\le
\Psi(Q,x,\lambda)-\lambda:D
+\frac{\beta^2}{4}
\left(|D-Q_{\rm end}|^2-\int x\,\dd|Q|^2\right).
\end{equation}
For fixed $D$ and $\lambda$, the bound passes to any sequence of finite
paths for which $Q_{\rm end}$, $\Psi(Q,x,\lambda)$, and the displayed
path integral converge.
\end{proposition}
\begin{proof}
Place independent copies of the single-site fields at all $n$ sites.
Interpolate from those fields to the common Hamiltonian disorder,
retaining the overlap constraint throughout. The terminal exponent is
\[
\sqrt t\,\beta\sum_{a=1}^d\widehat H_n(\sigma^a)
+\sqrt{1-t}\sum_{i=1}^n\sum_{j=0}^kg_{j,i}\cdot s_i
+\sum_{i=1}^ns_i^{\mathsf T}\lambda s_i,
\qquad s_i=(\sigma_i^1,\ldots,\sigma_i^d).
\]
Denote the expected recursive value, divided by $n$, by $\varphi(t)$.
For two terminal draws, define the cross-overlap matrix by
$R_{ab}=R(\sigma^a,\tau^b)$; this matrix need not be symmetric.
Equation~\eqref{static-eq:hamiltonian-covariance} gives Hamiltonian
covariance $n\beta^2|R|^2/2$, whereas the $j$th site-field increment
has covariance $n\beta^2(Q_j-Q_{j-1}):R$. The self-overlap is $D$.
Sum Lemma~\ref{static-lem:hierarchy-derivative} over these Gaussian
increments to obtain
\begin{align*}
\varphi'(t)
&=\frac{\beta^2}{4}\left[
|D|^2-2Q_k:D
-\sum_{l=0}^kw_l\E_l^{(2)}
\bigl(|R|^2-2Q_l:R\bigr)\right]\\
&\le\frac{\beta^2}{4}
\left(|D|^2-2Q_k:D+\sum_{l=0}^kw_l|Q_l|^2\right).
\end{align*}
The inequality discards the nonnegative squares $|R-Q_l|^2$ and
therefore does not require symmetry of $R$. Summation by parts gives
\[
|Q_k|^2-\sum_{l=0}^kw_l|Q_l|^2=\int x\,\dd|Q|^2.
\]
It remains to identify the endpoints. At $t=1$ the fields vanish and
$\varphi(1)=F_{d,n}(D)+\lambda:D$. At $t=0$, remove the overlap
constraint. This increases the terminal sum and hence every recursive
value, because each integration is monotone. The resulting recursion
factors over sites, giving $\varphi(0)\le\Psi(Q,x,\lambda)$.
Integration of the derivative bound proves the proposition. Singular
covariance increments are permitted by continuity, or directly by
representing a field through a square root of its covariance.
\end{proof}
\subsection{The scalar functional}
The scalar version of the hierarchy gives the variational quantity
against which we compare the unconstrained pressure. Let $\mu$ be a
probability measure on $[0,1]$ and put $\alpha(s)=\mu([0,s])$.
Define $\Phi_\alpha(s,z)$ by the terminal condition
$\Phi_\alpha(1,z)=\log(2\cosh z)$ and the backward equation
\begin{equation}\label{static-eq:scalar-pde}
-\partial_s\Phi_\alpha
=\frac{\beta^2}{2}
\left(\partial_{zz}\Phi_\alpha
+\alpha(s)(\partial_z\Phi_\alpha)^2\right).
\end{equation}
For a step function $\alpha$, the equation means the Gaussian
recursion already defined: over an interval of length $b-a$ with
constant mass $x$, its terminal function $f$ is mapped to
\[
z\longmapsto\frac1x\log\E
\exp\{x f(z+\beta\sqrt{b-a}\,G)\},
\]
with ordinary expectation at $x=0$. For general $\alpha$, take
$L^1$ limits of step functions; Lemma~\ref{static-lem:scalar-regularity}
proves that this definition exists, is independent of the approximation,
and has the stability properties we use. Define
\begin{equation}\label{static-eq:scalar-functional}
\cP_\beta(\mu)=\Phi_\alpha(0,0)
-\frac{\beta^2}{2}\int_0^1s\alpha(s)\dd s,
\qquad P_\beta=\inf_\mu\cP_\beta(\mu).
\end{equation}
The infimum is attained by compactness and continuity, as proved in
Proposition~\ref{static-prop:scalar-minimizer}. These scalar approximation
and compactness arguments are independent of the pressure comparison;
the lower comparison below will use only finitely supported trial measures. Apply
Proposition~\ref{static-prop:matrix-bound} with $d=1$,
$D=Q_{\rm end}=1$, and $\lambda=0$. First for step measures and
then by approximation, this gives the upper comparison
\begin{equation}\label{static-eq:scalar-upper-bound}
p_n(\beta):=\frac1n\E\log Z_{n,\beta}\le P_\beta.
\end{equation}
\begin{theorem}[Quantitative comparison]\label{static-thm:finite-size}
For every fixed $\beta>0$, there is $C_\beta<\infty$ such that for all
sufficiently large $n$,
\begin{equation}\label{static-eq:finite-size}
P_\beta-C_\beta n^{-1/24}\le p_n(\beta)\le P_\beta.
\end{equation}
\end{theorem}
The remaining task is the lower comparison. We will choose an
interpolation path from expectations of overlaps, using a hierarchical
backward system. Such expectation-dependent choices are also a feature
of the adaptive interpolation method of Barbier and
Macris~\cite{BarbierMacris19}. Here we first prove an explicit overlap
error estimate for the hierarchy, without assuming a prior description
of the Gibbs measure, and then use it along the chosen path.
Mourrat's enriched-free-energy and Hamilton--Jacobi viewpoint
\cite[Section~3, equations~(3.24)--(3.27)]{Mourrat2020} provides related
motivation for organizing errors by hierarchical overlaps. This is a
methodological connection only: the direct covariance and concentration
argument and the quantitative bound here do not invoke a Hamilton--Jacobi
theorem.
\subsection{Concentration along a hierarchy}
To obtain an error estimate that can be summed over the levels, we
use equally spaced masses for the remainder of the section:
\[
w_j=\frac1{k+1},\qquad x_j=\frac{j+1}{k+1}\quad(0\le j\le k).
\]
Fix $0\le t\le1$ and levels $0=q_{-1}<q_0<\cdots<q_k$.
Set $\Delta_j=\beta^2(q_j-q_{j-1})$ and use the terminal exponent
\begin{equation}\label{static-eq:adaptive-exponent}
U(\sigma)=\sqrt t\,\beta\widehat H_n(\sigma)
+\sum_{j=0}^k\sqrt{\Delta_j}\,z_j\cdot\sigma,
\end{equation}
where $z_0,\ldots,z_k$ are independent standard Gaussian vectors in
$\R^n$. Let $\mathcal A_j$ contain the prefix through level $j$,
including the quenched Hamiltonian disorder. Under the path law, put
\begin{equation}\label{static-eq:hierarchy-means}
m_j=\E[\sigma\mid\mathcal A_j],\qquad
M_j=\frac{\|m_j\|^2}{n},\qquad r_j=\E M_j.
\end{equation}
The conditional means $m_j$ form a martingale. Consequently
$0\le r_0\le\cdots\le r_k\le1$. They can also be obtained by
differentiating $F_j$ with respect to an additional deterministic field
coupled linearly to the spins. We begin by controlling the Gaussian
field energy under the same path law.
\begin{lemma}[Field-energy concentration]\label{static-lem:field-energy}
For every $0\le j\le k$, under the single-path law,
\begin{equation}\label{static-eq:field-energy-concentration}
\Var\left(\frac{z_j\cdot\sigma}{n}\right)\le\frac1n.
\end{equation}
\end{lemma}
\begin{proof}
Condition first on the terminal spin $\sigma$. The product of the
tilts in Equation~\eqref{static-eq:hierarchy-tilt} and the terminal
Gibbs density telescopes. Thus the conditional joint density of all
standard Gaussian disorder variables and fields is proportional to
their original Gaussian density multiplied by
\begin{equation}\label{static-eq:fixed-spin-density}
\exp\left\{U(\sigma)-\sum_{l=0}^kw_lF_l\right\}.
\end{equation}
Each $F_l$ is jointly convex in its prefix Gaussian variables. At the
terminal level this follows from log-sum-exp with linearly entering
variables. Backwards integration preserves convexity: expectation does
so at mass zero, and H\"older's inequality does so for a positive-mass
logarithmic expectation. Extend $F_l$ to all variables by making it
constant in its descendants; it is still convex. The weights $w_l$
are nonnegative, and $U(\sigma)$ is linear for fixed $\sigma$.
The negative logarithm of the conditional density therefore has Hessian
at least $I$. The linear specialization of
Lemma~\ref{static-lem:gaussian-tools} bounds the conditional variance
of $z_j\cdot\sigma/n$ by $1/n$.
To remove the conditioning, we must also check the conditional mean.
It is independent of $\sigma$, as follows from the gauge transformation
with signs $\eps_i=\sigma_i$:
\[
J_{ab}\longmapsto\eps_a\eps_bJ_{ab}\quad(a<b),
\qquad z_{l,i}\longmapsto\eps_i z_{l,i}\quad(0\le l\le k).
\]
Leave the diagonal variables fixed. This transformation preserves the
Gaussian law and carries the conditional law at $\sigma$ to the
conditional law at the all-plus spin. It also preserves
$z_j\cdot\sigma$. Thus the conditional means agree, and the law of
total variance proves Equation~\eqref{static-eq:field-energy-concentration}.
\end{proof}
\begin{lemma}[Average overlap concentration]\label{static-lem:average-overlap}
Let $\Delta_{\min}=\min_j\Delta_j$. If $k\ge1$ and
$k/\sqrt{n\Delta_{\min}}\le1$, then
\begin{equation}\label{static-eq:average-overlap-concentration}
\sum_{j=0}^kw_j\E_j^{(2)}(R-r_j)^2
\le C\left(\frac{k}{\sqrt{n\Delta_{\min}}}+k^{-1/2}\right),
\end{equation}
where $C$ is an absolute constant.
\end{lemma}
\begin{proof}
The law of total variance splits the overlap error into two terms:
the fluctuation of its conditional mean $M_j$, and the variance left
after the prefix is fixed. We estimate these terms in that order.
\par\medskip\noindent\textbf{Fluctuation of the conditional mean.}\enspace
Fix $j<k$. Sample three paths $a,b,c$ with common prefix through
$j+1$ and conditionally independent continuations. If
$B=B(\sigma^a,\sigma^b,\sigma^c)$ satisfies $|B|\le1$, integration
by parts in the two adjacent Gaussian fields gives
\begin{equation}\label{static-eq:tree-ibp}
\E B\left(\frac{z_j}{\sqrt{\Delta_j}}
-\frac{z_{j+1}}{\sqrt{\Delta_{j+1}}}\right)
\cdot\frac{\sigma^a}{n}
=-w_j\E B\frac{m_j\cdot\sigma^a}{n}.
\end{equation}
Here is the density calculation, including the common prefix. Relative
to the product Gaussian law, multiply $W_l$ along each common edge
$1\le l\le j+1$, along each of the three distinct descendant
branches, and then by the three terminal Gibbs probabilities. The
logarithm of this density is
\begin{align*}
&\sum_{l=1}^{j+1}x_{l-1}(F_l-F_{l-1})\\
&\quad+\sum_{v\in\{a,b,c\}}\left[
\sum_{l=j+2}^kx_{l-1}(F_l^v-F_{l-1}^v)
+U^v(\sigma^v)-F_k^v\right].
\end{align*}
Apply the derivative difference
$\Delta_j^{-1/2}\partial_{z_{j,i}}
-\Delta_{j+1}^{-1/2}\partial_{z_{j+1,i}}$ coordinate by coordinate.
At level $j+1$ and below, every term depends on these fields only
through $\sqrt{\Delta_j}z_j+\sqrt{\Delta_{j+1}}z_{j+1}$, so its
two derivatives cancel. Terms above $j$ depend on neither field.
Only $F_j$ remains, with coefficient $x_{j-1}-x_j=-w_j$; the same
calculation applies at $j=0$ because $x_{-1}=0$. Its normalized
derivative is $m_{j,i}$. Throughout this differentiation the spin
labels and $B$ are fixed. Integration by parts and summation over
$i$ now prove Equation~\eqref{static-eq:tree-ibp}.
Let $Y$ be the quantity multiplying $B$ on the left-hand side of
Equation~\eqref{static-eq:tree-ibp}. Every branch has the single-path
marginal. Lemma~\ref{static-lem:field-energy} and the $L^2$ triangle
inequality therefore imply
\[
\Var(Y)^{1/2}\le\frac{2}{\sqrt{n\Delta_{\min}}}.
\]
With $B=1$, Equation~\eqref{static-eq:tree-ibp} reads
$\E Y=-w_jr_j$. Subtract this mean, use $|B|\le1$, and recall
$w_j=1/(k+1)$ to obtain
\begin{equation}\label{static-eq:tree-centered}
\left|\E B\left(\frac{m_j\cdot\sigma^a}{n}-r_j\right)\right|
\le\frac{2(k+1)}{\sqrt{n\Delta_{\min}}}.
\end{equation}
Choose $B=R(\sigma^b,\sigma^c)$. Conditional on
$\mathcal A_{j+1}$ the three descendants are independent, so the
left-hand side becomes
\[
\left|\E\left(\frac{m_j\cdot m_{j+1}}n-r_j\right)M_{j+1}\right|.
\]
We now replace $m_{j+1}$ by $m_j$ in both factors. The change in the
expectation is at most $3\sqrt{\gamma_j}$, where
\[
\gamma_j=\frac1n\E\|m_{j+1}-m_j\|^2.
\]
This follows from the bounds $\|m_l\|\le\sqrt n$ and
$|M_j-r_j|\le1$. After replacement the expectation is
$\E(M_j-r_j)M_j=\Var(M_j)$. Martingale orthogonality gives
$\sum_{j<k}\gamma_j\le1$; averaging the errors and using
Cauchy--Schwarz proves
\begin{equation}\label{static-eq:mean-overlap-variance}
\sum_{j<k}w_j\Var(M_j)
\le C\left(\frac{k}{\sqrt{n\Delta_{\min}}}+k^{-1/2}\right).
\end{equation}
\par\medskip\noindent\textbf{Conditional variance.}\enspace
The conditional mean fluctuation is now controlled. To bound the
variance of the overlap of two descendants after their common prefix
is fixed, define
\[
C_j=\Cov(\sigma\mid\mathcal A_j),\qquad
D_j=\nabla^2F_j,\qquad
V_j=\Cov(m_{j+1}\mid\mathcal A_j),
\]
where all conditional quantities are given $\mathcal A_j$, and
$D_j$ is the Hessian in an additive deterministic spin field. The
law of total covariance and the differentiated recursion give
\[
C_j=\E[C_{j+1}\mid\mathcal A_j]+V_j,
\qquad
D_j=\E[D_{j+1}\mid\mathcal A_j]+x_jV_j.
\]
The terminal identity $D_k=C_k$, together with the nondecreasing
masses, allows backward induction and gives
\begin{equation}\label{static-eq:hessian-covariance-bound}
D_j\succeq x_jC_j.
\end{equation}
For the next field, abbreviate $\Delta=\Delta_{j+1}$ and set
$y=\sqrt\Delta z_{j+1}$. Conditional on $\mathcal A_j$, its
density is proportional to
$\exp\{-\|y\|^2/(2\Delta)+x_jF_{j+1}\}$.
Integration by parts, using $\nabla_yF_{j+1}=m_{j+1}$, yields
\begin{align}
\Cov(m_{j+1},y\mid\mathcal A_j)&=\Delta D_j,
\label{static-eq:conditional-cross-covariance}\\
\Cov(y\mid\mathcal A_j)&=\Delta I+\Delta^2x_jD_j.
\label{static-eq:conditional-field-covariance}
\end{align}
For the first identity, use $\E[y\mid\mathcal A_j]=\Delta x_jm_j$
and subtract the product of means from the integration-by-parts
expression for $\E[m_{j+1}y^{\mathsf T}\mid\mathcal A_j]$.
A further application to $yy^{\mathsf T}$ proves the second identity.
The joint covariance matrix of $(m_{j+1},y)$ is positive semidefinite,
so its Schur complement gives
\begin{equation}\label{static-eq:covariance-schur-bound}
V_j\succeq\Delta D_j(I+\Delta x_jD_j)^{-1}D_j.
\end{equation}
Order the eigenvalues of $C_j,D_j,V_j$ decreasingly and denote them
by $c_i,d_i,v_i$, respectively. Equation~\eqref{static-eq:hessian-covariance-bound}
implies $d_i\ge x_jc_i$. The function
$d\mapsto\Delta d^2/(1+\Delta x_jd)$ is increasing on $d\ge0$.
Apply the min--max principle to
Equation~\eqref{static-eq:covariance-schur-bound} to get
\[
v_i\ge\frac{\Delta d_i^2}{1+\Delta x_jd_i}
\ge\frac{\Delta x_j^2c_i^2}{1+\Delta x_j^2c_i}.
\]
At or above the threshold $c_i\ge(\Delta x_j^2)^{-1}$, the final
expression is at least $c_i/2$. Since also $c_i\le\tr C_j\le n$,
these eigenvalues contribute at most $2n\tr V_j$ to
$\sum_i c_i^2$. Those below the threshold contribute at most
$n/(\Delta x_j^2)$. Together the two bounds give
\begin{equation}\label{static-eq:conditional-square-trace}
\frac{\tr C_j^2}{n^2}
\le\frac{2\tr V_j}{n}
+\frac1{n\Delta_{j+1}x_j^2}.
\end{equation}
Martingale orthogonality again supplies the summable quantity:
$\sum_{j<k}\E\tr V_j\le n$. Sum over $j$, using
$x_j=(j+1)/(k+1)$, to obtain
\begin{equation}\label{static-eq:average-square-trace}
B_0:=\sum_{j<k}w_j\E\frac{\tr C_j^2}{n^2}
\le\frac Ck+\frac{Ck}{n\Delta_{\min}}.
\end{equation}
For two independent descendants of the level-$j$ prefix, the
conditional overlap has mean $M_j$ and variance
\[
\frac1{n^2}\left(\tr C_j^2+2m_j^{\mathsf T}C_jm_j\right).
\]
The contribution of the second term is at most
$2\sqrt{\tr C_j^2/n^2}$: use $\|m_j\|^2\le n$ and
$\|C_j\|_{\rm op}\le\sqrt{\tr C_j^2}$. Thus the averaged
conditional variance is at most $B_0+2\sqrt{B_0}$.
Under $k/\sqrt{n\Delta_{\min}}\le1$,
Equation~\eqref{static-eq:average-square-trace} gives $B_0\le C/k$.
Combine this estimate with the conditional-mean bound
\eqref{static-eq:mean-overlap-variance} through the law of total
variance. Finally, the level $k$ costs at most $4/(k+1)$ because
$|R-r_k|\le2$. These bounds prove
Equation~\eqref{static-eq:average-overlap-concentration}.
\end{proof}
\subsection{Adaptive interpolation and the finite-size error}
\begin{proof}[Proof of Theorem~\ref{static-thm:finite-size}]
The upper comparison was proved in
Equation~\eqref{static-eq:scalar-upper-bound}. To obtain the lower
comparison, we choose the field levels so that the pressure derivative
separates into a quadratic expression in the overlap means and the
overlap error controlled by Lemma~\ref{static-lem:average-overlap}.
After constructing that path, we will integrate the derivative and
identify its single-site endpoint with a scalar trial measure.
\par\medskip\noindent\textbf{Choosing the adaptive path.}\enspace
Consider provisionally a differentiable path $q(t)$ with positive gaps,
and set $f(t)=n^{-1}\E F_0(t,q(t))$. For a field gap in overlap units,
$a_j=q_j-q_{j-1}$, the hierarchy identity gives
\begin{equation}\label{static-eq:field-gap-derivative}
\frac{\partial}{\partial a_j}\frac1n\E F_0
=\frac{\beta^2}{2}
\left(1-\sum_{l=j}^kw_lr_l\right).
\end{equation}
The Hamiltonian contribution to the derivative is
$\frac{\beta^2}{4}(1-\sum_lw_l\E_l^{(2)}R^2)$.
Summing the field contributions and using
$\sum_{j=0}^la_j'=q_l'$ therefore gives
\[
f'(t)=\frac{\beta^2}{4}
\left(1-\sum_{l=0}^kw_l\E_l^{(2)}R^2\right)
+\frac{\beta^2}{2}
\left(q_k'-\sum_{l=0}^kw_lr_lq_l'\right).
\]
This calculation suggests the system
\begin{equation}\label{static-eq:adaptive-ode}
q_j'(t)=-r_j(t,q(t)).
\end{equation}
Indeed, since $\E_j^{(2)}R=r_j$, every positive-gap solution satisfies
\begin{equation}\label{static-eq:adaptive-pressure-derivative}
f'(t)=\frac{\beta^2}{4}\left[
1-2r_k+\sum_{j=0}^kw_jr_j^2
-\sum_{j=0}^kw_j\E_j^{(2)}(R-r_j)^2\right].
\end{equation}
The final term is exactly the error already estimated; the remaining
terms can be bounded after integration using the differential equation.
We now construct a solution with gaps large enough to apply that estimate.
Choose
\[
k=\lfloor n^{1/12}\rfloor,\qquad \eta=n^{-1/6},
\qquad q_j(1)=(j+1)\eta\quad(0\le j\le k),
\]
and solve Equation~\eqref{static-eq:adaptive-ode} backwards from $t=1$.
\par\medskip\noindent\textbf{Existence and bounds for the adaptive path.}\enspace
We verify that this system has a solution throughout $[0,1]$ and
keeps every gap positive. On the open region of positive gaps, the
finite Gaussian recursions make $r_j$ continuous in $t$ and locally
Lipschitz in $q$. The following domination justifies those assertions.
Fix a compact subset of this region; in this local argument constants
may depend on $n,k,\beta$ and that subset. Uniformly in the prefix and
in $t\in[0,1]$, we have
$\|\nabla_{z_h}F_h\|=\sqrt{\Delta_h}\|m_h\|\le L$.
Subtract the value of $F_h$ at $z_h=0$ from both numerator and
denominator in the normalized tilt. This gives
\[
W_h(z_h)\le C\exp(L\|z_h\|).
\]
The conditional first moments of subsequent fields are consequently
uniformly bounded. The derivative $\partial_{q_i}U$ is bounded by a
constant times $\|z_i\|+\|z_{i+1}\|$, with the second term omitted
at $i=k$. Differentiation through the recursions therefore yields
\[
|\partial_{q_i}F_l|\le C\left(1+\sum_{h\le l}\|z_h\|\right).
\]
To apply this to $r_j$, represent it as the expectation of $R$ on
a two-path tree. Since terminal Gibbs probabilities are at most one,
the preceding estimates dominate its density and the $q_i$ derivative
of that density by $C(1+S)e^{LS}$, where $S$ is the sum of the norms
of all fields in this finite tree. The bound is Gaussian integrable,
uniformly in $t$ and in the chosen compact set of $q$ values. It
justifies the differentiated recursions and gives the required local
Lipschitz property. At $t=0$, continuity follows from the same bound
and continuity in $\sqrt t$; no derivative of $\sqrt t$ is needed.
Along the local solution, $r_j\ge r_{j-1}$ implies that each gap
$q_j-q_{j-1}$ is nondecreasing as time runs backwards, with
$r_{-1}=0$. Each gap therefore stays at least $\eta$. The bound
$0\le r_j\le1$ also prevents escape to infinity and gives
\begin{equation}\label{static-eq:adaptive-path-bounds}
q_k(t)\le(k+1)\eta+1-t,
\qquad\Delta_j(t)\ge\beta^2\eta.
\end{equation}
These estimates continue the local solution across the full interval.
They also verify the field-gap hypothesis of
Lemma~\ref{static-lem:average-overlap}. Thus, for all sufficiently
large $n$, uniformly along the path,
\begin{equation}\label{static-eq:adaptive-overlap-error}
\sum_{j=0}^kw_j\E_j^{(2)}(R-r_j)^2
\le C_\beta n^{-1/24}.
\end{equation}
Indeed, the two terms in that lemma are controlled by
$k/\sqrt{n\beta^2\eta}=O_\beta(n^{-1/3})$ and
$k^{-1/2}=O(n^{-1/24})$.
\par\medskip\noindent\textbf{Integration of the pressure derivative.}\enspace
The adaptive path has now been constructed and its overlap error is
bounded, so we can integrate
Equation~\eqref{static-eq:adaptive-pressure-derivative}.
Write $\delta=(k+1)\eta=O(n^{-1/12})$ and $Q_j=q_j(0)$.
The derivative in Equation~\eqref{static-eq:field-gap-derivative}
lies between $0$ and $\beta^2/2$. Consequently, removing the fields present at $t=1$
costs at most $\beta^2\delta/2$, and
\[
f(1)\le p_n(\beta)+\frac{\beta^2}{2}\delta.
\]
The differential equation and Jensen's inequality also give
\[
\int_0^1r_j(t)\dd t=Q_j-q_j(1),\qquad
\int_0^1r_j(t)^2\dd t\ge\bigl(Q_j-q_j(1)\bigr)^2.
\]
The endpoint bounds are $0\le Q_j\le1+\delta$ and
$0\le q_j(1)\le\delta$. Integrate
Equation~\eqref{static-eq:adaptive-pressure-derivative}, apply these
bounds and the overlap error in
Equation~\eqref{static-eq:adaptive-overlap-error}, and conclude that
\begin{equation}\label{static-eq:adaptive-lower-bound}
p_n(\beta)\ge
f(0)+\frac{\beta^2}{4}
\left(1-2Q_k+\sum_{j=0}^kw_jQ_j^2\right)
-C_\beta n^{-1/24}.
\end{equation}
\par\medskip\noindent\textbf{Identification of a scalar trial measure.}\enspace
At $t=0$ there is no Hamiltonian coupling, so the recursion factors
over sites. Thus $f(0)$ is the scalar single-site value at levels
$Q_0,\ldots,Q_k$. The only possible obstruction to using these
levels as the support of a trial measure on $[0,1]$ is that some
may exceed one.
Replace each such level by $\overline Q_j=\min(Q_j,1)$.
Each gap can only decrease, and their total decrease is
$Q_k-\overline Q_k\le\delta$. Equation~\eqref{static-eq:field-gap-derivative}
bounds the change in the scalar value by $\beta^2\delta/2$.
The quadratic expression in
Equation~\eqref{static-eq:adaptive-lower-bound} changes by at most
$C_\beta\delta$ as well. We may therefore take $Q_k\le1$ within
the stated error, interpreting zero gaps by continuity.
Now define $\mu=\sum_{j=0}^kw_j\delta_{Q_j}$. Its distribution
function is zero up to $Q_0$, equals $x_{j-1}$ on each gap
$(Q_{j-1},Q_j)$, and equals one between $Q_k$ and $1$.
Over this final interval, the recursion maps $\log(2\cosh z)$ to
$\log(2\cosh z)+\beta^2(1-Q_k)/2$. Hence
\[
\Phi_\alpha(0,0)=f(0)+\frac{\beta^2}{2}(1-Q_k).
\]
The penalty is identified by summation by parts:
\[
\frac{\beta^2}{2}\int_0^1s\alpha(s)\dd s
=\frac{\beta^2}{4}\left(1-\sum_{j=0}^kw_jQ_j^2\right).
\]
The expression preceding the error in
Equation~\eqref{static-eq:adaptive-lower-bound} is therefore exactly
$\cP_\beta(\mu)$. It is at least $P_\beta$, which completes the
lower comparison and the proof.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/02-gaussian.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/03-scalar.tex ====
\section{Scalar regularity and properties of a minimizer}\label{static-sec:scalar}
We now establish the scalar identities used in the overlap comparisons
and the equilibrium energy estimate. The argument has three parts.
First, we justify the scalar equation and its stochastic evolution for
general distribution functions, including those with atoms. Next, we
derive first variations and use them to locate the support of an
arbitrary minimizing measure. Finally, the resulting stationarity
identity gives a strict energy bound when $\beta>1$. No further
description of the minimizer's support is assumed.
Regularity and step approximation for the Parisi evolution, together
with zero-field support properties of its minimizers, were developed
by Auffinger and Chen~\cite[Propositions~1--2 and Theorems~1, 5]{AC2015}.
The full first-variation characterization
appears in Chen~\cite[Theorem~2 and Proposition~1]{Chen2017} and,
independently, in Jagannath and Tobasco~\cite[Proposition~1.1,
Corollary~3.10 and Remark~1.3]{JT2017}. We prove the required statements
here with the normalization and atom conventions used by the later
pressure comparisons.
Throughout this section, $\alpha(s)=\mu([0,s])$ is the right-continuous
distribution function of a probability measure on $[0,1]$. Recall the
scalar terminal-value equation
\begin{equation}\label{static-eq:scalar-pde-recalled}
-\partial_s\Phi_\alpha
=\frac{\beta^2}{2}\bigl(\partial_{zz}\Phi_\alpha
+\alpha(s)(\partial_z\Phi_\alpha)^2\bigr),
\qquad \Phi_\alpha(1,z)=\log(2\cosh z).
\end{equation}
For a step distribution function this is the Gaussian recursion from
the preceding section. For general distribution functions, the scalar
values are defined by approximation, which the next lemma justifies.
The time derivative in the equation is understood almost everywhere;
no differentiability at a jump of $\alpha$ is required.
We denote the first two spatial derivatives by $u=\partial_z\Phi_\alpha$ and
$\chi=\partial_{zz}\Phi_\alpha$.
\subsection{Regularity and first variations}
\begin{lemma}[Regularity, stability, and scalar evolution]
\label{static-lem:scalar-regularity}
The step approximation defining $\Phi_\alpha$ is independent of the
approximating sequence. The functions $\Phi_\alpha,u,\chi$ are continuous
on $[0,1]\times\R$, with
\begin{equation}\label{static-eq:scalar-derivative-bounds}
|u|\le1,\qquad 0<\chi\le1,\qquad
|\chi(s,z)-\chi(s,z')|\le C_\beta|z-z'|.
\end{equation}
The function $\Phi_\alpha(s,\cdot)$ is even, so $u(s,0)=0$.
For two distribution functions $\alpha,\widetilde\alpha$,
\begin{equation}\label{static-eq:scalar-stability}
\sup_{s,z}\bigl(
|\Phi_\alpha-\Phi_{\widetilde\alpha}|
+|u_\alpha-u_{\widetilde\alpha}|
+|\chi_\alpha-\chi_{\widetilde\alpha}|
\bigr)
\le C_\beta\int_0^1|\alpha(s)-\widetilde\alpha(s)|\dd s.
\end{equation}
These functions are also continuous in $\beta>0$, uniformly in $(s,z)$
for their differences, locally in $\beta$.
The stochastic differential equation
\begin{equation}\label{static-eq:scalar-diffusion}
\dd X_s=\beta\dd B_s+\beta^2\alpha(s)u(s,X_s)\dd s,
\qquad X_0=0,
\end{equation}
has a unique strong solution. For a step distribution function, its
transitions across the constant-mass intervals are the normalized tilted
Gaussian kernels of the scalar recursion. Its bounded martingale
$U_s=u(s,X_s)$ satisfies
\begin{equation}\label{static-eq:scalar-martingale}
U_t-U_s=\beta\int_s^t\chi(v,X_v)\dd B_v.
\end{equation}
Moreover,
\begin{equation}\label{static-eq:scalar-chi-expectation}
\E\chi(t,X_t)-\E\chi(s,X_s)
=-\beta^2\int_s^t\alpha(v)\E\chi(v,X_v)^2\dd v.
\end{equation}
For $0\le s<t\le1$, the conditional form is
\begin{equation}\label{eq:scalar-evolution}
\E[\chi(t,X_t)\mid X_s]
=\chi(s,X_s)-\beta^2\int_s^t\alpha(v)
\E[\chi(v,X_v)^2\mid X_s]\dd v.
\end{equation}
Consequently $\Gamma(s):=\E u(s,X_s)^2$ is continuously differentiable,
with one-sided derivatives at the endpoints, and
\begin{equation}\label{static-eq:gamma-derivative}
\Gamma(0)=0,\qquad
\Gamma'(s)=\beta^2\E\chi(s,X_s)^2.
\end{equation}
The well-posedness statement and the evolution identities
\eqref{static-eq:scalar-martingale}--\eqref{static-eq:scalar-chi-expectation} also hold
for a diffusion started at any deterministic time and spatial value.
\end{lemma}
\begin{proof}
We prove estimates uniform over step distribution functions, use them
to pass to general $\alpha$, and then establish continuity in the
temperature parameter.
\par\medskip\noindent\textbf{Derivative bounds along a finite step path.}\enspace
Suppose first that $\alpha$ is a step function. On an interval of
constant mass $m$, its recursion is
\[
T_m f(z)=m^{-1}\log\E e^{m f(z+\beta\sqrt{\ell}\,Z)}
\]
where $\ell$ is the interval length and
$T_0f=\E f(z+\beta\sqrt{\ell}\,Z)$. To keep track of the successive
tilts, fix $(s,z)$ and list the remaining breakpoints as
$s=t_0<\cdots<t_r=1$. Let $m_j$ be the mass on
$[t_j,t_{j+1})$ and set $f_j=\Phi_\alpha(t_j,\cdot)$.
Starting with $Z_0=z$, sample $Z_{j+1}$ from
\[
K_j(v,\dd y)
=\exp\{m_j[f_{j+1}(y)-f_j(v)]\}
N(v,\beta^2(t_{j+1}-t_j))(\dd y),
\]
where the exponential equals one at $m_j=0$. Each kernel is normalized by
the scalar recursion on its own interval. Put
$\mathcal F_j=\sigma(Z_0,\ldots,Z_j)$ and $M_j=f_j'(Z_j)$.
Differentiating the recursion once and twice gives
\begin{align*}
\E[M_{j+1}\mid\mathcal F_j]&=M_j,\\
f_j''(Z_j)&=\E[f_{j+1}''(Z_{j+1})\mid\mathcal F_j]
+m_j\E[(M_{j+1}-M_j)^2\mid\mathcal F_j].
\end{align*}
Thus $M$ is a martingale for the concatenated kernels, even though
the tilts on successive intervals can differ. Its terminal value is
$M_r=\tanh Z_r$. Iterating the second identity yields
\[
\chi(s,z)=\E\operatorname{sech}^2(Z_r)
+\sum_{j=0}^{r-1}m_j\E(M_{j+1}-M_j)^2.
\]
The increments are orthogonal, so their unweighted squared sum has
expectation $\Var(\tanh Z_r)$. Since $0\le m_j\le1$, we obtain
\[
0<\E\bigl[1-\tanh^2(Z_r)\bigr]
\le\chi(s,z)
\le \E\bigl[1-\tanh^2(Z_r)\bigr]
+\Var(\tanh Z_r)
=1-u(s,z)^2\le1.
\]
Every recursion preserves evenness. Its gradient is a conditional
mean of the terminal gradient, which gives $|u|\le1$.
Before differentiating the equation, we record why all required
spatial derivatives exist and are bounded for a fixed finite path.
The $k$th derivative of one logarithmic Gaussian integral is a
polynomial in tilted moments of $f',\ldots,f^{(k)}$, with coefficients
bounded for $0\le m\le1$. At $m=0$, differentiation passes directly
through Gaussian expectation. All positive-order derivatives of
$\log(2\cosh z)$ are bounded, so finite induction gives bounded
derivatives at the breakpoints and within each interval. The scalar
values have at most linear growth, ensuring Gaussian integrability
at every step. These preliminary bounds may depend on the path; we
next obtain the uniform bound needed for approximation.
Within a constant-mass interval the recursion is smooth and solves
Equation~\eqref{static-eq:scalar-pde-recalled}. Its first two spatial
derivatives satisfy
\begin{align*}
-u_s&=\frac{\beta^2}{2}u_{zz}+\beta^2\alpha u u_z,\\
-\chi_s&=\frac{\beta^2}{2}\chi_{zz}
+\beta^2\alpha u\chi_z+\beta^2\alpha\chi^2.
\end{align*}
Set $w=\chi_z$. One more differentiation gives a linear equation:
\[
-w_s=\frac{\beta^2}{2}w_{zz}
+\beta^2\alpha u w_z+3\beta^2\alpha\chi w.
\]
The bounds already proved for $u$ and $\chi$ control this equation's
drift and potential uniformly over all step paths. Its terminal value
has absolute value at most $2$. Concatenate the diffusion
representations over the finitely many intervals to obtain
\[
\sup_{s,z}|w(s,z)|\le2e^{3\beta^2}.
\]
These differentiations may all be performed first inside the finite
Gaussian recursions. Their bounded derivatives justify passage under
the integrals, so the uniform estimate does not presuppose any
regularity of the limiting path.
\par\medskip\noindent\textbf{The scalar diffusion and its evolution identities.}\enspace
We next identify the tilted kernels with the diffusion in
Equation~\eqref{static-eq:scalar-diffusion}. On a constant-mass
interval $[a,b]$, start Brownian motion at $z$ by setting
$Y_s=z+\beta(B_s-B_a)$. Under this law, It\^o's formula identifies
\[
\exp\{m(\Phi(b,Y_b)-\Phi(a,z))\}
\]
as the exponential martingale with stochastic logarithm
$m\beta\int_a^b u(v,Y_v)\dd B_v$. Its bounded integrand justifies
the change of measure, which adds drift $\beta^2m u$. Under that
drift, a further application of It\^o's formula gives
\[
\dd u(s,X_s)=\beta\chi(s,X_s)\dd B_s,
\qquad
\dd\chi(s,X_s)=\beta\chi_z(s,X_s)\dd B_s
-\beta^2\alpha(s)\chi(s,X_s)^2\dd s.
\]
This establishes both evolution identities for step paths.
\par\medskip\noindent\textbf{Stability and passage to general distribution functions.}\enspace
The remaining task is to make these constructions independent of the
step approximation. Subtract the equations for two step paths, keeping
the same diffusion coefficient. The difference of the scalar values
has drift $\frac{\beta^2}{2}\alpha(u+\widetilde u)$ and source
$\frac{\beta^2}{2}(\alpha-\widetilde\alpha)\widetilde u^2$.
The equation for $u-\widetilde u$ has bounded drift and potential,
with source bounded by $C_\beta|\alpha-\widetilde\alpha|$.
For $\chi-\widetilde\chi$, the drift and potential remain bounded,
and the source is bounded by
\[
C_\beta\bigl(|\alpha-\widetilde\alpha|(s)
+\sup_z|u(s,z)-\widetilde u(s,z)|\bigr).
\]
To obtain the last bound, expand
$\alpha u\chi_z-\widetilde\alpha\widetilde u\widetilde\chi_z$
and use the uniform bound on $\widetilde\chi_z$. In the remaining
reaction term, factor the difference of squares. Each difference
has zero terminal value, so the diffusion representation of these
linear equations proves Equation~\eqref{static-eq:scalar-stability},
first for $\Phi$, then for $u$, and finally for $\chi$.
Approximate an arbitrary distribution function in $L^1$ by step
distribution functions. The stability estimates make $\Phi,u,\chi$
converge uniformly, with uniform convergence of differences for the
unbounded scalar value $\Phi$. The fundamental theorem of calculus
in $z$ identifies the limits of $u$ and $\chi$ with the first two
spatial derivatives of the limiting $\Phi$. Continuity, symmetry,
and all non-strict bounds follow by passage to the limit.
Passing to the limit in the time-integrated step equation also gives
Equation~\eqref{static-eq:scalar-pde-recalled} almost everywhere in time.
The limiting drift in Equation~\eqref{static-eq:scalar-diffusion} is
measurable in time, bounded, and uniformly Lipschitz in space. Hence
the equation has a strong solution and pathwise uniqueness. Couple
the solutions for two step approximations with the same Brownian
motion; Gronwall's inequality gives convergence uniformly in time.
We may therefore also pass the terminal lower bound for $\chi$ to
the limit:
\[
\chi(s,z)\ge\E_{s,z}\operatorname{sech}^2(X_1)>0.
\]
Since $X_1$ is finite almost surely, this lower bound is strictly
positive. It\^o isometry passes the martingale identity to the limit,
and bounded convergence does the same for the expectation identity.
Nothing in these arguments requires the starting point to be
$(0,0)$; they apply at every deterministic starting time and spatial
value. Applying the expectation identity to a diffusion started at
$(s,z)$ and then using the Markov property gives
Equation~\eqref{eq:scalar-evolution}. Finally,
Equation~\eqref{static-eq:scalar-martingale} and the
continuity of $\chi$ and $X$ give
Equation~\eqref{static-eq:gamma-derivative}.
\par\medskip\noindent\textbf{Continuity in the temperature parameter.}\enspace
Write $b=\beta^2$ and repeat the subtraction argument with different
values of $b$. In the equation for $\Phi$, the additional source is
$(\widetilde b-b)(\widetilde\chi+
\alpha\widetilde u^2)/2$, bounded by $C|\widetilde b-b|$.
The bound on $\widetilde\chi_z$ gives the same source bound for
$u$. Thus the differences of $\Phi$ and $u$ are
$O(|\widetilde b-b|)$ locally uniformly for $b>0$.
To obtain continuity of $\chi$, use its uniform spatial Lipschitz
bound and approximate it by a difference quotient of $u$. For
any $h>0$,
\[
\sup_{s,z}|\chi-\widetilde\chi|
\le\frac{2}{h}\sup_{s,z}|u-\widetilde u|+C h.
\]
Choose $h$ small first, then take $\widetilde b$ sufficiently close to
$b$. This proves the required uniform continuity of $\chi$.
The step approximation transfers these parameter estimates to
all distribution functions.
\end{proof}
We next differentiate the scalar functional in two directions: a
change of its distribution function and a change of temperature with
the distribution function fixed. We prove both formulas by differences,
so they require no differentiability of a minimizing measure.
For another distribution function $\widetilde\alpha$, set
$\alpha_\theta=\alpha+\theta(\widetilde\alpha-\alpha)$. The first
variation is
\begin{equation}\label{static-eq:scalar-mass-variation}
\left.\frac{\dd}{\dd\theta}\cP_\beta(\alpha_\theta)
\right|_{\theta=0+}
=\frac{\beta^2}{2}\int_0^1
(\widetilde\alpha-\alpha)(s)(\Gamma(s)-s)\dd s.
\end{equation}
We use $\cP_\beta(\alpha)$ to denote the functional of the measure
having distribution function $\alpha$. With $\alpha$ held fixed,
the temperature derivative of the scalar value is
\begin{equation}\label{static-eq:scalar-temperature-variation}
\frac{\partial}{\partial(\beta^2)}\Phi_\alpha(0,0)
=\frac12\int_0^1
\E\bigl[\chi(s,X_s)+\alpha(s)u(s,X_s)^2\bigr]\dd s.
\end{equation}
For each formula, start with step paths, where the equations are
classical between finitely many breakpoints. Subtraction gives an
integral representation, and that representation passes to general
paths by the stability lemma.
\par\medskip\noindent\textbf{Variation of the distribution function.}\enspace
Subtract the equations for $\Phi_{\alpha_\theta}$ and $\Phi_\alpha$.
The resulting drift is
$\beta^2\alpha(u_{\alpha_\theta}+u_\alpha)/2$, and the source is
$\theta\beta^2(\widetilde\alpha-\alpha)u_{\alpha_\theta}^2/2$.
Introduce $Y^\theta$ by $Y^\theta_0=0$ and
\[
\dd Y_s^\theta=\beta\dd B_s
+\frac{\beta^2}{2}\alpha(s)
(u_{\alpha_\theta}+u_\alpha)(s,Y_s^\theta)\dd s.
\]
The diffusion representation of the difference equation is the
exact identity
\begin{equation}\label{static-eq:scalar-mass-difference}
\frac{\Phi_{\alpha_\theta}(0,0)-\Phi_\alpha(0,0)}{\theta}
=\frac{\beta^2}{2}\E\int_0^1
(\widetilde\alpha-\alpha)(s)
u_{\alpha_\theta}(s,Y_s^\theta)^2\dd s.
\end{equation}
For general $\alpha,\widetilde\alpha$, choose step approximations
converging to each in $L^1$, and take their convex combinations at
a fixed $\theta>0$. Equation~\eqref{static-eq:scalar-stability}
shows that the mixed drifts converge in integrated spatial supremum
norm. They also have a common spatial Lipschitz bound. A coupling
with the same Brownian motion and Gronwall's inequality therefore
gives uniform convergence of the corresponding $Y^\theta$
processes. The sources converge in integrated expectation by
$|u|\le1$, the spatial Lipschitz bound for $u$, its uniform
convergence, and the $L^1$ convergence of
$\widetilde\alpha-\alpha$. Hence
Equation~\eqref{static-eq:scalar-mass-difference} holds for arbitrary
distribution functions.
We may now send $\theta\downarrow0$. The same estimates give
$Y^\theta\to X$ and uniform convergence of $u_{\alpha_\theta}$ to
$u_\alpha$. The right-hand side converges to
$\frac{\beta^2}{2}\int(\widetilde\alpha-\alpha)\Gamma$.
Subtract the derivative of the linear penalty to obtain
Equation~\eqref{static-eq:scalar-mass-variation}.
\par\medskip\noindent\textbf{Variation of the temperature.}\enspace
Set $b=\beta^2$ and use a superscript $b$ for scalar functions with
parameter $\sqrt b$. For $\widetilde b\ne b$, let
$Y^{b,\widetilde b}_0=0$ and define its evolution by
\[
\dd Y_s^{b,\widetilde b}=\sqrt b\dd B_s
+\frac b2\alpha(s)(u^b+u^{\widetilde b})
(s,Y_s^{b,\widetilde b})\dd s.
\]
Subtract the two step-path equations with diffusion coefficient
$b/2$ to obtain
\begin{equation}\label{static-eq:scalar-temperature-difference}
\frac{\Phi^{\widetilde b}(0,0)-\Phi^b(0,0)}{\widetilde b-b}
=\frac12\E\int_0^1
[\chi^{\widetilde b}+\alpha(u^{\widetilde b})^2]
(s,Y_s^{b,\widetilde b})\dd s.
\end{equation}
For fixed $b,\widetilde b$, pass to general $\alpha$ by the same
step approximation and Brownian coupling used above. The additional
term involving $\chi$ is controlled by its uniform spatial
Lipschitz bound. Next let $\widetilde b\to b$. Parameter continuity
in Lemma~\ref{static-lem:scalar-regularity} and the coupling give
$Y^{b,\widetilde b}\to X$. The integrands remain bounded and
converge in integrated expectation. The limit in
Equation~\eqref{static-eq:scalar-temperature-difference} therefore
proves Equation~\eqref{static-eq:scalar-temperature-variation} for
approach from either side.
\subsection{Minimizers and the energy gap}
The mass variation converts minimization into a condition on the
support. We will use that condition both to identify the expected
scalar susceptibility at every time and to show that the minimizing
measure is nontrivial when $\beta>1$.
\begin{proposition}[Support and stationarity identities]
\label{static-prop:scalar-minimizer}
The scalar functional has a minimizer $\mu$. For every such minimizer,
with its associated scalar process, the following assertions hold:
\begin{enumerate}[label=\textup{(\roman*)}]
\item $0\in\supp\mu$, and $\supp\mu\subset[0,1-\eps_0]$ for some
$\eps_0>0$.
\item At every $q\in\supp\mu\cap(0,1)$,
$\Gamma(q)=q$ and $\Gamma'(q)\le1$. At $q=0$,
$\Gamma(0)=0$ and $\Gamma'(0)\le1$.
\item Define
\begin{equation}\label{static-eq:scalar-A-d-definitions}
d_*(s)=1-\int_{(s,1]}t\,\mu(\dd t),\qquad
A(s)=\int_s^1\alpha(t)\dd t=d_*(s)-s\alpha(s).
\end{equation}
Then, at every $s\in[0,1]$,
\begin{equation}\label{static-eq:scalar-stationarity-identity}
\E\bigl[\chi(s,X_s)+\alpha(s)u(s,X_s)^2\bigr]=d_*(s).
\end{equation}
Equivalently, since $d_*(s)=A(s)+s\alpha(s)$,
\begin{equation}\label{eq:scalar-stationarity}
\E\chi(s,X_s)=A(s)+\alpha(s)(s-\Gamma(s))
\qquad(0\le s\le1).
\end{equation}
In particular
\begin{equation}\label{static-eq:scalar-susceptibility}
\chi(0,0)=A(0),\qquad \beta A(0)\le1.
\end{equation}
Furthermore $\alpha(s)>0$ for every $s\in(0,1]$,
$A(h)<A(0)$ for every $h\in(0,1]$, and $d_*(0+)=A(0)$.
\end{enumerate}
\end{proposition}
\begin{proof}
\par\medskip\noindent\textbf{Existence and variational support condition.}\enspace
Weak convergence of probability measures on $[0,1]$ implies
pointwise convergence of their distribution functions at continuity
points of the limit. Bounded convergence then gives convergence in
$L^1$. By Lemma~\ref{static-lem:scalar-regularity}, the functional
$\cP_\beta$ is continuous for this topology. The space of probability
measures on $[0,1]$ is compact, so a minimizer exists.
Fix any minimizer and define
\[
G(q)=\int_q^1(\Gamma(s)-s)\dd s.
\]
Fubini's theorem and Equation~\eqref{static-eq:scalar-mass-variation}
give $\int G\dd\widetilde\mu\ge\int G\dd\mu$ for every
probability measure $\widetilde\mu$. Testing with point masses shows
that $\mu$ is supported on the set of global minima of $G$.
The function $G$ is twice continuously differentiable and satisfies
\[
G'(q)=q-\Gamma(q),\qquad G''(q)=1-\Gamma'(q).
\]
The first- and second-derivative tests now give the claimed identities
and inequalities at interior support points. At the right endpoint,
$\Gamma(1)=\E\tanh^2(X_1)<1$, and hence $G'(1)>0$.
Thus $1$ does not minimize $G$. Its minimum set is compact, so
$\supp\mu$ stays a positive distance below $1$.
\par\medskip\noindent\textbf{The left endpoint of the support.}\enspace
Let $q_0=\min\supp\mu$. We claim that $q_0=0$. Suppose instead
that $q_0\in(0,1)$. Then $\alpha=0$ on $[0,q_0)$, so
$X_s=\beta B_s$ there and the equation for $\chi$ is the backward
heat equation. Consequently $\chi(s,X_s)$ is a bounded martingale
on $0\le s\le q_0$. One can see this either from the heat
semigroup or by choosing step approximations identically zero below
$q_0$. Its second moment is nondecreasing, as is $\Gamma'(s)$.
But $\Gamma(q_0)=q_0$, $\Gamma(0)=0$, and
$\Gamma'(q_0)\le1$. Thus the average of $\Gamma'$ on this
interval is $1$, whereas its maximum is at most $1$. Continuity
forces $\Gamma'\equiv1$ on $[0,q_0]$.
To contradict this constancy, observe that $\chi(q_0,\cdot)$ is
positive everywhere and cannot be constant: otherwise
$u(q_0,\cdot)$ would have a positive constant derivative, contrary
to $|u|\le1$. The function $\chi(q_0,\cdot)$ is continuous, and
$X_{q_0}$ is a nondegenerate Gaussian with full support. Therefore
$\Var(\chi(q_0,X_{q_0}))>0$, while $\chi(0,X_0)$ is
deterministic. The martingale's second moment strictly increases
between these endpoints, which contradicts the constancy of
$\Gamma'$. Hence $q_0=0$. At this endpoint, $G$ has a minimum
and symmetry gives $G'(0)=0$. The one-sided second-derivative test
then yields $\Gamma'(0)\le1$.
\par\medskip\noindent\textbf{Stationarity with atoms included.}\enspace
We have located the support and established $\Gamma(t)=t$ on it.
It remains to turn these support identities into an identity at every
time. Use Equation~\eqref{static-eq:scalar-chi-expectation}, the
terminal relation $\chi(1,z)=1-u(1,z)^2$, and
Equation~\eqref{static-eq:gamma-derivative} to write
\[
\E\chi(s,X_s)=1-\Gamma(1)+\int_s^1\alpha(t)\Gamma'(t)\dd t.
\]
Stieltjes integration by parts with the right-continuous
$\alpha$ gives the following formula, with every atom retained:
\[
\int_s^1\alpha(t)\Gamma'(t)\dd t
=\Gamma(1)-\alpha(s)\Gamma(s)
-\int_{(s,1]}\Gamma(t)\,\mu(\dd t).
\]
In particular, the integral uses $(s,1]$, while the endpoint term
uses $\alpha(s)$. Substituting $\Gamma(t)=t$ on $\supp\mu$,
including at zero, proves
Equation~\eqref{static-eq:scalar-stationarity-identity}. Fubini's
theorem gives the expression for $A$ in
Equation~\eqref{static-eq:scalar-A-d-definitions}. At $s=0$ the
stationarity identity becomes $\chi(0,0)=A(0)$. Finally,
$\Gamma'(0)=\beta^2\chi(0,0)^2\le1$ proves
Equation~\eqref{static-eq:scalar-susceptibility}.
Because $0\in\supp\mu$, each interval $[0,s]$ with $s>0$
has positive $\mu$-mass. Thus
$A(0)-A(h)=\int_0^h\alpha(s)\dd s>0$ for every $h\in(0,1]$.
The identity $d_*(s)=A(s)+s\alpha(s)$, together with continuity
of $A$ at zero, also gives $d_*(0+)=A(0)$.
\end{proof}
For $\beta>1$, the susceptibility bound excludes the measure
$\delta_0$. This makes the temperature derivative at a fixed
minimizer strictly smaller than $\beta/2$. A finite-volume secant
estimate transfers that strict inequality to the equilibrium energy.
\begin{lemma}[Strict bound on the equilibrium energy]\label{static-lem:energy-gap}
For each fixed $\beta>1$, there are $e_0=e_0(\beta)<\beta/2$ and
$N=N(\beta)$ such that
\begin{equation}\label{static-eq:energy-gap}
p_n'(\beta)=\frac1n\E\pi_{n,\beta}^J(H_n^J)\le e_0
\qquad(n\ge N).
\end{equation}
\end{lemma}
\begin{proof}
Choose a minimizing measure $\mu$ at the specified $\beta$ and
hold that measure fixed during the temperature variation. Combine
Equations~\eqref{static-eq:scalar-temperature-variation} and
\eqref{static-eq:scalar-stationarity-identity}, and differentiate
the penalty, to obtain
\begin{equation}\label{static-eq:fixed-minimizer-temperature-derivative}
\left.\frac{\dd}{\dd\gamma}\cP_\gamma(\mu)\right|_{\gamma=\beta}
=\beta\int_0^1\bigl(d_*(s)-s\alpha(s)\bigr)\dd s
=\beta\int_0^1 A(s)\dd s
=\frac{\beta}{2}\left(1-\int_0^1t^2\,\mu(\dd t)\right).
\end{equation}
The last equality follows from two applications of Fubini's theorem.
If $\mu=\delta_0$, then $A(0)=1$, contradicting
$\beta A(0)\le1$ for $\beta>1$. The second moment of $\mu$ is
therefore positive. Denote the derivative in
Equation~\eqref{static-eq:fixed-minimizer-temperature-derivative}
by $L$; we have proved $L<\beta/2$.
Choose $e_0$ with $L<e_0<\beta/2$. Then fix $\eps>0$
sufficiently small that
\[
\frac{\cP_{\beta+\eps}(\mu)-\cP_\beta(\mu)}{\eps}<e_0.
\]
For every $n$, the finite-volume pressure is differentiable and
convex in $\beta$. Bound its derivative by the forward secant,
use the scalar upper trial bound at $\beta+\eps$, and apply
Theorem~\ref{static-thm:finite-size} at $\beta$. This gives
\[
p_n'(\beta)
\le\frac{p_n(\beta+\eps)-p_n(\beta)}{\eps}
\le\frac{\cP_{\beta+\eps}(\mu)-P_\beta}{\eps}
+\frac{C_\beta}{\eps}n^{-1/24}.
\]
Since $\cP_\beta(\mu)=P_\beta$, the fixed secant slope is
strictly below $e_0$. Its strict margin absorbs the error once $n$
is sufficiently large. Finally, differentiate the finite partition
sum and then take Gaussian expectation to obtain the energy identity
in Equation~\eqref{static-eq:energy-gap}. The integrable bound
$\max_{\sigma\in\Sig}|H_n^J(\sigma)|$ justifies this passage.
The argument has varied only the temperature in the functional of
one fixed measure; it requires no differentiability of $P_\beta$
or of a choice of minimizer as a function of $\beta$.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/03-scalar.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/04-locking.tex ====
\section{Locking small overlaps across a large overlap}
\label{static-sec:locking}
When two configurations have a positive overlap bounded away from zero,
we will show that their small overlaps with a third configuration are
close, including their signs. This is the fixed-scale estimate used by
the crossing construction. Spin reversal will then give the
absolute-overlap version needed in Appendix~\ref{fixed-sec:crossings}.
Throughout the section, fix $\beta>1$ and a minimizer $\mu$ of the scalar
functional. We use the notation of Section~\ref{static-sec:scalar} and
write $\pi=\pi_{n,\beta}^J$. The scalar inputs are
\begin{equation}
\label{static-eq:locking-scalar-identities}
\begin{gathered}
0\leq\chi\leq1,\qquad |u|\leq1,\qquad
|\chi(s,z)-\chi(s,z')|\leq C_\beta|z-z'|,\\
\E\chi(s,X_s)=A(s)+\alpha(s)\bigl(s-\Gamma(s)\bigr),\qquad
\E\bigl[\chi(s,X_s)+\alpha(s)u(s,X_s)^2\bigr]=d_*(s),\\
\beta A(0)\leq1,\qquad A(h)<A(0)\quad(0<h\leq1),\qquad
d_*(s)\longrightarrow A(0)\quad(s\downarrow0).
\end{gathered}
\end{equation}
We also use the scalar martingale evolution from
Lemma~\ref{static-lem:scalar-regularity} and the bounds
$\alpha(s)>0$ for $s>0$ and $0\le d_*(s)\le1$ from
Proposition~\ref{static-prop:scalar-minimizer}.
For a nonempty Gram constraint $D$, write $Z_{d,n}(D)$ for the
unnormalized partition sum in~\eqref{static-eq:constrained-pressure}.
Thus $F_{d,n}(D)=n^{-1}\E\log Z_{d,n}(D)$, and division by
$Z_{n,\beta}^d$ gives the probability of $D$ under $d$ independent
Gibbs draws.
\begin{lemma}[Signed fixed-scale locking]\label{lem:fixed-locking}
Let $e_*=1/100$. For every fixed $h\in(0,1)$, there is $b>0$ such that,
for all sufficiently large $n$,
\begin{equation}\label{eq:fixed-locking}
\E\pi^{\otimes3}\!\left\{
\begin{gathered}
R(x,y)\ge h,\qquad |R(v,x)|,|R(v,y)|\le b,\\
|R(v,x)-R(v,y)|>2n^{-e_*}
\end{gathered}\right\}\le e^{-n^{9/10}}.
\end{equation}
The same conclusion holds after decreasing $b$.
\end{lemma}
We will prove a pressure deficit for each forbidden overlap matrix and
then sum probability bounds. The conversion below will also apply to
the shrinking gaps in Section~\ref{sec:narrow}.
\begin{lemma}[From a pressure gap to a probability bound]
\label{lem:pressure-prob}
Fix a replica number $d$. If a feasible overlap matrix $D$ satisfies
\[
F_{d,n}(D)\le dP_\beta-\eta,
\qquad \eta\ge2dC_\beta n^{-1/24},
\]
where $C_\beta$ is the constant in
Theorem~\ref{static-thm:finite-size}, then, for all sufficiently large $n$,
\begin{equation}
\label{static-eq:pressure-gap-probability}
\E\frac{Z_{d,n}(D)}{Z_{n,\beta}^d}
\le \exp(-n\eta/4)+\exp(-c_{\beta,d}n\eta^2).
\end{equation}
The constants are uniform over $D$ and over $n$-dependent gaps satisfying
the displayed lower bound.
\end{lemma}
\begin{proof}
For fixed $d$, the derivative of $\log Z_{d,n}(D)$ with respect to any
$J_{ij}$ has absolute value at most $\beta d/\sqrt n$. Both this function
and $d\log Z_{n,\beta}$ therefore have Gaussian Lipschitz constant at most
$\beta d\sqrt{(n-1)/2}$. Their difference
\[
T_D:=\log Z_{d,n}(D)-d\log Z_{n,\beta}
\]
has Lipschitz constant at most $C_{\beta,d}\sqrt n$ and satisfies
$T_D\le0$, since $e^{T_D}$ is a Gibbs probability.
Theorem~\ref{static-thm:finite-size} and the pressure-gap hypothesis give
$\E T_D\le-n\eta/2$. Gaussian concentration bounds
$\Prob\{T_D>-n\eta/4\}$ by $\exp(-c_{\beta,d}n\eta^2)$.
On its complement $e^{T_D}\le e^{-n\eta/4}$, and on the event itself
$e^{T_D}\le1$. Summing proves the claim. This calculation uses the
original off-diagonal disorder, so no auxiliary diagonal Gaussian
appears in the probability bound.
\end{proof}
Only $d=2,3$ will be needed. There are at most
$(n+1)^{d(d-1)/2}$ feasible overlap matrices, so a union over all
constraints costs only a polynomial factor.
\subsection{A pair constraint}
The three-replica comparison will use a large overlap $q$ for which
$\Gamma(q)$ is close to $q$. We first show that the remaining pair
overlaps already have an exponentially small Gibbs probability.
\begin{lemma}[Pair pressure bound]
\label{static-lem:pair-bound}
For every nonempty pair constraint with overlap $q\in[0,1]$,
\begin{equation}
\label{static-eq:pair-pressure}
F_{2,n}\!\left(\begin{smallmatrix}1&q\\q&1\end{smallmatrix}\right)
\leq 2P_\beta-\frac12\bigl(\Gamma(q)-q\bigr)^2.
\end{equation}
Consequently, for every fixed $\tau>0$, the averaged Gibbs probability
of pairs satisfying
$\bigl|\Gamma(|R(x,y)|)-|R(x,y)|\bigr|>\tau$
is at most $C\exp(-cn)$ for all large $n$.
\end{lemma}
\begin{proof}
Use Proposition~\ref{static-prop:matrix-bound} with two coordinates
sharing their covariance increments up to scalar time $q$ and receiving
independent increments thereafter. Give the common increments masses
$\alpha(s)/2$ and the independent increments masses $\alpha(s)$.
The endpoint is the prescribed Gram matrix.
At zero terminal multiplier this path has recursive value
$2\Phi(0,0)$. Indeed, integrating the independent portion first leaves
$\Phi(q,z_x)+\Phi(q,z_y)$. On the earlier common portion $z_x=z_y$;
integration of twice a scalar value at mass $\alpha/2$ is twice the
scalar integration at mass $\alpha$. The penalty has the same factor
two: before the split the squared norm is $4s^2$ with mass $\alpha/2$,
and after the split its varying part is $2s^2$ with mass $\alpha$.
Now add $L s_xs_y$ to the terminal exponent and denote the recursive
value by $\Psi(L)$. In the matrix convention this sets
$\lambda_{xy}=\lambda_{yx}=L/2$, so $\lambda:D=Lq$. Differentiating at
$L=0$ transports the terminal observable $s_xs_y$ backwards by
conditional expectation under the unperturbed tilted path. Given the
common field at $q$, the two later branches are independent and each
has conditional spin mean $u(q,X_q)$. Hence
\[
\Psi'(0)=\E u(q,X_q)^2=\Gamma(q).
\]
We also need the bound $\Psi''(L)\leq1$ for every $L$. For a terminal
observable $O\in[-1,1]$, the terminal Hessian is its thermal variance.
Backward integration at mass $m\in[0,1]$ averages the existing Hessian
and adds $m$ times the conditional variance of the gradient. Iterating
this formula and using the total-variance decomposition bounds the
result by $\Var(O)\leq1$ under the tilted path law. An outer quenched
expectation preserves the bound. Taylor's inequality therefore gives
\[
F_{2,n}(q)\leq2P_\beta+L\bigl(\Gamma(q)-q\bigr)+\frac{L^2}{2}.
\]
The choice $L=q-\Gamma(q)$ yields~\eqref{static-eq:pair-pressure}.
These computations are first made with step approximations of $\alpha$
whose segment endpoints include $q$. Lemma~\ref{static-lem:scalar-regularity}
gives convergence of the scalar values, derivatives, and tilted
transitions, so the inequality passes to the minimizing $\alpha$.
Finally, reversing one spin configuration preserves its Gibbs weight
and changes the sign of its overlap. The same pressure estimate thus
applies with the absolute overlap in place of $q$. There are at most
$n+1$ pair overlaps; summing~\eqref{static-eq:pressure-gap-probability}
with the fixed deficit $\tau^2/2$ proves the probability assertion.
\end{proof}
\subsection{A three-coordinate path with an asymmetric perturbation}
\label{static-subsec:locking-base}
Fix $q\in[h,1]$ and a nonempty constraint
\begin{equation}
\label{static-eq:locking-gram}
D=\begin{pmatrix}
1&r+\delta&r-\delta\\
r+\delta&1&q\\
r-\delta&q&1
\end{pmatrix},\qquad r\geq0.
\end{equation}
We will choose upper bounds for $r$ and $|\delta|$ depending only on
$\beta,h$. Write $D_0$ for the same matrix with $\delta=0$. Starting
from a hierarchy ending at $D_0$, we will add opposite field increments
to the $x$ and $y$ coordinates. This realizes the desired first-order
change in their overlaps with $v$. Keeping the increments positive
semidefinite also adds a second-order covariance term. The free-endpoint
matrix bound charges only its squared size, so that correction has a
fourth-order cost.
The reason to perturb an early increment is the strict inequality
$\beta^2A(h)A(0)<1$, supplied by
\eqref{static-eq:locking-scalar-identities}. At a small branching time
$r$, the two scalar branches are nearly independent. Their
susceptibility product is then bounded above by $A(h)A(0)$, up to
small errors.
The Gaussian variance part of the perturbation will cancel the
corresponding penalty up to the pair deviation $q-\Gamma(q)$; this
strict product inequality controls what remains.
The base hierarchy itself is valid for every $0\le r\le q\le1$;
only the later perturbation will require $r$ to be small. We use the
three-replica construction in Talagrand's discussion following
\cite[Proposition~4.4]{Talagrand07}, with masses divided by the size
of each common block. Its three stages, in local scalar time $s$, are:
\begin{enumerate}[label=\textup{(\roman*)}]
\item From $0$ to $r$, all three coordinates share each increment,
integrated at mass $\alpha(s)/3$.
\item From $r$ to $q$, the coordinate $v$ follows an independent branch
at mass $\alpha(s)$, while $x,y$ share a branch at mass
$\alpha(s)/2$.
\item From $q$ to $1$, all three coordinates have independent increments
at mass $\alpha(s)$.
\end{enumerate}
In the middle stage each branch has its own clock. If these clocks are
$a$ for $v$ and $b$ for the common $x,y$ branch, the current covariance is
\[
Q(a,b)=\begin{pmatrix}a&r&r\\r&b&b\\r&b&b\end{pmatrix},
\qquad r\le a,b\le q.
\]
Thus one branch can advance while the other is held fixed. For a step
function $\alpha$, merge the two finite lists of increments in nondecreasing
order of mass, retaining the order within each list. Each list is
nondecreasing, so such a merge exists; ties may be ordered in either
way consistent with the two local orders. The first-stage masses do
not exceed either middle starting mass, and the middle masses do not
exceed the starting mass of the last stage. Thus the entire schedule
is a valid nondecreasing hierarchy. Insert a dummy initial covariance
$Q_0=0$ to ensure that, even when $\alpha(0)>0$, the first genuine field
increment has its prescribed positive mass and the quenched initial
level has zero covariance.
The endpoint is $D_0$, and the recursive value minus the penalty equals
$3\cP_\beta(\mu)$. To check both assertions about the value, observe
that a common block of size $\ell$ has scalar value $\ell\Phi$.
Its mass $\alpha/\ell$ gives the ordinary scalar tilted transition,
and its squared-norm differential $\ell^2\dd(s^2)$ contributes
$\ell$ times the scalar penalty after multiplication by that mass.
Independent branches have additive values. When one branch is
integrated, the value of a frozen branch is independent of that
increment and passes unchanged through the logarithm. Applying these
facts at each split proves the claimed value and penalty for every
allowed interleaving. It also identifies the base tilted law. Denote the $v$ path and the
shared $x,y$ path by $X^v$ and $X^x$, and their common field at the
first split by $X_r$. Write $K_{r,z}$ for the scalar path law started at $(r,z)$ and run to time $q$.
If $\mathcal X_0$ denotes the common prefix through $r$, then
\begin{equation}\label{static-eq:locking-product-law}
\mathcal L(X^v_{[r,q]},X^x_{[r,q]}\mid\mathcal X_0)
=K_{r,X_r}\otimes K_{r,X_r}.
\end{equation}
Indeed, the frozen branch cancels from each normalized kernel, while
the factor three or two in the common-block value cancels the divided
mass. The two future branches are therefore independent conditional on
$X_r$, for every allowed merge. We will use this same base hierarchy
in Section~\ref{sec:narrow}.
For the fixed-scale perturbation, now fix $0<s_1<s_2<h$ and suppose
$r<s_1$. Define
\[
k(s)=\frac{\1_{[s_1,s_2]}(s)}{s_2-s_1},\qquad
g(s)=\int_0^s k(t)\dd t,\qquad
G(s)=\int_0^s k(t)^2\dd t,\qquad
K=G(s_2)=\frac1{s_2-s_1}.
\]
Let $L_{vx}=L_{xv}=1$, $L_{vy}=L_{yv}=-1$, with all other entries
of $L$ zero. Let $L_2$ vanish outside its $(x,y)$ block and give that
block the value
$\left(\begin{smallmatrix}1&-1\\-1&1\end{smallmatrix}\right)$.
Then
\[
|L|^2=|L_2|^2=4,\qquad L:L_2=0.
\]
During each $v$-increment with local clock in $[s_1,s_2]$, add
$\delta k$ times that increment to $x$ and its negative to $y$.
At the covariance level, this replaces $\dd s\,e_ve_v^{\mathsf T}$ by
\[
\dd s\,
\bigl(e_v+\delta k(e_x-e_y)\bigr)
\bigl(e_v+\delta k(e_x-e_y)\bigr)^{\mathsf T}.
\]
This rank-one form is positive semidefinite. The accumulated change
is $\delta gL+\delta^2GL_2$ until the modification ends, and is then
held fixed. The new endpoint is consequently
\[
\widetilde D=D_0+\delta L+\delta^2KL_2
=D+\delta^2KL_2.
\]
Apply Proposition~\ref{static-prop:matrix-bound} with zero terminal
multiplier and this endpoint. Its mismatch contribution is
\begin{equation}
\label{static-eq:locking-mismatch}
\frac{\beta^2}{4}|D-\widetilde D|^2
=\beta^2\delta^4K^2.
\end{equation}
We next isolate the beneficial increase in the penalty. Along the
base path, $Q:L=0$. Also $Q:L_2=0$ until the split at $q$, by which
time $g=1$ and $G=K$. In the final stage, $Q:L_2=2(s-q)$. Thus
\[
|\widetilde Q|^2-|Q|^2
=4\delta^2g^2+2\delta^2G(Q:L_2)+4\delta^4G^2.
\]
The last term has nonnegative increments. The first changes only
during the $v$-increments, whose masses are $\alpha(s)$; the middle
term changes only in the independent final stage. The penalty
increase is therefore at least
\begin{equation}
\label{static-eq:locking-penalty}
\beta^2\delta^2\left(
\int_0^1\alpha(s)\dd(g(s)^2)+K A(q)\right).
\end{equation}
\subsection{The recursive value under the perturbation}
To compare this penalty gain with the recursive value, first integrate
out the unchanged independent final stage. Let $z$ be the common base
field of $x,y$ at the end of the middle stage, and let $Y$ be the sum
of the raw $v$-increments weighted by $k$. The boundary value changes by
\begin{equation}
\label{static-eq:locking-boundary-increment}
\Delta=\Phi(q,z+\delta Y)+\Phi(q,z-\delta Y)-2\Phi(q,z).
\end{equation}
Convexity, the bound $\chi\leq1$, and its spatial Lipschitz bound yield
\begin{equation}
\label{static-eq:locking-taylor}
0\leq\Delta\leq\delta^2Y^2,
\qquad
\bigl|\Delta-\delta^2\chi(q,z)Y^2\bigr|
\leq C_\beta|\delta Y|^3.
\end{equation}
The remaining backward integrations are nonlinear, so a bound on the
boundary change must be propagated through their normalized kernels.
If the future value changes by $U$, an integration at positive mass
$m$ changes by
\[
\frac1m\log\E_{\rm base}
[e^{mU}\mid\text{current prefix}],
\]
where the expectation uses the unperturbed tilted kernel. At zero
mass, the change is its conditional expectation of $U$. For
$0<m\leq1$, Jensen's inequality and the power-mean inequality put
the displayed expression between
$\E_{\rm base}[U\mid\text{prefix}]$ and
$\log\E_{\rm base}[e^U\mid\text{prefix}]$.
Here is the resulting induction explicitly. Let $K_i$ be the
normalized base kernel at level $i$, and $U_i$ the difference of
recursive values at that prefix. Taking the ratio of the two defining
integrals gives $U_i=m_i^{-1}\log K_i e^{m_iU_{i+1}}$ for $m_i>0$;
this identity uses the base kernel only. Put
\[
A_i=\E_{\rm base}[\Delta\mid\text{prefix }i],\qquad
B_i=\log\E_{\rm base}[e^\Delta\mid\text{prefix }i].
\]
At the boundary these equal $U_i=\Delta$. If
$A_{i+1}\leq U_{i+1}\leq B_{i+1}$, the kernel inequalities and tower
property give
\[
U_i\geq K_iU_{i+1}\geq A_i,\qquad
U_i\leq\log K_ie^{U_{i+1}}
\leq\log K_ie^{B_{i+1}}=B_i.
\]
Jensen's inequality gives the same conclusions at zero mass. Applying
this induction through the hierarchy proves
\begin{equation}
\label{static-eq:locking-recursive-sandwich}
\E_{\rm base}\Delta
\ \leq\ \widetilde\Psi-\Psi
\ \leq\ \log\E_{\rm base}e^\Delta.
\end{equation}
Jensen's inequality also preserves these bounds under an outer
quenched expectation.
Under the base tilted law, take scalar diffusions $X^v,X^x$ that agree
through time $r$ and have independent Brownian noises thereafter.
The identified path law gives
\[
z=X_q^x,\qquad
Y=\int_{s_1}^{s_2} k(s)\dd X_s^v,
\qquad
\dd X_s^v=\beta\dd B_s^v
+\beta^2\alpha(s)u(s,X_s^v)\dd s.
\]
Thus $Y$ is a centered Gaussian increment of variance $\beta^2K$
plus a possibly dependent random term of absolute value at most
$\beta^2$. In particular, there are $c,C>0$, depending only on
$\beta,s_1,s_2$, such that
\begin{equation}
\label{static-eq:locking-Y-moment}
\E_{\rm base}e^{cY^2}\leq C.
\end{equation}
This estimate is uniform in $q\geq h$, $r<s_1$, and the step
approximations of $\alpha$. For small enough $|\delta|$,
\eqref{static-eq:locking-taylor} and this exponential moment imply
\[
\E\bigl(e^\Delta-1-\Delta\bigr)
\leq\tfrac12\E\bigl[\Delta^2e^\Delta\bigr]
\leq C\delta^4.
\]
Combining with~\eqref{static-eq:locking-recursive-sandwich} bounds the
recursive change by
\begin{equation}
\label{static-eq:locking-recursive-expansion}
\widetilde\Psi-\Psi
\leq\delta^2\E\bigl[\chi(q,X_q^x)Y^2\bigr]+C|\delta|^3.
\end{equation}
At this point all matrix paths can still be taken finite. Include
$r,q,s_1,s_2$ in their segment endpoints and choose nondecreasing step
approximations converging to $\alpha$ in $L^1([0,1])$.
Lemma~\ref{static-lem:scalar-regularity} gives convergence of the scalar
values, derivatives, and diffusions. Couple the diffusions with the
same Brownian noises, keeping the noises independent after $r$.
The uniform exponential moment~\eqref{static-eq:locking-Y-moment}
then justifies passage to the limit in the expectation in
\eqref{static-eq:locking-recursive-expansion}. The penalty integrals
converge as well: $\dd(g^2)/\dd s$ is bounded and supported on
$[s_1,s_2]$, and $A(q)$ is an integral of $\alpha$.
We have therefore applied the matrix comparison only to finite
hierarchies and passed their upper bounds to the limit. No infinite
hierarchy has been differentiated. In particular, convergence at a
possible atom $q$ is unnecessary; only $L^1$ convergence of the
masses is used. The minimizing identities will be applied after this
limit.
Subtracting~\eqref{static-eq:locking-penalty} from
\eqref{static-eq:locking-recursive-expansion} and including the
fourth-order cost~\eqref{static-eq:locking-mismatch} gives
\begin{equation}
\label{static-eq:locking-three-pressure-expansion}
\begin{split}
F_{3,n}(D)\leq 3P_\beta+\delta^2\mathcal B(q,r)+C|\delta|^3,\\
\mathcal B(q,r)=
\E\bigl[\chi(q,X_q^x)Y^2\bigr]
-\beta^2KA(q)-\beta^2\int_0^1\alpha(s)\dd(g(s)^2).
\end{split}
\end{equation}
The task is now to choose the perturbation interval so that this
quadratic coefficient is strictly negative.
\subsection{Choosing an interval with a strict quadratic gain}
The strict inequality $A(h)<A(0)$ will supply the gain. To expose it,
define
\[
V_t=\chi(t,X_t^v)+\alpha(t)u(t,X_t^v)^2,
\qquad Y_s=\int_{s_1}^{s}k(t)\dd X_t^v
\quad(s\in[s_1,s_2]).
\]
The martingale identity
$\dd u(s,X_s^v)=\beta\chi(s,X_s^v)\dd B_s^v$ and It\^o's formula,
conditional on the common starting field $X_r$, give for almost every
$s$
\[
\frac{\dd}{\dd s}\E[Y_su(s,X_s^v)\mid X_r]
=\beta^2 k(s)\E[V_s\mid X_r].
\]
Since $Y_{s_1}=0$, apply It\^o's formula once more to $Y_s^2$ and use
conditional independence of the $v$ and $x$ branches. This yields
\begin{equation}
\label{static-eq:locking-Ito}
\begin{split}
\E\bigl[\chi(q,X_q^x)Y^2\bigr]
={}&\beta^2K\E\chi(q,X_q)\\
&+2\beta^4\int_{s_1}^{s_2}\alpha(s)k(s)
\int_{s_1}^{s}k(t)
\E\bigl[\chi(q,X_q^x)V_t\bigr]\dd t\dd s.
\end{split}
\end{equation}
All factors except $Y$ are bounded, and its required moments follow
from~\eqref{static-eq:locking-Y-moment}. The conditioning and integrations
are therefore justified.
Only the common field at the small time $r$ prevents the product
expectation in this formula from factoring. Its effect is bounded by
\begin{equation}
\label{static-eq:locking-decoupling}
\left|\E\bigl[\chi(q,X_q^x)V_t\bigr]
-\E\chi(q,X_q)\,d_*(t)\right|
\leq C_\beta\sqrt r,
\qquad s_1\leq t\leq s_2,\quad h\leq q\leq1.
\end{equation}
To prove this estimate, restart the $x$ branch at an independent copy
$X_r'$ of $X_r$, independent also of the whole $v$ branch, and retain
its post-$r$ Brownian noise. The drift $\beta^2\alpha u$ is uniformly
Lipschitz in space, so the two $x$ solutions at time $q$ differ by at
most $e^{\beta^2}|X_r-X_r'|$. Use the spatial Lipschitz bound for
$\chi$, the bound $0\leq V_t\leq2$, and
$\E|X_r-X_r'|\leq C_\beta\sqrt r$, the latter following from the
bounded drift. The replacement has the same marginal law and is
independent of $V_t$; its expectation therefore factors, with
$\E V_t=d_*(t)$ by~\eqref{static-eq:locking-scalar-identities}.
The two inequalities $\beta A(0)\leq1$ and $A(h)<A(0)$ imply
\[
\beta^2 A(h)A(0)<1.
\]
Choose $s_2\in(0,h)$ and $\gamma>0$ small enough that
\begin{equation}
\label{static-eq:locking-margin}
\beta^2A(h)\sup_{0<t\leq s_2}d_*(t)\leq1-4\gamma,
\end{equation}
and set $s_1=s_2/2$. The functions $k,g$ and the constant $K$ are now
fixed. Also set
\[
W=2\beta^2\int_{s_1}^{s_2}\alpha(s)k(s)
\int_{s_1}^{s}k(t)\dd t\dd s>0.
\]
The strict positivity follows from $\alpha(s)>0$ for every $s>0$.
Let $\bar\chi(q)=\E\chi(q,X_q)$. For $q\geq h$ with
$|q-\Gamma(q)|\leq\tau$, the all-time scalar identity gives
\[
\bar\chi(q)=A(q)+\alpha(q)(q-\Gamma(q))\leq A(h)+\tau.
\]
First choose $\tau>0$ small enough and then
$r_0\in(0,s_1)$ small enough. Equations
\eqref{static-eq:locking-decoupling} and
\eqref{static-eq:locking-margin} give
\begin{equation}
\label{static-eq:locking-negative-factor}
\beta^2\E\bigl[\chi(q,X_q^x)V_t\bigr]-1\leq-2\gamma
\end{equation}
uniformly for $q\in[h,1]$ with $|q-\Gamma(q)|\leq\tau$,
$0\leq r\leq r_0$, and $t\in[s_1,s_2]$. More explicitly, since
$0\leq d_*(t)\leq1$, the left side is bounded above by
$-4\gamma+\beta^2\tau+C_\beta\sqrt r$.
We can now compare the entire quadratic term with this strict margin.
Using $\dd(g^2)=2gk\dd s$, substitute~\eqref{static-eq:locking-Ito}
into~\eqref{static-eq:locking-three-pressure-expansion}. The result is
the exact identity
\begin{equation}
\label{static-eq:locking-bracket}
\begin{split}
\mathcal B(q,r)
={}&\beta^2K\alpha(q)\bigl(q-\Gamma(q)\bigr)\\
&+2\beta^2\int_{s_1}^{s_2}\alpha(s)k(s)
\int_{s_1}^{s}k(t)
\left\{\beta^2\E\bigl[\chi(q,X_q^x)V_t\bigr]-1\right\}
\dd t\dd s.
\end{split}
\end{equation}
Decrease $\tau$ further until $\beta^2K\tau\leq\gamma W$. The first
term in~\eqref{static-eq:locking-bracket} is then at most $\gamma W$, and
\eqref{static-eq:locking-negative-factor} bounds the double integral
by $-2\gamma W$. Hence
\[
\mathcal B(q,r)\leq\gamma W-2\gamma W=-\gamma W.
\]
The cubic remainder in~\eqref{static-eq:locking-three-pressure-expansion}
is uniform over these $q,r$. Choose $\delta_0>0$ small enough to
absorb it. For some $c_0>0$ we have obtained
\begin{equation}
\label{static-eq:locking-three-gap}
F_{3,n}(D)\leq3P_\beta-c_0\delta^2
\end{equation}
whenever the constraint is nonempty,
$0\leq r\leq r_0$, $|\delta|\leq\delta_0$, $q\in[h,1]$, and
$|q-\Gamma(q)|\leq\tau$. In particular, this estimate places no
restriction $r\geq|\delta|$ on the two small signed overlaps.
\subsection{From the pressure gap to locking}
\begin{proof}[Proof of Lemma~\ref{lem:fixed-locking}]
Choose $b\le\min(r_0,\delta_0)$. For a triple in
\eqref{eq:fixed-locking}, set $q=R(x,y)\ge h$ and reverse $v$ if
necessary to make
\[
r=\frac{R(v,x)+R(v,y)}2\ge0,
\qquad \delta=\frac{R(v,x)-R(v,y)}2.
\]
This reversal preserves the Gibbs weight, the large overlap $q$, and
$|\delta|$. The event gives $r,|\delta|\le b$ and
$|\delta|>n^{-e_*}$. In particular, we do not need the two small
overlaps to have the same sign.
Lemma~\ref{static-lem:pair-bound} bounds pairs with
$|q-\Gamma(q)|>\tau$ by $C\exp(-cn)$ in averaged probability.
Adding an independent Gibbs draw does not change that bound.
For every remaining feasible triple,
\eqref{static-eq:locking-three-gap} gives a pressure deficit
$\eta=c_0\delta^2\ge c_0n^{-2e_*}$. Since
$2e_*=1/50<1/24$, this dominates the finite-size error in
Lemma~\ref{lem:pressure-prob}, uniformly over the constraints.
Their averaged probabilities are bounded by
\[
\exp(-c n^{49/50})+\exp(-c n^{24/25}).
\]
There are at most $(n+1)^3$ overlap triples and two choices of the
reversal. The resulting bound is at most
$C(n+1)^3\exp(-c n^{24/25})$, which is smaller than
$\exp(-n^{9/10})$ for all sufficiently large $n$.
\end{proof}
\begin{proposition}[Absolute-overlap locking]
\label{static-prop:locking}
For every fixed $h\in(0,1)$ there is $b>0$ such that, for all
sufficiently large $n$,
\begin{equation}\label{static-eq:locking-probability}
\E\pi^{\otimes3}\!\left\{
\begin{gathered}
|R(x,y)|\ge h,\qquad |R(v,x)|,|R(v,y)|\le b,\\
\bigl||R(v,x)|-|R(v,y)|\bigr|>2n^{-1/100}
\end{gathered}\right\}\le \exp(-n^{9/10}).
\end{equation}
The same conclusion holds after decreasing $b$.
\end{proposition}
\begin{proof}
Take $b$ from Lemma~\ref{lem:fixed-locking}. Reverse $y$ if needed to
make $R(x,y)\ge h$. This preserves the Gibbs weight and all absolute
overlaps in the event. The inequality
\[
\bigl||R(v,x)|-|R(v,y)|\bigr|
\le |R(v,x)-R(v,y)|
\]
then puts the transformed triple in the signed event. The two possible
reversals multiply its probability bound by at most two. The stronger
bound $C(n+1)^3\exp(-c n^{24/25})$ obtained in the preceding proof
absorbs this factor and gives the claimed exponent.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/04-locking.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/05-energy.tex ====
\section{Energy and disorder rotation}\label{sec:energy}
The strict energy gap from Lemma~\ref{static-lem:energy-gap} will force
overlap coverage on both stretched-exponential and polynomial mass scales.
This section supplies the two estimates behind that implication. We first
bound the upper tail of the energy of a Gibbs sample, then bound the
change of the partition function under a Gaussian rotation. We work with
the off-diagonal Hamiltonian of Equation~\eqref{eq:model} throughout;
the auxiliary diagonal variables used for interpolation are absent.
Write $H^J=H_n^J$, $Z^J=Z_{n,\beta}^J$ and $\pi^J=\pi_{n,\beta}^J$.
All disorder vectors in this section are indexed by $i<j$.
\begin{lemma}[An upper energy tail]\label{static-lem:coverage-energy}
There are constants $e_1\in(0,\beta/2)$ and $c_E>0$ such that, for all
sufficiently large $n$,
\begin{equation}\label{static-eq:coverage-energy-good}
\Prob_J\!\left(\pi^J\{H^J>ne_1\}>e^{-c_E n}\right)
\le e^{-c_E n}.
\end{equation}
\end{lemma}
\begin{proof}
The proof separates concentration from the location of the mean.
Lemma~\ref{static-lem:energy-gap} gives $e_0<\beta/2$ with
$p_n'(\beta)\le e_0$ once $n$ is sufficiently large.
Choose $e_1\in(\max\{0,e_0\},\beta/2)$ and put
$c_E=(e_1-e_0)^2/2$.
Sample the disorder $J$ and subsequently a configuration $V\sim\pi^J$.
Fix $v\in\Sig$. Bayes' formula expresses the conditional disorder
density given $V=v$ as proportional to
\[
\exp\!\left\{-\frac{|J|^2}{2}+\beta H^J(v)-\log Z^J\right\}.
\]
The log partition function is convex in the disorder. The negative
log-density therefore has Hessian $I+\nabla^2\log Z^J\succeq I$.
Adding a linear tilt $sH^J(v)$, for any $s\in\R$, leaves that lower
Hessian bound unchanged.
Write $H^J(v)=h_v\cdot J$, where
\[
|h_v|^2=\frac1n\sum_{i<j}v_i^2v_j^2=\frac{n-1}{2}.
\]
If $\varphi_v(s)=\log\E[e^{sH^J(v)}\mid V=v]$, differentiation under
the linearly tilted conditional law and the Brascamp--Lieb bound from
Lemma~\ref{static-lem:gaussian-tools} give
\[
\varphi_v''(s)=\Var_{v,s}(H^J(v))\le\frac{n-1}{2},\qquad s\in\R.
\]
Strong convexity ensures integrability under every tilt used here. Two
integrations of the variance bound, subtracting the conditional mean,
give
\begin{equation}\label{static-eq:coverage-posterior-mgf}
\E\!\left[
e^{s(H^J(v)-\E[H^J(v)\mid V=v])}\mid V=v
\right]\le \exp\!\left\{\frac{s^2(n-1)}4\right\}.
\end{equation}
To identify that conditional mean, use the gauge transformation
$J_{ij}\mapsto J_{ij}v_iv_j$, which preserves the Gaussian law, preserves $Z^J$,
and carries $H^J(v)$ to $H^J(\boldsymbol{+})$, where
$\boldsymbol{+}=(1,\ldots,1)$. More generally, these transformations act
transitively on the configurations. Thus $V$ has a uniform marginal
under the joint law, and $\E[H^J(v)\mid V=v]$ has the same value for
every $v$. Its average over $V$ is
\[
\E\pi^J(H^J)=np_n'(\beta).
\]
Differentiating the finite-volume pressure gives this identity; Gaussian
integrability permits interchange with expectation. The conditional mean
is therefore at most $ne_0$. Chernoff's inequality in
\eqref{static-eq:coverage-posterior-mgf}, followed by averaging, yields
\[
\E_J\pi^J\{H^J>ne_1\}
\le\exp\!\left\{-\frac{n^2(e_1-e_0)^2}{n-1}\right\}
\le e^{-2c_E n}.
\]
Finally, Markov's inequality turns this averaged Gibbs bound into the
disorder estimate~\eqref{static-eq:coverage-energy-good}.
\end{proof}
\begin{lemma}[A disorder rotation estimate]\label{static-lem:coverage-rotation}
Let $J,G$ be independent standard Gaussian vectors and, for
$\theta\in(0,\pi/2)$, set $J_\theta=\cos\theta J+\sin\theta G$. Then
for every $\lambda\in\R$,
\begin{equation}\label{static-eq:coverage-rotation-mgf}
\E\exp\!\left\{\lambda(\log Z^{J_\theta}-\log Z^J)\right\}
\le\exp\!\left\{\frac{\lambda^2\theta^2\beta^2(n-1)}4\right\}.
\end{equation}
In particular, for every $u>0$,
\begin{equation}\label{static-eq:coverage-rotation-tail}
\Prob\!\left(\log Z^{J_\theta}-\log Z^J\ge u\right)
\le\exp\!\left\{-\frac{u^2}{\theta^2\beta^2(n-1)}\right\}.
\end{equation}
\end{lemma}
\begin{proof}
We interpolate by rotation through independent Gaussian directions.
First put $F(J)=\log Z^J$. The coordinate derivative
$\partial_{J_{ij}}F=\beta n^{-1/2}\pi^J(\sigma_i\sigma_j)$ gives
\[
|\nabla F(J)|^2\le\frac{\beta^2(n-1)}2=:L_F^2.
\]
For $0\le s\le\theta$, let
$J_s=\cos sJ+\sin sG$ and $K_s=-\sin sJ+\cos sG$.
At a fixed $s$, orthogonal invariance makes $J_s$ and $K_s$
independent standard Gaussian vectors. Differentiation along the
rotation gives
\[
F(J_\theta)-F(J)=\int_0^\theta\nabla F(J_s)\cdot K_s\,\dd s.
\]
Apply Jensen's inequality to the uniform measure on $[0,\theta]$.
Inside the resulting integral, condition on $J_s$ and integrate the
independent Gaussian $K_s$:
\begin{align*}
\E e^{\lambda(F(J_\theta)-F(J))}
&\le\frac1\theta\int_0^\theta
\E e^{\lambda\theta\nabla F(J_s)\cdot K_s}\,\dd s\\
&\le\frac1\theta\int_0^\theta
e^{\lambda^2\theta^2L_F^2/2}\,\dd s
=e^{\lambda^2\theta^2L_F^2/2}.
\end{align*}
This is~\eqref{static-eq:coverage-rotation-mgf}. Exponential Markov
and optimization over $\lambda>0$ prove the tail bound. The argument
holds for every permitted angle, so it also applies to angles depending
on $n$.
\end{proof}
For coverage we parametrize the rotation by the variance of its fresh
disorder component. If $0<t<1/2$ and
$J'=\sqrt{1-t}\,J+\sqrt t\,G$, take
$\theta=\arcsin\sqrt t$. Since $\theta^2\le2t$, the preceding lemma gives,
for every $z>0$,
\begin{equation}\label{eq:rotation-tail}
\PP\{\log Z^{J'}-\log Z^J\ge z\}
\le\exp\!\left(-c_\beta\frac{z^2}{nt}\right),
\qquad c_\beta=\frac1{2\beta^2}.
\end{equation}
This estimate is uniform in $t$, including rotations whose variance
decreases with $n$.
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/05-energy.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/07-coverage.tex ====
\section{Coverage at stretched-exponential scales}
\label{sec:coverage}
The energy and rotation estimates now supply references for the dynamical
construction. Given a target $x$ in a subset $S$ of the cube, we seek a
reference $z\in S$ with $|R(x,z)|\ge q_*$ among independent Gibbs samples.
The required estimate has two parts: a small total Gibbs mass of targets
may be excluded, and every remaining target must have a neighborhood of
sufficient Gibbs mass. Both masses are measured by the original Gibbs
law, without conditioning on $S$. Thus the neighborhood mass is the
success probability of one sample from a reference bank.
We prove the estimate with both mass bounds on the scale $L=n^\kap$.
If a set has too few overlapping pairs, fresh Gaussian disorder produces
a gain in its restricted partition sum. The energy upper tail limits
the loss from decreasing the original disorder, leaving a net increase
that the rotation estimate makes unlikely. To make this argument uniform
over all subsets, we select a violating set using the original disorder
before sampling the perturbation.
\begin{proposition}[Stretched-exponential coverage]
\label{prop:coverage}
There exists $q_*\in(0,1)$, depending only on the fixed inverse
temperature $\beta>1$, with the following property. For each fixed
$D_1>0$, there exists a fixed $D_3=D_3(\beta,D_1)>0$ and a sequence of
events $\mathcal C_n(D_1)$, measurable with respect to the disorder,
such that $\PP_J(\mathcal C_n(D_1))\longrightarrow1$ and, on
$\mathcal C_n(D_1)$,
\begin{equation}
\pi^J\!\left(
\left\{x\in S:
\pi^J\{z\in S:|R(x,z)|\ge q_*\}<e^{-D_3L}
\right\}\right)
<e^{-D_1L}
\qquad\text{for every }S\subset\Sig.
\label{eq:coverage}
\end{equation}
\end{proposition}
\begin{proof}
We first control conditional pair mass and then pass to individual targets.
Let $e_1\in(0,\beta/2)$ and $c_E>0$ be given by
Lemma~\ref{static-lem:coverage-energy}, and define
\[
\mathcal E_n
=\left\{\pi^J\{x:H^J(x)>ne_1\}\le e^{-c_En}\right\}.
\]
Thus $\PP_J(\mathcal E_n)\to1$. Define
\[
g=\frac{\beta^2}{4}-\frac{\beta e_1}{2}>0,
\qquad
0<q_*<1,\qquad \beta^2q_*<\frac g8.
\]
Fix this threshold once for both the present proposition and
Appendix~\ref{poly-sec:coverage}; the choice precedes $D_1$.
Given $D_1>0$, choose fixed constants
\begin{equation}
M>\frac{8(D_1+1)}g,
\qquad
D_2>\frac{\beta^2M}{2}+1,
\qquad
t=\frac{ML}{n}.
\label{eq:coverage-parameters}
\end{equation}
For all sufficiently large $n$, we have $0<t<1/2$;
moreover, $t\to0$ and $nt=ML\to\infty$.
\par\medskip\noindent\textbf{Pair mass in every sufficiently large set.}\enspace
Our first claim is that, with disorder probability tending to one,
\begin{equation}
\bigl(\pi^J(\,\cdot\mid S)\bigr)^{\otimes2}
\{|R(x,y)|\ge q_*\}
\ge e^{-D_2L}
\quad\text{for every }S\subset\Sig
\text{ with }\pi^J(S)\ge e^{-D_1L}.
\label{eq:coverage-pair}
\end{equation}
Write $\mathcal V_n$ for the failure event. Order the subsets of $\Sig$
deterministically and, for $J\in\mathcal E_n\cap\mathcal V_n$, select
the first violating member $S(J)$. The Gibbs quantities for each fixed
subset are measurable in $J$, so the selection is measurable. Once we
condition on $J$, the chosen set is fixed and cannot depend on the
independent disorder used later.
Fix such a disorder, write $\pi=\pi^J$, and remove the upper-energy
tail from the selected set:
\[
T=S\cap\{x:H^J(x)\le ne_1\},
\qquad \nu=\pi(\,\cdot\mid T).
\]
Since $L=o(n)$, for all sufficiently large $n$,
\[
e^{-c_En}\le\frac12e^{-D_1L},
\qquad
\pi(T)\ge\frac12\pi(S)\ge\frac12e^{-D_1L}.
\]
The retained mass is at least half the original mass. Comparing the
conditional pair laws gives
\begin{equation}
\nu^{\otimes2}\{|R(x,y)|\ge q_*\}
\le
\left(\frac{\pi(S)}{\pi(T)}\right)^2
\bigl(\pi(\,\cdot\mid S)\bigr)^{\otimes2}
\{|R(x,y)|\ge q_*\}
<4e^{-D_2L}.
\label{eq:coverage-trimming}
\end{equation}
\par\medskip\noindent\textbf{Gain from independent disorder.}\enspace
Sample an independent Gaussian disorder $G$ and set
\[
W=\nu\!\left(e^{\beta\sqrt t H^G}\right),
\qquad J'=\sqrt{1-t}\,J+\sqrt t\,G.
\]
The subscript $G$ means that the original disorder and its selected set
are held fixed. The off-diagonal normalization gives
\[
\EE_G H^G(x)^2=\frac{n-1}{2},
\qquad
\EE_G[H^G(x)H^G(y)]=\frac{nR(x,y)^2-1}{2}.
\]
The Gaussian exponential moment formula consequently yields
\begin{align}
\EE_GW&=e^{\beta^2t(n-1)/4},\notag\\
\frac{\EE_GW^2}{(\EE_GW)^2}
&=\nu^{\otimes2}\!\left(
e^{\beta^2t(nR(x,y)^2-1)/2}\right)\notag\\
&\le e^{\beta^2ntq_*^2/2}
+4e^{(\beta^2M/2-D_2)L}
\le 2e^{\beta^2ntq_*^2/2}
\label{eq:coverage-moments}
\end{align}
for all sufficiently large $n$. Split the pair sum at $|R(x,y)|=q_*$. On the large-overlap part use
\eqref{eq:coverage-trimming}, then the choice of $D_2$ in
\eqref{eq:coverage-parameters}. Paley--Zygmund consequently yields
\begin{equation}
\PP_G\!\left(W\ge\frac12\EE_GW\right)
\ge\frac18e^{-A_0},
\qquad A_0=\frac{\beta^2ntq_*^2}{2}.
\label{eq:coverage-pz}
\end{equation}
Concentration raises this lower probability to one half after a controlled
loss in the threshold. For $f(G)=\log W$, differentiation gives
\[
\left|\frac{\partial f}{\partial G_{ij}}\right|
\le\beta\sqrt{\frac tn},
\qquad
\norm{\nabla f}_2^2\le\frac{\beta^2t(n-1)}2=:\sigma^2.
\]
Thus Gaussian concentration applies to $f$ with Lipschitz constant
at most $\sigma$. Write
\[
a=\frac{\beta^2t(n-1)}4-\log2,
\qquad \overline f=\EE_Gf.
\]
When $a>\overline f$, compare the upper concentration tail with
\eqref{eq:coverage-pz}:
\[
\frac18e^{-A_0}
\le\PP_G(f\ge a)
\le\exp\!\left(-\frac{(a-\overline f)^2}{2\sigma^2}\right),
\]
so $\overline f\ge a-\sigma\sqrt{2(A_0+\log8)}$.
If $a\le\overline f$, the same lower bound holds immediately.
In either case the lower concentration tail yields
\[
\PP_G\!\left(
f\ge a-\sigma\sqrt{2(A_0+\log8)}
-\sigma\sqrt{2\log2}
\right)\ge\frac12.
\]
Since $\sigma\le\beta\sqrt{nt/2}$, we have
\[
\sigma\sqrt{2(A_0+\log8)}+\sigma\sqrt{2\log2}
\le\frac{\beta^2ntq_*}{\sqrt2}+C_\beta\sqrt{nt}
\le\beta^2ntq_*+C_\beta\sqrt{nt}.
\]
Consequently, with conditional probability at least one half,
\begin{equation}
\log W\ge
\frac{\beta^2t(n-1)}4-\beta^2ntq_*
-C_\beta\sqrt{nt}-\log2.
\label{eq:coverage-logw}
\end{equation}
The constants and the large-$n$ threshold are independent of the selected
disorder and its violating set.
\par\medskip\noindent\textbf{Comparison with the rotation cost.}\enspace
The gain from $G$ must be compared with the change in the original energy.
Linearity in the disorder gives
\begin{align*}
\frac{Z^{J'}}{Z^J}
&=\pi\!\left(
e^{\beta(\sqrt{1-t}-1)H^J+\beta\sqrt t H^G}\right)\\
&\ge\pi(T)e^{-\beta ne_1(1-\sqrt{1-t})}W.
\end{align*}
On $T$, the original energy is at most $ne_1$. Its negative coefficient
$\sqrt{1-t}-1$ gives the last lower bound. Combining this with
\eqref{eq:coverage-logw}, we obtain
\begin{align*}
\log\frac{Z^{J'}}{Z^J}
&\ge -D_1L-\log2-\beta ne_1(1-\sqrt{1-t})+\log W\\
&\ge (g-\beta^2q_*)nt-D_1L
-C_\beta\sqrt{nt}-C_\beta(nt^2+1).
\end{align*}
Here we used $1-\sqrt{1-t}=t/2+O(t^2)$, uniformly for
$0<t<1/2$. Dividing the final lower bound by $nt$, its limit
inferior is at least
\[
g-\beta^2q_* -\frac{D_1}{M}>\frac{3g}{4}.
\]
It follows that, for all sufficiently large $n$ and every
$J\in\mathcal E_n\cap\mathcal V_n$,
\begin{equation}
\PP_G\!\left(
\log Z^{J'}-\log Z^J\ge\frac g2nt
\right)\ge\frac12.
\label{eq:coverage-gain}
\end{equation}
Average the conditional gain over $J$ and use the unconditional rotation
bound \eqref{eq:rotation-tail}:
\begin{align*}
\PP_J(\mathcal E_n\cap\mathcal V_n)
&\le2\PP_{J,G}\!\left(
\log Z^{J'}-\log Z^J\ge\frac g2nt\right)\\
&\le2e^{-c'_\beta nt}
=2e^{-c'_\beta ML}\longrightarrow0
\end{align*}
for a fixed $c'_\beta>0$. Adding
$\PP_J(\mathcal E_n^c)=o(1)$ proves
$\PP_J(\mathcal V_n)=o(1)$, and hence the simultaneous pair
statement in Equation~\eqref{eq:coverage-pair}. Since $\mathcal V_n$ is the event that any violating subset exists, this
proves the simultaneous claim. No union bound over subsets is needed.
\par\medskip\noindent\textbf{From pairs to individual targets.}\enspace
Choose a fixed $D_3>D_1+D_2$, and set
$\mathcal C_n(D_1)=\mathcal V_n^c$. Fix $J$ in this event and
any $S\subset\Sig$, and let
\[
B=\{x\in S:\pi^J\{z\in S:|R(x,z)|\ge q_*\}<e^{-D_3L}\}.
\]
Suppose that $\pi^J(B)\ge e^{-D_1L}$. Since $B\subset S$,
\begin{align*}
\bigl(\pi^J(\,\cdot\mid B)\bigr)^{\otimes2}
\{|R(x,y)|\ge q_*\}
&=\frac{1}{\pi^J(B)^2}
\sum_{x\in B}\pi^J(x)
\pi^J\{y\in B:|R(x,y)|\ge q_*\}\\
&<\frac{e^{-D_3L}}{\pi^J(B)}
\le e^{-(D_3-D_1)L}
<e^{-D_2L}.
\end{align*}
The set $B$ itself satisfies the mass hypothesis of
\eqref{eq:coverage-pair}, so this is a contradiction. Thus
$\pi^J(B)<e^{-D_1L}$ for every $S$, proving the target-wise statement.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/07-coverage.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/08-narrow-locking.tex ====
\section{Locking on a narrow annulus}\label{sec:narrow}
Lemma~\ref{lem:fixed-locking} applies when one pair overlap exceeds a
fixed positive cutoff and the other two are close to zero. The path
argument also needs a different regime: the latter overlaps are near a
positive level $r$, while the first overlap exceeds $r$ by an amount
that may tend to zero. We obtain a uniform pressure loss for that regime.
Fix a minimizing measure $\mu$ and use its scalar data from
Section~\ref{static-sec:scalar}. Recall
$\Gamma(s)=\EE u(s,X_s)^2$, and write
$\chi_s=\chi(s,X_s)$. We work in the case where
$\Gamma(s)=s$ on a fixed interval; if the identity fails there, the pair
pressure gap supplies the forbidden interval used in
Section~\ref{sec:dynamics}.
Fix $t_0>0$ with $3t_0<q_*$, where $q_*$ is the constant in
Proposition~\ref{prop:coverage}. Recall $\rho=n^{-\kap}$ with
$\kap=1/10000$, and set
\begin{equation}\label{eq:narrow-scales}
h=\rho^8,\qquad \varepsilon=\rho^{20}.
\end{equation}
Constants below may depend on $\beta,t_0$ and the fixed minimizing
measure. They do not depend on $n$ or on the overlap parameters.
The comparison will alter the fields by a contrast on the two sides
of a branching time, together with a short increment on the other
branch. When the surrounding interval has small Parisi mass, a tent
integration by parts shows that the contrast contributes little beyond
its Gaussian variance. The mixed term with the other branch then
provides the strict pressure loss.
\begin{lemma}[Narrow-annulus locking]\label{lem:narrow-locking}
Suppose that
\begin{equation}\label{eq:narrow-identity}
\Gamma(s)=s\qquad\text{for }t_0/2<s<3t_0.
\end{equation}
For three independent Gibbs configurations $v,x,y$, conditionally on
the disorder, define
\[
q=R(x,y),\qquad
r=\frac{R(v,x)+R(v,y)}2,\qquad
\delta=\frac{R(v,x)-R(v,y)}2.
\]
Then, for all sufficiently large $n$,
\begin{equation}\label{eq:narrow-locking-probability}
\EE\pi^{\otimes3}\left\{
\begin{array}{c}
t_0/2<r<3t_0,\quad \mu([r-h,r+h])\le\rho^4,\\
q\ge r+\rho,\quad \varepsilon\le|\delta|\le2\varepsilon
\end{array}\right\}
\le \exp(-n^{4/5}).
\end{equation}
\end{lemma}
\begin{proof}
Fix a feasible overlap matrix, in the coordinate order $v,x,y$,
\begin{equation}\label{eq:narrow-overlap-matrix}
D=
\begin{pmatrix}
1&r+\delta&r-\delta\\
r+\delta&1&q\\
r-\delta&q&1
\end{pmatrix},
\end{equation}
satisfying the event in~\eqref{eq:narrow-locking-probability}. We may
first impose the additional restriction
\begin{equation}\label{eq:narrow-pair-deviation}
|q-\Gamma(q)|\le\rho^{13}.
\end{equation}
For an excluded pair overlap, \eqref{static-eq:pair-pressure} already gives a
pressure deficit of at least $\rho^{26}/2$. At the end of the proof
we will apply Lemma~\ref{lem:pressure-prob} to these deficits and sum
over the at most $n+1$ pair overlaps. The independent third draw does
not change their total probability.
For the minimizing scalar data, the source of the strict loss is already
visible. The identity
$\Gamma'(r)=1$ normalizes $\beta^2\EE\chi_r^2$ to one, whereas the
conditional susceptibility evolution decreases the mixed moment:
\[
\beta^2\EE[\chi_q\chi_r]\le1-c(q-r).
\]
Step~8 will verify a uniform $c>0$. We construct a covariance
perturbation whose quadratic pressure change contains this deficit.
Its much larger Gaussian variance contribution will cancel against the
interpolation penalty.
The proof of the three-replica deficit has three stages. In
Steps~1--4 we construct an admissible perturbed hierarchy and reduce
the comparison to one quadratic moment. Steps~5--6 estimate that
moment using the tent on the first branch and a ramp on the second.
In Steps~7--9 we pass from finite hierarchies to the minimizing scalar
data, establish the negative quadratic coefficient, and sum the
resulting probability bounds.
\par\medskip\noindent\textbf{Step 1. Finite scalar data and the base hierarchy.}\enspace
For all large $n$, the overlap restrictions imply
\[
0<r-h<r<r+h<q\le1.
\]
Take increasingly fine partitions of $[0,1]$ containing
$r-h,r,r+h,q$. On each half-open segment, approximate $\alpha$ by
its actual value at the left endpoint, and set its value at $1$ to
$1$. These approximations are nondecreasing step distribution
functions. They converge in $L^1$ and preserve the actual endpoint
values at the four indicated points. In particular, each satisfies
\begin{equation}\label{eq:narrow-step-oscillation}
\alpha(r+h)-\alpha(r-h)\le\rho^4.
\end{equation}
During the finite calculations, we omit the approximation index:
$\alpha,\Phi,u,\chi,X,A$ mean the step data, and $\cP_\beta(\alpha)$ is
the scalar functional of the corresponding measure. These step
measures need not minimize the functional. We will use stationarity
only after passing to the minimizing limit.
We index only the integrated covariance increments: $Q_0=0$,
$Q_\ell-Q_{\ell-1}\succeq0$, with nondecreasing masses
$m_\ell\in[0,1]$ and field covariances
$\beta^2(Q_\ell-Q_{\ell-1})$. For this indexing write
\[
Q_{\rm end}=Q_N,\qquad
\int m\,\dd|Q|^2
=\sum_{\ell=1}^N m_\ell
(|Q_\ell|^2-|Q_{\ell-1}|^2).
\]
Prepending a zero-covariance quenched level gives precisely the
convention of Proposition~\ref{static-prop:matrix-bound}. We always use
zero terminal multiplier, so the terminal function is the sum of the
three scalar $\log(2\cosh)$ functions.
Build the base covariance path $Q$ in three stages:
\begin{enumerate}
\item From scalar time $0$ to $r$, give all three fields the same
increment of variance $\beta^2\,\dd s$, integrated at mass
$\alpha(s)/3$.
\item From $r$ to $q$, give $v$ its own scalar branch at mass
$\alpha(s)$ and give $x,y$ a common scalar branch at mass
$\alpha(s)/2$. The branches use independent Gaussian
increments. Merge their finite schedules in nondecreasing
mass order, preserving each branch's local order.
\item From $q$ to $1$, give the three coordinates independent
scalar increments simultaneously on each segment, all
integrated at mass $\alpha(s)$.
\end{enumerate}
This is the base hierarchy of
Section~\ref{static-subsec:locking-base}, now with the specified finite
approximation. In the middle stage, scalar time is local to each branch:
we merge by mass, preserving each branch's order. The end masses of the
first stage are at most the starting middle masses, and the end middle
masses are at most the starting third-stage masses. Thus the complete
schedule is nondecreasing, and every increment is positive semidefinite.
Write $\Psi_0$ for its recursive value. The block-size cancellation
proved in Section~\ref{static-subsec:locking-base} gives
\begin{equation}\label{eq:narrow-base-value}
\Psi_0-\frac{\beta^2}{4}\int m\,\dd|Q|^2
=3\Phi(0,0)-\frac{3\beta^2}{2}\int_0^1s\alpha(s)\,\dd s
=3\cP_\beta(\alpha).
\end{equation}
Its endpoint is $D$ with $\delta=0$. The same calculation identifies
the base tilted law: after the common path through $r$, the $x,y$
branch and the $v$ branch each have the scalar transition law. Denote
their scalar paths by $X^x$ and $X^v$; they coincide through $r$ and
then run to $q$ on their separate clocks. A frozen
branch's additive value cancels from every active normalized kernel,
so merging the schedules does not change their conditional independence.
To record the conditional independence that will be used below, let
$\mathcal X_0$ be the common prefix through $r$. If $K_{r,z}$ denotes
the scalar path law starting at $(r,z)$, then
\[
\mathcal L(X^x_{[r,q]},X^v_{[r,q]}\mid\mathcal X_0)
=K_{r,X_r}\otimes K_{r,X_r}.
\]
In particular, knowing the entire $x$ branch reveals no additional
future noise of the $v$ branch.
\par\medskip\noindent\textbf{Step 2. A perturbation preserving positive semidefiniteness.}\enspace
Set
\[
\theta=\rho^2,\qquad a=1-\theta,
\]
and form the following linear combinations of field increments:
\begin{equation}\label{eq:narrow-Y}
Y_0=\frac{(X^x_r-X^x_{r-h})-(X^x_{r+h}-X^x_r)}h,\qquad
Y_1=\frac{X^v_{r+h}-X^v_r}h,\qquad
Y=aY_0+\theta Y_1.
\end{equation}
Under the raw law these are Gaussian linear expressions. Under the
base tilted law the same expressions use the scalar paths just
identified. The contrast $Y_0$ includes the common increment before
the split and the $x$ increment after the split; $Y_1$ uses the
post-split increment on the $v$ branch. Figure~\ref{fig:narrow-increments}
shows these three increments and their covariance kernels.
\input{figures/narrow-increments}
At each level of the first two stages, add $\delta$ times that
level's contribution to $Y$ to the $x$ field, and subtract it from
the $y$ field. Leave $v$ unchanged. Each modified increment is a
deterministic linear image of its original Gaussian increment.
Consequently its covariance is positive semidefinite, and increments
at different levels remain independent. This gives an admissible
path $\widetilde Q$ with the original masses. At the end of the
second stage the added fields are $0,\delta Y,-\delta Y$.
For any prefix, write $K_c$ for the accumulated raw variance of
$Y$, divided by $\beta^2$, and write $g_v,g_x$ for its raw
covariances with the base $v,x$ fields in the same units. Its
covariance with the base $y$ field equals $g_x$. The coefficients
on the three active intervals are $a/h,-a/h,\theta/h$, so
\begin{equation}\label{eq:narrow-covariance-end}
K=\frac{2a^2+\theta^2}{h},\qquad
g_v^{\rm end}=a+\theta=1,\qquad g_x^{\rm end}=a-a=0.
\end{equation}
Let $e_v,e_x,e_y$ be the coordinate vectors and set
$L_2=(e_x-e_y)(e_x-e_y)^{\mathsf T}$. At every prefix,
\begin{equation}\label{eq:narrow-covariance-expansion}
\widetilde Q=Q+\delta L_1+\delta^2K_cL_2,
\qquad
L_1=
\begin{pmatrix}
0&g_v&-g_v\\
g_v&2g_x&0\\
-g_v&0&-2g_x
\end{pmatrix}.
\end{equation}
For the Frobenius product $U:V=\sum_{i,j}U_{ij}V_{ij}$, both $Q$
and $L_2$ are orthogonal to $L_1$. Also
$|L_1|^2=4g_v^2+8g_x^2$ and $|L_2|^2=4$. Expanding gives
\begin{equation}\label{eq:narrow-norm-expansion}
|\widetilde Q|^2-|Q|^2
=\delta^2\bigl(4g_v^2+8g_x^2+2K_c\,Q:L_2\bigr)
+4\delta^4K_c^2.
\end{equation}
The endpoint is
\begin{equation}\label{eq:narrow-modified-end}
\widetilde Q_{\rm end}=D+\delta^2K L_2.
\end{equation}
Its diagonal need not be one. We therefore use precisely the
arbitrary-endpoint comparison~\eqref{static-eq:matrix-bound}, proved
in Proposition~\ref{static-prop:matrix-bound}. It assigns the
endpoint discrepancy the cost
\begin{equation}\label{eq:narrow-mismatch}
\frac{\beta^2}{4}|D-\widetilde Q_{\rm end}|^2
=\beta^2\delta^4K^2.
\end{equation}
\par\medskip\noindent\textbf{Step 3. The penalty gain.}\enspace
We next compute the quadratic terms in the interpolation penalty.
The base $x,y$ fields coincide before the third stage, so $Q:L_2=0$
there. In the third stage, $Q:L_2=2(s-q)$ and $K_c=K$ is fixed.
The term $2K_c\,Q:L_2$ in~\eqref{eq:narrow-norm-expansion}
therefore contributes $\beta^2\delta^2K A(q)$ to the penalty.
For the other quadratic terms put
\[
I=\int m\,\bigl(\dd(g_v^2)+2\dd(g_x^2)\bigr).
\]
The three active pieces have separate roles. Before the split, on
$[r-h,r]$, both covariances grow from $0$ to $a$ at mass
$\alpha(s)/3$; this contributes $\alpha(r)a^2+O(\rho^4)$.
On the active $v$ steps, $g_v$ grows from $a$ to $a+\theta$
and $g_x$ is fixed; this contributes
$\alpha(r)(2a\theta+\theta^2)+O(\rho^4)$.
On the active shared $x,y$ steps, $g_x$ decreases from $a$ to zero
and $g_v$ is fixed. The factor $2$ in $I$ cancels the factor
$1/2$ in their masses, giving $-\alpha(r)a^2+O(\rho^4)$.
The error in replacing each mass by $\alpha(r)$ is controlled by
\eqref{eq:narrow-step-oscillation} times the total variation of the
relevant covariance square. These total variations are uniformly
bounded, independently of the partition and its interleaving. Thus
\begin{equation}\label{eq:narrow-I}
I=\alpha(r)(2a\theta+\theta^2)+O(\rho^4).
\end{equation}
Finally, $K_c$ is a nondecreasing cumulative variance and all masses
are nonnegative. The fourth-order contribution
$\int m\,\dd(K_c^2)$ is therefore nonnegative. Combining the terms
proves
\begin{equation}\label{eq:narrow-penalty}
\frac{\beta^2}{4}\int m\,
\dd\bigl(|\widetilde Q|^2-|Q|^2\bigr)
\ge
\beta^2\delta^2
\bigl(KA(q)+\alpha(r)(2a\theta+\theta^2)-C\rho^4\bigr).
\end{equation}
\par\medskip\noindent\textbf{Step 4. The recursive-value cost.}\enspace
Let $\widetilde\Psi$ denote the modified one-site value. Integrating
the unchanged third stage backwards leaves the boundary difference
\begin{equation}\label{eq:narrow-U}
U=\Phi(q,X_q^x+\delta Y)+\Phi(q,X_q^x-\delta Y)
-2\Phi(q,X_q^x).
\end{equation}
Initially view this as a function of the full raw Gaussian prefix
through the second stage. Let $\EE_0$ denote expectation under its
base tilted law. We claim that
\begin{equation}\label{eq:narrow-recursive-comparison}
\widetilde\Psi-\Psi_0\le\log\EE_0 e^U.
\end{equation}
Apply the normalized-kernel comparison
\eqref{static-eq:locking-recursive-sandwich} with boundary change $U$.
Here the state at each level is the entire raw Gaussian prefix: $Y$
uses several increments, so the current field alone need not determine
the modified future value. If $K_i$ is the normalized base kernel and
$\Delta_i$ is the change in recursive value, the ratio of the defining
integrals still gives
\[
\Delta_i=\frac1{m_i}\log K_i e^{m_i\Delta_{i+1}}
\quad(m_i>0),\qquad
\Delta_i=K_i\Delta_{i+1}\quad(m_i=0).
\]
The kernel induction used for
\eqref{static-eq:locking-recursive-sandwich} therefore applies without
change and proves~\eqref{eq:narrow-recursive-comparison}. This argument
uses no inverse transformation of the fields, so singular covariance
increments cause no difficulty. Every kernel is from the base hierarchy,
whose tilted prefix law was identified in Step~1.
Write $\chi_q=\chi(q,X_q^x)$. By Lemma~\ref{static-lem:scalar-regularity},
\begin{equation}\label{eq:narrow-U-bounds}
0\le U\le\delta^2Y^2,\qquad
|U-\delta^2\chi_qY^2|\le C_\beta|\delta Y|^3.
\end{equation}
The diffusion representation under $\EE_0$ writes $Y=G+B_*$,
where $G$ is a centered Gaussian linear combination of Brownian
increments with variance $\beta^2K$, and $|B_*|\le C_\beta$.
Indeed, each drift in~\eqref{eq:narrow-Y} is bounded by
$\beta^2$, integrated for time $h$, and then divided by $h$.
No independence between $G$ and $B_*$ is required. Since
$\delta^2/h=O(\rho^{32})$, Gaussian exponential moments and
$|Y|\le|G|+C_\beta$ imply, for all large $n$,
\[
\EE_0|Y|^3\le C_\beta h^{-3/2},\qquad
\EE_0\!\left[Y^4e^{\delta^2Y^2}\right]\le C_\beta h^{-2}.
\]
Use $e^u\le1+u+\tfrac12u^2e^u$ for $u\ge0$, followed by
$\log(1+z)\le z$, and then~\eqref{eq:narrow-U-bounds}. We obtain
\begin{equation}\label{eq:narrow-exponential-error}
\log\EE_0e^U
\le \delta^2\EE_0[\chi_qY^2]
+C_\beta\bigl(|\delta|^3h^{-3/2}+\delta^4h^{-2}\bigr).
\end{equation}
We have now reduced the comparison to estimating
$\EE_0[\chi_qY^2]$. Its Gaussian variance term will cancel the
$KA(q)$ part of the penalty up to the pair deviation. The strict
gain must come from the mixed moment of $Y_0$ and $Y_1$.
\par\medskip\noindent\textbf{Step 5. The tent on the first branch.}\enspace
We will establish the quadratic-moment bound
\begin{equation}\label{eq:narrow-quadratic}
\EE_0[\chi_qY^2]
\le\beta^2K\,\EE_0\chi_q
+2a\theta\beta^4\alpha(r)\EE_0[\chi_q\chi_r]
+C_\beta(\rho^4+\theta^2),
\qquad \chi_r=\chi(r,X_r^x).
\end{equation}
Begin with the scalar $x$ path through $q$. Write its partition as
$0=s_0<\cdots<s_N=q$ and its scalar masses as
$m_j^{\rm sc}=\alpha(s_j)$. Its normalized tilted density $D_x$
relative to the raw Gaussian increments has logarithm
\begin{equation}\label{eq:narrow-density}
\ell:=\log D_x
=\sum_{j=0}^{N-1}m_j^{\rm sc}
\bigl(\Phi(s_{j+1},X^x_{s_{j+1}})
-\Phi(s_j,X^x_{s_j})\bigr).
\end{equation}
The scalar masses appear here because the base-law factorization
in Step~1 has already canceled the divided matrix masses.
Let $T$ be the continuous piecewise linear tent supported on
$[r-h,r+h]$, zero at both endpoints and equal to one at $r$.
For functions of the raw cumulative coordinates, denote by
$\partial$ the derivative obtained by replacing $X^x_s$ with
$X^x_s+tT(s)$ and differentiating at $t=0$. Under the raw law,
\[
{\rm Cov}_{\rm raw}(Y_0,X^x_s)=\beta^2T(s),\qquad
{\rm Var}_{\rm raw}(Y_0)=\frac{2\beta^2}{h}.
\]
Finite-dimensional Gaussian integration by parts gives
\begin{equation}\label{eq:narrow-tent-ibp}
\EE_{\rm raw}[Y_0f]=\beta^2\EE_{\rm raw}[\partial f],\qquad
\EE_{\rm raw}[Y_0^2f]
=\frac{2\beta^2}{h}\EE_{\rm raw}f
+\beta^4\EE_{\rm raw}[\partial^2f].
\end{equation}
For a direct verification, multiply the Gaussian density by
$\exp\{tY_0-t^2(2\beta^2/h)/2\}$. This shifts every cumulative
coordinate by $t\beta^2T(s)$; one and two differentiations give
the two formulas. These derivatives act on raw coordinates, not
on the solution map of the tilted stochastic differential equation.
Telescope~\eqref{eq:narrow-density}. Since $T(0)=T(q)=0$,
only internal coefficients $m_{j-1}^{\rm sc}-m_j^{\rm sc}$ remain:
\[
\partial\ell
=\sum_{j=1}^{N-1}(m_{j-1}^{\rm sc}-m_j^{\rm sc})
T(s_j)u(s_j,X^x_{s_j}),\qquad
\partial^2\ell
=\sum_{j=1}^{N-1}(m_{j-1}^{\rm sc}-m_j^{\rm sc})
T(s_j)^2\chi(s_j,X^x_{s_j}).
\]
Only points strictly between $r-h$ and $r+h$ contribute. The sum
of the absolute coefficients at those points is at most
$\alpha(r+h)-\alpha(r-h)$. Thus
\begin{equation}\label{eq:narrow-density-derivatives}
|\partial\ell|\le\rho^4,\qquad
|\partial^2\ell|\le\rho^4.
\end{equation}
This estimate includes an atom at $r$. Atoms at $r-h$ or $r+h$
contribute zero because the tent vanishes there. No continuity of
$\alpha$ at these points is assumed.
The endpoint $q$ lies outside the tent, so $\partial\chi_q=0$.
Apply the second identity of~\eqref{eq:narrow-tent-ibp} to
$f=D_x\chi_q$:
\begin{align}
\EE_0[\chi_qY_0^2]
&=\frac{2\beta^2}{h}\EE_0\chi_q
+\beta^4\EE_0\!\left[
\chi_q\bigl((\partial\ell)^2+\partial^2\ell\bigr)\right]
\notag\\
&=\frac{2\beta^2}{h}\EE_0\chi_q+O_\beta(\rho^4).
\label{eq:narrow-Y0-square}
\end{align}
At the tent's peak, put $u_r=u(r,X_r^x)$; then
$\partial u_r=\chi_r$. The first identity applied to
$f=D_x\chi_qu_r$ gives
\begin{align}
\EE_0[\chi_qY_0u_r]
&=\beta^2\EE_0[\chi_q\chi_r]
+\beta^2\EE_0[\chi_qu_r\partial\ell]
\notag\\
&=\beta^2\EE_0[\chi_q\chi_r]+O_\beta(\rho^4).
\label{eq:narrow-Y0-linear}
\end{align}
Both integrations by parts are legitimate for every fixed finite
partition. Scalar values grow at most linearly, their needed
derivatives are bounded, and the density grows at most
exponential-linearly in the finite Gaussian vector. Gaussian
integrability therefore justifies differentiation directly, or
after truncation and passage to the limit. The error estimates
above are uniform over the partitions.
\par\medskip\noindent\textbf{Step 6. The ramp on the other branch and the mixed moment.}\enspace
Condition now on the entire path $X^x$ through $q$. The product law
from Step~1 leaves the future $v$ branch as a scalar diffusion
starting at the fixed value $X_r$. On $[r,r+h]$, its conditional
tilted density has the form~\eqref{eq:narrow-density}, with starting
time $r$. Use the ramp $S(s)=(s-r)/h$ and denote the corresponding
raw-coordinate derivative by $\partial_v$. The raw covariance of
$Y_1$ with the cumulative coordinate at $s$ is $\beta^2S(s)$,
and its raw variance is $\beta^2/h$.
Let $\ell_v$ be this branch's conditional log density. Telescoping
its derivative produces the terminal term
$m_{\rm last}u(r+h,X^v_{r+h})$ and internal terms with total
absolute coefficient at most
$m_{\rm last}-m_{\rm first}\le1$. The initial value is fixed and
the ramp is zero there. Using $|u|\le1$ and $\chi\le1$ gives
\[
|\partial_v\ell_v|\le2,\qquad
|\partial_v^2\ell_v|\le2.
\]
The conditional second Gaussian integration-by-parts identity
therefore yields
\begin{equation}\label{eq:narrow-Y1-square}
\EE_0[Y_1^2\mid X^x]
=\frac{\beta^2}{h}
+\beta^4\EE_0\!\left[
(\partial_v\ell_v)^2+\partial_v^2\ell_v\mid X^x\right]
=\frac{\beta^2}{h}+O_\beta(1).
\end{equation}
The error is uniform in the starting field $X_r$. Multiplying by
the $X^x$-measurable variable $\chi_q$ and averaging gives
\[
\EE_0[\chi_qY_1^2]
=\frac{\beta^2}{h}\EE_0\chi_q+O_\beta(1).
\]
For the mixed moment we need the conditional first moment of $Y_1$.
The Brownian increment has conditional mean zero. The arbitrary-start
martingale identity~\eqref{static-eq:scalar-martingale} gives
$\EE_0[u(z,X_z^v)\mid X^x]=u_r$, since the independent branch
starts at $X_r$. Hence its drift representation gives exactly
\begin{equation}\label{eq:narrow-Y1-mean}
\EE_0[Y_1\mid X^x]
=\frac{\beta^2}{h}\int_r^{r+h}
\alpha(z)\EE_0[u(z,X^v_z)\mid X^x]\,\dd z
=\beta^2\bar\alpha\,u_r,\qquad
\bar\alpha=\frac1h\int_r^{r+h}\alpha(z)\,\dd z.
\end{equation}
Conditioning on the whole $x$ path introduces no additional
conditioning on the $v$ branch's future noise. Moreover,
\eqref{eq:narrow-step-oscillation} implies
$|\bar\alpha-\alpha(r)|\le\rho^4$. Since $Y_0,\chi_q,u_r$ are
$X^x$-measurable, combine~\eqref{eq:narrow-Y0-linear} with
\eqref{eq:narrow-Y1-mean} to obtain
\begin{equation}\label{eq:narrow-mixed}
\EE_0[\chi_qY_0Y_1]
=\beta^2\bar\alpha\,\EE_0[\chi_qY_0u_r]
=\beta^4\alpha(r)\EE_0[\chi_q\chi_r]+O_\beta(\rho^4).
\end{equation}
Finally expand $Y^2=a^2Y_0^2+2a\theta Y_0Y_1+\theta^2Y_1^2$.
Equations~\eqref{eq:narrow-covariance-end},
\eqref{eq:narrow-Y0-square}, \eqref{eq:narrow-Y1-square}, and
\eqref{eq:narrow-mixed} prove~\eqref{eq:narrow-quadratic}.
The distinction between the two errors matters: the tent contributes
$O_\beta(\rho^4)$, while the ramp's $O_\beta(1)$ error is multiplied
by $\theta^2$.
\par\medskip\noindent\textbf{Step 7. Passing the pressure comparison to the minimizer.}\enspace
We can now apply~\eqref{static-eq:matrix-bound} to the modified finite
hierarchy. Combine the base value, mismatch, penalty, recursive
comparison, and moment estimates in
\eqref{eq:narrow-base-value}, \eqref{eq:narrow-mismatch},
\eqref{eq:narrow-penalty}, \eqref{eq:narrow-recursive-comparison},
\eqref{eq:narrow-exponential-error}, and
\eqref{eq:narrow-quadratic}. Since $K=O(h^{-1})$, the endpoint
cost is part of the $\delta^4h^{-2}$ error. Discard the additional
nonpositive term $-\beta^2\delta^2\alpha(r)\theta^2$ to get
\begin{align}
F_{3,n}(D)\le 3\cP_\beta(\alpha)
+\delta^2\bigl\{&
\beta^2K(\EE_0\chi_q-A(q))
+2a\theta\beta^2\alpha(r)
(\beta^2\EE_0[\chi_q\chi_r]-1)
+C_\beta(\rho^4+\theta^2)\bigr\}
\notag\\
&\hspace{-1em}
+C_\beta\bigl(|\delta|^3h^{-3/2}+\delta^4h^{-2}\bigr).
\label{eq:narrow-step-pressure}
\end{align}
The constants do not depend on the partition.
Keep $n,r,q,\delta$ fixed and send the mesh size to zero.
Lemma~\ref{static-lem:scalar-regularity} gives uniform convergence
of the scalar derivatives and $L^1$ stability of the values. Couple
the diffusions with the same Brownian motion. Their drifts are
uniformly bounded and Lipschitz in space, so $L^1$ convergence of
$\alpha$ together with uniform convergence of $u$ gives path
convergence by Gronwall's inequality. The bounded expectations in
\eqref{eq:narrow-step-pressure} consequently converge. So do $A(q)$
and the scalar functional, the latter to $P_\beta$. By construction,
every approximation uses the actual value $\alpha(r)$.
This limit is taken at fixed $h$, so it requires no approximation
rate uniform as $h\downarrow0$. Returning to the minimizing data,
and writing $\EE$ for its scalar-process expectation, we have
\begin{align}
F_{3,n}(D)\le 3P_\beta
+\delta^2\bigl\{&
\beta^2K(\EE\chi_q-A(q))
+2a\theta\beta^2\alpha(r)
(\beta^2\EE[\chi_q\chi_r]-1)
+C_\beta(\rho^4+\theta^2)\bigr\}
\notag\\
&\hspace{-1em}
+C_\beta\bigl(|\delta|^3h^{-3/2}+\delta^4h^{-2}\bigr).
\label{eq:narrow-pressure}
\end{align}
In the product $\chi_q\chi_r$, both susceptibilities belong to the
same scalar path. It remains to show that their evolution makes
the quadratic coefficient negative by more than all displayed
errors.
\par\medskip\noindent\textbf{Step 8. The strict quadratic margin and the scale choices.}\enspace
The assumed identity~\eqref{eq:narrow-identity} and the scalar
evolution imply
\[
\beta^2\EE\chi_r^2=\Gamma'(r)=1.
\]
Multiply the conditional susceptibility identity
\eqref{eq:scalar-evolution} by $\chi_r$ and average. It gives
\begin{equation}\label{eq:narrow-susceptibility-loss}
\beta^2\EE[\chi_q\chi_r]
=1-\beta^4\int_r^q\alpha(z)\EE[\chi_z^2\chi_r]\,\dd z
\le1-c(q-r)\le1-c\rho
\end{equation}
with a uniform $c>0$ over the allowed $r,q$. To justify uniformity,
including $q=1$, use the drift bound to obtain
$\sup_{s\le1}|X_s|\le\beta\sup_{s\le1}|B_s|+\beta^2$.
Choose a fixed $M$ so that the event that the right side is at most
$M$ has positive probability $p_M$. Continuity and strict positivity
of $\chi$ imply
\[
c_M:=\min_{0\le s\le1,\ |z|\le M}\chi(s,z)>0.
\]
Thus $\EE[\chi_z^2\chi_r]\ge p_Mc_M^3$ for all the times under
consideration. Also $\alpha(z)\ge\alpha(t_0/2)>0$ when $z\ge r$.
We may take $c=\beta^4\alpha(t_0/2)p_Mc_M^3$ in
\eqref{eq:narrow-susceptibility-loss}. This argument uses positivity
on a compact set, not a lower bound for $\chi$ on the whole real
line.
The all-time stationarity identity~\eqref{eq:scalar-stationarity}
and the pair restriction~\eqref{eq:narrow-pair-deviation} give
\[
\EE\chi_q-A(q)=\alpha(q)(q-\Gamma(q))\le\rho^{13}.
\]
Since $K=O(\rho^{-8})$, the first term in braces in
\eqref{eq:narrow-pressure} is at most $C_\beta\rho^5$.
For all large $n$, $a\ge1/2$; using
$\theta=\rho^2$, $\alpha(r)\ge\alpha(t_0/2)>0$, and
\eqref{eq:narrow-susceptibility-loss}, the second term is at most
$-c_1\rho^3$. The remaining brace term is $O_\beta(\rho^4)$.
The last two errors, after division by $\delta^2$, satisfy
\[
C_\beta|\delta|h^{-3/2}\le C_\beta\rho^8,
\qquad
C_\beta\delta^2h^{-2}\le C_\beta\rho^{24}.
\]
Every positive error is therefore smaller than the negative
$\rho^3$ contribution for all large $n$. Uniformly over the
feasible matrices in question, this proves
\begin{equation}\label{eq:narrow-pressure-gap}
F_{3,n}(D)\le3P_\beta-c_2\delta^2\rho^3
\le3P_\beta-c_2\rho^{43}.
\end{equation}
\par\medskip\noindent\textbf{Step 9. From pressure deficits to the claimed probability.}\enspace
The choice $\kap=1/10000$ satisfies $43\kap<1/24$, so the deficit
in~\eqref{eq:narrow-pressure-gap} dominates the finite-size error
required by Lemma~\ref{lem:pressure-prob}. For each feasible triple
matrix that lemma gives two exponential bounds, with exponents at
least constant multiples of $n\rho^{43}$ and $n\rho^{86}$.
The latter equals $n^{1-86\kap}=n^{0.9914}$. Sum these bounds over
the at most $(n+1)^3$ triple overlap matrices.
For the pairs removed in~\eqref{eq:narrow-pair-deviation},
\eqref{static-eq:pair-pressure} and the same lemma give exponents at least
constant multiples of $n\rho^{26}$ and
$n\rho^{52}=n^{0.9948}$. There are at most $n+1$ pair overlaps
to sum over. The fixed constants and polynomial counts in both
sums are absorbed by $\exp(-n^{4/5})$ for all sufficiently large
$n$. This proves~\eqref{eq:narrow-locking-probability}.
\end{proof}
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/08-narrow-locking.tex ====
==== BEGIN Stretched-exponential-barriers-for-typical-SK-initial-states//sections/09-dynamics.tex ====
\section{Reference banks, crossings and slow mixing}\label{sec:dynamics}
We have established all static inputs. It remains to use them along a path:
coverage selects references, fixed-scale locking controls incompatible
references, and narrow-annulus locking preserves constraints during
backwards searches. We first reduce Theorem~\ref{thm:main} to a sign
passage, then construct the required references, and finally bound the
probability of all resulting threshold constraints. As before,
$L=n^\kap$, $\rho=n^{-\kap}$, and constants may depend on the fixed
inverse temperature but not on $n$.
\subsection{Stationary paths and a sign-passage criterion}
Given $J$, start the discrete heat-bath chain $(x_k)_{k\ge0}$ at
$x_0\sim\pi^J$. Each step chooses a uniform site and resamples its spin
from the conditional Gibbs law. Stationarity gives marginal $\pi^J$
at every deterministic index. If $N$ is an independent rate-$n$ Poisson
process, $x_{N(t)}$ has conditional transition kernel $P_t^J$ from the
realized $x_0$. We work up to the deterministic index
\begin{equation}\label{eq:dyn-time-horizon}
m=\lceil e^{2L}\rceil.
\end{equation}
Markov's inequality gives
\begin{equation}\label{eq:dyn-poisson-tail}
\PP\{N(e^L)>m\}\leq \frac{ne^L}{m}\leq ne^{-L}=o(1).
\end{equation}
For every fixed configuration $v$, the overlap $R(v,x_k)$ changes by at
most $2/n$ at a single step.
A \emph{bank} is an ordered list of independent $\pi^J$ samples. Given
$J$, all raw bank entries are mutually independent and independent of
the complete path. Their sizes are deterministic functions of $n$.
A chosen reference may be reversed in sign at selection; that
orientation is thereafter fixed.
Suppose a first bank is used to select a reference $v_1$ satisfying
$R(v_1,x_0)\geq q_*$, with the choice and orientation depending only on
$J,x_0$ and this bank. Write $\mathcal S$ for successful selection, and
define its sign-passage event by
\[
\mathcal A=\mathcal S\cap
\{R(v_1,x_k)\leq0\text{ for some }0\leq k\leq m\}.
\]
Also write
\[
\mathcal M=\left\{
\norm{P_{e^L}^J(x_0,\cdot)-\pi^J}_{\TV}\leq\frac14\right\}.
\]
Spin-flip symmetry implies, for every oriented $v$,
\[
\pi^J\{z:R(v,z)\leq0\}\geq\frac12.
\]
Condition on $J,x_0$ and successful first-bank selection. Since that
selection uses no future updates, the path and clock still have their
original conditional laws. On $\mathcal M$, total variation therefore
puts conditional probability at least $1/4$ on the opposite-sign
half-space at time $e^L$. Removing clock overruns gives
\begin{equation}\label{eq:dyn-tv-bridge}
\PP(\mathcal A)\geq
\frac14\PP(\mathcal M\cap\mathcal S)
-\PP\{N(e^L)>m\}.
\end{equation}
Thus the remaining task is concrete: make the first reference available
with probability tending to one, and make its sign passage have
probability tending to zero.
Choose a fixed $b\in(0,1)$ no larger than the small-overlap cutoff in
Lemma~\ref{lem:fixed-locking} with large-overlap cutoff $q_*$. Next choose
$b'\in(0,b)$ no larger than that lemma's small-overlap cutoff with
large-overlap cutoff $b$. Fix $t_0>0$ so small that
\begin{equation}\label{eq:dyn-t0-choice}
4t_0<\min\{q_*,b',b^2/2\}.
\end{equation}
We use the notation
\begin{equation}\label{eq:dyn-fixed-error}
\eta_n=n^{-1/100}=\rho^{100}
\end{equation}
for the scale in fixed-scale locking.
\subsection{The case with a fixed forbidden overlap interval}
There are two possible scalar regimes. In the first,
$\Gamma(z)=z$ fails at some point of $(t_0/2,3t_0)$. Continuity then
supplies fixed constants
\[
t_0/2<a_0<a_1<3t_0,\qquad d_0>0,
\]
such that $|q-\Gamma(q)|\geq d_0$ for all $q\in[a_0,a_1]$.
The pair gap~\eqref{static-eq:pair-pressure},
Lemma~\ref{lem:pressure-prob}, and the polynomial number of feasible pair
overlaps imply, for some $c>0$ and all sufficiently large $n$,
\begin{equation}\label{eq:dyn-fixed-forbidden}
\EE\big[(\pi^J)^{\otimes2}
\{(v,x):R(v,x)\in[a_0,a_1]\}\big]\leq e^{-cn}.
\end{equation}
Apply Proposition~\ref{prop:coverage} with $D_1=3$, let $D_3$ be its
constant, and take a single bank with
$K=\lceil\exp((D_3+1)L)\rceil$ entries. Select its first entry with
$|R(v,x_0)|\geq q_*$, oriented to give positive overlap. On the disorder
event of coverage for the whole configuration space, the Gibbs mass of
starting states whose qualifying neighborhood has mass below $e^{-D_3L}$
is at most $e^{-3L}$. For every other starting state, the independent bank
misses that neighborhood with probability at most $e^{-e^L}$. Hence
$\PP(\mathcal S)\to1$.
At a deterministic $k\le m$, the path point and a fixed signed bank
entry are independent Gibbs draws conditional on $J$. Sum
\eqref{eq:dyn-fixed-forbidden} over entries, signs and indices:
\[
\PP\{R(\sigma v,x_k)\in[a_0,a_1]
\text{ for some bank entry }v,\ \sigma\in\{-1,1\},\ k\leq m\}
\leq 2K(m+1)e^{-cn}=o(1).
\]
For large $n$, a sequence beginning at overlap at least $q_*>a_1$ and
reaching nonpositive overlap, with jumps at most $2/n<a_1-a_0$, must visit
$[a_0,a_1]$. Therefore $\PP(\mathcal A)\to0$. Together with
\eqref{eq:dyn-tv-bridge}, this proves the theorem in this case.
This completes the forbidden-interval case. In the second regime we have
\begin{equation}\label{eq:dyn-identity-interval}
\Gamma(z)=z\qquad(t_0/2<z<3t_0),
\end{equation}
and put
\begin{equation}\label{eq:dyn-local-scales}
h=\rho^8,\qquad \varepsilon=\rho^{20}.
\end{equation}
\subsection{Thresholds and simultaneous bank properties}
Define the fixed integers and the number of levels by
\begin{equation}\label{eq:dyn-block-parameters}
H_0=\lceil 2/b^2\rceil,\qquad B=H_0+2,\qquad
p=B\left\lfloor\frac{t_0}{5B\rho}\right\rfloor.
\end{equation}
For large $n$, $p\ge B$. Divide the $p$ levels into consecutive blocks
of $B$ levels each. At the first level of every block after the first,
we \emph{reset} the selection cell to the whole space. Older retained
references are kept unless they meet the conflict rule specified below.
Independently of the disorder, banks and path, draw independent
thresholds $r_i$, uniform on
\begin{equation}\label{eq:dyn-threshold-intervals}
I_i=[t_0+4(i-1)\rho,\ t_0+(4(i-1)+1)\rho].
\end{equation}
These intervals lie in $[t_0,2t_0]$, and
\begin{equation}\label{eq:dyn-threshold-separation}
r_j-r_i\geq3\rho\qquad(i<j).
\end{equation}
Fubini's theorem gives
\[
\EE\big[\mu([r_i-2h,r_i+2h])\big]\leq\frac{4h}{\rho}.
\]
Consequently Markov's inequality and a union bound show that the event
\begin{equation}\label{eq:dyn-threshold-good}
\mathcal T=\{\mu([r_i-2h,r_i+2h])\leq\rho^4
\text{ for every }1\leq i\leq p\}
\end{equation}
satisfies
\begin{equation}\label{eq:dyn-threshold-good-prob}
\PP(\mathcal T^c)\leq\frac{4ph}{\rho^5}=O(\rho^2).
\end{equation}
The Fubini calculation includes atoms; no regularity of the distribution
of the minimizing measure is imposed here.
There will be a bank $\mathcal B_j$ at every level. Its size must
compensate for the number of earlier reference tuples that can define
the selection cell. Using all preceding levels would give a recursion
of growing depth $p$; Proposition~\ref{prop:coverage} does not control
its constants uniformly in $n$. The reset confines this recursion to
$B$ fixed steps.
Bank sizes can then depend only on the position within a block, keeping
the total number of samples at $\exp(O(L))$. Older retained references
will still enter the crossing constraints below.
We require coverage for every possible earlier tuple in the current
block. For entries $u_i\in\mathcal B_i$ from those earlier levels, define
\begin{equation}\label{eq:dyn-cell}
S=\{z:|R(u_i,z)|\leq3t_0
\text{ for every index in this tuple}\}.
\end{equation}
The empty tuple gives the whole space. We require
\begin{equation}\label{eq:dyn-bank-property}
x_k\in S,\quad 0\leq k\leq m
\quad\Longrightarrow\quad
\text{some }w\in\mathcal B_j\cap S
\text{ satisfies }|R(w,x_k)|\geq q_*.
\end{equation}
Absolute values make this property insensitive to orientations.
Choose the bank sizes recursively over the fixed positions
$s=1,\ldots,B$. Once the constants at smaller positions are fixed, set
\[
C_s=\sum_{u<s}(D_{3,u}+2),\qquad D_{1,s}=C_s+5.
\]
Let $D_{3,s}$ be a constant supplied by
Proposition~\ref{prop:coverage} for $D_{1,s}$, and use
\[
K_s=\lceil\exp((D_{3,s}+1)L)\rceil
\]
entries at position $s$. For large $n$, the number of earlier tuples in
the block is at most $\exp(C_sL)$. Only $B$ fixed applications of coverage
are needed, so their disorder events have an intersection $\mathcal D$
with $\PP(\mathcal D)\to1$.
Fix a level at position $s$, a realization $J\in\mathcal D$, and its
earlier bank lists. Each cell $S$ has a bad subset
\[
B_S=\{x\in S:\pi^J\{z\in S:|R(z,x)|\geq q_*\}<e^{-D_{3,s}L}\}
\]
of Gibbs mass less than $e^{-D_{1,s}L}$. The path is independent of those
lists given $J$, and each deterministic-index marginal is $\pi^J$.
A union bound over tuples and deterministic indices therefore bounds the
probability that any $x_k$ lies in its cell's bad subset by
\[
(m+1)e^{(C_s-D_{1,s})L}=O(e^{-3L}).
\]
After excluding these bad targets, condition also on the complete path.
The bank at the present level is still fresh. Its probability of missing
a qualifying neighborhood inside the cell is at most
\[
(1-e^{-D_{3,s}L})^{K_s}\leq e^{-e^L}.
\]
Union over the same tuples and indices, and then over the $p=O(L)$
levels, shows that \eqref{eq:dyn-bank-property} holds simultaneously with
probability $1-o(1)$. The recursion has only $B$ steps, so the total number
of entries across all $p$ banks is $\exp(O(L))$.
We also impose simultaneous avoidance of the exceptional triples in
Lemma~\ref{lem:fixed-locking}, for the two cutoff pairs $(q_*,b)$ and
$(b,b')$, and in Lemma~\ref{lem:narrow-locking}. We do so for every
ordered triple of signed entries from distinct banks, and for every
ordered triple consisting of two signed entries from distinct banks and
one path point $x_k$, $0\leq k\leq m$. For each deterministic choice,
the conditional law given $J$ is $(\pi^J)^{\otimes3}$: bank independence,
stationarity at that deterministic index, and spin-flip symmetry suffice.
The sampled configurations themselves need not be different.
There are $\exp(O(L))$ such choices, including all orders and signs.
The exceptional probability for each is bounded by a constant times
$e^{-n^{9/10}}+e^{-n^{4/5}}$. A union bound thus makes all these locking
assertions simultaneous with probability $1-o(1)$. Let $\mathcal G$ be
the intersection of this event, the simultaneous bank property, and
$\mathcal T$. We have proved
\begin{equation}\label{eq:dyn-good-event-prob}
\PP(\mathcal G^c)=o(1).
\end{equation}
We have made all these estimates simultaneously before choosing references
or crossing times. Later application at a backwards crossing therefore
uses this deterministic-index event, not an assertion of stationarity
at a random time.
\subsection{Backwards searches and deletion of conflicts}
Select $v_1$ from $\mathcal B_1$ as the first entry whose absolute overlap
with $x_0$ is at least $q_*$, and orient it positively. If it exists, let
$\tau_1$ be the first $k\leq m$ with $R(v_1,x_k)\leq r_1$. On success
initialize the ordered retained list as $A_1=\{1\}$.
At a subsequent level $j$, assuming all earlier levels have succeeded,
form the cell \eqref{eq:dyn-cell} using the selected references in earlier
positions of the current block. Choose as $v_j$ the first entry of
$\mathcal B_j$ lying in this cell and having absolute overlap at least
$q_*$ with $x_{\tau_{j-1}}$. Orient it so that
$R(v_j,x_{\tau_{j-1}})\geq q_*$. From $A_{j-1}$ delete the conflicts,
namely those indices $i$ with
\begin{equation}\label{eq:dyn-conflict}
|R(v_i,v_j)|>b.
\end{equation}
The remaining old indices are called compatible. Search backwards from
$\tau_{j-1}$, stopping at the first point encountered with overlap at
most $r_j$ with $v_j$. Equivalently, if the set is nonempty, define
\begin{equation}\label{eq:dyn-backward-time}
\tau_j=\max\{0\leq k\leq\tau_{j-1}:R(v_j,x_k)\leq r_j\}.
\end{equation}
Add $j$ to the compatible list to obtain $A_j$.
To use independence of each fresh threshold, define $v_j$ on failure
realizations as well. A missing candidate or missing crossing ends
the procedure. A candidate already selected before its crossing fails is
kept as that level's reference. At a level with no candidate, and at all
levels after an earlier failure, use the first entry of the corresponding
bank without sign reversal. These defaults serve the final conditional-probability calculation.
Selection and orientation of $v_j$ use only thresholds earlier than
$r_j$; its own threshold is used solely for the subsequent crossing. The first reference, when selected successfully, uses only
$J,x_0$ and $\mathcal B_1$ as required by
\eqref{eq:dyn-tv-bridge}.
We now work deterministically on $\mathcal G\cap\mathcal A$.
For all sufficiently large $n$, we will show that every level succeeds,
that reference $1$ survives, and that
\begin{align}
|R(v_i,x_{\tau_j})-r_i|&\leq6j\varepsilon
&&(i\in A_j),\label{eq:dyn-induction-path}\\
|R(v_i,v_\ell)-r_i|&\leq6j\varepsilon
&&(i,\ell\in A_j,\ i<\ell).
\label{eq:dyn-induction-references}
\end{align}
The induction also keeps the current block complete: after level $j$,
all references from that block's first level through $j$ remain on the
list. At the next reset, the preceding block is therefore intact.
Older blocks may already have gaps, but the complete preceding block
will prevent a new deletion from reaching them.
\input{figures/protected-block}
Deletions occur only at the first level of a new block, and at most $H_0$
indices, forming a suffix of the old list, are deleted at each such level.
At level $1$, coverage of the whole space supplies the reference.
Sign passage forces the first crossing, where the one-step overlap
bound puts the value in $[r_1-2/n,r_1]$. This starts the induction. In the following induction we
use the scale relations
\begin{equation}\label{eq:dyn-scale-relations}
\frac1n\ll\eta_n=\rho^{100}\ll\varepsilon=\rho^{20},\qquad
p\varepsilon=O(\rho^{19})\ll h=\rho^8\ll\rho.
\end{equation}
Suppose the assertions hold through level $j-1$. All earlier references
in the current block are still retained, since there have been no
deletions within that block after its first level. Their overlaps with
$x_{\tau_{j-1}}$ have absolute value at most
$2t_0+6p\varepsilon<3t_0$. Thus this target belongs to the required cell,
and the new selection $w=v_j$ exists by
\eqref{eq:dyn-bank-property}.
For each compatible old index $i$, the two small overlaps in the triple
$(v_i,w,x_{\tau_{j-1}})$ have absolute value at most $b$, whereas
$R(w,x_{\tau_{j-1}})\geq q_*$. Fixed-scale locking gives
\begin{equation}\label{eq:dyn-target-transfer}
|R(v_i,w)-R(v_i,x_{\tau_{j-1}})|\leq2\eta_n.
\end{equation}
In particular,
\begin{equation}\label{eq:dyn-reference-transfer}
|R(v_i,w)-r_i|\leq6(j-1)\varepsilon+2\eta_n
\qquad(i\text{ compatible}).
\end{equation}
\par\medskip\noindent\textbf{Conflicts occupy a bounded final segment.}\enspace
A conflicting old index $i$ cannot precede a compatible old index $\ell$. Indeed, by
\eqref{eq:dyn-induction-references} and
\eqref{eq:dyn-reference-transfer}, the magnitudes of
$R(v_\ell,v_i)$ and $R(v_\ell,w)$ are within
$6(j-1)\varepsilon$ and $6(j-1)\varepsilon+2\eta_n$ of $r_i$ and
$r_\ell$, respectively.
Both magnitudes are less than $b'$, by
\eqref{eq:dyn-t0-choice}. Their difference in magnitude is at least
\[
3\rho-12p\varepsilon-2\eta_n>2\eta_n.
\]
On the other hand $|R(v_i,w)|>b$. Reverse one of this pair if necessary
to make their mutual overlap positive. The signed difference of the two
small overlaps has absolute value at least their difference in magnitude.
This contradicts fixed-scale locking with cutoff pair $(b,b')$, which
holds for all signs on $\mathcal G$. The conflicts consequently form a
suffix of the old retained list.
The number of conflicts is bounded by $H_0$. Indeed, the induction
bounds all absolute overlaps between retained old references by
$2t_0+6p\varepsilon<3t_0$. If $d\geq1$ of them conflict with $w$,
normalize them to unit vectors and reverse each sign to give inner
product greater than $b$ with $w/\sqrt n$. Denote these vectors by
$u_1,\ldots,u_d$. Then
\[
d^2b^2<\left\langle\sum_{a=1}^d u_a,\frac{w}{\sqrt n}\right\rangle^2
\leq\norm{\sum_{a=1}^d u_a}^2
\leq d+3t_0d(d-1).
\]
Since $3t_0<b^2/2$, this implies
\[
d<\frac{1-3t_0}{b^2-3t_0}<\frac{2}{b^2}\leq H_0.
\]
In particular $d\leq H_0$.
If $j$ is not the first level of its block, the most recent old index
$j-1$ is compatible: the cell condition gives
$|R(v_{j-1},w)|\leq3t_0<b$. A suffix of conflicts is therefore empty.
At the first level of a new block, the additional invariant says that
all $B$ entries selected in the preceding block are still present. Removing at most $H_0<B$ final indices
cannot reach any older block, or the first index of the preceding block.
In particular index $1$ survives, including at the first reset. Adding the new index starts the current block as a retained singleton;
at a nonreset it extends the complete current block because there is no
deletion. This establishes the additional invariant and the deletion
claims before proving that the new crossing exists.
\par\medskip\noindent\textbf{The backwards search preserves old constraints.}\enspace
Follow the search for $\tau_j$, including the crossing step when there
is one. The new reference satisfies throughout this search
\begin{equation}\label{eq:dyn-search-large-overlap}
R(w,x_k)\geq r_j-2/n.
\end{equation}
If there is no crossing, this follows with the stronger strict bound
$R(w,x_k)>r_j$ at every index down to zero. If there is a crossing, all
indices strictly after it have overlap greater than $r_j$, and the jump
bound gives \eqref{eq:dyn-search-large-overlap} at the crossing itself.
For each compatible old reference $v_i$,
\eqref{eq:dyn-reference-transfer} places the fixed value $R(v_i,w)$
near $r_i$. The larger threshold $r_j$ separates the new overlap from
that old level throughout the search. We will use narrow-annulus
locking to prove
\begin{equation}\label{eq:dyn-search-locking}
|R(v_i,x_k)-R(v_i,w)|<2\varepsilon
\end{equation}
on this entire searched portion. Equation~\eqref{eq:dyn-target-transfer} proves the claim at the target.
At a first violation while moving backwards, the one-step bound puts
the left side in $[2\varepsilon,2\varepsilon+2/n]$. For this triple
use the parameters of Lemma~\ref{lem:narrow-locking}:
\[
r=\frac{R(v_i,x_k)+R(v_i,w)}2,\qquad
\delta=\frac{R(v_i,x_k)-R(v_i,w)}2,\qquad q=R(w,x_k).
\]
Then
\[
\varepsilon\leq|\delta|\leq\varepsilon+1/n\leq2\varepsilon,
\qquad
|r-r_i|\leq6(j-1)\varepsilon+2\eta_n+\varepsilon+1/n<h.
\]
It follows that $t_0/2<r<3t_0$ and
$[r-h,r+h]\subseteq[r_i-2h,r_i+2h]$. Thus $\mathcal T$ gives the
required bound $\mu([r-h,r+h])\leq\rho^4$. Also, by
\eqref{eq:dyn-threshold-separation},
\eqref{eq:dyn-search-large-overlap}, and
\eqref{eq:dyn-scale-relations},
\[
q-r\geq3\rho-6(j-1)\varepsilon-2\eta_n-\varepsilon-3/n
\geq\rho.
\]
All conditions of the forbidden narrow-annulus event hold, contradicting
its simultaneous avoidance on $\mathcal G$. This proves
\eqref{eq:dyn-search-locking}.
The crossing must exist. Otherwise the search reaches index zero, and
\eqref{eq:dyn-reference-transfer}--\eqref{eq:dyn-search-locking}
apply there to the surviving compatible reference $1$. They would give
\[
R(v_1,x_0)<r_1+6(j-1)\varepsilon+2\eta_n+2\varepsilon<q_*,
\]
contradicting the first selection. Hence the crossing exists. Its new
reference overlap belongs to $[r_j-2/n,r_j]$. For each retained old
reference, the error at the new crossing is at most
$6(j-1)\varepsilon+2\eta_n+2\varepsilon\leq6j\varepsilon$.
The old-new reference errors satisfy the same claimed bound by
\eqref{eq:dyn-reference-transfer}, and old-old reference errors are
unchanged. This proves
\eqref{eq:dyn-induction-path}--\eqref{eq:dyn-induction-references} and
completes the induction.
There are at most $p/B$ reset occasions, so the final list has size at least
\begin{equation}\label{eq:dyn-retained-count}
|A_p|\geq p-H_0p/B\geq p/B.
\end{equation}
Consequently $\mathcal G\cap\mathcal A$ implies that, for some
index $k$ in $\{0,\ldots,m\}$ and some subset of $p/B$
indices, the inequalities
\begin{equation}\label{eq:dyn-small-target-windows}
|R(v_i,x_k)-r_i|\leq6p\varepsilon
\end{equation}
hold simultaneously. We have reduced sign passage to this existence event. The probability
bound that follows will not condition on $\mathcal G$ or on the
random retained list.
\subsection{Predictable thresholds and completion of the proof}
Let $\mathcal F_0$ be generated by $J$, all raw bank lists, and the path
$(x_0,\ldots,x_m)$, and set
\[
\mathcal F_i=\sigma(\mathcal F_0,r_1,\ldots,r_i),
\qquad 1\leq i\leq p.
\]
Every $r_i$ is uniform on $I_i$ and independent of $\mathcal F_{i-1}$.
The failure conventions ensure that each oriented $v_i$ is
$\mathcal F_{i-1}$-measurable. Here is the complete induction. At level $1$ its
selection does not use any threshold. Inductively, whether a prior level
failed, the previously selected references, and the preceding crossing
time are all determined by earlier thresholds. Selection at level $i$
uses only that information and the bank and path data in $\mathcal F_0$.
The subsequent crossing may use $r_i$, but never changes that selected
reference. A default caused by an earlier failure likewise depends only
on earlier thresholds. This proves the claim, including on failure
events.
For each deterministic $k\leq m$, let
\[
E_i(k)=\{|R(v_i,x_k)-r_i|\leq6p\varepsilon\},\qquad
\gamma_n=\frac{12p\varepsilon}{\rho}.
\]
The interval defining $E_i(k)$ has length at most $12p\varepsilon$,
while $I_i$ has length $\rho$. Thus, for all sufficiently large $n$,
\begin{equation}\label{eq:dyn-conditional-window}
\PP(E_i(k)\mid\mathcal F_{i-1})\leq\gamma_n<1.
\end{equation}
For a deterministic subset $\{i_1<\cdots<i_s\}$, the earlier events
$E_{i_1}(k),\ldots,E_{i_{s-1}}(k)$ are
$\mathcal F_{i_s-1}$-measurable. Conditioning first at the largest
selected index and using \eqref{eq:dyn-conditional-window} yields
\[
\PP\left(\bigcap_{a=1}^s E_{i_a}(k)\right)
\leq\gamma_n\,
\PP\left(\bigcap_{a=1}^{s-1}E_{i_a}(k)\right)
\leq\gamma_n^s.
\]
The elimination order is essential: the latest selected threshold is
integrated out first. Later centers may depend on earlier thresholds,
but the displayed conditioning remains valid. Neither the random
retained subset nor the good event has been conditioned on.
Taking $s=p/B$, then taking a union over all deterministic subsets and all
deterministic indices $k$, proves
\begin{equation}\label{eq:dyn-threshold-union}
\PP\{\text{\eqref{eq:dyn-small-target-windows} holds for some }k
\text{ and some subset of size }p/B\}
\leq(m+1)2^p\left(\frac{12p\varepsilon}{\rho}\right)^{p/B}
=o(1).
\end{equation}
Indeed $p=(t_0/5)L+O(1)$, $\rho=L^{-1}$, and
$12p\varepsilon/\rho=O(\rho^{18})$. The logarithm of the right-hand
side is therefore bounded above by
\[
2L+O(p)-\frac{18p}{B}\log(1/\rho)
=-\Omega(L\log L).
\]
The deterministic induction and \eqref{eq:dyn-good-event-prob} now imply
\[
\PP(\mathcal A)
\leq\PP(\mathcal G^c)
+\PP(\mathcal G\cap\mathcal A)=o(1).
\]
The first selection succeeds on the simultaneous bank event at level $1$,
so $\PP(\mathcal S^c)=o(1)$ as well. Finally,
\eqref{eq:dyn-tv-bridge} and \eqref{eq:dyn-poisson-tail} give
\[
\PP(\mathcal M)
\leq\PP(\mathcal S^c)+4\PP(\mathcal A)
+4\PP\{N(e^L)>m\}=o(1).
\]
This proves Theorem~\ref{thm:main} also under
\eqref{eq:dyn-identity-interval}, and the two cases exhaust the
possibilities.
Finally take $T_n=\lfloor e^L\rfloor$ discrete update attempts.
This deterministic index lies below $m$ for large $n$. If its transition
law is within $1/4$ of equilibrium, the opposite-sign half-space has
conditional probability at least $1/4$ at that index. Thus the same
sign-passage inequality holds with no clock-overrun term. All bank,
locking and threshold estimates already concern the discrete path,
so they prove the second assertion of Theorem~\ref{thm:main} at exactly
$\lfloor e^L\rfloor$ attempts.
==== END Stretched-exponential-barriers-for-typical-SK-initial-states//sections/09-dynamics.tex ====