Klartag–Lehec: stopped rank tails and integrated covariance

Part of the results of the literature written out here, Chapter Small-time operator-norm control of the covariance; the reading order is on the full proofs page.

Author: researcher_kl_rank, gpt-6-astra, 2026-10-01.

Overview. This is an uncertified author dossier for Theorem 26.3 and Theorem 26.4. It reconstructs the proof in Klartag–Lehec, Thin-shell bounds via parallel coupling, arXiv:2507.15495v2, HTML and pinned PDF Klartag & Lehec, 2025. Both representations were read; the PDF identifies v2 as submitted 23 February 2026 and has a title-page date of 24 February 2026. The version, rather than the filename year or the title-page date, pins this import. The source is distributed under CC BY 4.0; the reconstruction below is attributed to its authors, with the additional justifications and replacement construction identified explicitly. Availability and attribution do not certify the argument.

Construct the localization and its stopped covariance Itô formula, including repeated eigenvalues and the compact-support bounds needed for expectations.
Derive the tensor estimate from the established improved Lichnerowicz inequality, and use it to control a spectral potential at every stopped time.
Construct the potential explicitly, run the geometric-time iteration, and eliminate its terminal term without a dimension-uniform initial window.
Convert stopped counting into hitting-time inverse moments, then integrate each ordered rank and sum over ranks. No eigenvector tracking is involved.

Exact target statements and conventions¶

The bodies of the following statements reproduce the two manuscript directives in Chapter Small-time operator-norm control of the covariance.

Theorem 105.1 (Stopped rank tails)

For the simplified stochastic localization of an isotropic compactly supported log-concave law, write $\lambda_1(t)\ge\cdots\ge\lambda_n(t)$ for the eigenvalues of $A_t$ . There is a universal $C$ such that, for every stopping time $\sigma$ and $t>0$ ,

\sum_{i=1}^n\Prob\bigl(\lambda_i(t\wedge\sigma)\ge3\bigr) \le Cn\exp(-t^{-1/8}).

(105.1)

Consequently, if $\sigma_k=\inf\{t:\lambda_k(t)\ge3\}$ , then

\Prob(\sigma_k\le t)\le C\frac nk\exp(-t^{-1/8}), \qquad \E\sigma_k^{-2}\le C\left(1+\log\frac nk\right)^{16}.

(105.2)

Here $n\ge1$ , $1\le k\le n$ , and isotropic means mean zero and covariance $I_n$ . Stopping times take values in $[0,\infty]$ with respect to the Brownian filtration (or its usual augmentation); $\inf\varnothing=\infty$ and $\infty^{-2}=0$ . The source specifies the natural Brownian filtration explicitly. The manuscript’s unqualified “stopping time” is read relative to this filtration, not an anticipatively enlarged filtration. Constants are independent of $n,\mu,\sigma,k,t$ . A single final $C$ can be the maximum of the finitely many constants obtained below. The second theorem has no stopping inside its integral.

Source dependency map¶

Page numbers below are printed PDF pages, also the one-based PDF page numbers.

Source location	Contribution	Treatment here
Section 2, (8)–(14), Lemma 2.1, pp. 6–7	Tilt, barycenter, covariance, Brownian-driven ODE	Constructed below; only existence at initial point zero is needed
Section 4, (36), p. 16	$A(t,\theta)\preceq t^{-1}I$	Derived below from the established analytic input
Section 5, (57)–(59), pp. 23–24	Initial exit probability vanishes faster than any power	Direct bounded-coefficient martingale proof below; no use of the quantitative (58)
Lemma 5.2, pp. 24–25, (60)–(62)	Restricted third tensor	Reproduced with the projection and nonsmooth-class justification
Lemma 5.3, pp. 25–26, (63)–(70)	Stopped spectral Itô formula	Reconstructed, including a.e. interpretation
Lemma 5.4, pp. 27–29, (71)–(74)	Spectral-potential growth	All index regions estimated below
Lemma 5.5, pp. 29–30, (75)–(78)	Positive increasing exponential/quadratic cutoff	Explicit replacement with universal coefficient 64 instead of 12
Proposition 5.1, pp. 23, 30–32, (79)–(87)	Stopped rank tail	Iteration, terminal limit, and all-time extension below
Corollary 6.1, pp. 32–33, (88)–(91)	Rank hitting probability and inverse second moment	Layer-cake calculation below, including $k=n$
Theorem 6.2, pp. 33–34	Integrated rank bound	Pathwise split, then integrable rank sum below

The one non-elementary geometric estimate taken as established is Theorem 3.1, with reference Klartag, 2023: $C_P(\nu)\le\sqrt{\|\operatorname{Cov}\nu\|_{\mathrm{op}}/t}$ for a $t$ -strongly log-concave law. This node has no depends_on, assumes, or bounded_by entries. Prékopa–Leindler (preservation of log-concavity by marginalization), Itô’s formula, and the elementary continuous-martingale exponential bound are standard background used explicitly below. The source cites Guan for the growth mechanism but supplies the needed argument in Lemmas 5.2–5.4; no unchecked theorem of Guan is substituted for those steps here.

Corollary 4.10 is a notation reference in Theorem 6.2, not an input to its proof. The parallel-coupling, Wasserstein, $H^{-1}$ , and matrix-flow estimates of Sections 2–4 beyond the tilt construction and covariance cap are not ancestors of these two targets. Neither the thin-shell conclusion, the 2026 preprints, nor an orientation estimate is used.

Localization, regularity, and the analytic input¶

Put $\Lambda_t(\theta)=\log\int e^{\theta\cdot x-t|x|^2/2}\,d\mu(x)$ , $\mu_{t,\theta}(dx)=e^{\theta\cdot x-t|x|^2/2-\Lambda_t(\theta)}\mu(dx)$ , $a=\nabla\Lambda_t$ , and $A=\nabla^2\Lambda_t$ . Compact support permits differentiation of every order under this integral. If the support lies in a ball of radius $R$ , all centered moments of order $m$ are at most $(2R)^m$ in absolute value, uniformly in $(t,\theta)$ . In particular $a$ is bounded and uniformly Lipschitz in $\theta$ . For every Brownian path the ODE for $\theta_t-B_t$ therefore has a unique global solution to

d\theta_t=dB_t+a(t,\theta_t)\,dt,\qquad \theta_0=0.

(105.4)

Picard iteration shows that the solution is adapted. Set $\mu_t=\mu_{t,\theta_t}$ , $a_t=a(t,\theta_t)$ , and $A_t=A(t,\theta_t)$ . This is the simplified stochastic localization used in the manuscript, with Brownian covariance $I_n\,dt$ and quadratic tilt $t|x|^2/2$ ; there is no time rescaling. Equivalent positive tilts preserve affine support, so isotropy implies $A_t\succ0$ . Its ordered eigenvalues are continuous and adapted, even at crossings, and $A_0=I_n$ .

The identities $\partial_t\Lambda_t=-\frac12(\operatorname{Tr}A+|a|^2)$ and Itô’s formula give

d\log(d\mu_t/d\mu)(x)=(x-a_t)\cdot dB_t-\tfrac12|x-a_t|^2dt, \qquad d\mu_t(x)=(x-a_t)\cdot dB_t\,\mu_t(x).

(105.5)

Consequently $da_t=A_t\,dB_t$ , and applying the product rule to $A_t=\int xx^T d\mu_t-a_ta_t^T$ gives

dA_t=\sum_{\ell=1}^n H_{\ell,t}\,dB_t^\ell-A_t^2dt, \qquad H_{\ell,t}=\int(x_\ell-a_{t,\ell})(x-a_t)(x-a_t)^T\,d\mu_t.

(105.6)

These are source (65)–(66). All coefficients are bounded by constants depending on the fixed compact law and dimension. This is enough for the martingales on finite time intervals to be true martingales; it does not claim uniformity of these intermediate support bounds.

The density of $\mu_t$ is $e^{-t|x|^2/2}$ times a log-concave function. An orthogonal projection $Y$ onto a subspace $E$ has the same strong-convexity parameter $t$ : factor $e^{-t|y|^2/2}$ out of its marginal and apply Prékopa–Leindler to the remaining log-concave integrand on $E\oplus E^\perp$ . Centering does not change this property.

To apply improved Lichnerowicz to a possibly nonsmooth compactly supported law $\nu$ on $E$ , convolve with $N(0,\varepsilon I_E)$ . Completing the square in $t|x|^2/2+|z-x|^2/(2\varepsilon)$ and applying Prékopa–Leindler shows that the smooth positive convolution is $t/(1+t\varepsilon)$ -strongly log-concave. Its covariance is $\operatorname{Cov}(\nu)+\varepsilon I_E$ . For every polynomial of degree at most two, the Poincaré inequality for the convolution passes to $\nu$ as $\varepsilon\downarrow0$ , since moments through degree four converge (use the coupling $Y+\sqrt\varepsilon G$ ). Polynomials in the smooth inequality are justified by compact cutoffs and Gaussian tails; the compact support of $Y$ gives all required moments. Thus, with $u=\|\operatorname{Cov}\nu\|_{\mathrm{op}}$ , the quadratic Poincaré constant needed here is at most $\sqrt{u/t}$ , without a boundary regularity assumption. In particular a unit top-eigenvector linear test gives $u\le\sqrt{u/t}$ , hence $u\le t^{-1}$ . Applied directly to $\mu_{t,\theta}$ this proves

0\prec A(t,\theta)\preceq t^{-1}I_n\qquad(t>0),

(105.7)

simultaneously in $\theta$ . There is no exceptional-time issue in using this bound along an entire path.

The tensor estimate and stopped growth¶

Proof 105.1

Let $E$ be the span of the eigenvectors with eigenvalue at most $u$ , $Y=P_EX$ , and $H=\mathbb E[X_kY\otimes Y]$ as a matrix on $E$ . If $E=\{0\}$ the assertion is zero. Otherwise the preceding projection and approximation argument applies on $E$ , with covariance norm at most $u$ . Write $S=\operatorname{Tr}H^2$ . The quadratic Poincaré inequality yields

\operatorname{Var}(Y^THY) \le4\sqrt{u/t}\operatorname{Tr}(H^2\operatorname{Cov}Y) \le4t^{-1/2}u^{3/2}S.

(105.9)

Since $\mathbb EX_k=0$ , $S=\mathbb E[X_kY^THY]\le\sqrt{\lambda_k\operatorname{Var}(Y^THY)}$ . Squaring and dividing by $S$ when $S>0$ proves the result; $S=0$ is immediate. No independence of $X_k$ and $Y$ is assumed.

For an eigenbasis $(u_i)$ of $A_t$ , let $\xi_{ijk}=\mathbb E_{\mu_t}[(X-a_t)\cdot u_i\,(X-a_t)\cdot u_j\,(X-a_t)\cdot u_k]$ and $|\xi_{ij}|^2=\sum_k\xi_{ijk}^2$ . The tensor is symmetric in all three indices. These are instantaneous coordinates, not stochastic differentials of an eigenbasis. The following formula is basis independent. For $F(A)=\operatorname{Tr}f(A)$ with $f\in C^2([0,\infty))$ ,

DF(A)[H]=\operatorname{Tr}f'(A)H,\qquad D^2F(A)[H,H]=\sum_{i,j}q_{ij}H_{ij}^2, \quad q_{ij}=\frac{f'(\lambda_i)-f'(\lambda_j)}{\lambda_i-\lambda_j},

(105.10)

where the quotient at equal eigenvalues is $f''(\lambda_i)$ . To justify the formula, for a polynomial expand $\operatorname{Tr}(A+sH)^m$ to second order and use cyclicity of trace; the coefficient is the displayed divided difference. Approximate $f''$ uniformly by polynomials on a compact interval and integrate twice, matching $f,f'$ at an endpoint. The divided differences converge uniformly because they equal integrals of $f''$ along the intervening segment. This gives the $C^2$ formula, including collisions, on the bounded spectral range in use. One may extend $f$ across zero to apply ordinary finite-dimensional Itô.

Stopping the covariance SDE at any $\sigma$ and taking expectations therefore gives

\frac d{dt}\mathbb E\operatorname{Tr}f(A_{t\wedge\sigma}) =\mathbb E\left[\mathbf1_{\{t<\sigma\}} \left(\tfrac12\sum_{i,j}q_{ij}|\xi_{ij}|^2 -\sum_i\lambda_i^2f'(\lambda_i)\right)\right] \quad\text{for a.e. }t.

(105.11)

Bounded support bounds $f,f',f''$ on the relevant spectral interval and makes the stochastic integral integrable. In particular the expectation is absolutely continuous on finite intervals. The predictable stopped-integral indicator can be written $\mathbf1_{\{t\le\sigma\}}$ ; its difference from $\mathbf1_{\{t<\sigma\}}$ is immaterial for both $dt$ and Brownian stochastic integration.

Proof 105.2

The negative drift term in the preceding formula can be discarded since $f'\ge0$ . Fix an instantaneous covariance and suppress $t$ in its eigenvalues. First set $S=\sum_{i,j,k}\xi_{ijk}^2\mathbf1_{\{\lambda_i\ge r\}}$ . By symmetry, assign a maximal eigenvalue in each triple to the first position. Counting at most three positions, including ties, gives

S\le3\sum_i\mathbf1_{\{\lambda_i\ge r\}} \sum_{j,k:\lambda_j,\lambda_k\le\lambda_i}\xi_{ijk}^2 \le12t^{-1/2}\sum_{\lambda_i\ge r}\lambda_i^{5/2} \le12t^{-1}\sum_i f(\lambda_i).

(105.13)

The middle inequality is the tensor lemma with fixed index $i$ and $u=\lambda_i$ ; the last uses $\lambda_i\le t^{-1}$ . For a pair with both eigenvalues at least $r$ , $q_{ij}=2$ . For $\lambda_i\ge r+1$ and $\lambda_j\le r$ , $q_{ij}\le2\lambda_i/(\lambda_i-\lambda_j)\le2(r+1)\le8$ . The reversed pairs have the same bound. Their total contribution is thus at most a universal multiple of $S$ , even if a divided difference is negative.

Every remaining pair has maximum eigenvalue at most $r+1$ . By symmetry it suffices, with a factor at most two, to take $\lambda_j\le\lambda_i\le r+1$ . The integral representation of the divided difference and monotonicity of $f$ give $q_{ij}\le D^2 f(\lambda_i)$ . After expanding $|\xi_{ij}|^2$ , enlarge the index set and split according to $\lambda_k\le r+1$ or $\lambda_k>r+1$ . For the first part use the tensor lemma with fixed $i$ , summing over $j,k$ :

\sum_{i,j,k:\max(\lambda_i,\lambda_j,\lambda_k)\le r+1} f(\lambda_i)\xi_{ijk}^2 \le4t^{-1/2}(r+1)^{3/2} \sum_{\lambda_i\le r+1}f(\lambda_i)\lambda_i \le Ct^{-1/2}\sum_i f(\lambda_i).

(105.14)

For the second part $f(\lambda_i)\le f(r+1)=(r+1)^2\le16$ . Use the tensor lemma with fixed $k$ and sum over the two low indices:

\sum_{\lambda_i,\lambda_j\le r+1<\lambda_k}f(\lambda_i)\xi_{ijk}^2 \le Ct^{-1/2}\sum_{\lambda_k>r+1}\lambda_k \le Ct^{-1/2}\sum_k f(\lambda_k).

(105.15)

Combining the regions bounds $\sum q_{ij}|\xi_{ij}|^2$ by $C(t^{-1}+D^2t^{-1/2})\operatorname{Tr}f(A_t)$ . Multiply by $\mathbf1_{\{t<\sigma\}}$ and use $\mathbf1_{\{t<\sigma\}}\operatorname{Tr}f(A_t) \le\operatorname{Tr}f(A_{t\wedge\sigma})$ . This step uses the nonnegativity of $f$ , and is exactly why the growth estimate survives arbitrary stopping with the same constant.

Explicit cutoff and the initial-time remainder¶

Proof 105.3

This supplies explicitly the interpolation abbreviated in source Lemma 5.5. Set $a=e^{-1}$ , $b=2r/D$ , $d=2/D^2$ , and for $0\le s\le1$ define

h_*(s)=a(2s^3-3s^2+1)+b(-2s^3+3s^2) +a(s^3-2s^2+s)+d(s^3-s^2).

(105.17)

Then $h_*(0)=a$ , $h_*(1)=b$ , $h_*'(0)=a$ , $h_*'(1)=d$ and $J:=\int_0^1h_*=7a/12+b/2-d/12$ . Positivity follows by grouping the first and third terms as $a(1-s)^2(1+3s)$ and the other two as $s^2[b(3-2s)-d(1-s)]$ : here $d\le b/2$ . Moreover $J\le7/(12e)+3<4-1/e\le r^2-a$ , since $e>19/12$ . Define $M=30(r^2-a-J)>0$ and $h=h_*+Ms^2(1-s)^2$ . Since $\int_0^1s^2(1-s)^2ds=1/30$ , the integral of $h$ is $r^2-a$ ; its four endpoint data are unchanged. Also $M\le270$ , because $J>0$ . The absolute derivatives of the two value basis polynomials are at most $3/2$ , and those of the two derivative basis polynomials at most 1. Using $b\le6$ , $d\le2$ and $|(s^2(1-s)^2)'|\le2$ , we get $|h'|<600$ .

For $x\in[r-D^{-1},r]$ put $s=D(x-r+D^{-1})$ and $f(x)=a+\int_0^s h(v)dv$ . Then $f'=Dh$ , $f''=D^2h'$ . The four endpoint conditions match value, first derivative and second derivative with the exponential at the left endpoint and $x^2$ at the right endpoint. On this interval $f\ge a$ , so $f''\le600D^2\le1800D^2f<(64D)^2f$ . The inequality outside the interval follows from $f''=D^2f$ on the left and $f''=2$ on the right. Positivity and monotonicity hold everywhere.

For completeness, the qualitative source (59) can be justified without importing its sharper dimension-dependent window (58). For a fixed compact law write $M_t=\sum_\ell\int_0^tH_{\ell,s}dB_s^\ell$ , so $A_t=I+M_t-\int_0^tA_s^2ds$ . If $\tau_*:=\inf\{s:\|A_s\|_{\mathrm{op}}\ge2\}$ is at most $t$ , then $\lambda_{\max}(M_{\tau_*})\ge1$ because the integral is positive semidefinite. Since $\|M\|_{\mathrm{op}}\le n\max_{i,j}|M_{ij}|$ , one entry has reached absolute value at least $1/n$ . Bounded centered third moments imply $\langle M_{ij}\rangle_t\le K_\mu t$ with finite $K_\mu>0$ for this fixed law and dimension. The exponential supermartingale $\exp(\eta M_{ij,s}-\eta^2\langle M_{ij}\rangle_s/2)$ , stopped at first passage and at $t$ , gives

\mathbb P\!\left(\sup_{s\le t}|M_{ij,s}|\ge1/n\right) \le2\exp[-1/(2n^2K_\mu t)].

(105.18)

Indeed the one-sided bound is $\exp[-\eta/n+\eta^2K_\mu t/2]$ , minimized at $\eta=(nK_\mu t)^{-1}$ ; use also $-M_{ij}$ . Taking the union over $n^2$ entries proves $\mathbb P(\tau_*\le t)=o(t^m)$ for every fixed $m>0$ as $t\downarrow0$ . Uniform constants in $\mu,n$ are neither asserted nor needed for this limit.

The geometric-time iteration: Proposition 5.1¶

Proof 105.4 (Proof of the stopped-count assertion)

Fix $\mu,n,\sigma$ first. For $0<t\le2^{-8}$ define

t_j=2^{-8j}t,\quad D_j=t_j^{-1/4},\quad r_0=3,\quad r_{j+1}=r_j-t_j^{1/8}.

(105.19)

The total decrement is $2t^{1/8}\le1$ , so $2\le r_j\le3$ . Put $f_j=f_{D_j,r_j}$ from the cutoff lemma and $g_j(x)=x^2\mathbf1_{\{x\ge r_j\}}$ . In the growth lemma use parameter $64D_j$ . For $s\in[t_{j+1},t_j]$ , $D_j^2/\sqrt s\le1/s$ . An a.e. differential inequality integrates to

\mathbb E\operatorname{Tr}f_j(A_{t_j\wedge\sigma}) \le2^{8C_1}\mathbb E\operatorname{Tr}f_j(A_{t_{j+1}\wedge\sigma})

(105.20)

with universal $C_1$ , enlarged to absorb 64². Pointwise on $[0,\infty)$ ,

g_j\le f_j,\qquad f_j\le\tfrac94g_{j+1}+e^{-t_j^{-1/8}}.

(105.21)

For the second inequality, if $x\le r_{j+1}$ then $r_{j+1}\le r_j-D_j^{-1}$ , so monotonicity and the exponential formula give $f_j(x)\le e^{-D_jt_j^{1/8}}=e^{-t_j^{-1/8}}$ . If $x\ge r_j$ , $f_j(x)=g_{j+1}(x)=x^2$ . On the intervening interval $f_j(x)\le r_j^2\le(9/4)x^2$ and $g_{j+1}(x)=x^2$ .

Writing $F_j=\mathbb E\operatorname{Tr}g_j(A_{t_j\wedge\sigma})$ , choose universal $b=8C_1+2>0$ to obtain

F_0\le2^{b m}F_m+n\sum_{j=0}^{m-1}2^{b(j+1)}e^{-2^jt^{-1/8}}.

(105.22)

Because $t^{-1/8}\ge2$ , the sum is at most

ne^{-t^{-1/8}}\sum_{j=0}^{\infty}2^{b(j+1)}e^{-2(2^j-1)} =C_b n e^{-t^{-1/8}},\qquad C_b<\infty.

(105.23)

It remains to justify discarding the first term. Compact support gives $\sup_s\operatorname{Tr}A_s^2\le L_\mu<\infty$ , also at stopped times. As $r_m\ge2$ and eigenvalues are continuous from $A_0=I$ ,

F_m\le L_\mu\mathbb P(\|A_{t_m\wedge\sigma}\|_{\mathrm{op}}\ge2) \le L_\mu\mathbb P(\tau_*\le t_m).

(105.24)

The previous qualitative estimate is faster than every power. Since $2^{bm}\le t_m^{-b}$ , the remainder tends to zero for this fixed law, uniformly over the choice of $\sigma$ in this bound. The constants $L_\mu,K_\mu$ vanish with the terminal term and do not enter $C_b$ . Finally $\mathbf1_{\{x\ge3\}}\le g_0(x)$ , proving the claimed rank count for $t\le2^{-8}$ . For $t\ge2^{-8}$ the count is at most $n$ and $e^{-t^{-1/8}}\ge e^{-2}$ ; increasing $C$ to at least $e^2$ proves the same statement for every $t>0$ . This also covers $\sigma=0$ and $\sigma=\infty$ .

Hitting times and time integration: Corollary 6.1 and Theorem 6.2¶

Proof 105.5 (Proof of the two hitting-time consequences)

Continuity and adaptation make $\sigma_k$ a stopping time for the usual Brownian filtration. Pathwise $\sigma_k>0$ , because $\lambda_k(0)=1$ . On $\{\sigma_k\le t\}$ continuity gives $\lambda_k(\sigma_k)=3$ and $\lambda_i(t\wedge\sigma_k)\ge3$ for all $i\le k$ . Thus

k\mathbb P(\sigma_k\le t) \le\sum_{i=1}^n\mathbb P(\lambda_i(t\wedge\sigma_k)\ge3) \le Cn e^{-t^{-1/8}}.

(105.25)

This uses the stopped estimate, not merely a bound at the deterministic time $t$ ; an eigenvalue can fall below threshold after its first hit.

Set $q=n/k\ge1$ , $\alpha=1/8$ , and $x_0=(2\log q)^{1/\alpha}$ . Tonelli’s layer-cake identity, with the convention $\infty^{-1}=0$ , yields

\mathbb E\sigma_k^{-2} =\int_0^\infty2x\,\mathbb P(\sigma_k^{-1}>x)\,dx \le x_0^2+Cq\int_{x_0}^\infty2x e^{-x^\alpha}\,dx.

(105.26)

Use $qe^{-x_0^\alpha/2}=1$ and split the exponent into two halves:

\mathbb E\sigma_k^{-2} \le(2\log q)^{16}+C\int_0^\infty2x e^{-x^\alpha/2}\,dx \le C'(1+\log q)^{16}.

(105.27)

The integral is finite by the substitution $y=x^\alpha/2$ , giving a constant times $\int_0^\infty y^{15}e^{-y}dy$ . When $k=n$ , $x_0=0$ and the same calculation applies. Strict versus non-strict hitting events does not affect the upper bound. This completes all assertions of the first target.

Proof 105.6 (Proof of the integrated rank covariance assertion)

Each $\lambda_k$ is the $k$ th largest eigenvalue at that time, with no choice of a persistent eigenvector. Before $\sigma_k$ it is below 3, and at every positive time it is at most $1/t$ . Hence, pathwise, with $u_k=\sigma_k\wedge1\in(0,1]$ ,

\int_0^1\lambda_k(t)dt \le3u_k+\int_{u_k}^1\frac{dt}{t} \le3-\log u_k, \qquad \exp\left(2\int_0^1\lambda_k(t)dt\right) \le e^6(1+\sigma_k^{-2}).

(105.28)

There is no integration of $1/t$ at zero: the initial interval uses the threshold bound. If $\sigma_k=\infty$ , $u_k=1$ and the right side is $e^6$ . Taking expectations and using the inverse moment bound reduces the claim to $\sum_{k=1}^n(1+\log(n/k))^{16}\le Cn$ . For the decreasing nonnegative function $h(x)=(1+\log(1/x))^{16}$ on $(0,1]$ , every interval $((k-1)/n,k/n]$ has $h(x)\ge h(k/n)$ . Therefore

\frac1n\sum_{k=1}^n h(k/n) \le\int_0^1h(x)dx =\int_0^\infty(1+y)^{16}e^{-y}dy<\infty.

(105.29)

The finite sum and expectation can be interchanged by linearity (or Tonelli before integrability has been shown). The pathwise integrals exist by covariance continuity. This proves the second target with a universal constant, without interchanging exponential and expectation or integrating a fixed-time tail.

Hypothesis usage, scope, and reviewer handoff¶

Hypothesis or convention	Exact use	What is not inferred
Compact support	Uniform bounds on all tilt moments; global ODE; true martingales; polynomial approximation; vanishing terminal term	No extension of these two stochastic statements to noncompact laws is asserted
Isotropy	$A_0=I$ , full-dimensional support, separation from thresholds $2,3$	No arbitrary-covariance formulation or hidden rescaling
Log-concavity	Strong log-concavity after tilting; projected quadratic Poincaré bound and $1/t$ cap	No assertion for all compact isotropic distributions
Brownian-filtration stopping	Stopped Itô formula and first-hit argument	No anticipative random-time estimate
Decreasing eigenvalue order	At least $k$ eigenvalues at a rank- $k$ hit	No eigenvector or source alignment
Positive increasing cutoff	Discard negative drift; dominate the stopped integrand	No need for convexity of the interpolating cutoff
$t>0$ , $n\ge1$ , $1\le k\le n$	Growth away from zero; rank sum and $k=n$ endpoint	No dimension restriction hidden in a logarithmic window

Mechanism and bottleneck. For stopped rank tails: improved Lichnerowicz controls a low-eigenvalue slice of the third tensor; tensor symmetry bounds the spectral-potential drift; the shrinking thresholds and geometric times give a summable exponential error. The growth estimate, not the initial window, governs the exponents obtained by this iteration. For integrated covariance: a first hit controls the number of large ranks; its inverse moment pays for the logarithmic post-hit integral; summation of the logarithmic rank profile costs only $n$ . This last step sees eigenvalue ranks and loses all eigenspace orientation.

Fences respected. Neither target has a bounded_by, depends_on, or assumes edge in the ledger read for this mission, so there are no explicit node fences to discharge. The brief’s thin-shell/KLS distinction and the manuscript’s orientation warning remain in force: no claim is made about Conjecture 29.1, Conjecture 21.1, or Conjecture 29.3. The proof uses the established Theorem 3.1 as an input, not an open antecedent. The source dependency map above records the internal dependence of the integrated bound on the stopped one even though the present ledger does not encode it.

Remark 105.3 (Source-audit boundary and certification boundary)

No source-access gap remained for the pinned HTML/PDF sections used here. The pointwise derivative formulation is replaced by an a.e. identity; the source cutoff’s coefficient 12 is not established by this dossier and is replaced by the explicitly proved universal coefficient 64. Neither difference leaves a step of the two target arguments conditional on that unverified formulation or constant. The established improved Lichnerowicz theorem and standard Prékopa–Leindler inequality are external inputs, not newly proved here. No independent certification is claimed. In particular the regularization, cutoff algebra, tensor index accounting, and elimination of the nonuniform terminal term still require a cold review of this reconstruction.

Next for the reviewer. Independently check both exact manuscript directives against the statements above and v2 Proposition 5.1, Corollary 6.1, and Theorem 6.2. Audit every row of the dependency map, in particular the nonsmooth projected Poincaré passage, spectral Hessian at collisions, all regions of Lemma 5.4, the explicit replacement cutoff, the a.e. Gronwall interpretation, the fixed-law terminal limit and surviving universal constants, and the inverse-moment/rank integration endpoints. Distinguish the published analytic input from the preprint assertions being reviewed. Neither target has an explicit ledger fence; retain the compact/isotropic/log-concave class and Brownian-filtration scope, and exclude every orientation or KLS claim. Run the full checker and fingerprint the stabilized dossier and relevant statements before any review record. The author proposes no status or proof-record delta.

References¶

Klartag, B., & Lehec, J. (2025). Thin-Shell Bounds via Parallel Coupling. https://arxiv.org/abs/2507.15495
Klartag, B. (2023). Logarithmic Bounds for Isoperimetry and Slices of Convex Sets. Ars Inveniendi Analytica, 2023(4), 1–17. 10.15781/jsjy-0b06