BKL: the tilt criterion and Appell duality - The KLS theorem and its methods

Part of the Bizeul–Klartag–Lehec proof, Chapter Bizeul–Klartag–Lehec: cumulants and suspension; the reading order is on the full proofs page.

Overview. This reconstructs the criterion in Section 3 of Bizeul et al., 2026, version arXiv:2610.05474v1. A sequence of normalized inverse-gradient tensors spends the Dirichlet energy of a first eigenfunction. Its failure of symmetry is paid by the Bochner defects. Taylor coefficients transfer along the sequence; a finite dyadic sum then bounds the total energy loss. The final section proves the exact duality with the existing Appell coefficients, including its factorial and tensor-norm conventions.

Dependencies. The criterion uses Definition 8.1, Lemma 8.1, and Lemma 8.2. The Appell comparison uses the polynomial convention in Theorem 7.1, but does not use its quantitative bound. We do not invoke Proposition 7.1 to prove the criterion, nor assume KLS or the cumulant estimate. The exponential Taylor bound is the stated premise of the criterion, not a conclusion of this dossier.

Criterion and analytic conventions¶

Theorem 131.1 (Poincaré inequality from the full Taylor hierarchy)

There is a universal $C<\infty$ with the following property. Let $\mu=e^{-V}dx$ be centered and regular in the sense of Definition 8.1, with $\operatorname{Cov}(\mu)\preceq I$ . For $f\in L^2(\mu)$ put

F_f(z)=\frac{\int f(x)e^{z\cdot x}\,d\mu(x)} {\int e^{z\cdot x}\,d\mu(x)},\qquad \mathcal T_df=\frac{\nabla^dF_f(0)}{d!}.

(131.1)

Suppose $R\ge1$ and $|\mathcal T_df|\le R^d\|f\|_2$ for every integer $d\ge1$ and every $f\in L^2(\mu)$ . Then $C_P(\mu)\le CR^2$ . This proves Theorem 8.1.

We prove this statement below, after deriving the estimates used in the finite summation. All quantities in that proof belong to this one fixed measure. Write $L=\Delta-\nabla V\cdot\nabla$ , $\lambda=C_P(\mu)^{-1}>0$ , and $\nabla_0u=\nabla u-\int\nabla u\,d\mu$ . The space $\mathcal A$ and its centered subspace are those of Lemma 8.1. Derivatives add a new first tensor index; Taylor indices precede the indices of the input. Full tensor norms sum all ordered indices. The scalar Taylor premise extends to every tensor-valued $u$ componentwise:

|\mathcal T_du|^2 =\sum_J|\mathcal T_du_J|^2 \le R^{2d}\sum_J\|u_J\|_2^2 =R^{2d}\|u\|_2^2. \tag{T1}

(131.2)

Strong convexity gives Gaussian tails. Cauchy--Schwarz therefore permits every derivative under the integral defining $F_f$ , locally uniformly in $z\in\mathbb R^n$ , even for arbitrary $f\in L^2(\mu)$ . For instance its differentiated integrands are bounded by $|f(x)|\,|x|^k e^{M|x|}$ on $|z|\le M$ , which is integrable by Cauchy--Schwarz and Gaussian decay.

The inverse-gradient sequence and its energy¶

Choose a real $f\in\mathcal A_0$ with $\|f\|_2=1$ and $-Lf=\lambda f$ , as furnished by Lemma 8.1. Put $h_1=f$ and $w_i=\nabla_0h_i$ . If $w_i=0$ , set $h_{i+1}=0$ and $\alpha_{i+1}=0$ . Otherwise set

u_{i+1}=(-L)^{-1}w_i,\qquad \alpha_{i+1}=\frac{\|w_i\|_2}{\|\nabla u_{i+1}\|_2}, \qquad h_{i+1}=\alpha_{i+1}u_{i+1}. \tag{T2}

(131.3)

All components are centered and in $\mathcal A$ . The denominator is positive when $w_i\ne0$ : a zero gradient would make centered $u_{i+1}$ zero, contradicting $-Lu_{i+1}=w_i$ . The inverse bound gives $\alpha_{i+1}^2\ge\lambda$ whenever it is positive. The identity $-Lh_{i+1}=\alpha_{i+1}w_i$ holds also in the zero case. In particular

v_i:=\|\nabla h_i\|_2^2,\qquad v_{i+1}=\|w_i\|_2^2,\qquad p_i:=v_i-v_{i+1}=\left|\int\nabla h_i\,d\mu\right|^2\ge0, \tag{T3}

(131.4)

and $v_1=\lambda$ . Define the nonnegative defect

\chi_i=\|\nabla^2h_i-\alpha_{i+1}\nabla h_{i+1}\|_2^2. \tag{T4}

(131.5)

Integration by parts gives

\langle\nabla^2h_i,\nabla h_{i+1}\rangle_{L^2} =\langle w_i,-Lh_{i+1}\rangle_{L^2} =\alpha_{i+1}v_{i+1}.

(131.6)

Expanding (T4) now shows

\|D^2h_i\|_2^2=\alpha_{i+1}^2v_{i+1}+\chi_i =\|Lh_{i+1}\|_2^2+\chi_i.

(131.7)

If $aI\preceq D^2V$ , Bochner’s identity therefore implies

\|Lh_i\|_2^2\ge \|Lh_{i+1}\|_2^2+\chi_i+a v_i. \tag{T5}

(131.8)

Sum through any finite index. Because $\|Lh_1\|_2^2=\lambda^2$ , this proves

\sum_{i\ge1}\chi_i\le\lambda^2,\qquad a\sum_{i\ge1}v_i\le\lambda^2,\qquad v_i\longrightarrow0. \tag{T6}

(131.9)

Let $N\ge1$ be the first index such that $v_{N+1}<\lambda/2$ . It exists by (T6). For $2\le j\le N$ we have $v_j\ge\lambda/2$ , so $\alpha_j>0$ and

\lambda\le\alpha_j^2 =\frac{\|Lh_j\|_2^2}{v_j}\le2\lambda. \tag{T7}

(131.10)

The upper bound follows by telescoping (T5). The two budgets required below are

\sum_{i=1}^N p_i=\lambda-v_{N+1}>\frac{\lambda}{2}, \qquad \sum_{i=1}^N\chi_i\le\lambda^2. \tag{T8}

(131.11)

Symmetry defects¶

Let $\tau_\ell$ exchange slots $\ell$ and $\ell+1$ , and let $\mathcal S_q$ average permutations of the first $q$ slots. These are isometries and orthogonal projections, respectively, on the full tensor space.

If $\alpha_{i+1}>0$ , the tensor $\nabla_0(\nabla h_i)/\alpha_{i+1}$ is symmetric in its first two slots. Centering (T4) and using $\alpha_{i+1}^2\ge\lambda$ gives

\left\|w_{i+1}- \frac{\nabla_0(\nabla h_i)}{\alpha_{i+1}}\right\|_2^2 \le\frac{\chi_i}{\lambda}.

(131.12)

If $\alpha_{i+1}=0$ then $w_{i+1}=0$ , so the same distance assertion holds with the zero tensor. It follows in every case that

\|w_j-\tau_1w_j\|_2^2\le\frac{4\chi_{j-1}}{\lambda} \quad(j\ge2). \tag{T9}

(131.13)

For $\ell\ge2$ , a permutation of slots $\ell,\ell+1$ in $w_j=\alpha_j\nabla_0(-L)^{-1}w_{j-1}$ acts on slots $\ell-1,\ell$ of its input. Therefore

w_j-\tau_\ell w_j =\alpha_j\nabla_0(-L)^{-1} (w_{j-1}-\tau_{\ell-1}w_{j-1}).

(131.14)

The input is centered componentwise. By the analytic inverse bound and (T7), this operator has norm at most $\sqrt2$ whenever $2\le j\le N$ . Consequently, for $2\le q\le i\le N$ and $1\le\ell<q$ , iteration down to (T9) gives

\|w_i-\tau_\ell w_i\|_2^2 \le\frac{2^{\ell+1}}{\lambda}\chi_{i-\ell}. \tag{T10}

(131.15)

A permutation of $q$ objects is a product of at most $q^2$ adjacent exchanges. Telescoping along such a product and applying Cauchy--Schwarz bounds its squared displacement of $w_i$ by $q^4\sum_{\ell=1}^{q-1}\|w_i-\tau_\ell w_i\|_2^2$ . Average this inequality over permutations and use convexity of the squared norm. Since $\ell+1\le q$ and $q^4\le16^q$ , we obtain the explicit bound

\|w_i-\mathcal S_qw_i\|_2^2 \le\frac{32^q}{\lambda} \sum_{j=i-q+1}^{i-1}\chi_j \qquad(2\le q\le i\le N). \tag{T11}

(131.16)

Thus the defects in (T5) pay for all the permutations needed below.

Moving Taylor coefficients along the sequence¶

For a scalar or tensor-valued $u\in\mathcal A$ , integrate by parts against the tilted density. Its potential is $V-z\cdot x$ and its generator is $L_z=L+z\cdot\nabla$ . Polynomial growth and Gaussian tails justify the integration, giving

F_{-Lu}(z)=z\cdot F_{\nabla u}(z).

(131.17)

Differentiate at zero. Exactly one derivative hits the explicit factor $z$ ; the other $d$ derivatives hit $F_{\nabla u}$ . The factorial normalization therefore gives

\mathcal T_1(-Lu)=\int\nabla u\,d\mu,\qquad \mathcal T_{d+1}(-Lu) =\mathcal S_{d+1}\mathcal T_d(\nabla u)\quad(d\ge1). \tag{T12}

(131.18)

In the second identity the symmetrization acts on the $d$ Taylor slots and the newly added gradient slot, leaving pre-existing tensor slots unchanged. Constants have zero Taylor tensors of positive order. Apply (T12) to $h_{i+1}$ and use (T2):

\mathcal S_{d+1}\mathcal T_dw_{i+1} =\alpha_{i+1}\mathcal T_{d+1}w_i. \tag{T13}

(131.19)

We now derive a uniform doubling estimate. If $2d\le i\le N$ and $T=\mathcal T_dw_i$ , repeated use of (T13) yields

\mathcal S_{d+k}T =\left(\prod_{j=i-k+1}^{i}\alpha_j\right) \mathcal T_{d+k}w_{i-k}\quad(1\le k\le d). \tag{T14}

(131.20)

This follows inductively because $\mathcal S_{r+1}\mathcal S_r=\mathcal S_{r+1}$ : averaging over the larger permutation group absorbs averaging over its subgroup. All indices of the product satisfy $2\le j\le N$ . Taking $k=d$ and using (T7) gives

|\mathcal S_{2d}T|^2 \le(2\lambda)^d|\mathcal T_{2d}w_{i-d}|^2. \tag{T15}

(131.21)

Set $\widetilde T=\mathcal T_d(\mathcal S_{2d}w_i)$ . This tensor has at least $3d$ slots; its first $d$ are symmetric Taylor slots and its next $2d$ are symmetric input slots. By (T1) and (T11),

\delta^2:=|T-\widetilde T|^2 \le\frac{32^{2d}R^{2d}}{\lambda} \sum_{j=i-2d+1}^{i-1}\chi_j. \tag{T16}

(131.22)

Apply Lemma 8.2 to $\widetilde T$ . Since $\mathcal S_{2d}$ is a contraction, the triangle inequality gives

|T|\le\delta+4^d|\mathcal S_{2d}\widetilde T| \le4^d|\mathcal S_{2d}T|+(1+4^d)\delta.

(131.23)

After squaring and using (T15)--(T16), this implies

|\mathcal T_dw_i|^2 \le(C_0\lambda)^d|\mathcal T_{2d}w_{i-d}|^2 +\frac{C_0^dR^{2d}}{\lambda} \sum_{j=i-2d+1}^{i-1}\chi_j, \qquad C_0=2^{17}. \tag{T17}

(131.24)

Indeed the first coefficient is at most $2\cdot32^d\le64^d$ , and the second coefficient is at most $2(1+4^d)^2\,32^{2d}\le8\cdot16384^d\le(2^{17})^d$ . This gives one universal constant valid for every degree; it is not optimized.

The finite dyadic argument¶

Proof 131.1 (Proof of the tilt criterion)

First express the lost gradient means using Taylor tensors. Testing $-Lf=\lambda f$ against the coordinate functions gives $\int\nabla f\,d\mu=\lambda\int xf\,d\mu$ . Since $\mu$ is centered and $\operatorname{Cov}(\mu)\preceq I$ , duality and Cauchy--Schwarz give $|\int xf\,d\mu|\le1$ . Thus $p_1\le\lambda^2$ . For $2\le i\le N$ , (T12), (T2), and (T7) give

p_i=|\mathcal T_1(-Lh_i)|^2 =\alpha_i^2|\mathcal T_1w_{i-1}|^2 \le2\lambda|\mathcal T_1w_{i-1}|^2.

(131.25)

Define, for every positive integer $d$ ,

S_d=\sum_{i=1}^{N-d}|\mathcal T_dw_i|^2,

(131.26)

with an empty sum equal to zero. In particular $S_d=0$ for $d\ge N$ . The preceding bounds imply

\sum_{i=1}^Np_i\le\lambda^2+2\lambda S_1. \tag{T18}

(131.27)

For each term of $S_d$ , (T1) and (T3) give $|\mathcal T_dw_i|^2\le R^{2d}v_{i+1}\le\lambda R^{2d}$ . There are at most $2d-1$ terms with $i<2d$ . For the others, sum (T17) over $2d\le i\le N-d$ . Its higher-order terms are a sub-sum of $S_{2d}$ , since $d\le i-d\le N-2d$ . Every $\chi_j$ occurs in at most $2d$ of the intervals $[i-2d+1,i-1]$ . Thus (T8) gives

\begin{aligned} S_d &\le(2d-1)\lambda R^{2d} +(C_0\lambda)^dS_{2d} +\frac{2dC_0^dR^{2d}}{\lambda}\sum_{j=1}^N\chi_j\\ &\le(C_0\lambda)^dS_{2d} +4dC_0^d\lambda R^{2d}. \end{aligned} \tag{T19}

(131.28)

This remains true when any of the sums is empty. Suppose, for a contradiction, that $\lambda R^2\le(100C_0^2)^{-1}$ . Multiplication of (T19) by $(C_0\lambda)^{d-1}$ gives

(C_0\lambda)^{d-1}S_d \le(C_0\lambda)^{2d-1}S_{2d} +4dC_0^{2d-1}(\lambda R^2)^d \le(C_0\lambda)^{2d-1}S_{2d}+\frac{4d}{100^d}.

(131.29)

Choose $m\ge1$ with $2^m\ge N$ and sum over $d=1,2,\ldots,2^{m-1}$ restricted to powers of two. The left and right weighted sums telescope, and $S_{2^m}=0$ . Consequently

S_1\le4\sum_{k=0}^{m-1}\frac{2^k}{100^{2^k}} \le4\sum_{d=1}^{\infty}\frac d{100^d} =\frac{400}{9801}<\frac1{20}.

(131.30)

The equality is the derivative of the geometric series at $1/100$ . Our supposition also implies $\lambda\le1/100$ , because $R,C_0\ge1$ . By (T18),

\sum_{i=1}^Np_i <\lambda\left(\frac1{100}+\frac1{10}\right) <\frac{\lambda}{2},

(131.31)

contradicting (T8). We conclude $C_P(\mu)=\lambda^{-1}\le100C_0^2R^2$ , proving the theorem with a universal constant. The summation is finite at each fixed measure; there is no exchange of a spectral limit with an approximation limit.

Exact comparison with Appell coefficients¶

Theorem 131.2 (Tilt--Appell duality with ordered-index norms)

Let $\mu$ be a full-dimensional log-concave probability on $\mathbb R^n$ , let $d\ge1$ , and use the Appell convention of Theorem 7.1. For every symmetric $d$ -tensor $T$ and $f\in L^2(\mu)$ ,

\langle T,\nabla^dF_f(0)\rangle =\int fP_d^\mu[T]\,d\mu. \tag{T20}

(131.32)

If $K_d(\mu)=\sup_{T=T^{\rm sym},\,|T|=1} \operatorname{Var}_\mu P_d^\mu[T]$ , then

\|\mathcal T_d:L^2(\mu)\longrightarrow \operatorname{Sym}^d(\mathbb R^n)\| =\frac{\sqrt{K_d(\mu)}}{d!}=c_d(\mu). \tag{T21}

(131.33)

This proves Proposition 8.1.

Proof 131.2

A finite-dimensional full-dimensional log-concave probability has an exponential moment $\int e^{\eta|x|}\,d\mu<\infty$ for some $\eta>0$ . Recall the elementary reason: an integrable log-concave density has a bounded convex superlevel set with nonempty interior; concavity of its logarithm along rays from an interior point bounds that logarithm above by a negative linear function outside a sufficiently large ball. Radial integration then gives an exponential moment after decreasing its exponent. In particular all polynomial moments are finite. For $|z|<\eta/4$ , differentiation under the integral defining $F_f$ is valid to every fixed order: Cauchy--Schwarz reduces it to finiteness of integrals of $|x|^{2k}e^{2|z||x|}$ , and these are dominated locally by a constant times $e^{\eta|x|}$ . The denominator is positive for real $z$ near zero.

The normalized exponential has formal expansion

\frac{e^{z\cdot x}}{\int e^{z\cdot y}\,d\mu(y)} =\sum_{k\ge0}\frac{\langle\mathcal A_k^\mu(x), z^{\otimes k}\rangle}{k!},\qquad P_k^\mu[T]=\langle T,\mathcal A_k^\mu\rangle.

(131.34)

Its derivative tensor of order $d$ at zero is exactly $\mathcal A_d^\mu(x)$ , with no additional symmetrization factor. The preceding integrable domination permits differentiating its integral against $f$ , proving (T20). The integral of the normalized exponential against $\mu$ is the constant one; taking derivatives shows $\int P_d^\mu[T]\,d\mu=0$ for $d\ge1$ . Thus its squared $L^2$ norm is its variance.

Consider the linear operator $B_d:\operatorname{Sym}^d(\mathbb R^n)\to L^2(\mu)$ , $B_dT=P_d^\mu[T]/d!$ . Its finite-dimensional domain makes it bounded, and its norm is $\sqrt{K_d(\mu)}/d!$ by the definition of $K_d$ . Equation (T20) says that its Hilbert-space adjoint is $B_d^*f=\mathcal T_df$ . An operator and its adjoint have equal norms, proving (T21). In particular this is the norm on the full symmetric tensor space, not a supremum restricted to tensors $u^{\otimes d}$ .

Source concordance. Equations (T2)--(T8) reconstruct the normalized sequence, Bochner decay and exit index in BKL Section 3. (T9)--(T11) give its Lemmas 3.2--3.3; (T12)--(T17) give Lemmas 3.6--3.8; (T18)--(T19) and the finite summation give Lemma 3.9 and Theorem 3.1. The duality identifies the result with the Appell convention already used in this repository.

Scope and fences. The premise must hold simultaneously for all degrees and all $L^2$ tests for the fixed measure. Different constants at different degrees do not meet it. The coefficient condition in Proposition 7.1 is not assumed as an unconditional theorem. The tensor operations respect Remark 25.4. No covariance occupation, sharp CMH bound or posterior-profile estimate is inferred, so Remark 33.6, Remark 33.5, and Remark 32.6 are unaffected. No additional bounded-by edge is proposed.

References¶

Bizeul, P., Klartag, B., & Lehec, J. (2026). Presenting a Proof of the Kannan–Lovasz–Simonovits Conjecture. https://arxiv.org/abs/2610.05474v1