Song–Zhang v2: the common coefficient radius - The KLS theorem and its methods

Part of the second version of Song–Zhang, Chapter Song–Zhang, second version: repeated refinement with summable losses; the reading order is on the full proofs page.

Overview. The coefficient radius is one scalar assigned to a fixed measure, independent of polynomial degree, block length, and block order. Its finiteness follows from the bounded inverse diffusion operator. Polynomial testing relates it to every finite block, and the already certified curvature comparison bounds the Poincaré constant by this radius. This proves Proposition 9.1, with the normalization of Definition 9.1. These are the interfaces of Lemmas 8.17–8.18 of Song & Zhang, 2026, reconstructed below without a uniform KLS input.

Dependencies. The proof uses Lemma 7.1, the Appell conventions in Theorem 7.1, and Theorem 7.2. It does not use the improved iteration, BKL, any consequence of BKL, or a uniform bound on the Poincaré constant. The finiteness of the inverse operator is used for one regular measure at a time, with no claim of uniformity in that measure.

The common radius¶

Let $\mu=e^{-W}dx$ be centered, with $W\in C^\infty$ , $aI\preceq D^2W\preceq bI$ for $0<a\le b<\infty$ , and $\operatorname{Cov}(\mu)\preceq I$ . Put $\mathscr H=L^2_0(\mu)$ , let $H$ be its nonnegative diffusion operator, and write $B=H^{-1}$ on $\mathscr H$ . Let $P_+$ subtract componentwise means, and set

\mathcal T=P_+\nabla B,\qquad R=\|\mathcal T\|_{\mathrm{op}}^2.

(139.1)

The codomain of $\mathcal T$ is $\mathscr H\otimes\mathbb R^n$ . Its iterates apply it componentwise and append an ordered output slot. All tensor norms below sum over ordered indices. In particular, finite orthogonal amplification preserves the operator norm of $\mathcal T$ . For the Appell coefficients $c_d=\sqrt{K_d}/d!$ , define

\mathcal A(\mu)=\max\left\{1,\sup_{d\ge2}c_d^{2/(d-1)}\right\}.

(139.2)

Theorem 139.1 (A finite radius shared by every degree and block)

For every measure in the stated regular class,

C_P(\mu)-1\le R\le C_P(\mu),\qquad 1\le\mathcal A(\mu)\le\max\{1,R\},\qquad c_d\le\mathcal A(\mu)^{(d-1)/2}\quad(d\ge1).

(139.3)

If $G:\mathbb N_{\ge1}\to[1,\infty)$ is nondecreasing and $c_k\le G(k)^{k-1}$ for every $k\ge1$ , then, for every integer $m\ge1$ ,

\mathcal A(\mu)\le \max\{G(m)^2,\|\mathcal T^m\|_{\mathrm{op}}^{2/m}\}.

(139.4)

Moreover,

C_P(\mu)\le2^{85}\mathcal A(\mu).

(139.5)

Each assertion concerns the same fixed measure. The constant in the last inequality is independent of its dimension and curvature bounds.

Proof 139.1

The restricted operator. Spectral calculus from Lemma 7.1 gives a bounded positive inverse $B$ on $\mathscr H$ , with $\|B\|=C_P(\mu)<\infty$ . For $f\in\mathscr H$ , $Bf\in\operatorname{Dom}H$ and

\|\nabla Bf\|_2^2=\langle Bf,HBf\rangle=\langle f,Bf\rangle.

(139.6)

The coordinate $x_i$ belongs to the form domain; testing the form against it gives $\mathbb E\partial_i Bf=\mathbb E[X_i f]$ . Write $Lf=\mathbb E[Xf]$ . By duality and covariance normalization,

|Lf|=\sup_{|v|=1}|\mathbb E[f\langle v,X\rangle]| \le\|f\|_2.

(139.7)

Consequently

\mathcal T^*\mathcal T=B-L^*L, \qquad B-I\preceq\mathcal T^*\mathcal T\preceq B.

(139.8)

Taking suprema of the quadratic forms over unit vectors proves the claimed bounds on $R$ , without any uniform estimate on $C_P$ .

The testing identity. For $d\ge1$ let $Q_df=\mathbb E[f\mathcal A_d^\mu]$ . This bounded map from $\mathscr H$ to symmetric $d$ -tensors is adjoint to $T\mapsto P_d^\mu[T]$ ; therefore $\|Q_d\|=\sqrt{K_d}=d!c_d$ . All these polynomials and their derivatives are square integrable because the positive lower Hessian bound gives Gaussian tails. They belong to the form domain by cutoff approximation.

For $d\ge2$ , integration by parts in the form sense gives

\mathbb E[fP_d[T]] =\mathbb E\langle\nabla Bf,\nabla P_d[T]\rangle.

(139.9)

Differentiating the formal generating identity shows $\partial_iP_d[T]=dP_{d-1}[T_i]$ , where $T_i$ is the slice with one index fixed at $i$ . Since $\mathbb E P_{d-1}[T_i]=0$ , replacing $\nabla Bf$ by $P_+\nabla Bf$ leaves this integral unchanged. It follows, as an identity of bounded maps, that

\frac{Q_d}{d!} =\mathsf S_d\left(\frac{Q_{d-1}}{(d-1)!}\otimes I\right)\mathcal T. \tag{H1}

(139.10)

Here $\mathsf S_d$ averages all permutations of the $d$ tensor slots. The displayed identity follows by pairing against every symmetric tensor; its right side is symmetric and hence this determines it uniquely. Symmetrization is an orthogonal projection. Thus $c_d\le\sqrt R\,c_{d-1}$ . The covariance bound gives $c_1\le1$ , and finite induction yields

c_d\le R^{(d-1)/2}\quad(d\ge2).

(139.11)

This also covers $R<1$ . If $R=0$ , all $c_d$ with $d\ge2$ vanish by the same identity. The definition of $\mathcal A$ now proves its finiteness and its first stated bounds. In particular no supremum is taken before obtaining one finite bound valid at every degree.

Finite blocks. Iterating (H1) exactly $m$ times, for $d\ge m+1$ , gives

\frac{Q_d}{d!} =\mathsf S_d\left(\frac{Q_{d-m}}{(d-m)!}\otimes I\right)\mathcal T^m. \tag{H2}

(139.12)

To justify the intermediate symmetrizations, $\mathcal T$ acts on scalar components and thus commutes with any permutation of existing output slots. The final average absorbs every intermediate permutation average. This proves (H2) by finite induction, with no commutation of a spatial derivative and $B$ . Taking norms gives $c_d\le\|\mathcal T^m\|c_{d-m}$ . For each $d\ge1$ write $d=jm+r$ , where $j\ge0$ and $1\le r\le m$ . Finite iteration, including the case $j=0$ , gives

c_d\le \|\mathcal T^m\|^j c_r \le \|\mathcal T^m\|^jG(m)^{r-1} \le \max\{G(m)^2,\|\mathcal T^m\|^{2/m}\}^{(d-1)/2}.

(139.13)

When the block norm is zero and $j\ge1$ , the first inequality already gives zero; no ambiguous zeroth power is needed. The maximum is at least one, so taking the defining supremum for $\mathcal A$ proves the assertion.

The conversion to Poincaré. Set $\mathcal R=2^{40}\sqrt{\mathcal A(\mu)}$ , which meets the threshold of Theorem 7.2 with $\epsilon=1$ and $\ell\equiv1$ . Since $(k+1)^2\le4^k$ for $k\ge1$ and $\mathcal A\ge1$ ,

c_k\le\mathcal A^{(k-1)/2} \le\frac{\mathcal R^k}{(k+1)^2}.

(139.14)

For every finite dyadic integer $d\ge2$ , that certified comparison gives

C_P(\mu)\le32\mathcal R^2\max\{1,a^{-1/(d+1)}\} =2^{85}\mathcal A(\mu)\max\{1,a^{-1/(d+1)}\}.

(139.15)

The curvature parameter $a>0$ is fixed throughout these inequalities. Letting $d$ increase along dyadic integers removes the last factor. Only scalar inequalities are passed to the limit; no infinite family of inverse-gradient iterates is constructed.

Fences respected. The common radius is finite for each regular law, but no dimension-free upper bound on this radius has been proved here. Thus the coefficient/KLS equivalence is respected, and no conclusion about uniform conditional initialization or summable block losses is inferred from pointwise finiteness. No bounded_by node is proposed for this interface.

Reconstruction boundary¶

References¶

Song, Z., & Zhang, X. (2026). An O(1) Bound for the KLS Constant. https://arxiv.org/abs/2610.01447v2