BK: uniform operator powers and the quadratic seed - The KLS theorem and its methods

Part of the Balasubramanian–Kasiviswanathan proof, Chapter Balasubramanian–Kasiviswanathan: compatible integration; the reading order is on the full proofs page.

Overview. The operator comparison first bounds the spectral radius. A recurrence along each adjoint orbit then puts its largest normalized squared norm among the first finitely many iterates. Summing second differences gives one prefactor for all powers. Two applications give integration powers and the quadratic seed. This reconstructs Lemma 5.1, Appendix C, and Corollaries 5.2–5.3 of Balasubramanian & Kasiviswanathan, 2026, pinned at commit 4837c33649ba2271f43c9684e9350ecbdd725f95.

Dependencies. The abstract lemma uses only bounded Hilbert-space operators and spectral calculus. Its applications take Proposition 11.1 as an explicit input: bounded graded operators $J,L$ , the comparison $J^*J\preceq g(JJ^*+LL^*)$ , observations $\|J^{j-1}L\|\le c_j(\nu)$ , the block identity for $J_0\cdots J_{q-1}$ , and $C_P(\nu)\le1+\|J\|^2$ . They also use the normalizations in Definition 11.2 and Definition 11.1. The quadratic seed alone uses Theorem 25.1. No BKL theorem, SZ v2 theorem, or KLS conclusion is an input.

Theorem 153.1 (Finitely observed powers)

Let $R:X\to X$ and $L:Y\to X$ be bounded operators on real or complex Hilbert spaces and let $a>0$ . Suppose

R^*R\preceq g(RR^*+LL^*),\qquad g(u)=u/(1+au).

(153.1)

Let $D\ge2$ be an integer, and let finite nonnegative numbers $\gamma_2,\ldots,\gamma_D$ satisfy $\|R^{j-1}L\|\le\gamma_j$ for $2\le j\le D$ . For every $B>0$ satisfying $aB^{D+1}>\gamma_D^2$ , and every integer $k\ge0$ ,

\|R^k\|^2\le B^k\left(1+\sum_{j=1}^{D-2} \frac{j\gamma_{j+1}^2}{B^{j+1}}\right).

(153.2)

This is Lemma 11.3.

Proof 153.1

Complexification preserves the operator inequality and all norms, so work over complex Hilbert spaces. For every bounded nonnegative operator $H$ and unit vector $v$ , its scalar spectral measure and concavity of $g$ give

\langle v,g(H)v\rangle\le g(\langle v,Hv\rangle).

(153.3)

We first show $r(R)^2\le(\gamma_D^2/a)^{1/(D+1)}$ . If $r(R)=0$ there is nothing to show. Otherwise take $\lambda\in\sigma(R^*)$ with $|\lambda|=r(R)>0$ . There exist unit vectors $v_h$ with $(R^*-\lambda)v_h\to0$ . Here is a direct verification: the points $z_h=(1+h^{-1})\lambda$ lie in the resolvent set; the Neumann-series test at $\lambda$ gives $\|(R^*-z_h)^{-1}\|\ge|z_h-\lambda|^{-1}$ . Choose unit vectors on which the resolvent norm is achieved within a factor of two and normalize their resolvent images. The resulting vectors satisfy

\|(R^*-\lambda)v_h\|\le 2/\|(R^*-z_h)^{-1}\|+|z_h-\lambda|\longrightarrow0.

(153.4)

s\le g(s+p)<a^{-1},\qquad p\ge\frac{as^2}{1-as}.

(153.5)

Factoring the difference of powers shows $((R^*)^{D-1}-\lambda^{D-1})v_h\to0$ . Applying $L^*$ and the last observation bound gives $s^{D-1}p\le\gamma_D^2$ . Consequently

\frac{as^{D+1}}{1-as}\le\gamma_D^2,

(153.6)

which proves the asserted spectral-radius bound.

Fix a unit vector $f$ and define

x_m=\|(R^*)^mf\|^2,\quad b_m=\|L^*(R^*)^mf\|^2, \quad y_m=x_m/B^m.

(153.7)

For $m\ge1$ with $x_{m-1},x_m>0$ , Cauchy–Schwarz gives

x_m^2\le x_{m-1}\|R(R^*)^mf\|^2.

(153.8)

Applying the comparison and Jensen to $(R^*)^mf/\sqrt{x_m}$ yields

\frac{x_m}{x_{m-1}}\le g\left(\frac{x_{m+1}+b_m}{x_m}\right).

(153.9)

In particular $ax_m/x_{m-1}<1$ . Since $g^{-1}(v)=v/(1-av)\ge v+av^2$ on $[0,a^{-1})$ , inversion gives

y_{m+1}\ge\frac{y_m^2}{y_{m-1}}+ aB\frac{y_m^3}{y_{m-1}^2}-\frac{b_m}{B^{m+1}}.\tag{1}

(153.10)

The strict inequality for $B$ and the spectral-radius bound give $r(R)<\sqrt B$ . Choose $r(R)<\rho<\sqrt B$ ; the spectral-radius formula implies $\|R^m\|\le\rho^m$ for all sufficiently large $m$ . Thus $y_m\to0$ , and, since $y_0=1$ , the sequence attains a positive global maximum $Y$ at some finite index $h$ . If $h\ge D-1$ , then

b_h\le\gamma_D^2x_{h-D+1}\le\gamma_D^2B^{h-D+1}Y.

(153.11)

Every iterate preceding the positive $x_h$ is positive. Using (1) at $h$ and $y_{h-1}\le y_h=Y$ gives

y_{h+1}\ge Y+(aB-\gamma_D^2/B^D)Y>Y,

(153.12)

a contradiction. Therefore $h\le D-2$ .

If $h=0$ , the conclusion is immediate. Otherwise put $\Delta_j=y_j-y_{j-1}$ . For $1\le j\le h$ , we have $b_j\le\gamma_{j+1}^2$ and all denominators in (1) are positive. Dropping the nonnegative curvature term, and using $v^2/u-2v+u=(v-u)^2/u\ge0$ , gives

\Delta_{j+1}\ge\Delta_j-\gamma_{j+1}^2/B^{j+1}.

(153.13)

Since $\Delta_{h+1}\le0$ , backwards summation and then summation over $j$ give

Y-1=\sum_{j=1}^h\Delta_j \le\sum_{j=1}^h\sum_{i=j}^h\frac{\gamma_{i+1}^2}{B^{i+1}} =\sum_{i=1}^h\frac{i\gamma_{i+1}^2}{B^{i+1}}\le Q_D(B)-1.

(153.14)

This argument allows $x_{h+1}=0$ . Thus $y_k\le Q_D(B)$ for all $k$ . Taking the supremum over unit $f$ and using $\|(R^*)^k\|=\|R^k\|$ proves the claim. The zero Hilbert space causes no exception, since all its operator norms are zero.

Theorem 153.2 (The integration consequences)

Assume the integration-calculus input stated above. Let $\nu$ be centered, regular and log-concave, with covariance at most $I$ and curvature at least $aI$ , $a>0$ . If $D\ge2$ , $\rho\ge1$ and $c_j(\nu)\le\rho^j/(j+1)^4$ for $2\le j\le D$ , put $B=\rho^2\max\{1,a^{-1/(D+1)}\}$ . Then for every integer $q\ge1$ ,

\|J^q\|^2\le2B^q,\qquad\|J_0\cdots J_{q-1}\|^2\le2B^q, \qquad C_P(\nu)\le1+2B.

(153.15)

Without any higher coefficient assumption,

\|J\|^2\le(2/a)^{1/3}.

(153.16)

These are Corollary 11.1 and Corollary 11.2.

Proof 153.2

For the first assertion apply the preceding theorem with $R=J$ , the constant-field operator $L$ , and $\gamma_j=\rho^j/(j+1)^4$ . Indeed $aB^{D+1}\ge\rho^{2D+2}>\gamma_D^2$ , and

Q_D(B)-1\le\sum_{j=1}^\infty\frac{j}{(j+2)^8} \le\sum_{j=1}^\infty(j+2)^{-7} \le\int_2^\infty x^{-7}\,dx=\frac1{384}<1.

(153.17)

The block of $J^q$ from grade $q$ to grade zero is $J_0\cdots J_{q-1}$ , so its norm is no greater. The Poincaré conclusion uses the stated calculus input.

For the seed, Letwin’s inequality gives $\operatorname{Var}(X^TTX)\le8\|T\|_{\mathrm{HS}}^2$ for isotropic log-concave $X$ . Since the normalized quadratic Appell polynomial is $(X^TTX-\mathbb EX^TTX)/2$ , its coefficient is at most $\sqrt2$ . If covariance is at most $I$ , write the law as $AZ$ on its linear support, where $Z$ is isotropic and $\|A\|\le1$ . Replace $T$ by $A^TTA$ ; its Hilbert–Schmidt norm is no larger, proving $c_2(\nu)\le\sqrt2$ . (The point mass case is zero.) Use $D=2$ , $\gamma_2=\sqrt2$ . The sum in $Q_2$ is empty, so for every $B>(2/a)^{1/3}$ the theorem gives $\|J\|^2\le B$ . Taking the infimum over such $B$ proves the seed.

Fences respected. No bounded_by fence is proposed for the abstract operator statement. The integration assertions keep positive curvature and regularity, and their coefficient hypotheses are restricted to the stated finite degrees. They do not assert that separate one-step bounds retain a uniform prefactor under multiplication. The quadratic estimate is precisely Letwin’s constant-eight variance normalization, not a sharper third-moment or moment-map claim.

References¶

Balasubramanian, K., & Kasiviswanathan, S. (2026). A Dimension-Free Bound on the Poincaré Constant of Isotropic Log-Concave Measures. https://arxiv.org/abs/2610.07728v1