Carleson control implies centroid control - The KLS theorem and its methods

Part of the fixed-cut archive, Chapter The fixed cut: the mass martingale and the Carleson estimate; the reading order is on the full proofs page.

Overview. We first use the certified directional dissipation estimate to obtain a finite, dimension-dependent total source budget. Localizing the scalar Riccati identity and using Fatou then establishes all the finiteness needed for absorption. Only after removing this auxiliary stopping do we apply the assumed Carleson estimate on deterministic prefixes. Gronwall gives a dimension-free bound and hence the stopped centroid estimate and KLS.

Setup. Fix an isotropic log-concave probability measure $\mu$ on $\mathbb R^n$ and a measurable cut $E$ with $p_0=\mu(E)\in[2/5,3/5]$ . Use the two-color localization quantities of Theorem 24.1:

s_t=p_t(1-p_t),\quad B_t=s_t\delta_t\delta_t^T,\quad r_t=\operatorname{Tr}B_t, \quad R_t=A_t-B_t,

(101.1)

S_t=s_t\|G_t\|_{\mathrm{HS}}^2,\qquad D_t=2s_t\delta_t^TA_t\delta_t-r_t^2.

(101.2)

Here $p_t$ is the mass martingale, $A_t$ the full covariance, and $G_t$ the difference of the two conditional covariances. Covariance decomposition gives $0\preceq B_t\preceq A_t$ , $R_t\succeq0$ , and the scalar Riccati theorem gives

dr_t=dM_t+(S_t-D_t)\,dt,\qquad D_t\ge r_t^2\ge0,

(101.3)

where $M$ is a continuous local martingale with $M_0=0$ . At time zero, $B_0\preceq I_n$ and $B_0$ has rank at most one, so $r_0\le1$ . Put $\tau=\inf\{t:p_t\notin[1/3,2/3]\}$ , using continuous exit.

Proof 101.1

Apply Corollary 24.1 to the deterministic coordinate vectors and sum. Since $S_t=\sum_{j=1}^n s_t|G_te_j|^2$ , Tonelli gives

\mathbb E\int_0^\infty S_t\,dt \le\sum_{j=1}^n e_j^TR_0e_j =\operatorname{Tr}R_0\le n.

(101.5)

To justify the expectation inequality for $r$ , take an increasing sequence of localizing times $\sigma_k\uparrow\infty$ for the Riccati identity. They can be chosen to bound the stopped local martingale and the accumulated integrals of $S+D$ as well as $r$ , by intersecting the usual localizers with their level hitting times and with the deterministic time $k$ . The continuous semimartingale identity is valid on bounded horizons, and its drift integrals are locally finite; hence these extra level times also tend to infinity. At $T\wedge\tau\wedge\sigma_k$ the martingale has zero expectation and all terms are integrable, so

\mathbb E r_{T\wedge\tau\wedge\sigma_k} +\mathbb E\int_0^{T\wedge\tau\wedge\sigma_k}D_t\,dt =r_0+\mathbb E\int_0^{T\wedge\tau\wedge\sigma_k}S_t\,dt.

(101.6)

Continuity gives almost sure convergence of the terminal $r$ values, without asserting they are monotone. Apply Fatou to the nonnegative left-hand side and monotone convergence to the source integrals on the right. The resulting inequality is the first assertion; its right side is bounded by $1+n$ by the source budget above. In particular $u(T)\le1+n$ for every $T$ . The stopped process is continuous and adapted, so its expectation is measurable; this bound makes $u$ locally integrable. Finally, nonnegativity and the pointwise inequality $\mathbf1_{\{t<\tau\}}r_t\le r_{t\wedge\tau}$ give the occupation claim by Tonelli. Endpoints do not affect time integrals.

Proof 101.2

Fix $\mu,E$ as in the setup. For each deterministic $0<T\le T_0$ , the finiteness lemma establishes that all three occupation expectations in the Carleson premise are finite. Apply that premise with $I=[0,T]$ , after the auxiliary localization has already been removed. Combining the premise with the lemma gives

u(T)+(1-\alpha)\mathbb E\int_0^{T\wedge\tau}D_t\,dt \le r_0+C_0T+C_1\mathbb E\int_0^{T\wedge\tau}r_t\,dt \le1+C_0T+C_1\int_0^T u(t)\,dt.

(101.8)

Because $1-\alpha>0$ and the damping expectation is finite and nonnegative, we may discard it. Integral Gronwall applies to the nonnegative, locally integrable function $u$ and yields

u(T)\le(1+C_0T)e^{C_1T}\le C_*:=(1+C_0T_0)e^{C_1T_0}.

(101.9)

For completeness, set $v(T)=1+C_0T+C_1\int_0^T u(t)\,dt$ . Then $u\le v$ , $v(0)=1$ , and $v'\le C_0+C_1v$ almost everywhere. Multiplication by $e^{-C_1T}$ and integration gives $v(T)\le e^{C_1T}(1+C_0\int_0^T e^{-C_1t}\,dt) \le e^{C_1T}(1+C_0T)$ , including the case $C_1=0$ .

On $t<\tau$ , $s_t\ge2/9$ and $r_t=s_t|\delta_t|^2$ . Hence for every $0<T\le T_0$ ,

\mathbb E\int_0^{T\wedge\tau}|\delta_t|^2\,dt \le\frac92\mathbb E\int_0^{T\wedge\tau}r_t\,dt \le\frac92 C_*T.

(101.10)

The constants are independent of dimension, the measure, and the cut; the argument covered the entire initial range $p_0\in[2/5,3/5]$ . This is the asserted stopped centroid assumption.

To conclude KLS without importing another conditional implication, restrict to cuts with $p_0=1/2$ . The localization identity for their mass martingale is $dp_t=s_t\delta_t\cdot dW_t$ , since $\operatorname{Cov}_{\mu_t}(\mathbf1_E,X)=s_t\delta_t$ . The stopped martingale $p_{t\wedge\tau}-1/2$ is bounded. Its quadratic variation and the weak maximal inequality for its square give

\mathbb P(\tau\le T) \le36\mathbb E(p_{T\wedge\tau}-1/2)^2 =36\mathbb E\int_0^{T\wedge\tau}s_t^2|\delta_t|^2\,dt \le36CT.

(101.11)

Continuity supplies the displacement $1/6$ on exit. Choose $T_* =\min\{T_0,[72(C+1)]^{-1}\}>0$ . Then with probability at least $1/2$ the cut survives, and its masses at $T_*$ are at least $1/3$ . Lemma 30.1 now yields the universal Cheeger lower bound.

Dependencies and fences. The proof uses Theorem 24.1, Corollary 24.1, and Lemma 30.1, with Assumption 28.1 as an antecedent. The target has no bounded_by edge. The dimension-dependent bound $1+n$ is used only to establish finiteness before absorption; it is never asserted to be universal. The operator-to-trace difficulty is not discharged. No estimate on an additional random interval, no uniform integrability of the terminal Riccati process, and no transfer of Carleson through an approximation are required.