Proof of The Tracked Energy Bound for the Block Cascade under Local Joint Coercivity

Throughout we use linearity and monotonicity of the integral freely; all random variables appearing are bounded, hence integrable, so every expectation exists and is finite. Integrals over compact subintervals are Lebesgue integrals. Write $\mathcal{T}_{k}$ for the tracked slice of block $k$, that is, the set of pairs $(t,\omega)$ with $\omega\in G_{k}$, $t\in[t_{k},t_{k+1}]$ and $t<\sigma^{(k)}(\omega)$.

Step 1: proof of claim 1. By claim 6 of the anchored clocks lemma each $\Delta_{k}\mathcal{E}$ is a random variable with values in $[0,4R^{2}h_{k}]$; since $\mathbf{1}_{G_{k}}\le1$ and $\sum_{k}h_{k}=T$, monotonicity gives $0\le\mathcal{Z}\le N\sum_{k}4R^{2}h_{k}=4R^{2}TN$, finite.

The function $Q$ is a nonnegative random variable with $\mathbb{E}[Q^{4}]\le c_{Q}\kappa_{0}N^{-2}$ by claim 2 of the pre-stopping envelope lemma, as adopted in the anchored envelope lemma; hence $\mathcal{N}=\{Q>q_{0}\}$ is an event and, since $q_{0}^{4}\mathbf{1}_{\mathcal{N}}\le Q^{4}$ pointwise, monotonicity gives $q_{0}^{4}P(\mathcal{N})\le\mathbb{E}[Q^{4}]\le c_{Q}\kappa_{0}N^{-2}$, that is, $P(\mathcal{N})\le c_{Q}\kappa_{0}q_{0}^{-4}N^{-2}$. Finally $\mathbf{1}_{G_{k}\cap\mathcal{N}}\Delta_{k}\mathcal{E}\le4R^{2}h_{k}\mathbf{1}_{\mathcal{N}}$ pointwise, so summing and multiplying by $N$,

Step 2: proof of claim 2. Let $k$, $\omega\in G_{k}\setminus\mathcal{N}$ and $t\in[t_{k},T]$ with $t<\sigma^{(k)}(\omega)$ be as stated; then $\omega\in\Omega_{0}$, since $G_{k}\subseteq\Omega_{0}$ by claim 2 of the cascade lemma.

(2a) The state deviation. By claim 3 of the cascade lemma $Y_{t_{k}}(\omega)<L_{k}$, so claim 1 of the anchored envelope lemma — applied for the instance with anchor $t_{k}$ and level $L_{k}$, at this $t$, where $\min(t,\sigma^{(k)}(\omega))=t$ — gives

using $L_{k}\le\varepsilon_{1}$ (claim 1 of the cascade lemma) and $Q(\omega)\le q_{0}$ (as $\omega\notin\mathcal{N}$). Equivalently $|\Sigma_{t}(\omega)-S_{t}|\le\varepsilon_{1}+q_{0}$.

(2b) The clipped case. Suppose first $|\hat{\alpha}(t,\omega)-A_{t}|\le\delta$. Since $\omega\in\Omega_{0}$ we have $\hat{\alpha}(t,\omega)=\alpha_{t}(\omega)$ by claim 2 of the realized-control lemma, so $|\alpha_{t}(\omega)-A_{t}|\le\delta$ and, with (2a),

by the first part of (SM); as both sides are nonnegative this gives $\rho_{t}(\omega)\le\rho^{*}$. Claim 3 of the localized coercivity lemma therefore applies and yields

(2c) The clipped-out case. Suppose instead $|\hat{\alpha}(t,\omega)-A_{t}|>\delta$, so that $|\mathfrak{a}_{t}(\omega)|=\sqrt{N}|\alpha_{t}(\omega)-A_{t}|>\sqrt{N}\delta$ and hence $N\delta^{2}<|\mathfrak{a}_{t}(\omega)|^{2}$. Hypothesis (JC) supplies (H1) with $r=c_{J}$ by claim 5 of the localized coercivity lemma, so with (A) and (U) the constant $r_{0}>0$ of conclusion (d) of the quadratic growth lemma exists and part (d) of the first-order expansion lemma applies at every point, giving $N\mathcal{D}_{t}\ge\frac{r_{0}}{2}|\mathfrak{a}_{t}|^{2}-C_{3}|\mathfrak{s}_{t}|^{2}$. By (2a) and the second part of (SM),

so $N\mathcal{D}_{t}(\omega)\ge\bigl(\frac{r_{0}}{2}-\frac{r_{0}}{4}\bigr)|\mathfrak{a}_{t}(\omega)|^{2}=\frac{r_{0}}{4}|\mathfrak{a}_{t}(\omega)|^{2}\ge c_{\star}|\mathfrak{a}_{t}(\omega)|^{2}$. In both cases claim 2 follows.

Step 3: proof of claim 3. Let $\omega\in G_{K}$; then $\omega\in\Omega_{0}$ and $\sigma^{(k)}(\omega)\ge t_{k+1}$ for every $k$, so $\Delta_{k}\mathcal{E}(\omega)=\mathcal{E}_{t_{k+1}}(\omega)-\mathcal{E}_{t_{k}}(\omega)$ and, telescoping, $\sum_{k=0}^{K-1}\Delta_{k}\mathcal{E}(\omega)=\mathcal{E}_{T}(\omega)$.

By claim 4 of the extended good-set stopping-time lemma, whose hypotheses are the standing hypothesis on $S^{*}$, $Y_{T}(\omega)\le C_{S}\,\mathcal{E}_{T}(\omega)^{1/2}$, that is, $|\Phi_{T}(\omega)-S_{T}|\le C_{S}\mathcal{E}_{T}(\omega)^{1/2}$. Also, by claim 2 of the pathwise tracking lemma and claim 4 of the flow stability lemma (comparing the flows from $\Sigma_{0}(\omega)$ and from $x_{0}$ under the same control $\hat{\alpha}(\omega)$, for which the control-perturbation functionals vanish), $|\Sigma_{T}(\omega)-\Phi_{T}(\omega)|\le|M_{T}(\omega)|+\Lambda_{b}e^{\Lambda_{b}T}\mathcal{M}_{T}(\omega)+e^{\Lambda_{b}T}|\Sigma_{0}(\omega)-x_{0}|$, and this majorant is at most $Q(\omega)$ by the chain of inequalities displayed in claim 1 of the anchored envelope lemma, invoked for the instance with anchor $t_{K-1}$ and level $L_{K-1}=\varepsilon_{1}$ at the time $t=T$ — legitimate because $G_{K}\subseteq G_{K-1}\subseteq\{Y_{t_{K-1}}<L_{K-1}\}$ by claim 3 of the cascade lemma and because $\min(T,\sigma^{(K-1)}(\omega))=T$ on $G_{K}$ — and because there $|\Sigma_{0}-x_{0}|=N^{-1/2}|\mathfrak{s}_{0}|$, since $S_{0}=x_{0}$. Hence $|\Sigma_{T}-S_{T}|\le C_{S}\mathcal{E}_{T}^{1/2}+Q$ and, by the elementary inequality $(u+v)^{2}\le2u^{2}+2v^{2}$,

Multiplying by $\mathbf{1}_{G_{K}}$ and taking expectations, and using $\mathbf{1}_{G_{K}}\le\mathbf{1}_{G_{k}}$ for every $k$ (the good sets being nested) together with the telescoping above,

Finally $\mathbb{E}[Q^{2}]\le\mathbb{E}[Q^{4}]^{1/2}\le(c_{Q}\kappa_{0})^{1/2}N^{-1}$ by claim 4 of the toolkit applied on the probability space (the Cauchy-Schwarz inequality with $f=Q^{2}$, $g=1$) and the multiplicativity of the nonnegative square root, so $2N\mathbb{E}[Q^{2}]\le2c_{Q}^{1/2}\kappa_{0}^{1/2}$, giving claim 3.

Step 4: proof of claim 4. By part (c) of the first-order expansion lemma,

By claim 2 of the cascade lemma the sets $D_{0},\dots,D_{K-1},G_{K}$ are pairwise disjoint with union $\Omega_{0}$, so $\mathbf{1}_{\Omega_{0}}=\sum_{k}\mathbf{1}_{D_{k}}+\mathbf{1}_{G_{K}}$ pointwise. Moreover, for each $k$, on $G_{k}$ the intervals $[t_{k},\min(\sigma^{(k)},t_{k+1}))$ over $k$ together with the post-exit interval on each $D_{k}$ and the whole of $[0,T]$ on $G_{K}$ decompose $[0,T]$; concretely, fix $\omega\in\Omega_{0}$ and let $j$ be the unique index with $\omega\in D_{j}$ if $\omega\notin G_{K}$, no such index existing when $\omega\in G_{K}$. If $\omega\in G_{K}$ then $\sigma^{(k)}(\omega)\ge t_{k+1}$ for every $k$, so the closed intervals $[t_{k},\min(\sigma^{(k)},t_{k+1})]=[t_{k},t_{k+1}]$ cover $[0,T]$ and overlap only in the finitely many endpoints $t_{k}$; if $\omega\in D_{j}$ then $\sigma^{(k)}(\omega)\ge t_{k+1}$ for $k<j$ while $\sigma^{(j)}(\omega)<t_{j+1}$, so the intervals $[t_{k},\min(\sigma^{(k)},t_{k+1})]$ for $k\le j$ together with $[\sigma^{(j)}(\omega),T]$ cover $[0,T]$ and again overlap only in finitely many points, the blocks $k>j$ contributing nothing because $\omega\notin G_{k}$ for such $k$. In either case the covering is exact up to a finite set, hence up to a set of Lebesgue measure zero, so the corresponding indicators sum to $1$ for Lebesgue-almost every $t\in[0,T]$; summing them, using the Tonelli and Fubini theorems (all integrands being product-measurable and bounded, by part (a) of the first-order expansion lemma and the progressive measurability of the pre-stopping indicators of the stopped-integral lemma) to exchange the time integral and the expectation, we obtain the decomposition

the first sum running over the tracked slices, the second over the post-exit segments (each $\mathbf{1}_{D_{j}}$ contributing its own terminal term since $D_{j}$ and $G_{K}$ are disjoint), and the endpoint $t=\min(\sigma^{(k)},t_{k+1})$ being a single point, hence of Lebesgue measure zero, by claim 1 of the null-set lemma.

(4a) Tracked terms. Fix $k$. Splitting $\mathbf{1}_{G_{k}}=\mathbf{1}_{G_{k}\setminus\mathcal{N}}+\mathbf{1}_{G_{k}\cap\mathcal{N}}$ and applying claim 2 on the first part and the crude bound $N\mathcal{D}_{t}\ge-NC_{\mathcal{D}}$ (part (a) of the first-order expansion lemma) on the second,

Since $\int_{[t_{k},\min(\sigma^{(k)},t_{k+1})]}|\mathfrak{a}_{t}|^{2}dt=N\,\Delta_{k}\mathcal{E}$ at every point of $\Omega_{0}$ (the integrand being $N|\hat{\alpha}(t)-A_{t}|^{2}$ there, by the realized-control lemma), summing over $k$ and using claim 1,

(4b) Post-exit terms. Fix $j$. The event $\mathcal{N}$ need not be $\mathcal{F}_{\sigma^{(j)}}$-measurable, so it cannot itself be used to split $D_{j}$; instead let $q^{\bullet}$ be the majorant $q^{\bullet}_{t}=|M_{t}|+\Lambda_{b}e^{\Lambda_{b}T}\mathcal{M}_{t}+e^{\Lambda_{b}T}N^{-1/2}|\mathfrak{s}_{0}|$ appearing in claim 1 of the anchored envelope lemma. The family $q^{\bullet}$ is progressively measurable: $\mathbf{1}_{\Omega_{0}}M^{\gamma}$ is progressively measurable by claim 2 of the stopped covariation lemma, hence so is $|M|$ on $\Omega_{0}$, the indefinite integrals $\mathcal{M}_{t}$ are progressively measurable by claim 4 of the progressive measurability toolkit, and $|\mathfrak{s}_{0}|$ is $\mathcal{F}^{\mathrm{sys}}_{0}$-measurable and constant in $t$. Invoking claim 1 of the anchored envelope lemma for the instance with anchor $t_{j}$ and level $L_{j}$ at the time $t=\sigma^{(j)}(\omega)$ — legitimate because $D_{j}\subseteq G_{j}\subseteq\{Y_{t_{j}}<L_{j}\}$ by claim 3 of the cascade lemma — gives both $|\mathfrak{s}_{\sigma^{(j)}}|\le\sqrt{N}(L_{j}+q^{\bullet}_{\sigma^{(j)}})$ and $q^{\bullet}_{\sigma^{(j)}}\le Q$ there. Put $D_{j}'=D_{j}\cap\{q^{\bullet}_{\sigma^{(j)}}\le q_{0}\}$ and $D_{j}''=D_{j}\setminus D_{j}'$. The sampled function $q^{\bullet}_{\sigma^{(j)}}$ is $\mathcal{F}_{\sigma^{(j)}}$-measurable by claim 4(ii) of the stopping-time toolkit, and $D_{j}\in\mathcal{F}_{\sigma^{(j)}}$ by claim 2 of the cascade lemma, so $D_{j}'\in\mathcal{F}_{\sigma^{(j)}}$ and $D_{j}'\subseteq\Omega_{0}$. On $D_{j}'$ the envelope of (2a), applied at $t=\sigma^{(j)}$, gives $|\Sigma_{\sigma^{(j)}}-S_{\sigma^{(j)}}|\le L_{j}+q_{0}\le\varepsilon_{1}+q_{0}\le\varepsilon_{tg}$ by (SM). Hence claim 3 of the post-exit comparison lemma applies with $\varsigma=\sigma^{(j)}$ and $D=D_{j}'$:

On $D_{j}''$ we use the crude bound: $|\int_{[\sigma^{(j)},T]}N\mathcal{D}_{t}dt+N\mathcal{D}_{G}|\le N(C_{\mathcal{D}}T+C_{\mathcal{D}G})$ pointwise, and $D_{j}''\subseteq\{q^{\bullet}_{\sigma^{(j)}}>q_{0}\}\subseteq\mathcal{N}$ since $q^{\bullet}_{\sigma^{(j)}}\le Q$; the $D_{j}''$ being disjoint, their total probability is at most $P(\mathcal{N})$.

By claim 3 of the anchored envelope lemma applied with $D=D_{j}'\subseteq\{Y_{t_{j}}<L_{j}\}$ (claim 3 of the cascade lemma), $\mathbb{E}[\mathbf{1}_{D_{j}'}|\mathfrak{s}_{\sigma^{(j)}}|^{2}]\le2NL_{j}^{2}P(D_{j}')+2c_{Q}^{1/2}\kappa_{0}^{1/2}P(D_{j}')^{1/2}$. Summing over $j$, using $P(D_{j}')\le P(D_{j})$, claim 6 of the cascade lemma with $C_{\dagger}=\mathcal{Z}$ — legitimate, hypothesis (EB) holding with that value by the definition of $\mathcal{Z}$ and claim 1 — for $\sum_{j}NL_{j}^{2}P(D_{j})\le\Lambda_{\star}\mathcal{Z}$ and for $\sum_{j}\sqrt{N}P(D_{j})^{3/4}\le(\Lambda_{\star}\mathcal{Z})^{3/4}N^{-1/4}\Xi_{K}$, and the bound $\sum_{j}P(D_{j})^{1/2}\le\sqrt{K}$ — which holds because for each $j$ the inequality $(u-v)^{2}\ge0$ with $u=(\sqrt{K}P(D_{j}))^{1/2}$ and $v=K^{-1/4}$ gives $P(D_{j})^{1/2}=uv\le\tfrac{1}{2}(u^{2}+v^{2})=\tfrac{1}{2}\bigl(\sqrt{K}P(D_{j})+K^{-1/2}\bigr)$, and summing over the $K$ values of $j$ and using $\sum_{j}P(D_{j})\le1$ (the $D_{j}$ being disjoint) yields $\tfrac{1}{2}(\sqrt{K}+K\cdot K^{-1/2})=\sqrt{K}$ — the post-exit terms total at least

(4c) Terminal term. On $G_{K}\setminus\mathcal{N}$ the envelope of (2a) at $t=T$ (with $\sigma^{(K-1)}\ge t_{K}=T$, so the stopped time is $T$) gives $|\Sigma_{T}-S_{T}|\le\varepsilon_{1}+q_{0}\le\rho^{*}_{G}$, so claim 4 of the localized coercivity lemma gives $N\mathcal{D}_{G}\ge-\epsilon|\mathfrak{s}_{T}|^{2}$ there; on $G_{K}\cap\mathcal{N}$ we use $N\mathcal{D}_{G}\ge-NC_{\mathcal{D}G}$. With claim 3,

Adding (4a), (4b) and (4c) into the decomposition of $\mathcal{J}_{N}$, rearranging, and bounding $N\,C_{\mathcal{D}}T\,P(\mathcal{N})+N(C_{\mathcal{D}}T+C_{\mathcal{D}G})P(\mathcal{N})+NC_{\mathcal{D}G}P(\mathcal{N})$ by $N(C_{\mathcal{D}}T+C_{\mathcal{D}G})\cdot2\,P(\mathcal{N})$ and then by the corresponding term of $\mathcal{R}_{N}$ using claim 1 gives claim 4.

Step 5: proof of claim 5. Write $\mathcal{Z}\in[0,\infty)$ by claim 1. By claim 4 and the absorption condition,

using the multiplicativity of $x\mapsto x^{3/4}$ recorded in the restricted moments lemma. Put $a=C_{ns}\Lambda_{\star}^{3/4}\Xi_{K}N^{-1/4}$, a nonnegative real, and note $a\le\bar{a}:=C_{ns}\Lambda_{\star}^{3/4}\Xi_{K}$ because $N^{-1/4}\le1$ for $N\ge1$.

An elementary Young-type inequality. For all reals $a\ge0$ and $z\ge0$,

Indeed, if $a\,z^{3/4}\le\frac{c_{\star}}{4}z$ the inequality holds, the second term being nonnegative. Otherwise $a\,z^{3/4}>\frac{c_{\star}}{4}z$; here $z>0$ (else both sides vanish), so dividing by $z^{3/4}>0$ gives $a>\frac{c_{\star}}{4}z^{1/4}$, that is, $z^{1/4}<\frac{4a}{c_{\star}}$, whence, raising to the third power (which is nondecreasing on $[0,\infty)$), $z^{3/4}=(z^{1/4})^{3}<\bigl(\frac{4a}{c_{\star}}\bigr)^{3}$ and therefore $a\,z^{3/4}<a\bigl(\frac{4a}{c_{\star}}\bigr)^{3}=\frac{64a^{4}}{c_{\star}^{3}}$. (No upper bound on $c_{\star}$ is needed; the constant $64$ is not optimal.)

Applying this with $z=\mathcal{Z}$, and then $a\le\bar{a}$ together with the monotonicity of $x\mapsto x^{4}$ on $[0,\infty)$, we obtain

Since $\mathcal{Z}$ is finite by claim 1, the two terms $\frac{c_{\star}}{4}\mathcal{Z}$ may be subtracted, leaving $\frac{c_{\star}}{2}\mathcal{Z}\le\mathcal{J}^{\sharp}+\mathcal{R}_{N}+\frac{64\bar{a}^{4}}{c_{\star}^{3}}$, whence, multiplying by $2/c_{\star}>0$,

which is exactly the constant named in the statement, since $\bar{a}=C_{ns}\Lambda_{\star}^{3/4}\Xi_{K}$. Hypothesis (EB) of the cascade lemma is the inequality $\mathcal{Z}\le C_{\dagger}$ just proved. $\blacksquare$