Proof of The Block Cascade of Anchored Good-Set Clocks: Adapted Good Sets, Matched Escape Bounds, and the Energy Ledger

Throughout we use linearity and monotonicity of the integral freely, on $[0,T]$ and on $(\Omega,\mathcal{F},P)$; every random variable appearing below is bounded, hence integrable on the probability space, so all expectations exist and are finite. We write $f_{\omega}(s)=|\hat{\alpha}(s,\omega)-A_{s}|^{2}$, a nonnegative measurable function of $s$ with $f_{\omega}\le4R^{2}$ (each of $\hat{\alpha}(s,\omega)$ and $A_{s}$ lying in $\mathcal{A}$ and $R=\sup_{a\in\mathcal{A}}|a|$), so that $\mathcal{E}_{t}(\omega)=\int_{[0,t]}f_{\omega}(s)\,ds$.

Step 0: elementary facts about the powers used below. Write $x^{1/2}$ for the nonnegative square root of $x\ge0$ and $x^{1/4}=(x^{1/2})^{1/2}$, as in the preamble of the restricted moments lemma, which records that $x\mapsto x^{1/4}$, and hence also $x\mapsto x^{1/2}$, is multiplicative and nondecreasing on $[0,\infty)$; by definition $(x^{1/2})^{2}=x$ and $(x^{1/4})^{2}=x^{1/2}$. We use:

(P1) For $x>0$ one has $x^{1/2}>0$ and $x^{1/4}>0$ (a nonnegative square root of a positive number is nonzero, its square being positive), and $(x^{-1})^{1/2}=(x^{1/2})^{-1}$, $(x^{-1})^{1/4}=(x^{1/4})^{-1}$: indeed $(x^{-1})^{1/2}x^{1/2}=(x^{-1}x)^{1/2}=1$ by multiplicativity, and likewise for the fourth root.

(P2) The map $u\mapsto u^{3}$ is multiplicative, nondecreasing on $[0,\infty)$, and satisfies $u^{3}>1$ whenever $u>1$; consequently $x\mapsto x^{3/4}=(x^{1/4})^{3}$ (the convention of the restricted moments lemma) and $x\mapsto x^{3/2}=(x^{1/2})^{3}$ are multiplicative and nondecreasing on $[0,\infty)$.

(P3) For $x>0$ and an integer $n$, $(x^{n})^{1/2}=(x^{1/2})^{n}$: for $n\ge0$ this follows from multiplicativity of $x\mapsto x^{1/2}$ by induction on $n$ (the case $n=0$ reading $1^{1/2}=1$), and for $n<0$ by combining that case with (P1).

(P4) For $y>0$, $(y^{2})^{1/4}=y^{1/2}$: both sides are nonnegative and the square of the left-hand side is $(y^{2})^{1/2}=y$ by definition of the fourth root and since the nonnegative square root of $y^{2}$ is $y$; so the two agree by uniqueness of the nonnegative square root of $y$.

Step 1: proof of claim 1. Since $0\cdot T_{0}=0<T$, the number $0$ does not satisfy the defining inequality, so $K\ge1$. By minimality $(K-1)T_{0}<T$, hence $kT_{0}\le(K-1)T_{0}<T$ and $t_{k}=kT_{0}$ for every $k\in\{0,\dots,K-1\}$; also $t_{0}=0$ and $t_{K}=\min(KT_{0},T)=T$ because $KT_{0}\ge T$. For $k\in\{0,\dots,K-2\}$ we get $h_{k}=(k+1)T_{0}-kT_{0}=T_{0}$, and $h_{K-1}=T-(K-1)T_{0}$, which is positive by minimality and at most $KT_{0}-(K-1)T_{0}=T_{0}$ since $T\le KT_{0}$; so $0<h_{k}\le T_{0}$ and $t_{k}<t_{k+1}$ for every $k\in\{0,\dots,K-1\}$.

By the solution definition $l\ge2$, so $\sqrt{l}\ge\sqrt{2}>1$ by monotonicity of the nonnegative square root; and $\Lambda_{b}\ge0$ by its definition in the affine rate-family lemma, so $e^{\sqrt{l}\Lambda_{b}T}\ge1$ by the basic properties of the exponential. Hence $C_{a}\ge1$ and $2C_{a}\ge2>1$. Since $\varepsilon_{1}>0$ we get $L_{k}>0$; the exponent $k+1-K$ is at most $0$ for $k\le K-1$, so $(2C_{a})^{k+1-K}\le1$ and $L_{k}\le\varepsilon_{1}$, with equality at $k=K-1$; and $L_{k}/L_{k-1}=(2C_{a})^{(k+1-K)-(k-K)}=2C_{a}$ for $k\in\{1,\dots,K-1\}$.

Step 2: proof of claim 2. That each $\sigma^{(k)}$ is a stopping time of both filtrations with values in $[t_{k},T]$ is claim 1 of the anchored clocks lemma for the instance with anchor $t_{k}$ and level $L_{k}$.

Good sets. We show $G_{k}\in\mathcal{F}^{\mathrm{sys}}_{t_{k}}$ by induction on $k$. For $k=0$: $G_{0}=\Omega_{0}$, the regular event of the solution, which belongs to $\mathcal{F}^{\mathrm{sys}}_{0}$ by the solution definition. Assume $G_{k}\in\mathcal{F}^{\mathrm{sys}}_{t_{k}}$ for some $k\le K-1$. Then $G_{k}\in\mathcal{F}^{\mathrm{sys}}_{t_{k+1}}$ because $t_{k}\le t_{k+1}$ and a filtration is nondecreasing, while $\{\sigma^{(k)}\ge t_{k+1}\}\in\mathcal{F}^{\mathrm{sys}}_{t_{k+1}}$ by claim 1 of the stopping-time toolkit; a $\sigma$-algebra is closed under intersection, so $G_{k+1}\in\mathcal{F}^{\mathrm{sys}}_{t_{k+1}}$. Nesting is immediate from the definition, and $D_{k}\subseteq G_{k}\subseteq G_{0}=\Omega_{0}$.

Partition. For each $k\le K-1$ the sets $\{\sigma^{(k)}\ge t_{k+1}\}$ and $\{\sigma^{(k)}<t_{k+1}\}$ are complementary, so $G_{k}$ is the disjoint union of $G_{k+1}$ and $D_{k}$. Applying this for $k=0,1,\dots,K-1$ in turn gives $\Omega_{0}=G_{0}=D_{0}\cup G_{1}=D_{0}\cup D_{1}\cup G_{2}=\dots=D_{0}\cup\dots\cup D_{K-1}\cup G_{K}$, the union being disjoint at each stage because $D_{j}\subseteq G_{j}$ while $G_{j+1}\supseteq G_{j+2}\supseteq\dots$ are disjoint from $D_{j}$.

$D_{k}\in\mathcal{F}_{\sigma^{(k)}}$. By the definition of the prior $\sigma$-algebra we must check $D_{k}\cap\{\sigma^{(k)}\le s\}\in\mathcal{F}^{\mathrm{sys}}_{s}$ for every $s\in[0,T]$. If $s<t_{k}$ then $\sigma^{(k)}\ge t_{k}>s$ everywhere, so the set is empty and belongs to $\mathcal{F}^{\mathrm{sys}}_{s}$. If $t_{k}\le s<t_{k+1}$ then $\{\sigma^{(k)}\le s\}\subseteq\{\sigma^{(k)}<t_{k+1}\}$, so $D_{k}\cap\{\sigma^{(k)}\le s\}=G_{k}\cap\{\sigma^{(k)}\le s\}$, which lies in $\mathcal{F}^{\mathrm{sys}}_{s}$ because $G_{k}\in\mathcal{F}^{\mathrm{sys}}_{t_{k}}\subseteq\mathcal{F}^{\mathrm{sys}}_{s}$ and $\{\sigma^{(k)}\le s\}\in\mathcal{F}^{\mathrm{sys}}_{s}$ by the definition of a stopping time. If $s\ge t_{k+1}$ then $\sigma^{(k)}<t_{k+1}\le s$ at every point of $D_{k}$, so $D_{k}\cap\{\sigma^{(k)}\le s\}=D_{k}$, and $D_{k}=G_{k}\cap\{\sigma^{(k)}<t_{k+1}\}\in\mathcal{F}^{\mathrm{sys}}_{t_{k+1}}\subseteq\mathcal{F}^{\mathrm{sys}}_{s}$, using claim 1 of the stopping-time toolkit for $\{\sigma^{(k)}<t_{k+1}\}$.

Step 3: proof of claim 3. First, $Y_{0}(\omega)=0$ for every $\omega$: by claim 2 of the flow stability lemma the flow satisfies $S_{0}(x_{0},\xi)=x_{0}$, so $\Phi_{0}(\omega)=x_{0}$, while $S^{*}_{0}=x_{0}$ by the standing hypothesis; hence $Y_{0}(\omega)=|x_{0}-x_{0}|=0$.

We prove $G_{k}\subseteq\{Y_{t_{k}}\le L_{k}/(2C_{a})\}$ by induction on $k\in\{0,\dots,K-1\}$; the second inclusion of claim 3 then follows at once, since $2C_{a}>1$ and $L_{k}>0$ give $L_{k}/(2C_{a})<L_{k}$. For $k=0$ the claim holds because $Y_{0}=0\le L_{0}/(2C_{a})$ everywhere. Suppose it holds for some $k\le K-2$ and let $\omega\in G_{k+1}$. Then $\omega\in G_{k}$, so $Y_{t_{k}}(\omega)\le L_{k}/(2C_{a})<L_{k}$, and $\sigma^{(k)}(\omega)\ge t_{k+1}$. Apply claim 2 of the anchored clocks lemma for the instance with anchor $t_{k}$ and level $L_{k}$, at the time $t=t_{k+1}\in[t_{k},T]$: there $u=\min(t_{k+1},\sigma^{(k)}(\omega))=t_{k+1}$, and since $Y_{t_{k}}(\omega)<L_{k}$ that claim gives $Y_{t_{k+1}}(\omega)\le L_{k}$. By claim 1, $L_{k}=L_{k+1}/(2C_{a})$, so $Y_{t_{k+1}}(\omega)\le L_{k+1}/(2C_{a})$, completing the induction.

Step 4: proof of claim 4. Fix $k\in\{0,\dots,K-1\}$ and consider the instance of the anchored clocks lemma with anchor $t_{k}$, level $L_{k}$, energy threshold $c^{(k)}_{\mathcal{E}}$ and clipped-out threshold $\theta^{(k)}_{\mathrm{out}}$. Its claim 6 applies with $h=h_{k}$: indeed $0<h_{k}$ by claim 1 and $h_{k}=t_{k+1}-t_{k}\le T-t_{k}$ because $t_{k+1}\le T$; the hypotheses of claim 4 of that lemma are the standing hypothesis on $S^{*}$; and $L_{k}>0$. Write $m^{(k)}$ for the constant $m^{*,h}_{b}$ of that claim for this instance. Taking there the event $D=G_{k}$, admissible by claim 3 above (which gives $G_{k}\subseteq\{Y_{t_{k}}\le L_{k}/(2C_{a})\}$, the required inclusion with the instance's level $L_{k}$ in place of $\varepsilon_{1}$), and using $t_{k}+h_{k}=t_{k+1}$, claim 6 yields

$$P(D_{k})=P\bigl(G_{k}\cap\{\sigma^{(k)}<t_{k+1}\}\bigr)\ \le\ \frac{\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\bigl(\mathcal{E}_{\min(\sigma^{(k)},\,t_{k+1})}-\mathcal{E}_{t_{k}}\bigr)\bigr]}{m^{(k)}}=\frac{\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\bigr]}{m^{(k)}} .$$

Now $\delta^{2}\theta^{(k)}_{\mathrm{out}}=L_{k}^{2}/\lambda_{o}$ and $c^{(k)}_{\mathcal{E}}=L_{k}^{2}/\lambda_{c}$. If $K_{2}>0$ then

$$m^{(k)}=\min\Bigl(\frac{L_{k}^{2}}{4C_{a}^{2}K_{2}^{2}h_{k}},\ \frac{L_{k}^{2}}{\lambda_{c}},\ \frac{L_{k}^{2}}{\lambda_{o}}\Bigr)=\frac{L_{k}^{2}}{\max\bigl(4C_{a}^{2}K_{2}^{2}h_{k},\ \lambda_{c},\ \lambda_{o}\bigr)},$$

using that for positive reals the minimum of the reciprocals is the reciprocal of the maximum; hence $L_{k}^{2}/m^{(k)}=\max(4C_{a}^{2}K_{2}^{2}h_{k},\lambda_{c},\lambda_{o})\le\Lambda_{\star}$, because $h_{k}\le T_{0}$ by claim 1. If $K_{2}=0$ then $m^{(k)}=\min(c^{(k)}_{\mathcal{E}},\delta^{2}\theta^{(k)}_{\mathrm{out}})=L_{k}^{2}/\max(\lambda_{c},\lambda_{o})$ and again $L_{k}^{2}/m^{(k)}=\max(\lambda_{c},\lambda_{o})\le\Lambda_{\star}$. In both cases $m^{(k)}>0$, and multiplying the displayed inequality by $L_{k}^{2}>0$ gives

$$L_{k}^{2}\,P(D_{k})\ \le\ \frac{L_{k}^{2}}{m^{(k)}}\,\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\bigr]\ \le\ \Lambda_{\star}\,\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\bigr],$$

the last step because $\mathbb{E}[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}]\ge0$, the integrand being nonnegative by claim 6 of the anchored clocks lemma.

Step 5: proof of claim 5. The energy is nondecreasing in time: for $0\le r\le t\le T$ and every $\omega$, the zero extensions of the restrictions of $f_{\omega}$ to $[0,r]$ and to $[0,t]$ (claim 2 of the integral toolkit) satisfy the pointwise inequality $\le$ between them, both being nonnegative and the first vanishing outside $[0,r]$; so monotonicity of the integral gives $\mathcal{E}_{r}(\omega)\le\mathcal{E}_{t}(\omega)$. Also $\mathcal{E}_{0}=0$ by the convention for the degenerate interval.

Fix $k$. Since $\sigma^{(k)}\ge t_{k}$ everywhere, $t_{k}\le\min(\sigma^{(k)}(\omega),t_{k+1})\le t_{k+1}$, so by the monotonicity just proved

$$0\ \le\ \Delta_{k}\mathcal{E}(\omega)\ \le\ \mathcal{E}_{t_{k+1}}(\omega)-\mathcal{E}_{t_{k}}(\omega)\qquad(\omega\in\Omega).$$

Since $G_{k}\subseteq\Omega_{0}$ we have $\mathbf{1}_{G_{k}}\le\mathbf{1}_{\Omega_{0}}$ pointwise, so $\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\le\mathbf{1}_{\Omega_{0}}(\mathcal{E}_{t_{k+1}}-\mathcal{E}_{t_{k}})$ pointwise, both sides being bounded random variables ($\Delta_{k}\mathcal{E}$ by claim 6 of the anchored clocks lemma, and $\mathcal{E}_{t}$ being a random variable bounded by $4R^{2}T$). Taking expectations and summing over $k\in\{0,\dots,K-1\}$, then using linearity and the telescoping of $\sum_{k=0}^{K-1}(\mathcal{E}_{t_{k+1}}-\mathcal{E}_{t_{k}})=\mathcal{E}_{t_{K}}-\mathcal{E}_{t_{0}}=\mathcal{E}_{T}$,

$$\sum_{k=0}^{K-1}\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\bigr]\ \le\ \mathbb{E}\Bigl[\mathbf{1}_{\Omega_{0}}\sum_{k=0}^{K-1}\bigl(\mathcal{E}_{t_{k+1}}-\mathcal{E}_{t_{k}}\bigr)\Bigr]=\mathbb{E}\bigl[\mathbf{1}_{\Omega_{0}}\mathcal{E}_{T}\bigr].$$

Step 6: proof of claim 6. Each $\Delta_{k}\mathcal{E}$ is nonnegative by claim 6 of the anchored clocks lemma, so the left-hand side of (EB) is nonnegative and hypothesis (EB) forces $C_{\dagger}\ge0$. Multiplying claim 4 by $N>0$, summing over $k$, and applying (EB),

$$\sum_{k=0}^{K-1}N\,L_{k}^{2}\,P(D_{k})\ \le\ \Lambda_{\star}\,N\sum_{k=0}^{K-1}\mathbb{E}\bigl[\mathbf{1}_{G_{k}}\Delta_{k}\mathcal{E}\bigr]\ \le\ \Lambda_{\star}\,C_{\dagger},$$

using $\Lambda_{\star}\ge\lambda_{c}>0$. This is the first assertion.

Every summand on the left is nonnegative, so each satisfies $N L_{k}^{2}P(D_{k})\le\Lambda_{\star}C_{\dagger}$, that is,

$$P(D_{k})\ \le\ \Lambda_{\star}C_{\dagger}\,N^{-1}L_{k}^{-2}.$$

Write $b_{k}=\Lambda_{\star}C_{\dagger}\,N^{-1}L_{k}^{-2}$, so $0\le P(D_{k})\le b_{k}$. By (P2) the map $x\mapsto x^{3/4}$ is nondecreasing and multiplicative on $[0,\infty)$, so

$$P(D_{k})^{3/4}\ \le\ b_{k}^{3/4}=\bigl(\Lambda_{\star}C_{\dagger}\bigr)^{3/4}\,\bigl(N^{-1}\bigr)^{3/4}\,\bigl(L_{k}^{-2}\bigr)^{3/4}.$$

By (P1) and (P2), $(N^{-1})^{3/4}=\bigl((N^{-1})^{1/4}\bigr)^{3}=\bigl((N^{1/4})^{-1}\bigr)^{3}=(N^{1/4})^{-3}$. Similarly $(L_{k}^{-2})^{1/4}=\bigl((L_{k}^{2})^{1/4}\bigr)^{-1}=(L_{k}^{1/2})^{-1}$ by (P1) and (P4), so $(L_{k}^{-2})^{3/4}=(L_{k}^{1/2})^{-3}=L_{k}^{-3/2}$ in the notation of the statement. Finally $\sqrt{N}=(N^{1/4})^{2}$ because $(N^{1/4})^{2}=N^{1/2}$, so $\sqrt{N}\,(N^{1/4})^{-3}=(N^{1/4})^{-1}=N^{-1/4}$. Multiplying the display by $\sqrt{N}$ therefore gives

$$\sqrt{N}\,P(D_{k})^{3/4}\ \le\ \bigl(\Lambda_{\star}C_{\dagger}\bigr)^{3/4}\,N^{-1/4}\,L_{k}^{-3/2},$$

and summing over $k$ gives the second assertion, with $\Xi_{K}=\sum_{k=0}^{K-1}L_{k}^{-3/2}$.

Finally we evaluate $\Xi_{K}$. Put $\rho=(2C_{a})^{1/2}$ and $r=\rho^{3}=(2C_{a})^{3/2}$. Since $2C_{a}>1$ by claim 1 and $x\mapsto x^{1/2}$ is nondecreasing with $1^{1/2}=1$, we have $\rho\ge1$; in fact $\rho>1$, for $\rho=1$ would give $2C_{a}=\rho^{2}=1$. Hence $r>1$ by (P2). From $L_{k}=(2C_{a})^{k+1-K}\varepsilon_{1}$, multiplicativity of $x\mapsto x^{1/2}$ and (P3) give $L_{k}^{1/2}=\rho^{\,k+1-K}\,\varepsilon_{1}^{1/2}$; cubing and taking reciprocals (using (P1), (P2) and $L_{k}>0$) yields $L_{k}^{-3/2}=\varepsilon_{1}^{-3/2}\,r^{K-k-1}$, so, substituting $j=K-1-k$ (a bijection of $\{0,\dots,K-1\}$ onto itself),

$$\Xi_{K}=\varepsilon_{1}^{-3/2}\sum_{j=0}^{K-1}r^{j}=\varepsilon_{1}^{-3/2}\,\frac{r^{K}-1}{r-1},$$

the last equality because $(r-1)\sum_{j=0}^{K-1}r^{j}=\sum_{j=1}^{K}r^{j}-\sum_{j=0}^{K-1}r^{j}=r^{K}-1$ by telescoping, and $r-1\neq0$. Since $r^{K}=(2C_{a})^{3K/2}$, this is the stated expression. $\blacksquare$