Proof. Using \(L_{ui}=L_uX_i(u)\), independence of the level-\(n\) descendants from \(\mathcal{F}_n^X\), and the mean-one condition \(\mathbb{E}(X_0+X_1)=1\), we get \(\mathbb{E}[M_{n+1}\mid\mathcal{F}_n^X]=M_n\). ◻

Proof. Work on the full-probability event on which \(M_n\to M<\infty\) and \(M_m^{(u)}\to M^{(u)}\) for every \(u\in\mathcal{T}\). Then \[\mu_n(I_u)=L_uM_{n-|u|}^{(u)}\longrightarrow L_uM^{(u)}\] for every dyadic cylinder \(I_u\). Thus the integrals of dyadic step functions against \(\mu_n\) have limits. Since \(\sup_n\mu_n([0,1])<\infty\), uniform approximation of continuous functions by dyadic step functions extends these limits to all \(g\in C([0,1])\). The resulting linear functional is positive and bounded by \(\|g\|_\infty\sup_n\mu_n([0,1])\). By the Riesz representation theorem, this positive bounded functional is integration against a finite Borel measure \(\mu\). Hence \(\mu_n\xrightarrow{\mathrm{w}}\mu\), and evaluating the convergence at the constant function \(g\equiv1\) gives \(\mu([0,1])=M\). ◻

Proof. For \(0<q\le1\), \(x^q\le1+x\), and hence \(\rho(q)<\infty\). Writing \[\rho(q)=\int x^q\,d\nu_X(x),\qquad \nu_X=\mathbb{E}(\delta_{X_0}+\delta_{X_1}),\] Hölder’s inequality implies the log-convexity of \(\rho\) on its finiteness intervals.

The balanced assumption implies \(\rho(1)=1\). Since \(H^+(X)<\infty\), the family \(x^q\log_2 x\), \(q\uparrow1\), is dominated by \(x\log_2^+x+C\mathbf{1}_{\{0<x\le1\}}\), and hence \[\left.\frac{d}{dq}\right|_{q=1-}\log_2\rho(q) =\mathbb{E}[X_0\log_2X_0+X_1\log_2X_1] =\gamma(X)<0 .\] Convexity then implies \[\log_2\rho(q)\ge \gamma(X)(q-1)>0,\] so \(\rho(q)>1\).

If \(q_0\in I_X\), convexity between \(1\) and \(q_0\) implies \(\rho(q)<1\) for all \(1<q\le q_0\). Hence \(I_X\) is an initial interval of \((1,2)\), with the stated endpoint alternatives. ◻

Proof. Let \(0<s<1\). By Lemma 4, \(2^{qs/2}\rho(q)>1\) for \(0<q\le1\). Hence only \(q\in(1,2)\) can satisfy \(2^{qs/2}\rho(q)<1\). For such \(q\), this inequality is equivalent to \(\rho(q)<1\) and \(s<-\frac{2}{q}\log_2\rho(q)\). Taking the supremum over \(s\) and \(q\) implies the formula and the equivalence \[D_E(X)>0\iff \rho(q)<1 \quad\text{for some}\quad q\in(1,2).\]

If \(\rho(q)<1\), then \(\rho(q)<\infty\). Conversely, if \(\rho(q_0)<\infty\) for some \(q_0>1\), choose \(r\in(1,2)\) with \(r\le q_0\). The right derivative of \(\log_2\rho\) at \(1\) along \([1,r]\) equals \(\gamma(X)<0\), so \(\rho(q)<1\) for all \(q>1\) sufficiently close to \(1\). ◻

Proof. Equation 24 follows immediately from the identity \(C_{uv}=L_uC_v^{(u)}\). The asserted independence and equality in law follow from the independence of the descendant environments below distinct level-\(m\) vertices. ◻

Proof. We work on the event where all \(C_u\) are defined and consistent. Iterating \(C_u=C_{u0}+C_{u1}\) implies \(\sum_{\lvert u \rvert=n}C_u=M\). Cauchy’s inequality implies 25 . On \(\{M>0\}\), taking logarithms and passing to the liminf yields the asserted bound. ◻

Proof. For every dyadic cylinder \([u]_\partial\), the measures \(\mu_n\) assign asymptotic mass \(C_u\) to \(I_u\), while \(\mu_\partial([u]_\partial)=C_u\). Thus \(\mu_n\) and \(\pi_\#\mu_\partial\) have the same limits on dyadic step functions. Uniform approximation of \(C([0,1])\) by dyadic step functions, together with \(\sup_n\mu_n([0,1])<\infty\), implies \(\mu=\pi_\#\mu_\partial\). ◻

Proof. The second equality is the definition of \(\dim_2(\mu)\). It remains to compare the tree and Euclidean square sums.

If \(\mu\) has an atom of mass \(a>0\), then by Lemma 7, \(\mu_\partial(\pi^{-1}\{x\})=a\) for some \(x\). Since \(\pi^{-1}\{x\}\) has at most two points, there exists a boundary path \(\omega\) such that \(\mu_\partial(\{\omega\})\ge a/2\). Thus \[C_{\omega|n}\ge a/2 \qquad(n\ge0),\] and hence \(\Sigma_n\ge a^2/4\). Together with \(\Sigma_n\le M^2\), this yields \[\liminf_{n\to\infty}\frac{\log_2\Sigma_n}{-n}=0.\] By Remark 14, \(\dim_2(\mu)=0\).

If \(\mu\) is non-atomic, then so is \(\mu_\partial\); since \(\pi^{-1}(I_u)\) and \([u]_\partial\) differ only by finitely many endpoint codings, \(\mu(I_u)=C_u\) for every \(u\). Hence \(\Sigma_n(\mu)=\Sigma_n\) for all \(n\). ◻

Proof. Set \(S=X_0+X_1\). We first treat \(1<q\le2\). Since \(S^q\le 2^{q-1}(X_0^q+X_1^q)\), it follows that \(\mathbb{E}|S-1|^q<\infty\). Let \(D_{n+1}=M_{n+1}-M_n\). Writing \(S(u)=X_0(u)+X_1(u)\), \[D_{n+1} = \sum_{|u|=n}L_u\bigl(S(u)-1\bigr).\] Conditionally on \(\mathcal{F}_n^X\), the variables \(L_u(S(u)-1)\) \((|u|=n)\), are independent and centered. Hence the conditional von Bahr–Esseen inequality yields \[\mathbb{E}\bigl[|D_{n+1}|^q\mid\mathcal{F}_n^X\bigr] \le 2\mathbb{E}|S-1|^q\sum_{|u|=n}L_u^q .\] Taking expectations and using Lemma 3, we obtain \[\mathbb{E}|D_{n+1}|^q \le 2\mathbb{E}|S-1|^q\,\rho(q)^n, \qquad \|D_{n+1}\|_q \le \bigl(2\mathbb{E}|S-1|^q\bigr)^{1/q}\rho(q)^{n/q}.\] Since \(M_n=1+\sum_{k=0}^{n-1}D_{k+1}\), Minkowski’s inequality implies \[\sup_{n\ge0}\|M_n\|_q \le 1+ \frac{\bigl(2\mathbb{E}|S-1|^q\bigr)^{1/q}}{1-\rho(q)^{1/q}}.\] Thus \(\sup_{n\ge0}\mathbb{E}[M_n^q]<\infty\). Since \(M_n\to M\) a.s., Fatou’s lemma implies \(\mathbb{E}[M^q]<\infty\).

It remains to consider \(q>2\). We first note that \(\rho(q)<1\) implies \(\rho(p)<1\) for every \(1<p\le q\). Indeed, if \(p=(1-\theta)+\theta q\), \(0<\theta\le1\), then Hölder’s inequality yields \[\rho(p) = \mathbb{E}[X_0^p+X_1^p] \le \bigl(\mathbb{E}[X_0+X_1]\bigr)^{1-\theta} \bigl(\mathbb{E}[X_0^q+X_1^q]\bigr)^\theta = \rho(q)^\theta <1.\]

We argue by induction over \((m,m+1]\), \(m\ge2\). Assume the assertion holds at exponent \(q-1\). Let \(M_n^{(0)}\) and \(M_n^{(1)}\) be independent copies of \(M_n\), independent of \(X=(X_0,X_1)\). By the branching construction, \[M_{n+1} \stackrel{d}= X_0M_n^{(0)}+X_1M_n^{(1)}.\] Using \[(x+y)^q \le x^q+y^q+C_q\bigl(x^{q-1}y+xy^{q-1}\bigr), \qquad x,y\ge0,\] together with independence and \(\mathbb{E}M_n=1\), we obtain \[\mathbb{E}[M_{n+1}^q] \le \rho(q)\mathbb{E}[M_n^q] + C_q\mathbb{E}\!\left[X_0^{q-1}X_1+X_0X_1^{q-1}\right]\mathbb{E}[M_n^{q-1}].\] Young’s inequality implies \[\mathbb{E}\!\left[X_0^{q-1}X_1+X_0X_1^{q-1}\right] \le \rho(q)<\infty .\] Since \(\rho(q-1)<1\), the induction hypothesis yields \[\sup_{n\ge0}\mathbb{E}[M_n^{q-1}]<\infty .\] Consequently, for some finite \(C_q'\), \[\mathbb{E}[M_{n+1}^q] \le \rho(q)\mathbb{E}[M_n^q]+C_q', \qquad n\ge0 .\] Since \(\rho(q)<1\), iteration yields \[\sup_{n\ge0}\mathbb{E}[M_n^q]<\infty .\] Finally, Fatou’s lemma and \(M_n\to M\) a.s. imply \[\mathbb{E}[M^q] \le \liminf_{n\to\infty}\mathbb{E}[M_n^q] <\infty.\] ◻

Proof. Apply Lemma 9 to the binary vector \((X_0,X_1)\); descendant cascades have the same law as the original one. ◻

Proof. By the preceding observation, \(\mathbb{E}[M^q]<\infty\). Using \(C_u=L_uM^{(u)}\), independence, \(M^{(u)}\stackrel d=M\), and subadditivity of \(x^p\) for \(p=q/2<1\), we get \(\mathbb{E}\Sigma_n^p\le \mathbb{E}[M^q]\rho(q)^n\). Multiplying by \(2^{snp}=2^{snq/2}\), we obtain \[\mathbb{E}\left[(2^{sn}\Sigma_n)^p\right] \le \mathbb{E}[M^q]\bigl(2^{sq/2}\rho(q)\bigr)^n = \mathbb{E}[M^q]m_s(q)^n.\]

Since \(m_s(q)<1\), Markov’s inequality implies, for every \(\varepsilon>0\), \[\sum_{n=0}^\infty \mathbb{P}(2^{sn}\Sigma_n>\varepsilon) \le \varepsilon^{-p}\mathbb{E}[M^q]\sum_{n=0}^\infty m_s(q)^n <\infty.\] By Borel–Cantelli, applied with \(\varepsilon=1/k\), it follows that \(2^{sn}\Sigma_n\to0\) almost surely. ◻

Proof. If \(D_E(X)=0\), the final bound follows from Lemma 6. Assume that \(D_E(X)>0\), and let \(0<s<D_E(X)\). By Proposition 11, there exists \(q\in(1,2)\) such that \(m_s(q)<1\). Hence Theorem 15 implies \(2^{sn}\Sigma_n\to0\) almost surely. Thus, almost surely, \(\Sigma_n\le 2^{-sn}\) for all sufficiently large \(n\). On \(\{M>0\}\), Lemma 6 ensures that \(\Sigma_n>0\), and therefore \[\liminf_{n\to\infty} \frac{\log_2\Sigma_n}{-n} \ge s.\] Letting \(s\uparrow D_E(X)\) along a countable sequence yields 30 . ◻

Proof. By Lemma 8 and Corollary 3, \[\dim_2(\mu) = \liminf_{n\to\infty}\frac{\log_2\Sigma_n}{-n} \ge D_E(X)\] almost surely on \(\{M>0\}\). Since \(\mu\) is then a nonzero finite Borel measure, Proposition 13 applies and implies \(\dim_{\mathrm E}(\mu)=\dim_2(\mu)\). ◻

Proof. Let \(\mathcal{T}_+=\{u:L_u>0\}\). Its offspring law is \(N_X\). Since energy-admissibility implies \(\mathbb{P}(M>0)>0\), the alive tree survives with positive probability.

If \(\mathbb{E}N_X<1\), then, for the alive population \(Z_n\), \[\mathbb{P}(Z_n>0)\le \mathbb{E}Z_n=(\mathbb{E}N_X)^n\to0,\] so the survival probability is zero. If \(\mathbb{E}N_X=1\), positive survival probability forces the critical Galton–Watson law to be degenerate, namely \(N_X=1\) almost surely. Thus exactly one coordinate of \(X\) is positive almost surely. With \(S=X_0+X_1\), balance implies \(\mathbb{E}S=1\), and \[X_0\log_2X_0+X_1\log_2X_1=S\log_2S.\] By Jensen’s inequality, \[\mathbb{E}[S\log_2S]\ge \mathbb{E}S\log_2\mathbb{E}S=0,\] contradicting the strict entropy condition in energy-admissibility. Hence \(\mathbb{E}N_X>1\). ◻

Proof. For \(0<q\le1\), one has \(X_0^q+X_1^q\le 2+X_0+X_1,\) which is integrable. Since \(X_i^q\to\mathbf{1}_{\{X_i>0\}}\) as \(q\downarrow0\), dominated convergence and Lemma 11 imply \[\rho(q)\to \mathbb{E}N_X>1.\] Moreover, \(2^{qs/2}\to1\) uniformly for \(s\in K\). Hence \(m_s(q)=2^{qs/2}\rho(q)\to\mathbb{E}N_X>1\) uniformly on \(K\), establishing the claimed gap. ◻

Proof. Suppose that \(D_E(X)<s<1\) and \(\inf_{0<q<2}m_s(q)=1\). Set \[s_0=\frac{D_E(X)+s}{2}.\] Then \(D_E(X)<s_0<s\), and by the definition of \(D_E(X)\), \(\inf_{0<q<2}m_{s_0}(q)\ge1\). By Lemma 12, there exist \(q_0>0\) and \(\eta>0\) such that \[m_s(q)\ge1+\eta \qquad(0<q\le q_0).\] Thus there exists a sequence \(q_j\in[q_0,2)\) such that \(m_s(q_j)\to1\). Since \(m_{s_0}(q)=2^{-q(s-s_0)/2}m_s(q)\), we obtain \[\inf_{0<q<2}m_{s_0}(q) \le \liminf_j m_{s_0}(q_j) \le 2^{-q_0(s-s_0)/2}<1,\] contradicting the preceding lower bound. Hence \(\inf_{0<q<2}m_s(q)>1\). ◻

Proof. For \(0<\theta<1\), setting \(q=2\theta\) yields \(\kappa_s(\theta)=m_s(q)\), so the interior bound follows from Lemma 13. The endpoint \(\theta=0\) follows from Lemma 11, and the endpoint \(\theta=1\) follows by taking \(q\uparrow2\). Hence \[c:=\inf_{0\le\theta\le1}\kappa_s(\theta)>1.\]

Choose \(\lambda\in(0,1)\) sufficiently close to \(1\) so that \(\lambda c>1\). For \(0<\theta\le1\), \[\kappa_{s,\lambda}(\theta) = \lambda^\theta\kappa_s(\theta) \ge \lambda c>1,\] while \[\kappa_{s,\lambda}(0)=\kappa_s(0)\ge c.\] This proves the discounted condition. ◻

Proof. If \(Y_s<\infty\) almost surely, then \(Z_n^{(s)}\) is eventually bounded almost surely; since \(\lambda<1\), the discounted series is finite almost surely.

Let \(Z_{s,\lambda}^{(0)}\) and \(Z_{s,\lambda}^{(1)}\) be the corresponding discounted series in the two descendant subtrees. They are independent copies of \(Z_{s,\lambda}\), independent of \(X\). Using 31 , \[Z_{s,\lambda} = Z_0^{(s)} + \lambda A_0^{(s)}Z_{s,\lambda}^{(0)} + \lambda A_1^{(s)}Z_{s,\lambda}^{(1)}.\] Since \(Z_0^{(s)}=M^2\ge0\), \[Z_{s,\lambda} \ge \lambda A_0^{(s)}Z_{s,\lambda}^{(0)} + \lambda A_1^{(s)}Z_{s,\lambda}^{(1)}.\] Taking \(e^{-t(\cdot)}\), using monotonicity, and conditioning on \(A_0^{(s)},A_1^{(s)}\), we obtain 32 . ◻

Proof. Let \[\alpha_0=\inf_{0\le\theta\le1}\kappa(\theta)>1.\] Choose \(\delta_0>0\) such that \[1+2\delta_0<\alpha_0 .\]

For \(m\ge2\), set \[A_i^{[m]} = A_i\mathbf{1}_{\{m^{-1}\le A_i\le m\}}, \qquad i=0,1,\] and define \[\kappa_m(\theta) = \begin{cases} \displaystyle \mathbb{E}\left[ \mathbf{1}_{\{A_0^{[m]}>0\}}+\mathbf{1}_{\{A_1^{[m]}>0\}} \right], & \theta=0,\\[8pt] \displaystyle \mathbb{E}\left[ \bigl(A_0^{[m]}\bigr)^\theta+ \bigl(A_1^{[m]}\bigr)^\theta \right], & 0<\theta\le1. \end{cases}\] For each fixed \(\theta\in[0,1]\), monotone convergence yields \(\kappa_m(\theta)\uparrow\kappa(\theta)\), with the case \(\theta=0\) understood through the indicators of \(\{A_i>0\}\). Since \(\inf_{0\le\theta\le1}\kappa(\theta)>1+2\delta_0\), continuity of the truncated functions \(\kappa_m\) and compactness of \([0,1]\) allow us to choose a single \(m_0\) such that \[\inf_{0\le\vartheta\le1}\kappa_{m_0}(\vartheta)>1+\delta_0.\]

We claim that it is enough to prove the lemma for the truncated pair \((A_0^{[m_0]},A_1^{[m_0]}).\) Indeed, for each vertex \(v\in\mathcal{T}\), set \[A_i^{[m_0]}(v) = A_i(v)\mathbf{1}_{\{m_0^{-1}\le A_i(v)\le m_0\}}, \qquad i=0,1.\] Then \(\{(A_0^{[m_0]}(v),A_1^{[m_0]}(v)):v\in\mathcal{T}\}\) is an i.i.d. family of copies of \((A_0^{[m_0]},A_1^{[m_0]})\). If \[A_u^{[m_0]} = \prod_{j=1}^{\lvert u \rvert}A_{u_j}^{[m_0]}(u|_{j-1}),\] then \[0\le A_u^{[m_0]}\le A_u\] for every \(u\in\mathcal{T}\). Consequently, for every \(\tau>0\), \[\{u:A_u^{[m_0]}\ge\tau\} \subseteq \{u:A_u\ge\tau\}.\] Thus any lower bound for the expected number of truncated products exceeding \(\tau\) is also a lower bound for the corresponding original products.

Therefore, from now on, replace \((A_0,A_1)\) by the truncated pair \((A_0^{[m_0]},A_1^{[m_0]})\), and write it again as \((A_0,A_1)\). We also write \(\kappa\) for the corresponding function associated with this new pair. Then \[A_i\in\{0\}\cup[m_0^{-1},m_0], \qquad i=0,1,\] and, by the preceding display, \[\label{eq:truncated-uniform-supercritical-paper} \inf_{0\le\theta\le1}\kappa(\theta)>1.\tag{33}\]

Define a finite measure \(\nu\) on \(\mathbb{R}\) by \[\nu(B) = \mathbb{E}\sum_{i=0}^1 \mathbf{1}_{\{A_i>0,\;\ln A_i\in B\}}, \qquad B\in\mathcal{B}(\mathbb{R}).\] Since \(A_i\in\{0\}\cup[m_0^{-1},m_0]\), the measure \(\nu\) is supported on \([-\ln m_0,\ln m_0]\). For \(\theta\in\mathbb{R}\), define \[\Lambda(\theta) = \ln\int e^{\theta x}\,\nu(dx).\] This is well-defined because \(\nu\) has compact support and positive total mass. Moreover, for \(0\le\theta\le1\), \[\int e^{\theta x}\,\nu(dx) = \mathbb{E}\sum_{i=0}^1 \mathbf{1}_{\{A_i>0\}}e^{\theta\ln A_i}.\] For \(0<\theta\le1\), this equals \(\mathbb{E}\left[A_0^\theta+A_1^\theta\right]\), while for \(\theta=0\) it equals \(\mathbb{E}\left[\mathbf{1}_{\{A_0>0\}}+\mathbf{1}_{\{A_1>0\}}\right]\).

By 33 , \[\Lambda(\theta)=\ln\kappa(\theta)>0, \qquad 0\le\theta\le1.\]

Since \(\nu\) has compact support and positive total mass, \(\Lambda\) is finite, convex, and \(C^\infty\) on \(\mathbb{R}\).

We next use a simple geometric consequence of convexity. Since \(\Lambda>0\) on \([0,1]\), one can choose \(\theta_0\in(0,1)\) such that the tangent line to \(\Lambda\) at \(\theta_0\) is positive at both endpoints \(0\) and \(1\). Indeed, this is immediate by taking a minimizer of \(\Lambda\) on \([0,1]\): if the minimizer lies in the interior, the tangent is horizontal, while if it lies at an endpoint, one moves slightly into the interval.

Set \[r=\Lambda'(\theta_0), \qquad \gamma=\Lambda(\theta_0)-\theta_0\Lambda'(\theta_0).\] The values of the tangent line \[t\mapsto \Lambda(\theta_0)+(t-\theta_0)\Lambda'(\theta_0)\] at \(t=0\) and \(t=1\) are respectively \(\gamma\) and \(\gamma+r\). Thus \(\gamma>0\), \(\gamma+r>0\). Define a probability measure \(Q\) on \(\mathbb{R}\) by \[Q(dx) = e^{\theta_0x-\Lambda(\theta_0)}\,\nu(dx).\] Indeed, \[Q(\mathbb{R}) = e^{-\Lambda(\theta_0)} \int e^{\theta_0x}\,\nu(dx) = 1.\] Let \(Y_1,Y_2,\ldots\) be i.i.d. random variables with law \(Q\), and write \(S_N=Y_1+\cdots+Y_N\). Since \(Q\) has compact support, \(Y_1\) is integrable, and by the strong law of large numbers, \[\frac{S_N}{N}\longrightarrow \mathbb{E}_QY_1 \qquad Q\text{-a.s.}\] Moreover, \[\mathbb{E}_QY_1 = \int x\,Q(dx) = e^{-\Lambda(\theta_0)}\int xe^{\theta_0x}\,\nu(dx) = \Lambda'(\theta_0) = r.\] Therefore \[\frac{S_N}{N}\longrightarrow r \qquad Q\text{-a.s.}\]

Choose \(\eta>0\) sufficiently small that \[\gamma-\theta_0\eta>0, \qquad \gamma+r-\eta-\theta_0\eta>0.\]

Then choose \(\delta>0\) sufficiently small that \[\label{eq:delta-choice-finite-block-paper} \gamma-\theta_0\eta-2\delta>0, \qquad \gamma+r-\eta-\theta_0\eta-2\delta>0.\tag{34}\]

For \(N\ge1\), define \[Z_N = \#\left\{ u\in\{0,1\}^N: A_u\ge e^{N(r-\eta)} \right\}.\] Equivalently, \[Z_N = \sum_{\lvert u \rvert=N} \mathbf{1}_{\{A_u>0\}} \mathbf{1}_{[N(r-\eta),\infty)}(\ln A_u).\]

We shall use the following many-to-one identity: for every nonnegative Borel function \(f:\mathbb{R}\to[0,\infty]\) and every \(N\ge1\), \[\label{eq:many-to-one-nu-convolution-paper} \mathbb{E}\sum_{\lvert u \rvert=N} \mathbf{1}_{\{A_u>0\}}f(\ln A_u) = \int f(x)\,\nu^{*N}(dx).\tag{35}\] Indeed, the case \(N=1\) is precisely the definition of \(\nu\), and the general case follows by induction on \(N\), using the independence of the weights on different vertices and the definition of convolution.

Taking \[f(x)=\mathbf{1}_{[N(r-\eta),\infty)}(x)\] in 35 , we obtain \[\mathbb{E}Z_N = \nu^{*N}\bigl([N(r-\eta),\infty)\bigr).\]

Let \(I_N=[N(r-\eta),N(r+\eta)]\). Then \[\mathbb{E}Z_N \ge \nu^{*N}(I_N).\] We next estimate \(\nu^{*N}(I_N)\). Since \(Q(dx)=e^{\theta_0x-\Lambda(\theta_0)}\,\nu(dx)\), we have \[\nu(dx)=e^{\Lambda(\theta_0)-\theta_0x}\,Q(dx).\] Hence, by the definition of convolution and by the fact that \((Y_1,\ldots,Y_N)\) has law \(Q^{\otimes N}\), \[\begin{align} \nu^{*N}(I_N) &= e^{N\Lambda(\theta_0)} \mathbb{E}_Q\left[ e^{-\theta_0S_N}\mathbf{1}_{\{S_N\in I_N\}} \right] \\ &\ge e^{N\Lambda(\theta_0)} e^{-\theta_0N(r+\eta)} Q(S_N\in I_N). \end{align}\] By the strong law, \[Q(S_N\in I_N)\longrightarrow1.\] Consequently, for all sufficiently large \(N\), \(Q(S_N\in I_N)\ge e^{-N\delta}\). For such \(N\), \[\mathbb{E}Z_N \ge \exp\left(N(\Lambda(\theta_0)-\theta_0r-\theta_0\eta-\delta)\right).\] Since \(\gamma=\Lambda(\theta_0)-\theta_0r\), this becomes \[\mathbb{E}Z_N \ge \exp\left(N(\gamma-\theta_0\eta-\delta)\right).\]

Choose such an \(N\), and set \[\tau=e^{N(r-\eta)}, \qquad \beta=\exp\left(N(\gamma-\theta_0\eta-2\delta)\right).\] By 34 , \(\beta>1\). Moreover, \(\mathbb{E}Z_N>\beta\). Since \[Z_N=\#\{u\in\{0,1\}^N:A_u\ge\tau\},\] we have \[\mathbb{E}\#\{u\in\{0,1\}^N:A_u\ge\tau\}>\beta.\] Finally, \[\beta\tau = \exp\left( N(\gamma+r-\eta-\theta_0\eta-2\delta) \right)>1\] by 34 . This proves the lemma for the truncated pair, and hence for the original pair. ◻

Proof. Since \(\xi\) is bounded, \(\mathbb{E}[\xi\log^+\xi]<\infty\). By the Kesten–Stigum theorem for supercritical Galton–Watson processes, \[\frac{Z_k}{m^k}\longrightarrow Y_\infty\] almost surely, where \(Y_\infty>0\) almost surely on the survival event \(S\); see [42], or the original theorem of Kesten and Stigum [43]. Hence, on \(S\), \[\frac{Z_k}{\beta^k} = \frac{Z_k}{m^k}\left(\frac{m}{\beta}\right)^k \longrightarrow \infty.\] Therefore \(Z_k\ge \beta^k\) for all sufficiently large \(k\), almost surely on \(S\). ◻

Proof. Recall that \(A_i^{(s)}=2^sX_i^2\), \(i=0,1\), and, for \(u\in\mathcal{T}\), \[A_u^{(s)} = \prod_{j=1}^{\lvert u \rvert} A_{u_j}^{(s)}(u|_{j-1}).\] Thus \(A_u^{(s)}=2^{s\lvert u \rvert}L_u^2\), and \[\label{eq:scaled-square-sum-as-A-sum-paper} 2^{sn}\Sigma_n = \sum_{\lvert u \rvert=n} A_u^{(s)}\bigl(M^{(u)}\bigr)^2.\tag{36}\]

By Lemma 14, \[\inf_{0\le\theta\le1} \mathbb{E}\left[ \bigl(A_0^{(s)}\bigr)^\theta+ \bigl(A_1^{(s)}\bigr)^\theta \right] >1,\] with \(a^0=\mathbf{1}_{\{a>0\}}\). Applying Lemma 16 to \(\bigl(A_0^{(s)},A_1^{(s)}\bigr)\), we obtain \(N\ge1\), \(\tau>0\), and \(\beta>1\) such that \[\beta\tau>1, \qquad m:= \mathbb{E}\#\{u\in\{0,1\}^N:A_u^{(s)}\ge\tau\}>\beta.\]

Define an embedded \(N\)-block Galton–Watson process. The root is good; if \(u\) is good, then \(uv\), \(v\in\{0,1\}^N\), is good if and only if \[\frac{A_{uv}^{(s)}}{A_u^{(s)}}\ge\tau.\] Let \(\mathcal{G}_k\) be the set of good vertices in generation \(k\). The offspring distribution has mean \(m>\beta>1\) and is bounded by \(2^N\); hence the process survives with positive probability. Denote its survival event by \(S_{\mathrm{good}}\). By Lemma 17, on \(S_{\mathrm{good}}\), \[\label{eq:good-tree-growth-eventual-paper} \#\mathcal{G}_k\ge\beta^k\tag{37}\] for all sufficiently large \(k\), almost surely. Moreover, every \(u\in\mathcal{G}_k\) satisfies \[\lvert u \rvert=kN, \qquad A_u^{(s)}\ge\tau^k.\]

Since \(\mathbb{P}(M>0)>0\), choose \(a>0\) with \[p_a:=\mathbb{P}(M^2\ge a)>0.\] For each \(k\), conditionally on \(\mathcal{F}^X_{kN}\), the variables \[\mathbf{1}_{\{(M^{(u)})^2\ge a\}}, \qquad u\in\mathcal{G}_k,\] are independent Bernoulli variables with parameter \(p_a\). Hence, by the binomial Chernoff bound, there exists \(c_a>0\) such that \[\mathbb{P}(B_k\mid\mathcal{F}^X_{kN}) \le \exp(-c_a\#\mathcal{G}_k),\] where \[B_k = \left\{ \#\left\{ u\in\mathcal{G}_k: (M^{(u)})^2\ge a \right\} < \frac{p_a}{2}\#\mathcal{G}_k \right\}.\]

For \(L\ge1\), set \[C_L = \left\{ \#\mathcal{G}_k\ge\beta^k \text{ for every } k\ge L \right\}.\] For \(k\ge L\), \[\mathbb{P}(B_k\cap C_L) \le \mathbb{E}\left[ \mathbf{1}_{\{\#\mathcal{G}_k\ge\beta^k\}} \mathbb{P}(B_k\mid\mathcal{F}^X_{kN}) \right] \le \exp(-c_a\beta^k).\] Thus \(\sum_{k=L}^{\infty}\mathbb{P}(B_k\cap C_L)<\infty\), and Borel–Cantelli implies \[\mathbb{P}(B_k\;\text{i.o.}\;\cap C_L)=0.\] Since, by 37 , \[S_{\mathrm{good}} \subseteq \bigcup_{L=1}^{\infty}C_L\] up to a null event, we have \[\mathbb{P}(B_k\;\text{i.o.}\;\cap S_{\mathrm{good}})=0.\] Consequently, on \(S_{\mathrm{good}}\), almost surely, for all sufficiently large \(k\), \[\#\left\{ u\in\mathcal{G}_k: (M^{(u)})^2\ge a \right\} \ge \frac{p_a}{2}\#\mathcal{G}_k \ge \frac{p_a}{2}\beta^k.\]

Along \(n=kN\), from 36 , we obtain \[\begin{align} 2^{skN}\Sigma_{kN} &= \sum_{\lvert u \rvert=kN} A_u^{(s)}\bigl(M^{(u)}\bigr)^2 \\ &\ge a\tau^k \#\left\{ u\in\mathcal{G}_k: (M^{(u)})^2\ge a \right\} \ge \frac{a p_a}{2}(\beta\tau)^k . \end{align}\] Since \(\beta\tau>1\), the right-hand side tends to \(+\infty\). Therefore \[\limsup_{n\to\infty}2^{sn}\Sigma_n=\infty\] on \(S_{\mathrm{good}}\), up to a null event. Since \(\mathbb{P}(S_{\mathrm{good}})>0\), ?? follows. ◻

Proof. Each alive vertex has offspring law \[N_X=\mathbf{1}_{\{X_0>0\}}+\mathbf{1}_{\{X_1>0\}},\] and offspring numbers at distinct vertices are independent. Hence \(\mathcal{T}_+\) is a Galton–Watson tree. By Lemma 11, \(\mathbb{E}N_X>1\); thus it is supercritical, and its population tends to infinity almost surely on its survival event.

If the alive tree becomes extinct, then \(L_u=0\) for all sufficiently deep vertices \(u\). Hence \(M_n=0\) eventually, and therefore \(M=0\). Thus \[\{M>0\}\subseteq\{\mathcal{T}_+\text{ survives}\}\] up to null events. ◻

Proof. Set \[Y_s=\limsup_{n\to\infty}2^{sn}\Sigma_n .\] By Proposition 16, \[p_s:=\mathbb{P}(Y_s=\infty)>0, \qquad q_s:=\mathbb{P}(Y_s<\infty)<1.\]

Fix \(m\ge0\). The branching identity yields \[2^{s(m+n)}\Sigma_{m+n} = \sum_{\lvert u \rvert=m}2^{sm}L_u^2 \left(2^{sn}\Sigma_n^{(u)}\right).\] Hence, if \(L_u>0\) and \(Y_s^{(u)}=\infty\), then \(Y_s=\infty\). Therefore \[\{Y_s<\infty\} \subseteq \bigcap_{\substack{\lvert u \rvert=m\\ L_u>0}} \{Y_s^{(u)}<\infty\}.\] Conditionally on \(\mathcal{F}_m^X\), the variables \(\{Y_s^{(u)}:\lvert u \rvert=m\}\) are independent copies of \(Y_s\), and hence \[\mathbb{P}(Y_s<\infty\mid\mathcal{F}_m^X) \le q_s^{Z_m^+}.\]

Let \(A=\{Y_s<\infty\}\) and \(B=\{M>0\}\). For \(k\ge1\), \[\mathbb{P}(A\cap B) \le \mathbb{P}(A\cap\{Z_m^+\ge k\}) + \mathbb{P}(B\cap\{Z_m^+<k\}) \le q_s^k+\mathbb{P}(B\cap\{Z_m^+<k\}).\] By Lemma 18, \(Z_m^+\to\infty\) almost surely on \(B\). Letting \(m\to\infty\), we obtain \[\mathbb{P}(A\cap B)\le q_s^k.\] Finally, let \(k\to\infty\). Since \(q_s<1\), \[\mathbb{P}(Y_s<\infty,\;M>0)=0.\] Thus \(Y_s=\infty\) almost surely on \(\{M>0\}\). ◻

Proof. If \(D_E(X)=1\), the claim follows from Lemma 6. Assume that \(D_E(X)<1\), and let \(s\in(D_E(X),1)\). By Proposition 17, \[\limsup_{n\to\infty}2^{sn}\Sigma_n=\infty \qquad\text{almost surely on }\{M>0\}.\] Hence, on this event, infinitely many \(n\) satisfy \(\Sigma_n\ge2^{-sn}\), and therefore \(\liminf_{n\to\infty} \frac{\log_2\Sigma_n}{-n} \le s\). Letting \(s\downarrow D_E(X)\) along a countable sequence proves the claim. ◻

Proof. The result follows by combining the lower and upper bounds on the intersection of their full-probability events. ◻

Proof. Endpoint transfer and Theorem 18 give \(\dim_2(\mu)=D_E(X)\); since \(\mu\) is nonzero and finite, Proposition 13 yields \(\dim_{\mathrm E}(\mu)=\dim_2(\mu)\). ◻

Proof. This follows immediately from \(\dim_{\mathrm E}(\mu)=D_E(X)\), the definition of energy dimension, and the monotonicity of \(I_t(\mu)\) in \(t\). ◻

Proof. For a continuous function \(e^{-2\pi i\xi x}\), the descendant decomposition of the limiting measure implies \[\int e^{-2\pi i\xi x}\,d\mu(x) = \sum_{\lvert u \rvert=m} L_u\int e^{-2\pi i\xi(a_u+2^{-m}y)}\,d\mu^{(u)}(y),\] which is precisely 40 .

The identity 41 follows from the same computation applied to Lebesgue measure, using the deterministic decomposition \[\mathcal{L}=\sum_{\lvert u \rvert=m}2^{-m}(S_u)_\#\mathcal{L}.\] Subtracting 41 from 40 with \(\xi=2^n\eta\), we arrive at 42 . ◻

Proof. Work on an event where \(M<\infty\), where 43 holds for all \(n,\eta,\eta'\), and where 44 holds eventually. Let \(2^n\le \xi\le 2^{n+1}\), write \(\xi=2^n\eta\) with \(\eta\in[1,2]\), and choose \(\eta_n\in\Gamma_n(1+\beta+\varepsilon)\) with \[\lvert\eta-\eta_n\rvert\le 2^{-(1+\beta+\varepsilon)n}.\] By Lemma 21, \[\lvert H_n(\eta)-H_n(\eta_n)\rvert \le C(M+1)2^n2^{-(1+\beta+\varepsilon)n} = C(M+1)2^{-(\beta+\varepsilon)n}.\] Hence, eventually, \[\lvert H_n(\eta) \rvert \le A_\beta2^{-\beta n} + C(M+1)2^{-(\beta+\varepsilon)n} \le C_\beta2^{-\beta n}.\] Moreover, \[\lvert\widehat\mu(2^n\eta)\rvert \le \lvert H_n(\eta) \rvert+\lvert\widehat{\mathcal{L}}(2^n\eta)\rvert \le C_\beta2^{-\beta n}+C2^{-n} \le C_\beta2^{-\beta n},\] where the last inequality uses \(\beta<1\). This proves the bound for positive frequencies, while negative frequencies follow from \(\widehat\mu(-\xi)=\overline{\widehat\mu(\xi)}\). ◻

Proof. For \(p\ge2\), the scalar Khintchine inequality implies \[\mathbb{E}_\varepsilon \left| \sum_{j\in J}\varepsilon_j c_j \right|^p \le (C\sqrt p)^p \left(\sum_{j\in J}\lvert c_j\rvert^2\right)^{p/2}\] for every finite complex family \((c_j)\). Applying this with \(p=r\), summing over \(g\in G\), and using Minkowski’s inequality in \(\ell^{r/2}(G)\), we obtain \[\mathbb{E}_\varepsilon \left\| \sum_{j\in J}\varepsilon_jz_j \right\|_{\ell^r(G)}^r \le (C\sqrt r)^r \left(\sum_{j\in J}\lVert z_j \rVert_{\ell^r(G)}^2\right)^{r/2}.\] Since \(q\le r\), monotonicity of \(L^p\)-norms and the assumption \(q<2\) yield \[\mathbb{E}_\varepsilon \left\| \sum_{j\in J}\varepsilon_jz_j \right\|_{\ell^r(G)}^q \le C_q r^{q/2} \left(\sum_{j\in J}\lVert z_j \rVert_{\ell^r(G)}^2\right)^{q/2} \le C_q r^{q/2} \sum_{j\in J}\lVert z_j \rVert_{\ell^r(G)}^q.\] This proves 45 . ◻

Proof. Condition on \(\mathcal{A}\). The coefficients \(a_j\) are then deterministic, and the conditional law of the independent mean-zero vectors \(Z_j\) is their original law. The standard symmetrization argument yields \[\mathbb{E}\left\|\sum_j a_jZ_j\right\|_{\ell^r(G)}^q \le \mathbb{E}\mathbb{E}_\varepsilon \left\|\sum_j\varepsilon_j a_j(Z_j-Z_j')\right\|_{\ell^r(G)}^q,\] where \((Z_j')\) is an independent copy of \((Z_j)\). Lemma 23 and \[\|a_j(Z_j-Z_j')\|_{\ell^r(G)}^q \le 2^{q-1}|a_j|^q \bigl(\|Z_j\|_{\ell^r(G)}^q+\|Z_j'\|_{\ell^r(G)}^q\bigr)\] then yield ?? . ◻

Proof. Apply Proposition 21 with \(a_u=L_u\) and \(\mathcal{A}=\mathcal{F}^X_m\) to obtain 46 . Taking expectations and using \(\mathbb{E}\sum_{\lvert u \rvert=m}L_u^q=\rho(q)^m\) from Lemma 3, we obtain 47 . ◻

Proof. For each \(n\), by balance, \(\mathbb{E}\mu_n=\mathcal{L}\), hence \(\mathbb{E}[\widehat{\mu_n}(\xi)]=\widehat{\mathcal{L}}(\xi)\). By Theorem 6, \(\mu_n\xrightarrow{\mathrm{w}}\mu\) almost surely, so \(\widehat{\mu_n}(\xi)\to\widehat\mu(\xi)\) almost surely. Energy-admissibility and Theorem 8 imply uniform integrability of the total mass martingale and \(\mathbb{E}M=1\). Since \(\lvert\widehat{\mu_n}(\xi)\rvert\le M_n\), the family \(\{\widehat{\mu_n}(\xi):n\ge0\}\) is uniformly integrable. Therefore convergence holds in \(L^1\), and \[\mathbb{E}[\widehat\mu(\xi)] = \lim_{n\to\infty}\mathbb{E}[\widehat{\mu_n}(\xi)] = \widehat{\mathcal{L}}(\xi).\] Thus \(\mathbb{E}[H_n(\eta)]=0\) by \(H_n(\eta)=\widehat\mu(2^n\eta)-\widehat{\mathcal{L}}(2^n\eta)\). The descendant statement follows because each descendant cascade has the same law as the original one. ◻

Proof. By Lemma 20, \(\lvert\widehat{\mathcal{L}}(2^{k-m}\eta)\rvert\le C2^{-(k-m)}\) on \([1,2]\), and \[\left\lvert \sum_{\lvert u \rvert=m} \bigl(L_u-2^{-m}\bigr)e^{-2\pi i a_u2^k\eta} \right\rvert \le \sum_{\lvert u \rvert=m}L_u+\sum_{\lvert u \rvert=m}2^{-m} = M_m+1.\] Thus \[\lVert R_{k,m} \rVert_{\ell^r(G)} \le C2^{-(k-m)}(M_m+1)(\#G)^{1/r},\] and hence \[\mathbb{E}\lVert R_{k,m} \rVert_{\ell^r(G)}^q \le C2^{-q(k-m)}(\#G)^{q/r}\mathbb{E}[(M_m+1)^q].\] By Lemma 9 and \(\rho(q)<1\), \(\sup_{m\ge0}\mathbb{E}[M_m^q]<\infty\), hence also \(\sup_{m\ge0}\mathbb{E}[(M_m+1)^q]<\infty\). This proves 50 . ◻

Proof. Since \(\rho(q)<1\), the preceding moment estimate yields \[\mathbb{E}(M+1)^q=O_{\alpha,q}(1).\] Together with the pointwise bound \(|H_k(\eta)|\le M+1\) for \(\eta\in[1,2]\), this yields \[\mathbb{E}\|H_k\|_{\ell^{r_n}(G)}^q \le (\#G)^{q/r_n}\mathbb{E}(M+1)^q =O_{\alpha,q}(1).\] ◻

Proof. Fix \(G\in\mathcal{G}_n(\alpha)\). By the mesoscopic decomposition, \[H_k=U_{k,m}+R_{k,m}\] on \(G\). Hence \[\lVert H_k\rVert_{\ell^{r_n}(G)}^q \le 2^{q-1} \lVert U_{k,m}\rVert_{\ell^{r_n}(G)}^q + 2^{q-1} \lVert R_{k,m} \rVert_{\ell^{r_n}(G)}^q.\] We estimate the two terms separately.

For the random descendant term \(U_{k,m}\), define \[Z_u(\eta) = e^{-2\pi i a_u2^k\eta}H_{k-m}^{(u)}(\eta), \qquad \eta\in G.\] By Lemma 25, conditional on \(\mathcal{F}^X_m\), the family \((Z_u)_{\lvert u \rvert=m}\) is independent, mean zero, and independent of \(\mathcal{F}^X_m\). Moreover, \[\lVert Z_u\rVert_{\ell^{r_n}(G)} = \lVert H_{k-m}^{(u)}\rVert_{\ell^{r_n}(G)}.\] Applying Corollary 7, with coefficients \(L_u\), we obtain \[\mathbb{E}\left[ \lVert U_{k,m}\rVert_{\ell^{r_n}(G)}^q \middle| \mathcal{F}^X_m \right] \le C_q r_n^{q/2} \left(\sum_{\lvert u \rvert=m}L_u^q\right) \mathbb{E}\lVert H_{k-m}\rVert_{\ell^{r_n}(G)}^q.\] Since \(G\in\mathcal{G}_n(\alpha)\), \[\mathbb{E}\lVert H_{k-m}\rVert_{\ell^{r_n}(G)}^q \le A_{k-m,n}.\] Taking expectations and using Lemma 3, \[\mathbb{E}\lVert U_{k,m}\rVert_{\ell^{r_n}(G)}^q \le C_q r_n^{q/2} A_{k-m,n} \mathbb{E}\left[\sum_{\lvert u \rvert=m}L_u^q\right] = C_q r_n^{q/2}\rho(q)^m A_{k-m,n}.\]

For the forcing term, Lemma 26 yields \[\mathbb{E}\lVert R_{k,m} \rVert_{\ell^{r_n}(G)}^q \le C_q2^{-q(k-m)}(\#G)^{q/r_n}.\] By 51 and \(r_n\ge n\) up to finitely many small \(n\)’s, \[(\#G)^{q/r_n}\le C_{q,\alpha}.\] Thus \[\mathbb{E}\lVert R_{k,m} \rVert_{\ell^{r_n}(G)}^q \le C_{q,\alpha}2^{-q(k-m)}.\]

Combining the estimates for \(U_{k,m}\) and \(R_{k,m}\), and then taking the supremum over \(G\in\mathcal{G}_n(\alpha)\), proves \[A_{k,n} \le C_{q,\alpha}r_n^{q/2}\rho(q)^mA_{k-m,n} + C_{q,\alpha}2^{-q(k-m)}.\] This is ?? . ◻

Proof. Choose \(\gamma\in(\beta,1)\), and set \(k'=\lceil\gamma k\rceil\). Since \(\rho(q)=2^{-q\beta_X(q)}\), Proposition 23 yields a constant \(C_0<\infty\) such that, whenever \(k-k'\ge1\), \[\label{eq:ladder-recursion-proof-paper} A_{k,n} \le C_0 r_n^{q/2}2^{-q\beta_X(q)(k-k')}A_{k',n} + C_0 2^{-qk'}.\tag{53}\] Here \(A_{k,n}=A_{k,n}^{(q,\alpha_{\mathrm{gr}})}\), and we use \(r_n^{q/2}\le Cn^{q/2}\), with \(C\) absorbed into \(C_0\).

Since \[k-k'=k-\lceil\gamma k\rceil\ge (1-\gamma)k-1,\] we have \(k-k'\ge c_\gamma k\) for all sufficiently large \(k\), where \(c_\gamma=(1-\gamma)/2\). Choose \(K>0\) such that \[\label{eq:K-choice-ladder-paper} q(\beta_X(q)-\beta)c_\gamma K>\frac{q}{2}+2.\tag{54}\] Then, for \(k\ge K\log_2 n\) and all sufficiently large \(n\), \(k-k'\ge c_\gamma K\log_2 n\).

We prove ?? by induction on \(k\), with \(n\) fixed and sufficiently large. If \(0\le k\le K\log_2 n\), then Lemma 27 implies \(A_{k,n}\le C_1\). Choosing \(\tau>q\beta K\), and then increasing \(A\) and \(n_0\), we obtain \[A_{k,n}\le A n^\tau2^{-q\beta k} \qquad (0\le k\le K\log_2 n,\;n\ge n_0).\]

Assume now that \(k>K\log_2 n\) and that the bound holds at all smaller scales. For \(n\) sufficiently large, \(k-k'\ge1\) and \(k'<k\). By 53 and the induction hypothesis at \(k'\), \[\begin{align} A_{k,n} &\le C_0A n^{\tau+q/2} 2^{-q\beta_X(q)(k-k')}2^{-q\beta k'} + C_0 2^{-qk'} \\ &= C_0A n^{\tau+q/2} 2^{-q\beta k}2^{-q(\beta_X(q)-\beta)(k-k')} + C_0 2^{-qk'} . \end{align}\] Moreover, by 54 , \[n^{q/2}2^{-q(\beta_X(q)-\beta)(k-k')} \le n^{q/2-q(\beta_X(q)-\beta)c_\gamma K} \le n^{-2}\] for all sufficiently large \(n\). Thus, after increasing \(n_0\), the recursive term is at most \[C_0A n^\tau2^{-q\beta k}n^{-2} \le \frac{1}{2} A n^\tau2^{-q\beta k}.\] Also, \[2^{-qk'}\le 2^{-q\gamma k}\le 2^{-q\beta k}, \qquad C_0 2^{-qk'} \le \frac{1}{2} A n^\tau2^{-q\beta k},\] provided \(A\) is chosen sufficiently large. Combining the two bounds closes the induction and proves ?? . The terminal estimate ?? is the special case \(k=n\). ◻

Proof. Choose \(\beta<\beta'<\beta_X(q)\). By Proposition 25 and the fact that \(\Gamma_n(\alpha_{\mathrm{gr}})\in\mathcal{G}_n(\alpha_{\mathrm{gr}})\), \[\mathbb{E}\left[ \lVert H_n \rVert_{\ell^{r_n}(\Gamma_n(\alpha_{\mathrm{gr}}))}^q \right] \le A n^\tau 2^{-q\beta' n}\] for all sufficiently large \(n\). If \(\max_{\eta\in\Gamma_n(\alpha_{\mathrm{gr}})}\lvert H_n(\eta) \rvert>2^{-\beta n}\), then \(\lVert H_n \rVert_{\ell^{r_n}(\Gamma_n(\alpha_{\mathrm{gr}}))}^q>2^{-q\beta n}\). By Markov’s inequality, \[\mathbb{P}\left( \max_{\eta\in\Gamma_n(\alpha_{\mathrm{gr}})} \lvert H_n(\eta) \rvert>2^{-\beta n} \right) \le A n^\tau 2^{-q(\beta'-\beta)n},\] which is summable in \(n\). The Borel–Cantelli lemma proves 55 . ◻

Proof. By Lemma 28, \(\beta<1\). Choose \(\varepsilon>0\) and set \(\alpha_{\mathrm{gr}}=1+\beta+\varepsilon\). By Corollary 8, almost surely, \[\max_{\eta\in\Gamma_n(\alpha_{\mathrm{gr}})} \lvert H_n(\eta) \rvert \le 2^{-\beta n}\] eventually. Lemma 22 upgrades this bound to 56 . ◻

Proof. If \(D_E(X)=0\), there is nothing to prove. Fix \(0<\sigma<D_E(X)\). By Proposition 11, there exists \(q\in(1,2)\) such that \[\rho(q)<1 \qquad \text{and} \qquad -\frac{2}{q}\log_2\rho(q)>\sigma,\] equivalently \(2\beta_X(q)>\sigma\). Choose \(\sigma/2<\beta<\beta_X(q)\). By Theorem 26, almost surely there is \(C_\beta(\omega)<\infty\) such that, for all sufficiently large \(\lvert\xi\rvert\), \[\lvert\widehat\mu(\xi)\rvert \le C_\beta(\omega)\lvert\xi\rvert^{-\beta} \le C_\beta(\omega)\lvert\xi\rvert^{-\sigma/2}.\] Thus 57 holds with \(C_\sigma(\omega)=C_\beta(\omega)\).

Intersecting the full-probability events over rational \(\sigma\in(0,D_E(X))\), we obtain \(\dim_{\mathrm F}(\mu)\ge D_E(X)\) on \(\{M>0\}\). ◻

Proof. On \(\{M>0\}\), Theorem 19 yields \[\dim_{\mathrm E}(\mu)=\dim_2(\mu)=D_E(X),\] while Theorem 27 yields \(\dim_{\mathrm F}(\mu)\ge D_E(X)\). Since \(\mu\) is then a nonzero finite Borel measure, Proposition 5 yields \(\dim_{\mathrm F}(\mu)\le\dim_{\mathrm E}(\mu)\). Hence \[\dim_{\mathrm F}(\mu)=\dim_{\mathrm E}(\mu)=\dim_2(\mu)=D_E(X)\] almost surely on \(\{M>0\}\). ◻

Proof. The vector law is balanced. Moreover, \[\mathbb{E}\bigl[X_0\log_2^+X_0+X_1\log_2^+X_1\bigr]<\infty\] is equivalent to \[\mathbb{E}[W\log_2^+W]<\infty,\] and \[\mathbb{E}[X_0\log_2 X_0+X_1\log_2 X_1] = \mathbb{E}[W\log_2 W]-1.\] Thus the scalar assumptions imply energy-admissibility of the vector law.

By Corollary 2, \[\rho(q)=2^{1-q}\mathbb{E}[W^q], \qquad D_E(X) = \sup_{1<q<2} \max\left\{ 0,\, 2-\frac{2}{q}\bigl(1+\log_2\mathbb{E}[W^q]\bigr) \right\} = D^+(W).\] Theorem 29 then yields \[\dim_{\mathrm F}(\mu)=\dim_{\mathrm E}(\mu)=\dim_2(\mu)=D_E(X)=D^+(W)\] almost surely on \(\{M>0\}\), and its decay formulation implies the asserted decay estimate. ◻

Proof. If \(\mathbb{E}[W^q]=\infty\) for every \(q>1\), then every term in the definition 61 is interpreted as \(0\). Hence \(D^+(W)=0\), and the conclusion follows from Corollary 10. ◻

Proof. The equivalences are contained in Proposition 11. If \(D_E(X)>0\), then Theorem 29 implies that \(\dim_{\mathrm F}(\mu)=D_E(X)>0\) almost surely on \(\{M>0\}\). ◻

Proof. Choose \(q>1\) such that \(\mathbb{E}[W^q]<\infty\) and \(\tau(q)/q>\beta\). Then \(\tau(q)>0\), equivalently \(2^{1-q}\mathbb{E}[W^q]<1\), and the standard terminal-mass moment criterion implies \(\mathbb{E}[Y^q]<\infty\). By 64 , the independence of \(Q_v\) and \(Y^{(v)}\), and the identity \(Y^{(v)}\stackrel{d}=Y\), \[\mathbb{E}[\mu_\circ(\mathcal{I}_v)^q] = 2^{-q\lvert v \rvert}\mathbb{E}[Q_v^q]\mathbb{E}[Y^q] = 2^{-q\lvert v \rvert}\bigl(\mathbb{E}[W^q]\bigr)^{\lvert v \rvert}\mathbb{E}[Y^q].\] Thus \[\mathbb{E}\sum_{\lvert v \rvert=n}\mu_\circ(\mathcal{I}_v)^q = \mathbb{E}[Y^q]\,2^{-n\tau(q)}.\] Therefore \[\mathbb{P}\left(\max_{\lvert v \rvert=n}\mu_\circ(\mathcal{I}_v)>2^{-\beta n}\right) \le 2^{\beta q n} \mathbb{E}\sum_{\lvert v \rvert=n}\mu_\circ(\mathcal{I}_v)^q \le C2^{-n(\tau(q)-\beta q)}.\] Since \(\tau(q)-\beta q>0\), the right-hand side is summable. The Borel–Cantelli lemma implies \(\max_{\lvert v \rvert=n}\mu_\circ(\mathcal{I}_v)\le2^{-\beta n}\) for all sufficiently large \(n\), almost surely. Enlarging the random constant to cover the finitely many initial generations proves 65 . ◻

Proof. If \(2^{-(n+1)}<r\le2^{-n}\), then \(B(x,r)\cap\mathbb{S}^1\) meets at most a bounded number of level-\(n\) dyadic arcs. Hence \[\nu(B(x,r))\le C'2^{-\beta n}\le C''r^\beta.\] For \(x\in\operatorname{spt}\nu\), divide \(\log_2\nu(B(x,r))\le \log_2 C''+\beta\log_2 r\) by \(\log_2 r<0\) and let \(r\downarrow0\) to obtain the claim. ◻

Proof. Apply Biggins’ Chernoff theorem for branching random walks [13] to the displacement \(-\ln W\), with edges for which \(W=0\) treated as killed. The one-generation transform is \[m(0)=2\mathbb{P}(W>0),\qquad m(\theta)=2\mathbb{E}(W^\theta)\quad(\theta>0),\] with the value \(+\infty\) allowed. By Lemma 31, \(m(0)>1\), and \(m(1)=2\mathbb{E}W=2<\infty\). Since \(N_n(a)\) counts the particles with total displacement at most \(-an\ln2\), Biggins’ theorem yields \[\lim_{n\to\infty} \bigl(\mathbb{E} N_n(a)\bigr)^{1/n} = \inf_{\theta\ge0} e^{-a\theta\ln2}m(\theta).\] Taking \(\log_2\) yields \[\Lambda(a)= \inf_{q\ge0} \left\{ 1+\log_2\mathbb{E}(W^q)-qa \right\},\] with \(W^0=\mathbf{1}_{\{W>0\}}\) and with the stated extended-value convention.

For \(q>0\), define \[\Psi(q)=\frac{1+\log_2\mathbb{E}(W^q)}{q},\] with value \(+\infty\) if \(\mathbb{E}(W^q)=+\infty\). We first show that \[\inf_{q>0}\Psi(q)=1-A_{\mathrm{loc}}(W).\] For \(0<q\le1\), put \(\varphi(q)=\log_2\mathbb{E}(W^q)\). Then \(\varphi\) is convex, \(\varphi(1)=0\), and \(\varphi'_-(1)=\mathbb{E}(W\log_2 W)\). Hence, for \(0<q<1\), \[\varphi(q)\ge(q-1)\mathbb{E}(W\log_2 W).\] Since the minimal Kahane–Peyrière regime ensures \(\mathbb{E}(W\log_2 W)<1\), it follows that \(\Psi(q)>1\) for \(0<q<1\), while \(\Psi(1)=1\). Thus \[\inf_{0<q\le1}\Psi(q)=1.\] Together with the definition of \(A_{\mathrm{loc}}(W)\), this implies \[1-A_{\mathrm{loc}}(W) = \min\left\{1,\inf_{q>1}\Psi(q)\right\} = \inf_{q>0}\Psi(q).\]

Fix \(a<1-A_{\mathrm{loc}}(W)\), and set \(\eta=1-A_{\mathrm{loc}}(W)-a>0\). For \(q>0\), \[1+\log_2\mathbb{E}(W^q)-qa = q(\Psi(q)-a) \ge q\eta\] in the extended sense. Also, \(\mathbb{E}(W^q)\to\mathbb{P}(W>0)\) as \(q\downarrow0\), and Lemma 31 ensures that \(1+\log_2\mathbb{P}(W>0)>0\). Hence there exist \(q_0,c_0>0\) such that \[1+\log_2\mathbb{E}(W^q)-qa\ge c_0 \qquad(0\le q\le q_0),\] whereas for \(q\ge q_0\) the previous bound yields the lower bound \(q_0\eta\). Therefore \[\Lambda(a)\ge\min\{c_0,q_0\eta\}>0.\]

Finally, for each \(q\ge0\), the map \(a\mapsto 1+\log_2\mathbb{E}(W^q)-qa\) is affine; hence \(\Lambda\) is concave. On \((-\infty,1-A_{\mathrm{loc}}(W))\), the positivity just proved and the \(q=0\) term imply \[0<\Lambda(a)\le1+\log_2\mathbb{P}(W>0)<\infty.\] Thus \(\Lambda\) is finite and concave on this open interval, and is therefore continuous there. ◻

Proof. Fix \(0<\delta<\beta-A_{\mathrm{loc}}(W)\), and set \(a=1-\beta+\delta<1-A_{\mathrm{loc}}(W)\). By the descendant construction in Subsection 7.1, \(Y^{(v)}=\mu_\circ^{(v)}(\mathbb{S}^1)\stackrel{d}{=}\mu_\circ(\mathbb{S}^1)\), and \[Y^{(v)} = \frac{1}{2} W_{v0}Y^{(v0)} + \frac{1}{2} W_{v1}Y^{(v1)}, \qquad \mu_\circ(\mathcal{I}_v)=2^{-\lvert v \rvert}Q_vY^{(v)}\] by the first-step descendant decomposition and 64 .

By Theorem 32, \(\mathbb{P}(Y^{(u)}>0)>0\), and since \(W_u\) is independent of \(Y^{(u)}\) with \(\mathbb{P}(W_u>0)>0\), choose \(c>0\) such that \[p_c:=\mathbb{P}(W_uY^{(u)}\ge 2c)>0,\] independently of \(u\).

Let \(0<\varepsilon<\Lambda(a)/2\). Choose \(h\ge1\) such that \[\frac{1}{h}\log_2\mathbb{E} N_h(a)>\Lambda(a)-\varepsilon, \qquad \frac{1}{h}\log_2\left(\frac{p_c}{2}\right)>-\varepsilon.\] Then \[m_h:=\frac{p_c}{2}\mathbb{E} N_h(a) > 2^{h(\Lambda(a)-2\varepsilon)} > 1.\]

For \(v,u\in\mathcal{T}\), set \[Q_u^{(v)}=\prod_{j=1}^{\lvert u \rvert}W_{v(u|j)}, \qquad Q_{\varnothing}^{(v)}=1.\] If \(Q_v>0\), then \(Q_u^{(v)}=Q_{vu}/Q_v\). Define \[\mathcal{C}(v) = \left\{ v0u: u\in\{0,1\}^{h-1},\; W_{v1}Y^{(v1)}\ge 2c,\; Q_{0u}^{(v)}\ge 2^{ah} \right\}, \qquad \xi_v=\#\mathcal{C}(v).\] The side-witness event \(\{W_uY^{(u)}\ge 2c\}\) is independent of the \(v0\)-subtree and has probability \(p_c\). By symmetry, \[\mathbb{E}\xi_v = p_c\, \mathbb{E}\#\{u\in\{0,1\}^{h-1}:Q_{0u}^{(v)}\ge 2^{ah}\} = \frac{p_c}{2}\mathbb{E} N_h(a) = m_h>1.\]

Starting from \(v\), set \[Z_0^{(v)}=\{v\}, \qquad Z_{k+1}^{(v)}=\bigcup_{w\in Z_k^{(v)}}\mathcal{C}(w).\] Then \((\#Z_k^{(v)})_{k\ge0}\) is a Galton–Watson process with offspring law \(\xi_v\): different current particles use disjoint subtrees; if \(z\in\mathcal{C}(w)\), the event \(z\in\mathcal{C}(w)\) uses only the finite path from \(w\) to \(z\) and the side subtree \(w1\), and no weights strictly below \(z\); the laws are identical by the tree-indexed i.i.d. construction. Its mean is \(m_h>1\). Therefore \[\rho_h := \mathbb{P}\left( Z_k^{(\varnothing)}\neq\varnothing \text{ for every }k\ge0 \right) >0.\]

We next show that, on non-extinction, some alive vertex starts a surviving witnessed block process. Let \[\mathcal{F}_n=\sigma(W_w:1\le \lvert w \rvert\le n), \qquad H_v=\bigcap_{k=0}^{\infty}\{Z_k^{(v)}\neq\varnothing\}, \quad v\in\{0,1\}^{n}.\] Conditional on \(\mathcal{F}_n\), the events \(H_v\) are independent and \(\mathbb{P}(H_v\mid\mathcal{F}_n)=\rho_h\). Put \[B_n = \bigcap_{v\in\{0,1\}^{n}} \left( \{Q_v=0\}\cup H_v^c \right), \qquad G=\bigcap_{n=0}^{\infty}B_n.\] Then \[\mathbb{P}(B_n\mid\mathcal{F}_n)=(1-\rho_h)^{A_n}, \qquad \mathbb{E}(\mathbf{1}_G\mid\mathcal{F}_n) \le (1-\rho_h)^{A_n}.\] By Lemma 31, \(A_n\to\infty\) almost surely on \(\{\mu_\circ(\mathbb{S}^1)>0\}\), while \(\mathbb{E}(\mathbf{1}_G\mid\mathcal{F}_n)\to\mathbf{1}_G\) almost surely by martingale convergence. Hence \(G\) has probability zero on the non-extinction event. Thus, almost surely on the non-extinction event, there is an alive vertex \(v_0\) such that \(Q_{v_0}>0\) and \(Z_k^{(v_0)}\neq\varnothing\) for every \(k\ge0\).

Define \[\partial\mathcal{T}_{v_0}^{\mathrm{wit}} := \left\{ (w_k)_{k=0}^{\infty}: w_0=v_0,\; w_{k+1}\in\mathcal{C}(w_k)\;\text{for all }k\ge0 \right\},\] with \(h\)-block ultrametric \[d_{\mathrm{tree}}(\gamma,\gamma') = 2^{-h\lvert\gamma\wedge\gamma'\rvert_{\mathrm{GW}}},\] where \(\lvert\gamma\wedge\gamma'\rvert_{\mathrm{GW}}\) denotes the common initial block length. Let \(\Pi_{v_0}\) be the dyadic coding map \[\Pi_{v_0}(\gamma) \in \bigcap_{k=0}^{\infty}\overline{\mathcal{I}_{w_k}}.\] If \(D\) is the countable set of dyadic endpoints on \(\mathbb{S}^1\), set \[K_{v_0} := \Pi_{v_0}\bigl(\partial\mathcal{T}_{v_0}^{\mathrm{wit}}\bigr)\setminus D.\] By the Galton–Watson boundary dimension theorem and the metric-change remark [45], on survival, \[\dim_{\mathrm H}K_{v_0} = \frac{\log_2 m_h}{h} = \frac{1}{h}\log_2\mathbb{E} N_h(a) + \frac{1}{h}\log_2\left(\frac{p_c}{2}\right) > \Lambda(a)-2\varepsilon.\] Here deleting \(D\) does not change Hausdorff dimension, and the symbolic \(h\)-block metric and the Euclidean metric are dimensionally equivalent because each \(h\)-block cylinder projects to a dyadic arc of comparable length, while each arc at that scale intersects only \(O_h(1)\) such arcs.

It remains to show that \(K_{v_0}\subseteq E_{\beta,C}(\mu_\circ)\). Let \(x\in K_{v_0}\). Then \(x\notin D\) corresponds to a chain \[v_0\prec v_1\prec v_2\prec\cdots, \qquad v_{k+1}\in\mathcal{C}(v_k), \qquad \lvert v_k \rvert=\lvert v_0 \rvert+kh,\] with \(\mathcal{I}_{\lvert v_k \rvert}(x)=\mathcal{I}_{v_k}\). The relative-product condition yields \[\frac{Q_{v_{k+1}}}{Q_{v_k}}\ge 2^{ah}, \qquad Q_{v_k}\ge Q_{v_0}2^{akh},\] and the side witness ensures \[Y^{(v_k)} = \frac{1}{2}W_{v_k0}Y^{(v_k0)} + \frac{1}{2}W_{v_k1}Y^{(v_k1)} \ge c.\] Therefore \[\mu_\circ(\mathcal{I}_{v_k}) = 2^{-\lvert v_k \rvert}Q_{v_k}Y^{(v_k)} \ge c\,2^{-\lvert v_k \rvert}Q_{v_0}2^{akh}.\] Since \(\lvert v_k \rvert=\lvert v_0 \rvert+kh\) and \(a=1-\beta+\delta\), \[\frac{\mu_\circ(\mathcal{I}_{v_k})}{2^{-\beta\lvert v_k \rvert}} \ge cQ_{v_0}2^{(\beta-1)\lvert v_0 \rvert}2^{\delta kh} \to+\infty.\] As \(C(\omega)<\infty\) on non-extinction, for all large \(k\), \[\mu_\circ(\mathcal{I}_{\lvert v_k \rvert}(x)) = \mu_\circ(\mathcal{I}_{v_k}) \ge C(\omega)2^{-\beta\lvert v_k \rvert}.\] The cylinders \(\mathcal{I}_{v_k}\) shrink to \(x\) and have positive \(\mu_\circ\)-mass for all large \(k\), so \(x\in\operatorname{spt}\mu_\circ\). Thus \(K_{v_0}\subseteq E_{\beta,C}(\mu_\circ)\).

Consequently, \[\dim_{\mathrm H}E_{\beta,C}(\mu_\circ) \ge \dim_{\mathrm H}K_{v_0} > \Lambda(1-\beta+\delta)-2\varepsilon.\] Letting \(\varepsilon\downarrow0\), then \(\delta\downarrow0\), and using the continuity from Lemma 32, we obtain \[\dim_{\mathrm H}E_{\beta,C}(\mu_\circ) \ge \Lambda(1-\beta)>0.\] ◻

Proof. The lower bound follows from Theorem 35. We prove the reverse inequality.

Fix \(\gamma>A_{\mathrm{loc}}(W)\), and set \[E_{\gamma}(\mu_\circ) := \left\{ x\in\operatorname{spt}\mu_\circ: \mu_\circ(\mathcal{I}_{n_k}(x)) \ge 2^{-\gamma n_k} \text{ for some }n_k\uparrow\infty \right\}.\] For \(n\ge1\), let \[H_n(\gamma) = \left\{ v\in\{0,1\}^{n}: \mu_\circ(\mathcal{I}_v)\ge 2^{-\gamma n} \right\}.\] Then \[E_\gamma(\mu_\circ) \subseteq \limsup_{n\to\infty} \bigcup_{v\in H_n(\gamma)}\mathcal{I}_v.\]

Let \[\mathcal{A} = (0,1] \cup \left\{ q>1: \mathbb{E}(W^q)<\infty,\; 2^{1-q}\mathbb{E}(W^q)<1 \right\}.\] For \(q\in\mathcal{A}\), the \(L^q\)-boundedness theorem for Mandelbrot cascade martingales implies that \(\mathbb{E}(Y^q)<\infty\), where \(Y=\mu_\circ(\mathbb{S}^1)\). Put \[\Phi_\gamma(q)=1+\log_2\mathbb{E}(W^q)-q(1-\gamma).\] Since \[\mu_\circ(\mathcal{I}_v)=2^{-n}Q_vY^{(v)},\] with \(Y^{(v)}\) an independent copy of \(Y\), independent of \(Q_v\), \[\mathbb{E}\sum_{\lvert v \rvert=n}\mu_\circ(\mathcal{I}_v)^q = \mathbb{E}(Y^q)\, 2^{n(1-q)} \bigl(\mathbb{E}(W^q)\bigr)^n.\] For \(s>0\), \[\mathbf{1}_{\{\mu_\circ(\mathcal{I}_v)\ge 2^{-\gamma n}\}} \le 2^{q\gamma n}\mu_\circ(\mathcal{I}_v)^q, \qquad \operatorname{diam}(\mathcal{I}_v)\le C_1 2^{-n},\] and therefore \[\mathbb{E} \sum_{v\in H_n(\gamma)} \operatorname{diam}(\mathcal{I}_v)^s \le C_1^s\mathbb{E}(Y^q)\, 2^{-n(s-\Phi_\gamma(q))}.\] If \(s>\Phi_\gamma(q)\), summing over \(n\) and applying Fubini yields \[\sum_{n=1}^{\infty} \sum_{v\in H_n(\gamma)} \operatorname{diam}(\mathcal{I}_v)^s <\infty\] almost surely. The limsup cover then implies \[\dim_{\mathrm H}E_\gamma(\mu_\circ)\le \Phi_\gamma(q), \qquad q\in\mathcal{A}.\]

It remains to remove the restriction \(q\in\mathcal{A}\). If \(q>1\), \(\mathbb{E}(W^q)<\infty\), and \(2^{1-q}\mathbb{E}(W^q)\ge1\), then \[\Phi_\gamma(q) = q\gamma+1-q+\log_2\mathbb{E}(W^q) \ge q\gamma>\gamma=\Phi_\gamma(1),\] so such \(q\) cannot decrease the infimum; if \(\mathbb{E}(W^q)=+\infty\), the value is \(+\infty\). Since \(q=1\in\mathcal{A}\) and \(\mathbb{E}(W^q)\to\mathbb{P}(W>0)\) as \(q\downarrow0\), the endpoint \(q=0\) may also be included. Hence \[\inf_{q\in\mathcal{A}}\Phi_\gamma(q) = \inf_{q\ge0} \left\{ 1+\log_2\mathbb{E}(W^q)-q(1-\gamma) \right\}.\] By Lemma 32, the last infimum is \(\Lambda(1-\gamma)\). Thus \[\dim_{\mathrm H}E_\gamma(\mu_\circ)\le \Lambda(1-\gamma)\] almost surely.

Now let \(C\) be as in the statement. For each rational \(\eta>0\), on \(\{0<C(\omega)<\infty\}\), \[C(\omega)2^{-\beta n}\ge 2^{-(\beta+\eta)n}\] eventually, and hence \[E_{\beta,C}(\mu_\circ)\subseteq E_{\beta+\eta}(\mu_\circ).\] The deterministic upper bound with \(\gamma=\beta+\eta\) yields \[\dim_{\mathrm H}E_{\beta,C}(\mu_\circ) \le \Lambda(1-\beta-\eta)\] almost surely. Taking a countable intersection over rational \(\eta>0\), letting \(\eta\downarrow0\), and using the continuity of \(\Lambda\) from Lemma 32, we obtain \[\dim_{\mathrm H}E_{\beta,C}(\mu_\circ)\le\Lambda(1-\beta).\] Together with the lower bound, this proves the equality. The variational formula follows from Lemma 32. ◻

Proof. We first prove the lower bound. If \(A_{\mathrm{loc}}(W)=0\), the claim is immediate, since, for every finite positive measure \(\nu\), \[\inf_{x\in\operatorname{spt}\nu} \liminf_{r\downarrow0} \frac{\log_2\nu(B(x,r))}{\log_2 r} \ge0.\] Assume \(A_{\mathrm{loc}}(W)>0\), and choose \(0<\beta_j<A_{\mathrm{loc}}(W)\) with \(\beta_j\uparrow A_{\mathrm{loc}}(W)\). For each \(j\), Lemma 29 and Lemma 30 imply, on a full-probability event, that \[\liminf_{r\downarrow0} \frac{\log_2\mu_\circ(B(x,r))}{\log_2 r} \ge \beta_j \qquad \text{for every }x\in\operatorname{spt}\mu_\circ.\] Intersecting these countably many full-probability events yields \(\alpha_{\min}(\mu_\circ)\ge\beta_j\) for all \(j\), hence \(\alpha_{\min}(\mu_\circ)\ge A_{\mathrm{loc}}(W)\) almost surely on \(\mathcal{S}_\circ\).

For the upper bound, choose \(\beta_j>A_{\mathrm{loc}}(W)\) with \(\beta_j\downarrow A_{\mathrm{loc}}(W)\). By Theorem 36, \[\dim_{\mathrm H}E_{\beta_j,1}(\mu_\circ)=\Lambda(1-\beta_j)>0\] almost surely on \(\mathcal{S}_\circ\). Hence, after a countable intersection, for every \(j\) there exist \(x_j\in\operatorname{spt}\mu_\circ\) and \(n_{j,k}\uparrow\infty\) such that \[\mu_\circ(\mathcal{I}_{n_{j,k}}(x_j)) \ge 2^{-\beta_j n_{j,k}}.\] For fixed \(j\), \(\mathcal{I}_{n_{j,k}}(x_j)\) is contained in a Euclidean ball \(B(x_j,C2^{-n_{j,k}})\) with deterministic \(C<\infty\). Set \(r_{j,k}=C2^{-n_{j,k}}\). Then \[\mu_\circ(B(x_j,r_{j,k})) \ge 2^{-\beta_j n_{j,k}} = C^{-\beta_j}r_{j,k}^{\beta_j},\] and hence \[\log_2\mu_\circ(B(x_j,r_{j,k})) \ge \beta_j\log_2 r_{j,k}-\beta_j\log_2 C.\] Since \(\log_2 r_{j,k}<0\), \[\frac{\log_2\mu_\circ(B(x_j,r_{j,k}))}{\log_2 r_{j,k}} \le \beta_j-\frac{\beta_j\log_2 C}{\log_2 r_{j,k}}.\] Letting \(k\to\infty\) yields \[\liminf_{r\downarrow0} \frac{\log_2\mu_\circ(B(x_j,r))}{\log_2 r} \le \beta_j.\] Since \(x_j\in\operatorname{spt}\mu_\circ\), \(\alpha_{\min}(\mu_\circ)\le\beta_j\). Letting \(j\to\infty\) yields \(\alpha_{\min}(\mu_\circ)\le A_{\mathrm{loc}}(W)\) almost surely on \(\mathcal{S}_\circ\). Combining the two bounds proves 66 . ◻

Proof. Let \(x_0=f(t_0)\), and pull \(\nu\) back to \([0,1)\); we denote the resulting measure again by \(\nu\). Set \[a(t)=f(t)\cdot x_0=\cos 2\pi(t-t_0).\] For \(t,u\) sufficiently close to \(t_0\), \[\lvert a(t)-a(t_0)\rvert\le C\lvert t-t_0\rvert^2, \qquad \lvert a(u)-a(t_0)\rvert\le C\lvert u-t_0\rvert^2.\] Let \(\psi(R)=e^{-\pi R^2}\). Its Fourier transform is again \(e^{-\pi\rho^2}\), which is positive and bounded below in a neighborhood of the origin. Expanding the square and applying Fubini, \[\begin{align} \int_{\mathbb{R}}\lvert\widehat\nu(Rx_0)\rvert^2\psi(R/\Lambda)\,dR &= \int_{\mathbb{R}}\iint e^{-2\pi iR(a(t)-a(u))}\,d\nu(t)d\nu(u)\,\psi(R/\Lambda)\,dR\\ &= \Lambda\iint e^{-\pi\Lambda^2(a(t)-a(u))^2}\,d\nu(t)d\nu(u). \end{align}\] If \(t,u\) both lie in a sufficiently small arc of length \(cr\) around \(t_0\), then \[\Lambda \lvert a(t)-a(u)\rvert\le C r^{-2}r^2\le1\] after choosing \(c\) sufficiently small. On this set the Gaussian factor is bounded below by a positive absolute constant. Restricting the double integral to the square of this arc, and using the comparability of chord distance and arclength at small scales, proves 67 . ◻

Proof. Let \(t\) be an admissible Fourier decay exponent for \(\nu\), and assume \(0<t<1\). Thus \[\lvert\widehat\nu(\xi)\rvert\le C_t(1+\lvert\xi\rvert)^{-t/2} \qquad(\xi\in\mathbb{R}^2).\] We prove that \(\alpha_{\min}(\nu)\ge t\). Fix \(0<s<t\) and \(x_0\in\operatorname{spt}\nu\). For small \(r>0\), set \(\Lambda=r^{-2}\). Since \(x_0\in\mathbb{S}^1\), \[\begin{align} \int_{\mathbb{R}}\lvert\widehat\nu(Rx_0)\rvert^2e^{-\pi(R/\Lambda)^2}\,dR &\le C_t^2\int_{\mathbb{R}}(1+\lvert R\rvert)^{-t}e^{-\pi(R/\Lambda)^2}\,dR\\ &\le C_s\Lambda^{1-s}. \end{align}\] On the other hand, Lemma 33 implies \[\int_{\mathbb{R}}\lvert\widehat\nu(Rx_0)\rvert^2e^{-\pi(R/\Lambda)^2}\,dR \ge c\Lambda\,\nu(B(x_0,cr))^2.\] Combining the two estimates, \[\nu(B(x_0,cr))^2\le C_s\Lambda^{-s}=C_sr^{2s},\] and hence \(\nu(B(x_0,cr))\le C_sr^s\). After changing constants, \(\nu(B(x_0,\rho))\le C_s\rho^s\) for all sufficiently small \(\rho>0\). Since \(\log_2\rho<0\), \[\liminf_{\rho\downarrow0} \frac{\log_2\nu(B(x_0,\rho))}{\log_2\rho} \ge s.\] As \(x_0\in\operatorname{spt}\nu\) was arbitrary, \(\alpha_{\min}(\nu)\ge s\). Letting \(s\uparrow t\) yields \(\alpha_{\min}(\nu)\ge t\). Thus every admissible Fourier decay exponent \(t<1\) is bounded above by \(\alpha_{\min}(\nu)\).

It remains to exclude admissible exponents larger than \(1\). Since \(\nu\) is supported on \(\mathbb{S}^1\), one has \(\dim_{\mathrm F}(\nu)\le1\). Taking the supremum over admissible Fourier decay exponents and using the preceding conclusion for all \(t<1\), we obtain the desired claim. ◻

Proof. On \(\mathcal{S}_\circ\), the measure \(\mu_\circ\) is a nonzero finite Borel measure on \(\mathbb{S}^1\). Proposition 39 and Theorem 38 imply the desired claim. ◻

Proof. Choose \(0<\delta_1<\min\{\delta,1-s\}\), and set \(a=s+\delta_1\). Then \(s<a<1\), and since \(\delta_1<\delta\), \[\beta_{\mathrm{comp}} = s+r\delta-(r-1)\delta_1>s, \qquad \gamma_{\mathrm{comp}}=\min\{\beta_{\mathrm{comp}},1\}>s.\] For \(\varepsilon=0\), \(\vartheta_0=1-s/(2a)>1/2\), so \[\beta_{\mathrm{comp}}\vartheta_0>\frac{s}{2}, \qquad \gamma_{\mathrm{comp}}\vartheta_0>\frac{s}{2}.\] By continuity, choose \(\varepsilon>0\) sufficiently small so that the same strict inequalities hold with \[\vartheta=1-\frac{s}{2a}-\frac{8\varepsilon}{a},\] while also ensuring that \(\vartheta>0\) and \(20\varepsilon<\delta_1/2\). Finally, choose \[0<\kappa<\frac{\delta_1}{2}-20\varepsilon.\] Then \(20\varepsilon+\kappa<\delta_1/2\), as required. ◻

Proof. The scalar cascade can be viewed as a dyadic vector cascade with normalized child coefficients \(U_0/2\) and \(U_1/2\), where \(U_0\) and \(U_1\) are independent copies of \(U\). By the finite-\(r\) witnessed hypothesis, \[\mathbb{E}\left[\left(\frac{U_0}{2}\right)^r+\left(\frac{U_1}{2}\right)^r\right] = 2^{1-r}\mathbb{E}[U^r] \le 2^{-r(s+\delta)} <1.\] In addition, \[\mathbb{E}\left[\left(\frac{U_0+U_1}{2}\right)^r\right] \le \frac{\mathbb{E}U_0^r+\mathbb{E}U_1^r}{2} <\infty.\] Therefore Lemma 9 implies \[\sup_{\ell\ge0}\mathbb{E}[\widetilde{Y}_\ell^r]<\infty.\] Since \(\widetilde{Y}_\ell\to\widetilde{Y}\) almost surely, Fatou’s lemma yields \[\mathbb{E}[\widetilde{Y}^r]\le\liminf_{\ell\to\infty}\mathbb{E}[\widetilde{Y}_\ell^r]<\infty.\] ◻

Proof. Fix \(J_v\in\mathcal{D}_k^\circ\) with \(\lvert v \rvert=k\). For \(\ell\ge k\), \[\widetilde{\nu}_\ell(J_v) = 2^{-k}Q_v\widetilde{Y}_{\ell-k}^{(v)},\] where \(\widetilde{Y}_{\ell-k}^{(v)}\) is independent of \(Q_v\) and has the same law as \(\widetilde{Y}_{\ell-k}\). Hence \[\mathbb{E}[\widetilde{\nu}_\ell(J_v)^r] = 2^{-kr}\mathbb{E}[Q_v^r]\mathbb{E}[(\widetilde{Y}_{\ell-k})^r] \le C_r2^{-kr}(\mathbb{E}[U^r])^k.\] Summing over the \(2^k\) intervals \(J_v\), we obtain \[\mathbb{E}\left[ \sum_{J\in\mathcal{D}_k^\circ} \widetilde{\nu}_\ell(J)^r \right] \le C_r(2^{1-r}\mathbb{E}[U^r])^k \le C_r2^{-r(s+\delta)k}.\] This proves 72 . For the limiting measure, \[\widetilde{\nu}(J_v)=2^{-k}Q_v\widetilde{Y}^{(v)},\] where \(\widetilde{Y}^{(v)}\) is an independent copy of \(\widetilde{Y}\). The same computation, now using \(\mathbb{E}[\widetilde{Y}^r]<\infty\), implies 73 . ◻

Proof. Fix \(k\ge0\) and \(J\in\mathcal{D}_k^\circ\). Since \((\widetilde{\nu}_\ell(J))_{\ell\ge k}\) is a nonnegative martingale, Doob’s \(L^r\) maximal inequality implies, for every \(L\ge k\), \[\mathbb{P}\left( \max_{k\le \ell\le L}\widetilde{\nu}_\ell(J)>T_{k,n} \right) \le \left(\frac{r}{r-1}\right)^r T_{k,n}^{-r}\mathbb{E}[\widetilde{\nu}_L(J)^r].\] After summing over \(J\in\mathcal{D}_k^\circ\), applying 72 , and letting \(L\to\infty\), we get \[\mathbb{P}\left( \exists J\in\mathcal{D}_k^\circ,\;\exists \ell\ge k: \widetilde{\nu}_\ell(J)>T_{k,n} \right) \le C T_{k,n}^{-r}2^{-r(s+\delta)k}.\] Using \(T_{k,n}=2^{\varepsilon n}2^{-ak}\) and \(a=s+\delta_1\), the right-hand side becomes \(C2^{-r\varepsilon n}2^{-r(\delta-\delta_1)k}\).

Summing over \(k\ge0\) yields \[\mathbb{P}((\mathcal{G}^{\mathrm{pre}}_n)^c)\le C2^{-r\varepsilon n}.\]

For the limiting event, Markov’s inequality together with 73 implies \[\mathbb{P}\left( \exists J\in\mathcal{D}_k^\circ: \widetilde{\nu}(J)>T_{k,n} \right) \le C2^{-r\varepsilon n}2^{-r(\delta-\delta_1)k}.\] Summing over \(k\ge0\), we obtain \(\mathbb{P}((\mathcal{G}^{\mathrm{lim}}_n)^c)\le C2^{-r\varepsilon n}\). Since \[\mathcal{G}_n^c \subset (\mathcal{G}^{\mathrm{pre}}_n)^c \cup (\mathcal{G}^{\mathrm{lim}}_n)^c,\] we obtain \(\mathbb{P}(\mathcal{G}_n^c)\le C2^{-r\varepsilon n}\), which implies ?? after decreasing \(c>0\), if necessary. Summability is immediate. ◻

Proof. For (i), choose \(k\ge0\) such that \(2^{-(k+1)}<\lvert B\rvert\le2^{-k}\). Up to the wrap-around convention, the interval \(B\) can be covered by at most two intervals in \(\mathcal{D}_k^\circ\). On \(\mathcal{G}^{\mathrm{lim}}_n\), each such interval has \(\widetilde{\nu}\)-mass at most \(T_{k,n}=2^{\varepsilon n}2^{-ak}\). Hence \[\widetilde{\nu}(B)\le C2^{\varepsilon n}2^{-ak}\le C2^{\varepsilon n}\lvert B\rvert^a.\]

For (ii), set \(\rho=\lvert B\rvert+2^{-\ell}\). If \(\rho>1\), the estimate is immediate from the case \(k=0\) and the choice of the constant \(C\). Otherwise, choose \(0\le k\le\ell\) such that \(2^{-(k+1)}<\rho\le2^{-k}\). The union of the level-\(\ell\) intervals meeting \(B\) is contained in an interval of length \(O(\rho)\), and hence in a bounded number of level-\(k\) intervals. On \(\mathcal{G}^{\mathrm{pre}}_n\), \[\widetilde{\nu}_\ell(B)\le C2^{\varepsilon n}2^{-ak}\le C2^{\varepsilon n}\rho^a.\] This proves 75 .

For (iii), on \(\mathcal{G}^{\mathrm{pre}}_n\), every level-\(\ell\) interval \(I\) satisfies \(\widetilde{\nu}_\ell(I)\le T_{\ell,n}=2^{\varepsilon n}2^{-a\ell}\). Hence \[\sum_{\substack{I\in\mathcal{D}_\ell^\circ\\ I\cap B\neq\varnothing}} \widetilde{\nu}_\ell(I)^2 \le 2^{\varepsilon n}2^{-a\ell} \sum_{\substack{I\in\mathcal{D}_\ell^\circ\\ I\cap B\neq\varnothing}} \widetilde{\nu}_\ell(I).\] The last sum is bounded by the prelimit mass of an enlarged interval of length \(O(\lvert B\rvert+2^{-\ell})\). Applying (ii) to this enlarged interval implies 76 . ◻

Proof. Taking \(k=0\) and \(J=[0,1)\) in \(\mathcal{G}^{\mathrm{lim}}_n\), we obtain \(\widetilde{\nu}([0,1))\le T_{0,n}=2^{\varepsilon n}\). Since \(\nu_\circ=f_\#\widetilde{\nu}\), it follows that \(\nu_\circ(\mathbb{S}^1)\le2^{\varepsilon n}\). Also, \[\begin{align} \lvert\widehat{\nu_\circ}(\xi)-\widehat{\nu_\circ}(\eta)\rvert &\le \int_{\mathbb{S}^1} \left\lvert e^{-2\pi i x\cdot \xi} - e^{-2\pi i x\cdot \eta} \right\rvert\,d\nu_\circ(x)\\ &\le 2\pi\lvert\xi-\eta\rvert\,\nu_\circ(\mathbb{S}^1) \le 2\pi 2^{\varepsilon n}\lvert\xi-\eta\rvert, \end{align}\] using \(\lvert e^{-iu}-e^{-iv}\rvert\le \lvert u-v\rvert\) and \(\lvert x \rvert=1\) on \(\mathbb{S}^1\). ◻

Proof. We first treat the low-frequency case. If \(\Lambda\le \Lambda_0\), where \(\Lambda_0\) is a sufficiently large absolute constant, we simply take \[\chi_{\xi,\mathrm{stat}}\equiv 1, \qquad \mathfrak D_\xi=\varnothing.\] Then all assertions are immediate after enlarging the absolute constant \(C\).

Assume from now on that \(\Lambda>\Lambda_0\). Let \(h_\xi(t)=\sin^2(2\pi t-\theta)\). Since \(Z_\xi\) is the zero set of \(\sin(2\pi t-\theta)\), we have the uniform comparison \[h_\xi(t)^{1/2}\asymp \operatorname{dist}(t,Z_\xi) \qquad(t\in\mathbb{R}/\mathbb{Z}).\] Choose a nonnegative \(C^\infty\) cutoff \(\psi\) on \([0,\infty)\) such that \[\psi(u)=1\quad(0\le u\le1), \qquad \psi(u)=0\quad(u\ge4),\] and set \[\chi_{\xi,\mathrm{stat}}(t) = \psi\bigl(\Lambda h_\xi(t)\bigr).\] Then \[\operatorname{spt}\chi_{\xi,\mathrm{stat}} \subset \{t:h_\xi(t)\le 4\Lambda^{-1}\} \subset \{t:\operatorname{dist}(t,Z_\xi)\le C\Lambda^{-1/2}\}.\]

Next choose a smooth dyadic partition of unity on \((0,\infty)\). Let \(\rho:[0,\infty)\to[0,1]\) be nonincreasing and \(C^\infty\), with \[\rho(u)=1\quad(0\le u\le1), \qquad \rho(u)=0\quad(u\ge4),\] and set \(\zeta(u)=\rho(u)-\rho(4u)\). Then \(\zeta\ge0\), \(\operatorname{spt}\zeta\subset[1/4,4]\), and, writing \(d=2^j\), \[\sum_{d\in2^{\mathbb{Z}}}\zeta\left(\frac{u}{d^2}\right) = \sum_{j\in\mathbb{Z}} \left[ \rho\left(\frac{u}{4^j}\right) - \rho\left(\frac{u}{4^{j-1}}\right) \right] =1, \qquad u>0,\] by telescoping. This provides the required \(C^\infty\) dyadic partition of unity.

Let \(\mathfrak D_\xi\) be the finite set of dyadic numbers \(d\) for which \[\zeta\left(\frac{h_\xi(t)}{d^2}\right)\] can be nonzero for some \(t\) with \(1-\chi_{\xi,\mathrm{stat}}(t)\neq0\). For \(d\in\mathfrak D_\xi\), define \[\chi_{\xi,d}(t) = (1-\chi_{\xi,\mathrm{stat}}(t)) \zeta\left(\frac{h_\xi(t)}{d^2}\right).\] Then, for every \(t\), \[\chi_{\xi,\mathrm{stat}}(t) + \sum_{d\in\mathfrak D_\xi}\chi_{\xi,d}(t) = 1.\]

The support condition for the derivative bands follows from the support of \(\zeta\). If \(t\in\operatorname{spt}\chi_{\xi,d}\), then \(h_\xi(t)\asymp d^2\), and hence \(\operatorname{dist}(t,Z_\xi)\asymp d\). Thus \[\operatorname{spt}\chi_{\xi,d} \subset \{t:C^{-1}d\le \operatorname{dist}(t,Z_\xi)\le Cd\}.\] Moreover, since \(1-\chi_{\xi,\mathrm{stat}}\) vanishes inside a stationary tube of radius \(\asymp\Lambda^{-1/2}\), any such \(d\) satisfies \[C^{-1}\Lambda^{-1/2}\le d\le C.\] It follows that the number of possible dyadic scales is bounded by \[\#\mathfrak D_\xi \le C(1+\log_2(2+\Lambda)).\]

We next verify the derivative size. Since \[\lvert\phi_\xi'(t)\rvert = 4\pi^2\Lambda\lvert\sin(2\pi t-\theta)\rvert = 4\pi^2\Lambda h_\xi(t)^{1/2},\] and since \(h_\xi(t)^{1/2}\asymp d\) on \(\operatorname{spt}\chi_{\xi,d}\), we get \[C^{-1}\Lambda d \le \lvert\phi_\xi'(t)\rvert \le C\Lambda d \qquad(t\in\operatorname{spt}\chi_{\xi,d}).\] It remains to prove the cutoff derivative bounds. We have \[\lvert h_\xi'(t)\rvert\le C h_\xi(t)^{1/2}.\] On the support of \(\zeta(h_\xi/d^2)\), we obtain \[\left\lvert \frac{d}{dt}\zeta\left(\frac{h_\xi(t)}{d^2}\right) \right\rvert \le C\frac{\lvert h_\xi'(t)\rvert}{d^2} \le Cd^{-1}.\] Also, \[\lvert\chi_{\xi,\mathrm{stat}}'(t)\rvert \le C\Lambda \lvert h_\xi'(t)\rvert.\] This derivative is supported where \(h_\xi(t)\asymp\Lambda^{-1}\), and hence where \(h_\xi(t)^{1/2}\asymp\Lambda^{-1/2}\). If this region intersects the support of \(\zeta(h_\xi/d^2)\), then \(d\asymp\Lambda^{-1/2}\), so \[\lvert\chi_{\xi,\mathrm{stat}}'(t)\rvert \le C\Lambda h_\xi(t)^{1/2} \le C\Lambda^{1/2} \le Cd^{-1}.\] Combining the two derivative estimates, we obtain \(\lvert\chi_{\xi,d}'(t)\rvert\le Cd^{-1}\).

Finally, the bounded overlap follows from the dyadic construction: for each fixed \(t\), the condition \[\zeta\left(\frac{h_\xi(t)}{d^2}\right)\neq0\] can hold for only \(O(1)\) dyadic values of \(d\). Hence \[\sum_{d\in\mathfrak D_\xi}\mathbf{1}_{\operatorname{spt}\chi_{\xi,d}}(t)\le C.\] This proves all the stated properties. ◻

Proof. The absolute value is at most \(\widetilde{\nu}(\operatorname{spt}\chi_{\xi,\mathrm{stat}})\). By Lemma 39, this support is contained in the union of two intervals, each of length at most \(C\Lambda^{-1/2}\). By 74 , \[\widetilde{\nu}(\operatorname{spt}\chi_{\xi,\mathrm{stat}}) \le C2^{\varepsilon n}\Lambda^{-a/2}.\] Since \(\Lambda\ge2^n\), \[2^{\varepsilon n}\Lambda^{-a/2} \le 2^{-sn/2}2^{-(\delta_1/2-\varepsilon)n}.\] The parameter choice ensures \(\delta_1/2-\varepsilon>0\), so 79 holds with \(c_{\mathrm{stat}}=\delta_1/2-\varepsilon\). ◻

Proof. For a fixed \(d\), the support of \(\chi_{\xi,d}\) is contained in a bounded number of intervals of length at most \(Cd\). By 74 , \[\widetilde{\nu}(\operatorname{spt}\chi_{\xi,d}) \le C2^{\varepsilon n}d^a.\] If \(d\) is mass-only, then \(d^a\le2^{-sn/2}2^{-8\varepsilon n}\), hence \[\left| \int e^{i\phi_\xi(t)}\chi_{\xi,d}(t)\,d\widetilde{\nu}(t) \right| \le C2^{-sn/2}2^{-7\varepsilon n}.\] There are \(O(n)\) derivative bands in the annulus, and \(n2^{-7\varepsilon n}\le C2^{-6\varepsilon n}\).

This proves 80 . ◻

Proof. Since \(d\) is non-mass, \[d^a>2^{-sn/2}2^{-8\varepsilon n}, \qquad d>2^{-sn/(2a)}2^{-8\varepsilon n/a}.\] Using \(\Lambda\ge2^n\), we get \(\Lambda d\ge2^{\vartheta n}\). By Definition 11, \(m_{\xi,d}\ge\vartheta n-C_0\). For the upper bound, derivative bands satisfy \(d\le C\), and \(\Lambda\le2^{n+1}\), hence \(\Lambda d\le C2^n\), so \(m_{\xi,d}\le n+C_0\) after increasing \(C_0\). ◻

Proof. The second estimate is immediate from \(\lvert c_J\rvert\le2^\ell\lvert J\rvert=1/2\).

Assume \(0\le\ell<m_{\xi,d}\). It suffices to prove \[\left\lvert \int_J e^{i\phi_\xi(t)}\chi_{\xi,d}(t)\,dt \right\rvert \le C(\Lambda d)^{-1},\] since multiplying by \(2^\ell\) yields \(\lvert c_J\rvert\le C2^\ell(\Lambda d)^{-1}\le C2^{\ell-m_{\xi,d}}\).

Let \(G(t)=\chi_{\xi,d}(t)\). On \(\operatorname{spt}G\), \[\lvert\phi_\xi'(t)\rvert\ge C^{-1}\Lambda d, \qquad \lvert G'(t)\rvert\le Cd^{-1}, \qquad \lvert\phi_\xi''(t)\rvert\le C\Lambda.\] The set \(J\cap\operatorname{spt}G\) has a bounded number of connected components. On each component, integration by parts yields \[\int e^{i\phi_\xi(t)}G(t)\,dt = \left[ \frac{e^{i\phi_\xi(t)}G(t)}{i\phi_\xi'(t)} \right]_{\mathrm{endpoints}} - \int e^{i\phi_\xi(t)} \left( \frac{G'(t)}{i\phi_\xi'(t)} - \frac{G(t)\phi_\xi''(t)}{i(\phi_\xi'(t))^2} \right)\,dt.\] The boundary term is \(O((\Lambda d)^{-1})\). Also, \[\int_{\operatorname{spt}G}\frac{\lvert G'(t)\rvert}{\lvert\phi_\xi'(t)\rvert}\,dt \le C\int_{\operatorname{spt}G}\frac{d^{-1}}{\Lambda d}\,dt \le C(\Lambda d)^{-1},\] because \(\operatorname{spt}G\) has length \(O(d)\), and \[\int_{\operatorname{spt}G}\frac{\lvert G(t)\rvert\,\lvert\phi_\xi''(t)\rvert}{\lvert\phi_\xi'(t)\rvert^2}\,dt \le C\int_{\operatorname{spt}G}\frac{\Lambda}{(\Lambda d)^2}\,dt \le C(\Lambda d)^{-1}.\] Combining these bounds proves the prefix estimate. ◻

Proof. The integration-by-parts estimate from Lemma 43 yields \[\lvert F^{(0)}_{\xi,d}\rvert\le C(\Lambda d)^{-1}.\] Since \(d\) is non-mass, \(d^{-1}<2^{sn/(2a)}2^{8\varepsilon n/a}\). Using \(\Lambda\ge2^n\), \[(\Lambda d)^{-1} \le 2^{-n}2^{sn/(2a)}2^{8\varepsilon n/a} = 2^{-\vartheta n}= 2^{-sn/2}2^{-(\vartheta-s/2)n}.\] By 78 , \(\vartheta>s/2\). This proves the first estimate with \(c_{\mathrm{arc}}=\vartheta-s/2>0\). The \(O(n)\) derivative bands are absorbed into the exponential decay after decreasing \(c_{\mathrm{arc}}>0\), if necessary. ◻

Proof. Let \(G_{\xi,d}(t)=e^{i\phi_\xi(t)}\chi_{\xi,d}(t)\). For \(L\ge1\), \[\int G_{\xi,d}\,d\widetilde{\nu}_L = \int G_{\xi,d}(t)\,dt + \sum_{\ell=0}^{L-1} \int G_{\xi,d}\,d(\widetilde{\nu}_{\ell+1}-\widetilde{\nu}_\ell).\] The first term is \(F^{(0)}_{\xi,d}\). For a parent \(I\in\mathcal{D}_\ell^\circ\) and child \(J\subset I\), the densities of \(\widetilde{\nu}_\ell\) on \(I\) and \(\widetilde{\nu}_{\ell+1}\) on \(J\) are \(2^\ell M_I\) and \(2^\ell M_IU_J\), respectively. Hence \[\int_JG_{\xi,d}\,d(\widetilde{\nu}_{\ell+1}-\widetilde{\nu}_\ell) = M_I(U_J-1)\,2^\ell\int_JG_{\xi,d}(t)\,dt= M_I(U_J-1)\,c_J(\xi,d,\ell).\] Summing over children and generations, we obtain, for \(L>m_{\xi,d}\), \[\int G_{\xi,d}\,d\widetilde{\nu}_L = F^{(0)}_{\xi,d} + F^{\mathrm{pre}}_{\xi,d} + F^{\mathrm{post}}_{\xi,d,L}.\] Since \(G_{\xi,d}\) is continuous and \(\widetilde{\nu}_L\xrightarrow{\mathrm{w}}\widetilde{\nu}\), the left side converges to \(\int G_{\xi,d}\,d\widetilde{\nu}\). Thus the post-bin partial sums converge and satisfy 81 . ◻

Proof. By Lemma 39, \[1=\chi_{\xi,\mathrm{stat}}+\sum_{d\in\mathfrak D_\xi}\chi_{\xi,d}.\] Therefore \[\widehat{\nu_\circ}(\xi) = \int e^{i\phi_\xi(t)}\chi_{\xi,\mathrm{stat}}(t)\,d\widetilde{\nu}(t) + \sum_{d\in\mathfrak D_\xi} \int e^{i\phi_\xi(t)}\chi_{\xi,d}(t)\,d\widetilde{\nu}(t).\] The stationary term is bounded by Lemma 40. The sum over mass-only bands is bounded by Lemma 41. We include these terms in \(E^{\mathrm{safe}}_{\xi,n}\).

For each non-mass band \(d\in\mathfrak D_{\xi,n}^{\mathrm{osc}}\), Lemma 45 provides the decomposition \[\int e^{i\phi_\xi(t)}\chi_{\xi,d}(t)\,d\widetilde{\nu}(t) = F^{(0)}_{\xi,d} + F^{\mathrm{pre}}_{\xi,d} + F^{\mathrm{post}}_{\xi,d}.\] The arclength term \(F^{(0)}_{\xi,d}\) is bounded by Lemma 44; summing over all non-mass bands still contributes a term of size \(C2^{-sn/2}2^{-c n}\) for some \(c>0\). These arclength terms are also included in \(E^{\mathrm{safe}}_{\xi,n}\).

Combining the stationary term, the mass-only band terms, and the arclength forcing terms, we obtain ?? with \[\lvert E^{\mathrm{safe}}_{\xi,n}\rvert \le C2^{-sn/2} \left( 2^{-c_{\mathrm{stat}}n} + 2^{-6\varepsilon n} + 2^{-c_{\mathrm{arc}}n} \right).\] After decreasing the exponent and increasing \(C\), this is \[\lvert E^{\mathrm{safe}}_{\xi,n}\rvert \le C2^{-sn/2}2^{-c_{\mathrm{safe}}n}.\] The coefficient estimates are exactly Lemma 43. ◻

Proof. For fixed \(\ell\), the event \(\{\tau_n\le \ell\}\) is determined by the collection of masses \(\widetilde{\nu}_j(J)\), \(0\le j\le\ell\), \(J\in\mathcal{D}_k^\circ\), \(0\le k\le j\), all of which are \(\mathcal{F}_\ell\)-measurable. Hence \(\tau_n\) is a stopping time. The identity \(\mathcal{G}_n^{\mathrm{pre}}=\{\tau_n=\infty\}\) follows directly from the definitions. ◻

Proof. The centeredness follows immediately from the definition of \(\overline{U}_{I,n}\). If \(M_I=0\), then \(X_{J,n}^{\mathrm{cap}}=0\). We may therefore assume that \(M_I>0\). Since \(0\le U_{J,n}^{\mathrm{cap}}\le C_{I,n}\) and \(C_{I,n}\) is \(\mathcal{F}_\ell\)-measurable, \[0\le\overline{U}_{I,n}\le C_{I,n}, \qquad \lvert U_{J,n}^{\mathrm{cap}}-\overline{U}_{I,n}\rvert\le C_{I,n}.\] Therefore \[\lvert X_{J,n}^{\mathrm{cap}}\rvert \le \lvert c_J\rvert M_IC_{I,n} = 2\lvert c_J\rvert L_nT_{\ell+1,n},\] which proves 82 . For the variance, \[\operatorname{Var}(U_{J,n}^{\mathrm{cap}}\mid\mathcal{F}_\ell) \le \mathbb{E}[(U_{J,n}^{\mathrm{cap}})^2\mid\mathcal{F}_\ell].\] Since \((U_{J,n}^{\mathrm{cap}})^2\le C_{I,n}U_{J,n}^{\mathrm{cap}}\le C_{I,n}U_J\), \[\mathbb{E}[(U_{J,n}^{\mathrm{cap}})^2\mid\mathcal{F}_\ell]\le C_{I,n}.\] Thus \[\mathbb{E}[\lvert X_{J,n}^{\mathrm{cap}}\rvert^2\mid\mathcal{F}_\ell] \le \lvert c_J\rvert^2M_I^2C_{I,n} = 2L_n\lvert c_J\rvert^2M_IT_{\ell+1,n},\] which establishes 83 . ◻

Proof. On \(\mathcal{G}_n^{\mathrm{pre}}\), \(\widetilde{\nu}_{\ell+1}(J)\le T_{\ell+1,n}\). Since \(\widetilde{\nu}_{\ell+1}(J)=M_IU_J/2\),

if \(M_I>0\), then \[U_J\le \frac{2T_{\ell+1,n}}{M_I}\le L_n\frac{2T_{\ell+1,n}}{M_I}=C_{I,n}.\] Thus, on \(\mathcal{G}_n^{\mathrm{pre}}\), we have \(U_{J,n}^{\mathrm{cap}}=U_J\), and \[c_JM_I(U_J-1) = c_JM_I(U_{J,n}^{\mathrm{cap}}-\overline{U}_{I,n}) + c_JM_I(\overline{U}_{I,n}-1).\] This proves 84 . Finally, since \(\mathbb{E}U_J=1\), \[\overline{U}_{I,n} = \mathbb{E}[U_J-(U_J-C_{I,n})_+\mid\mathcal{F}_\ell] = 1-\mathbb{E}[(U_J-C_{I,n})_+\mid\mathcal{F}_\ell].\] ◻

Proof. Only intervals meeting a fixed \(C2^{-\ell}\)-neighborhood \(S_{\xi,d}^{(\ell)}\) of \(S_{\xi,d}\) can contribute. For \(0\le\ell<m\), Lemma 43 yields \(\lvert c_J\rvert\le C2^{\ell-m}\), and 75 implies \[\sum_{\substack{I\in\mathcal{D}_\ell^\circ\\ I\cap S_{\xi,d}^{(\ell)}\neq\varnothing}} M_I \le C2^{\varepsilon n}(d+2^{-\ell})^a.\] Since \(T_{\ell+1,n}\le C2^{\varepsilon n}2^{-a\ell}\), the prefix sum is at most \[C2^{2\varepsilon n} \sum_{\ell=0}^{m-1} 2^{2(\ell-m)}2^{-a\ell}(d+2^{-\ell})^a.\] We claim that the last sum is \(O(\lvert\xi\rvert^{-a})\). We split the sum into the two ranges \(2^{-\ell}\ge d\) and \(2^{-\ell}<d\). In the first range, \[(d+2^{-\ell})^a\le C2^{-a\ell},\] so the summand is bounded by \(C2^{-2m}2^{(2-2a)\ell}\). Since \(a<1\), \[\sum_{2^{-\ell}\ge d} 2^{-2m}2^{(2-2a)\ell} \le C2^{-2m}d^{-(2-2a)}.\] Using \(2^{-m}\asymp(\lvert\xi\rvert d)^{-1}\) and \(d\gtrsim\lvert\xi\rvert^{-1/2}\), this is bounded by \(C\lvert\xi\rvert^{-a}\). In the second range, \[(d+2^{-\ell})^a\le Cd^a,\] so \[C2^{-2m}d^a\sum_{\ell<m}2^{(2-a)\ell} \le Cd^a2^{-am} \asymp C\lvert\xi\rvert^{-a}.\] This proves 86 .

For \(\ell\ge m\), we have \(\lvert c_J\rvert\le C\). Since \(2^{-\ell}\le2^{-m}\lesssim(\lvert\xi\rvert d)^{-1}\lesssim d\), we have \(d+2^{-\ell}\le Cd\). Therefore the level-\(\ell\) contribution is bounded by \(C2^{2\varepsilon n}2^{-a\ell}d^a\). After summing over \(\ell\ge m\), we obtain \[C2^{2\varepsilon n}d^a2^{-am} \asymp C2^{2\varepsilon n}\lvert\xi\rvert^{-a},\] which proves 87 . ◻

Proof. Set \(X_j=\xi_j/R\). Then \((X_j)\) is a martingale difference sequence with \(\lvert X_j\rvert\le1\), and for \(0\le n\le N\), \[\sum_{j=1}^n X_j=\frac{M_n}{R}, \qquad \sum_{j=1}^n\operatorname{Var}(X_j\mid\mathcal{H}_{j-1}) \le \frac{V}{R^2}.\] If \(\max_{0\le j\le N}M_j>u\), then for some \(1\le n\le N\), \[\sum_{j=1}^nX_j>\frac{u}{R}, \qquad \sum_{j=1}^n\operatorname{Var}(X_j\mid\mathcal{H}_{j-1})\le\frac{V}{R^2}.\] Applying Theorem 44 with \(a=u/R\) and \(b=V/R^2\), we get \[\mathbb{P}\left( \max_{0\le j\le N}M_j>u \right) \le \exp\left( -\frac{(u/R)^2}{2(u/R+V/R^2)} \right) = \exp\left( -\frac{u^2}{2(V+Ru)} \right).\] ◻

Proof. Write \(S_j=X_j+iY_j\), where \(X_j=\operatorname{Re}S_j\) and \(Y_j=\operatorname{Im}S_j.\) Then \(X_j\) and \(Y_j\) are real-valued martingales with differences \(\operatorname{Re}\Delta_j\) and \(\operatorname{Im}\Delta_j,\) respectively. Moreover, \(\lvert\operatorname{Re}\Delta_j\rvert,\;\lvert\operatorname{Im}\Delta_j\rvert \le \lvert\Delta_j\rvert\le R,\) and \[\sum_{j=1}^N \mathbb{E}\bigl((\operatorname{Re}\Delta_j)^2\mid\mathcal{H}_{j-1}\bigr) \le V, \qquad \sum_{j=1}^N \mathbb{E}\bigl((\operatorname{Im}\Delta_j)^2\mid\mathcal{H}_{j-1}\bigr) \le V.\] The same bounds also hold for \(-X_j\) and \(-Y_j\). Therefore, applying Lemma 50 to \(X_j,-X_j,Y_j,-Y_j\) with \(u=t/2\), and then using the union bound, \[\mathbb{P}\left( \max_{0\le j\le N}\lvert S_j\rvert>t \right) \le 4\exp\left( -\frac{(t/2)^2}{2(V+Rt/2)} \right) \le 4\exp\left( -\frac{t^2}{8(V+Rt)} \right).\] This proves the lemma. ◻

Proof. We prove the pre-bin assertion; the post-bin assertion follows in the same way. Set \[\Delta_r^{\mathrm{pre}} := \mathbf{1}_{\{\tau_n>\ell_r\}} X_{J_r,n}^{\mathrm{cap}}(\xi,d,\ell_r).\] It is enough to prove \(\mathbb{E}(\Delta_r^{\mathrm{pre}}\mid\mathcal{H}_{r-1}^{\mathrm{pre}})=0\). By definition, \[\Delta_r^{\mathrm{pre}} = \mathbf{1}_{\{\tau_n>\ell_r\}} c_{J_r}(\xi,d,\ell_r)M_{I_r} \left( U_{J_r}\wedge C_{I_r,n} - \overline{U}_{I_r,n} \right).\] At the stage when \(U_{J_r}\) is revealed, the factors \[\mathbf{1}_{\{\tau_n>\ell_r\}}, \quad c_{J_r}(\xi,d,\ell_r), \quad M_{I_r}, \quad C_{I_r,n}, \quad \overline{U}_{I_r,n}\] are \(\mathcal{H}_{r-1}^{\mathrm{pre}}\)-measurable. Since \(U_{J_r}\) is independent of the refined past and \(C_{I_r,n}\) is already measurable with respect to that past, \[\mathbb{E}\left( U_{J_r}\wedge C_{I_r,n} \mid \mathcal{H}_{r-1}^{\mathrm{pre}} \right) = \overline{U}_{I_r,n}.\] Thus \(\mathbb{E}(\Delta_r^{\mathrm{pre}}\mid\mathcal{H}_{r-1}^{\mathrm{pre}})=0\), and \(S_r^{\mathrm{pre}}\) is a complex-valued martingale. Its terminal value is \[S_N^{\mathrm{pre}} = \sum_{\ell=0}^{m-1} \mathbf{1}_{\{\tau_n>\ell\}} \sum_{I\in\mathcal{D}_\ell^\circ} \sum_{\substack{J\in\mathcal{D}_{\ell+1}^\circ\\ J\subset I}} X_{J,n}^{\mathrm{cap}}(\xi,d,\ell) = \widehat F^{\mathrm{pre,cap}}_{\xi,d,n}.\] The same conditional-expectation computation applied to the post-bin ordering proves the second assertion. ◻

Proof. By 82 , \[\lvert X_{J,n}^{\mathrm{cap}}\rvert\le C\lvert c_J\rvert L_nT_{\ell+1,n}.\] If \(0\le\ell<m\), then \(\lvert c_J\rvert\le C2^{\ell-m}\), and since \(a<1\), \[\lvert c_J\rvert T_{\ell+1,n} \le C2^{\ell-m}2^{\varepsilon n}2^{-a\ell} \le C2^{\varepsilon n}2^{-am}.\] If \(\ell\ge m\), then the bound \(\lvert c_J\rvert\le C\) yields the same estimate. Therefore \[\lvert X_{J,n}^{\mathrm{cap}}(\xi,d,\ell)\rvert \le CL_n2^{\varepsilon n}2^{-am}.\] Using \(L_n=2^{\kappa n}\), \(2^{-m}\asymp(\lvert\xi\rvert d)^{-1}\), \(d^{-a}<2^{sn/2}2^{8\varepsilon n}\), \(\lvert\xi\rvert\ge2^n\), and \(a=s+\delta_1\), we get \[\lvert X_{J,n}^{\mathrm{cap}}(\xi,d,\ell)\rvert \le C2^{-sn/2}2^{-(\delta_1-\kappa-9\varepsilon)n}.\] By the parameter choice, \(\delta_1-\kappa-9\varepsilon>0\), which proves 88 .

For the quadratic variation, Lemmas 47 and 49 give, uniformly for the prefix and all finite post-bin cutoffs, \[V_{\xi,d,n} \le CL_n2^{2\varepsilon n}\lvert\xi\rvert^{-a} \le C2^{-sn}2^{-(\delta_1-\kappa-2\varepsilon)n}.\] Since \(\delta_1-\kappa-2\varepsilon>0\) by the parameter choice, this proves 89 . ◻

Proof. We prove the prefix estimate; the post-bin estimate is obtained in the same way. Set \[t_n=2^{-sn/2}2^{-3\varepsilon n}.\] By Lemmas 52 and 53, Lemma 51 applies with \[R\le C2^{-sn/2}2^{-(\delta_1-\kappa-9\varepsilon)n}, \qquad V\le C2^{-sn}2^{-(\delta_1-\kappa-2\varepsilon)n}.\] Since \(t_n^2=2^{-sn}2^{-6\varepsilon n}\), \[V \le Ct_n^2\,2^{-(\delta_1-\kappa-8\varepsilon)n}, \qquad Rt_n \le Ct_n^2\,2^{-(\delta_1-\kappa-12\varepsilon)n}.\] By the parameter choice, the exponents are positive. Hence, for some \(\eta>0\), \[V+Rt_n \le Ct_n^2\,2^{-\eta n}, \qquad \frac{t_n^2}{V+Rt_n}\ge c2^{\eta n}.\] Lemma 51 then yields ?? . Applying the same argument to the stopped post-bin martingales implies ?? . Finally, on \(\mathcal{G}_n^{\mathrm{pre}}\), Lemma 46 implies \(\tau_n=\infty\), and hence the stopped and unstopped centered capped arrays agree. ◻

Proof. On \(\mathcal{G}_n^{\mathrm{pre}}\), the stopped and unstopped arrays agree. Proposition 45 bounds both the prefix and post-bin bad events by \(C\exp(-c2^{\eta n})\). The union bound, together with \(2\cdot2^{-3\varepsilon n}\le2^{-2\varepsilon n}\) for all sufficiently large \(n\), implies the claim after adjusting \(C\). ◻

Proof. If \(M_I=0\), then \(D_{J,n}=0\). It remains to consider the case \(M_I>0\). For every \(K>0\), \[(U-K)_+\le U\mathbf{1}_{\{U>K\}}\le K^{1-r}U^r.\] Since \(C_{I,n}\) is \(\mathcal{F}_\ell\)-measurable and \(U_J\) is independent of \(\mathcal{F}_\ell\), \[\mathbb{E}[(U_J-C_{I,n})_+\mid\mathcal{F}_\ell] \le C_{I,n}^{1-r}\mathbb{E}[U^r].\] Hence \[\lvert D_{J,n}\rvert \le \lvert c_J\rvert M_IC_{I,n}^{1-r}\mathbb{E}[U^r].\] Substituting \(C_{I,n}=2L_nT_{\ell+1,n}/M_I\), we obtain \[M_IC_{I,n}^{1-r} = 2^{1-r}L_n^{1-r}T_{\ell+1,n}^{1-r}M_I^r,\] which proves 90 . ◻

Proof. We use \[L_n^{1-r}=2^{-\kappa(r-1)n}, \qquad T_{\ell+1,n}^{1-r} \le C2^{-\varepsilon(r-1)n}2^{a(r-1)\ell},\] together with 72 : \[\mathbb{E}\left[\sum_{I\in\mathcal{D}_\ell^\circ}M_I^r\right] \le C2^{-r(s+\delta)\ell}.\] It follows that \[\mathbb{E}[R_{\ell,n}] \le C2^{-(\varepsilon+\kappa)(r-1)n} 2^{(r-1)a\ell} 2^{-r(s+\delta)\ell}.\] Since \(a=s+\delta_1\) and \(r(s+\delta)-(r-1)(s+\delta_1)=\beta_{\mathrm{comp}}\), this is 91 . ◻

Proof. For a non-mass band \(d\), write \(m=m_{\xi,d}\). By Lemmas 43 and 54, \[\mathfrak C^{\mathrm{pre}}_{\xi,d,n} \le C\sum_{\ell=0}^{m-1}2^{\ell-m}R_{\ell,n}.\] For fixed \(\xi\), only boundedly many dyadic bands \(d\) have the same value of \(m_{\xi,d}\). Hence \[\mathfrak C^{\mathrm{pre}}_{\xi,n} \le C\sum_{m=m_n^-}^{m_n^+} \sum_{\ell=0}^{m-1}2^{\ell-m}R_{\ell,n}.\] Let the right-hand side be \(E^{\mathrm{pre}}_n\). Taking expectations and using Lemma 55, \[\mathbb{E}[E^{\mathrm{pre}}_n] \le C2^{-(\varepsilon+\kappa)(r-1)n} \sum_{m=m_n^-}^{m_n^+} \sum_{\ell=0}^{m-1}2^{\ell-m}2^{-\beta_{\mathrm{comp}}\ell}.\] For fixed \(m\), \[\sum_{\ell=0}^{m-1}2^{\ell-m}2^{-\beta_{\mathrm{comp}}\ell} \le C(1+m)2^{-\gamma_{\mathrm{comp}}m}, \qquad \gamma_{\mathrm{comp}}=\min\{\beta_{\mathrm{comp}},1\}.\] Thus \[\mathbb{E}[E^{\mathrm{pre}}_n] \le C2^{-(\varepsilon+\kappa)(r-1)n} \sum_{m=m_n^-}^{m_n^+}(1+m)2^{-\gamma_{\mathrm{comp}}m}.\] Since there are \(O(n)\) terms and \(m_n^-\ge\vartheta n-C_0-1\), \[\mathbb{E}[E^{\mathrm{pre}}_n] \le Cn^2 2^{-(\varepsilon+\kappa)(r-1)n} 2^{-\gamma_{\mathrm{comp}}\vartheta n}.\] By Lemma 34, \(\gamma_{\mathrm{comp}}\vartheta>s/2\), so the last expression is bounded by \(C2^{-sn/2}2^{-cn}\) after absorbing the polynomial factor. ◻

Proof. For \(\ell\ge m_{\xi,d}\), we have \(\lvert c_J\rvert\le C\). Hence Lemma 54 implies \[\mathfrak C^{\mathrm{post}}_{\xi,n,L} \le C\sum_{m=m_n^-}^{m_n^+}\sum_{\ell=m}^{\infty}R_{\ell,n}.\] Let the right-hand side be \(E^{\mathrm{post}}_n\). Taking expectations and using Lemma 55, \[\mathbb{E}[E^{\mathrm{post}}_n] \le C2^{-(\varepsilon+\kappa)(r-1)n} \sum_{m=m_n^-}^{m_n^+}\sum_{\ell=m}^{\infty}2^{-\beta_{\mathrm{comp}}\ell}.\] Since \(\beta_{\mathrm{comp}}>0\), \[\sum_{\ell=m}^{\infty}2^{-\beta_{\mathrm{comp}}\ell} \le C2^{-\beta_{\mathrm{comp}}m}.\] Thus \[\mathbb{E}[E^{\mathrm{post}}_n] \le Cn2^{-(\varepsilon+\kappa)(r-1)n} 2^{-\beta_{\mathrm{comp}}m_n^-}.\] By Lemma 34, \(\beta_{\mathrm{comp}}\vartheta>s/2\), and since \(m_n^-\ge\vartheta n-C_0-1\), 92 follows after reducing \(c>0\). ◻

Proof. By Lemmas 56 and 57, \[\sup_{2^n\le\lvert\xi\rvert\le2^{n+1}} \left( \mathfrak C^{\mathrm{pre}}_{\xi,n} + \sup_{L\ge1}\mathfrak C^{\mathrm{post}}_{\xi,n,L} \right) \le E^{\mathrm{pre}}_n+E^{\mathrm{post}}_n,\] and \[\mathbb{E}[E^{\mathrm{pre}}_n+E^{\mathrm{post}}_n] \le C2^{-sn/2}2^{-cn}.\] Markov’s inequality therefore yields \[\mathbb{P}\left( E^{\mathrm{pre}}_n+E^{\mathrm{post}}_n > 2n^{-2}2^{-sn/2} \right) \le Cn^2 2^{-cn} \le C2^{-cn}\] after reducing \(c\). Summability follows. ◻

Proof. On \(\mathcal{G}_n^{\mathrm{pre}}\), Lemma 48 yields the identity \[c_JM_I(U_J-1)=X_{J,n}^{\mathrm{cap}}+D_{J,n}.\] Summing over \(0\le\ell<m_{\xi,d}\) proves the prefix identity, while summing over \(m_{\xi,d}\le\ell<L\) proves the finite post-bin identity. The absolute-value bounds are immediate from the definitions of \(\mathfrak C^{\mathrm{pre}}\) and \(\mathfrak C^{\mathrm{post}}\). If the absolute post-bin compensator is finite, the signed compensator converges absolutely, so passing to the limit as \(L\to\infty\) proves the limiting identity. ◻

Proof. Let \(\rho_n=(8\pi)^{-1}2^{-(s/2+\varepsilon)n}\). Take a square lattice with mesh comparable to \(\rho_n\), and retain the points needed to cover the annulus. Since the annulus is contained in a square of side length \(O(2^n)\), \[\#\mathcal{N}_n \le C\left(\frac{2^n}{\rho_n}\right)^2 \le C2^{(2+s+2\varepsilon)n}.\] ◻

Proof. For \(2^n\le\lvert\xi\rvert\le2^{n+1}\), choose \(\eta\in\mathcal{N}_n\) satisfying 93 . By 77 , \[\lvert\widehat{\nu_\circ}(\xi)-\widehat{\nu_\circ}(\eta)\rvert \le 2\pi2^{\varepsilon n}\lvert\xi-\eta\rvert \le \frac{1}{4}\,2^{-sn/2}.\] Thus \(\lvert\widehat{\nu_\circ}(\xi)\rvert\le (A+1)2^{-sn/2}\). ◻

Proof. For fixed \(\xi\) and \(d\), Corollary 15 yields the bound \(C\exp(-c2^{\eta n})\). For each \(\xi\), the number of derivative bands is \(O(n)\), and by 94 , \[\#\mathcal{N}_n\le C2^{(2+s+2\varepsilon)n}.\] The union bound therefore yields \[\mathbb{P}(\mathcal{E}_n^{\mathrm{cap}}) \le Cn2^{(2+s+2\varepsilon)n}\exp(-c2^{\eta n}).\] The stretched-exponential factor dominates the ordinary exponential prefactor, and the claim follows after decreasing \(c\) and \(\eta\), if necessary. ◻

Proof. By Proposition 42, Lemma 61, and Proposition 46, \[\mathbb{P}(\mathcal{G}_n^c)\le C2^{-cn}, \qquad \mathbb{P}(\mathcal{E}_n^{\mathrm{cap}})\le C\exp(-c2^{\eta n}), \qquad \mathbb{P}(\mathcal{E}_n^{\mathrm{comp}})\le C2^{-cn}.\] It remains to prove that, on \[\mathcal{G}_n\cap(\mathcal{E}_n^{\mathrm{cap}})^c\cap(\mathcal{E}_n^{\mathrm{comp}})^c,\] one has \(\max_{\xi\in\mathcal{N}_n}\lvert\widehat{\nu_\circ}(\xi)\rvert\le C2^{-sn/2}.\)

Fix \(\xi\in\mathcal{N}_n\). By Proposition 43, \[\widehat{\nu_\circ}(\xi) = E^{\mathrm{safe}}_{\xi,n} + \sum_{d\in\mathfrak D_{\xi,n}^{\mathrm{osc}}} \left( F^{\mathrm{pre}}_{\xi,d} + F^{\mathrm{post}}_{\xi,d} \right),\] with \(\lvert E^{\mathrm{safe}}_{\xi,n}\rvert\le C2^{-sn/2}2^{-c_{\mathrm{safe}}n}.\) On \(\mathcal{G}_n^{\mathrm{pre}}\), Lemma 58 decomposes each raw array into a centered capped part and a compensator. Since \((\mathcal{E}_n^{\mathrm{cap}})^c\) holds, for every non-mass band \(d\), \[\lvert F^{\mathrm{pre,cap}}_{\xi,d,n}\rvert + \sup_{L>m_{\xi,d}}\lvert F^{\mathrm{post,cap}}_{\xi,d,n,L}\rvert \le 2^{-sn/2}2^{-2\varepsilon n}.\] The raw post-bin limit exists by Lemma 45; on \((\mathcal{E}_n^{\mathrm{comp}})^c,\) the post-bin compensator converges absolutely, and hence the limiting centered capped post-bin term exists. Therefore, \[\lvert F^{\mathrm{pre,cap}}_{\xi,d,n}\rvert + \lvert F^{\mathrm{post,cap}}_{\xi,d,n}\rvert \le 2^{-sn/2}2^{-2\varepsilon n}.\] Since there are \(O(n)\) derivative bands, \[\sum_{d\in\mathfrak D_{\xi,n}^{\mathrm{osc}}} \left( \lvert F^{\mathrm{pre,cap}}_{\xi,d,n}\rvert + \lvert F^{\mathrm{post,cap}}_{\xi,d,n}\rvert \right) \le Cn2^{-2\varepsilon n}2^{-sn/2} \le C2^{-sn/2}.\] On \((\mathcal{E}_n^{\mathrm{comp}})^c,\) \[\sum_{d\in\mathfrak D_{\xi,n}^{\mathrm{osc}}} \left( \lvert D^{\mathrm{pre}}_{\xi,d,n}\rvert + \lvert D^{\mathrm{post}}_{\xi,d,n}\rvert \right) \le 2n^{-2}2^{-sn/2}.\] Combining the safe term, the centered capped terms, and the compensators, we obtain \[\lvert\widehat{\nu_\circ}(\xi)\rvert \le C2^{-sn/2}.\] Taking the union of the exceptional probabilities proves 95 . ◻

Proof of Theorem 40. By Lemma 62, \[\mathbb{P}\left( \max_{\xi\in\mathcal{N}_n} \lvert\widehat{\nu_\circ}(\xi)\rvert > C2^{-sn/2} \right) \le C\exp(-c2^{\eta n})+C2^{-cn}.\] By Proposition 42, \[\mathbb{P}\bigl((\mathcal{G}_n^{\mathrm{lim}})^c\bigr) \le C2^{-cn}.\] On \(\mathcal{G}_n^{\mathrm{lim}}\), Lemma 60 upgrades the grid estimate to the full annulus. Therefore, after increasing \(C\), \[\mathbb{P}\left( \sup_{2^n\le\lvert\xi\rvert\le2^{n+1}} \lvert\widehat{\nu_\circ}(\xi)\rvert > C2^{-sn/2} \right) \le C\exp(-c2^{\eta n})+C2^{-cn}.\] This is 69 . The right-hand side is summable in \(n\). Hence, by the Borel–Cantelli lemma, almost surely, for all sufficiently large \(n\), \[\sup_{2^n\le\lvert\xi\rvert\le2^{n+1}} \lvert\widehat{\nu_\circ}(\xi)\rvert \le C2^{-sn/2}.\] This completes the proof. ◻

Proof. We prove the lower bound. If \(A_{\mathrm{loc}}(W)=0\), there is nothing to prove. Assume \(A_{\mathrm{loc}}(W)>0\), and fix \(0<\sigma<A_{\mathrm{loc}}(W)\). By the definition of \(A_{\mathrm{loc}}(W)\), there exists \(r>1\) such that \[\mathbb{E}[W^r]<\infty, \qquad \frac{r-1-\log_2\mathbb{E}[W^r]}{r}>\sigma.\] Choose \(\delta>0\) such that \[\sigma+\delta < \frac{r-1-\log_2\mathbb{E}[W^r]}{r}.\] Equivalently, \[1-r+\log_2\mathbb{E}[W^r]<-r(\sigma+\delta),\] and therefore \[2^{1-r}\mathbb{E}[W^r]\le 2^{-r(\sigma+\delta)}.\]

Apply Theorem 40 with \(U=W\) and \(s=\sigma\). The interval cascade generated by \(U\), pushed forward by \(f(t)=(\cos 2\pi t,\sin 2\pi t)\), is precisely \(\mu_\circ\). Hence Theorem 40 provides a finite random constant \(C_\sigma(\omega)<\infty\) such that \(\lvert\widehat{\mu_\circ}(\xi)\rvert \le C_\sigma(\omega)\lvert\xi\rvert^{-\sigma/2}\) for all sufficiently large \(\lvert\xi\rvert\), almost surely. This is exactly the Fourier decay required for the exponent \(\sigma\); no endpoint estimate at \(\sigma=A_{\mathrm{loc}}(W)\) is needed. Consequently, \(\dim_{\mathrm F}(\mu_\circ)\ge \sigma\) almost surely on \(\mathcal{S}_\circ\).

Applying the preceding argument to every rational \(\sigma\in\mathbb{Q}\cap(0,A_{\mathrm{loc}}(W))\) and intersecting the corresponding probability-one events, we obtain \(\dim_{\mathrm F}(\mu_\circ)\ge A_{\mathrm{loc}}(W)\) almost surely on \(\mathcal{S}_\circ\). The countability of the chosen exponents is the only measurability point in the passage to the endpoint. The opposite inequality is Corollary 13. Hence Theorem 40 completes the proof. ◻

Proof. For (i), the assumption ensures that \(q-1-\log_2\mathbb{E}[W^q]>0\) for some \(q>1\), hence \(A_{\mathrm{loc}}(W)>0\). For (ii), every \(q>1\) term in the definition of \(A_{\mathrm{loc}}(W)\) is interpreted as \(0\), so \(A_{\mathrm{loc}}(W)=0\). Both conclusions follow from Theorem 47. ◻