Article Title]Inverse of the Gaussian multiplicative chaos: Decoupling and Multipoint moments

[1]Ilia<EMAIL_ADDRESS>
[1]Tomas<EMAIL_ADDRESS>
*[1]Math Department, University of Toronto, 40 St. George Street, Toronto, M5S 2E4, Ontario, Canada

In this article we study the decoupling structure and multipoint moment of the inverse of the Gaussian multiplicative chaos. It is also the second part of preliminary work for extending the work in "Random conformal weldings" <cit.> to the existence of Lehto welding for the inverse. In particular, we prove that the dilatation of the inverse homeomorphism on the positive real line is in $L^{1}([0,1]\times[0,2])$.

[MSC Classification]3Primary 60G57; Secondary 60G15

In this article we continue our systematic study of the inverse of 1d-GMC from <cit.>. We consider a Gaussian field $U(t)$ on the real line $[0,\infty)$ with logarithmic covariance $\Expe{U(t)U(s)}=-\ln\abs{t-s}$, the corresponding GMC measure $\eta$

η(I):=∫_Ie^γU_(x)-γ^2/2U_(x)^2, I⊂[0,∞)

and its inverse $Q=Q_{\eta}$ i.e. $Q(\eta(x))=x=\eta(Q(x))$ (see precise definitions below) and let $Q(a,b):=Q(b)-Q(a)$. In <cit.>, we studied the moments $\Expe{(Q(a,b))^{p}}$ for $p\in\para{-\frac{(1+\frac{\gamma^{2}}{2})^{2}}{2\gamma^{2}}, \infty}$. In this article, we study the behaviour of the multipoint moments

\begin{equation}
\Expe{\prod_{k}\para{Q^{k}(a_{k},b_{k})}^{p_{k}} } \tand \Expe{\prod_{k}\para{\frac{Q^{k}(a_{k},b_{k})}{Q^{k}(c_{k},d_{k})}}^{p_{k}} } ,  \end{equation}

with $d_{k+1}<a_{k}<b_{k}<c_{k}<d_{k}$. This requires an in-depth understanding of the correlation between two intervals $Q([a,b])$ and $Q([c,d])$. This article is also the second part of the project extending the work in <cit.> to the case of inverse conformal welding, see <cit.> for the precise statement and motivation of the problem.

§.§ Outline

Here is an outline of the paper.

* In <ref> we study the decoupling structure for the inverse.

* In <ref> we study the moments of the ratio $\frac{Q(a,a+x)}{Q(b,b+x)}$ because of its relevance to the dilatation in the <cit.>-framework.

* In <ref> we use the decoupling structure to study the multipoint moments

\begin{equation}
\Expe{\prod_{k}\para{Q^{k}(a_{k},b_{k})}^{p_{k}} } \tand \Expe{\prod_{k}\para{\frac{Q^{k}(a_{k},b_{k})}{Q^{k}(c_{k},d_{k})}}^{p_{k}} },
\end{equation}

with $d_{k+1}<a_{k}<b_{k}<c_{k}<d_{k}$.

* In the appendix, we prove GMC estimates that we need for this work.

§ MAIN RESULTS

The main results are the moments for the ratio $\frac{Q(a,b)}{Q(c,d)}$ and the product of the ratios.

§.§ Single-point Ratio

For the moments of the ratio, we go over the cases of decreasing numerator, equal-length intervals for $Q$ and equal-length intervals for $Q_{H}$, the inverse corresponding to the field $H$ in <ref>.

We go over the cases of decreasing numerator, equal-length intervals for $Q$ and the same for $Q_{H}$.

* Fix  $\beta\in  (0,0.152)$ and $\delta\leq 1$ . Assume $\abs{I}=r\delta$ for some $r\in (0,1)$ and $\abs{J}\to 0$. Then there exists $\e_{1},\e_{2}>0$, such that for each $p=1+\e_{1}$, one can find $q=1+\e_{2}\tand \tilde{q}>0$ such that

Q^δJ/Q^δI^p⪅J/δ^q r^-q̃.

* Fix $\gamma<\frac{2}{\sqrt{3}}$ and $\delta\leq 1$. Assume we have two equal-length intervals $J=(a,a+x),I=(b,b+x)\subset [0,1]$ with $b-a=c_{b-a}x$ for $c_{b-a}>1$  and $a=c_{1}\delta,b=c_{2}\delta$ and $x< \delta$.Then we have for all $p\in [1,1+\e_{1}]$ with small enough $\e_{1}>0$, a bound of the form

Q(a,a+x)/Q(b,b+x)^p ,Q(b,b+x)/Q(a,a+x)^p ≤c_a,b,p,δx/δ^-_ratio(p),

where $\e_{ratio}(p)\in (0,1)$ can be made arbitrarily small at the cost of larger comparison constant $c_{a,b,p,\delta }$. The constant $c_{a,b,p,\delta }$ is uniformly bounded in $\delta$.

* Fix $\gamma<\frac{2}{\sqrt{3}}$.  For the inverse $Q_{H}$ corresponding to the field $H$ in <Ref>, we have the same singular estimate as above for $\delta=1$

Q_H(a,a+x)/Q_H(b,b+x)^p ,Q_H(b,b+x)/Q_H(a,a+x)^p ≤c_a,b,p,δx^-_ratio(p).

One major corollary of this proposition is that the dilatation of the inverse homeomorphism on the positive real line is in $L^{1}_{loc}$. This is the analogous result to <cit.>. There they studied the dilation $K$ for the Ahlfors-Beurling extension of the homeomorphism $h(x)=\frac{\tau(0,x)}{\tau(0,1)}$. Here we study the dilatation $K_{Q}$ for the Ahlfors-Beurling extension of the inverse map $Q^{1}(0,x)$ mapping $Q^{1}:\Rplus\to \Rplus$.

We have finiteness

\begin{equation}
\int_{[0,1]\times [0,2]}K_{Q}(x+iy)\dx\dy\leq \frac{c}{2(1-\epsilon_{ratio}(1))}<\infty.
\end{equation}

§.§ Decoupling

In the decoupling the goal is to estimate the multipoint estimates

\begin{equation}
\Expe{\prod_{k}\para{\frac{Q^{k}(a_{k},b_{k})}{Q^{k}(c_{k},d_{k})}}^{q_{k}} } ,
\end{equation}

with $d_{k+1}<a_{k}<b_{k}<c_{k}<d_{k}$, using a conditional independence result for the inverse. The ratios $\frac{Q^{1}(a_{1},b_{1})}{Q^{1}(c_{1},d_{1})},\frac{Q^{2}(a_{2},b_{2})}{Q^{2}(c_{2},d_{2})}$ can be decoupled into separate conditional factors  given the event that they are well-separated $G_{1,2}:=\set{Q^{2}(d_{2})+\delta_{2} <Q^{1}(a_{1})}$

\begin{equation}
\Expe{\frac{Q^{1}(a_{1},b_{1})}{Q^{1}(c_{1},d_{1})}\frac{Q^{2}(a_{2},b_{2})}{Q^{2}(c_{2},d_{2})}\ind{G_{1,2}}  }=\Expe{\Expe{\frac{Q^{1}(a_{1},b_{1})}{Q^{1}(c_{1},d_{1})}\ind{G_{1,2}} \mid \mathcal{F}}\frac{Q^{2}(a_{2},b_{2})}{Q^{2}(c_{2},d_{2})} },
\end{equation}

see notations for detail on the filtration $\mathcal{F}$. These estimates are needed because they are the analogous of the decoupling

∏_kη^k(a_k,b_k)/η^k(c_k,d_k)^q_k =∏_kη^k(a_k,b_k)/η^k(c_k,d_k)^q_k ,

showing up in estimating the dilatation in <cit.>. Starting with a collection of increments $\set{Q^{k}(a_{k},b_{k})}_{k\geq 1}$,we aim to extract a subsequence of them that will satisfy the gap events $Q_{a}^{k}-Q_{b}^{k+m}\geq \delta_{k+m}$ that we need in order to do decoupling i.e. obtaining with high probability a subsequence $S=\set{i_{1},\cdots, i_{\abs{S}}}\subset \set{1,\cdots,N}$ of size $\abs{S}\geq c N$, $c\in (0,1)$, such that for $i_{k}\in S$

We turn this existence into a deviation estimate using graph theory and the concept of maximally independent subset. For $k,m\in [N]$ we consider the complement gap event i.e. the overlap event

\begin{equation}
O_{k,m}:=G_{k,m}^{c}:=\ind{Q^{k\wedge m}(a_{k\wedge m})-Q^{m\vee k}(b_{m\vee k}) \leq \delta_{k\vee m}   }.
\end{equation}

We then consider the random graph $G:=G_{Q}(N)$ with $N$ vertices labeled by $v_{k}:=\set{(Q^{k}(a_{k}),Q^{k}(b_{k})}$ and we say that $v_{k},v_{k+m}$  are connected by edges only when this event happens $G_{k,k+m}^{c}=1$. So the question of having a subsequence of increments that don't intersect is the same as obtaining the largest clique of vertices $S\subset G$ that do not connect to each other, this is called the independence number $\alpha(G)$ of $G$.

We show that the independence number being less than $c_{gap}N$ decays exponentially in $N$

\begin{equation}
\Proba{\alpha(G)< c_{gap}N}\lessapprox c \rho_{*}^{(1+\e_{*})N},
\end{equation}

for some $\e_{*}>0$.

This theorem shows that for each $N$, we can obtain such a subset $S\subset G$  with probability at least $1- c \rho_{*}^{(1+\e_{*})N}$.

§.§ Multipoint moments and ratios

From the above theorem for the independence number, for each $N\geq 1$ we are given a subset $S:=\set{i_{1},\cdots,i_{M}}\subset \set{1,\cdots,N}$ with $M\geq \alpha N$ such that we have large enough gaps

\begin{equation}
G_{S}:=G_{i_{1},\cdots,i_{M}}:=\bigcap_{k\in S}\set{Q_{a_{i_{k}}}^{i_{k}}-Q_{b_{i_{k+1}}}^{i_{k+1}}\geq \delta_{i_{k+1}}}.
\end{equation}

If we intersect with this event, we get the following estimate on the multipoint-moments.

We fix $p_{k},k\in S$ then we have

∏_k∈S Q^k(J_k)^p_k G_S≤c^S ∏_k∈S x_k^q_k,

for $q_{k}>0$ satisfying

p_k>-ζ(-q_k)+1, k∈S.

In the following proposition we study the first term to get moment estimates as in the single-ratio case.

We fix set $S\subset [N]$.

* (Equal length)  Fix $\beta<\frac{2}{3}$. Suppose we have equal length intervals $x_{k}=\abs{J_k}=\abs{I_k}<\delta_k$ for each $k\in S$. Fix $\delta_k\leq 1$ and intervals $J=(c_k,c_k+x_k),I=(d_k,d_k+x_k)\subset [0,1]$ with $d_k-c_k=c_{d-c,k}x_k$ for $c_{d-c,k}>1$  and $c_k=\lambda_{1,k}\delta_k,d_k=\lambda_{2,k}\delta_k$. Then we have for all $p_k\in [1,1+\e_{k}]$ with small enough $\e_{k}>0$, a bound of the form

∏_k∈S Q^k(J_k)/Q^k(I_k)^p_k G_S≤c^S∏_k∈S x_k /δ_k^-_ratio(p_k),

where $\e_{ratio}(p_k)>0$ can be made arbitrarily small at the cost of a larger proportionality constant $c$. The constants are uniformly bounded in $x_k$ and also in $\delta_k$.

* (Decreasing numerator) Fix $\beta\in \para{0,0.152}.$
Suppose that $\abs{I_{k}}=r_{k}\delta_{k}$ for $r_{k}>0$ and $\abs{J_{k}}\to 0$ for each $k\in S$. Then we have for each $p_{k}\in [1,1+\e_{ratio}]$ some  $q_{k}>1$ such that

∏_k∈S Q^k(J_k)/Q^k(I_k)^p_k G_S≤∏_k∈S J_k/δ_k^q_k.

§.§ Acknowledgements

First and foremost, we thank Eero Saksman and Antti Kupiainen. We had numerous useful discussions over many years. Eero asked us to compute the moments of the inverse and that got the ball rolling seven years ago. We also thank J.Aru, J.Junnila and V.Vargas for discussions and comments.

§ NOTATIONS AND GMC RESULTS

§.§ White noise expansion on the unit circle and real line

Gaussian fields with logarithmic covariance indexed over time or even functions $f\in \mathbb{H}$, for some Hilbert space $\mathbb{H}$, have appeared in many places in the literature: for the Gaussian free field which is a Gaussian field indexed over the $L^{2}$ Hilbert space equipped with the Dirichlet inner product <cit.> and for the Gaussian multiplicative chaos which is about Gaussian fields with logarithmic covariance plus some continuous bounded function $\ln\frac{1}{\abs{x-y}}+g(x,y)$ <cit.>. In this work we will work with the Gaussian field indexed over sets in the upper half-plane found in the works <cit.> and used in the random welding context in <cit.>. For the hyperbolic measure $\dlambda(x,y):=\frac{1}{y^{2}}\dx \dy$ in the upper half-plane $\uhp$ we consider a Gaussian process $\set{W(A)}_{A\in B_{f}(\uhp)}$ indexed by Borel sets of finite hyperbolic area:

B_f():={A⊂: λ(A)<∞; (x,y),(x',y')∈A|x-x'|<∞}

with covariance

Its periodic version $W_{per}$ has covariance

W_per(A_1)W_per(A_2):=λ(A_1∩⋃_n∈ (A_2+n)).

§.§.§ Logarithmic field on the real line
Consider the triangular region

𝒮:={(x,y)∈: x∈[-1/2,1/2]y>2x}

and define the field on the real line

It has logarithmic covariance (<cit.>)

This is the main field we will work with throughout this article. Because of the divergence along the diagonal, we upper and lower truncate by considering the field evaluated over shifts of the region

𝒮_^δ(x_0):={(x,y)∈: x∈[x_0-δ/2,x_0+δ/2]2x_0-x∨<y≤δ}.

and $\mathcal{S}_{\e}^{\delta}=\mathcal{S}_{\e}^{\delta}(0)$. The covariance of $U_{\e}^{\delta}(x)=W(\mathcal{S}_{\e}^{\delta}+x)$ is the following (<cit.>).

The truncated covariance satisfies for $\delta\geq \e$ and all $x_{1},x_{2}\in \mathbb{R}$:

[U_ ε^  δ(x_1 )U_ ε^  δ(x_2 )  ]=R_ ε^  δ(x_1-x_2):={
ln(δ/ε )-1/-1/δx_2-x_1    x_2-x_1≤ε

ln(δ/x_2-x_1) +x_2-x_1/δ-1   ≤x_2-x_1≤δ

0   δ≤x_2-x_1.
.    .

In the case of $\e=0$, we shorthand write $R^{  \delta }(\abs{x_{1}-x_{2}})=R_{0}^{  \delta }(\abs{x_{1}-x_{2}})$.

This implies that we have a bound for its difference:

U_ε_1^δ(x_1)-U_ε_2^δ(x_2)^2≤2 |x_2-x_1|+|ε_1-ε_2|/ε_2∧_1.

Therefore, being Gaussian this difference bound is true for any $a>1$

|U_ε_1^δ(x_1)-U_ε_2^δ(x_2)|^a≤c |x_2-x_1|+|ε_1-ε_2|/ε_2∧_1     ^a/2.

and so we can apply the Kolmogorov-Centsov lemma from <cit.>:

|U_ε_1(x_1)-U_ε_2(x_2)|≤M  (ln(1/ε_2))^ζ(|x_2-x_1|+|ε_1-ε_2|)^γ/ ε_2^γ+ε ,

where $\frac{1}{2}< \frac{\e_{1}}{\e_{2}}<2$, $0<\gamma<\frac{1}{2}$, $\varepsilon,\zeta>0 $ and $M=M(\varepsilon,\gamma,\zeta)$.

Also, we can consider strictly increasing larger heights i.e. $U^{h}_{0}$ denotes the field of height $h\geq \delta_{1}$. The infinitely large field  $U^{\infty}_{0}$ does not exist because it has infinite variance.

This field is needed because it allows us to work on the entire real line and not worry about the boundary effects of working on the unit circle $[0,1]/ 0\sim 1$.

§.§.§ Infinite cone field

For the region

𝒜_^δ:={(x,y)∈: x∈[-δ/2,δ/2]2x∨<y}

we let $U_{\omega,\e}^{\delta}:=\omega_{\e}^{\delta}(x):=W(\mathcal{A}+x)$. This field was considered in <cit.>. It has the covariance

[ω_ ε^  δ(x_1 )ω_ ε^  δ(x_2 )  ]={
ln(δ/ε )+1-x_2-x_1 /   x_2-x_1≤ε

ln(δ/x_2-x_1)    ≤x_2-x_1≤δ

0   δ≤x_2-x_1
.    .

This field is interesting because it has "exact scaling laws" as we will see below. And so it helps in estimating the moments for the GMC corresponding to $U$ from above.

§.§.§ Trace of the Gaussian free field on the unit circle

For the wedge shaped region

H:={(x,y)∈: x∈[-1/2,1/2] and y>2/πtan(πx)},

we define the trace GFF $H$ to be

H(x):=W_per(H+x), x∈/.

A regularized version of $H$ comes from truncating all the lower scales. Let $A_{r,\varepsilon}:=\{(x,y)\in \uhp: r>y>\varepsilon\}$ and $H_{\varepsilon}^{r}:=H\cap A_{r,\varepsilon}$ then we set:

and $H_{\varepsilon}:=h^{\infty}_{\varepsilon}$. Its covariance is as follows: For points $x,\xi\in [0,1)$, with 0 and 1 identified,  and $\varepsilon\leq r$ we find for $y:=|x-\xi|$:

[H_ε(x)H_r(ξ)]:=   2log(2)+log1/2sin(πy)     y> 2/π arctan(π/2)
log(1/)+(1/2)log(π^2ε^2+4)+2/πarctan(π/2)/ε   y≤2/π arctan(π/2)

-log(π)- y ε^-1-log(cos(π/2y))   .

From the above covariance computations, we find the difference bound for $1/2\leq \frac{\e_{1}}{\e_{2}}\leq 2$:

[|H_ε_1(x_1)-H_ε_2(x_2)|^2]⪅ |x_2-x_1|+|ε_1-ε_2|/ε_2∧_1

Therefore, by Kolmogorov-Centsov <cit.>, there is a continuous modification with the modulus:

|H_ε_1(x_1)-H_ε_2(x_2)|≤M (ln(1/ε_1))^ζ(|x_2-x_1|+|ε_1-ε_2|)^γ/ ε_1^γ+ε ,

where $0<\gamma<\frac{1}{2}$, $\varepsilon,\zeta>0 $ and $M=M(\varepsilon,\gamma,\zeta)$. A similar modulus is true for the field $U(x)$ on the real line constructed above.

This field is important because of the original problem being the conformal welding on the unit circle. So all our estimates need to return to the GMC with the field $H$.

§.§.§ Measure and Inverse notations

For a field $X$, let us define the corresponding Gaussian multiplicative chaos (GMC) as

η_X(I):=∫_Ie^X_(x),  X_:=γX_-γ^2/2E[X_^2],  I⊂[0,1) ,

provided the limit exists.

For the real-line field $U^{\delta}$ we use the following notation to match <cit.>:

η^δ(A):=η_^δ(A):=∫_Ae^γU_^δ(x)-γ^2/2ln1/,   A⊂ℝ.

Its inverse $Q^{\delta}:\Rplus\to \Rplus$ is defined as

Q^δ(x)=Q^δ_x=:=t≥0 : η^δ0,t≥x

and we will also consider increments of the inverse over intervals $I:=(y,x)$

For the inverse of the lower-truncated GMC $\eta_{\e}^{\delta}$, we will write

In the ensuing articles we will work with a sequence of truncations $\delta_{n}\to 0$, and so we will shorthand denote $U^{n}:=U^{\delta_{n}}$, $\eta^{n}$and $Q^{n}$. This is because as in <cit.>, we need to decouple nearby measures: to have the following two objects

be independent with $a_{n+1}<b_{n+1}<a_{n}<b_{n}$ decreasing to zero, we simply require $b_{n+1}+\delta_{n+1}<a_{n}$. So we see that as $\delta_{n}\to 0$, this becomes easier to achieve. Whereas if we kept $\delta_{n}=\delta$, this would be false as $a_{n}\to 0$.

Similarly, when lower truncating $U_{n}:=U_{\delta_{n}}$, we will write $\eta_{n}$ and $Q_{n}$. By slight abuse of notation, we denote

U(s)∩U(t):=W(U+s∩U+t)    U(s)∖U(t):=W(U+s∖U+t).

§.§.§ Semigroup formula

In the spirit of viewing $Q_{a}=Q(a)$ as a hitting time we have the following semigroup formula and notation for increments

Q^δ(y,x)=Q^δ(x)-Q^δ(y)=t≥0: Q^δ(y),Q^δ(y)+tδ≥x-y =:Q_x-y^δ∙Q_y^δ.

Generally for any nonnegative $T\geq 0$ we use the notation

Q_x^δ ∙T:=t≥0: T,T+tδ≥x .

To be clear the notation $Q_{x}^{\delta }\bullet Q_{y}^{\delta}$ is not equal to the composition $Q^{\delta }\circ Q_{y}^{\delta}=Q^{\delta }(Q_{y}^{\delta})=Q^{\delta }(a),$ which is the first time that the GMC $\eta$ hits the level $a:=Q_{y}^{\delta}$.

§.§.§ Scaling laws

Let $\omega^{\delta,\lambda}_{\e}(x)$ be the field with the following covariance
ln(δ/ε )+1-1/x_2-x_1   x_2-x_1≤ε

ln(δ/x_2-x_1)    ≤x_2-x_1≤δ/λ

0   δ/λ≤x_2-x_1
.    .

In our study of the scaling laws we will use the following statement (<cit.>, <cit.>.)

For $\lambda\in (0,1)$ and fixed $x_{0}$ and all Borel sets $A\subset B_{\delta/2}(x_0)$ we have

η^δ_ω(λA)_A⊂B_δ/2(x_0)λe^Ω_λη^δ,λ_ω(A)_A⊂B_δ/2(x_0) ,

where  $\overline{\Omega_{\lambda}}:=\gamma\sqrt{ \ln(\frac{1}{\lambda})}N(0,1)-\frac{\gamma^{2}}{2}\ln(\frac{1}{\lambda})$ and the measure $\eta_{\omega_{\e}}^{\delta,\lambda}$ has the underlying field $\omega^{\delta,\lambda}_{\e}(x)$

In the rest of the article, we let

We will also use the field $U_{ \varepsilon}^{\delta, \lambda}$ with covariance $R_{\e}^{\delta,\lambda}(\abs{x_{1}-x_{2}}):=\Expe{U_{ \varepsilon}^{  \delta,\lambda }(x_{1} )U_{ \varepsilon}^{  \delta,\lambda }(x_{2} )  }$

ln(δ/ε )-1/-1/δx_2-x_1+(1-λ)(1-x_2-x_1/δ)   x_2-x_1≤ε

ln(δ/x_2-x_1)-1+x_2-x_1/δ+(1-λ)(1-x_2-x_1/δ)    ≤x_2-x_1≤δ/λ

0   δ/λ≤x_2-x_1
.    .

We note that after $\abs{x_{2}-x_{1}}\geq \delta$ this covariance is negative and thus discontinuous at $\abs{x_{2}-x_{1}}=\frac{\delta}{\lambda}$ and so the GMC $\eta^{\delta,\lambda}$ cannot be defined (so far as we know in the literature, GMC has been built for positive definite covariances <cit.>). So we are forced to evaluate those fields only over sets $A$ with length $\abs{A}\leq \delta$ (see <cit.> where they also need this restriction to define the scaling law.).

The $U_{ \varepsilon}^{\delta, \lambda}$ is related to $U^{\delta}_{\e}$ in the following way

U^δ_λ(λx)Z_λ+  U_ ε^  δ,λ(x),

where $Z_{\lambda}:=N(0,\ln\frac{1}{\lambda}-1+\lambda)$ is a Gaussian independent of $ U_{ \varepsilon}^{  \delta,\lambda }$. The corresponding lognormal is denoted by

c_λ:=   1/λ-Z_λ.

We can now restate (<ref>):
[Scaling transformation for $U^{\delta}_{\e}$]
For $\lambda\in (0,1)$, $x_{0}\in\mathbb{R}$ and all Borel sets $A\subset B_{\delta/2}(x_0)$ we have

η^δ_U(λA)_A⊂B_δ/2(x_0)λe^Z_λη^δ,λ_U(A)_A⊂B_δ/2(x_0) ,    η_U^δ,λ:=η_U^δ,λ_.

For any $p\in \R$ we have the moment formula

pZ_λ  =pβr_λ (p-1).

More generally, for positive continuous bounded function $g_{\delta}:\Rplus\to \Rplus$ with $g_{\delta}(x)=0$ for all $x\geq \delta$, we let

U_ ε^δ,g (x_1 )U_ ε^  δ,g (x_2 )  ={
ln(δ/ε )-1 /-1/δx_2-x_1+g_δ(x_2-x_1)   x_2-x_1≤ε

ln(δ/x_2-x_1)-1+x_2-x_1/δ+g_δ(x_2-x_1)    ≤x_2-x_1≤δ

0   δ≤x_2-x_1
.  .

§.§.§ Singular integral

Let the singular/fusion integrals for GMC measure be defined by (<cit.>)

As before, let $Q_{R}(x)$ denote its inverse.

Using the Girsanov/tilting-lemma (<cit.>) we have

§.§.§ Filtration notations

Following <cit.>, for a Borel set $S\subset \mathbb{H}$ we define $\mathcal{U}_{S}=\mathcal{U}(S)$ to be the $\sigma$-algebra generated by the randoms variables $U(A)$, where $A$ runs over Borel subsets $A\subseteq S$. For short if $X\in \CU_{S}$ we will call such a variable measurable. Moreover, since we will be using the triangular region $\mathcal{S}$ over intervals $[a,b]$, we define

We fix decreasing sequence $(\delta_{n}=\rho_{*}^{n})_{n\geq 1}$ for some $\rho_{*}<1$. We denote the $\sigma-$algebra of upper truncated fields

All the lower truncations are measurable with respect to the filtration of the larger scales

U^n(s)=U^δ_n_0(s)∈^k([a,b]) n≥k,s∈[a,b],

and the same is true for the measure and its inverse

We define measurability with respect to random times in a standard fashion. Namely, for a single time $Q^{k}(b)$ we have the usual notion of the stopping time filtration

ℱ[0,Q^k(b)]:={A∈⋃_t≥0  ^k([0,t]): A∩{Q^k(b) ≤t}∈^k([0,t]), ∀t≥0}.

For example, $\CU^{n}\para{[0,Q^{k}(b)-s]}\subset \mathcal{F}\para{[0,Q^{k}(b)]}$ for all $n\geq k$ and $s\in [0,Q^{k}(b)]$. We also have, for $n\geq k$,

Q^k(b)≥Q^n(a)+s ∈ℱ[0,Q^k(b)].

This perspective helps us achieve a decoupling. For two interval increments $Q^{k}(a_{k},b_{k}),Q^{n}(a_{n},b_{n})$ from possibly different scales $n\geq k$ , we we will need to consider the gap event

G_k,n:=Q^k(a_k)-Q^n(b_n)≥δ_n ∈ℱ[0,Q^k(a_k)].

In this part we study the conditional decoupling for the increments of the inverse $Q^{\delta}(x)$ (for either field $U^{\delta}$ or $\omega^{\delta}$ because we only use their short-range correlations). We start by briefly discussing in a heuristic manner the main idea of decoupling the inverse.  This was inspired by the Brownian motion setting where increments of hitting times $T_{d}-T_{c}$ and $T_{b}-T_{a}$ are independent iff $c>b$ or $a>d$ <cit.>. It was also inspired by the following conditional independence for Markov chains $S_{n}=\sum_{k=1}^{n} X_{k}$ (<cit.> )

\begin{equation}\label{eq:randomwalkcondinde}
\Proba{S_{1}=i, S_{3}=j\conditional S_{2}=k}=\Proba{S_{1}=i\conditional S_{2}=k}\Proba{S_{3}=j\conditional S_{2}=k}.
\end{equation}

In words the past $S_{1}$ and the future $S_{3}$ are conditionally independent given the present $S_{2}$.

We start with two inverse increments $Q^{k}(a_{k},b_{k}),Q^{n}(a_{n},b_{n})$ from possibly different scales $n\geq k$ and an event about their gap

\begin{equation}\label{gapeventkndelta}
G_{k,n}:=\set{Q^{k}(a_{k})-Q^{n}(b_{n})\geq \delta_{n}}.
\end{equation}

In analogy with the random walk, this gap is thought of as the current step: $\set{S_{2}=k}$ and so conditional on the the gap $G_{k,n}$ we study the decoupling of $S_{1}:=Q^{n}(a_{n},b_{n})$ and $S_{3}:=Q^{k}(a_{k},b_{k})$. Here is a way we will use this idea. We start with applying the tower property with respect to $ \CF\para{[0,Q^{n}(b_{n})]}$:

=   Q^k(a_k,b_k)∈A_1,G_k,n|[0,Q^n(b_n)] Q^n(a_n,b_n)∈A_2 ,

where $A_{1},A_{2}$ are Borel subsets of $\Rplus$. Next we further clarify the source of the correlation between $S_{1}$ and $S_{3}$ via the gap $G_{k,n}$. We take discrete dyadic approximations $T_{N}(x)\downarrow Q^{k}(x)$:

\begin{equation}
T_{N}(x):=\frac{m+1}{2^{N}}\twhen \frac{m}{2^{N}}\leq Q^{k}(x) < \frac{m+1}{2^{N}}, m\geq 0.
\end{equation}

By fixing $T_{N}(a_{k})=\ell$, we have the gap event $G_{k,n}^{N}:=\set{\ell-Q^{n}(b_{n})\geq \delta_{n}}\in \CF\para{[0,Q^{n}(b_{n})]}$ and so we are left with

∑_ℓ∈D_NQ^k_b_k-a_k∙ℓ∈A_1,T_N(a_k)=ℓ|[0,Q^n(b_n)] G_k,n^N,Q^n(a_n,b_n)∈A_2 ,

where  we used the semigroup notation

\begin{equation}
Q^{k}(x)-Q^{k}(y)=\infp{t\geq 0: \etamu{Q^{k}(y),Q^{k}(y)+t}{k}\geq x-y }=:Q_{x-y}^{k }\bullet Q_{y}^{k}.
\end{equation}

But here because of the gap event $\set{\ell\geq Q^{n}(b_{n})+ \delta_{n}}$, we get that

\begin{equation}
Q^{k}_{b_{k}-a_{k}}\bullet \ell=\infp{t\geq 0: \etamu{\ell,\ell+t}{k}\geq b_{k}-a_{k} }
\end{equation}

is independent of the filtration $ \CF\para{[0,Q^{n}(b_{n})]}$. So the correlation got localized through the decomposed event $\set{T_{N}(a_{k})=\ell}$. This is the essential decoupling that we will use. It comes at the cost of obtaining the gap events with high probability and so we next study the error of not having such gaps.

§ DECAY OF OVERLAP

In this section we estimate the probability that the gap between $Q^{k+m}(b_{k+m})$ and $Q^{k}(a_{k})$ is small:

\begin{equation}
\Proba{Q^{k}(a_{k})-Q^{k+m}(b_{k+m})< \delta_{k+m}}\to 0\tas m\to +\infty.
\end{equation}

This says that the decoupling from above becomes asymptotically true in smaller scales. It shows that the complement of the gap event from <ref>

\begin{equation}
G_{k,k+m}^{c}=\set{Q^{k}(a_{k})-Q^{k+m}(b_{k+m})\leq \delta_{k+m}}
\end{equation}

decays fast in probability and so it is reasonable that the intervals $Q^{k}(a_{k},b_{k})$ and $Q^{k+m}(a_{k+m},b_{k+m})$ decorrelate for large $m$. This kind of asymptotic independence was crucial in <cit.>, where even though the annuli were correlated, after removing the upper-scales, the lower scales exhibited decoupling.

We fix $k,m\geq 1$. We let $\delta_{k}:=\rho_{\delta}^{n}$ for some $\rho_{\delta}\leq \rho_{*}$. Then the probability for a small gap can be bounded by:

Q^k(a_k)-Q^k+m(b_k+m)<δ_k+m⪅     a_k-g_k,m-b_k+m^-2  δ_k^p_1b_k+m^2-p_1

+a_k-g_k,m-b_k+m^-p_2δ_k^p_2 δ_k+m/δ_k^ζ(p_2)-1

=:   e^gap_k,k+m,

where $p_{1}\in (0,1),p_{2}\in [1,\beta^{-1})$, and the $\e$'s are arbitrarily small and sequence $g=g_{k,m}:=\rho_{g}\rho_{1}^{k}\rho_{2}^{m}$ for some $\rho_{1}\in [\rho_{\delta},\rho_{*}],\rho_{2}\in (0,1)$. Using the def:exponentialchoiceparam for general $\rho_{\delta}\leq \rho_{*}$, we bound

⪅      ρ_a- ρ_1/ρ_*^kρ_2^m-ρ_bρ_*^m^-2 ρ_δ/ρ_*^p_1kρ_*^(2-p_1)m+1/ρ_g^p_2 ρ_δ/ρ_1^p_2kρ_δ^/ρ_2^p_2^m.

The main choice in this article is $\rho_{\delta}=\rho_{*}$. And the above estimate cannot be used directly to obtain a subset of decoupled intervals. So we just include it to showcase the phenomenon of the exponential decay of having overlaps.

For simplicity we let

f_k,m:=ρ_δ/ρ_*^p_1kρ_*^(2-p_1)m,ρ_δ/ρ_1^p_2k ρ_δ^ζ(p_2)-1/ρ_2^p_2^m.

We see that $f_{k,m}$ can be naturally represented as $f_{1}^{k} f_{2}^{m}$ for some $f_{1}\in (0,1],f_{2}\in (0,1)$.

A recurring way to write the overlap event in terms of shifted GMCs is as follows

Q^k(a_k)-Q^k+m(b_k+m)< δ_k+m=a_k-η^k(Q^k+m(b_k+m)≤∫_0^δ_k+me^U^k(s+Q^k+m(b_k+m).

Here we remove the upper bound

(<ref>)≤   a_k-g_k,m-b_k+m≤η^k(Q^k+m(b_k+m)-b_k+m

for our choice of $g_{k,m}$. For the first term we use <ref> to bound it by

1/a_k-g_k,m-b_k+m^2η^k(Q^k+m(b_k+m)-b_k+m^2⪅    1/a_k-g_k,m-b_k+m^2δ_k^p_1b_k+m^2-p_1,

for $p_{1}\in (0,1)$. For the second term we use scaling by $\delta_{k}$ and <ref> for some $p_{2}\in [1,\beta^{-1})$ to upper bound

\begin{equation}
\Proba{g_{k,m}\leq \int_{0}^{\delta_{k+m}}e^{U^{k}(s+Q^{k+m}(b_{k+m})}\ds}\lessapprox g_{k,m}^{-p_{2}} \delta_{k}^{p_{2}} \para{\frac{\delta_{k+m}}{\delta_{k}}}^{\zeta(p_{2})-1}.
\end{equation}

Using the def:exponentialchoiceparam

the first term is

1/ρ_a- ρ_1/ρ_*^kρ_2^m-ρ_bρ_*^m ^2 ρ_δ/ρ_*^p_1kρ_*^(2-p_1)m,

and the second term is

1/ρ_g^p_2 ρ_δ/ρ_1^p_2kρ_δ^ζ(p_2)-1/ρ_2^p_2^m.

One possible room for sharpening this estimate is to write

=   g-δ_k+m≤∫_0^δ_k+me^U^k(s+Q^k+m(b_k+m)-δ_k+m

≤   1/g-δ_k+m^2 ∫_0^δ_k+me^U^k(s+Q^k+m(b_k+m)-δ_k+m^2.

However, it is unclear to us how to estimate this difference  because $U^{k}(s+Q^{k+m}(b_{k+m})$ is not necessarily Gaussian anymore. If one uses the supremum, then we can do a chaining argument as in maximum estimates in <cit.> but including the indicator

\begin{equation}
X_{T}:=\sup_{T}\ind{\eta(T,T+x)-x\geq g}\para{\eta(T,T+x)-x}
\end{equation}

in order to preserve positivity.

§ THE INDEPENDENCE NUMBER OF GRAPHS

In this section we study the family of increments $\set{Q^{k}(a_{k},b_{k})}_{k\geq 1}$ and aim to extract a subsequence of them that will satisfy the gap events $Q_{a}^{k}-Q_{b}^{k+m}\geq \delta_{k+m}$ that we need in order to do decoupling i.e. obtaining with high probability a subsequence $S=\set{i_{1},\cdots, i_{\abs{S}}}\subset \set{1,\cdots,N}$ of size $\abs{S}\geq c N$, $c\in (0,1)$, such that for $i_{k}\in S$

We turn this existence into a deviation estimate using graph theory and the concept of maximally independent subset.

For $k,m\in [N]$ we consider the complement gap event i.e. the overlap event

\begin{equation}
O_{k,m}:=G_{k,m}^{c}:=\ind{Q^{k\wedge m}(a_{k\wedge m})-Q^{m\vee k}(b_{m\vee k}) \leq \delta_{k\vee m}   }.
\end{equation}

We then consider the random graph $G:=G_{Q}(N)$ with $N$ vertices labeled by $v_{k}:=\set{(Q^{k}(a_{k}),Q^{k}(b_{k})}$ and we say that $v_{k},v_{k+m}$  are connected by edges only when this event happens $G_{k,k+m}^{c}=1$. So the question of having a subsequence of increments that don't intersect is the same as obtaining the largest clique of vertices $S\subset G$ that do not connect to each other, this is called the independence number $\alpha(G)$ of $G$. The independence number has the following lower bounds <cit.>

\begin{equation}
\alpha(G)\geq \frac{N}{1+\thickbar{d}}, \alpha(G)\geq \frac{N}{\Delta_{N}+1}\tand    \alpha(G)\geq\sum_{k=1}^{N}\frac{1}{d(v_{k})+1},
\end{equation}

where the kth-degree is $d(v_k):=\sum_{m\in[N]\setminus \set{k} } O_{k,m}$, the average degree is $\thickbar{d}:=\frac{\sum_{k=1}^{N}d(v_{k})}{N}$ and the maximum degree is $\Delta_{N}:=\maxl{k\in[N] } d(v_{k})$. The sharpest bound is the third one and is called Caro-Wei bound. So returning to our decoupling, by showing that the event $\alpha(G)\geq c N$, for some constant $c\in (0,1)$, has high probability as $N\to +\infty$,  we obtain $cN$ large enough gaps $\set{Q_{a_{i_{k}}}^{i_{k}}-Q_{b_{i_{k+m}}}^{i_{k+m}}\geq \delta_{i_{k+m}}}$ for $i_{j}\in S$, in particular

\begin{equation}\label{eq:decayofsmallindependennumber}
\Proba{\alpha(G)\leq c N}\leq c_{1} e^{-c_{2}N},
\end{equation}

where $c_{2}$ can be arbitrarily large by adjusting the parameters in def:exponentialchoiceparamannul.

§.§ Maximum degree

For independent interest, here we will study the tail of the maximum $\Delta_{N}:=\maxl{k\in[N] } d(v_{k})$ using a tail for sum of indicators <cit.>.

Let $X_1,\ldots,X_n$ be indicator random variables, $0 < \beta < 1$ and $ 0 < k < \beta n $.

\begin{equation}\label{main}
\Pr \left( \sum_{i=1}^n{X_i}\geq \beta n\right) \leq
\frac{1}{\binom{\beta n}{k}}\sum_{|S|=k}{\Pr\left(\wedge_{i \in S} (X_i = 1) \right)}.
\end{equation}

In particular, if $\Pr\left(\wedge_{i \in S} (X_i = 1) \right) \leq \alpha^k$ for every $S$ of size $k = \left(\frac{\beta-\alpha}{1-\alpha}\right)n$, where $0 < \alpha < \beta$ then

\begin{equation}
\Pr \left( \sum_{i=1}^n{X_i}\geq \beta n\right)  \leq e^{-D(\beta||\alpha) n},
\end{equation}

where $D(\beta||\alpha):=\beta\ln\frac{\beta}{\alpha}+(1-\beta)\ln\frac{1-\beta}{1-\alpha}$ is the Kullback-Leibler distance between $\alpha,\beta$.

By the union bound we study

\begin{equation}
\Proba{\maxl{k\in[N] } d(v_{k})\geq \beta N}\leq \sum_m\Proba{d(v_{m})\geq \beta N}.
\end{equation}

For every set $S\subset [N]$ with large enough cardinality $\abs{S}=k>\frac{N}{2}$, there exists an index $m_{*}\in S$ that is the most distant scale $m_{*}(m,S):=\arg\max\limits_{i\in S}\abs{m-i}$ such that $\abs{m-m_{*}}\geq r \abs{S}=rk$ for some uniform in $S$ fixed constant $r<\frac{1}{2}$. Moreover, as explained in <ref>, if we use exponential choices we get uniform exponential bound $e^{Q}_{k,k+m}\leq cf_{1}^{k}f_{2}^{m}$ and so we further bound

\begin{equation}
\Pr\left(\wedge_{i \in S} (G_{m,i}^{c} = 1) \right)\leq c f_{2}^{\abs{m-m_{*}}} \leq cf_{2}^{rk}.
\end{equation}

Therefore, the <ref> implies

\begin{equation}
\Proba{\max_{m}d(v_{m})\geq \beta N}\leq cNe^{-rD(\beta||f_{2}) N}.
\end{equation}

However, we will need a stronger estimate to get <ref>. This is because to get a linearly-growing independent-clique we need for the maximum degree to behave almost as a constant

\begin{equation}
\Proba{\alpha(G)\leq c_{N}N}\leq \Proba{\maxl{k\in[N] } d(v_{k})\geq \frac{1}{c_{N}}-1},
\end{equation}

for some sequence $ c_{N}$. So for this probability to decay exponentially, the maximum needs to behave almost as a constant or the sequence $c_{N}$ needs to go to zero and hence we lose the linear growth for $\alpha(G)$.

§.§ Average degree and mean

Another bound could come from the average degree. In the event

\begin{equation}
\thickbar{d}=\frac{\sum_{k=1}^{N}d(v_{k})}{N}\geq \alpha \doncl 2\sum_{k=1}^{N}\sum_{m>k}^{N}\ind{Q^{k}(a_{k})-Q^{m}(b_{m})\leq \delta_{m}}\geq \alpha N
\end{equation}

the lower bound is linear in $N$ and the sum has order $N^{2}$, so we would need a large number $N^{2}-\alpha N=N(N-\alpha)$ of indicator terms to be zero. This is reasonable because its mean is uniformly bounded.

For $L\leq N$ we have the formula

∑_k=L^N∑_m>k^NQ^k(a_k)-Q^m(b_m)≤δ_m   ≤C f_1^Lf_2/(1-f_1)(1-f_2),

for the $f_{i}\in (0,1)$ from <ref>. So for $L=1$ we have
∑_k=1^N∑_m>k^NQ^k(a_k)-Q^m(b_m)≤δ_m   ≤C f_1f_2/(1-f_1)(1-f_2).

Once again we use the exponential choice in <ref> that bound the gap

\begin{equation}
\Proba{Q^{k}(a_{k})-Q^{k+m}(b_{k+m})\leq \delta_{k+m}}\lessapprox f_{1}^{k}f_{2}^{m}.
\end{equation}

So we get

\begin{equation}
\sum_{k=L}^{N}\sum_{m=k+1}^{N}f_{1}^{k}f_{2}^{m-k}= f_{2}f_{1}^{L}\frac{-f_{2}\para{1-f_{2}^{N-L}}+f_{1}\para{1-f_{2}^{N+1-L}}-\para{1-f_{2}}f_{1}^{N+1-L}}{(1-f_{1})(1-f_{2})(f_{1}-f_{2})}.
\end{equation}

We take $N$ to infinity to be left with

\begin{equation}
\frac{f_{1}^{L}f_{2}}{(1-f_{1})(1-f_{2})}\tand \frac{f_{1}f_{2}}{(1-f_{1})(1-f_{2})}.
\end{equation}

So we see here that if we restricted to the graph $G_{L,N}$ with vertices in $[L,N]$ and set $L:=\alpha_{0}N$ for some $\alpha_{0}\in (0,1)$, then by Markov we get

\begin{equation}
\Proba{\frac{\sum_{k=\alpha_{0}N}^{N}d(v_{k})}{N}\geq \alpha } \leq \frac{1}{\alpha N} \frac{f_{1}^{\alpha_{0}N}f_{2}}{(1-f_{1})(1-f_{2})},
\end{equation}

which indeed it decays exponentially in $N$. This means that we get with high probability

\begin{equation}
\alpha(G_{L,N})\geq \frac{N-L}{1+\alpha}=\frac{(1-\alpha_{0})N}{1+\alpha} .
\end{equation}

We tested getting a large deviation result

\begin{equation}
\Proba{2\sum_{k=1}^{N}\sum_{m>k}^{N}\ind{Q^{k}(a_{k})-Q^{m}(b_{m})\leq \delta_{m}}\geq \alpha N }\lessapprox e^{-cN}
\end{equation}

using projective operators <cit.> but we got negative results because those estimates often assume a $L_{\infty}$-bounded object whereas here if we use filtrations we end up with semi-random GMC objects and thus lognormals which are unbounded.

§ EXISTENCE OF GAP EVENT ESTIMATE

In this section we obtain the exponential decay of the independent number being small.

We show that the independence number being less than $c_{gap}N$ decays exponentially in $N$

\begin{equation}
\Proba{\alpha(G)< c_{gap}N}\leq c \rho_{*}^{(1+\e_{*})N},
\end{equation}

for some $\e_{*}>0$, given the following constraint

(β+1)^2/4β-(1+_*)/1-c_gap-β^-1+1/2r_a> (1+_*)/1-c_gap.

The constraint <ref> is a quadratic inequality in $\beta$ that has a solution for $\beta<0.171452$  and some small enough $\e_{*},c_{gap}$.

Case $\rho_{\delta}<\rho_{*}$
Recall $\delta_{m}:=\rho_{\delta}^{m}$. Even though we are not using $\rho_{\delta}<\rho_{*}$ in this article, we do this case because the proof is very short and so it gives an idea of the underlying phenomenon being the decay of the overlap. We study the following event and its complement going over all totally isolated subsets of the graph

Since this event gives the existence of an independent set with cardinality larger than $c_{gap}N$, we have

α(G)< c_gapN=    E_α(G)^c

=   S⊂[1,N]

Here we only keep the leftmost subset

\begin{equation}\label{eq:sstarset}
\end{equation}

and then use union bound and <ref>

(<ref>) ≤k,k+m∈S_*Q^k(a_k)-Q^k+m(b_k+m)≤δ_k+m

⪅   f_1^(1-c_gap)Nf_2/(1-f_1)(1-f_2).

Case $\rho_{\delta}=\rho_{*}$
We start with splitting over all possibilities of positive/zero for the sums of edge-weights with $m>k$

α(G)< c_gapN   =∑_S⊂[N]    α(G)< c_gapN, ⋂_i∈SY_i≥1 , ⋂_i∈S^cY_i= 0   ,

If we have $\abs{S^{c}}\geq c_{gap}N$ , then we obtain that the vertices corresponding to $S^{c}:=\set{j_{1},...,j_{R}}$ do not overlap with each other. This is because the zero sum $Y_{j_{1}}=0$ implies that the node $v_{j_{1}}$ doesn't have any edges with $v_{j_{2}},...,v_{j_{R}}$; the zero sum $Y_{j_{2}}=0$ similarly implies that $v_{j_{2}}$ doesn't have any edges with $v_{j_{3}},...,v_{j_{R}}$ and so on and so forth. But this not allowed due to the event $\alpha(G)< c_{gap}N$. Therefore, we reduce the sum over only subsets that satisfy $\abs{S^{c}}\leq  c_{gap}N\doncl \abs{S}\geq (1-c_{gap}) N$

S≥(1-c_gap) N    α(G)<c_gapN, ⋂_i∈SY_i≥1 , ⋂_i∈S^cY_i= 0   .

So it remains to study the probability term

\begin{equation}\label{eq:Ykdecompseventprob}
\Proba{\bigcap_{i\in S} \set{\sum_{m>i, m\in S }\ind{O_{i,m}} \geq 1}}=\Proba{\bigcap_{i\in S} \bigcup_{m_{i}>i, m\in S } O_{i,m}},
\end{equation}

for  $S:=\set{s_{1},...,s_{M}}\subset [N]$ with $s_{i}<s_{i+1}$ and $M=  \ceil{(1-c_{gap}) N}$.

Step 1: Independent and Shrinking intervals

Here we use a similar strategy as done for the terms $L_{n,m}$ in <cit.>. Namely we invert back to GMC and split into shrinking intervals. We insert the deviation estimate

\begin{equation}
\set{Q^{m}(b_{m})\leq g_{m,k}},
\end{equation}

for sequence $g_{m,k}:=\rho_{g}^{m-k}\rho_{*}^{k}$ with $\rho_{g}>\rho_{*}$, to get the inclusion

Q^k(a_k)-Q^m(b_m)≤δ_m⊆   Q^k(a_k)≤g_m,k+δ_m∪Q^m(b_m)≥g_m,k

=   a_k≤k0,g_m,k+ δ_m∪b_m≥m0,g_m,k.

So we reduced to studying the tail and small-ball of GMC respectively. We decompose

\begin{equation}
\spara{0,g_{m,k}+ \delta_{m}}=\bigcup_{\ell\geq 1}I_{m,k,\ell}\tand   \spara{0,g_{m,k}}=\bigcup_{\ell\geq 1}J_{m,k,\ell},
\end{equation}

for sequences of partition intervals $I_{m,k,\ell}=\spara{x_{m,k,\ell},x_{m,k,\ell}+y_{m,k,\ell}}$ and $J_{m,k,\ell}=\spara{z_{m,k,\ell},z_{m,k,\ell}+w_{m,k,\ell}}$. In particular, we let

x_m,k,ℓ:=   ρ_*(g_m,k+δ_m)ρ_*^ℓ=ρ_*ρ_g^m-kρ_*^k+ρ_*^mρ_*^ℓ,

y_m,k,ℓ:=   (1-ρ_*) ρ_g^m-kρ_*^k+ρ_*^mρ_*^ℓ,

z_m,k,ℓ:=   ρ_*ρ_g^m-kρ_*^kρ_*^ℓ,

w_m,k,ℓ:=   (1-ρ_*) ρ_g^m-kρ_*^k+ℓ

to obtain the partition constraints

x_m,k,ℓ+1+y_m,k,ℓ+1=x_m,k,ℓ since  ρ_*^2+(1-ρ_*)ρ_* =ρ_*,

∑_ℓ≥0y_m,k,ℓ=  g_m,k+δ_m,

and same for $J_{m,k,\ell}$

z_m,k,ℓ+1+w_m,k,ℓ+1=z_m,k,ℓ since  ρ_*^2+(1-ρ_*)ρ_* =ρ_*,

∑_ℓ≥0w_m,k,ℓ=  g_m,k.

u_k,ℓ:=   (1-ρ_u)/2a_kρ_u^ℓ    ũ_m,ℓ:=1/ρ_bb_mρ_ũ^ℓ=δ_mρ_ũ^ℓ,

for $\rho_{u}:=\rho_{*}^{r_{u}},\rho_{\tilde{u}}:=\rho_{*}^{r_{\tilde{u}}}$, with $r_{u},r_{\tilde{u}}>0$, to get

∑_ℓ≥0u_k,ℓ=a_k/2<a_k ∑_ℓ≥0ũ_m,ℓ=b_m/ρ_b(1-ρ_ũ)>b_m.

In summary,  we upper bound by the following union over $\ell$

(<ref>)≤   j∈S⋃_m_j=j+1,m_j∈SE_j,m_j=j∈SE_j,

E_a,b:=   ⋃_ℓ≥0u_a,ℓ≤aI_b,a,ℓ∪⋃_ℓ̃≥0ũ_b,ℓ̃≥bJ_b,a,ℓ̃ ,

E_j:=   ⋃_m_j=j+1,m_j∈SE_j,m_j.

Step 2: Decoupling
As done at step <cit.>

we pull the first scale $i_{1}$ for which $r_{i_{1}}>0$ using the union bound

(<ref>)≤   ∑_m_i_1=i_1+1,m_i_1∈S ∑_ℓ_1≥0u_i_1,ℓ_1≤i_1I_m_i_1,i_1,ℓ_1∩j∈S∖{i_1}E_j

+   ∑_m_i_1=i_1+1,m_i_1∈S ∑_ℓ̃_1≥0ũ_m_i_1,ℓ̃_1≥m_i_1J_m_i_1,i_1,ℓ̃_1∩j∈S∖{i_1}E_j.

For the first sum we pick the smallest scale $i_{2}\geq i_{1}$ such that

and if none exists then we have $x_{m_{i_{1}},\ell_{1}}\leq g_{s_{M}}+\delta_{s_{M}+1}+\delta_{s_{M}}$. We have the analogous choice for the second sum, namely the smallest scale  $i_{2}\geq m_{i_{1}}$ with

and if none exists then we have $w_{m_{i_{1}},i_{1},\tilde{\ell}_{1}}\leq g_{s_{M}}+\delta_{s_{M}+1}+\delta_{s_{M}}$. By repeating the process, for each $L\in [N]$,  we get a subsequence $S_{L}:=\set{i_{1},...,i_{L}}\subset S$ such that we have the bound

(<ref>)≤   L=1^M  m_i=i+1,m_i∈S,
i∈S_L^NE⊂S_L  ℓ_j≥0
j∈E  ℓ̃_j≥0

where we sum over all subsets $E\subset S_{L}$ and the $m_{i_{j}},\ell_{j},\tilde{\ell}_{j}$ and $i_{i_{j+1}}$ satisfy

depending on whether we study tail or small-ball respectively. Next we translate this condition into concrete inequalities for $m,\ell,\tilde{\ell}$.

Step 3: Covering condition
First we study the $x-$terms. Using the def:exponentialchoiceparam, we write

g_i_2,1+δ_i_2+1+δ_i_2≤x_m_i_1,i_1,ℓ_1    ρ_gρ_*^i_2+ρ_*^i_2+1+ρ_*^i_2≤ρ_* ρ_g^m_i_1-i_1ρ_*^i_1+ρ_*^m_i_1ρ_*^ℓ_1

c_1,ρ:=   ln1/ρ_g/ln1/ρ_*=r_gc_2,ρ:=ln1/ρ_*^-1ln1+ρ_*/ρ_g^m_i_1-i_1^-1 +lnρ_g+ρ_*+1^-1  .

Conversely, if $r\in S$ and $r\leq i_{2}$, since $i_{2}$ is the smallest-choice,  we have the reverse inequality

g_r,1+δ_r+1+δ_r> x_m_i_1,i_1,ℓ_1   i_1+ℓ_1+(m_i_1-i_1)r_g+c_2,ρ≥r.

For the $w-$terms we similarly have

By taking $\rho_{*}$ small enough, we get

\begin{equation}
c_{2,\rho},c_{3,\rho}\in (0,\frac{1}{2})
\end{equation}

and so since $i_{2},r$ are integers, we obtain

r≤i_1+ℓ_1+(m_i_1-i_1)r_g< i_2.

By repeating this process , we get a subsequence $S_{L}:=\set{i_{1},...,i_{L}}\subset S$ that satisfy covering condition for $S$

\begin{equation}
S\subset [i_{1},C_{1}]\cup [i_{2},C_{2}]\cup...\cup      [i_{L},C_{L}],
\end{equation}

\begin{equation}
\end{equation}

Therefore, due to the covering condition we obtain

Step 4: Estimates For the product over $E$ in <ref> we simply use Markov for $p>1$

u_i_j,ℓ_j≤i_jI_m_i_j,i_j,ℓ_j≤   c I_m_i_j,i_j,ℓ_j/δ_i_j^ζ(p)/u_i_j ,ℓ_j/δ_i_j^p

≤   c

Here we need to extract a bound $\rho_{*}^{\frac{1}{1-\alpha}(1+\e_{*})(\ell_{j}+(m_{i_{j}}-i_{j})r_{g}+1)}$. As mentioned in <ref>, the $\ell_{j}\geq 0$ and $m_{i_{j}}-i_{j}\geq 1$, so in order to extract the term $+1$ we need to use the second inequality $m_{i_{j}}-i_{j}\geq 1$. Assuming $\ell_{j}=0$, we are left with

and so we require

r_g(m_i_j-i_j)ζ(p)-pr_a> 1/1-c_gap (1+_*)(r_g(m_i_j-i_j)+1).

Secondly, for $\ell_{j}\geq 1$ we similarly require

ζ(p)-p r_u>1/1-c_gap(1+_*).

We get

We discuss those constraints at the compatibility section below.

For the product over $E^{c}$ in <ref> we apply the small ball estimate <ref> for the case $n>k$ and $\ell\geq 0$, we need to take the parameters as follows:

ρ_1=ρ_ũ=ρ_*^r_ũ, ρ_2=ρ_*,ρ_3=ρ_g=ρ_*^r_g

a_1=r_ũ,a_2=1, a_3=r_g,

n=m_i_j, k=i_j,ℓ=ℓ̃_j,:=ln(1 - ρ_*)/ln1/ρ_*.

To apply this corollary, we also need

r_ũ>1>r_g r_ũ>(1+)(1+β).

We study these conditions later in <ref>. Therefore, we get the bound

ũ_m_i_j,ℓ̃_j≥m_i_jJ_m_i_j,i_j,ℓ̃_j≤cρ_*^ℓ_j c_2      ℓ_j>c_g(m_i_j-i_j)

ρ_*^ℓ_j c_3,1+(m_i_j-i_j) c_3,2      0≤ℓ_j≤c_g(m_i_j-i_j),

for $c_{g}:=1-r_{g}$ and

c_2:=   1/32r_ũ-(1+)-β(1+)/√(β1+  )      ^2,

c_3,1:=   (r_ũ-(1+))ln1/ρ_*^α_0-1c_3,2:=(1-r_g)ln1/ρ_*^α_0-1,

for $\e_{0}\approx 1$ and $\alpha_{0}>1$. To use the covering condition <ref> we again need to extract a bound $\rho_{*}^{(1+\e_{*})(\ell_{j}+(m_{i_{j}}-i_{j})r_{g}+1)}$. For the second range $0\leq\ell_{j}\leq c_{g}(m_{i_{j}}-i_{j})$  we simply require

where the $2$ is needed as above in order to give the term $+1$. For the first range $\ell_{j}>c_{g}(m_{i_{j}}-i_{j})$, we need to use the summation in order to extract the exponent $c_{g}(m_{i_{j}}-i_{j})$. By requiring

we split this exponent to bound

\begin{equation}
\rho_{*}^{\ell_{j} c_{2}  }\leq \rho_{*}^{\ell_{j} \frac{c_{2}}{2}+\ell_{j} \frac{c_{2}}{2}  }\leq \rho_{*}^{ \frac{1}{1-c_{gap}}(1+\e_{*})\para{\ell_{j}+r_{g} (m_{i_{j}}-i_{j})+1} }.
\end{equation}

Step 5: Conclusion
Here we can finally use the covering condition <ref>. We take out an $\e_{*}(1-\lambda)$-part, for small $\lambda\in (0,1)$, in order to extract exponent $M$

(<ref>)≤   ρ_*^1/1-c_gap(1+_**(1-λ))ML=1^Mc^L  m_i=i+1,m_i∈S
i∈S_L^NE⊂S_L  ℓ_j≥0
j∈E  ℓ̃_j≥0

The $\ell$-sums give a factor $\para{\frac{1}{1-\rho_{*}^{\e_{*}\lambda}}}^{L}$. The  $m_{i}$ sums give a factor $\para{\frac{\rho_{*}^{r_{g}\e_{*}\lambda}}{1-\rho_{*}^{r_{g}\e_{*}\lambda}}}^{L}$, and so we are left with

(<ref>)≤   ρ_*^(1+_**(1-λ))NL=1^Mc^Lρ_*^(r_g_*(1-λ))L≤c̃ ρ_*^(1+_**(1-λ))M.

Step 6: Compatibility of constraints
In summary we have the constraints

1)     r_g(m_i_j-i_j)ζ(p)-pr_a> 1/1-c_gap (1+_*)(r_g(m_i_j-i_j)+1),

2)     ζ(p)-p r_u>1/1-c_gap(1+_*),

3)     r_ũ>1>r_g r_ũ>(1+)(1+β),

4)     c_3,1=(r_ũ-(1+))ln1/ρ_*^α_0-1>1/1-c_gap(1+_*),

5)     c_3,2=(1-r_g)ln1/ρ_*^α_0-1>1/1-c_gap2(1+_*),

6)     c_2=1/32r_ũ-(1+)-β(1+)/√(β1+  )      ^2>1/1-c_gap2(1+_*)r_g/1-r_g,1.

The (4,5) constraints are immediate by taking small enough $\rho_{*}$. The constraints (3,6) also follow by taking large $r_{\tilde{u}}$. So it remains to study the first two constraints. We maximize $\zeta(p)$ by setting $p=\frac{\beta^{-1}+1}{2}$ (allowed since $\beta<1$) to get $\zeta(p)=\frac{(\beta+1)^{2}}{4\beta}$. For the second constraint we simply take small enough $r_{u}$. Since $m_{i_{j}}-i_{j}\geq 1$, the first constraint simplifies to

r_g(β+1)^2/4β-(1+_*)/1-c_gap-β^-1+1/2r_a> (1+_*)/1-c_gap.

To further simplify we take $r_{g}=1-\e$ for small $\e>0$ (at the cost of larger $r_{\tilde{u}}$ in (6))

(1-)(β+1)^2/4β-(1+_*)/1-c_gap-β^-1+1/2r_a> (1+_*)/1-c_gap.

Ratios of increments
In this section we study the moments of the ratios of $Q_{\eta}$. There is an analogous result for $\eta$ in <cit.>, where using the lognormal-scaling law one can bound

\begin{equation}
\Expe{\para{\frac{\eta^{\delta}(I)}{\eta^{\delta}((J)}}^{p}  }\leq c\para{\frac{\abs{I}}{\abs{J}}}^{\zeta(p)},
\end{equation}

for all small enough intervals $I,J$ ie. $\abs{I},\abs{J}<\delta$ that are close and for $1<p<\beta^{-1}$.

In the conformal welding part for the inverse, the dilatation $K$ is controlled by ratios $\frac{Q^{\delta}(J)}{Q^{\delta}(I)}$. The case $\abs{I}=\abs{J}$ represents the most singular behaviour of the dilatation as it approaches the real axis. In that case we return to the scaling law for GMC.

In the following proposition we will study moments of the analogous ratio of the inverse increments.

As in <cit.> we have two separate statements.

We go over the cases of decreasing numerator, equal-length intervals for $Q$ and the same for $Q_{H}$.

* Fix  $\beta\in  (0,0.152)$ and $\delta\leq 1$ . Assume $\abs{I}=r\delta$ for some $r\in (0,1)$ and $\abs{J}\to 0$. Then there exists $\e_{1},\e_{2}>0$, such that for each $p=1+\e_{1}$, one can find $q=1+\e_{2}\tand \tilde{q}>0$ such that

Q^δJ/Q^δI^p⪅J/δ^q r^-q̃.

* Fix $\gamma<\frac{2}{\sqrt{3}}$ and $\delta\leq 1$. Assume we have two equal-length intervals $J=(a,a+x),I=(b,b+x)\subset [0,1]$ with $b-a=c_{b-a}x$ for $c_{b-a}>1$  and $a=c_{1}\delta,b=c_{2}\delta$ and $x< \delta$.Then we have for all $p\in [1,1+\e_{1}]$ with small enough $\e_{1}>0$, a bound of the form

Q(a,a+x)/Q(b,b+x)^p ,Q(b,b+x)/Q(a,a+x)^p ≤c_a,b,p,δx/δ^-_ratio(p),

where $\e_{ratio}(p)\in (0,1)$ can be made arbitrarily small at the cost of larger comparison constant $c_{a,b,p,\delta }$. The constant $c_{a,b,p,\delta }$ is uniformly bounded in $\delta$.

* Fix $\gamma<\frac{2}{\sqrt{3}}$.  For the inverse $Q_{H}$ corresponding to the field $H$ in <Ref>, we have the same singular estimate as above for $\delta=1$

Q_H(a,a+x)/Q_H(b,b+x)^p ,Q_H(b,b+x)/Q_H(a,a+x)^p ≤c_a,b,p,δx^-_ratio(p).

§ DILATATION IS IN $L^{1}_{LOC}$

One major corollary of this proposition is that the dilatation of the inverse homeomorphism on the positive real line is in $L^{1}_{loc}$. This is the analogous result to <cit.>. In particular, here we study the dilatation $K_{Q}$ corresponding to the inverse of scale one $Q^{1}:\Rplus\to \Rplus$.

We have finiteness

\begin{equation}
\int_{[0,1]\times [0,2]}K_{Q}(x+iy)\dx\dy\leq \frac{c}{2(1-\epsilon_{ratio}(1))}<\infty.
\end{equation}

We use the dilatation bound that depends linearly on the homeomorphism <cit.>: for each $I\in \mathcal{D}:=\bigcup_{n\geq 0}\mathcal{D}_{n}$ and Whitney square $C_{I}:=\set{(x,y)\in I: x\in I\tand y\in [2^{-n-1},2^{-n}]}$, we have

\begin{equation}
\sup_{z\in C_{I}}K(z)\leq \sum_{\mathbf{J}\in j_{\ell}(I) } \frac{Q(J_{1})}{Q(J_{2})}+ \frac{Q(J_{2})}{Q(J_{1})},
\end{equation}

\begin{equation}
j_{\ell}(I):=\set{ \mathbf{J}=(J_{1},J_{2}): J_{1},J_{2}\in \mathcal{D}_{n+5}\tand  J_{1},J_{2}\subset j_{0}(I)}.
\end{equation}

and $j_{0}(I)$ be the union of $I$ and its neighbors in $\mathcal{D}_n$. So by covering the rectangle $[0,1]\times [0,2]$ by Whitney squares we get similarly to <cit.> the bound

\begin{equation}
\int_{[0,1]\times [0,2]}K_{Q}(x+iy) \dx\dy \leq c\sum_{n=1}^{\infty}\sum_{k=1}^{2^{n}}\int_{\frac{k-1}{2^{n}}}^{\frac{k}{2^{n}}}\frac{1}{x^{\e_{ratio}(1)}}\frac{1}{2^{n+1}}\dx =\frac{c}{2(1-\epsilon_{ratio}(1))}<\infty.
\end{equation}

Since we have the bound for the unit circle case too in <Ref>, we have the above corollary there as well. For the inverse homeomorphism $Q_{H}(x\eta_{H}(1))$ is a bit more tricky because one has to first scale out the total mass.

One seemingly major research problem are extending the constraints on $\gamma$. It is unclear how to remove these given the current techniques. One source of them are in <Ref> which are because in the the maximum modulus estimates we get singular factors. So it seems that one would have  to avoid the modulus techniques all together. We talk about various routes in the further directions of <cit.>.

§ MOMENTS OF THE RATIO: PROOF OF <REF>

The proof is tricky for the following three reasons.

* lack of prefactor scaling law  The lognormal scaling law shows up as a prefactor in <Ref> and so there is no cancelation of these lognormals as in the GMC-case <cit.>. And even then one still needs the constraint $Q(a,a+x)\leq x$ which an event that goes to zero in probability. Even for $a=0$ using the log-normal scaling law we actually have

Q_x≥δx = x≥η^δxδ=√(βln1/x)+ln1/η^δ(δ)/√(βln1/x)≥N(0,1)  →1 x→0,

for independent Gaussian $N(0,1)$. This means that $Q_{x}$ decays a lot slower than linear.

* $Q_{a}$ is random In the GMC-case $\frac{\eta(a,a+x))}{\eta(b,b+x)}$, one can apply <Ref> at $x_{0}=a$. However, here $Q_{a}$ is random. In fact, the distribution of $U(s+Q_{a})$ will likely be assymmetric and non-Gaussian.  In <cit.> for the case of Brownian motion $U_{s}=B_{s}$ , they show that $\expo{\sigma B_{T_{a}}-(\frac{\sigma^{2}}{2}-\mu)T_{a}}\eqdis\frac{\sigma^{2}}{4} X_{a}^{2}$ for a Bessel process $X_{a}$ at location $a$ and index $\nu=\frac{2\mu}{\sigma^{2}}-1$. So in this work we decompose $Q_{a}$ and then extract a factor $x^{-p}$. If there was a way to deal with this issue, then perhaps it would also remove that singular factor.

* $Q_{b-a}\bullet Q_{a}$ is random In the GMC-case, one can restrict $b-a\leq c x$ so that the scaling law will go through. However, the analogous object to $b-a$ here is $Q_{b-a}\bullet Q_{a}$ and that can be arbitrarily large.
We turn the ratio into

Q(a,a+x)/Q(b,b+x)^p≤Q(a,a+x)/Q_b-a∙Q_a^p_1^1/p_1 Q_b-a∙Q_a/Q(b,b+x)^p_2 ^1/p_2 ,

in order to match scales and heuristically apply the scaling law wrt $\tilde{x}\approx Q_{b-a}\bullet Q_{a}$. This fits the perspective of truly working in the image and thus needing to do all the operations as if $Q$ is the domain i.e. keep reverting to shifted-GMC.

To understand the overall strategy, we start with a heuristic argument. We first approximate over the dyadics: $Q^{\delta,n}(a)=\ell\in D_{n}(0,\infty)$. Using the semigroup formula

Q^δ(x)-Q^δ(y)=   t≥0: Q^δ(y),Q^δ(y)+tδ≥x-y

=   :Q_x-y^δ∙Q_y^δ,

and the approximation $Q^{\delta,n}(a)$ for $ Q^{\delta}(a)$ we replace $Q^{\delta}(J)=Q^{\delta}(a,a+x)=Q^{\delta}_{x}\bullet Q^{\delta}_{a}$ by

Q_x∙Q^δ,n(a)=  Q_x∙ℓ=t≥0:  ℓ,ℓ+tδ≥x

and we replace $Q^{\delta}(I)=Q^{\delta}(b,b+x)$ by

Q_x∙ Q_b-a∙ℓ+ℓ=t≥0: Q_b-a∙ℓ+ℓ,Q_b-a∙ℓ+ℓ+tδ≥x .

So by applying we get

∑_ℓ Q^δ(J)/Q^δ(I)>R,Q^δ,n(a)=ℓ,Q_b-Q_a≤ρδ

≤∑_ℓQ_x∙ℓ/Q_x∙Q_b-a∙ℓ+ℓ >R,Q_b-a∙ℓ≤ρδ^1/b_1 Q^δ,n(a)=ℓ^1/b_2,

for $b_{i}$ conjugates. The first factor is perfectly set up for the log-scaling law for $\lambda=x$.  Now the sum is proportional to $2^{n}$ and thus diverges when we take the limit to approximate the original $Q^{\delta}(a)$. So instead we will decompose as in the proof of <Ref>.

§.§ Proof of $\abs{I}=r\delta$ for $r>0$ and $\abs{J}\to 0$

Here we simply scale by $\delta$ and apply :

Q^1 J/δ^pQ^1 I/δ^-p

≤    Q^1 J/δ^pp_1^1/p_1Q^1 I/δ^-pp_2^1/p_2

for $\frac{1}{p_{1}}+\frac{1}{p_{2}}=1$ and then just apply the inverse increment moments <Ref>

\begin{equation}
\leq \para{c_{11}\para{\frac{J}{\delta}}^{q_{11}}+c_{12}\para{\frac{J}{\delta}}^{q_{12}}}^{1/p_{1}}\para{c_{21}r^{-q_{21}}+c_{22}r^{-q_{22}}}^{1/p_{2}}.
\end{equation}

Next we further explore the constraints for getting $\frac{q_{11}}{p_{1}},\frac{q_{12}}{p_{1}}\geq 1+\e_{ratio}$ for some $\e_{ratio}>0$. The constraint from <Ref> is

\begin{equation}
\zeta(-q_{11})+pp_{1}>1  \tand q_{12}>pp_{1},
\end{equation}

where we simplified and relabelled to match the current setting. The second inequality is cleared by taking large $q_{12}>0$. We set $\frac{q_{11}}{p_{1}}=1+\e_{ratio}$ and obtain a constrain for $\e_{ratio}$ in terms of $\beta$. This choice translates second inequality to

Since $pp_{2}\in [1,\beta^{-1})$, we further take

and so we study

(<ref>)   β^-1p_1-1/p_1>(1+_ratio)1+βp_1(1+_ratio)+1+1/p_1.

Since the inverse has all its positive moments we are free to take arbitrarily large $p_{1}>1$. This has a solution $p_{1}>1, \e_{ratio}\in (0,1)$ when we further require $\beta\in (0,0.152)$.

§.§ Proof of $\abs{J}=\abs{I}<\delta$ and ratio of the form $\frac{Q(a,a+x)}{Q(b,b+x)}$ for $a<b$

Here we will use a similar argument as in the computation of moment increments in <cit.>.

Now we proceed to the proof. For simplicity we assume $\delta=1$ by scaling

\begin{equation}
\Expe{\para{\frac{Q^{\delta}(a,a+x)}{Q^{\delta}(b,b+x)}}^{p}} = \Expe{\para{\frac{Q(\frac{a}{\delta},\frac{a+x}{\delta})}{Q(\frac{b}{\delta},\frac{b+x}{\delta})}}^{p}}
\end{equation}

and instead just abuse notation to write $\Expe{\para{\frac{Q(a,a+x)}{Q(b,b+x)}}^{p}}$ whereas in the proposition we write the fractions. We fix $\rho_{b-a}\in (0,1)$ that will be taken to be small enough below given the various constraints at the cost of large comparison constants. We will go over all the possibilities of $Q\para{b,b+x},Q\para{a,a+x}\tand Q_{b-a}\bullet Q_{a}$ depending on them being less or greater than one or $\rho_{b-a}$ respectively.

§.§.§ Case $Q(b,b+x)\in [1,\infty)$

This is the easiest case, since we can simply lower bound $Q(b, b+x)$ by $1$

\begin{equation}
\Expe{\para{\frac{Q(a,a+x)}{Q(b,b+x)}}^{p}\ind{Q\para{b,b+x}\geq 1}}\leq \Expe{\para{Q\para{a,a+x}}^{p}}.
\end{equation}

So below we assume $Q(b,b+x)\in [0,1]$.

§.§.§ Case $Q(b,b+x)\in [0,1]$ and $Q_{b-a}\bullet Q_{a}\geq \rho_{b-a}$

Here we use the upper bound $x\leq b-a$ and just study

Step 1: Bounding the ratio $\para{\frac{Q_{b-a}\bullet Q_{a}}{Q(b,b+x)}}^{p}$
Here we will decompose the numerator event $E_{b-a}^{[\rho_{b-a},\infty)}$ with $E_{b-a}^{k}:=\set{Q_{b-a}\bullet Q_{a}\in \spara{b_{k},b_{k+1}}=:I_{b-a}^{k}}$ for $k\geq 1$ and the sequence  $b_{k}:=\rho_{b-a}k^{\lambda_{2}}$ for some $\lambda_{2}\in (0,1)$ and $k\geq 1$

Q_b-a∙Q_a/Q(b,b+x)^pE_b-a^[ρ_b-a,∞),E_b,b+x^[0,1]    =∑_kQ_b-a∙Q_a/1∧Q(b,b+x)^pE_b-a^k.

Then we use the upper bound and apply layercake

=   ∑_kb_k+1^p∫_1^∞η(Q_b,Q_b+t^-1/p)≥x ,E_b-a^k.

Here we need to further decompose $Q_{a}$

D_ℓ^a:=     Q(a)∈[a_ℓ,a_ℓ+1]=:I_ℓ,

for the $a_{\ell}:=\rho_{1}\ell^{\lambda_{1}},\ell\geq 0$ and $\rho_{1},\lambda_{1}\in (0,1)$. Using the semigroup formula $Q(b)= Q_{b-a}\bullet Q(a)    +Q(a) $, we further upper bound by

(<ref>)≤   ∑_kb_k+1^p∫_1^∞∑_ℓ

max_u∈I_ℓ,ũ∈I_b-a^kη(ũ+u,ũ+u+t^-1/p)≥x, c_b-ax≥min_u∈I_ℓη(u,u+b_k) ,D_ℓ^a,

where we only kept the upper bound in $E_{b-a}^{k}$. Now that we used the decomposition events $D_{\ell}^{a}$ that we will study later, we bound by the minimum

\begin{equation}
\minp{\Proba{\max_{u\in I_{\ell},\tilde{u}\in I_{b-a}^{k}}\eta(\tilde{u}+u,\tilde{u}+u+t^{-1/p})\geq x, c_{b-a}x\geq \min_{u\in I_{\ell}}\eta(u,u+b_{k})},\Proba{D_{\ell}^{a}}}.
\end{equation}

For the first two events, we further apply the FKG inequality <Ref>

max_u∈I_ℓ,ũ∈I_b-a^kη(ũ+u,ũ+u+t^-1/p)≥x, c_b-ax≥min_u∈I_ℓη(u,u+b_k)

≤   max_u∈I_ℓ,ũ∈I_b-a^kη(ũ+u,ũ+u+t^-1/p)≥xc_b-ax≥min_u∈I_ℓη(u,u+b_k).

For the first event depending on $t$, we simply use the positive moments in <Ref> for $t^{-1}\leq 1$ and $p_{1}\in [1,\beta^{-1})$

\begin{equation}\label{eq:reallinelinemaxbound}
\Proba{\max_{u\in I_{\ell},\tilde{u}\in I_{b-a}^{k}}\eta(\tilde{u}+u,\tilde{u}+u+t^{-1/p})\geq x}\lessapprox x^{-p_{1}} t^{-\frac{\zeta(p_{1})-1}{p}} ,
\end{equation}

where the constant is uniformly bounded since $\abs{I_{\ell}}+\abs{I_{b-a}^{k}}\leq 10$ (*)conditionone, which will be shown below. So for the t-integral to converge we require $\zeta(p_{1})>p+1$. In summary returning to <Ref> we will study

(<ref>)≤   ∫_[1,∞)∑_k∑_ℓb_k+1^px^-p_1 t^-ζ(p_1)-1/pc_b-ax≥min_u∈I_ℓη(u,u+b_k), D_ℓ^a,

where we also applied translation invariance. We let $\alpha_{1}:=\frac{\zeta(p_{1})-1}{p}$ and

R_1,k:=   c_b-ax≥min_u∈I_ℓη(u,u+b_k) R_2,ℓ:=D_ℓ^a.

Step 2: The t-integral
When $t\geq 1$, we have

x^-p_1 t^-α_1R_1,k ≤R_2,ℓt≥x^-p_1 R_1,kR_2,ℓ^-1^1/α_1 =:N_1.

Therefore, all together, by integrating in $t$ we have

(<ref>)≤   ∑_kb_k+1^p∑_ℓN_1  R_2,ℓ+N_1^1-α_1 x^p_1 R_1,k

=   ∑_kb_k+1^p(∑_ℓx^-p_1 R_1,kR_2,ℓ^-1^1/α_1   R_2,ℓ

+x^-p_1 R_1,kR_2,ℓ^-1^(1-α_1)/α_1  x^-p_1R_1,k)

≤   2∑_kb_k+1^px^-p_1 R_1,k^1/α_1 ∑_ℓR_2,ℓ^1-1/α_1.

Step 3: The $\ell$-sum for $R_{2,\ell}$
For the event $D_{\ell}^{a}:=\set{  Q(a)\in [a_{\ell},a_{\ell+1}]=:I_{\ell}}$, we choose $a_{\ell}:=\rho_{1}\ell^{\lambda_{1}},\ell\geq 0$ and $\rho_{1},\lambda_{1}\in (0,1)$. For $\ell=0$, we bound by 1 and for $\ell\geq 1$ we apply Markov for $r_{1}>0$

\begin{equation}
R^{2}_{\ell}= \Proba{D_{\ell}^{a}}\leq  \Proba{a\geq \eta(a_{\ell})}\lessapprox (a/\rho_{1})^{r_{1}}\ell^{-\lambda_{1}r_{1} }=:c\ell^{-\lambda_{1}r_{1} },
\end{equation}

where for finiteness we require $\lambda_{1}r_{1}>1$. To get summability in the  $\ell$-sum <Ref> we require

\begin{equation}
\lambda_{1}r_{1}\para{1-\frac{1}{\alpha_{1}}}  >1.
\end{equation}

Here we can take large $r_{1}$ to get this as long as again $\alpha_{1}>1\doncl \zeta(p_{1})>p+1$. Since $p=1+\e_{1}$, we require

which forces $\beta<\frac{2}{3}$.

Step 4: The $k$-sum and $x$
We choose $b_{k}:=\rho_{b-a}k^{\lambda_{2}}$ for some $\lambda_{2}\in (0,1)$ and $k\geq 1$ in order to cover the $[\rho_{b-a},\infty)$ image. Together with above constraint on $\lambda_{1}$, we also get the constraint <ref>. We apply the negative moments <Ref> for some $p_{2}>0$: when $b_{k}\leq 1$ we bound by

\begin{equation}\label{eq:sumkevent1}
R_{1,k}=\Proba{c_{b-a}x\geq \min_{u\in I_{\ell}}\eta(u,u+b_{k})}\lessapprox x^{p_2} b_{k}^{\zeta(-p_{2})-1}=x^{p_2}\rho_{b-a}^{\zeta(-p_{2})-1}k^{(\zeta(-p_{2})-1)\lambda_{2}}.
\end{equation}

when $b_{k}\geq 1$ we bound by

\begin{equation}\label{eq:sumkevent2}
R_{1,k}=\Proba{c_{b-a}x\geq \min_{u\in I_{\ell}}\eta(u,u+b_{k})}\lessapprox x^{p_2} b_{k}^{-p_{2}-1}=x^{p_2}\rho_{b-a}^{-p_{2}-1}k^{-(p_{2}-1)\lambda_{2}}.
\end{equation}

So to get finiteness in the $k$-sum we require

λ_2p_2-1/α_1-p=λ_2pp_2-1/ζ(p_1)-1 >1,

which is possible for large enough $p_{2}$.
For $x$ we have the exponent

where again we take large enough $p_{2}$ to get a positive exponent.

This case was simple because we truncated away from zero for $Q_{b-a}\bullet Q_{a}\in [\rho_{b-a},\infty)$ which then came at the price of a very large constant in $\rho_{b-a}^{-p_{2}}$ that doesn't affect overall finiteness.

§.§ Case $Q_{b-a}\bullet Q_{a}\in [0,\rho_{b-a}]$ and $Q(b,b+x)\in [0,1]$

Here we finally study the singular case of both $Q_{b-a}\bullet Q_{a}$ and $Q(b,b+x)$ going to zero and thus we need to use the lognormal scaling law. As mentioned in <Ref>, a key part is the difference $b-a$ which is required to decay proportionally to $x$ i.e. $b-a\leq c_{b-a}x$ for the scaling law to go through. Here we deal with a random variable $Q_{b-a}\bullet Q_{a}$ and as mentioned in <cit.> even for $a=0$ using the log-normal scaling law we actually have

Q_x≥δx = x≥η^δxδ=√(βln1/x)+ln1/η^δ(δ)/√(βln1/x)≥N(0,1)  →1 x→0,

for independent Gaussian $N(0,1)$. This means that $Q_{x}$ decays a lot slower than linear. So here the scaling law needs to be done with respect to a different variable, not $x$. Heuristically by turning to the shifted GMC, we will do the scaling law with respect to $\tilde{x}\approx Q_{b-a}\bullet Q_{a}$.

Step 1: Decomposing $Q_{b-a}\bullet Q_{a}$ and $Q_{a}$ We decompose the numerator

\begin{equation}
E_{b-a}^{k}(Q_{a}):=\set{Q_{b-a}\bullet Q_{a}\in I_{k}:=\spara{a_{k+1},a_{k}}\subset [0,\rho_{b-a}]},
\end{equation}

for the fixed small constant $\rho_{b-a}\in (0,1)$ and the sequence $a_{k}:=\rho_{b-a}k^{-\lambda_{b-a}}$ for $k\geq 1$ and $\lambda_{b-a}>0$, to obtain as above

≤   1+∑_k∫_1^∞ηQ_b,Q_b+ a_kt^-1/p ≥x,  ηQ_a,Q_a+ a_k+1≤b-a≤ηQ_a,Q_a+ a_k  ,

where for convenience we removed the small ratio part i.e. $\para{\frac{Q_{b-a}\bullet Q_{a}}{Q(b,b+x)}}^{p}\leq 1$. So we again write $b-a=c_{b-a}x$ for $c_{b-a}>1$. In preparation of scaling law, we will use the lower bound in $E_{b-a}^{k}(Q_{a})$ to form a ratio and use the semigroup formula $Q(b)= Q_{b-a}\bullet Q(a)    +Q(a) $ to bound

ηQ_b,Q_b+ a_kt^-1/p /ηQ_a,Q_a+ a_k+1≥c_b-a^-1,E_b-a^k(Q_a)

≤   max_u∈I_kηQ_a+u,Q_a+u+ a_kt^-1/p /ηQ_a,Q_a+ a_k+1≥c_b-a^-1,E_b-a^k(Q_a) .

For each $k\geq 1$, we also decompose the variable $Q(a)$ with events $D_{\ell,k}^{a}:=\set{  Q(a)\in [c_{\ell,k},c_{\ell+1,k}]=:I_{\ell,k}}$ to overall bound

(<ref>)≤   ∑_k∫_1^∞∑_ℓ

u∈I_k,ũ∈I_ℓ,kηũ+u,ũ+u+ a_kt^-1/p/ηũ,ũ+ a_k+1≥c_b-a^-1 ,E_b-a^k(ũ)D_ℓ,k^a .

Now that we used the event $D_{\ell,k}^{a}$, we bound by a inequality

∑_ℓ∑_kD_ℓ,k^a ^1/b_12

·   ∫_1^∞u∈I_k,ũ∈I_ℓ,kηũ+u,ũ+u+ a_kt^-1/p/ηũ,ũ+ a_k+1≥c_b-a^-1 ,E_b-a^k(ũ) ^1/b_11
for $b_{11}^{-1}+b_{12}^{-1}=1$. So now we study

u∈I_k,ũ∈[0, I_ℓ,k]ηũ+u,ũ+u+ a_kt^-1/p/ηũ,ũ+ a_k+1≥c_b-a^-1,E_b-a^k(ũ) ,

where we also used translation invariance.

Step 2: Scaling law step Here we are finally in the position to apply the lognormal-scaling law <ref> for $\lambda=\frac{a_{k+1}}{\rho_{b-a}}\in (0,1)$ by requiring the condition

ρ_b-a/a_k+1 I_ℓ,k+a_k+a_k t^-1/p≤1/2.

We let $L_{1}:=\frac{\rho_{b-a}\abs{I_{\ell,k}}}{a_{k+1}}, L_{2}:=\frac{\rho_{b-a} a_{k}}{a_{k+1}}$ and we study this condition in <Ref>. Therefore, by the scaling law we get

S_k,ℓ(u,ũ) ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ_b-a≥c_b-a^-1,E_b-a^k(ũ,N(0,1)) ,

where $S_{k,\ell}(u,\tilde{u}):=\set{u\in \frac{\rho_{b-a}}{a_{k+1}}I_{k},\tilde{u}\in\spara{0, \frac{\rho_{b-a}}{a_{k+1}}\abs{I_{\ell,k}}}}$ and

E_b-a^k(ũ,N(0,1))   :=ηũ,ũ+ ρ_b-a≤ρ_b-a(b-a)/a_k+1e^-Z̅_λ≤ηũ,ũ+ L_2

=ln1/ηũ,ũ+ L_2≤σ_kN(0,1)-σ_k^2/2 +ln a_k+1/ρ_b-a(b-a)≤ln1/ηũ,ũ+ ρ_b-a,

and the independent $\bar{Z}_{\lambda}:=\sigma_{k}N(0,1)-\frac{\sigma_{k}^{2}}{2}$ (defined in <Ref>) and $\sigma_{k}^{2}:=2\beta\para{\ln\frac{1}{\lambda}-1+\lambda}$. Here we further integrate out this independent lognormal
∫_S_k,ℓ(u,ũ) ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ_b-a≥c_b-a^-1,E_b-a^k(ũ,y) g_N(0,1)(y),

where $g_{N(0,1)}(y):=\frac{1}{\sqrt{2\pi}}\expo{-y^{2}/2}$. Now that we used the event $E_{b-a}^{k}$, we again bound by another for $b_{21}^{-1}+b_{22}^{-1}=1$. So in summary we are studying

∑_ℓ∑_kD_ℓ,k^a ^1/b_12

·∫_1^∞{∫_⋃_S_k,ℓ(u,ũ) ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ_b-a≥c_b-a^-1^1/b_21
·⋃_S_k,ℓ(u,ũ) E_b-a^k(ũ,y)^1/b_22 g_N(0,1)(y)}^1/b_11.

Step 3: Ratio modulus
Now, we apply Markov inequality for $q_{0}\in [1,\beta^{-1})$ and inequality for $q_{1}^{-1}+q_{2}^{-1}=1$ with $q_{0}q_{1}\in \para{0,\frac{2}{\gamma^{2}}}$ to separate numerator and denominator. We then use <Ref> to get

max_S_k,ℓ(u,ũ)ηu+ũ,u+ũ+L_2/t^1/p  ^q_0q_1^1/q_1⪅    L_2/t^1/p ^ζ(q_0q_1)-1/q_1≈t^-1/q_1pζ(q_0q_1)-1 ,

and the <Ref> to get

max_ũ∈0, L_1ηũ,ũ+ρ_b-a ^-q_0q_2^1/q_2⪅ ρ_b-a^(ζ(-q_0q_2)-1)/q_2,

which is uniformly bounded since we fix $\rho_{b-a}$.

Step 4: Integral over $t$ We return to the t-integral in  <Ref>. Here we have

∫_1^∞{⋃_S_k,ℓ(u,ũ) ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ_b-a≥c_b-a^-1^1/b_21 }^1/b_11

≤   c_ρ_b-a ∫_1^∞t^-1/b_11b_21q_1pζ(q_0q_1)-1 .

So in order for the t-integral to be finite we require

\begin{equation}\label{eq:tintegralcosnstraint}
\frac{\zeta(q_{0}q_{1})-1}{b_{11}b_{21}}>  pq_{1}.
\end{equation}

We take $p=1+\e$. We also take $q_{0}=\beta^{-1}-\e$ by taking $q_{1}=1+\e_{4}$ and thus $q_{1}$'s conjugate $q_{2}$ to be arbitrarily large. Finally, we take $b_{11}=1+\e_{5}$  (see <Ref> where we take $b_{12}$ arbitrarily large) and $b_{21}=1+\e_{5}$ (we take $b_{22}$ arbitrarily large). So all together we get the constraint

\begin{equation}\label{eq:constraintongamma}
\beta^{-1}>2+o(\e_{i})\doncl \gamma<\sqrt{\frac{2}{2+o(\e_{i})}}  \doncl \gamma<1.
\end{equation}

Step 5: Event $D_{\ell,k}^{a}$
Here we study $D_{\ell,k}^{a}:=\set{  Q(a)\in [c_{\ell,k},c_{\ell+1,k}]=:I_{\ell,k}}$.  For $\ell=0$, we bound by $\Proba{D_{\ell,k}^{a}}\leq 1$ and for $\ell\geq 1$ we bound by a minimum as we will see below.  We set $c_{\ell,k}:=\rho_{a}a_{k}c_{\ell}$ for some small enough $\rho_{a}\in (0,1)$ and $c_{\ell}:=\ell^{\lambda_{a}}$ for $\lambda_{a}\in (0,1)$ so that indeed $c_{\ell+1}-c_{\ell}\to 0$ as $\ell\to +\infty$. Then we have the following bounds by applying Markov for $p_{1,a}>0$ and $p_{2,a}\in [1,\beta^{-1})$ by <Ref>

\begin{equation}
\Proba{ \eta^{1}(c_{\ell,k})\leq a}\leq \branchmat{\para{\rho_{a}a_{k}c_{\ell}}^{\zeta(-p_{1,a})}  & c_{\ell,k}\leq 1 \\\para{\rho_{a}a_{k}c_{\ell}}^{-p_{1,a}}  & c_{\ell,k}\geq 1    }   =:B_{1,\ell,k}
\end{equation}

\begin{equation}
\Proba{ a\leq \eta^{1}(c_{\ell+1})}\leq \branchmat{\para{\rho_{a}a_{k}c_{\ell+1}}^{\zeta(p_{2,a})}  & c_{\ell+1}\leq 1 \\ \para{\rho_{a}a_{k}c_{\ell+1}}^{p_{2,a}}  & c_{\ell+1}\geq 1    } =:B_{2,\ell,k}
\end{equation}

and bound by their minimum $ \Proba{D_{\ell,k}^{a}}\leq\min_{i=1,2}B_{i,\ell,k}$. When $1\geq c_{\ell+1,k}\geq c_{\ell,k}$, we have

\begin{equation}
B_{1,\ell,k}\leq B_{2,\ell,k}    \doncl\ell \geq \para{\frac{c_{\ell}}{c_{\ell+1}}}^{\alpha(1)/\lambda_{a}}\para{\rho_{a}a_{k}}^{-1/\lambda_{a}},
\end{equation}

where $\alpha(1):=\frac{\zeta(p_{2,a})}{\zeta(p_{2,a})-\zeta(-p_{1,a})}$. When $c_{\ell+1,k}\geq 1\geq c_{\ell,k}$, we have

\begin{equation}
B_{1,\ell,k}\leq B_{2,\ell,k}    \doncl \ell\geq \para{\frac{c_{\ell}}{c_{\ell+1}}}^{\alpha(2)/\lambda_{a}}\para{\rho_{a}a_{k}}^{-1/\lambda_{a}},
\end{equation}

where $\alpha(2):=\frac{p_{2,a}}{p_{2,a}-\zeta(-p_{1,a})}$. Finally, when $c_{\ell+1,k}\geq c_{\ell,k}\geq 1$, we have

\begin{equation}
B_{1,\ell,k}\leq B_{2,\ell,k}    \doncl \ell\geq \para{\frac{c_{\ell}}{c_{\ell+1}}}^{\alpha(3)/\lambda_{a}}\para{\rho_{a}a_{k}}^{-1/\lambda_{a}},
\end{equation}

where $\alpha(3):=\frac{p_{2,a}}{p_{2,a}+p_{1,a}}$. The ratio is uniformly bounded $\frac{c_{\ell}}{c_{\ell+1}}\in [\frac{1}{2},1]$ and so we upper bound by terms of the form

∑_ℓ≥1 D_ℓ,k^a ^1/b_12  ⪅     ρ_aa_k^-p_1,a/b_12(ρ_aa_k)^-1/λ_a^1-p_1,aλ_a/b_12

=   c 2ρ_aa_k^-1/b_12λ_a  =:c_aa_k^-1/b_12λ_a  ,

where for finiteness of the $\ell$-sum we require $\frac{p_{1,a}\lambda_{a}}{b_{12}}>1$ since as $\ell\to +\infty$ we eventually get $c_{\ell}\geq 1$. Due to existence of negative moments of GMC, here we can take $p_{1,a}>0$ arbitrarily large so that we allow $\lambda_{a}=1-\e_{a}$ and $b_{12}$ arbitrarily large which in turn allows $b_{11}=1+\e_{5}$ in <Ref>.

In summary the <Ref> is upper bounded by

∑_ka_k^-1/b_12λ_a  {∫_⋃_S_k,ℓ(u,ũ) E_b-a^k(ũ,y)^1/b_22 g_N(0,1)(y)}^1/b_11.
Step 6: Requirement for scaling
We set $a_{k}:=\rho_{b-a}k^{-\lambda_{b-a}}$ for $k\geq 1$ and $\lambda_{b-a}>0$. We verify the constraints in <Ref>. We required that $\frac{a_{k+1}}{\rho_{b-a}}\in (0,1)$ and

ρ_b-a/a_k+1 I_ℓ,k+a_k+a_k t^-1/p≤1/2,

where $t\geq 1$. For $k\geq 1$ we have that $\frac{a_{k}}{a_{k+1}}\leq 2^{\lambda_{b-a}} $ and for $\ell\geq 0$ we have

\begin{equation}
\abs{I_{\ell,k}}=\rho_{a}a_{k}((\ell+1)^{\lambda_{a}}-(\ell)^{\lambda_{a}})\leq    \rho_{a}a_{k}
\end{equation}

and so we ask for $\frac{a_{k+1}}{\rho_{b-a}}\leq  \frac{\rho_{b-a}}{ 2^{\lambda_{b-a}}\rho_{b-a}}<1$ and

ρ_b-a/a_k+1 I_ℓ,k+a_k+a_k t^-1/p≤ρ_b-a 2^λ_b-a ρ_a+2<1/2.

Step 7: The event $E^{k}_{a,a+x}$
Here we are studying

·(∫_⋃_u∈[0,ρ_b-a] ln1/ηu,u+ L_2≤σ_ky-σ_k^2/2 +ln a_k+1/ρ_b-a(b-a)≤ln1/ηu,u+ ρ_b-a ^1/b_22

where we had set $a_{k}=\rho_{b-a}k^{-\lambda_{b-a}}$ for $k\geq 1$ and $\lambda_{b-a}>0$ and $g_{N(0,1)}$ is the standard normal density,

σ_k   :=√(2βlnρ_b-a/a_k+1-1+a_k+1/ρ_b-a)

M_k,x   :=σ_k^2/2+ln ρ_b-a(b-a)/a_k+1=lnx/a_k+1^1+β+ β-1+a_k+1/ρ_b-a +ln ρ_b-a c_b-a .

For the probability term, one upper bound follows by taking a minimum and using the <Ref> and <Ref> for $z^{0}_{2}>0$ and $z_{1}^{0}\in (0,\beta^{-1})$

⋃_u∈[0,ρ_b-a] ln1/ηu,u+ L_2≤σ_ky-M_k,x≤ln1/ηu,u+ ρ_b-a ^1/b_22

≤   e^σ_ky-M_k,x≤max_u∈[0,ρ_b-a]1/ηu,u+ ρ_b-a ,e^-(σ_ky-M_k,x)≤max_u∈[0,ρ_b-a]ηu,u+ L_2 ,1^1/b_22

≤    c_ρ_b-a,L_2 e^-(σ_ky-M_k,x)z_2^0/b_22,e^(σ_ky-M_k,x)z_1^0/b_22,1,

for $z_{2}:=\frac{z_{2}^{0}}{b_{22}}>0$ and $z_{1}:=\frac{z_{1}^{0}}{b_{22}}$. Therefore, we are left with

∑_k≥1a_k^-1/b_12λ_a  ∫_ e^-(σ_ky-M_k,x)z_2,e^(σ_ky-M_k,x)z_1,1   g_N(0,1)(y)^1/b_11,

where we also brought in the power $b_{11}$ because we take $b_{11}=1+\e$ for small $\e>0$. Since it is almost one, we will ignore $b_{11}$ below. We actually only need to study

Step 8: Constraints from the minimum
Next we identify how the minimum behaves

σ_k^2/2+ln ρ_b-a(b-a)/a_k+1   -σ_ky>0

lnx/a_k+1^1+β+ β-1+a_k+1/ρ_b-a +ln ρ_b-a c_b-a -y√(2βlnρ_b-a/a_k+1-1+a_k+1/ρ_b-a)>0

(1+β)λ_b-aln(k+1)+ lnx+C_1(k)-y√(2βλ_b-aln(k+1)+C_2(k))>0

ln(k+1)+ lnx+C_1(k)/(1+β)λ_b-a-y/(1+β)λ_b-a√(2βλ_b-aln(k+1)+C_2(k))>0,

C_1(k):=    β-1+a_k+1/ρ_b-a    +ln ρ_b-a c_b-a C_2(k):=   2β-1+a_k+1/ρ_b-a.

For all other variables fixed, as $k$ grows to infinity this inequality becomes true. Next we focus on getting concrete lower bounds for $k$. The complication happens as $x\to 0$, so to simplify analysis we just focus on the case

The reverse inequality of larger $x$ is simpler because $\ln x$ can be treated as a constant that can be ignored by taking large $\lambda_{b-a}$ and so we leave it. For $y\leq 0$, the inequality <Ref> holds for:

ln(k+1)≥     ln1/x-C_1(k)/(1+β)λ_b-a=:N_x,k

and for $y> 0$ the inequality <Ref> holds for:

ln(k+1)≥     ln1/x-C_1(k)/(1+β)λ_b-a+y/(1+β)λ_b-a^2βλ_b-a

+1/2y/(1+β)λ_b-a√(4 ln1/x-C_1(k)/(1+β)2β+y/(1+β)2β^2+4C_2(k))

=:   N_x,k+g_y+p(x,y,k),

N_x,k:=   ln1/x-C_1(k)/(1+β)λ_b-a,     g_y:=y/(1+β)λ_b-a^2βλ_b-a,

p(x,y,k):=   1/2y/(1+β)λ_b-a√(4 ln1/x-C_1(k)/(1+β)2β+y/(1+β)2β^2+4C_2(k)).

Whereas the opposite inequality $ M_{k,x}-\sigma_{k}y\leq 0$ holds  for

y≤0 ln(k+1)≤     N_x,k+g_y-p(x,y,k)

and for

y>0ln(k+1)≤     N_x,k+g_y+p(x,y,k).

We go over each the four cases of the signs of $ M_{k,x}-\sigma_{k}y\tand y$ to bound the $k-$sum and the $y-$integral and extract a singular factor for $x$.

Case $M_{k,x}-\sigma_{k}y>0$ and $y\leq 0$
Here we just study the tail-sum $k\geq \expo{N_{x,k}}$ for $e^{(\sigma_{k}y-M_{k,x})z_{1}}$

∑_k≥N_x,ka_k^-1/b_12λ_a  -z_1M_k,x+σ_kyz_1   ⪅    x^-z_1∑_k≥N_x,kk^-(z_1(1+β)-1/b_12λ_a )λ_b-a-C_1(k)

⪅    x^-z_1 x^(z_1(1+β)-1/b_12λ_a )λ_b-a-1/(1+β)λ_b-a

⪅    x^-1/(1+β)λ_b-a-1/b_12λ_a

for $z_{1}(1+\beta)\lambda_{b-a}$. For summability we require $(z_{1}(1+\beta)-\frac{1}{b_{12}\lambda_{a} })\lambda_{b-a}>1$, which is possible by taking large enough $\lambda_{b-a},b_{12}$. So the exponent of $x$ is of the form $x^{-\e}$ for small $\e>0$.

Case $M_{k,x}-\sigma_{k}y>0$ and $y> 0$
We start by extracting the $y$-part in the summand $a_{k}^{-\frac{1}{b_{12}\lambda_{a} }}e^{(\sigma_{k}y-M_{k,x})z_{1}}$. We ignore the first part $a_{k}^{-\frac{1}{b_{12}\lambda_{a} }}$ since we can take arbitrarily small power. For the second part we factor out the leading asymptotic

=   y√(2βλ_b-aln(k+1)+C_2(k))+ln1/x-C_1(k)-(1+β)λ_b-aln(k+1)

=    ln1/x-C_1(k)-ln(k+1)(1+β)λ_b-a-y/√(ln(k+1))√(2βλ_b-a+C_2(k)/ln(k+1)).

Then we use the $k-$ lower bound $G:=\expo{N_{x,k}+g_{y}+p(x,y,k)}$ to bound by

(<ref>)≤   ln1/x-C_1(k)-ln(k+1)(1+β)λ_b-a-y/√(ln(G))√(2βλ_b-a+C_2(k)/ln(G)).

Since the focus in the $y-$integral is for large $y$, we see that for large $y$ the $G$ term is asymptotically equal to

G≈N_x,k+ y^22β/(1+β)^2λ_b-a.

Therefore, the $y-$part in <Ref> behaves like

\begin{equation}
\frac{y}{\sqrt{\ln (G+1)}}\sqrt{2\beta\lambda_{b-a}+\frac{C_{2}(k)}{\ln (G+1)}}\approx    \frac{y}{\ln\frac{1}{x}+y^{2}}.
\end{equation}

However, outside we are multiplying by the standard normal density $e^{-y^{2}}$ and so we still have integrability in the $y-$integral in <Ref>. Also the effect of small $x$ only decreases this bound and so it doesn't contribute a singular factor to $x$. Finally, for simplicity we also drop the $y-$part from the lower bound and thus return to the same tail sum $k\geq N_{x,k}$ as in the previous case in <Ref>.

Case $M_{k,x}-\sigma_{k}y\leq 0$ and $y\leq 0$
Here we study the bulk-sum $ k\leq \expo{ N_{x,k}+g_{y}-p(x,y,k)}=:\tilde{G}$. The part $e^{-(\sigma_{k}y-M_{k,x})z_{2}}$ actually grows in $k$, so we simply keep the $1$ and thus get the upper bound

∑_1≤k≤G̃ 1⪅1/(1+β)λ_b-aln1/x+y/(1+β)λ_b-a^2βλ_b-a.

Here again we get a singular power $x^{-\e}$. We also get integrability in the $y$-integral by taking large $\lambda_{b-a}$ and using the normal density $e^{-y^{2}/2}$.

Case $M_{k,x}-\sigma_{k}y\leq 0$ and $y>0$
Here we study the bulk-sum $ k\leq \expo{ N_{x,k}+g_{y}+p(x,y,k)}=:\bar{G}$. For the $g(y)+p(x,y,k)$ we have an upper bound without $k$ since $C_{1}(k),C_{2}(k)$ are bounded in $k$

≤   y/(1+β)λ_b-a^2βλ_b-a+1/2y/(1+β)λ_b-a√(4 ln1/x+1/(1+β)2β+y/(1+β)2β^2+4):=R_x,y.

So we get

\begin{equation}
\sum_{1\leq k\leq \bar{G} } \expo{\frac{1}{(1+\beta)\lambda_{b-a}}\ln\frac{1}{x}+R_{x,y}}.
\end{equation}

We are left with the $y-$integral

=   ∫_0^∞1/λ_b-ac_1y^2+c_2y√(c_3ln1/x+c_4y^2+4) -y^2/2,

for some positive constants $c_{i}$. Integrability is clear by taking large $\lambda_{b-a}$. Next we study the singular factor $x$. Using the triangle inequality $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ we are left with

∫_0^∞-c̃_11/2-1/λ_b-ay^2+c̃_2/λ_b-ay√(ln1/x) ,

for some positive $\tilde{c}_{i}$. Here we use the general formula

\begin{equation}\label{eq:gaussianintegralpolynomial}
\int_{0}^{\infty}e^{ay-y^{2}/2}dy=\sqrt{\frac{\pi}{2}}e^{a^{2}/2}Erf(\frac{a}{\sqrt{2}}+1),
\end{equation}

where $Erf(z)$ is the error function $\int_{0}^{z}e^{-t^{2}}\dt$ and so for $a:=\frac{\tilde{c}_{2}}{\lambda_{b-a}}\sqrt{\ln\frac{1}{x}}$ we get the singular factor

\begin{equation}
\expo{\frac{1}{2}\para{\frac{\tilde{c}_{2}}{\lambda_{b-a}}}^{2}\ln{\frac{1}{x}}}=x^{-\frac{1}{2}\para{\frac{\tilde{c}_{2}}{\lambda_{b-a}}}^{2}}.
\end{equation}

§.§ Proof of $\abs{J}=\abs{I}<\delta$ and ratio of the form $\frac{Q(b,b+x)}{Q(a,a+x)}$ for $a<b$

We will try to keep find the similarities with the previous case of $\frac{Q(a,a+x)}{Q(b,b+x)}$ as much as possible to shorten it.

§.§.§ Case $Q(a,a+x)\in [1,\infty)$

This is again the easiest case, since we can simply upper bound the denominator

\begin{equation}
\Expe{\para{\frac{Q(b,b+x)}{Q(a,a+x)}}^{p}\ind{Q\para{a,a+x}\geq 1}}\leq  1^{-p}\Expe{\para{Q\para{b,b+x}}^{p}}.
\end{equation}

§.§.§ Case $Q(a,a+x)\in [0,1]$ and $E_{\rho,b-a}:=\set{Q_{b-a+x}\bullet Q_{a}\geq \rho_{b-a+x}}$

Because the ratio is flipped here we need to instead divide and multiply by $Q(a,b+x)$. That leaves us with studying

\begin{equation}\label{eq:mainsingularatio}
\Expe{\para{\frac{Q(a,b+x)}{Q(a,a+x)}}^{p}\ind{Q\para{a,a+x}\geq 1}}
\end{equation}

since $Q(a,b+x)\geq Q(b,b+x)$. Here we repeat the steps up to <Ref> to get

(<ref>)≤∑_kb_k+1^p∫_1^∞∑_ℓ max_u∈I_ℓη(u,u+t^-1/p)≥x, (c_b-a+1)x≥min_u∈I_ℓη(u,u+b_k) ,D_ℓ^a,

for $D_{\ell}^{a}:=\set{  Q(a)\in [a_{\ell},a_{\ell+1}]=:I_{\ell}}$. From here we proceed the same way we did with <Ref> to get a non-singular power of $x$ as in <Ref>.

§.§.§ Case $Q_{b-a+x}\bullet Q_{a}\in [0,\rho_{b-a+x}]$ and $Q(a,a+x)\in [0,1]$

We decompose the numerator

\begin{equation}
E_{b-a+x}^{k}(Q_{a}):=\set{Q_{b-a+x}\bullet Q_{a}\in I_{k}:=\spara{a_{k+1},a_{k}}\subset [0,\rho_{b-a+x}]},
\end{equation}

for arbitrarily small $\rho_{b-a+x}\in (0,1)$ and $a_{k}:=\rho_{b-a+x}k^{-\lambda_{b-a}}$ as before, to obtain

≤    1+∑_k∫_1^∞ηQ_a,Q_a+ a_kt^-1/pp_1 ≥x,  ηQ_a,Q_a+ a_k+1≤b-a+x≤ηQ_a,Q_a+ a_k  .

Here we again need to apply the scaling law and so we write it as a ratio

ηQ_a,Q_a+ a_kt^-1/pp_1 /ηQ_a,Q_a+ a_k+1≥(c_b-a+1)^-1,  ηQ_a,Q_a+ a_k+1≤(c_b-a+1)x ≤ηQ_a,Q_a+ a_k  .

This is a slightly easier version of the singular case that we already studied in <Ref>. It is a bit easier because here the ratio has the same shift $Q_{a}$. So from here we again get the same singular power of $x^{-\e}$.

Inverse on the unit circle
Here we study the ratios

\begin{equation}
\Expe{\para{\frac{Q_{H}(a,a+x)}{Q_{H}(b,b+x)}}^{p}}   \tand \Expe{\para{\frac{Q_{H}(b,b+x)}{Q_{H}(a,a+x)}}^{p}}.
\end{equation}

We will go over all the possibilities of $Q_{H}\para{b,b+x},Q_{H}\para{a,a+x}$ depending on them being less or greater than one. The strategy here is to switch back from $\eta_{H}$ to the measure $\eta=\eta_{U}$ as described in <Ref> via $\eta_{H}=G\eta$ and then use the results for $\eta$.

§.§ Case $Q_{H}(b,b+x)\in [1,\infty)$

This is again the easiest case as before.

§.§ Case $Q_{H}(b,b+x)\in [0,1]$ and $Q_{H}(b-a)\bullet Q_{H}(a)\geq \rho_{b-a}$

Again we use $x\leq b-a$ to bound

Here we will decompose the numerator event and $Q_{H}(a)$

E_b-a^k:=   Q_H(b-a)∙Q_H(a)∈b_k,b_k+1=:I_b-a^k, k≥1

D_ℓ^a:=     Q_H(a)∈[a_ℓ,a_ℓ+1]=:I_ℓ,ℓ≥0,

for the same sequence $b_{k}\to +\infty,a_{\ell}\to +\infty$ as before $b_{k}:=\rho_{b-a}k^{\lambda_{2}}$ for some $\lambda_{2}\in (0,1)$ and  $a_{\ell}:=\rho_{1}\ell^{\lambda_{1}},\ell\geq 0$ and $\rho_{1},\lambda_{1}\in (0,1)$. Thus as before, we obtain an analogous upper bound to <Ref>

max_u∈[0,I_ℓ],ũ∈[0,I_b-a^k]η_H(ũ+u,ũ+u+t^-1/p)≥xc_b-ax≥min_u∈[0,I_ℓ]η_H(u,u+b_k), D_ℓ^a,

where we also applied the periodicity and the translation invariance. Here the only difference with <Ref> is that we switch back to the measure $\eta$ as described in <Ref> via $\eta_{H}=G\eta$ and then use the results for $\eta$. So for the first event depending on $t$, we first apply Markov for $p_{0}>0$

≤    x^-p_0max_T∈[0,ρ_b-a+ρ_1]η_H(T,T+t^-1/p)^p_0

≤    x^-p_0e^p_0sup_s∈[0,1]ξ(s)max_T∈[0,ρ_b-a+ρ_1]η(T,T+t^-1/p)^p_0

then apply for pair $p_{0,1},p_{0,2}$ to separate the $\xi$-factor and finally use the positive moments in <Ref>

\begin{equation}
\eqref{eq:unitcirclemaxxi}\leq  x^{-p_{0}}\para{\Expe{e^{p_{0}p_{0,1}\sup_{s\in [0,1]}\xi(s)}}}^{1/p_{0,1}} t^{-\frac{\zeta(p_{0}p_{0,2})-1}{p_{0,2}}}.
\end{equation}

We use that the supremum of the $\xi$-field from <Ref> has finite exponential moments over any closed interval $I$

\begin{equation}
\Expe{\expo{\alpha\sup_{s\in I}\xi(s)}}<\infty \tfor \alpha>0.
\end{equation}

So here we can take $p_{0,2}=1+\e$ by pushing its conjugate $p_{0,1}$ to be arbitrarily large. We do the same for the second factor with the $b_{k}$. So for the first two factors we return to the previous setting in <ref>. It remains to study the $\Proba{D_{\ell}^{a}}$.

Event $D_{\ell}^{a}$ for the $Q_{H}(a)$
Here we have the same result as before because as shown in <Ref>, the exponents for the q-moments of $\eta_{H}(t)$ and $t\in [0,\infty)$ are the same those of $\eta(t)$ as long as they are in $q\in (-\infty,0)\cup [1,\beta^{-1})$ with the only difference of bounding by $\floor{t}^{q}$ when $t\geq 1$ and $q<0$.

§.§ Case $Q_{H}(b-a)\bullet Q_{H}(a)\in [0,\rho_{b-a}]$ and $Q_{H}(b,b+x)\in [0,1]$

Here we again study the singular case. We follow the same overall strategy. We mainly switch back to $\eta$ at the cost of having to control a $\xi$-factor as above.

Step 1: Decomposing $Q_{H}(b-a)\bullet Q_{H}(a)$ and $Q_{H}(a)$ We repeat similar steps to get to the analogue of <Ref>

∑_ℓ∑_kD_ℓ,k^a ^1/b_12
·   ∫_1^∞u∈I_k,ũ∈0,I_ℓ,kη_Hũ+u,ũ+u+ a_kt^-1/p/η_Hũ,ũ+ a_k+1≥c_b-a^-1 ,E_b-a^k(ũ) ^1/b_11 ,

D_ℓ,k^a:=  Q_H(a)∈[c_ℓ,k,c_ℓ+1,k]=:I_ℓ,k,

Now we make the transition to $\eta$. In the event $E_{b-a}^{k}(\tilde{u})$ we use <Ref> to write

\begin{equation}
G_{\tilde{u}}^{1}= \xi(\tilde{u}+\theta_{[0,b-a]}) ,
\end{equation}

and $\theta_{[0,b-a]}\in [0,Q_{H}(b-a)\bullet T]\subseteq [0,\rho_{b-a}]$. For the ratio event we pull a $\xi$ field from numerator and denominator and combine them

η_Hũ+u,ũ+u+ a_kt^-1/p/η_Hũ,ũ+ a_k+1≤e^G_u,ũ^2-G_ũ^3    ηũ+u,ũ+u+ a_kt^-1/p/ηũ,ũ+ a_k+1,

G_u,ũ^2   :=sup_θ∈[0,ρ_b-a]ξ(ũ+u+θ) ,

G_ũ^3   :=sup_θ∈[0,ρ_b-a]ξ(ũ+θ).

In order to progress to the scaling law, we need to remove these two suprema. Since they have all their exponential moments, we decompose them and separate them out by . We use

D^G^1_m_1(ũ):   =    G_ũ^1∈[d_m_1,d_m_1+1]=:I_m^G^1m_1∈_0,

D^G^2_m_2(u,ũ):   =    G_u,ũ^2∈[d_m_2,d_m_2+1]=:I_m_2^G^2m_2∈_0,

D^G^3_m_3(ũ):   =    G_ũ^3∈[d_m_3,d_m_3+1]=:I_m_3^G^3m_3∈_0,

where $\Z_0:=\Z\setminus \set{0}$ and $d_{m}$ is a bi-infinite sequence going to minus and plus infinity to cover the image of those constants defined in <Ref>: we set

\begin{equation}\label{eq:sequencesdefinition}
d_{k}:=\branchmat{\ln k^{\lambda}& k\geq 1 \\ \ln \abs{k}^{-\lambda}& k\leq -1 \\},
\end{equation}

for $k\in \Z_0$ and $\lambda>0$. So finally, we take yet another supremum and apply triple to separate these events

(<ref>)   ≤∑_ℓ,k,m_iD_ℓ,k^a ^1/b_12∫_1^∞

Se^u_2-u_3ηũ+u,ũ+u+ a_kt^-1/p/ηũ,ũ+ a_k+1≥c_b-a^-1 ,E_b-a^k(ũ,u_1) ^1/b_11r_ratio

·∏_i=1,3max_ũ∈0,ρ_aD^G^i_m_i(ũ)^1/r_G^i max_u∈[0,ρ_b-a],ũ∈0,ρ_aD^G^2_m_2(u,ũ)^1/r_G^2,

where $r_{ratio}^{-1}+\sum_i r_{G^{i}}^{-1}=1$ and

\begin{equation}
S:=\set{u\in I_{k},\tilde{u}\in \spara{0,\abs{I_{\ell,k}}} ,\set{u_{i}\in I_{m_i}^{G^i}}_{1,2,3}}
\end{equation}

and we also used the bounds on $\abs{I_{k}},\abs{I_{\ell,k}}$ from the previous section to decouple the factors $G^{i}$ from the $k,\ell$-sums. Here we take $r_{ratio}=1+\e$ for arbitrarily small $\e>0$, at the cost of taking arbitrarily large $r_{G^{i}}$.

Step 2: Scaling law step This step proceed as before in <Ref> with the only difference that we now include the $e^{u_{i}}$ factors

⋃_S̃e^u_2-u_3ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ≥c_b-a^-1^1/b_21b_11r_ratio

·   {∫_⋃_S̃ E_b-a^k(ũ,y,u_1)^1/b_22 g_N(0,1)(y)}^1/b_11r_ratio,

where the maximum-domain changed to

\begin{equation}
\tilde{S}:=\set{u\in \frac{\rho}{a_{k+1}}I_{k},\tilde{u}\in\spara{0, \frac{\rho}{a_{k+1}}\abs{I_{\ell,k}}} ,\set{u_{i}\in I_{m_i}^{G^i}}_{1,2,3} }.
\end{equation}

and the decomposition event changed to

E_b-a^k(ũ,y,u_1)   =ln1/ηũ,ũ+ L_2≤σ_kN(0,1)-σ_k^2/2 +ln a_k+1/e^u_1ρ(b-a)≤ln1/ηũ,ũ+ ρ,

with the difference being that we added the $e^{u_{1}}$-factor.

Step 3: Ratio modulus
Now, we apply Markov inequality for $q_{0}\in [1,\beta^{-1})$ and inequality for $q_{1}^{-1}+q_{2}^{-1}=1$ with $q_{0}q_{1}\in \para{1,\beta^{-1}}$ to separate numerator and denominator

⋃_S̃e^u_2-u_3ηũ+u,ũ+u+ L_2t^-1/p/ηũ,ũ+ ρ≥c_b-a^-1^1/b_21b_11r_ratio

≤   max_S̃e^q_0d_m_2+1-d_m_3ηu+ũ,u+ũ+L_2/t^1/p  ^q_0q_1^1/q_1b_21b_11r_ratio

We then use <Ref> to get

max_S̃e^q_0d_m_2+1-d_m_3ηu+ũ,u+ũ+L_2/t^1/p  ^q_0q_1^1/q_1b_21b_11r_ratio

⪅    e^q_0d_m_2+1-d_m_3L_2/t^1/p ^ζ(q_0q_1)-1/q_1b_21b_11r_ratio

⪅   e^q_0d_m_2+1-d_m_3t^-ζ(q_0q_1)-1/pq_1b_21b_11r_ratio

and we use the <Ref> to get

max_S̃ηu+ũ,u+ũ+ρ^-q_0q_2^1/q_2⪅ ρ^ζ(-q_0q_2)-1/q_2b_21b_11r_ratio.

Step 4: Integral over $t$ and $\para{\frac{Q_{b-a}\bullet Q_{a}}{Q(b,b+x)}}^{p}$ Here the only difference in the exponent of $t$ in <Ref> is that we divide by $r_{ratio}$, which we can take it to be $r_{ratio}=1+\e$ by taking $r_{G},r_{G^{2}}$ arbitrarily large. So we get the same constraint

\begin{equation}
\beta^{-1}>2+o(\e_{i})\doncl \gamma<\sqrt{\frac{2}{2+o(\e_{i})}}  \doncl \gamma<1.
\end{equation}

Step 5: Sum over $\ell$ and event $D_{\ell,k}^{a}$
Here the only difference in <Ref> is that we restrict $p_{2}\in [1,\beta^{-1})$ in order that we have the convexity-condition that we need in <Ref>. So we again get the singular factor $c_{a}a_{k}^{-\frac{1}{b_{12}\lambda_{a} } }.$

Step 6: Sum over $k$ and event $E^{k}_{a,a+x}$
Here we are now studying

∫_⋃_u_1∈I_m_1^G_1 ,u∈[0,ρ] ln1/ηũ,ũ+ L_2≤σ_ky-σ_k^2/2 +ln a_k+1/e^u_1ρ(b-a)≤ln1/ηũ,ũ+ ρ ^1/b_22

where now we have an extra maximum for $u_{1}$. For the probability term, one upper bound follows by taking a minimum and using the <Ref>,<Ref>

⋃_u_1∈I_m_1^G_1 ,u∈[0,ρ] ln1/ηu,u+ L_2≤y-u_1≤ln1/ηu,u+ ρ ^1/b_22

≤c_ρ,L_2 e^-yz_2^0/b_22e^d_m_1+1z_2^0/b_22,e^yz_1^0/b_22e^-d_m_1z_1^0/b_22,

for $z_{2}:=\frac{z_{2}^{0}}{b_{22}}>0$ and $z_{1}:=\frac{z_{1}^{0}}{b_{22}}\twith z_{1}^{0}\in (0,\beta^{-1})$. Here reverse the change of variables again to be left with

∫_ e^-(σ_ky-M_k,x)z_2e^d_m_1+1z_2,e^(σ_ky-M_k,x)z_1e^-d_m_1z_1,1   g_N(0,1)(y).

Here we returned to a similar integral as in <Ref>. The only difference is that we picked up the factors $e^{d_{m_{1}+1}z_{2}},e^{-d_{m_{1}}z_{1}}$. We will need to extract the $e^{d_{m_{1}+1}}$ factors for the summations. We again only study

So we instead split over

ln(k+1)+ ln(xe^d_m_1)++C_1(k)/(1+β)λ_b-a-y/(1+β)λ_b-a√(2βλ_b-aln(k+1)+C_2(k))>0,

where for convenience we grouped together the $x$ and $e^{d_{m_{1}}}$ and let $\tilde{x}:=xe^{d_{m_{1}}}$.  As defined in <Ref>, this $\tilde{x}$ can be arbitrarily small and arbitrarily large and so we have to slightly modify the proof. In the case of $m_{1}\leq -1$, we get that $\tilde{x}\leq x$ and we again satisfy <Ref> i.e.

In this case we proceed as before writing $\tilde{x}$ instead of $x$, in order to be left with

So it remains to study the case $m_{1}\geq 1$ where we have $\tilde{x}\geq x$ and arbitrarily large as $m_{1}$ grows. For the opposite inequality

we get similar lower bounds. For all $y\in (-\infty,\infty)$, we get a solution if we require the same $N_{x,k}$ as before but now replacing $x$ by $\tilde{x}$

ln(k+1)≥     ln1/x̃-C_1(k)/(1+β)λ_b-a+1/2y/(1+β)λ_b-a^2βλ_b-a

+1/2y/(1+β)λ_b-a√(4 ln1/x-C_1(k)/(1+β)2β+y/(1+β)2β^2+4C_2(k))

=:   N_x̃,k+g_y+p(x̃,y,k),

N_x̃,k:=   ln1/x̃-C_1(k)/(1+β)λ_b-a,     g_y:=y/(1+β)λ_b-a^2βλ_b-a,

p(x̃,y,k):=   1/2y/(1+β)λ_b-a√(4 ln1/x̃-C_1(k)/(1+β)2β+y/(1+β)2β^2+4C_2(k)).

If we follow the same steps, we again get the singular factor

Now it finally remains to study the decomposition events for $\xi$.

Step 7: Events $D^{G^{i}}$
Here we study

∑_m_i      e^q_0d_m_2+1-d_m_3e^-d_m_1z_1
·∏_i=1,3max_ũ∈0,ρ_aD^G^i_m_i(ũ)^1/r_G^i max_u∈[0,ρ_b-a],ũ∈0,ρ_aD^G^2_m_2(u,ũ)^1/r_G^2,

D^G^1_m_1(ũ):   =    G_ũ^1∈[d_m_1,d_m_1+1]=:I_m_1^G^1m_1∈_0,

D^G^2_m_2(u,ũ):   =    G_u,ũ^2∈[d_m_2,d_m_2+1]=:I_m_2^G^2m_2∈_0,

D^G^3_m_3(ũ):   =    G_ũ^3∈[d_m_3,d_m_3+1]=:I_m_3^G^3m_3∈_0,

G_ũ^1   :=ξ(ũ+θ_[0,b-a]),

G_u,ũ^2   :=sup_θ∈[0,ρ_b-a]ξ(ũ+u+θ),

G_ũ^3   :=sup_θ∈[0,ρ_b-a]ξ(ũ+θ).

These sums are all decoupled and so we study them separately. We mainly use the finiteness of exponential moments <Ref>. Since the three events are handled similarly, we just focus on the on the event $D^{G^{2}}$.

Step 8: Event $D^{G^{2}}$
We set

\begin{equation}
e_{k}=\branchmat{\ln k^{\lambda}& k\geq 1 \\ \ln \abs{k}^{-\lambda}& k\leq -1 \\},
\end{equation}

for $k\in \Z_0$ and $\lambda>0$. When we have $d_{m_{2}}<0$, we keep only the upper bound and apply Chernoff inequality for $\alpha_{2,1}>0$

≤    e^α_2,1 d_m_2+1/r_G^2 s∈[0,ρ_a+2ρ_b-a]α_2,1γH_(s)-U_(s)^1/r_G^2.

Here we take large enough $\alpha_{2,1}$ to satisfy the finiteness constraint $(q_{0}+\frac{\alpha_{2,1}}{r_{G^{2}}})\lambda>1$ for the $k$-sum. Similarly, when $d_{m_{2}}>0$, we keep the lower bound and apply Chernoff inequality for $\alpha_{2,2}>0$

≤    e^-α_2,2 d_m_2 /r_G^2s∈[0,ρ_a+2ρ_b-a]α_2,2γH_(s)-U_(s)^1/r_G^2.

§.§ Case $\abs{J}=\abs{I}<\delta$ and ratio of the form $\frac{Q_{H}(b,b+x)}{Q_{H}(a,a+x)}$ for $a<b$

Bounding this ratio follows the same strategy as for the case on the real line. In particular for the singular cases, we have the same situation.  The $q_{0}$ being arbitrarily small didn't affect it either since in <Ref> we can again take $\alpha_{2,1}$ large enough.

Multipoint estimates
In this part we study multipoint estimates that appear in the <cit.>-framework (eg. <cit.>).

§ MULTIPOINT MOMENTS

We fix $N\geq 1$ and generic intervals $[a_{k},b_{k}]$ and $J_{k}:=[c_{k},c_{k}+x_{k}]\subseteq [a_{k},b_{k}]$ for $k\in [N]$. We consider the gap event for some subset $S:=\set{i_{1},\cdots,i_{M}}\subset \set{1,\cdots,N}$ with $M\geq \alpha N$, for some $\alpha>0$, such that we have large enough gaps

\begin{equation}
G_{S}:=G_{i_{1},\cdots,i_{M}}:=\bigcap_{k\in S}\set{Q_{a_{i_{k}}}^{i_{k}}-Q_{b_{i_{k+1}}}^{i_{k+1}}\geq \delta_{i_{k+1}}}.
\end{equation}

If we intersect with this event, we get the following estimate on the multipoint-moments.

We fix $p_{k},k\in S$ then we have

∏_k∈S Q^k(J_k)^p_k G_S≤c^S ∏_k∈S x_k^q_k,

for $q_{k}>0$ satisfying

p_k>-ζ(-q_k)+1, k∈S.

The natural follow-up question is whether the gap event $G_{S}$ has high enough probability. If we start with the joint tail of inverses $Q^{k}(J_{k})$ with $J_{k}:=[c_{k},c_{k}+x_{k}]\subseteq [a_{k},b_{k}]$ , we upper bound by the decoupled ratio plus the error $g_{Gap}(N)$

⋂_k∈[N]  Q^k(J_k)> R_k ≤   ⋂_k∈[N]  Q^k(J_k)> R_k∩α(G)≥αN +g_Gap(N)

≤   S⊂[N]
S≥αN⋂_k∈S  Q^k(J_k)> R_k∩G_S +g_Gap(N).

So here to get decay $g_{Gap}(N)\to 0$, this will impose constraints on $a_{k},b_{k},\delta_{k}$ as described in <ref>.

To ease on the notation we simply write $1,....,M$ as opposed to $i_{1},...,i_{M}$ since for now the argument only requires them to have gaps and it is not number-specific. The strategy is to peel-off one moment at a time by successively decomposing the $\para{Q^1(a_{1}),Q^{2}(b_{2})},\para{Q^{2}(a_{2}),Q^{3}(b_{3})},...$. We partially follow the proof-logic in the single-point moments of $Q([a,b])$ from <cit.>.

Step 1: $\delta$-Strong Markov
We start with the largest scale $\para{Q^{1}(J_{1})}^{p_{1}}$ for $J_{1}:=[c_{1},c_{1}+x_{1}]$ and denote the product of other scales as

\begin{equation}
Y_{[2, M]}:= \prod\limits_{k\in [2,M]} \para{Q^{k}(J_{k})}^{p_{k}}   \ind{ G_{[3,M]}}
\end{equation}

We write

∏_k∈S Q^k(J_k)^p_k G_S=   ∫_0^B_δx_1≥η^1Q_c_1^1,Q_c_1^1+ t  G_1,2Y_[2, M] p_1t^p_1-1

+∫_B_δ^∞x_1≥η^1Q_c_1^1,Q_c_1^1+ t  G_1,2Y_[2, M] p_1t^p_1-1,

for $B_{\delta}:=\minp{2,\maxp{2\delta,1}}\in [1,2]$ which satisfies $B_{\delta}-\delta\geq 1/2$. For the second integral, we simply lower bound

η^1Q_c_1^1,Q_c_1^1+ t   ≥η^1Q_c_1^1+δ_1,Q_c_1^1+ t

and use <ref>. This gives the bound

∫_B_δ^∞x_1≥η^1Q_c_1^1,Q_c_1^1+ t  G_1,2Y_[2, M] p_1t^p_1-1≤cx^q_1,2Y_[2, M]

for $q_{1,2}>p_{1}$.

Step 2: Decomposition
For the first integral we have to do the decomposition argument as done in <cit.>.  We fix sequence $a_{\ell}:=\rho\delta_{1} \ell^{\lambda},\ell\geq 0$ for $\rho,\lambda\in (0,1)$. We see that $a_{\ell+1}-a_{\ell}\leq \rho\delta_{1}$. Using the gap event $G_{1,2}$, we consider the decomposition

D^1_ℓ:=a_ℓ+Q^2(b_2)+δ_2  ≤Q^1(c_1)≤a_ℓ+1+Q^2(b_2)+δ_2, ℓ≥0.

Therefore, we decompose and bound using an infimum

x_1≥η^1Q_c_1^1,Q_c_1^1+ t
=   ∑_ℓ≥0x_1≥η^1Q_c_1^1,Q_c_1^1+ t ,D^1_ℓ

≤   ∑_ℓ≥0x_1≥inf_T∈I_ℓη^1T,T+ t ,D^1_ℓ,

Step 3: Tower property
Now here for each $\ell$-term we upper bound to remove the gap event and apply the tower property for $\CF\para{[0,Q^{2}(b_{2})]}$

x_1≥inf_T∈I_ℓη^1T,T+ t ,D^1_ℓ G_1,2Y_[2, M]

≤   x_1≥inf_T∈I_ℓη^1T,T+ t ,D^1_ℓ Y_[2, M]

=    x_1≥inf_T∈I_ℓη^1T,T+ t ,D^1_ℓ|[0,Q^2(b_2)] Y_[2, M] ,

where we used that $Y_{[2, M]}\in \CF\para{[0,Q^{2}(b_{2})]}$. Then we apply conditional-to separate the decomposition event

x_1≥inf_T∈I_ℓη^1T,T+ t ,D^1_ℓ|[0,Q^2(b_2)]
≤   x_1≥inf_T∈I_ℓη^1T,T+ t [0,Q^2(b_2)]^1/b_11

·   D^1_ℓ [0,Q^2(b_2)]^1/b_12,

for $1/b_{11}+1/b_{12}=1$.

Step 4: Decoupling
For the first factor

x_1≥inf_T∈I_ℓη^1T,T+ t [0,Q^2(b_2)]^1/b_11

the supremum is over a shifted interval $[a_{\ell},a_{\ell+1}]$, so we study

inf_T∈[a_ℓ,a_ℓ+1]η^1T+Q^2(b_2)+δ_2,T+Q^2(b_2)+δ_2+ t .

So due to <ref>, this variable is independent of events in $\CF\para{[0,Q^{2}(b_{2})]}$

(<ref>)=x_1≥inf_T∈[0,ρδ_1]η^1T,T+ t ^1/b_11.
Now that we decoupled, we use <ref> for $q_{1,1}>0$

(<ref>)≤x_1^q_1,1 t^α_2δ_1(q_1,1)/b_11,

where $\alpha_{2\delta}(p)=\zeta(-p)-1$ if $\frac{x}{2}\leq \delta$ and $\alpha_{2\delta}(p)=-p$ if $\frac{x}{2}\geq \delta$. Returning to the $t-$integral in <ref> over $[0,B_{\delta_{1}}]$, we obtain the constraint

ζ(-q_1,1)-1/b_11+p_1-1>-1   p_1> -ζ(-q_1,1)+1/b_11.

Step 5: Event $D^{1}_{\ell}$
Here we study

D^1_ℓ=   η^10,a_ℓ+Q^2(b_2)+δ_2 ≤c_1   ≤η^10,a_ℓ+1+Q^2(b_2)+δ_2

=   η^1Q^2(b_2)+δ_2,a_ℓ+Q^2(b_2)+δ_2 ≤c_1-η^10,Q^2(b_2)+δ_2   ≤η^1Q^2(b_2)+δ_2,a_ℓ+1+Q^2(b_2)+δ_2 .

We again apply <ref> to decouple

(<ref>)η^10,a_ℓ ≤c_1-η^10,Q^2(b_2)+δ_2   ≤η^10,a_ℓ+1 ,

where $\wt{\eta}^{1}$ is an independent copy of $\eta^{1}$. To avoid the coupled term $\eta^{1}\para{0,Q^{2}(b_{2})+\delta_{2}}$
, we keep only the lower bound

D^1_ℓ≤η^10,a_ℓ ≤c_1

and so

D^1_ℓ [0,Q^2(b_2)]≤η^10,a_ℓ ≤c_1 .

For $\ell=0$, we bound by zero.  When $c_{\ell}\leq \delta_{1}\tor c_{\ell}> \delta_{1}$ we use Markov with $r_{1}>0\tand r_{2}>0$ respectively,to upper bound by

∑_ℓ≥1 D^1_ℓ [0,Q^2(b_2)]^1/b_12≤    ∑_ℓ≥1  η^10,a_ℓ ≤c_1 ^1/b_12

≤   ∑_ℓ≥1   c_1^r_1/b_12a_ℓ^ζ(-r_1)/b_12+ c_1^r_2/b_12a_ℓ^-r_2/b_12.

So to get finiteness we require both $-\zeta(-r_{1})\lambda>b_{12},r_{2}\lambda>b_{12}$, which is possible since GMC has all its negative moments and so we take arbitrarily large $r_{1},r_{2}$. This allows to take $b_{12}$ arbitrarily large too so that we get $b_{11}=1+\e$. This procedure didn't leave any terms that are coupled with the other scales and so we proceed the same for them.

§ MOMENTS OF MULTIPOINT RATIOS

In the following proposition we study the first term to get moment estimates as in the single-ratio case in <ref>.

We fix set $S\subset [N]$.

* (Equal length)  Fix $\beta<\frac{2}{3}$. Suppose we have equal length intervals $x_{k}=\abs{J_k}=\abs{I_k}<\delta_k$ for each $k\in S$. Fix $\delta_k\leq 1$ and intervals $J=(c_k,c_k+x_k),I=(d_k,d_k+x_k)\subset [0,1]$ with $d_k-c_k=c_{d-c,k}x_k$ for $c_{d-c,k}>1$  and $c_k=\lambda_{1,k}\delta_k,d_k=\lambda_{2,k}\delta_k$. Then we have for all $p_k\in [1,1+\e_{k}]$ with small enough $\e_{k}>0$, a bound of the form

∏_k∈S Q^k(J_k)/Q^k(I_k)^p_k G_S≤c^S∏_k∈S x_k /δ_k^-_ratio(p_k),

where $\e_{ratio}(p_k)>0$ can be made arbitrarily small at the cost of a larger proportionality constant $c$. The constants are uniformly bounded in $x_k$ and also in $\delta_k$.

* (Decreasing numerator) Fix $\beta\in \para{0,0.152}.$
Suppose that $\abs{I_{k}}=r_{k}\delta_{k}$ for $r_{k}>0$ and $\abs{J_{k}}\to 0$ for each $k\in S$. Then we have for each $p_{k}\in [1,1+\e_{ratio}]$ some  $q_{k}>1$ such that

∏_k∈S Q^k(J_k)/Q^k(I_k)^p_k G_S≤∏_k∈S J_k/δ_k^q_k.

Equal length intervals
To ease on the notation we simply write $1,....,M$ as opposed to $i_{1},...,i_{M}$ since for now the argument only requires them to have gaps and it is not number-specific. The strategy is to peel-off one ratio at a time by successively decomposing the $\para{Q^1(a_{1}),Q^{2}(b_{2})},\para{Q^{2}(a_{2}),Q^{3}(b_{3})},...$. We will follow some of the steps from the proof of <ref>. Here we only cover the singular case involving the scaling law and the other cases are handled similarly.

Step 1: Decomposition
We start with the largest scale $\para{\frac{Q^{1}(J_{1})}{Q^{1}(I_1)}}^{p_{1}}$ for $J_{1}:=[c_{1},c_{1}+x_{1}],I_{1}:=[d_{1},d_{1}+y_{1}]$ and denote the product of other scales as

\begin{equation}
Y_{[2, M]}:= \prod\limits_{k\in [2,M]} \para{\frac{Q^{k}(J_{k})}{Q^{k}(I_{k})}}^{p_{k}}   \ind{ G_{[3,M]}}
\end{equation}

We decouple this first ratio using the gap $G_{1,2}$. We only cover the details of the particular case $c_{1}< d_{1}$ and $x_{1}=d_{1}-c_{1}$ and

\begin{equation}
Q^{1}(J_{1})\in [0,\rho] \tand Q^{1}(I_{1})\in [0,1]
\end{equation}

and so we study

Q^1(J_1)/Q^1(I_1)^p_1Q^1(J_1)∈[0,ρ] Q^1(I_1)∈[0,1]G_1,2Y_[2, M].

This corresponds to the singular case and the other cases are easier to handle, for details see the proof of <ref>. As done there we first decompose the numerator

\begin{equation}
D^{x_{1}}_{k}:=\set{ Q^{1}(J_{1})\in [r_{k+1},r_{k}]},  \end{equation}

for the same decaying sequence $r_{k}:=\rho k^{-\lambda}$ for $k\geq 1$ and $\lambda>0$. We then apply layercake to upper bound by

(<ref>)≤   G_1,2Y_[2, M]

+∑_k∫_[1,∞)η^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k G_1,2Y_[2, M] .

Next for each k-term we decompose $Q^{1}(c_{1})$.  We set $c_{\ell,k}:=\rho_{a}\delta_{1}r_{k}c_{\ell}$ for some small enough $\rho_{a}\in (0,1)$ and $c_{\ell}:=\ell^{\lambda_{a}}$ for $\lambda_{a}\in (0,1)$ so that indeed $c_{\ell+1}-c_{\ell}\to 0$ as $\ell\to +\infty$. So we consider

D^c_1_ℓ,k:=  Q^1(c_1)∈[c_ℓ,k,c_ℓ+1,k]+Q^2(b_2)+δ_2=:I_ℓ,k^c_1+Q^2(b_2)+δ_2 ,

where we included the information from the gap event $G_{1,2}$ and so for the probability term in <ref> we write

η^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k G_1,2Y_[2, M]

=   ∑_ℓη^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k ,D^c_1_ℓ,k G_1,2Y_[2, M]

≤   ∑_ℓη^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k ,D^c_1_ℓ,k Y_[2, M] ,

where in the last step we upper bounded to remove the gap event $\ind{G_{1,2}}$ now that we used it.

Step 2: Tower property
Now here for each $\ell$-term we apply the tower property for $\CF\para{[0,Q^{2}(b_{2})]}$

η^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k ,D^c_1_ℓ,k Y_[2, M]

=   η^1Q_d_1^1,Q_d_1^1+ r_kt^-1/p ≥y_1, D^x_1_k ,D^c_1_ℓ,k[0,Q^2(b_2)] Y_[2, M] .

Next we proceed as in <ref> inside the conditional expectation, we take supremum over $u\in I_{\ell,k}^{c_{1}}+Q^{2}(b_{2})+\delta_{2}$

[   u∈I_ℓ,k^c_1+Q^2(b_2)+δ_2η^1u+Q_d_1-c_1^1∙u,u+Q_d_1-c_1^1∙u+ r_kt^-1/p ≥y_1, D^x_1_k(u)

·D^c_1_ℓ,k [0,Q^2(b_2)]],

\begin{equation}
D^{x_{1}}_{k}(u):= \set{ Q^{1}_{x_{1}}\bullet u\in [r_{k+1},r_{k}]}.
\end{equation}

So now that we used the event $D^{c_{1}}_{\ell,k}$, as in the proof of <ref>  we apply  conditional-to remove it out

(u∈I_ℓ,k^c_1+Q^2(b_2)+δ_2η^1u+Q_d_1-c_1^1∙u,u+Q_d_1-c_1^1∙u+ r_kt^-1/p ≥y_1,D^x_1_k(u) [0,Q^2(b_2)])^1/b_11

·   D^c_1_ℓ,k [0,Q^2(b_2)]^1/b_12,

for $b_{11}^{-1}+b_{12}^{-1}=1$.

Step 3: Decoupling
For the first factor, we simply use <ref> to decouple

u∈I_ℓ,k^c_1+Q^2(b_2)+δ_2η^1u+Q_d_1-c_1^1∙u,u+Q_d_1-c_1^1∙u+ r_kt^-1/p ≥y_1,D^x_1_k(u) [0,Q^2(b_2)]

=   u∈I_ℓ,k^c_1η^1u+Q_d_1-c_1^1∙u,u+Q_d_1-c_1^1∙u+ r_kt^-1/p ≥y_1,D^x_1_k(u)  .

Now that we decoupled this ratio, we proceed as in the proof of the unconditional version <ref>. Next we study the event $D^{c_{1}}_{\ell,k}$.

Step 4: The event $D^{c_{1}}_{\ell,k}$
Here we study

=   η^10,c_ℓ,k+Q^2(b_2)+δ_2 ≤c_1   ≤η^10,c_ℓ+1,k+Q^2(b_2)+δ_2

=   η^1Q^2(b_2)+δ_2,c_ℓ,k+Q^2(b_2)+δ_2 ≤c_1-η^10,Q^2(b_2)+δ_2   ≤η^1Q^2(b_2)+δ_2,c_ℓ+1,k+Q^2(b_2)+δ_2 .

We again apply <ref> to decouple

(<ref>)η^10,c_ℓ,k ≤c_1-η^10,Q^2(b_2)+δ_2   ≤η^10,c_ℓ+1,k ,

where $\wt{\eta}^{1}$ is an independent copy of $\eta^{1}$. To avoid the coupled term $\eta^{1}\para{0,Q^{2}(b_{2})+\delta_{2}}$, we keep only the lower bound

D^c_1_ℓ,k [0,Q^2(b_2)]≤η^10,c_ℓ,k ≤c_1 .

Now that we decoupled we proceed as in proof of <ref>. For $\ell=0$, we bound by zero.  When $c_{\ell,k}\leq \delta_{i_{1}}\tor c_{\ell,k}> \delta_{i_{1}}$ we use Markov with $p_{1}>0\tand p_{2}>0$ respectively,to upper bound by

∑_ℓ≥1 D^c_1_ℓ,k [0,Q^2(b_2)]^1/b_12

≤    ∑_ℓ≥1  η^10,c_ℓ,k ≤c_1 ^1/b_12

≤   ∑_ℓ≥1   c_1^p_1/b_12c_ℓ,k^ζ(-p_1)/b_12+ c_1^p_2/b_12c_ℓ,k^-p_2/b_12.

So from here to get finiteness we require both $-\zeta(-p_{1})\lambda_{a}>b_{12},p_{2}\lambda_{a}>b_{12}$. Therefore, we get the singular power $r_{k}^{-\frac{1+\e}{\lambda_{a}}}$ for arbitrarily small $\e>0$. Here we are completely back to the unconditional setting of <ref>. This procedure didn't leave any terms that are coupled with the other scales and so we proceed the same for them. So we bootstrap the same for the next scales.

Decreasing numerator case
We assume $c_{1}<d_{1}$ because the other case is symmetric. Here we use the semigroup formula to write

\begin{equation}
\frac{Q^{1}(J_{1})}{Q^{1}(I_1)}=\frac{Q_{x_{1}}^{1}\bullet Q_{c_{1}}^{1}}{Q^{1}_{y_{1}}\bullet \para{Q_{d_{1}-c_{1}}^{1}\bullet Q_{c_{1}}^{1}+Q_{c_{1}}^{1}}}.
\end{equation}

Here the decay only comes from the numerator. So there is no need for scaling law and so in the decomposition event <ref>, we only index over $\ell\geq 0$ without any presence of $r_{k}$

D^c_ℓ:=  Q^1(c_1)∈[ℓ^λ,(ℓ+1)^λ]+Q^2(b_2)+δ_2.

That means there won't be any singular factor $r_{k}^{-1/\lambda}$. As above we take the maximum and use the conditional-$b_{11}^{-1}+b_{12}^{-1}=1$ to separate out the event $D^{c}_{\ell}$

≤   ∑_ℓ max_u∈[0,2]Q_x_1^1∙u/Q^1_y_1∙Q_d_1-c_1^1∙u+u^b_11p^1/b_11·D^c_1_ℓ [0,Q^2(b_2)]^1/b_12.

We deal with the second term as we dealt with $D^{c_{1}}_{\ell,k}$in the proof of the first part. So we are left with studying

\begin{equation}\label{eq:maximumratiodecnum}
\para{ \Expe{\max_{u\in [0,2]}\para{\frac{Q_{x_{1}}^{1}\bullet u}{Q^{1}_{y_{1}}\bullet \para{Q_{d_{1}-c_{1}}^{1}\bullet u+u}}}^{b_{11}p}}}^{1/b_{11}}.
\end{equation}

The denominator is just a constant since we already scaled by $\delta_{i_{1}}$ and we can use the <Ref>. So we simply apply again to separate numerator and denominator

\begin{equation}\label{eq:holderseparatnumde}
\para{\Expe{\max_{u\in [0,2]}\para{Q_{x_{1}}^{1}\bullet u}^{b_{11}pp_{1}}}}^{1/b_{11}p_{1}}\para{\Expe{\max_{u\in [0,2]}\para{\frac{1}{Q^{1}_{y_{1}}\bullet \para{Q_{d_{1}-c_{1}}^{1}\bullet u+u}}}^{b_{11}pp_{2}}}}^{1/b_{11}p_{2}},
\end{equation}

for $p_{1}^{-1}+p_{2}^{-1}=1$. For the first factor we apply layercake, Markov for for $q_{1},q_{2}>0$ and then <Ref>

max_u∈[0,2]Q_x_1^1∙u^pb_11p_1   =∫_0^∞x_1≥max_u∈[0,2]ηu,u+t^1/pb_11p_1

≤   x_1^q_1∫_0^1t^ζ(-q_1)-1/b_11pp_1 +x_1^q_2∫_1^∞t^-q_2/b_11pp_1.

For finiteness we require

Next we get the exponents for $x$ to be $\frac{q_{1}}{p_{1}}>1+\e$ and $\frac{q_{2}}{p_{1}}>1+\e$ for small $\e>0$. The second inequality is cleared by taking large $q_{2}>0$. For the first inequality we have that an arbitrarily large $q_{1}$ will break the first integral constraint. In the unconditional work we took $b_{11}=1+\e$ and we do the same here. So for the exponent of we ask $\frac{q_{1}}{p_{1}}=1+\e$, which translates the integral constraint to

\begin{equation}\label{eq:integralconstraint}
1<p_{1}\tand \frac{(1+\e)(1+\beta(p_{1}(1+\e)+1))+\frac{1}{p_{1}}}{p}<1.
\end{equation}

Since $pp_{2}\in [1,\beta^{-1})$, we further take $p=\beta^{-1}\frac{p_{1}-1}{p_{1}}-o(\e)$, which makes the above into a single variable problem

\begin{equation}
1<p_{1}\tand \frac{(1+\e)(1+\beta(p_{1}(1+\e)+1))+\frac{1}{p_{1}}}{\beta^{-1}\frac{p_{1}-1}{p_{1}}}<1.
\end{equation}

This has a solution $p_{1}>1$ when we further require

\begin{equation}
\beta\in \para{0,0.152}.
\end{equation}

If we have $c_{1}>d_{1}$, we would have to use <Ref> instead of <Ref> and it has the same exponent and the same constraint for $\beta$.

§ MULTIPOINT AND UNIT CIRCLE

This is the relevant part for the Lehto-welding. Namely we study the multipoint estimate for the inverse $Q_{H}$ on the unit circle. In this section we will use the following notations. We consider the events

G_[1,M]:=   Q^i(a_i)≥Q^n_i+1(b_n_i+1)+δ_n_i+1, i∈[1,M],

Ξ_[1,M]:=   (s,t)∈[0,Q^i(b_i)]^2ξ(t)-ξ(s)≤u_i,ξ ,i∈[1,M],

U_[1,M]:=   (s,t)∈[0,Q^i(b_i)]^2U^1_i(t)-U^1_i(s)≤u_i,us, i∈[1,M],

where we use the shorthand notation $Q^{i}=Q^{n_{i}}$ and consider any sequences $u_{i,\xi},u_{i,us}\in (0,1)$ strictly decreasing to zero as $i\to+\infty$ and interval

We will bound the following product

∏_k∈[1,M]Q_H(d_k,d_k+y_k)/Q_H(c_k,c_k+x_k)^p_k    G_[1,M] ,Ξ_[1,M], U_[1,M],

where for set $S:=\set{n_{1},...,n_{M}}\subset [1,N]$. In our ensuing work on Lehto-welding we obtained the existence of this set $S$ for these events using deviation estimates.

We fix set $S:=\set{n_{1},...,n_{M}}\subset [N]$.

* (Equal length)  Fix $\gamma<\frac{2}{\sqrt{3}}$. Suppose we have equal length intervals $x_{k}=\abs{J_k}=\abs{I_k}\leq \delta_k$ for each $k\in S$. Fix $\delta_k\leq 1$ and intervals $J=(c_k,c_k+x_k),I=(d_k,d_k+x_k)\subset [0,1]$ with $d_k-c_k=c_{d-c,k}x_k$ or $c_k-d_{k}=c_{d-c,k}x_k$ for some $c_{d-c,k}>1$  and $c_k=\lambda_{1,k}\delta_k,d_k=\lambda_{2,k}\delta_k$ for some $\lambda_{i,k}>0$.  Then we have for all $p_k\in [1,1+\e_{k}]$ with small enough $\e_{k}>0$, a bound of the form

∏_k∈[1,M]Q_H(d_k,d_k+y_k)/Q_H(c_k,c_k+x_k)^p_k    G_[1,M] ,Ξ_[1,M], U_[1,M]≤c^S∏_k∈S x_k /δ_k^-_ratio(p_k),

where $\e_{ratio}(p_k)>0$ can be made arbitrarily small at the cost of a larger proportionality constant $c$. The constants are uniformly bounded in $x_k$ and also in $\delta_k$.

* (Decreasing numerator) Fix $\beta\in  (0,0.152).$ Suppose that $\abs{I_{k}}=r_{k}\delta_{k}$ for $r_{k}>0$ and $\abs{J_{k}}\to 0$ for each $k\in S$. Then we have for each $p_{k}\in [1,1+\e_{ratio,1}]$ some  $q_{k}=1+\e_{ratio,2}>1$ such that

∏_k∈[1,M]Q_H(d_k,d_k+y_k)/Q_H(c_k,c_k+x_k)^p_k    G_[1,M] ,Ξ_[1,M], U_[1,M]≤∏_k∈S J_k/δ_k^q_k.

§.§ Proof of <ref>: Equal lengths

In order to achieve any decoupling the $\xi$ field and upper scales need to be removed. If we take a maximum over their domains $P_{k}$,

∏_k∈[1,M]u_1,u_2∈P_kQ^ke^u_1 c_k,c_k+x_ke^u_2/u_3,u_4∈P_kQ^ke^u_3d_k,d_k+y_ke^u_4^p_k

then we can no longer use the trick of <ref> where we divided and multiplied by $Q_{b-a}\bullet Q_{a}$ and applied scaling law with respect to its deterministic approximation because now the denominator has "multiple" transitions $Q_{e^{u_{3}}(d_{k}-c_{k})}\bullet Q_{e^{u_{3}}c_{k}}$ for each $u_{3}$ and so it's no longer clear which factor to decompose and apply the scaling law on.

So instead we move back to the level of $Q_{H}$. Let $C_{1},C_{2}$ contain all the $k$ such that $d_{k}\geq c_{k}$ and $d_{k}<c_{k}$ respectively. We divide and multiply by the intermediate-intervals

M_k:=Q_H(c_k,c_k+d_k-c_k)    k∈C_1
Q_H(c_k,d_k+y_k)    k∈C_2.

For the $C_{1}$-group we use that $x_{k}<d_{k}-c_{k}$ to bound the first ratios by one:

∏_k∈C_1 Q_H(c_k,c_k+x_k)/Q_H(d_k,d_k+y_k)^p_k≤∏_k∈C_1Q_H(c_k,c_k+d_k-c_k)/Q_H(d_k,d_k+y_k)^p_k

and for the $C_{2}-$group we use $Q_{H}(d_{k},d_{k}+y_{k})\leq Q_{H}(c_{k},d_{k}+y_{k})$ to bound

\begin{equation}
\prod\limits_{k\in C_{2}} \para{\frac{Q_{H}(d_{k},d_{k}+y_{k})}{Q_{H}(c_{k},c_{k}+x_{k})}}^{p_{k}}    \leq \prod\limits_{k\in C_{2}} \para{\frac{Q_{H}(c_{k},d_{k}+y_{k})}{Q_{H}(c_{k},c_{k}+x_{k})}}^{p_{k}}.
\end{equation}

The strategy will be to first transition to $Q_{U}$, then decouple to reduce single-point estimates and then study the first one and bootstrap for the rest.

Decoupling part
Step 1: Decomposing the intermediate intervals for all the scales
As we did before in the single-point case <ref>, we decompose the intermediate interval $Q_{H}(c_{k},c_{k}+d_{k}-c_{k})$.

So we consider the following decompositions for $k\in [1,M]$

D^k_ℓ_k,M:=   M_k=Q_H(c_k,c_k+d_k-c_k)∈[a_ℓ_k,a_ℓ_k+1]=:I_ℓ_k^k    k∈C_1
M_k=Q_H(c_k,d_k+y_k)∈[a_ℓ_k,a_ℓ_k+1]=:I_ℓ_k^k    k∈C_2     ℓ_k∈,

S^k_m_k,c(ℓ_k):=   Q_H(c_k)∈[b_m_k(ℓ_k),b_m_k+1(ℓ_k)]=:I_m_k,c(ℓ_k)     m_k∈,

where the $a_{\ell_{k}},b_{m_{k}}(\ell_{k})$ are bi-directional sequences possibly going to zero and infinity; and the notation $b_{m_{k}}(\ell_{k})$ means that this sequence will also depend on $\ell_{k}$ (as required in the singular case of the proof in <ref>). In particular we let

a_k:=ρ_b-ak^-λ_neg    k≤-1

ρ_b-ak^λ_pos    k≥1

for $\lambda_{neg}>0$ and $\lambda_{pos}\in (0,1)$, and we will see $b_{m_{k}}(\ell_{k})$ in (<ref>).

In the next paragraph, we will also incorporate the gap-events in their definition so as to achieve decoupling. We will also split them over $\rho_{b-a}$ to get the various cases as in <ref>. So we write

∏_k∈C_1 D^k_ℓ_k,M,S^k_m_k,c(ℓ_k)M_k/B_k^p_k

· G_[1,M],Ξ_[1,M], U_[1,M].

Next we study the relation to $Q_{U}^{k}$.

Step 2: Relation with $Q_{U}^{k}$
We first transition to $Q=Q_{U}$ over general intervals $(z_{n,M},z_{n,M}+r_{n,M})$ for

z_n,M:=z_nξ(b_n)+U^1_n(θ_b_n) r_n,M:=  r_nξ(b_n)+U^1_n(θ_b_n),

for  $\theta_{b_{n}}\in [0,Q^{n}(b_{n})].$  We write

=    Qe^ξ(b_n)/e^ξ(z_n,M)e^U^1_n(θ_b_n)·p_n,p_n+r_nξ(z_n,M)-ξ(z_n,M,z_n,M+r_n,M)

=:   QA,A+B,

Then we transition to $Q^{n}$

We next insert the gap information into the event $D^{k}_{\ell_{k},M}$. By using the events $\Xi_{[1,M]}, U_{[1,M]}$, we identify the supersets for the mean-values

QA+B=   Qe^U^1_n(θ_b_n)e^ξ(b_n)/e^ξ(z_n,M)p_n+e^ξ(b_n)/e^ξ(z_n,M,z_n,M+r_n,M)r_n≤Qe^U^1_n(θ_b_n)b_n
= Q^n(b_n),

and so $\theta_{A},\theta_{A+B}\in [0,Q^{n}(b_{n})]$. Whereas for lower bound we have

and so we have the interval inclusion

Q_U^n(e^U̅_n^1(θ_A)A),Q_U^n(e^U̅_n^1(θ_A)(A+B)⊆[Q^n(a_n) ,Q^n(b_n) ].

Step 3: Using the Gap event
It is convenient to rewrite the decomposition <ref>

We have $Q_{H}(c_{n})\geq Q^{n}(a_{n})$ and so from the gap event

we have $Q_{H}(c_{n})>Q^{n+1}(b_{n+1})+\delta_{n+1}$. That means we rewrite the decomposition using the sequence

\begin{equation}\label{eq:usinggapQhQu3}
\end{equation}

for adeterministic bi-infinite sequence $\wt{b}_{m_{n}}(\ell_{n})$ going both to zero and plus infinity. In particular, we let

for some small enough $\rho_{a}\in (0,1)$ and $\lambda_{a}\in (0,1)$. We further denote their intervals as $\wt{I}_{m_{n},a}^{n}(\ell_{n}):=[\wt{b}_{m_{n}}(\ell_{n}),\wt{b}_{m_{n}+1}^{n}(\ell_{n})]$.

Removing the $\Xi_{[1,M]}, U_{[1,M]}$ via maximum
Now we are finally in the position to remove the $\xi$ and upper-scale $U^{1}_{n}$ fields. All the quantities in <ref> that involve $Q_{H}$ need to get switched to $Q^{k}_{U}$ in order for the decoupling argument to go through.

We start from the outside i.e. we temporarily leave $Q_{H}(c_{k}),M_{k}=Q_{H}(c_{k},c_{k}+d_{k}-c_{k})$ untouched. We also return to the scaled notation

\begin{equation}
\end{equation}

Step 4: Switching the $Q_{H}(x_{k})$ to $Q_{U}^{k}$
For the term $A_{k}=Q_{H}(x_{k})\bullet Q_{H}(c_{k})=Q_{H}(x_{k})\bullet T$, with $T:= Q_{H}(c_{k})$, using the transition formulas in <ref> we bound

min_u_1,u_2∈I_T Q_U^ke^U^1_k(θ_b_k)/e^U̅_k^1(u_1)e^ξ(b_k)/e^ ξ(u_2)x_k∙T  ≤Q_H(x_k,U,ξ)∙T≤max_u_1,u_2∈I_T Q_U^ke^U^1_k(θ_b_k)/e^U̅_k^1(u_1)e^ξ(b_k)/e^ ξ(u_2)x_k∙T,

where $I_{T}:= [T,T+Q_{H}(p)\bullet T]$. We now examine the relation of those the extremizers to $Q^{k}_{U}(b_{k})$. We have

u_i≤T+ Q_H(p)∙T=Q_H(c_k,U,ξ+x_k,U,ξ)≤Q^k_U(b_k),

where we applied the events $\Xi_{[1,M]}, U_{[1,M]}$.

Step 5: Switching the $Q_{H}(y_{k})$ to $Q_{U}^{k}$
For the term $B_{k}=Q_{H}(y_{k})\bullet \para{M_{k}+Q_{H}(c_{k})}=Q_{H}(y_{k})\bullet S_{k}$, with $S_{k}:=M_{k}+Q_{H}(c_{k})$, we again bound

where $I_{S}:= [S,S+Q_{H}(\tilde{p})\bullet S]\subset [0,Q^{k}_{U}(b_{k})]$ due to the events $\Xi_{[1,M]}, U_{[1,M]}$. As a result, we can take the following  maximum for the outer inverses over interval the $P_{k}$

∏_k∈C_1M_k/B_k^p_k·∏_k∈C_2M_k/A_k^p_k≤∏_k∈C_1u_k∈P_k M_k/B_k(u_k)^p_k·∏_k∈C_2u_k∈P_k M_k/A_k(u_k)^p_k,

A_k(u_k):=Q_U^ke^2u_kx_k∙Q_H(c_k)B_k(u_k):=    Q_U^ke^2u_ky_k∙M_k+Q_H(c_k).

Step 6: Switching the $M_{k}=Q_{H}(c_{k},c_{k}+d_{k}-c_{k})$ to $Q^{k}_{U}$
Next we repeat for $M_{k}$. Before we modify it we take a maximum over $M_{k}$ using the event $D^{k}_{\ell_{k},M}$

(<ref>)≤   ∏_k∈C_1 u_k∈P_k

for $B_{k}(u_{k},T_{k}):=Q_{U}^{k}\para{e^{2u_{k}}y_{k}}\bullet\para{T_{k}+Q_{H}(c_{k})}$. We repeat as above to upper bound the decomposition event $D^{k}_{\ell_{k},M}$ using the events $\Xi_{[1,M]}, U_{[1,M]}$ and maximum

D^k_ℓ_k,M≤u_k∈P_k   D^k_ℓ_k,M(u_k),

D^k_ℓ_k,M(u_k):=   Q_U^ke^2u_kz_k∙Q_H(c_k)∈[a_ℓ_k,a_ℓ_k+1]=I_ℓ_k^k,

z_k:=   d_k-c_k   k∈C_1
d_k-c_k+y_k    k∈C_2.

Step 7: Switching the $ Q_{H}(c_{k}) $ to $ Q_{U}^{k}$
Finally, we repeat for $ Q_{H}(c_{k}) $. We take maximum using the event $S^{k}_{m_{k},c}(\ell_{k})$ and bound that event by maximum too using the events $\Xi_{[1,M]}, U_{[1,M]}$ to get the following overall bound

(<ref>)≤(ℓ_1,m_1,⋯,ℓ_M,m_M)∏_k∈[1,M]E_k^DE_k^S R_k G_[1,M]
=   (ℓ_1,m_1,⋯,ℓ_M,m_M)∏_ku_k∈P_k
T_k∈I_m_k,c^k(ℓ_k)  D^k_ℓ_k,M(u_k,T_k)u_k∈P_k S^k_m_k,c(ℓ_k,u_k) ·

·   ∏_k∈C_1 u_k∈P_k
T_k∈I_m_k,c^k(ℓ_k) a_ℓ_k+1/Q_U^ke^2u_ky_k∙T_k+T_k+Q^k+1(b_k+1)+δ_k+1 ^p_k

·   ∏_k∈C_2u_k∈P_k
T_k∈I_m_k,c^k(ℓ_k)  a_ℓ_k+1/Q_U^ke^2u_kx_k∙T_k+Q^k+1(b_k+1)+δ_k+1  ^p_k · G_[1,M]
=:   (ℓ_1,m_1,⋯,ℓ_M,m_M)∏_kE_k^DE_k^SR_k G_[1,M],

T_k∈I_m_k,c^k(ℓ_k) a_ℓ_k+1/Q_U^ke^2u_ky_k∙T_k+T_k+Q^k+1(b_k+1)+δ_k+1 ^p_k   k∈C_1
T_k∈I_m_k,c^k(ℓ_k)  a_ℓ_k+1/Q_U^ke^2u_kx_k∙T_k+Q^k+1(b_k+1)+δ_k+1  ^p_k    k∈C_2

T_k∈I_m_k,c^k(ℓ_k)  D^k_ℓ_k,M(u_k,T_k)

E_k^S:=u_k∈P_k S^k_m_k,c(ℓ_k,u_k)

D^k_ℓ_k,M(u_k,T_k):=   Q_U^ke^2u_k(d_k-c_k)∙T_k+Q^k+1(b_k+1)+δ_k+1∈[a_ℓ_k,a_ℓ_k+1]=I_ℓ_k^k,

\begin{equation*}
S^{k}_{m_{k},c}(\ell_{k},u_{k}):=\{Q_{U}^{k}(e^{2u_{k}}c_{k})\in  [\wt{b}_{m_{k}}^{k}(\ell_{k}),\wt{b}_{m_{k}+1}^{k}(\ell_{k})]+Q^{k+1}(b_{k+1})+\delta_{k+1}=\wt{I}_{m_{k},c}^{k}(\ell_{k})+Q^{k+1}(b_{k+1})+\delta_{k+1}\}.
\end{equation*}

Step 8: Decoupling Now that we turned everything into $Q^{k}_{U}$, we can decouple. We will apply <ref> in order to remove the shifts $Q^{k+1}(b_{k+1})+\delta_{k+1}$ and so we will return to the single-point estimates. We will do this by bootsrapping starting from the first scale $R_{1}$. First as we did in the proof of <ref>, we scale all the terms by $\delta_{1}$ which means that the first scale is now $Q^{1}$ (i.e. $U$ of height one) but for the other scales to ease on notation we still just write $Q^{2},Q^{3}$ (as opposed to $Q^{\delta_{2}/\delta_{1}}$). We start with applying the tower property for the filtration $\CF(Q^{2}(b_{2}))$

(ℓ_1,m_1,⋯,ℓ_M,m_M)∏_k∈[1,M] E_k^DE_k^SR_k G_[1,M]

=   (ℓ_1,m_1,⋯,ℓ_M,m_M)E_1^DE_1^SR_1(Q^2(b_2)∏_k∈[2,M]E_k^DE_k^S R_k G_[1,M].

We separate the decomposition event $E_{1}^{S}$ by applying conditional
\begin{equation}
\Expe{E_{1}^{D}E_{1}^{S}R_{1}\conditional \CF(Q^{2}(b_{2}))}\leq \para{\Expe{\para{E_{1}^{D}R_{1}}^{b_{11}}\conditional \CF(Q^{2}(b_{2}))}}^{1/b_{11}}\para{\Proba{E_{1}^{S}\conditional \CF(Q^{2}(b_{2}))}}^{1/b_{12}},
\end{equation}

for $b_{11}^{-1}+b_{12}^{-1}=1$. So now we study those two factors. As explained in the previous section, for the first event we have a delta-gap and so we can apply <ref>

where we removed the shift from <ref>

R̃_k:=   u_k∈P_k
T_k∈I_m_k,c^k(ℓ_k) a_ℓ_k+1/Q_U^ke^2u_ky_k∙T_k+T_k ^p_k   k∈C_1
T_k∈I_m_k,c^k(ℓ_k)  a_ℓ_k+1/Q_U^ke^2u_kx_k∙T_k  ^p_k    k∈C_2

Ẽ_k^D:=   u_k∈P_k
T_k∈I_m_k,c^k(ℓ_k)  D^k_ℓ_k,M(u_k,T_k)

Ẽ_k^S:=   u_k∈P_k S^k_m_k,c(ℓ_k,u_k)

D̃^k_ℓ_k,M(u_k,T_k):=   Q_U^ke^2u_k(d_k-c_k)∙T_k∈[a_ℓ_k,a_ℓ_k+1]=I_ℓ_k^k,

S̃^k_m_k,c(ℓ_k,u_k):=   Q_U^k(e^2u_kc_k)∈[b_m_k^k(ℓ_k),b_m_k+1^k(ℓ_k)]=I_m_k,c^k(ℓ_k).

Since we decoupled the decomposition event $E_{1}^{S}$, we also translate

in order to remove the $\ell_1$-dependence and thus bring in the $\ell_{1}$-sum and we used the bound

\begin{equation}
\abs{\tilde{I}_{k,a}(\ell)}=\rho_{a}a_{k}((\ell+1)^{\lambda_{a}}-\ell^{\lambda_{a}})\leq    \rho_{a}a_{k}.
\end{equation}

Step 9: Event $E_{1}^{S}$
Here we study the event

⊂   η^10,c_ℓ,k+Q^2(b_2)+δ_2 ≤e^2u_1c_1

=   η^1Q^2(b_2)+δ_2,c_ℓ,k+Q^2(b_2)+δ_2 ≤e^2u_1c_1-η^10,Q^2(b_2)+δ_2

⊂    η^1Q^2(b_2)+δ_2,c_ℓ,k+Q^2(b_2)+δ_2 ≤e^2u_1c_1 ,

where in the context of <ref>, we have $c_{\ell_{1},m_{1}}=\wt{b}_{m_{1}}(\ell_{1})$
and then we again apply <ref> to decouple

η^1Q^2(b_2)+δ_2,c_ℓ,k+Q^2(b_2)+δ_2 ≤e^2u_1c_1η^10,c_ℓ,k ≤c_1 e^2u_1 ,

where $\wt{\eta}^{1}$ is an independent copy of $\eta^{1}$. So we have

\begin{equation}
\Expe{E_{1}^{S}\conditional \CF\para{[0,Q^{2}(b_{2})]}}\leq \Expe{\maxls{u_{1}\in P_{1} }\ind{\wt{\eta}^{1}\para{0,c_{\ell,k}} \leq e^{2u_{1}}c_{1}}}= \Proba{\wt{\eta}^{1}\para{0,c_{\ell,k}} \leq e^{4}c_{1}},
\end{equation}

where we use the $P_{1}\subset (-2,2)$ from <ref>.
Now that we decoupled we proceed as in <ref>.  Recall that $c_{\ell,k}=\rho_{a}a_{k}c_{\ell}$ for some small enough $\rho_{a}\in (0,1)$ and $c_{\ell}:=\ell^{\lambda_{a}}$ for $\lambda_{a}\in (0,1)$ so that indeed $c_{\ell+1}-c_{\ell}\to 0$ as $\ell\to +\infty$. For $\ell=0$, we bound by zero.  When $c_{\ell,k}\leq \delta_{1}\tor c_{\ell,k}> \delta_{1}$ we use Markov with $p_{1}>0\tand p_{2}>0$ respectively,to upper bound by

∑_ℓ≥1 E_1^S [0,Q^2(b_2)]^1/b_12≤∑_ℓ≥1  η^10,c_ℓ,k ≤e^4c_1 ^1/b_12

≤   ∑_ℓ≥1   c_1^p_1/b_12c_ℓ,k^ζ(-p_1)/b_12+ c_1^p_2/b_12c_ℓ,k^-p_2/b_12

≤   c a_k^-1+/λ_a.

So from here to get finiteness we require both $-\zeta(-p_{1})\lambda_{a},p_{2}\lambda_{a}>b_{12}$. Therefore, we get the singular power $a_{k}^{-\frac{1+\e}{\lambda_{a}}}$ for arbitrarily small $\e>0$.

Step 10: Bootstrap
Therefore, we removed all the conditional expectations and are left with

(<ref>)≤    ℓ_1,m_1a_m_1^-1+/λ_a Ẽ_1^DR̃_1^b_11^1/b_11 · ℓ_i,m_i
i∈[2,M]∏_k∈[2,M]E_k^DE_k^S R_k G_[1,M].

So that means that we decoupled the first scale at the cost of a summation that will give a singular factor in $x^{-\e_{ratio}}$. In particular, we next prove that

(<ref>)≤    cx^-_ratio  · ℓ_i,m_i
i∈[2,M]∏_k∈[2,M]E_k^DE_k^S R_k G_[1,M].

Repeating for the rest of the scales in the order given by the gap-event, will give us $M$ such singular factors.

Step 11: Ratio part for $1\in C_{1}$
Here we study the first factor

\begin{equation}
\para{\Expe{\para{\tilde{E}_{1}^{D}\tilde{R}_{1}}^{b_{11}}}}^{1/b_{11}}
\end{equation}

where recall from <ref>

Ẽ_1^D=   u_1∈P_1
T_1∈I_ρ,ℓ_1  D^1_ℓ_1,M(u_1,T_1),

D^1_ℓ_1,M(u_1,T_1)=   Q_U^1e^2u_1(d_1-c_1)∙T_1∈[a_ℓ_1^1,a_ℓ_1+1^1]=I_ℓ_1^1

R̃_1=   u_1∈P_1

We let

We go over each case as in the proof of <ref> for

Step 12: Case $B^{min}_{1}\in [1,\infty)$
In this case we are left with studying the sum

\begin{equation}
\sumls{\ell_{1}\in \Z\setminus\set{0}}a_{\ell_{1}}^{p_{1}-\frac{1+\e}{\lambda_{a}b_{12} }}\para{\Expe{\maxls{u_{1}\in P_{1}\\\wt{T}_{1}\in I_{\rho,\ell_{1}}  }\ind{Q_{U}^{1}\para{e^{2u_{1}}(d_{1}-c_{1})}\bullet\para{\wt{T}_{1}}\in [a_{\ell_{1}}^{1},a_{\ell_{1}+1}^{1}]}}}^{1/b_{11}}
\end{equation}

for bi-infinite sequence $a_{\ell_{1}}$ going to zero and infinity. In particular, we set

a_k:=ρ_b-ak^-λ_neg    k≤-1

ρ_b-ak^λ_pos    k≥1

for $\lambda_{neg}>0$ and $\lambda_{pos}\in (0,1)$. For $k\geq 1$, we only keep the lower bound

T_1∈I_ρ,ℓ_1  Q_U^1e^2u_1(d_1-c_1)∙T_1∈[a_ℓ_1^1,a_ℓ_1+1^1]

≤    T_1∈I_ρ,ℓ_1  η_U^1T_1,T_1+ρ_b-ak^λ_pos ≤e^2R_1^pert(d_1-c_1)

Here we use <ref> for $p_{min}>0$ to get the sum

∑_k≥1k^λ_posp_1-1+/λ_ab_12 +ζ(-p_min)-1/b_11.

This is finite by taking large enough $p_{min}$. For $k\leq -1$ we bound the probability by one and just study

k≥1k^-λ_negp_1-1+/λ_ab_12 .

Here as in the proof <ref> we take $p_{1}=1+\e$ and $b_{12}$ large enough so that the overall bracketed-exponent is positive $p_{1}-\frac{1+\e}{\lambda_{a}b_{12} }>0$ and then we just take $\lambda_{neg}$ arbitrarily large.

Step 13: Case $B^{min}_{1}\in [0,1]$ and $a_{\ell_{1}}\in [\rho_{b-a},\infty)$
Here we study the sum

ℓ_1≥1a_ℓ_1^p_1-1+/λ_ab_12 Ẽ_1^DB^min_1∈[0,1]u_1∈P_1

As in the proof of <ref>, we start with applying layercake to write the expectation term as

∫_1^∞Ẽ_1^D1/B^min_1 ≥t^1/p_1b_11

=   ∫_1^∞Ẽ_1^Dmax_u_1,T_1,T_1 ηT_1+T_1 ,T_1+T_1 +t^-1/p_1b_11≥e^2u_1y_1

≤   ∫_1^∞T_1∈I_ρ,ℓ_1  η_U^1T_1,T_1+ρ_b-ak^λ_pos ≤e^2R_1^pert(d_1-c_1)

·  max_T_1,T_1ηT_1+T_1 ,T_1+T_1 +t^-1/p_1b_11≥e^-2R_1^perty_1  ,

where in the last inequality we only kept the lower bound in $\tilde{E}_{1}^{D}$ and then applied FKGineq. From here the proof proceeds as in the section "Case $Q(b,b+x)\in [0,1]$ and $Q_{b-a}\bullet Q_{a}\geq \rho_{b-a}$" in the proof of <ref>.

Step 14: Case $B^{min}_{1}\in [0,1]$ and $a_{\ell_{1}}\in [0,\rho_{b-a}]$
Here we study the sum

ℓ_1≤-1a_ℓ_1^-1+/λ_ab_12 Ẽ_1^DB^min_1∈[0,1]u_1∈P_1

This is again the singular case that requires the use of scaling law. In the expectation we have

Ẽ_1^Dmax_T_1,T_1ηT_1+T_1 ,T_1+T_1 +a_ℓ_1t^-1/p_1b_11/T_1∈I_ρ,ℓ_1  η_U^1T_1,T_1+ρ_b-aa_ℓ_1+1 ≥e^-2R_1^pertc_d-c,y ,

where we did the same trick as in the singular case of <ref> of using the lower bound. From here we proceed the same because we have the same constraints on the shifts. The only difference is the bound $ e^{-2R_{1}^{pert}}$ which we treat as a constant.

Step 15: Ratio part for $1\in C_{2}$
Here we study

\begin{equation}
\para{\Expe{\para{\tilde{E}_{1}^{D}\tilde{R}_{1}}^{b_{11}}}}^{1/b_{11}},
\end{equation}

where recakk

Ẽ_1^D=   u_1∈P_1
T_1∈I_ρ,ℓ_1  D^1_ℓ_1,M(u_1,T_1),

D^1_ℓ_1,M(u_1,T_1)=   Q_U^1e^2u_1(d_1-c_1+y_1)∙T_1∈[a_ℓ_1^1,a_ℓ_1+1^1]=I_ℓ_1^1,

R̃_1=   u_1∈P_1
T_1∈I_ρ,ℓ_1  a_ℓ_1+1^1/A_1(u_1,T_1)^p_1=a_ℓ_1+1^1/A^min^p_1A^min:=u_1∈P_1
T_1∈I_ρ,ℓ_1  A_1(u_1,T_1)

As explained in the proof of <ref> this is an easier case of the previous because now they have matching shifts $Q_{c_{1}}$. So we leave it.

§.§ Proof of <ref>: Decreasing numerator

Here we again study the product

∏_k Q_H(c_k,c_k+x_k)/Q_H(d_k,d_k+y_k)^p_k G_[1,M],Ξ_[1,M], U_[1,M].

After we scale by $\delta$, the denominators will be  nonsingular and so there is no need for the log-normal scaling law. We assume $c_{k}<d_{k}$ because the other case is symmetric. We start with decomposing $Q_{H}(c_{k})$ using the analogous decomposition events from <ref>

S_m_k,c(u_k):=   Q_U^k(e^2u_kc_k)∈[b_m_k,b_m_k+1]+Q^k+1(b_k+1)+δ_k+1,

I_m_k,c:=   [b_m_k,b_m_k+1],

for the diverging sequence from <ref> but for $\ell=1$, i.e. $\wt{b}_{m}:=\wt{b}_{m}(1)=\rho_{a}a_{m}$. So using the the semigroup formula we upper bound by a maximum

(<ref>)≤[   ∏_k∑_m_kmax_u_k,T_kS^k_m_k,c(ℓ_k,u_k)Q_H(x_k)∙T_k/Q_H(y_k)∙Q_H(d_k-c_k)∙T_k+T_k^p_k

· G_[1,M],Ξ_[1,M], U_[1,M]],

where the maximum is over $u_{k}\in P_{k}$ and $T_{k}\in \wt{I}_{m_{k},c}^{k}(\ell_{k})+Q^{k+1}(b_{k+1})+\delta_{k+1}$. Here we need to further switch to $Q_{U}^{k}$ by taking maximum again

≤   max_T_kmax_u^i_k∈P_kQ_U^k(e^u^1_kx_k)∙T_k/Q_U^k(e^u^2_ky_k)∙Q_U^k(e^u^3_k(d_k-c_k)∙T_k+T_k^p_k=:R_k.

As we did in the proof of <ref>, we scale all the terms by $\delta_{1}$ which means that the first scale is now $Q^{1}$ (i.e. $U$ of height one) but for the other scales to ease on notation we still just write $Q^{2},Q^{3}$ (as opposed to $Q^{\delta_{2}/\delta_{1}}$). That also means that the denominator for $k=1$ is a constant term that does not depend on $\delta_{1}$ (since we set $y_{1}=r_{1}\delta_{1}$).

Next as done in the equal-length case, we apply the tower-property

∑_m_1 E_1^S  R_1Y_[2,M]
=∑_m_1 E_1^S  R_1(Q^2(b_2)) Y_[2,M],

where $E_{1}^{S}:=\max_{u_{1},T_{1}}\ind{S^{1}_{m_{1},a}(u_{k})}$ and $Y_{[2,M]}$ are the remaining factors. Since we used the decomposition event, we separate it via conditional-
\begin{equation}
\Expe{E_{1}^{S} R_{1}\conditional \CF(Q^{2}(b_{2})}\leq \para{\Expe{\para{R_{1}}^{b_{11}}\conditional \CF(Q^{2}(b_{2})}}^{1/b_{11}}\para{\Proba{E_{1}^{S}\conditional \CF(Q^{2}(b_{2})}}^{1/b_{12}},
\end{equation}

for $b_{11}^{-1}+b_{12}^{-1}=1$. For the event $E_{1}^{S}$ the situation is simpler than before since there is no $\ell$-dependence and so we can use the same method as before in <ref>. For the first factor, we can apply <ref> to be left with studying

By applying and bounding $e^{u^{1}_{1}}x_{1}\leq e^{2R_{1}^{pert}}x_{1}$, we almost return to the setting of <ref>. For the numerator we simply substitute $x_{1}$ by $e^{2R_{1}^{pert}}x_{1}$ to get same exponent for $x_{1}$. For the denominator we instead use a version with perturbation in <ref>. But in the above specific situation it gives the same exponent as the un-perturbed version in <ref>. So by boostrapping we get the desired bound in <ref>.

§ FURTHER RESEARCH DIRECTIONS

* Conditional independence Starting with variables $Q(b,c),Q_{a}$ and event $E:=\set{Q(b)\geq Q(a)+1}$, is there some sigma-algebra $\mathcal{G}$ that gives conditional independence

Q(a)≥t_1,Q(b,c)≥t_2,E =Q(a)≥t_1|𝒢Q(b,c)≥t_2|𝒢?

It seems unlikely because contrary to the case of random walks's conditional independence <ref>, we don't have any independent increments. So if we decompose $Q_{b}=\ell$, then this event is still highly correlated with the increment $Q_{c-b}\bullet Q_{b}=Q_{c-b}\bullet \ell$.

* Independence number

The independent interval graph with endpoints has been studied in <cit.>. It would be interesting if a similar construction of some correlated Poisson process could give exact growth of the independence number.

* Better bounds on the ratio moments. In the similar spirit for <ref>, it would be interesting to possible experiment getting sharper exponents for the factor $x$ (in the decreasing numerator case) or even bounds that are uniform in $x$ (in equal length case).

* Convergence to Lebesgue: The extension to

Q^n(a,b)-(b-a)>Δ η(Q_a,Q_a+x)-x≥Δ

is unclear. This is because the current proof using decoupling doesn't work as evaluating GMC over two distant subintervals $I_{1},I_{3}\subset \spara{Q_{a},Q_{a}+x} $ is still correlated.

§ COMPARISON INEQUALITIES

We will use the Kahane's inequality (eg. <cit.>).
(Kahane Inequality)
Let $\rho$ be a Radon measure on a domain $D\subset\R^{n}$, $X(\cdot)$ and $Y(\cdot)$ be two continuous centred Gaussian fields on $D$, and $F: \Rplus \to \R$ be some smooth function with at most polynomial growth at infinity. For $t \in [0,1]$, define $Z_t(x) = \sqrt{t}X(x) + \sqrt{1-t}Y_t(x)$ and

φ(t) := [ F(W_t)],     W_t := ∫_D e^Z_t(x) - 1/2[Z_t(x)^2] ρ(dx).

Then the derivative of $\varphi$ is given by

\begin{equation}\label{eq:Kahane_int}
\begin{split}
\varphi'(t) & = \frac{1}{2} \int_D \int_D \left(\EE[X(x) X(y)] - \EE[Y(x) Y(y)]\right) \\
& \qquad \qquad \times \EE \left[e^{Z_t(x) + Z_t(y) - \frac{1}{2}\EE[Z_t(x)^2] - \frac{1}{2}\EE[Z_t(y)^2]} F''(W_t) \right] \rho(dx) \rho(dy).
\end{split}
\end{equation}

In particular, if

[X(x) X(y)] ≤[Y(x) Y(y)]     ∀x, y ∈D,

then for any convex $F: \Rplus \to \R$

[F(∫_D e^X(x) - 1/2 [X(x)^2]ρ(dx))]
≤[F(∫_D e^Y(x) - 1/2 [Y(x)^2]ρ(dx))].

and the inequality is reversed if $F$ is concave instead. In the case of distributional fields $X,Y$, we use this result for the mollifications $X_{\e},Y_{\e}$ and pass to the limit due to the continuity of $F$.

The interpolation has implications for the max/min of Gaussian fields.
(Slepian's lemma)<cit.>
Let $\set{X_t}_{t\in T}$ and $\set{Y_t}_{t\in T}$ be two mean zero Gaussian processes. Assume that for all $t, s \in T$, in some index set $T$, we have

\begin{equation}
\Expe{X_{t}^{2}}=\Expe{Y_{t}^{2}}    \tand \Expe{X_{t}X_{s}}\leq \Expe{Y_{t}Y_{s}}.
\end{equation}

In particular, for every $\tau\in \R$, we have

\begin{equation}
\Proba{\sup_{t\in T}X_{t}\leq \tau }\leq \Proba{\sup_{t\in T}Y_{t}\leq \tau }\tand \Proba{\inf_{t\in T}X_{t}\geq -\tau }\leq \Proba{\inf_{t\in T}Y_{t}\geq -\tau }.
\end{equation}

We will also consider another comparison inequality for increasing/decreasing functionals which generalizes the Harris-inequality <cit.> that is called FKG inequality for Gaussian fields <cit.>,<cit.>.

Let $\set{Z(x)}_{x\in U}$ be an a.s. continuous centred Gaussian field on $U \subset \R^d$ with $\Expe{Z(x)Z(y)} \geq 0$ for all $x, y \in U$. Then, if $f, g$ are two bounded, tincreasing measurable functions,

\begin{equation}
\Expe{f\para{\set{Z(x)}_{x\in U}}g\para{\set{Z(x)}_{x\in U}}}    \geq \Expe{f\para{\set{Z(x)}_{x\in U}}}\Expe{g\para{\set{Z(x)}_{x\in U}}}
\end{equation}

and the opposite inequality if one function is increasing and the other one is decreasing

\begin{equation}
\Expe{f\para{\set{Z(x)}_{x\in U}}g\para{\set{Z(x)}_{x\in U}}}    \leq \Expe{f\para{\set{Z(x)}_{x\in U}}}\Expe{g\para{\set{Z(x)}_{x\in U}}}.
\end{equation}

§ MOMENT ESTIMATES FOR GMC

§.§ Moments bounds of GMC

For positive continuous bounded function $g_{\delta}:\Rplus\to \Rplus$ with $g_{\delta}(x)=0$ for all $x\geq \delta$ and uniform bound $M_{g}:=\sup_{\delta\geq 0}\norm{g_{\delta}}_{\infty}$,  we will need the following moment estimates for $\eta^{\delta}=\eta_{g}^{\delta}$ in order to compute the moments of the inverse $Q_{g}^{\delta}$. As mentioned in the notations we need to study more general fields such as $U^{\delta,\lambda}$ in <ref>

\begin{equation}
\Expe{U_{ \varepsilon}^{\delta,g }(x_{1} )U_{ \varepsilon}^{  \delta,g }(x_{2} )  }=\left\{\begin{matrix}
\ln(\frac{\delta }{\varepsilon} )-\para{\frac{1 }{\e}-\frac{1}{\delta}}\abs{x_{2}-x_{1}}+g_{\delta}(\abs{x_{2}-x_{1}})&\tifc \abs{x_{2}-x_{1}}\leq \varepsilon\\
\ln(\frac{\delta}{\abs{x_{2}-x_{1}}})-1+\frac{\abs{x_{2}-x_{1}}}{\delta}+g_{\delta}(\abs{x_{2}-x_{1}}) &\tifc \e\leq \abs{x_{2}-x_{1}}\leq \delta\\
0&\tifc \delta\leq \abs{x_{2}-x_{1}}
\end{matrix}\right.    .
\end{equation}

The following proposition shown in <cit.> is studying the moment bounds for this field.

We have the following estimates:

* For $q\in (-\infty,0)\cup [1,\frac{2}{\gamma^{2}})$ we have

\begin{equation}
\Expe{\para{\eta^{\delta}[0,  t]}^{q}}\leq c_{1}\branchmat{t^{\zeta(q)}\delta^{\frac{\gamma^{2}}{2}(q^{2}-q)} &t\in [0,\delta]\\ t^{q}& t\in [\delta,\infty] },
\end{equation}

where $\zeta(q):=q-\frac{\gamma^{2}}{2}(q^{2}-q)$ is the multifractal exponent and $c_{1}:=e^{\frac{\gamma^{2}}{2}(q^{2}-q)M_{g}}\Expe{\para{\int_{0}^{1}e^{ \overline{\omega}^{1}(x)}\dx}^{q}}$ and $\omega^1$ is the exact-scaling field of height $1$.

* When $t,\delta\in [0,1]$ then we have a similar bound with no $\delta$

\begin{equation}
\Expe{\para{\eta^{\delta}[0,  t]}^{q}}\leq c_{1} t^{\zeta(q)},
\end{equation}

but this exponent is not that sharp: if $\delta\leq t\leq 1$ and $q>1$, we have $q>\zeta(q)$ and so the previous bound is better.

* Finally, for all $t\in [0,\infty)$ and $q\in [0,1]$, we have

\begin{equation}
\Expe{\para{\eta^{\delta}[0,  t]}^{q}}\leq c_{2,\delta } t^{\zeta(q)}\delta^{\frac{\gamma^{2}}{2}(q^{2}-q)},
\end{equation}