Linear response and moderate deviations: hierarchical approach. II

TL;DR

This paper extends the Moderate Deviations Principle to a new class of random fields using a hierarchical, inductive approach, improving understanding of deviations in complex stochastic systems.

Contribution

It introduces a hierarchical, inductive method to establish upper bounds for moderate deviations in random fields, generalizing previous results beyond sums of independent variables.

Findings

Upper bound for moderate deviations in a new class of random fields.

Inductive approach applicable in multiple dimensions.

Refined formulations removing stationarity assumptions.

Abstract

The Moderate Deviations Principle (MDP) is well-understood for sums of independent random variables, worse understood for stationary random sequences, and scantily understood for random fields. An upper bound for a new class of random fields is obtained here by induction in dimension. Version 3. Sect 1. Stationarity, being not essential in the proofs, is removed from the definitions and the main result formulation. Sect. 2. $[C,2C]$ instead of $[C_1,2C_1]$ before Prop. 2.6; $ a\ge1$ instead of $a\ge C/C_1$ in the last proof; Remark 2.5 added; supremum over shifts in (2.2) (formerly (2.3)); "centered" instead of "CMS". Cosmetic changes: indexing of leaks; Remark 2.11. Sect. 3. Cosmetic change: semicolon after the second display of the proof of Lemma 3.9. References: "response", not "responce".

Equations

$$\begin{gathered} {\int }_{[\alpha ,\beta ]\times B}{X}_t\mathrm{d}t+{\int }_{[\beta ,\gamma ]\times B}{X}_t\mathrm{d}t= \\ ={\int }_{[\alpha ,\gamma ]\times B}{X}_t\mathrm{d}t\text{a.s. whenever }-\infty <\alpha <\beta <\gamma <\infty , \end{gathered}$$

$$\mathbb{E}\,\bigg{|}\int_{B}X_{t}\,\mathrm{d}t\bigg{|}<\infty\quad\text{and}\quad\mathbb{E}\,\int_{B}X_{t}\,\mathrm{d}t=0\quad\text{for all boxes }B\subset\mathbb{R}^{d}\,.$$

$$\begin{gathered} \text{the distribution of}{\int }_{[0,r_1]\times \cdots \times [0,r_d]}{X}_t\mathrm{d}t \\ \text{ is a Borel measurable function of }(r_1,\dots ,r_d). \end{gathered}$$

$$\begin{gathered} Y(t_2,\dots ,t_d)= \\ {\int }_a^r{X}^{-}(t_1,t_2,\dots ,t_d)\mathrm{d}{t}_1+{\int }_r^b{X}^{+}(t_1,t_2,\dots ,t_d)\mathrm{d}{t}_1-{\int }_a^b{X}^0(t_1,t_2,\dots ,t_d)\mathrm{d}{t}_1 \end{gathered}$$

$${\int }_{B}Y(t)\mathrm{d}t={\int }_{[a,r]\times B}{X}^{-}(t)\mathrm{d}t+{\int }_{[r,b]\times B}{X}^{+}(t)\mathrm{d}t-{\int }_{[a,b]\times B}{X}^0(t)\mathrm{d}t$$

$$\exists \epsilon >0\forall i\in I\mathbb{E}\exp \epsilon |X_i|\le 2,$$

$$\forall i\in I\;\>\forall s\in\mathbb{R}^{d}\;\;\mathbb{E}\,\exp\varepsilon\bigg{|}\int_{B+s}X_{i}(t)\,\mathrm{d}t\bigg{|}\leq 2\,,$$

$$\mathrm{vol}B=({\beta }_1-{\alpha }_1)\dots ({\beta }_d-{\alpha }_d),0ptB=\min ({\beta }_1-{\alpha }_1,\dots ,{\beta }_d-{\alpha }_d).$$

$$\text{if}C|\lambda |\le \frac{1}{{\log }^d\mathrm{vol}B}\text{ and }0ptB\ge C,\text{then}$$

$$\log \mathbb{E}\exp \lambda {\int }_B{X}_{t,i}\mathrm{d}t\le C(\mathrm{vol}B){\lambda }^2.$$

$$\log \mathbb{E}\exp \lambda {\int }_B{Y}_{k,r,a,b,i}(t)\mathrm{d}t\le C_1(\mathrm{vol}B){\lambda }^2$$

$$\log \mathbb{E}\exp \lambda Y_{1,r,a,b,i}\le C_1{\lambda }^2\text{whenever }{C}_1|\lambda |\le 1,$$

$$\text{if}C|\lambda |\le \frac{1}{v^{1/(2d)}{\log }^{d-1}v},\text{then}\log \mathbb{E}\exp \lambda {\int }_B{X}_i(t)\mathrm{d}t\le Cv{\lambda }^2.$$

$$S_{B,i}=\frac{1}{\sqrt{\mathrm{vol}B}}{\int }_B{X}_{t,i}\mathrm{d}t$$

$$f_{B,i}(\lambda )=\underset{s\in {\mathbb{R}}^d}{\sup }\log \mathbb{E}\exp \lambda S_{B+s,i}.$$

$$f_B(\lambda )=\underset{i\in I}{\sup }{f}_{B,i}(\lambda );$$

$$f_{v,C}(\lambda )=\underset{\mathrm{vol}(B)=v,0ptB\ge C}{\sup }{f}_B(\lambda )\text{for }v\ge C^d.$$

$$R(v)=v^{\frac{1}{d}}\text{and}S(v)=v^{\frac{d-1}{d}}$$

$$f_{v,C}(\lambda )\le C{\lambda }^2\text{whenever }C|\lambda |\le \frac{\sqrt{S(v)}}{{\log }^{d-1}v}\text{ and }v\ge C^d.$$

$$f_{B_{2}}(\lambda)\leq\frac{2}{p}f_{B_{1}}\Big{(}\frac{p\lambda}{\sqrt{2}}\Big{)}+C_{1}\frac{p}{p-1}\cdot\frac{\lambda^{2}}{2r}\,.$$

$$f_{B_0}(\lambda )\le a{\lambda }^2\text{whenever }|\lambda |\le \delta ,$$

$$f_B(\lambda )\le 2a{\lambda }^2\text{whenever }|\lambda |\le \Delta ,$$

$$\displaystyle\!\!\!\!\!\!f_{B_{k+1}}(\lambda)\leq\frac{2}{p}f_{B_{k}}\Big{(}\frac{p\lambda}{\sqrt{2}}\Big{)}+C_{1}\frac{p}{p-1}\cdot\frac{\lambda^{2}}{R(\operatorname{vol}B_{k+1})}$$

$$\text{for }{C}_1|\lambda |\le \frac{p-1}{p}\sqrt{\mathrm{vol}{B}_{k+1}}{\log }^{-(d-1)}S(\mathrm{vol}{B}_{k+1})$$

$$f_{B_{k+1}}(\lambda)\leq\frac{2}{p}A_{k}^{2}\Big{(}\frac{p\lambda}{\sqrt{2}}\Big{)}^{2}+qx^{2}\lambda^{2}=(A_{k}^{2}p+x^{2}q)\lambda^{2}$$

$${\Delta }_k$$

$$\frac{p_k}{\sqrt{2}}\frac{p_{k+1}}{\sqrt{2}}\dots \frac{p_{N-1}}{\sqrt{2}}\sqrt{\frac{1}{a}S(\mathrm{vol}{B}_N)}\le \sqrt{\frac{1}{a}S(\mathrm{vol}{B}_k)};$$

$$p_k\dots p_{N-1}\le 2^{\frac{N-k}{2}}\cdot 2^{-\frac{d-1}{2d}(N-k)}=2^{\frac{N-k}{2d}}.$$

$$f_{v,C}(\lambda )\le 2a{\lambda }^2\text{whenever }{C}_1|\lambda |\le \sqrt{\frac{1}{a}S(v)}{\log }^{-(d-1)}S(v).$$

$$f_B(\lambda )\le a{\lambda }^2\text{whenever }|\lambda |\le \delta .$$

$$f_B(\lambda )$$

Full text

1. 1 Definition, and main result formulated
2. 2 Far from large deviations
3. 3 Close to large deviations

1 Definition, and main result formulated

We examine a class of random fields $X=(X_{t})_{t\in\mathbb{R}^{d}}$, but we are interested only in integrals $\int_{B}X_{t}\,\mathrm{d}t$ over boxes $B=[\alpha_{1},\beta_{1}]\times\dots\times[\alpha_{d},\beta_{d}]\subset\mathbb{R}^{d}$ (where $\alpha_{1}<\beta_{1},\dots,\alpha_{d}<\beta_{d}$) rather than “individual” random variables $X_{t}$. Similarly to [2, Sect. 1] we merely deal with a box-indexed family of random variables, denoted (if only for convenience) by $\int_{B}X_{t}\,\mathrm{d}t$ and satisfying additivity:

[TABLE]

for all boxes $B\subset\mathbb{R}^{d-1}$; this being additivity w.r.t. the first coordinate, the same is required for each coordinate. We say that $X$ is centered, if

[TABLE]

Two more conditions, stationarity and mesurability, will be used in [3] when proving the moderate deviations principle. Stationarity means measure preserving time shifts that send $\int_{B}X_{t}\,\mathrm{d}t$ to $\int_{B+s}X_{t}\,\mathrm{d}t$. Measurability (under stationarity) is similar to [2, (1.2)]:

[TABLE]

But for now, our goal being upper bounds, we need only (1.1) and (1.2), and call such $X$ a centered random field (on $\mathbb{R}^{d}$). The notions “independent” and “identically distributed” are interpreted for such processes similarly to [2, Sect. 1]. Sometimes, when convenient, we write $X(t)$ rather than $X_{t}$.

Splittability, defined in [2, Def. 1.4] for $d=1$, will be defined here for all $d$.

First, given a centered random field $X=(X_{t})_{t\in\mathbb{R}^{d}}$, we define a split of $X$ as a triple of random fields $X^{0},X^{-},X^{+}$ (on some probability space) such that the two fields $X^{-},X^{+}$ are (mutually) independent and the four fields $X,X^{0},X^{-},X^{+}$ are identically distributed. (Informally, a split is useful when its leak defined below is small.) Clearly, $X^{0},X^{-},X^{+}$ are centered (since $X$ is).

Second, given $r,a,b\in\mathbb{R}$, $a<r<b$, and a split $(X^{0},X^{-},X^{+})$ of $X$, we define the leak (of this split) along the hyperplane $\{r\}\times\mathbb{R}^{d-1}$ on the strip $[a,b]\times\mathbb{R}^{d-1}$ as the random field $(Y_{t})_{t\in\mathbb{R}^{d-1}}$ where

[TABLE]

in the sense that

[TABLE]

for all boxes $B\subset\mathbb{R}^{d-1}$. Clearly, $Y$ is also a centered random field. Similarly, for each $k\in\{1,\dots,d\}$ we define the leak along the hyperplane $\mathbb{R}^{k-1}\times\{r\}\times\mathbb{R}^{d-k}$ on the coordinate strip $\mathbb{R}^{k-1}\times[a,b]\times\mathbb{R}^{d-k}$. (Different $k$ and $r$ need different splits, to be useful.)

Given $X$, we need splits $(X_{k,r}^{0},X_{k,r}^{-},X_{k,r}^{+})$ of $X$ for all $k=1,\dots,d$ and $r\in\mathbb{R}$; and for all $a\in(-\infty,r)$, $b\in(r,\infty)$ we consider the corresponding leak $Y_{k,r,a,b}$ of this split. For $d=1$ the leak is just a single random variable $Y=\int_{a}^{r}X^{-}(t)\,\mathrm{d}t+\int_{r}^{b}X^{+}(t)\,\mathrm{d}t-\int_{a}^{b}X^{0}(t)\,\mathrm{d}t$.

Accordingly, a family $(X_{i})_{i\in I}$ of random fields on $\mathbb{R}^{d}$ leads to a family $(Y_{k,r,a,b,i})_{k=1,\dots,d;a<r<b;i\in I}$ of centered random fields on $\mathbb{R}^{d-1}$.

Third, we consider a family of centered random fields (X_{i})_{i\in I}=\bigl{(}(X_{t,i})_{t\in\mathbb{R}^{d}}\bigr{)}\vphantom{)}_{i\in I} (the index set $I$ being arbitrary) and define uniform splittability of such family. Then splittability of a centered random field appears as the particular case of a single-element set $I$. We define uniform splittability by recursion in the dimension $d=0,1,2,\dots$, treating a centered random field on $\mathbb{R}^{0}$ as just a single random variable of zero mean.

1.4 Definition**.**

A family $(X_{i})_{i\in I}$ of centered random fields $X_{i}=(X_{t,i})_{t\in\mathbb{R}^{d}}$ is uniformly splittable, if either $d=0$ and

[TABLE]

or $d\geq 1$ and the following two conditions hold:

(a) there exist a box $B\subset\mathbb{R}^{d}$ and $\varepsilon>0$ such that

[TABLE]

(b) there exist splits $(X_{k,r,i}^{0},X_{k,r,i}^{-},X_{k,r,i}^{+})$ of each $X_{i}$, whose leaks are a uniformly splittable family $(Y_{k,r,a,b,i})_{k=1,\dots,d;a<r<b;i\in I}$.

We use volume and width of a box $B=[\alpha_{1},\beta_{1}]\times\dots\times[\alpha_{d},\beta_{d}]\subset\mathbb{R}^{d}$:

[TABLE]

The theorem below applies first of all to a single random field (that is, a single-element set $I$); the general formulation enables the proof by induction in the dimension.

1.5 Theorem**.**

For every uniformly splittable family $(X_{i})_{i\in I}$ of centered random fields $X_{i}=(X_{t,i})_{t\in\mathbb{R}^{d}}$ there exists $C\in(1,\infty)$ such that for every $i\in I$, every box $B\subset\mathbb{R}^{d}$ and every $\lambda\in\mathbb{R}$,

[TABLE]

(Of course, $\log^{d}\operatorname{vol}B$ is \bigl{(}\log(\operatorname{vol}B)\bigr{)}\vphantom{)}^{d}.) This theorem will be proved by induction in the dimension $d=1,2,\dots$ Throughout we assume that a uniformly splittable family $(X_{i})_{i\in I}$ of centered random fields $X_{i}=(X_{t,i})_{t\in\mathbb{R}^{d}}$ is given. We assume that the theorem holds in dimension $d-1$, unless $d=1$; in the latter case the proof needs trivial modifications.

According to Def. 1.4 we have splits (of the fields $X_{i}$) whose leaks $Y_{k,r,a,b,i}$ are a uniformly splittable family (in dimension $d-1$). Theorem 1.5, applied to this family, gives $C_{1}$ such that for all $k,r,a,b,i$

[TABLE]

whenever $B\subset\mathbb{R}^{d-1}$ is a box of width $\geq C_{1}$, and $C_{1}|\lambda|\leq\frac{1}{\log^{d-1}\operatorname{vol}B}$.

For $d=1$ the leaks $Y_{1,r,a,b,i}$ are just random variables; their uniform splittability means $\exists\varepsilon>0\;\forall r,a,b\;\forall i\in I\;\>\mathbb{E}\,\exp\varepsilon|Y_{1,r,a,b,i}|\leq 2$. By Lemma 1.8 below this implies $\log\mathbb{E}\,\exp\varepsilon\lambda Y_{1,r,a,b,i}\leq\lambda^{2}$ for $\lambda\in[-1,1]$. Taking C_{1}=\max\bigl{(}\frac{1}{\varepsilon^{2}},1\bigr{)}\vphantom{)} we get for all $r,a,b,i$

[TABLE]

to be used for $d=1$ instead of (1.6).

We borrow from [2, Lemma 2a8] a general fact.

1.8 Lemma**.**

If a random variable $Z$ satisfies $\mathbb{E}\,\exp|Z|\leq 2$ and $\mathbb{E}\,Z=0$, then $\log\mathbb{E}\,\exp\lambda Z\leq\lambda^{2}$ for all $\lambda\in[-1,1]$.

Theorem 1.5 is proved in Sect. 2 for \lambda=\mathcal{O}\bigl{(}(\operatorname{vol}B)^{-\frac{1}{2d}}\log^{-(d-1)}\operatorname{vol}B\bigr{)}\vphantom{)}; larger $\lambda$ are treated in Sect. 3. I still do not know, what happens when $\lambda$ tends to [math] slower than $\log^{-d}\operatorname{vol}B$. This logarithmic gap between moderate and large deviations, is it a phenomenon or a drawback of my approach?

2 Far from large deviations

2.1 Proposition**.**

There exists $C\in(1,\infty)$ such that for every $i\in I$, every box $B\subset\mathbb{R}^{d}$ of volume $v$ and width $\geq C$, and every $\lambda\in\mathbb{R}$,

[TABLE]

Similarly to [2, Sect. 2a] we consider random variables

[TABLE]

their cumulant generating functions $\lambda\mapsto\log\mathbb{E}\,\exp\lambda S_{B,i}$, and ensure shift invariance by taking the supremum over all shifts of a box:

[TABLE]

Still, $f_{B,i}(\lambda)\geq 0$, since $\mathbb{E}\,\exp\lambda S_{B,i}\geq\mathbb{E}\,(1+\lambda S_{B,i})=1$.

Further, we take the supremum over all $i$ and all boxes of a given volume and width $\geq C$:

[TABLE]

All these functions map $\mathbb{R}$ to $[0,\infty]$.

Denoting for convenience

[TABLE]

we rewrite (not proved yet) Prop. 2.1 as follows.

2.3 Proposition**.**

There exists $C\in(1,\infty)$ such that

[TABLE]

We generalize [2, Prop. 2a9(a)]. Given a box $B\subset\mathbb{R}^{d-1}$ and a number $r>0$, we consider two boxes $B_{1}=[0,r]\times B$ and $B_{2}=[-r,r]\times B$ in $\mathbb{R}^{d}$. Let $v=\operatorname{vol}B_{1}$ and $0ptB\geq C_{1}$. For $d=1$ we mean $B_{1}=[0,r]$ and $B_{2}=[-r,r]$; $B$ disappears, as well as the condition on $0ptB$; by convention, $\log^{-(d-1)}\operatorname{vol}B=1$, and (in the proof) $\operatorname{vol}B=1$.

2.4 Lemma**.**

For all $p\in(1,\infty)$ and $\lambda$ such that $C_{1}|\lambda|\leq\frac{p-1}{p}\sqrt{2v}\log^{-(d-1)}\operatorname{vol}B$,

[TABLE]

Proof.

Given $i\in I$, we use the split $(X^{0},X^{-},X^{+})$ of $X_{i}$ whose leak $Y=Y_{1,0,-r,r,i}$ on the strip $[-r,r]\times\mathbb{R}^{d-1}$ satisfies (1.6). Similarly to [2, 2a7 and 2a9], the random variables $U=\frac{1}{\sqrt{v}}\int_{[-r,0]\times B}X_{t}^{-}\,\mathrm{d}t$, $V=\frac{1}{\sqrt{v}}\int_{[0,r]\times B}X_{t}^{+}\,\mathrm{d}t$, $W=\frac{1}{\sqrt{2v}}\int_{[-r,r]\times B}X_{t}^{0}\,\mathrm{d}t$ and $Z=-\int_{B}Y_{t}\,\mathrm{d}t$ satisfy $Z=\sqrt{2v}W-\sqrt{v}U-\sqrt{v}V$ and, by (1.6), $\log\mathbb{E}\,\exp\lambda Z\leq C_{1}(\operatorname{vol}B)\lambda^{2}$ for $C_{1}|\lambda|\leq\log^{-(d-1)}\operatorname{vol}B$. Similarly to [2, Prop. 2a9], by Hölder’s inequality, f_{B_{2},i}(\lambda)=\log\mathbb{E}\,\exp\frac{\lambda}{\sqrt{2v}}\int_{B_{2}}X_{t,i}\,\mathrm{d}t=\log\mathbb{E}\,\exp\lambda W=\log\mathbb{E}\,\exp\lambda\bigl{(}\tfrac{U+V}{\sqrt{2}}+\frac{1}{\sqrt{2v}}Z\bigr{)}\vphantom{)}\leq\frac{1}{p}\cdot 2f_{B_{1}}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}+\frac{p-1}{p}\log\mathbb{E}\,\exp\frac{p}{p-1}\frac{\lambda Z}{\sqrt{2v}}. The second term does not exceed \frac{p-1}{p}C_{1}(\operatorname{vol}B)\bigl{(}\frac{p}{p-1}\frac{\lambda}{\sqrt{2v}}\bigr{)}\vphantom{)}^{2}=C_{1}\frac{\operatorname{vol}B}{2v}\frac{p}{p-1}\lambda^{2} for $\frac{p}{p-1}\frac{C_{1}|\lambda|}{\sqrt{2v}}\leq\log^{-(d-1)}\operatorname{vol}B$; and $\frac{\operatorname{vol}B}{2v}=\frac{1}{2r}$. It remains to take supremum in $i\in I$.

For $d=1$ the leak $Y=Y_{1,0,-r,r,i}$, being a random variable, satisfies (1.7); $Z=-Y$; $\log\mathbb{E}\,\exp\lambda Z\leq C_{1}\lambda^{2}$ for $C_{1}|\lambda|\leq 1$ (and $v=r$, of course). ∎

2.5 Remark**.**

Above, a box is halved (divided in two boxes) by the hyperplane $x_{1}=0$. More generally, the same holds when halving the box by another coordinate hyperplane $x_{k}=c$ for $k\in\{1,\dots,d\}$ and the appropriate $c$.

Now we consider two boxes in $\mathbb{R}^{d}$, $B_{0}=[0,r_{1}]\times\dots\times[0,r_{d}]$ and $B=[0,2^{n_{1}}r_{1}]\times\dots\times[0,2^{n_{d}}r_{d}]$ for arbitrary $r_{1},\dots,r_{d}\in[C,2C)$ for some $C\geq C_{1}$, and arbitrary $n_{1},\dots,n_{d}\in\{0,1,2,\dots\}$.

2.6 Proposition**.**

If $C$ is large enough, then for all $r_{1},\dots,r_{d}\in[C,2C)$, $\delta>0$ and $a\geq\frac{C}{R(\operatorname{vol}B_{0})}$ satisfying

[TABLE]

where $B_{0}=[0,r_{1}]\times\dots\times[0,r_{d}]$, there exists natural $N$ such that the following holds for all $n_{1},\dots,n_{d}\in\{0,1,2,\dots\}$ satisfying $n_{1}+\dots+n_{d}\geq N$:

[TABLE]

where $B=[0,2^{n_{1}}r_{1}]\times\dots\times[0,2^{n_{d}}r_{d}]$ and $\Delta=\frac{1}{C_{1}}\sqrt{\frac{1}{a}S(\operatorname{vol}B)}\log^{-(d-1)}S(\operatorname{vol}B)$.

(For $d=1$, by convention, $\log^{-(d-1)}S(\dots)=1$, notwithstanding that $S(\dots)=1$.)

Proof.

We take $n=n_{1}+\dots+n_{d}$, $B_{n}=B$, halve the longest side of $B_{n}$, denote the half of $B_{n}$ by $B_{n-1}$, and repeat this operation getting $B_{n-2},\dots,B_{0}$. For each $k=0,\dots,n-1$ we have $\operatorname{vol}B_{k+1}=2^{k+1}\operatorname{vol}B_{0}$ and, assuming $C\geq C_{1}$,

[TABLE]

by Lemma 2.4 (and Remark 2.5), since (recall $r$ and $B$ of 2.4) the longest side $2r$ of $B_{k+1}$ cannot be less than $R(\operatorname{vol}B_{k+1})$, and $\operatorname{vol}B=\frac{\operatorname{vol}B_{k+1}}{2r}\leq S(\operatorname{vol}B_{k+1})$. (For $d=1$ this is just $1=\frac{2r}{2r}\leq 1$.)

Given $A_{k}$ and $\Delta_{k}$ such that $f_{B_{k}}(\lambda)\leq A_{k}^{2}\lambda^{2}$ for $|\lambda|\leq\Delta_{k}$, we denote $q=\frac{p}{p-1}$, $x=\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}$ and get

[TABLE]

for |\lambda|\leq\min\bigl{(}\frac{\sqrt{2}}{p}\Delta_{k},\frac{1}{C_{1}q}\sqrt{\operatorname{vol}B_{k+1}}\log^{-(d-1)}S(\operatorname{vol}B_{k+1})\bigr{)}\vphantom{)}. Generally, the minimum of $A_{k}^{2}p+x^{2}q$ over $p,q$ such that $\frac{1}{p}+\frac{1}{q}=1$ is equal to $(A_{k}+x)^{2}$, and is reached at $p=1+\frac{x}{A_{k}}$, $q=1+\frac{A_{k}}{x}$. Thus, $f_{B_{k+1}}(\lambda)\leq A_{k+1}^{2}\lambda^{2}$ for $|\lambda|\leq\Delta_{k+1}$, provided that $A_{k+1}\geq A_{k}+x=A_{k}+\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}$ and \Delta_{k+1}\leq\min\Big{(}\frac{\sqrt{2}}{p_{k}}\Delta_{k},\frac{1}{C_{1}q_{k}}\sqrt{\operatorname{vol}B_{k+1}}\log^{-(d-1)}S(\operatorname{vol}B_{k+1})\Big{)}; here $p_{k}=1+\frac{x}{A_{k}}=1+\frac{1}{A_{k}}\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}$ and $q_{k}=1+\frac{A_{k}}{x}=1+\frac{A_{k}}{\sqrt{C_{1}}}\sqrt{R(\operatorname{vol}B_{k+1})}$.

We take $A_{k}=\sqrt{a}+\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{0})}}\sum_{i=1}^{k}2^{-\frac{i}{2d}}$ (thus $A_{k+1}=A_{k}+\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}\,$) and note that $A_{k}\uparrow A_{\infty}\leq\sqrt{a}+\sqrt{\frac{C_{1}}{C}a}\sum_{i=1}^{\infty}2^{-\frac{i}{2d}}\leq\sqrt{2a}$ if $C$ is large enough (since $a\geq\frac{C}{R(\operatorname{vol}B_{0})}$). Also, $q_{k}\leq 1+\frac{\sqrt{2a}}{\sqrt{C_{1}}}\sqrt{R(\operatorname{vol}B_{k+1})}\leq\frac{\sqrt{3a}}{\sqrt{C_{1}}}\sqrt{R(\operatorname{vol}B_{k+1})}$ for all $k$, if $C$ is large enough (since $a\geq\frac{C}{R(\operatorname{vol}B_{0})}$ again). Assuming also $C_{1}\geq 3$ (which is harmless) we introduce $M_{k}=\frac{1}{C_{1}}\sqrt{\frac{1}{a}S(\operatorname{vol}B_{k})}\log^{-(d-1)}S(\operatorname{vol}B_{k})$, note that $M_{k+1}\leq\frac{1}{C_{1}q_{k}}\sqrt{\operatorname{vol}B_{k+1}}\log^{-(d-1)}S(\operatorname{vol}B_{k+1})$ (since $q_{k}\leq\sqrt{aR(\operatorname{vol}B_{k+1})}$), and replace the condition on $\Delta_{k+1}$ given above with the stronger condition \Delta_{k+1}\leq\min\bigl{(}\frac{\sqrt{2}}{p_{k}}\Delta_{k},M_{k+1}\bigr{)}\vphantom{)}. Now we note that $p_{k}-1=\frac{1}{A_{k}}\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}\leq\frac{1}{\sqrt{a}}\sqrt{\frac{C_{1}}{R(\operatorname{vol}B_{k+1})}}\leq\sqrt{\frac{C_{1}R(\operatorname{vol}B_{0})}{CR(\operatorname{vol}B_{k+1})}}\leq 2^{-\frac{k+1}{2d}}$ if $C\geq C_{1}$, take integer $N$ such that $2^{-\frac{N+1}{2d}}\leq 2^{\frac{1}{2d}}-1$, and get $M_{k+1}\leq\frac{\sqrt{2}}{p_{k}}M_{k}$ for all $k\geq N$ (since $\frac{M_{k+1}}{M_{k}}\leq 2^{\frac{d-1}{2d}}$ and $p_{k}\leq 1+2^{-\frac{N+1}{2d}}\leq 2^{\frac{1}{2d}}$). We choose $\Delta_{k}$ as follows:

[TABLE]

Clearly, $\Delta_{k+1}\leq\frac{\sqrt{2}}{p_{k}}\Delta_{k}$ for all $k$.

In order to obtain $f_{B_{n}}(\lambda)\leq 2a\lambda^{2}$ for $|\lambda|\leq M_{n}$ when $n\geq N$ it is sufficient to ensure that $\Delta_{0}\leq\delta$ and $\Delta_{k}\leq M_{k}$ for $k=0,1,\dots,N-1$. We note that $p_{0}\dots p_{N-1}\leq\prod_{k=0}^{N-1}(1+2^{-\frac{k+1}{2d}})\leq\exp\sum_{k=0}^{\infty}2^{-\frac{k+1}{2d}}$ and $C_{1}M_{N}\leq\sqrt{\frac{1}{a}S(\operatorname{vol}B_{N})}\leq\sqrt{\frac{1}{C}R(\operatorname{vol}B_{0})S(\operatorname{vol}B_{N})}=2^{\frac{d-1}{2d}N}\sqrt{\frac{1}{C}\operatorname{vol}B_{0}}$, thus \Delta_{0}=p_{0}\dots p_{N-1}\cdot 2^{-\frac{N}{2}}M_{N}\leq\bigl{(}\exp\sum_{k}2^{-\frac{k+1}{2d}}\bigr{)}\vphantom{)}\cdot 2^{-\frac{N}{2d}}\frac{1}{C_{1}}\sqrt{\frac{1}{C}(2C)^{d}}; by increasing $N$ as needed we get $\Delta_{0}\leq\delta$.

It remains to ensure that $\Delta_{k}\leq M_{k}$ for $k=0,1,\dots,N-1$. We’ll get a bit more: $\Delta_{k}\leq\frac{1}{C_{1}}\sqrt{\frac{1}{a}S(\operatorname{vol}B_{k})}\log^{-(d-1)}S(\operatorname{vol}B_{N})$, that is,

[TABLE]

We may check it only for $k=0$ and $k=N$ due to the fact that $p_{k}\dots p_{N-1}$ is a logarithmically convex function of $k$ (since $p_{k}$ decrease). For $k=N$ it is just $1\leq 1$. For $k=0$ we need $p_{0}\dots p_{N-1}\leq 2^{\frac{N}{2d}}$, which holds for $N$ such that $\exp\sum_{k}2^{-\frac{k+1}{2d}}\leq 2^{\frac{N}{2d}}$. ∎

2.7 Remark**.**

In the proof of 2.6, the restriction on $C$ depends only on $d$ and $C_{1}$. For large $d$, roughly, $C\geq\mathcal{O}(d^{2})C_{1}$. Also, the restriction on $N$ depends only on $d$, $C$ and $\delta$; roughly, $N\geq\mathcal{O}(d^{2})\log C+\mathcal{O}(d)\log\frac{1}{\delta}$.

2.8 Remark**.**

In 2.6, $a$ and $\delta$ may depend on $r_{1},\dots,r_{d}$. Assume for a while that they do not; that is, the given $a$ and $\delta$ serve all $r_{1},\dots,r_{d}$ (for the given $C$). Then the conclusion (that $f_{B}(\lambda)\leq 2a\lambda^{2}$ whenever $|\lambda|\leq\Delta$) holds for all $B$ such that $0ptB\geq C$ and $\operatorname{vol}B$ is large enough (namely, $\operatorname{vol}B\geq 2^{N}(2C)^{d}$). We get, for all $v$ large enough,

[TABLE]

It appears (Lemma 2.12 below) that the assumption of 2.8 is satisfied always (that is, for every uniformly splittable family). Alternatively, the reader may just include that assumption into Def. 1.4 (replacing Item (a) there) and skip to the proof of Prop. 2.3 near the end of this section.

2.9 Lemma**.**

For every uniformly splittable family $(X_{i})_{i\in I}$ of centered random fields on $\mathbb{R}^{d}$ and every box $B\subset\mathbb{R}^{d}$ there exist $a,\delta>0$ such that

[TABLE]

This lemma will be proved by induction in the dimension $d=1,2,\dots$ As was noted near (1.6), the given family $X=(X_{i})_{i}$ on $\mathbb{R}^{d}$ leads to another family $Y=(Y_{k,r,a,b,i})$ on $\mathbb{R}^{d-1}$. Both families of random fields lead to box-indexed families of functions $\mathbb{R}\to[0,\infty]$; $X$ leads to $(f_{B})_{B\subset\mathbb{R}^{d}}$ as before; likewise, $Y$ leads to $(g_{B})_{B\subset\mathbb{R}^{d-1}}$. (For $d=1$, just a single function $g$.)

Similarly to [2, 2a9(b) and 2a10] we modify Lemma 2.4 as follows. Given a box $B_{0}\subset\mathbb{R}^{d-1}$ and numbers $r,s>0$, we consider three boxes $B_{1}=[-r,0]\times B_{0}$, $B_{2}=[0,s]\times B_{0}$ and $B=[-r,s]\times B_{0}$ in $\mathbb{R}^{d}$.

2.10 Lemma**.**

For all $p\in(1,\infty)$ and $\lambda\in\mathbb{R}$,

[TABLE]

Proof.

Similar to the proof of 2.4. Denoting $v=\operatorname{vol}B_{0}$, the random variables $U=\frac{1}{\sqrt{rv}}\int_{B_{1}}X_{t}^{-}\,\mathrm{d}t$, $V=\frac{1}{\sqrt{sv}}\int_{B_{2}}X_{t}^{+}\,\mathrm{d}t$, $W=\frac{1}{\sqrt{(r+s)v}}\int_{B}X_{t}^{0}\,\mathrm{d}t$ and $Z=-\int_{B_{0}}Y_{t}\,\mathrm{d}t$ satisfy $\sqrt{r+s}W=\sqrt{r}U+\sqrt{s}V+\frac{1}{\sqrt{v}}Z$. By Hölder’s inequality, f_{B,i}(\lambda)=\log\mathbb{E}\,\exp\lambda W=\log\mathbb{E}\,\exp\lambda\bigl{(}\sqrt{\frac{r}{r+s}}U+\sqrt{\frac{s}{r+s}}V+\frac{1}{\sqrt{(r+s)v}}Z\bigr{)}\vphantom{)}\leq\frac{1}{p}\log\mathbb{E}\,\exp p\lambda\bigl{(}\sqrt{\frac{r}{r+s}}U+\sqrt{\frac{s}{r+s}}V\bigr{)}\vphantom{)}+\frac{p-1}{p}\log\mathbb{E}\,\exp\frac{p}{p-1}\lambda\frac{1}{\sqrt{r+s}}\frac{1}{\sqrt{v}}Z\leq\linebreak\frac{1}{p}f_{B_{1}}\bigl{(}p\lambda\sqrt{\frac{r}{r+s}}\,\bigr{)}\vphantom{)}+\frac{1}{p}f_{B_{2}}\bigl{(}p\lambda\sqrt{\frac{s}{r+s}}\,\bigr{)}\vphantom{)}+\frac{p-1}{p}g_{B_{0}}\bigl{(}-\frac{p}{p-1}\frac{\lambda}{\sqrt{r+s}}\bigr{)}\vphantom{)}; supremum in $i$ gives the first inequality (the upper bound). The second inequality (the lower bound), being rewritten as f_{B_{1}}\bigl{(}\lambda\sqrt{\frac{r}{r+s}}\,\bigr{)}\vphantom{)}+f_{B_{2}}\bigl{(}\lambda\sqrt{\frac{s}{r+s}}\,\bigr{)}\vphantom{)}\leq\frac{1}{p}f_{B}(p\lambda)+\linebreak\frac{p-1}{p}g_{B_{0}}\bigl{(}\frac{p}{p-1}\frac{\lambda}{\sqrt{r+s}}\bigr{)}\vphantom{)}, follows by Hölder’s inequality from the relation $\sqrt{r}U+\sqrt{s}V=\sqrt{r+s}W-\frac{1}{\sqrt{v}}Z$. ∎

2.11 Remark**.**

Above, a box $B$ is divided in two boxes $B_{1},B_{2}$ by the hyperplane $x_{1}=0$. More generally, the same holds when $B$ is divided by another coordinate hyperplane $x_{k}=c$ for $k\in\{1,\dots,d\}$ and any appropriate $c$.

Let us call a box $B\subset\mathbb{R}^{d}$ good when there exist $a,\delta>0$ such that $f_{B}(\lambda)\leq a\lambda^{2}$ whenever $|\lambda|\leq\delta$. Similarly, a box $B\subset\mathbb{R}^{d-1}$ is good when \exists a,\delta>0\;\forall\lambda\;\bigl{(}|\lambda|\leq\delta\;\Longrightarrow\;g_{B}(\lambda)\leq a\lambda^{2}\bigr{)}\vphantom{)}.

Existence of (at least one) good box follows from Item (a) of Def. 1.4 and Lemma 1.8. (For $d=0$ the only “box” is good.)

In order to prove Lemma 2.9 we assume (the induction hypothesis) that all boxes in $\mathbb{R}^{d-1}$ are good, and prove that all boxes in $\mathbb{R}^{d}$ are good.

By 2.10, $B$ is good if and only if $B_{1},B_{2}$ are good. It follows that every box contained in some good box is good (turn from $B$ to $B_{1}$ or $B_{2}$, and iterate).

Thus, all boxes that are small enough are good. It follows that every box is good (divide it into small boxes).

Given a set of boxes, we say that these boxes are uniformly good when there exist $a,\delta>0$ such that for every box $B$ of the given set, every $\lambda\in[-\delta,\delta]$ satisfies the inequality $f_{B}(\lambda)\leq a\lambda^{2}$ (or $g_{B}(\lambda)\leq a\lambda^{2}$, for $B\subset\mathbb{R}^{d-1}$).

2.12 Lemma**.**

Let $0<c<C<\infty$. Then the boxes $[0,r_{1}]\times\dots\times[0,r_{d}]$ for all $r_{1},\dots,r_{d}\in[c,C]$ are uniformly good.

Proof.

Induction in the dimension $d$. The induction hypothesis gives $a_{0},\delta_{0}>0$ such that $g_{[0,r_{1}]\times\dots\times[0,r_{d-1}]}(\lambda)\leq a_{0}\lambda^{2}$ whenever $|\lambda|\leq\delta_{0}$ and $r_{1},\dots,r_{d-1}\in[c,C]$. (When $d=1$, this holds for the single function $g$.) The box $[0,C]^{d}=[0,C]\times\dots\times[0,C]$ being good by 2.9, we take $a,\delta>0$ such that $f_{[0,C]^{d}}(\lambda)\leq a\lambda^{2}$ whenever $|\lambda|\leq\delta$. We use the second inequality of 2.10 for $p=2$ (taking 2.11 into account):

[TABLE]

for |\lambda|\leq\min\bigl{(}\frac{\delta}{2}\sqrt{\frac{c}{C}},\frac{\delta_{0}}{2}\sqrt{c}\bigr{)}\vphantom{)}. Thus, the boxes $[0,r_{1}]\times[0,C]^{d-1}$ for $r_{1}\in[c,C]$ are uniformly good. Now we divide the box $[0,r_{1}]\times[0,C]^{d-1}$ by the hyperplane $x_{2}=r_{2}$, apply again the argument used above, and see that the boxes $[0,r_{1}]\times[0,r_{2}]\times[0,C]^{d-2}$ for $r_{1},r_{2}\in[c,C]$ are uniformly good. And so on. ∎

Proof of Prop. 2.3.

We take $C$ large enough according to 2.6. By 2.12 the boxes $B=[0,r_{1}]\times\dots\times[0,r_{d}]$ for all $r_{1},\dots,r_{d}\in[C,2C]$ are uniformly good. We take $a\geq 1$ and $\delta>0$ such that $f_{B}(\lambda)\leq a\lambda^{2}$ for all these $B$ and all $\lambda\in[-\delta,\delta]$. Now 2.8 gives $V$ such that all $v\in[V,\infty)$ satisfy

[TABLE]

We take $M>1$ such that $M\geq C$, $M^{d}\geq V$, $M\geq 2a$, $M\geq C_{1}\sqrt{a}$, and get

[TABLE]

since $v>1$, $\log^{d-1}S(v)\leq\log^{d-1}v$ (just $1\leq 1$ for $d=1$), $v\geq M^{d}\geq V$, $C_{1}|\lambda|\leq\frac{C_{1}}{M}\sqrt{S(v)}\log^{-(d-1)}v\leq\sqrt{\frac{1}{a}S(v)}\log^{-(d-1)}S(v)$, and $f_{v,M}(\lambda)\leq f_{v,C}(\lambda)\leq 2a\lambda^{2}\leq M\lambda^{2}$. ∎

Proposition 2.1 is thus proved.

3 Close to large deviations

In this section we denote by $C_{2}$ the constant given by Prop. 2.3, use the functions $f_{v,C}$ mostly for $C=C_{2}$, and denote $f_{v}=f_{v,C_{2}}$. By 2.3,

[TABLE]

We still use (1.6), (1.7) and $C_{1}$ therefrom; $C_{1}\leq C_{2}$. By convention, $\log^{0}x=1$ always (also if $x$ does not belong to $(1,\infty)$ or even is ill-defined).

3.2 Lemma**.**

For all $p\in(1,\infty)$,

[TABLE]

whenever $C_{1}|\lambda|\leq\frac{p-1}{p}\sqrt{2v}\log^{-(d-1)}S(2v)$ and $2v\geq(2C_{2})^{d}$.

Proof.

Given a box $B$ such that $\operatorname{vol}B=2v$ and $0ptB\geq C_{2}$, we halve the longest side $2r$ of $B$ and apply Lemma 2.4 (as we did in the proof of 2.6). Once again, $2r\geq R(\operatorname{vol}B)=R(2v)\geq 2C_{2}$, thus a half of $B$ is still of width $\geq C_{2}$, and $\frac{2v}{2r}\leq S(2v)$, thus 2.4 applies and gives f_{B}(\lambda)\leq\frac{2}{p}f_{v}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}+C_{1}\frac{p}{p-1}\cdot\frac{\lambda^{2}}{2r}. ∎

It follows that

[TABLE]

since $\frac{p}{p-1}C_{1}|\lambda|\leq\sqrt{2v}\log^{-(d-1)}S(2v)$.

For convenience we denote

[TABLE]

3.3 Corollary**.**

For all $p\in(1,\infty)$,

[TABLE]

whenever $0<C_{1}|\lambda|\leq\frac{p-1}{p}\sqrt{2v}\log^{-(d-1)}S(2v)$ and $2v\geq(2C_{2})^{d}$.

Proof.

\varphi_{2v}(\lambda)-\varphi_{v}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}=\frac{1}{|\lambda|\sqrt{2v}}f_{2v}(\lambda)-\frac{\sqrt{2}}{p|\lambda|\sqrt{v}}f_{v}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}\leq\frac{1}{|\lambda|\sqrt{2v}}\bigl{(}\frac{2}{p}f_{v}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}+|\lambda|\frac{\sqrt{2v}}{R(2v)}\log^{-(d-1)}S(2v)\bigr{)}\vphantom{)}-\frac{\sqrt{2}}{p|\lambda|\sqrt{v}}f_{v}\bigl{(}\frac{p\lambda}{\sqrt{2}}\bigr{)}\vphantom{)}=\frac{1}{R(2v)\log^{d-1}S(2v)}. ∎

3.4 Corollary**.**

[TABLE]

whenever $2v\geq(2C_{2})^{d}$, $\lambda_{0}\lambda_{1}>0$ and

[TABLE]

Proof.

We take $p=\frac{\sqrt{2}\lambda_{0}}{\lambda_{1}}$ and note that $\lambda_{0}=\frac{p\lambda_{1}}{\sqrt{2}}$ and $\frac{p-1}{p|\lambda_{1}|}=\frac{1}{|\lambda_{1}|}-\frac{1}{\sqrt{2}|\lambda_{0}|}\geq\frac{C_{1}}{\sqrt{2v}}\log^{d-1}S(2v)$, that is, $C_{1}|\lambda_{1}|\leq\frac{p-1}{p}\sqrt{2v}\log^{-(d-1)}S(2v)$, thus 3.3 applies. ∎

3.5 Corollary**.**

Let numbers $v_{0},\dots,v_{n}$ and $\lambda_{0},\dots,\lambda_{n}$ satisfy

[TABLE]

for $k=0,\dots,n-1$; and $2v_{0}\geq(2C_{2})^{d}$. Then

[TABLE]

Proof.

Just apply 3.4 $n$ times. ∎

We rewrite Theorem 1.5 in terms of $f_{v}$.

3.6 Proposition**.**

There exists $C\in(1,\infty)$ such that for every $\lambda\in\mathbb{R}$,

[TABLE]

Taking $C\geq\max(C_{2},\mathrm{e}^{1/d})$ we see that the case $C|\lambda|\leq\frac{\sqrt{S(v)}}{\log^{d-1}v}$ is covered by (3.1).

From now on we assume that $v\geq C^{d}$ and

[TABLE]

ultimately we’ll prove that $f_{v,C}(\lambda)\leq C\lambda^{2}$ provided that the constant $C$, dependent only on $d,C_{1},C_{2}$, is large enough.

We take integer $n$ such that

[TABLE]

(the constant $M_{d}$, dependent on $d$ only, will be chosen later).

3.9 Lemma**.**

If $M_{d}\geq(2d)^{d-1}$ and $C\geq\mathrm{e}$, then $n\geq 1$.

Proof.

Assume the contrary: $M_{d}^{2d}\frac{(C|\lambda|)^{2d}}{v^{d-1}}\log^{2d(d-1)}\frac{\sqrt{v}}{C|\lambda|}\leq 1$, that is,

[TABLE]

For $d=1$ it means $M_{1}C|\lambda|\leq 1$ in contradiction to $C|\lambda|>1$ and $M_{1}\geq 1$. Assume $d\geq 2$. Using (3.7),

[TABLE]

thus, $\frac{1}{2d}\log v>\log\frac{\sqrt{v}}{C|\lambda|}$, that is, $C|\lambda|>\sqrt{S(v)}$. Now 3.10 gives $M_{d}\log^{d-1}\frac{\sqrt{v}}{C|\lambda|}<1$, which implies $M_{d}\log^{d-1}\log^{d}v<1$ (since $C|\lambda|\leq\frac{\sqrt{v}}{\log^{d}v}$ by (3.7)), and $2d\log\log^{d}v<1$ (since $M_{d}\geq(2d)^{d-1}$). On the other hand, $v\geq C^{d}\geq\mathrm{e}^{d}$ implies $\log v\geq d$ and $\log\log^{d}v=d\log\log v\geq d\log d$, thus $2d\log\log^{d}v\geq 2d^{2}\log d\geq 8\log 2>1$; a contradiction. ∎

From now on we assume $M_{d}\geq(2d)^{d-1}$ and $C\geq\mathrm{e}$ (thus, $n\geq 1$). We define $v_{0},\dots,v_{n}$ by

[TABLE]

Below, “$y=\mathcal{O}(x)$” means that $y\leq\operatorname{const}\cdot x$ for some constant dependent on $d$ only.

3.11 Lemma**.**

$n\leq\frac{\sqrt{v}}{2C_{1}|\lambda|}\log^{-(d-1)}v$ provided that $C$ is large enough.

Proof.

It is sufficient to prove that $n=\mathcal{O}(\log v)$; then, increasing $C$ as needed, we get $n\leq\frac{C}{2C_{1}}\log v\leq\frac{\sqrt{v}}{2C_{1}|\lambda|}\log^{-(d-1)}v$ since $C|\lambda|\leq\frac{\sqrt{v}}{\log^{d}v}$ by (3.7).

We have $\frac{1}{C|\lambda|}=\mathcal{O}(1)$ (since by (3.7), $\frac{1}{C|\lambda|}\leq\frac{\log^{d-1}v}{\sqrt{S(v)}}$, the latter being bounded in $v\in(1,\infty)$). Thus, $\log\frac{\sqrt{v}}{C|\lambda|}=\mathcal{O}(\log v)$ (since $\log v\geq\log C^{d}\geq d\geq 1$). Also, $C|\lambda|\leq\frac{\sqrt{v}}{\log^{d}v}\leq\sqrt{v}$. Using (3.8),

[TABLE]

which implies $n=\mathcal{O}(\log v)$. ∎

Having $n\log^{d-1}v\leq\frac{\sqrt{v}}{2C_{1}|\lambda|}$ (ensured by Lemma 3.11) we define $\lambda_{0},\dots,\lambda_{n}$ (either all positive or all negative) by

[TABLE]

That is,

[TABLE]

the right-hand side is positive, since

[TABLE]

moreover, for $k=n$ we get

[TABLE]

3.13 Lemma**.**

If $C$ is large enough, then $2v_{0}\geq(2C_{2})^{d}$.

Proof.

For $C$ large enough we have $\frac{x}{\log^{d-1}x}\geq M_{d}\sqrt{2C_{2}}$ for all $x\in[(d\log C)^{d},\infty)$. We take $x=\frac{\sqrt{v}}{C|\lambda|}$; using (3.7), $x\geq\log^{d}v\geq\log^{d}C^{d}=(d\log C)^{d}$, thus $\frac{x}{\log^{d-1}x}\geq M_{d}\sqrt{2C_{2}}$. Using (3.8), 2v_{0}=2^{-(n-1)}v>M_{d}^{-2d}\frac{v^{d}}{(C|\lambda|)^{2d}}\log^{-2d(d-1)}\frac{\sqrt{v}}{C|\lambda|}=\bigl{(}\frac{1}{M_{d}}x\log^{-(d-1)}x\bigr{)}\vphantom{)}^{2d}\geq\bigl{(}\sqrt{2C_{2}}\bigr{)}\vphantom{)}^{2d}. ∎

From now on we assume that $C$ is large enough, so that $2v_{0}\geq(2C_{2})^{d}$. Now Corollary 3.5 applies:

[TABLE]

3.14 Lemma**.**

If $M_{d}\geq 2(2d)^{d-1}$, then $C|\lambda_{0}|\leq\frac{\sqrt{S(v_{0})}}{\log^{d-1}v_{0}}$.

Proof.

Assume the contrary. Using (3.12), $2C\cdot 2^{-n/2}|\lambda|\geq C|\lambda_{0}|>\frac{\sqrt{S(2^{-n}v)}}{\log^{d-1}(2^{-n}v)}=2^{-n/2}\cdot 2^{\frac{n}{2d}}v^{\frac{d-1}{2d}}\log^{-(d-1)}(2^{-n}v)$; using (3.8), $M_{d}^{2d}\frac{(C|\lambda|)^{2d}}{v^{d-1}}\log^{2d(d-1)}\frac{\sqrt{v}}{C|\lambda|}\leq 2^{n}<\frac{(2C|\lambda|)^{2d}}{v^{d-1}}\log^{2d(d-1)}(2^{-n}v)$ and \bigl{(}2d\log\frac{\sqrt{v}}{C|\lambda|}\bigr{)}\vphantom{)}^{d-1}\leq\frac{1}{2}M_{d}\log^{d-1}\frac{\sqrt{v}}{C|\lambda|}<\log^{d-1}(2^{-n}v). For $d=1$ it means $1<1$. Assume $d\geq 2$. We have $2d\log\frac{\sqrt{v}}{C|\lambda|}<\log(2^{-n}v)$ and \bigl{(}\frac{\sqrt{v}}{C|\lambda|}\bigr{)}\vphantom{)}^{2d}<2^{-n}v, that is, 2^{n}<v\bigl{(}\frac{C|\lambda|}{\sqrt{v}}\bigr{)}\vphantom{)}^{2d}=\frac{(C|\lambda|)^{2d}}{v^{d-1}}. Using (3.8), $M_{d}^{2d}\log^{2d(d-1)}\frac{\sqrt{v}}{C|\lambda|}<1$, that is, $M_{d}\log^{d-1}\frac{\sqrt{v}}{C|\lambda|}<1$, which cannot be true, as was shown in the proof of Lemma 3.9. ∎

From now on we assume that $M_{d}\geq 2(2d)^{d-1}$, so that $C|\lambda_{0}|\leq\frac{\sqrt{S(v_{0})}}{\log^{d-1}v_{0}}$. Also, $v_{0}\geq C_{2}^{d}$ by 3.13, and $C_{2}\leq C$. Now (3.1) applies: $f_{v_{0}}(\lambda_{0})\leq C_{2}\lambda_{0}^{2}$; and therefore,

[TABLE]

3.16 Lemma**.**

\sum_{k=0}^{n-1}\frac{1}{R(2v_{k})\log^{d-1}S(2v_{k})}=\mathcal{O}\bigl{(}\frac{C|\lambda|}{\sqrt{v}}\bigr{)}\vphantom{)}.

Proof.

We rewrite the sum as \sum_{k=1}^{n}(2^{-(n-k)}v)^{-1/d}\log^{-(d-1)}S(v_{k})\leq\linebreak\bigl{(}\log^{-(d-1)}S(2v_{0})\bigr{)}\vphantom{)}(2^{-n}v)^{-1/d}\sum_{k=1}^{\infty}2^{-k/d}, note that $\log^{-(d-1)}S(2v_{0})=\mathcal{O}(1)$ (since $C_{2}>1$ and $2v_{0}\geq(2C_{2})^{d}\geq 2$) and see that the given sum is \mathcal{O}\bigl{(}\bigl{(}\frac{2^{n}}{v}\bigr{)}\vphantom{)}^{1/d}\bigr{)}\vphantom{)}. By (3.8), \bigl{(}\frac{2^{n}}{v}\bigr{)}\vphantom{)}^{1/d}<2^{1/d}M_{d}^{2}\frac{(C|\lambda|)^{2}}{v}\log^{2d-2}\frac{\sqrt{v}}{C|\lambda|}=\linebreak\mathcal{O}\bigl{(}\bigl{(}\frac{C|\lambda|}{\sqrt{v}}\bigr{)}\vphantom{)}^{2}\log^{2d-2}\frac{\sqrt{v}}{C|\lambda|}\bigr{)}\vphantom{)}. It remains to note that $\frac{C|\lambda|}{\sqrt{v}}\log^{2d-2}\frac{\sqrt{v}}{C|\lambda|}=\mathcal{O}(1)$, since the function $x\mapsto\frac{\log^{2d-2}x}{x}$ is bounded on $[1,\infty)$ and, using (3.7), $\frac{C|\lambda|}{\sqrt{v}}\leq\frac{1}{\log^{d}v}\leq\frac{1}{(d\log C)^{d}}\leq\frac{1}{d^{d}}\leq 1$. ∎

Proof of Prop. 3.6.

By (3.15) and Lemma 3.16, $\frac{f_{v}(\lambda)}{|\lambda|\sqrt{v}}\leq\frac{C_{2}|\lambda_{0}|}{\sqrt{v_{0}}}+N_{d}C\frac{|\lambda|}{\sqrt{v}}$ for some constant $N_{d}$ dependent on $d$ only. By (3.12), $\frac{|\lambda_{0}|}{\sqrt{v_{0}}}\leq\frac{2|\lambda|}{2^{n/2}\sqrt{v_{0}}}=\frac{2|\lambda|}{\sqrt{v}}$. Thus, f_{v}(\lambda)\leq|\lambda|\sqrt{v}\bigl{(}C_{2}\frac{2|\lambda|}{\sqrt{v}}+N_{d}C\frac{|\lambda|}{\sqrt{v}}\bigr{)}\vphantom{)}\leq(2C_{2}+N_{d}C)\lambda^{2}. And finally, $f_{v,2C_{2}+N_{d}C}(\lambda)\leq f_{v,C_{2}}(\lambda)=f_{v}(\lambda)$. ∎

Theorem 1.5 is thus proved.