2 Application in the high-dimensional configuration space: square-root equilibrium: fast approach and long-term stability

In the off-equilibrium ideal gas model, $N$ point particles are moving around in a cubic container, say the unit cube $[0,1)^{3}$ , bouncing back and forth on the walls like billiard balls. To study the time evolution of such a large billiard system, we use the well-known geometric trick of unfolding that converts a (zigzag) billiard orbit into a torus line. For illustration the figure below shows the square billiard. Then unfolding simply means that we keep reflecting the square itself in the respective side and unfold the piecewise linear billiard path to a straight line.

Unfolding defines four reflected copies $S_{1},S_{2},S_{3},S_{4}$ of a test set $S$ , where each one of the four unit squares contains a reflected copy of the given test set $S$ . In the last step we shrink the underlying $2\times 2$ square to the unit square $I^{2}=[0,1)^{2}$ . Of course, the test set $S$ can be upgraded to any periodic test function  $f$ .

Note that the unfolding of the billiard path in the $h$ -dimensional unit cube $[0,1)^{h}$ with $h\neq 2$ can be defined in an analogous way. Formally, unfolding means the map

$$\begin{eqnarray}2\biggl\|\frac{x}{2}\biggr\|\rightarrow \biggl\{\frac{x}{2}\biggr\}\end{eqnarray}$$

applied to each coordinate, where $\Vert z\Vert$ denotes the distance of a real number $z$ from a nearest integer and $0\leqslant \{z\}<1$ denotes the fractional part of  $z$ .

It follows from unfolding that a billiard path in the $h$ -dimensional unit cube $[0,1)^{h}$ intersects a given test set $S\subset [0,1)^{h}$ at time $t$ precisely when the corresponding torus line in the $h$ -dimensional $2\times \cdots \times 2$ (hyper)cube intersects the union of the $2^{h}$ reflected copies of  $S$ . Note that each one of the $2^{h}$ unit (hyper)cubes contains a reflected copy of the given test set  $S$ . In the last step, we shrink the $h$ -dimensional $2\times \cdots \times 2$ (hyper)cube to the unit (hyper)cube $I^{h}=[0,1)^{h}$ .

A constant speed piecewise linear point billiard motion in $I^{h}$ is defined by the equation

(2.1) $$\begin{eqnarray}\mathbf{x}(t)=(x_{1}(t),\ldots ,x_{h}(t)),\qquad x_{i}(t)=2\biggl\|\frac{y_{i}+t\unicode[STIX]{x1D6FC}_{i}}{2}\biggr\|,\quad 1\leqslant i\leqslant h,\;0<t<\infty .\end{eqnarray}$$

Here $\mathbf{y}=(y_{1},\ldots ,y_{h})$ is the starting point and the non-zero vector $\unicode[STIX]{x1D736}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{k})$ is the initial direction.

Motivated by the central limit theorem, it is a natural intuition to visualize snapshot equilibrium in the particle space (or gas container) as a state where the system exhibits square-root size fluctuations in particle counting. More precisely, let $S\subset [0,1)^{3}$ be an arbitrary but fixed Lebesgue measurable subset of the unit cube with volume $0<\operatorname{vol}(S)<1$ , and consider the particle-counting function

(2.2) $$\begin{eqnarray}N_{S}(t)=\mathop{\sum }_{\substack{ 1\leqslant k\leqslant N \\ \mathbf{x}_{k}(t)\in S}}1,\end{eqnarray}$$

where $\mathbf{x}_{k}(t)$ is the orbit of the $k$ th billiard; see (2.1). The billiard system exhibits square-root fluctuation if the particle-counting function (2.2) differs from the expected value $N\operatorname{vol}(S)$ by $O(\sqrt{N})$ . In other words, it is a good intuition to visualize snapshot equilibrium as a square-root fluctuation equilibrium in the particle space, or simply square-root equilibrium.

Using square-root equilibrium as the definition of snapshot equilibrium, the statement once the system reaches (snapshot) equilibrium (in the particle space), it stays in (snapshot) equilibrium forever is certainly untrue for the unlimited time evolution of a typical trajectory of the system, i.e. $t\rightarrow \infty$ . Indeed, given an arbitrary initial configuration of starting points, if the initial velocity coordinates are linearly independent over the rationals (representing typical directions), then by Kronecker’s well-known density theorem, the time evolution of this individual trajectory of the system eventually violates square-root equilibrium in the worst possible way infinitely many times. We will clarify this in part II in the sequel.

We know that the $N$ -point billiard model in the cube can be reduced to the torus model via unfolding. We assume that the particles independently have Gaussian initial velocity distribution in the 3-space. In other words, the set of the $N$ particles in $I^{3}=[0,1)^{3}$ at time $t$ is

(2.3) $$\begin{eqnarray}\displaystyle {\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t) & = & \displaystyle {\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D70C}_{1},\mathbf{e}_{1},\ldots ,\unicode[STIX]{x1D70C}_{N},\mathbf{e}_{N};t)\nonumber\\ \displaystyle & = & \displaystyle \{\mathbf{y}_{1}+\unicode[STIX]{x1D70C}_{1}t\mathbf{e}_{1},\ldots ,\mathbf{y}_{N}+\unicode[STIX]{x1D70C}_{N}t\mathbf{e}_{N}\}\hspace{0.2em}{\rm mod}\hspace{0.2em}1\nonumber\\ \displaystyle & = & \displaystyle \{\{\mathbf{y}_{1}+\unicode[STIX]{x1D70C}_{1}t\mathbf{e}_{1}\},\ldots ,\{\mathbf{y}_{N}+\unicode[STIX]{x1D70C}_{N}t\mathbf{e}_{N}\}\},\end{eqnarray}$$

where ${\mathcal{Y}}=\{\mathbf{y}_{1},\ldots ,\mathbf{y}_{N}\}\subset [0,1)^{3}$ is the $N$ -element set of initial point configuration, and the initial velocities of the particles are independent random variables having the same speed distribution with density

(2.4) $$\begin{eqnarray}g(u)=\sqrt{\frac{2}{\unicode[STIX]{x1D70B}}}u^{2}\text{e}^{-u^{2}/2},\quad 0\leqslant u<\infty ,\end{eqnarray}$$

which is the density of the speed of the three-dimensional Gaussian velocity distribution. So the trajectory of the $k$ th particle is $\mathbf{y}_{k}+\unicode[STIX]{x1D70C}_{k}t\mathbf{e}_{k}\in \mathbf{R}^{3}\hspace{0.2em}{\rm mod}\hspace{0.2em}1$ , $1\leqslant k\leqslant N$ , where

$$\begin{eqnarray}\Pr [\unicode[STIX]{x1D70C}_{k}\leqslant u]=\sqrt{\frac{2}{\unicode[STIX]{x1D70B}}}\int _{0}^{u}z^{2}\text{e}^{-z^{2}/2}\,dz.\end{eqnarray}$$

On the other hand, the curve in the configuration space $I^{d}=[0,1)^{d}$ with $d=3N$ , representing the time evolution of the system (2.3), is the straight line modulo one in $\mathbf{R}^{d}$ given by

(2.5) $$\begin{eqnarray}\vec{{\mathcal{Y}}}(\text{Gauss};\unicode[STIX]{x1D714};t)=\vec{{\mathcal{Y}}}+t\mathbf{v}(\unicode[STIX]{x1D714})\hspace{0.2em}{\rm mod}\hspace{0.2em}1,\end{eqnarray}$$

where

(2.6) $$\begin{eqnarray}\unicode[STIX]{x1D714}=(\unicode[STIX]{x1D70C}_{1},\mathbf{e}_{1},\ldots ,\unicode[STIX]{x1D70C}_{N},\mathbf{e}_{N})\in \unicode[STIX]{x1D6FA}_{\text{Gauss}}=([0,\infty )\times \mathbf{S}^{2})^{N},\end{eqnarray}$$

and

(2.7) $$\begin{eqnarray}\mathbf{v}(\unicode[STIX]{x1D714})=(\unicode[STIX]{x1D70C}_{1}\mathbf{e}_{1},\ldots ,\unicode[STIX]{x1D70C}_{N}\mathbf{e}_{N}).\end{eqnarray}$$

The product space $\unicode[STIX]{x1D6FA}_{\text{Gauss}}$ is equipped with the product measure $\unicode[STIX]{x1D707}_{\text{Gauss}}$ , where the half-line $[0,\infty )$ has the probability density function (2.4), and the sphere $\mathbf{S}^{2}$ has the normalized surface area. Here $\vec{{\mathcal{Y}}}$ denotes the $3N$ -dimensional vector

(2.8) $$\begin{eqnarray}\vec{{\mathcal{Y}}}=(\mathbf{y}_{1},\ldots ,\mathbf{y}_{N})\end{eqnarray}$$

formed from the $N$ -element set of starting points of the particles ${\mathcal{Y}}\subset [0,1)^{3}$ .

We need the well-known fact from probability theory that $\unicode[STIX]{x1D707}_{\text{Gauss}}$ is the $d$ -dimensional standard Gaussian distribution with $d=3N$ , i.e. the multidimensional Gaussian distribution is rotation invariant.

Let $B\subset I^{3}=[0,1)^{3}$ be an arbitrary but fixed measurable test set in the gas container, where $\operatorname{vol}(B)$ denotes the three-dimensional Lebesgue measure. Assume that $N$ is large. Is it then true that, once a typical time evolution of the (Gaussian torus) system reaches square-root equilibrium in the particle space, then it stays in that state in the quantitative sense, of factor 30 say,

$$\begin{eqnarray}||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)N|\leqslant 30\sqrt{N},\end{eqnarray}$$

for an extremely long time (with the possible exception of a totally negligible set of values of  $t$ )? The factor 30 is accidental, and square-root equilibrium is the best that we can hope for.

By using Theorem 1.2, in particular (1.12), we give a positive answer to this question. We use Theorem 1.2 as a short-time ergodic theorem in the configuration space. Thus the configuration space average nearly equals the short-term time average. The good news is that the configuration space average can be easily computed with direct application of probability theory, since the configuration space is a product space with product measure; see the application of Bernstein’s large deviation inequality in (2.11) below. Note that Birkhoff’s ergodic theorem does not have an explicit error term and works only for typical initial condition, and a typical initial condition represents equilibrium, the trivial case, since we are studying off-equilibrium dynamics. In contrast, Theorem 1.2 has the advantage that it works for arbitrary off-equilibrium initial configuration. It also has an explicit error term, and we can use it to describe the time evolution in realistic time. The details go as follows.

The family of time evolutions ${\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)$ , $\unicode[STIX]{x1D714}\in \unicode[STIX]{x1D6FA}_{\text{Gauss}}$ , of the three-dimensional Gaussian torus model (in the particle space $I^{3}$ ) is represented by the family of torus lines (2.5) in the configuration space  $I^{d}$ , all starting from the same point $\vec{{\mathcal{Y}}}\in I^{d}$ ; see also (2.6)–(2.8).

For an arbitrary $\unicode[STIX]{x1D6FE}>0$ , define the (extremely complicated) set

(2.9) $$\begin{eqnarray}S(B;\unicode[STIX]{x1D6FE})=\{\vec{{\mathcal{Z}}}\in I^{d}:||{\mathcal{Z}}\cap B|-\operatorname{vol}(B)N|>\unicode[STIX]{x1D6FE}\sqrt{N}\}\end{eqnarray}$$

in the configuration space, where

$$\begin{eqnarray}\vec{{\mathcal{Z}}}=(z_{1},\ldots ,z_{3N})\quad \text{and}\quad {\mathcal{Z}}=\{\mathbf{z}_{1},\ldots ,\mathbf{z}_{N}\!\}\end{eqnarray}$$

with $\mathbf{z}_{k}=(z_{3k-2},z_{3k-1},z_{3k})$ , $1\leqslant k\leqslant N$ .

Recall the following large deviation inequality in probability theory; see [Reference Feller3].

Bernstein’s inequality. Let $X_{1},\ldots ,X_{n}$ be independent random variables with binomial distribution $\Pr [X_{i}=1]=p$ and $\Pr [X_{i}=0]=q=1-p$ . Then for every positive  $\unicode[STIX]{x1D6FE}$ ,

(2.10) $$\begin{eqnarray}\displaystyle \Pr \biggl[\biggl|\mathop{\sum }_{i=1}^{n}(X_{i}-p)\biggr|\geqslant \unicode[STIX]{x1D6FE}\sqrt{npq}\biggr] & = & \displaystyle \mathop{\sum }_{\substack{ 0\leqslant k\leqslant n \\ |k-pn|\geqslant \unicode[STIX]{x1D6FE}\sqrt{pqn}}}\binom{n}{k}p^{k}q^{n-k}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2\exp \biggl(-\frac{\unicode[STIX]{x1D6FE}^{2}/2}{1+\unicode[STIX]{x1D6FE}/3\sqrt{npq}}\biggr),\qquad \quad\end{eqnarray}$$

where $\sqrt{npq}$ is the standard deviation of the binomial distribution.

Using (2.10) with $p=\operatorname{vol}(B)$ , we have

(2.11) $$\begin{eqnarray}\displaystyle \operatorname{vol}_{d}(S(B;\unicode[STIX]{x1D6FE})) & {\leqslant} & \displaystyle 2\exp \biggl(-\frac{\unicode[STIX]{x1D6FE}^{2}(2p(1-p))^{-1}}{1+\unicode[STIX]{x1D6FE}/3p(1-p)\sqrt{N}}\biggr)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2\exp \biggl(-\frac{2\unicode[STIX]{x1D6FE}^{2}}{1+2\unicode[STIX]{x1D6FE}/\sqrt{N}}\biggr),\end{eqnarray}$$

where the last inequality comes from the simple fact that $p(1-p)\leqslant 1/4$ . The reason why we could apply Bernstein’s inequality is that $\operatorname{vol}_{d}$ is a product measure, and so the $d=3N$ -dimensional volume $\operatorname{vol}_{d}(S(B;\unicode[STIX]{x1D6FE}))$ represents a large deviation probability for $N$ independent random variables.

For example,

(2.12) $$\begin{eqnarray}\text{if}~\unicode[STIX]{x1D6FE}=30~\text{and}~N\geqslant 10^{6},\text{then (2.11) gives}~2\exp \biggl(-\frac{2\unicode[STIX]{x1D6FE}^{2}}{1+2\unicode[STIX]{x1D6FE}/\sqrt{N}}\biggr)<10^{-700},\end{eqnarray}$$

which is extremely small. The long-term stability of, say, 30-square-root equilibrium (in the particle space) is based on this striking numerical fact.

Since the configuration space $I^{d}$ is translation invariant, we can apply Theorem 1.2 with $f=\unicode[STIX]{x1D712}_{S}$ , where $S=S(B;\unicode[STIX]{x1D6FE})-\vec{{\mathcal{Y}}}$ is a translated copy of $S(B;\unicode[STIX]{x1D6FE})$ in the torus $I^{d}$ . Clearly $\operatorname{vol}_{d}(S)=\operatorname{vol}_{d}(S(B;\unicode[STIX]{x1D6FE}))$ . Using (2.11) in Theorem 1.2 with $W=2^{k}U$ , we obtain

(2.13) $$\begin{eqnarray}\displaystyle & & \displaystyle (2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}\biggl(\int _{U}^{2^{k}U}\unicode[STIX]{x1D712}_{S}(t\mathbf{v})\,dt-\operatorname{vol}_{d}(S)(2^{k}-1)U\biggr)^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad =(2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}(D_{S}(\mathbf{v};U,2^{k}U))^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\exp \biggl(-\frac{2\unicode[STIX]{x1D6FE}^{2}}{1+2\unicode[STIX]{x1D6FE}/\sqrt{N}}\biggr)10k((2^{k}-1)U+1),\end{eqnarray}$$

assuming of course that $U\geqslant 1$ and $\text{e}^{\unicode[STIX]{x1D70B}^{2}U^{2}/2}\geqslant 3\,dU$ . By (2.6), (2.9) and using $S=S(B;\unicode[STIX]{x1D6FE})-\vec{{\mathcal{Y}}}$ , we have

(2.14) $$\begin{eqnarray}\displaystyle & & \displaystyle (2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}\biggl(\int _{U}^{2^{k}U}\unicode[STIX]{x1D712}_{S}(t\mathbf{v})\,dt-\operatorname{vol}_{d}(S)(2^{k}-1)U\biggr)^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}}\biggl(\int _{U}^{2^{k}U}\unicode[STIX]{x1D712}_{S}(t\mathbf{v}(\unicode[STIX]{x1D714}))\,dt-\operatorname{vol}_{d}(S)(2^{k}-1)U\biggr)^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}}\!\!(\operatorname{length}\{U\,\leqslant \,t\,\leqslant \,2^{k}U:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\,\cap \,B|\,-\,\text{vol}(B)N|\,>\,\unicode[STIX]{x1D6FE}\sqrt{N}\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\text{vol}_{d}(S(B;\unicode[STIX]{x1D6FE}))(2^{k}-1)U)^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714}).\qquad \quad\end{eqnarray}$$

Combining (2.13) and (2.14), we obtain the following result.

Theorem 2.1. Let ${\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)$ , $\unicode[STIX]{x1D714}\in \unicode[STIX]{x1D6FA}_{\text{Gauss}}$ , be the three-dimensional Gaussian torus model, and let $B\subset [0,1)^{3}$ be a measurable test set with three-dimensional Lebesgue measure $\operatorname{vol}(B)$ . Assume that

$$\begin{eqnarray}U\geqslant 1\quad \text{and}\quad \text{e}^{\unicode[STIX]{x1D70B}^{2}U^{2}/2}\geqslant 3\,dU.\end{eqnarray}$$

Then for every $\unicode[STIX]{x1D6FE}>0$ and every integer $k\geqslant 1$ ,

(2.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}}(\operatorname{length}\{U\leqslant t\leqslant 2^{k}U:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)N|>\unicode[STIX]{x1D6FE}\sqrt{N}\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\operatorname{vol}_{d}(S(B;\unicode[STIX]{x1D6FE}))(2^{k}-1)U)^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\exp \biggl(-\frac{2\unicode[STIX]{x1D6FE}^{2}}{1+2\unicode[STIX]{x1D6FE}/\sqrt{N}}\biggr)10k((2^{k}-1)U+1),\end{eqnarray}$$

where

$$\begin{eqnarray}\operatorname{vol}_{d}(S(B;\unicode[STIX]{x1D6FE}))<2\exp \biggl(-\frac{2\unicode[STIX]{x1D6FE}^{2}}{1+2\unicode[STIX]{x1D6FE}/\sqrt{N}}\biggr).\end{eqnarray}$$

Probably the reader does not find Theorem 2.1 very pretty, but it is an extremely powerful result. To illustrate its power, let $\unicode[STIX]{x1D6FE}=30$ , $U=4$ , $k=100$ and $N=10^{27}$ , so that $d=3N=3\cdot 10^{27}$ . Then by (2.12) and (2.15),

(2.16) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}}(\operatorname{length}\{4\leqslant t\leqslant 4\cdot 2^{100}:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)N|>30\sqrt{N}\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\operatorname{vol}_{d}(S(B;30))(2^{100}-1)4)^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 10^{-700}\cdot 400\cdot 2^{100}\cdot 20<10^{-661}.\end{eqnarray}$$

Let $\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{bad})}$ be the set of those $\unicode[STIX]{x1D714}\in \unicode[STIX]{x1D6FA}_{\text{Gauss}}$ for which

(2.17) $$\begin{eqnarray}\operatorname{length}\{4\leqslant t\leqslant 4\cdot 2^{100}:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)N|>30\sqrt{N}\}\geqslant 10^{-220}.\end{eqnarray}$$

We claim that (2.16) implies

(2.18) $$\begin{eqnarray}\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{bad})})\leqslant 10^{-220}.\end{eqnarray}$$

Indeed, otherwise

$$\begin{eqnarray}\displaystyle & & \displaystyle \int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}}(\operatorname{length}\{4\leqslant t\leqslant 4\cdot 2^{100}:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)N|>30\sqrt{N}\}\nonumber\\ \displaystyle & & \displaystyle \qquad -\,\operatorname{vol}_{d}(S(B;30))(2^{100}-1)4)^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant \int _{\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{bad})}} (\operatorname{length}\!\{\!4\leqslant t\leqslant 4\cdot 2^{100}:||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|\nonumber\\ \displaystyle & & \displaystyle \qquad -\operatorname{vol}(B)N|>30\sqrt{N}\!\}-{\operatorname{vol}_{d}(S(B;30))(2^{100}-1)4)}^{2}\,d\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D714})\nonumber\\ \displaystyle & & \displaystyle \quad \geqslant 10^{-220}(10^{-220}-10^{-600})^{2}>10^{-661},\nonumber\end{eqnarray}$$

which contradicts (2.16). In the last step we have used the fact that

$$\begin{eqnarray}\operatorname{vol}_{d}(S(B;30))(2^{100}-1)4\leqslant 10^{-700}\cdot 10^{31}<10^{-600}.\end{eqnarray}$$

Note that the choice of $N=10^{27}$ was realistic, in the sense that there are roughly $10^{27}$ gas molecules in a cubic box of volume $1~\text{m}^{3}$ . In the classical Bernoulli gas model, the gas molecules are represented by point billiards. Using unfolding, we can reduce the billiard model to the torus model. The threshold $U=4$ here represents, roughly speaking, the relaxation distance, i.e. the necessary number of jumps per particle in the torus model, equal to half of the number of bounces in the billiard model, to reach square-root equilibrium (in the particle space) for the typical time evolution of the Gaussian system. Assume that the gas molecules have average speed $10^{3}~\text{m s}^{-1}$ . For this Gaussian system, it takes only a few milliseconds to reach square-root equilibrium. Now (2.17) and (2.18) have the following interpretation. Choosing an arbitrary (measurable) test set $B\subset [0,1)^{3}$ in the gas container (or particle space) and an arbitrary $N$ -element initial point configuration  ${\mathcal{Y}}$ , for the totally overwhelming majority of the initial velocities (Gaussian distribution), the number of particles in $B$ remains very close to the expected value $\operatorname{vol}(B)N$ for an extremely long time, with the possible exception of a totally negligible set of times  $t$ .

Indeed, for every (measurable) test set $B\subset [0,1)^{3}$ and every $N=10^{27}$ -element initial point configuration ${\mathcal{Y}}\subset [0,1)^{3}$ , there exists a subset $\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{good})}$ , where

$$\begin{eqnarray}\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{good})}=\unicode[STIX]{x1D6FA}_{\text{Gauss}}\setminus \unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{bad})}\end{eqnarray}$$

with

$$\begin{eqnarray}\unicode[STIX]{x1D707}_{\text{Gauss}}(\unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{good})})\geqslant 1-10^{-220},\end{eqnarray}$$

noting (2.18), representing a totally overwhelming majority such that for every $\unicode[STIX]{x1D714}\in \unicode[STIX]{x1D6FA}_{\text{Gauss}}^{(\text{good})}$ , the inequality

(2.19) $$\begin{eqnarray}||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\operatorname{vol}(B)10^{27}|\leqslant 30\cdot 10^{13.5}\end{eqnarray}$$

holds for every $4\leqslant t\leqslant 4\cdot 2^{100}$ with the possible exception of a set of times $t$ of total length less than $10^{-220}$ ; see (2.17). The latter actually represents less than $10^{-223}$ seconds, which is a ridiculously short time.

Note that $4\leqslant t\leqslant 4\cdot 2^{100}$ represents a time interval of about $10^{27}$ seconds, which is an incredibly long time, being roughly a billion times the age of the universe.

Finally, by (2.19),

$$\begin{eqnarray}\frac{1}{N}||{\mathcal{Y}}(\text{Gauss};\unicode[STIX]{x1D714};t)\cap B|-\text{vol}(B)|\leqslant 3\cdot 10^{-12.5}<10^{-12},\end{eqnarray}$$

which can be interpreted as almost constant density for an incredibly long time.

What happens if we want to prove long-term stability of square-root equilibrium with respect to a whole family of nice sets instead of a fixed measurable test set? Of course we cannot expect that a system stays in square-root equilibrium with respect to all measurable test sets simultaneously. What we may expect, however, is that, starting from an arbitrary but fixed initial configuration  ${\mathcal{Y}}$ , and after reaching configuration space equilibrium, the typical time evolution of the system stays in square-root equilibrium in the particle space with respect to all nice test sets simultaneously for a very, very long time, without any violator time instant  $t$ . And indeed, again by using Theorem 1.2, we are going to prove such a result in part II in the sequel. In fact, this paper is the first in a series of papers devoted to the applications of Theorem 1.2, and its extensions beyond the Gaussian case, to describe the time evolution of large off-equilibrium systems.

3 Proof of Theorem 1.2

For technical reasons, it is convenient to prove first a special case with an upper bound on the ratio  $W/U$ .

Theorem 3.1. Let $f\in L_{2}([0,1)^{d})$ be a test function. If $1\leqslant U<V\leqslant 2U$ and $\text{e}^{\unicode[STIX]{x1D70B}^{2}U^{2}/2}\geqslant 3\,dU$ , then

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};U,V)\leqslant \unicode[STIX]{x1D70E}_{0}^{2}(f)(9(V-U)+1).\end{eqnarray}$$

The technical restriction $U<V\leqslant 2U$ in Theorem 3.1 can be easily eliminated by a routine application of the Cauchy–Schwarz inequality; see the end of §3.

To prove Theorem 3.1, we shall use Fourier analysis in the configuration space $I^{d}=[0,1)^{d}$ . For high dimension $d$ , this leads to technical difficulties that are combinatorial in nature.

Let $f\in L_{2}(I^{d})$ be a Lebesgue square-integrable function in the $d$ -dimensional unit torus, i.e. we extend $f$ over the whole $d$ -space $\mathbf{R}^{d}$ periodically, and consider the Fourier expansion

(3.1) $$\begin{eqnarray}f(\mathbf{u})=\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}}a_{\mathbf{n}}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}\mathbf{n}\cdot \mathbf{u}},\end{eqnarray}$$

where

(3.2) $$\begin{eqnarray}a_{\mathbf{n}}=\int _{I^{d}}f(\mathbf{y})\text{e}^{-2\unicode[STIX]{x1D70B}\text{i}\mathbf{n}\cdot \mathbf{y}}\,d\mathbf{y},\quad \mathbf{n}\in \mathbf{Z}^{d},\end{eqnarray}$$

are the Fourier coefficients. Here $\mathbf{v}\cdot \mathbf{w}$ denotes the dot product of $\mathbf{v}$ and  $\mathbf{w}$ .

Clearly

(3.3) $$\begin{eqnarray}a_{\mathbf{0}}=\int _{I^{d}}f(\mathbf{y})\,d\mathbf{y}\quad \text{and so}\quad \mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}=\unicode[STIX]{x1D70E}_{0}^{2}(f),\end{eqnarray}$$

where we have used Parseval’s formula.

By (3.1), (3.2) and noting that $\mathbf{e}\in \mathbf{S}^{d-1}$ is a unit vector in the $d$ -space, we have

(3.4) $$\begin{eqnarray}f(t\sqrt{d}\mathbf{e})-\int _{I^{d}}f\,d\text{V}=\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}t\sqrt{d}\mathbf{n}\cdot \mathbf{e}}.\end{eqnarray}$$

Recall (1.9) and (1.10). Using (1.6) and (3.4), we have

(3.5) $$\begin{eqnarray}\displaystyle D_{f}(\unicode[STIX]{x1D70C}\mathbf{e};T_{1},T_{2}) & = & \displaystyle \int _{T_{1}}^{T_{2}}f(t\unicode[STIX]{x1D70C}\mathbf{e})\,dt-(T_{2}-T_{1})\int _{I^{d}}f\,d\text{V}\nonumber\\ \displaystyle & = & \displaystyle \int _{T_{1}}^{T_{2}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}t\mathbf{n}\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\,dt.\end{eqnarray}$$

We need a simple estimate.

Lemma 3.1. For every $d$ -dimensional vector $\mathbf{w}=(w_{1},\ldots ,w_{d})$ , we have

$$\begin{eqnarray}\int _{\mathbf{S}^{d-1}}\int _{0}^{\infty }\text{e}^{\text{i}\mathbf{w}\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\frac{\unicode[STIX]{x1D70C}^{d-1}\text{e}^{-\unicode[STIX]{x1D70C}^{2}/2}}{C_{d}}\,d\unicode[STIX]{x1D70C}\,d\unicode[STIX]{x1D708}_{d-1}^{\ast }(\mathbf{e})=\text{e}^{-|\mathbf{w}|^{2}/2}.\end{eqnarray}$$

Proof. For simplicity, denote the integral under investigation by  ${\mathcal{I}}$ . Note that the vector $\unicode[STIX]{x1D70C}\mathbf{e}=\mathbf{v}=(v_{1},\ldots ,v_{d})$ has $d$ -dimensional standard Gaussian (normal) distribution. Thus

$$\begin{eqnarray}\displaystyle {\mathcal{I}} & = & \displaystyle (2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}\text{e}^{\text{i}\mathbf{w}\cdot \mathbf{v}}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}=\mathop{\prod }_{j=1}^{d}\biggl(\frac{1}{\sqrt{2\unicode[STIX]{x1D70B}}}\int _{-\infty }^{\infty }\text{e}^{\text{i}w_{j}v_{j}}\text{e}^{-v_{j}^{2}/2}\,dv_{j}\biggr)\nonumber\\ \displaystyle & = & \displaystyle \mathop{\prod }_{j=1}^{d}\text{e}^{-w_{j}^{2}/2}=\text{e}^{-|\mathbf{w}|^{2}/2},\nonumber\end{eqnarray}$$

where in the argument, we have used the well-known facts that the coordinates $v_{1},\ldots ,v_{d}$ of $\mathbf{v}$ are independent random variables, each having standard normal distribution, and that the Fourier transform of $\text{e}^{-x^{2}/2}$ is itself.◻

Let us return to (1.9) and (3.5). Applying Lemma 3.1, it is easy to establish the following result.

Lemma 3.2. For every $-\infty \leqslant W^{\prime }<W^{\prime \prime }\leqslant \infty$ , we have

$$\begin{eqnarray}\unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};W^{\prime },W^{\prime \prime })=\mathop{\sum }_{\mathbf{n}_{1},\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}_{1}}\overline{a_{\mathbf{n}_{2}}}\int _{W^{\prime }}^{W^{\prime \prime }}\int _{W^{\prime }}^{W^{\prime \prime }}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{1}\,dt_{2}.\end{eqnarray}$$

Proof. It is easy to see that

$$\begin{eqnarray}|D_{f}(\unicode[STIX]{x1D70C}\mathbf{e};W^{\prime },W^{\prime \prime })|^{2}=\int _{W^{\prime }}^{W^{\prime \prime }}\int _{W^{\prime }}^{W^{\prime \prime }}\mathop{\sum }_{\mathbf{n}_{1},\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}_{1}}\overline{a_{\mathbf{n}_{2}}}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2})\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\,dt_{1}\,dt_{2}.\end{eqnarray}$$

Applying this in (3.5), we obtain

(3.6) $$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};W^{\prime },W^{\prime \prime })\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{\mathbf{S}^{d-1}}\int _{0}^{\infty }\biggl(\int _{W^{\prime }}^{W^{\prime \prime }}\int _{W^{\prime }}^{W^{\prime \prime }}\mathop{\sum }_{\mathbf{n}_{1},\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}_{1}}\overline{a_{\mathbf{n}_{2}}}\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2})\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\,dt_{1}\,dt_{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\frac{\unicode[STIX]{x1D70C}^{d-1}\text{e}^{-\unicode[STIX]{x1D70C}^{2}/2}}{C_{d}}\,d\unicode[STIX]{x1D70C}\,d\unicode[STIX]{x1D708}_{d-1}^{\ast }(\mathbf{e})\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\mathbf{n}_{1},\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}a_{\mathbf{n}_{1}}\overline{a_{\mathbf{n}_{2}}}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \int _{W^{\prime }}^{W^{\prime \prime }}\!\int _{W^{\prime }}^{W^{\prime \prime }}\!\biggl(\int _{\mathbf{S}^{d-1}}\int _{0}^{\infty }\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2})\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\frac{\unicode[STIX]{x1D70C}^{d-1}\text{e}^{-\unicode[STIX]{x1D70C}^{2}/2}}{C_{d}}\,d\unicode[STIX]{x1D70C}\,d\unicode[STIX]{x1D708}_{d-1}^{\ast }(\mathbf{e})\biggr)\,dt_{1}\,dt_{2}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

Lemma 3.1 applied to the inner integral on the right-hand side of (3.6) now gives

$$\begin{eqnarray}\int _{\mathbf{S}^{d-1}}\int _{0}^{\infty }\text{e}^{2\unicode[STIX]{x1D70B}\text{i}(t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2})\cdot \unicode[STIX]{x1D70C}\mathbf{e}}\frac{\unicode[STIX]{x1D70C}^{d-1}\text{e}^{-\unicode[STIX]{x1D70C}^{2}/2}}{C_{d}}\,d\unicode[STIX]{x1D70C}\,d\unicode[STIX]{x1D708}_{d-1}^{\ast }(\mathbf{e})=\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}.\end{eqnarray}$$

Substituting this into (3.6) completes the proof. ◻

We are now ready to prove Theorem 3.1. The proof is an elementary brute force combinatorial argument. For every $\mathbf{n}=(n_{1},\ldots ,n_{d})\in \mathbf{Z}^{d}\setminus \mathbf{0}$ , write

$$\begin{eqnarray}L(\mathbf{n})=\{1\leqslant i\leqslant d:n_{i}\neq 0\}.\end{eqnarray}$$

Applying the inequality $|a_{\mathbf{n}_{1}}\overline{a_{\mathbf{n}_{2}}}||\leqslant (|a_{\mathbf{n}_{1}}|^{2}+|a_{\mathbf{n}_{2}}|^{2})/2$ in Lemma 3.2, we have

(3.7) $$\begin{eqnarray}\displaystyle & & \displaystyle |\unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};U,V)|\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathop{\sum }_{\mathbf{n}_{1},\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}\frac{1}{2}(|a_{\mathbf{n}_{1}}|^{2}+|a_{\mathbf{n}_{2}}|^{2})\int _{U}^{V}\int _{U}^{V}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{1}\,dt_{2}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\mathop{\sum }_{\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}\int _{U}^{V}\int _{U}^{V}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{1}\,dt_{2}\nonumber\\ \displaystyle & & \displaystyle \quad =\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}\nonumber\\ \displaystyle & & \displaystyle \quad =\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \,\biggl(\mathop{\sum }_{L_{1,2}\subseteq L(\mathbf{n}_{1})}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=\max \{|L_{1,2}|,1\}}^{d}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}.\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

We fix $t_{1}\in [U,V]$ , $\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}$ , $L_{1,2}\subseteq L(\mathbf{n}_{1})$ and $\unicode[STIX]{x1D706}_{2}$ , and focus on the inner integral on the right-hand side of (3.7).

Write

$$\begin{eqnarray}\unicode[STIX]{x1D706}_{1,2}=|L_{1,2}|=|\{1\leqslant i\leqslant d:n_{1,i}\neq 0~\text{and}~n_{2,i}\neq 0\}|.\end{eqnarray}$$

Let $k_{i}(\mathbf{n}_{2})$ , $i=1,2,3\ldots \,$ , denote the number of coordinates $n_{2,i}=\pm i$ of $\mathbf{n}_{2}$ which also satisfy $n_{1,i}=0$ . Note that

(3.8) $$\begin{eqnarray}k_{1}(\mathbf{n}_{2})+k_{2}(\mathbf{n}_{2})+k_{3}(\mathbf{n}_{2})+\cdots =\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}.\end{eqnarray}$$

Let $h_{0}(t_{2};\mathbf{n}_{2})$ denote the number of coordinates $j\in L_{1,2}$ such that

$$\begin{eqnarray}|t_{1}n_{1,j}-t_{2}n_{2,j}|<\frac{U}{2}.\end{eqnarray}$$

Let $h_{i}(t_{2};\mathbf{n}_{2})$ , $i=1,2,3\ldots \,$ , denote the number of coordinates $j\in L_{1,2}$ such that

$$\begin{eqnarray}\frac{(2i-1)U}{2}\leqslant |t_{1}n_{1,j}-t_{2}n_{2,j}|<\frac{(2i+1)U}{2}.\end{eqnarray}$$

Note that

(3.9) $$\begin{eqnarray}h_{0}(t_{2};\mathbf{n}_{2})+h_{1}(t_{2};\mathbf{n}_{2})+h_{2}(t_{2};\mathbf{n}_{2})+h_{3}(t_{2};\mathbf{n}_{2})+\cdots =\unicode[STIX]{x1D706}_{1,2}.\end{eqnarray}$$

By definition,

$$\begin{eqnarray}\displaystyle |t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2} & = & \displaystyle \mathop{\sum }_{j\in L_{1,2}}(t_{1}n_{1,j}-t_{2}n_{2,j})^{2}\nonumber\\ \displaystyle & & \displaystyle +\,\mathop{\sum }_{j\in L(\mathbf{n}_{1})\setminus L_{1,2}}(t_{1}n_{1,j})^{2}+\mathop{\sum }_{j\in L(\mathbf{n}_{2})\setminus L_{1,2}}(t_{2}n_{2,j})^{2}.\nonumber\end{eqnarray}$$

Using this and the definitions of $k_{i}(\mathbf{n}_{2})$ and $h_{i}(t_{2};\mathbf{n}_{2})$ , we have

(3.10) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \exp \biggl(-2\unicode[STIX]{x1D70B}^{2}\mathop{\sum }_{j\in L(\mathbf{n}_{1})\setminus L_{1,2}}|n_{1,j}|^{2}U^{2}\biggr)\nonumber\\ \displaystyle & & \displaystyle \qquad \cdot \int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}{\mathcal{K}}(t_{2};\mathbf{n}_{2}){\mathcal{H}}(t_{2};\mathbf{n}_{2};U)\,dt_{2},\end{eqnarray}$$

where

$$\begin{eqnarray}{\mathcal{K}}(t_{2};\mathbf{n}_{2})=\mathop{\prod }_{i\geqslant 1}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}k_{i}(\mathbf{n}_{2})i^{2}t_{2}^{2}}\end{eqnarray}$$

and

$$\begin{eqnarray}{\mathcal{H}}(t_{2};\mathbf{n}_{2};U)=\mathop{\prod }_{i\geqslant 1}\text{e}^{-\unicode[STIX]{x1D70B}^{2}h_{i}(t_{2};\mathbf{n}_{2})((2i-1)U)^{2}/2}.\end{eqnarray}$$

We estimate the last sum on the right-hand side of (3.10). Again, using the definitions of $k_{i}(\mathbf{n}_{2})$ and $h_{i}(t_{2};\mathbf{n}_{2})$ , and noting (3.8) and (3.9), we obtain the upper bound

(3.11) $$\begin{eqnarray}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}{\mathcal{K}}(t_{2};\mathbf{n}_{2}){\mathcal{H}}(t_{2};\mathbf{n}_{2};U)\leqslant {\mathcal{K}}(d;U){\mathcal{H}}(U),\end{eqnarray}$$

where

$$\begin{eqnarray}{\mathcal{K}}(d;U)=\mathop{\sum }_{\substack{ r\geqslant 1 \\ k_{1},\ldots ,k_{r-1}\geqslant 0 \\ k_{r}\geqslant 1 \\ k_{1}+\cdots +k_{r}=\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}}\mathop{\prod }_{i=1}^{r}\binom{d-\unicode[STIX]{x1D706}_{1}-k_{1}-\cdots -k_{i-1}}{k_{i}}2^{k_{i}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}k_{i}i^{2}U^{2}}\end{eqnarray}$$

and

$$\begin{eqnarray}\displaystyle {\mathcal{H}}(U) & = & \displaystyle \mathop{\sum }_{\substack{ r\geqslant 0 \\ h_{0},\ldots ,h_{r-1}\geqslant 0 \\ h_{r}\geqslant 1 \\ h_{0}+\cdots +r_{r}=\unicode[STIX]{x1D706}_{1,2}}}\binom{\unicode[STIX]{x1D706}_{1,2}}{h_{0}}\mathop{\prod }_{i=1}^{r}\binom{\unicode[STIX]{x1D706}_{1,2}-h_{0}-\cdots -h_{i-1}}{h_{i}}\nonumber\\ \displaystyle & & \displaystyle \times \,2^{h_{i}}\text{e}^{-\unicode[STIX]{x1D70B}^{2}h_{i}((2i-1)U)^{2}/2}.\nonumber\end{eqnarray}$$

Note that (3.11) includes the pathological case $\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}=0$ , with the convention that the summation means the single term $(k_{1},\ldots ,k_{r})=(0)$ , and similarly, if $\unicode[STIX]{x1D706}_{1,2}=0$ , then $(h_{0},h_{1},\ldots ,h_{r})$ is just the single term  $(0)$ .

We have the trivial bound

(3.12) $$\begin{eqnarray}{\mathcal{K}}(d;U)\leqslant \mathop{\sum }_{\substack{ r\geqslant 1 \\ k_{1},\ldots ,k_{r-1}\geqslant 0 \\ k_{r}\geqslant 1 \\ k_{1}+\cdots +k_{r}=\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}}\mathop{\prod }_{i=1}^{r}(2d)^{k_{i}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}k_{i}i^{2}U^{2}}\leqslant \biggl(\mathop{\sum }_{i\geqslant 1}2d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}i^{2}U^{2}}\biggr)^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}.\end{eqnarray}$$

On the other hand, using the multinomial theorem, we obtain

(3.13) $$\begin{eqnarray}{\mathcal{H}}(U)=\biggl(1+\mathop{\sum }_{i\geqslant 1}2\text{e}^{-\unicode[STIX]{x1D70B}^{2}((2i-1)U)^{2}/2}\biggr)^{\unicode[STIX]{x1D706}_{1,2}}.\end{eqnarray}$$

Combining (3.11)–(3.13), we deduce that

(3.14) $$\begin{eqnarray}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\!\!{\mathcal{K}}(t_{2};\mathbf{n}_{2}){\mathcal{H}}(t_{2};\mathbf{n}_{2};U)\leqslant (3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}.\end{eqnarray}$$

Combining this with (3.10), we conclude that

(3.15) $$\begin{eqnarray}\displaystyle & & \displaystyle \int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \text{e}^{-2\unicode[STIX]{x1D70B}^{2}(\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2})U^{2}}(V-U)(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}.\qquad \;\end{eqnarray}$$

Let us now return to (3.7). We have the decomposition

(3.16) $$\begin{eqnarray}|\unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};U,V)|\leqslant \unicode[STIX]{x1D6EF}_{1}+\unicode[STIX]{x1D6EF}_{2}+\unicode[STIX]{x1D6EF}_{3}+\unicode[STIX]{x1D6EF}_{4},\end{eqnarray}$$

where

(3.17) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{1} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d-1}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\biggl(\mathop{\sum }_{L_{1,2}\subseteq L(\mathbf{n}_{1})}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=\unicode[STIX]{x1D706}_{1}+1}^{d}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

is characterized by the property $\unicode[STIX]{x1D706}_{1}<\unicode[STIX]{x1D706}_{2}$ , and

(3.18) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{2} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\biggl(\mathop{\sum }_{L_{1,2}\subseteq L(\mathbf{n}_{1})}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=\max \{|L_{1,2}|,1\}}^{\unicode[STIX]{x1D706}_{1}-1}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{2} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}\nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

is characterized by the property $\unicode[STIX]{x1D706}_{1}>\unicode[STIX]{x1D706}_{2}$ . Furthermore, we split the case $\unicode[STIX]{x1D706}_{1}=\unicode[STIX]{x1D706}_{2}$ into two subcases according as $L(\mathbf{n}_{1})\neq L(\mathbf{n}_{2})$ or $L(\mathbf{n}_{1})=L(\mathbf{n}_{2})$ , thus

(3.19) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{3} & = & \displaystyle \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & & \displaystyle \cdot \,\biggl(\mathop{\sum }_{\substack{ L_{1,2}\subset L(\mathbf{n}_{1}) \\ L_{1,2}\neq L(\mathbf{n}_{1})}}\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{2})|=\unicode[STIX]{x1D706}_{1} \\ L(\mathbf{n}_{2})\cap L(\mathbf{n}_{1})=L_{1,2}}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}\end{eqnarray}$$

and

(3.20) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}=\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ L(\mathbf{n}_{2})=L(\mathbf{n}_{1})}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}.\end{eqnarray}$$

To estimate the last term (3.20), we need a simple but important lemma. To state this, we first need a definition. Given real numbers $C$ and  $C^{\prime }$ , consider the set

$$\begin{eqnarray}{\mathcal{B}}_{U}(C;C^{\prime })=\{t\in [U,2U]:\text{there exists}~n\in \mathbf{Z}\setminus \{0\}~\text{such that}~|C-tn|\leqslant C^{\prime }\}.\end{eqnarray}$$

We give an upper bound on the one-dimensional Lebesgue measure, i.e. length, of the set ${\mathcal{B}}_{U}(C;C^{\prime })$ .

Lemma 3.3. For arbitrary real numbers $C,C^{\prime }$ with $|C|\geqslant U\geqslant 1$ and $0<C^{\prime }<U/2$ , we have

$$\begin{eqnarray}\operatorname{length}({\mathcal{B}}_{U}(C;C^{\prime }))<6C^{\prime }.\end{eqnarray}$$

Proof. We can assume without loss of generality that $C>0$ . Clearly

$$\begin{eqnarray}|C-tn|\leqslant C^{\prime }\quad \text{if and only if}~\frac{C-C^{\prime }}{n}\leqslant t\leqslant \frac{C+C^{\prime }}{n},\end{eqnarray}$$

so

$$\begin{eqnarray}\operatorname{length}({\mathcal{B}}_{U}(C;C^{\prime }))=\mathop{\sum }_{n}^{\ast }\frac{2C^{\prime }}{n},\end{eqnarray}$$

where the summation $\sum _{n}^{\ast }$ is extended over all integers $n$ such that

$$\begin{eqnarray}\frac{C}{t}-\frac{C^{\prime }}{t}\leqslant n\leqslant \frac{C}{t}+\frac{C^{\prime }}{t}.\end{eqnarray}$$

Note that

$$\begin{eqnarray}\frac{C+C^{\prime }}{t}\leqslant \frac{3C}{2t}\leqslant \frac{3C}{2U}\quad \text{and}\quad \frac{C-C^{\prime }}{t}\geqslant \frac{C}{2t}\geqslant \frac{C}{4U}.\end{eqnarray}$$

Thus we have

$$\begin{eqnarray}\displaystyle \operatorname{length}({\mathcal{B}}_{U}(C;C^{\prime })) & {\leqslant} & \displaystyle \mathop{\sum }_{C/4U\leqslant n\leqslant 3C/2U}\frac{2C^{\prime }}{n}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C^{\prime }\biggl(1+\log \frac{3C/2U}{C/4U}\biggr)=2C^{\prime }(1+\log 6),\nonumber\end{eqnarray}$$

where we have used the well-known fact that $\sum _{A\leqslant n\leqslant B}1/n\leqslant 1+\log (B/A)$ for all $0<A<B$ . Since $\log 6<2$ , the proof is complete.◻

Applying (3.15) in (3.17), we obtain

(3.21) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{1} & {\leqslant} & \displaystyle (V-U)^{2}\max _{\unicode[STIX]{x1D706}_{1,2}\leqslant d}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{d}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{\unicode[STIX]{x1D706}_{2}-1}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{1}}\binom{\unicode[STIX]{x1D706}_{1}}{\unicode[STIX]{x1D706}_{1,2}}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-\unicode[STIX]{x1D706}_{1,2}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}(\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2})U^{2}}\nonumber\\ \displaystyle & = & \displaystyle (V-U)^{2}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{d}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{\unicode[STIX]{x1D706}_{2}-1}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}((3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-1}+\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}}.\end{eqnarray}$$

By hypothesis,

(3.22) $$\begin{eqnarray}dU\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2}\leqslant {\textstyle \frac{1}{3}}.\end{eqnarray}$$

It follows that

(3.23) $$\begin{eqnarray}\displaystyle & & \displaystyle ((3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-1}+\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}}\nonumber\\ \displaystyle & & \displaystyle \quad =(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-\unicode[STIX]{x1D706}_{1}}(1+3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}/2}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-\unicode[STIX]{x1D706}_{1}}\biggl(1+\frac{1}{2d}\biggr)^{d}<2(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{-\unicode[STIX]{x1D706}_{1}}.\end{eqnarray}$$

Using (3.22) and (3.23) in (3.21), we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{1} & {\leqslant} & \displaystyle 2(V-U)^{2}\biggl(1+\frac{1}{2d}\biggr)^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{\unicode[STIX]{x1D706}_{2}-1}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1}}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 4(V-U)^{2}\,d\mathop{\sum }_{j=1}^{\infty }(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{j}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & = & \displaystyle 12(V-U)^{2}\frac{d^{2}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}}}{1-3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2},\nonumber\end{eqnarray}$$

where we have used the substitution $j=\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1}$ . Thus

(3.24) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{1}\leqslant 12(V-U)^{2}\frac{3^{-4}U^{-2}}{1-3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\leqslant \frac{1}{6}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2},\end{eqnarray}$$

since $V-U\leqslant U$ .

Next, applying (3.15) in (3.18), we obtain

(3.25) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{2} & {\leqslant} & \displaystyle (V-U)^{2}\max _{\unicode[STIX]{x1D706}_{1,2}\leqslant d}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{\unicode[STIX]{x1D706}_{1}-1}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{2}}\binom{\unicode[STIX]{x1D706}_{1}}{\unicode[STIX]{x1D706}_{1,2}}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}(\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2})U^{2}}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (V-U)^{2}\biggl(1+\frac{1}{2d}\biggr)^{d}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{\unicode[STIX]{x1D706}_{1}-1}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{2}}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{2}-\unicode[STIX]{x1D706}_{1,2}}(d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2(V-U)^{2}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{\unicode[STIX]{x1D706}_{1}-1}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{2}}(3d\text{e}^{-2\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}+\unicode[STIX]{x1D706}_{2}-2\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2},\qquad\end{eqnarray}$$

where in the last steps we have used (3.22) and the trivial upper bound

(3.26) $$\begin{eqnarray}\binom{\unicode[STIX]{x1D706}_{1}}{\unicode[STIX]{x1D706}_{1,2}}=\binom{\unicode[STIX]{x1D706}_{1}}{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}\leqslant \unicode[STIX]{x1D706}_{1}^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}\leqslant d^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}.\end{eqnarray}$$

Using (3.22) in (3.25), we have

(3.27) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{2} & {\leqslant} & \displaystyle 2(V-U)^{2}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{2}=1}^{\unicode[STIX]{x1D706}_{1}-1}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{2}}(3^{-3}d^{-1}U^{-2})^{\unicode[STIX]{x1D706}_{1}+\unicode[STIX]{x1D706}_{2}-2\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\boldsymbol{ n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2(V-U)^{2}\mathop{\sum }_{j=1}^{\infty }d(j+1)j(3^{-3}d^{-1}U^{-2})^{j}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & = & \displaystyle 4(V-U)^{2}\frac{3^{-3}U^{-2}}{(1-3^{-3}d^{-1}U^{-2})^{3}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & = & \displaystyle \frac{4}{27}(V-U)^{2}U^{-2}(1-3^{-3}d^{-1}U^{-2})^{-3}\mathop{\sum }_{\boldsymbol{ n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{1}{4}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2},\end{eqnarray}$$

where we have used the substitution $j=\unicode[STIX]{x1D706}_{1}+\unicode[STIX]{x1D706}_{2}-2\unicode[STIX]{x1D706}_{1,2}$ , the assumption $V-U\leqslant U$ , and the simple fact that

$$\begin{eqnarray}\mathop{\sum }_{j=1}^{\infty }j(j+1)x^{j}=\frac{2x}{(1-x)^{3}}\quad \text{for all}~|x|<1.\end{eqnarray}$$

Next, applying (3.15) in (3.19), we obtain

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{3} & {\leqslant} & \displaystyle (V-U)^{2}\max _{\unicode[STIX]{x1D706}_{1,2}\leqslant d}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{1}-1}\binom{\unicode[STIX]{x1D706}_{1}}{\unicode[STIX]{x1D706}_{1,2}}(3d\text{e}^{-(2+2)\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}.\nonumber\end{eqnarray}$$

Using (3.26) in this, we have

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{3} & {\leqslant} & \displaystyle (V-U)^{2}\max _{\unicode[STIX]{x1D706}_{1,2}\leqslant d}(1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{\unicode[STIX]{x1D706}_{1,2}}\nonumber\\ \displaystyle & & \displaystyle \cdot \mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\substack{ \mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ |L(\mathbf{n}_{1})|=\unicode[STIX]{x1D706}_{1}}}|a_{\mathbf{n}_{1}}|^{2}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{1}-1}(3d^{2}\text{e}^{-4\unicode[STIX]{x1D70B}^{2}U^{2}})^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}.\nonumber\end{eqnarray}$$

Then using (3.22) in this, we have

(3.28) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{3} & {\leqslant} & \displaystyle (V-U)^{2}(1+3(dU)^{-1})^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1}=1}^{d}\mathop{\sum }_{\unicode[STIX]{x1D706}_{1,2}=0}^{\unicode[STIX]{x1D706}_{1}-1}(3(3^{4}\,dU)^{-2})^{\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (V-U)^{2}\biggl(1+\frac{1}{2d}\biggr)^{d}d\mathop{\sum }_{j=1}^{\infty }(3^{7}d^{2}U^{2})^{-j}\mathop{\sum }_{\boldsymbol{ n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2d(V-U)^{2}\frac{(3^{7}d^{2}U^{2})^{-1}}{1-(3^{7}d^{2}U^{2})^{-1}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & = & \displaystyle 2(V-U)^{2}U^{-2}\frac{(3^{7}d)^{-1}}{1-(3^{7}d^{2}U^{2})^{-1}}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \frac{1}{100d}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2},\end{eqnarray}$$

where we have used the substitution $j=\unicode[STIX]{x1D706}_{1}-\unicode[STIX]{x1D706}_{1,2}$ and the hypothesis $V-U\leqslant U$ .

Finally we estimate (3.20). We have

(3.29) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}=\unicode[STIX]{x1D6EF}_{4}^{(1)}+\unicode[STIX]{x1D6EF}_{4}^{(2)},\end{eqnarray}$$

where

(3.30) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(1)}=\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ L(\mathbf{n}_{2})=L(\mathbf{n}_{1}) \\ h_{0}(t_{2};\mathbf{n}_{2})<|L(\mathbf{n}_{1})|}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1},\end{eqnarray}$$

and where

(3.31) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(2)}=\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ L(\mathbf{n}_{2})=L(\mathbf{n}_{1}) \\ h_{0}(t_{2};\mathbf{n}_{2})=|L(\mathbf{n}_{1})|}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}\,dt_{2}\biggr)\,dt_{1}.\end{eqnarray}$$

We start with the term $\unicode[STIX]{x1D6EF}_{4}^{(1)}$ given by (3.30), and use the notation $\unicode[STIX]{x1D706}_{1}=|L(\mathbf{n}_{1})|$ . Note that here $\unicode[STIX]{x1D706}_{1}=\unicode[STIX]{x1D706}_{2}=\unicode[STIX]{x1D706}_{1,2}$ . Write

$$\begin{eqnarray}{\mathcal{H}}^{(1)}(U)=\mathop{\sum }_{\substack{ r\geqslant 0 \\ h_{0},\ldots ,h_{r-1}\geqslant 0 \\ h_{r}\geqslant 1 \\ h_{0}+\cdots +r_{r}=\unicode[STIX]{x1D706}_{1}}}\binom{\unicode[STIX]{x1D706}_{1}}{h_{0}}\mathop{\prod }_{i=1}^{r}\binom{\unicode[STIX]{x1D706}_{1}-h_{0}-\cdots -h_{i-1}}{h_{i}}2^{h_{i}}\text{e}^{-\unicode[STIX]{x1D70B}^{2}h_{i}((2i-1)U)^{2}/2}.\end{eqnarray}$$

In this special case the argument of (3.10)–(3.15) simplifies to the upper bound

$$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{4}^{(1)} & {\leqslant} & \displaystyle \int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}(V-U){\mathcal{H}}^{(1)}(U)\,dt_{1}\nonumber\\ \displaystyle & = & \displaystyle \int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}(V-U)\nonumber\\ \displaystyle & & \displaystyle \cdot \,((1+2\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2}+2\text{e}^{-\unicode[STIX]{x1D70B}^{2}3^{2}U^{2}/2}+2\text{e}^{-\unicode[STIX]{x1D70B}^{2}5^{2}U^{2}/2}+\cdots )^{\unicode[STIX]{x1D706}_{1}}-1)\,dt_{1},\nonumber\end{eqnarray}$$

where the term $-1$ at the end is explained by the restriction $h_{0}<\unicode[STIX]{x1D706}_{1}$ , i.e. the term $-1$ is due to the fact that in the application of the multinomial theorem we have an almost complete sum, where the single missing case is $h_{0}=\unicode[STIX]{x1D706}_{1}$ . It now follows that

(3.32) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D6EF}_{4}^{(1)} & {\leqslant} & \displaystyle \mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}(V-U)^{2}((1+3\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})^{d}-1)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D70E}_{0}^{2}(f)(V-U)^{2}(\exp (3d\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2})-1)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \unicode[STIX]{x1D70E}_{0}^{2}(f)(V-U)^{2}6\,d\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2}\end{eqnarray}$$

since

(3.33) $$\begin{eqnarray}d\text{e}^{-\unicode[STIX]{x1D70B}^{2}U^{2}/2}\leqslant {\textstyle \frac{1}{3}}.\end{eqnarray}$$

Note that in (3.32), we have used Parseval’s formula and the simple inequalities $1+x\leqslant \text{e}^{x}\leqslant 1+2x$ for $0\leqslant x\leqslant 1$ ; and of course (3.33) follows from (3.22). If we now apply (3.22) in (3.32), then we conclude that

(3.34) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(1)}\leqslant \unicode[STIX]{x1D70E}_{0}^{2}(f)(V-U)^{2}\frac{6}{3U}\leqslant 2\unicode[STIX]{x1D70E}_{0}^{2}(f)(V-U),\end{eqnarray}$$

since $V-U\leqslant U$ .

To estimate $\unicode[STIX]{x1D6EF}_{4}^{(2)}$ given by (3.31), we make use of the fact that for every given triple $(t_{1},\mathbf{n}_{1},t_{2})$ with $t_{1},t_{2}\in [U,V]$ and $\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}$ , there is at most one term in the sum

$$\begin{eqnarray}\mathop{\sum }_{\substack{ \mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0} \\ L(\mathbf{n}_{2})=L(\mathbf{n}_{1}) \\ h_{0}(t_{2};\mathbf{n}_{2})=|L(\mathbf{n}_{1})|}}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}|^{2}}.\end{eqnarray}$$

Indeed, it follows easily from the definition of $h_{0}(t_{2};\mathbf{n}_{2})$ that for every given triple $(t_{1},\mathbf{n}_{1},t_{2})$ , the inequality $|t_{1}n_{1,j}-t_{2}n_{2,j}|<U/2$ has at most one integer solution $n_{2,j}$ , and since $h_{0}(t_{2};\mathbf{n}_{2})=|L(\mathbf{n}_{1})|$ and $L(\mathbf{n}_{2})=L(\mathbf{n}_{1})$ , there is therefore at most one $\mathbf{n}_{2}\in \mathbf{Z}^{d}\setminus \mathbf{0}$ satisfying the requirements.

Let $\mathbf{n}_{2}^{\ast }\in \mathbf{Z}^{d}\setminus \mathbf{0}$ , $\mathbf{n}_{2}^{\ast }=\mathbf{n}_{2}^{\ast }(t_{1},\mathbf{n}_{1},t_{2})$ , denote this single integer lattice point, if it exists. Then (3.31) can be rewritten in the form

$$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(2)}=\int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}\mathbf{n}_{1}-t_{2}\mathbf{n}_{2}^{\ast }|^{2}}\,dt_{2}\biggr)\,dt_{1}\end{eqnarray}$$

if $\mathbf{n}_{2}^{\ast }=\mathbf{n}_{2}^{\ast }(t_{1},\mathbf{n}_{1},t_{2})$ exists, and vanishes otherwise. Let $j_{0}=j_{0}(\mathbf{n}_{1})$ be an arbitrary but fixed element of the set $L(\mathbf{n}_{1})$ . Then we have trivially

(3.35) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(2)}\leqslant \int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\int _{U}^{V}\text{e}^{-2\unicode[STIX]{x1D70B}^{2}|t_{1}n_{1,j_{0}}-t_{2}n_{2,j_{0}}^{\ast }|^{2}}\,dt_{2}\biggr)\,dt_{1}.\end{eqnarray}$$

Given $(t_{1},\mathbf{n}_{1})$ , we decompose the interval $U\leqslant t_{2}\leqslant V$ , where $V\leqslant 2U$ , into sets

(3.36) $$\begin{eqnarray}I_{\ell }(t_{1},\mathbf{n}_{1})=\{t_{2}\in [U,V]:\ell -1\leqslant |t_{1}n_{1,j_{0}}-t_{2}n_{2,j_{0}}^{\ast }|<\ell \},\quad \ell =1,2,3,\ldots ,\end{eqnarray}$$

where $j_{0}=j_{0}(\mathbf{n}_{1})$ and $\mathbf{n}_{2}^{\ast }=\mathbf{n}_{2}^{\ast }(t_{1},\mathbf{n}_{1},t_{2})$ . Using the decomposition (3.36) in (3.35), we obtain the upper bound

(3.37) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(2)}\leqslant \int _{U}^{V}\mathop{\sum }_{\mathbf{n}_{1}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}_{1}}|^{2}\biggl(\mathop{\sum }_{\ell =1}^{\infty }\operatorname{length}(I_{\ell })\text{e}^{-2\unicode[STIX]{x1D70B}^{2}(\ell -1)^{2}}\biggr)\,dt_{1},\end{eqnarray}$$

where $I_{\ell }=I_{\ell }(t_{1},\mathbf{n}_{1})$ . Applying Lemma 3.3 with $C=t_{1}n_{1,j_{0}}$ and $C^{\prime }=\ell$ , we obtain the upper bound $\operatorname{length}(I_{\ell })<6\ell$ . Using this in (3.37), we conclude that

(3.38) $$\begin{eqnarray}\unicode[STIX]{x1D6EF}_{4}^{(2)}\leqslant \int _{U}^{V}\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}\biggl(\mathop{\sum }_{\ell =1}^{\infty }6\ell \text{e}^{-2\unicode[STIX]{x1D70B}^{2}(\ell -1)^{2}}\biggr)\,dt_{1}\leqslant 7(V-U)\mathop{\sum }_{\mathbf{n}\in \mathbf{Z}^{d}\setminus \mathbf{0}}|a_{\mathbf{n}}|^{2}.\end{eqnarray}$$

Finally, combining (3.3), (3.24), (3.27)–(3.29), (3.34) and (3.38) with (3.16), we conclude that

$$\begin{eqnarray}\displaystyle |\unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};U,V)| & {\leqslant} & \displaystyle \biggl(\frac{1}{6}+\frac{1}{4}+\frac{1}{100d}+2(V-U)+7(V-U)\biggr)\unicode[STIX]{x1D70E}_{0}^{2}(f)\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (9(V-U)+1)\unicode[STIX]{x1D70E}_{0}^{2}(f).\nonumber\end{eqnarray}$$

This completes the proof of Theorem 3.1.

To prove Theorem 1.2, we have to eliminate the technical restriction $U<V\leqslant 2U$ in Theorem 3.1.

Given $1\leqslant U<W$ , let $k\geqslant 1$ denote the integer such that $2^{k-1}U<W\leqslant 2^{k}U$ . It is easy to find a sequence $1\leqslant W_{0}=U<W_{1}<\cdots <W_{k}=W$ such that $W_{i}<W_{i+1}\leqslant 2W_{i}$ for all $0\leqslant i<k$ . By using the Cauchy–Schwarz inequality and Theorem 3.1, we have

$$\begin{eqnarray}\displaystyle & & \displaystyle \unicode[STIX]{x1D6E5}_{f}^{2}(\text{Gauss};U,W)\nonumber\\ \displaystyle & & \displaystyle \quad =(2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}(D_{f}(\mathbf{v};U,W))^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad =(2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}\biggl(\mathop{\sum }_{j=0}^{k-1}D_{f}(\mathbf{v};W_{j},W_{j+1})\biggr)^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant (2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}\biggl(\mathop{\sum }_{j=0}^{k-1}(D_{f}(\mathbf{v};W_{j},W_{j+1}))^{2}\biggr)\biggl(\mathop{\sum }_{j=0}^{k-1}1\biggr)\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad =k\mathop{\sum }_{j=0}^{k-1}(2\unicode[STIX]{x1D70B})^{-d/2}\int _{\mathbf{R}^{d}}(D_{f}(\mathbf{v};W_{j},W_{j+1}))^{2}\text{e}^{-|\mathbf{v}|^{2}/2}\,d\mathbf{v}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant k\unicode[STIX]{x1D70E}_{0}^{2}(f)\mathop{\sum }_{j=0}^{k-1}(9(W_{j+1}-W_{j})+1)\leqslant k\unicode[STIX]{x1D70E}_{0}^{2}(f)(10(W-U)+10).\nonumber\end{eqnarray}$$

Since $k=\lceil \log _{2}(W/U)\rceil$ , the proof of Theorem 1.2 is complete.