On the Discrete Approximation by the Mellin Transform of the Riemann Zeta-Function

Abstract

In the paper, it is obtained that there are infinite discrete shifts $\Xi (s+ikh)$, $h>0$, $k\in {\mathbb{N}}_0$ of the Mellin transform $\Xi \left(s\right)$ of the square of the Riemann zeta-function, approximating a certain class of analytic functions. For the proof, a probabilistic approach based on weak convergence of probability measures in the space of analytic functions is applied.

1. Introduction

As usual, $\zeta \left(s\right)$ is denoted by $s=\sigma +it$, the Riemann zeta-function, which, for $\sigma >1$, is defined by

$$\zeta \left(s\right)=\sum _{m=1}^{\infty }\frac{1}{m^s},$$

and has the meromorphic continuation of the whole complex plane with a unique simple pole at the point of $s=1$ with a residue of 1. In the theory of the function of $\zeta \left(s\right)$, the modified Mellin transforms ${\Xi }_k\left(s\right)$ play an important role. For $k⩾0$ and $\sigma >\sigma \left(k\right)>1$, the functions ${\Xi }_k\left(s\right)$ are defined by

$${\Xi }_k\left(s\right)={\int }_1^{\infty }{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2k}{x}^{-s}\mathrm{d}x.$$

The functions ${\Xi }_k\left(s\right)$ were introduced in [1,2] and are applied for the investigation of the moments

$${\int }_1^T{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^{2k}\mathrm{d}t.$$

In general, ${\Xi }_k\left(s\right)$ are attractive analytic functions and are widely studied; see, for example, [3,4,5,6].

In [7], the approximation properties of the function ${\Xi }_1\left(s\right)$ were studied. Let $G=\{s\in \mathbb{C}:1/2<\sigma <1\}$. $\mathcal{H}\left(G\right)$ is denoted by the space of analytic functions on G endowed with the topology of uniform convergence on compacta, and by $\mathrm{meas}A$ the Lebesgue measure of a measurable set $A\subset \mathbb{R}$. Then, in [7], the following theorem is proven.

Theorem 1. 

There exists a closed, non-empty set $F\subset \mathcal{H}\left(G\right)$, such that, for every compact set $K\subset G$, function $f\left(s\right)\in F$, and $\epsilon >0$,

$$\underset{T\to \infty }{lim\; inf}\frac{1}{T}\mathrm{meas}\left\{\tau \in [0,T]:\underset{s\in K}{sup}\left|{\Xi }_1(s+i\tau )-f\left(s\right)\right|<\epsilon \right\}>0.$$

Moreover, the limit

$$\underset{T\to \infty }{lim}\frac{1}{T}\mathrm{meas}\left\{\tau \in [0,T]:\underset{s\in K}{sup}\left|{\Xi }_1(s+i\tau )-f\left(s\right)\right|<\epsilon \right\}$$

exists and is positive for all but, at most, is a countable number $\epsilon >0$.

Theorem 1 is of continuous type, $\tau$ in the shifts ${\Xi }_1(s+i\tau )$ takes arbitrary real values. The aim of this paper is to obtain a discrete version of Theorem 1 with shifts ${\Xi }_1(s+ikh)$, where $h>0$ is a fixed number and $k\in \mathbb{N}\cup \left\{0\right\}\stackrel{def}{=}{\mathbb{N}}_0$.

$\#A$ denotes the cardinality of a set $A\subset \mathbb{R}$. For brevity, we write $\Xi \left(s\right)$ in place of ${\Xi }_1\left(s\right)$. Let N run over the set ${\mathbb{N}}_0$.

Theorem 2. 

For every $h>0$, there exists a closed non-empty set $F_h\subset \mathcal{H}\left(G\right)$ such that, for every compact set $K\subset G$, function $f\left(s\right)\in F_h$, and $\epsilon >0$,

$$\underset{N\to \infty }{lim\; inf}\frac{1}{N+1}\#\left\{0⩽k⩽N:\underset{s\in K}{sup}\left|\Xi (s+ikh)-f\left(s\right)\right|<\epsilon \right\}>0.$$

Moreover, the limit

$$\underset{N\to \infty }{lim}\frac{1}{N+1}\#\left\{0⩽k⩽N:\underset{s\in K}{sup}\left|\Xi (s+ikh)-f\left(s\right)\right|<\epsilon \right\}$$

exists and is positive for all but at most countably many $\epsilon >0$.

Theorem 2 shows that the set of discrete shifts $\Xi (s+ikh)$ approximating with a given accuracy the function $f\left(s\right)\in F_h$ is infinite.

We note that Theorem 2 has a certain advantage against Theorem 1 because it is easier to detect discrete approximating shifts.

Unfortunately, the sets F and $F_h$ in Theorems 1 and 2, respectively, are not identified; however, Theorems 1 and 2 show good approximation properties of the function $\Xi \left(s\right)$. In some sense, Theorems 1 and 2 recall universality theorems for the function $\zeta \left(s\right)$. In this case, F and $F_h$ are sets of non-vanishing analytic functions on G; see, for example, [8,9].

Here, we prove that the set $F_h$ is a support of a certain $\mathcal{H}\left(G\right)$-valued random element defined in terms of $\Xi \left(s\right)$. The distribution of that random element is the limit measure in a probabilistic discrete limit theorem for the function $\Xi \left(s\right)$. $\mathcal{B}\left(\mathcal{X}\right)$ denotes the Borel $\sigma$-field of the space $\mathcal{X}$, by $\underset{}{\overset{\mathrm{W}}{\to }}$ the weak convergence of probability measures, and, for $A\in \mathcal{B}\left(\mathcal{H}\right(G\left)\right)$, define

$$P_{N,h}\left(A\right)=\frac{1}{N+1}\#\left\{0⩽k⩽N:\Xi (s+ikh)\in A\right\}.$$

Theorem 3. 

For every fixed $h>0$, on $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$, there exists a probability measure $P_h$ such that $P_{N,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_h$.

2. Some Lemmas

Let $a>1$ be a fixed number. Define the set

$${\Omega }_a=\prod _{u\in [1,a]}{\gamma }_u,$$

where ${\gamma }_u=\{s\in \mathbb{C}:|s|=1\}$ for all $u\in [1,a]$. As a Cartesian product of compact sets, the torus ${\Omega }_a$ is a compact topological Abelian group. Let $\omega =\{{\omega }_u:u\in [1,a]\}$ be elements of ${\Omega }_a$.

For $A\in \mathcal{B}\left({\Omega }_a\right)$ and $h>0$, define

$$Q_{N,a,h}\left(A\right)=\frac{1}{N+1}\#\left\{0⩽k⩽N:\left(u^{-ikh}:u\in [1,a]\right)\in A\right\}.$$

Lemma 1. 

On $({\Omega }_a,\mathcal{B}\left({\Omega }_a\right))$, there exists a probability measure $Q_{a,h}$ such that $Q_{N,a,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{Q}_{a,h}$.

Proof. 

We apply the Fourier transform method. Let $F_{Q_{N,a,h}}\left(\underline{k}\right)$, $\underline{k}=(k_u:k_u\in \mathbb{Z},u\in [1,a])$, be the Fourier transform of $Q_{N,a,h}$, i.e.,

$$F_{Q_{N,a,h}}\left(\underline{k}\right)={\int }_{{\Omega }_a}\underset{u\in [1,a]}{{\prod }^{*}}{\omega }_u^{k_u}\mathrm{d}{Q}_{N,a,h},$$

where “*” shows that only a finite number of integers $k_u$ are non-zero. Thus, by the definition of $Q_{N,a,h}$,

$$F_{Q_{N,a,h}}\left(\underline{k}\right)=\frac{1}{N+1}\sum _{k=0}^N\underset{u\in [1,a]}{{\prod }^{*}}{u}^{-ikh{k}_u}=\frac{1}{N+1}\sum _{k=0}^{N}exp\left\{-ikh\underset{u\in [1,a]}{{\sum }^{*}}{k}_{u}logu\right\}.$$

(1)

If

$$\underset{u\in [1,a]}{{\sum }^{*}}{k}_{u}logu=\frac{2\pi r}{h},r\in \mathbb{Z},$$

(2)

then

$$F_{Q_{N,a,h}}\left(\underline{k}\right)=1.$$

(3)

If $\underline{k}=(k_u:u\in [1,a])$ does not satisfy (2), then using the formula of geometric progression gives

$$F_{Q_{N,a,h}}\left(\underline{k}\right)=\frac{1}{N+1}\frac{1-A^{N+1}(\underline{k},h)}{1-A(\underline{k},h)},$$

where

$$A(\underline{k},h)=exp\left\{-ih\underset{u\in [1,a]}{{\sum }^{*}}{k}_{u}logu\right\}.$$

Therefore, by (3),

$$\underset{N\to \infty }{lim}{F}_{Q_{N,a,h}}\left(\underline{k}\right)=\left\{\begin{array}{cc}1\hfill & \mathrm{if}\underline{k}\mathrm{satisfies}(2),\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(4)

This shows that $Q_{N,a,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{Q}_{a,h}$, where the Fourier transform of $Q_{a,h}$ is the right-hand side of (4). □

We apply Lemma 1 for the proof of a limit theorem for one integral sum. For $x,y\in [1,\infty ]$ and fixed $\theta >1/2$, define

$$v(x,y)=exp\left\{-{\left(\frac{x}{y}\right)}^{\theta }\right\},$$

and

$${\Xi }_{a,y}\left(s\right)={\int }_1^{a}g(x,y)x^{-s}\mathrm{d}x,$$

where

$$g(x,y)={\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2}v(x,y).$$

${\mathcal{Z}}_{n,a,y}\left(s\right)$ denotes the integral sum of the function $g(x,y)x^{-s}$ over the interval $[1,a]$, i.e.,

$${\mathcal{Z}}_{n,a,y}\left(s\right)=\frac{a-1}{n}\sum _{l=1}^{n}g({\xi }_l,y){\xi }_l^{-s},$$

where ${\xi }_l\in [x_{l-1},x_l]$ and $x_l=1+((a-1)/n)l$, $n\in \mathbb{N}$. For $A\in \mathcal{B}\left(\mathcal{H}\right(G\left)\right)$, define

$$P_{N,n,a,y,h}\left(A\right)=\frac{1}{N+1}\#\left\{0⩽k⩽N:{\mathcal{Z}}_{n,a,y}(s+ikh)\in A\right\}.$$

Lemma 2. 

On $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$, there exists a probability measure $P_{n,a,y,h}$ such that $P_{N,n,a,y,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_{n,a,y,h}$.

Proof. 

We apply the following simple remark on the preservation of weak convergence under continuous mappings. Let $w:{\mathcal{X}}_1\to {\mathcal{X}}_2$ be a $(\mathcal{B}\left({\mathcal{X}}_1\right),\mathcal{B}\left({\mathcal{X}}_2\right))$-measurable mapping. Then, every probability measure P on $({\mathcal{X}}_1,\mathcal{B}\left({\mathcal{X}}_1\right))$ induces the unique probability measure $P{w}^{-1}$ on $({\mathcal{X}}_2,\mathcal{B}\left({\mathcal{X}}_2\right))$ defined by $P{w}^{-1}\left(A\right)=P\left(w^{-1}A\right)$, $A\in \mathcal{B}\left({\mathcal{X}}_2\right)$. If the mapping w is continuous, then the weak convergence is preserved. Thus, if $P_n\underset{n\to \infty }{\overset{\mathrm{W}}{\to }}P$ in the space ${\mathcal{X}}_1$, then $P_n{w}^{-1}\underset{n\to \infty }{\overset{\mathrm{W}}{\to }}P{w}^{-1}$ in the space ${\mathcal{X}}_2$ as well [10].

Define the mapping $w_{n,a,y}:{\Omega }_a\to \mathcal{H}\left(G\right)$ by the formula

$$w_{n,a,y}\left(\omega \right)=\frac{a-1}{n}\sum _{l=1}^{n}g({\xi }_l,y){\xi }_l^{-s}{\omega }_{{\xi }_l}.$$

Since the above sum is finite, the mapping $w_{n,a}$ is continuous in the product topology. Moreover,

$$w_{n,a,y}\left(u^{-ikh}:u\in [1,a]\right)=\frac{a-1}{n}\sum _{l=1}^{n}g({\xi }_l,y){\xi }_l^{-s-ikh}={\mathcal{Z}}_{n,a,y}(s+ikh).$$

Hence, $P_{N,n,a,y,h}=Q_{N,a,h}{w}_{n,a,y}^{-1}$. Therefore, the above remark, continuity of $w_{n,a,y}$ and Lemma 1 show that $P_{N,n,a,y,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_{n,a,y,h}=Q_{a,h}{w}_{n,a,y}^{-1}$. □

The next step consists of the passage from ${\mathcal{Z}}_{n,a,y}\left(s\right)$ to ${\Xi }_{a,y}\left(s\right)$ in Lemma 2. For this, one statement on convergence in distribution $(\underset{}{\overset{\mathcal{D}}{\to }})$ of $\mathcal{H}\left(G\right)$-valued random elements is useful, and we recall it. There exists a sequence $\{K_l:l\in \mathbb{N}\}\subset G$ of compact embedded sets such that G is union of sets $K_l$, and every compact $K\subset G$ lies in some set $K_l$. Then

$$\rho (g_1,g_2)=\sum _{l=1}^{\infty }{2}^{-l}\frac{{sup}_{s\in K_l}\left|g_1\left(s\right)-g_2\left(s\right)\right|}{1+{sup}_{s\in K_l}\left|g_1\left(s\right)-g_2\left(s\right)\right|},g_1,g_2\in \mathcal{H}\left(G\right),$$

is a metric in $\mathcal{H}\left(G\right)$ which induces the topology of uniform convergence on compacta.

Lemma 3. 

Suppose that X, $Y_N$ and $Y_{Nl}$ are $\mathcal{H}\left(G\right)$-valued random elements defined on the same probability space with measure P such that, for $l\in \mathbb{N}$,

$$X_{Nl}\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}{X}_l,$$

and

$$X_l\underset{l\to \infty }{\overset{\mathcal{D}}{\to }}X.$$

Moreover, let, for every $\epsilon >0$,

$$\underset{l\to \infty }{lim}\underset{N\to \infty }{lim\; sup}P\left\{\rho (X_{Nl},Y_N)⩾\epsilon \right\}=0.$$

Then, $Y_N\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}X$.

Proof. 

Since the space $\mathcal{H}\left(G\right)$ is separable, the lemma is a particular case of a general theorem on convergence in distribution; see, for example, Theorem 4.2 of [10]. □

An application of Lemma 3 requires the following statement:

Lemma 4. 

The equality

$$\underset{n\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\rho \left({\mathcal{Z}}_{n,a,y}(s+ikh),{\Xi }_{a,y}(s+ikh)\right)=0$$

holds for every fixed $h>0$.

Proof. 

In view of the definition of the metric $\rho$, it is suffice to show that, for arbitrary compact set $K\subset G$,

$$\underset{n\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|{\mathcal{Z}}_{n,a,y}(s+ikh)-{\Xi }_{a,y}(s+ikh)\right|=0.$$

(5)

Let L be a simple closed contour lying in G and enclosing a compact set $K\subset G$. Then, by the integral Cauchy formula,

$$\underset{s\in K}{sup}\left|{\mathcal{Z}}_{n,a,y}(s+ikh)-{\Xi }_{a,y}(s+ikh)\right|{\ll }_L{\int }_L\left|{\mathcal{Z}}_{n,a,y}(z+ikh)-{\Xi }_{a,y}(z+ikh)\right|\left|\mathrm{d}z\right|,$$

where $a{\ll }_{\xi }b$, $b>0$, means that there exists a constant $c=c\left(\xi \right)>0$ such that $\left|a\right|⩽cb$. Hence,

$$\begin{array}{cc}\hfill \frac{1}{N+1}& \sum _{k=0}^N\underset{s\in K}{sup}\left|{\mathcal{Z}}_{n,a,y}(s+ikh)-{\Xi }_{a,y}(s+ikh)\right|\hfill \\ & {\ll }_L{\int }_L\left|\mathrm{d}z\right|\left(\frac{1}{N+1}\sum _{k=0}^N\left|{\mathcal{Z}}_{n,a,y}(z+ikh)-{\Xi }_{a,y}(z+ikh)\right|\right).\hfill \end{array}$$

(6)

By the Cauchy–Schwarz inequality,

$$\begin{array}{cc}\hfill \frac{1}{N+1}& \sum _{k=0}^N\left|{\mathcal{Z}}_{n,a,y}(z+ikh)-{\Xi }_{a,y}(z+ikh)\right|\hfill \\ & ⩽{\left(\frac{1}{N+1}\sum _{k=0}^N{\left|{\mathcal{Z}}_{n,a,y}(z+ikh)-{\Xi }_{a,y}(z+ikh)\right|}^2\right)}^{1/2}.\hfill \end{array}$$

(7)

Obviously,

$$\begin{array}{cc}\hfill {\left|{\mathcal{Z}}_{n,a,y}(z+ikh)-{\Xi }_{a,y}(z+ikh)\right|}^2=& {\mathcal{Z}}_{n,a,y}(z+ikh)\overline{{\mathcal{Z}}_{n,a,y}(z+ikh)}\hfill \\ & -{\mathcal{Z}}_{n,a,y}(z+ikh)\overline{{\Xi }_{a,y}(z+ikh)}\hfill \\ & -\overline{{\mathcal{Z}}_{n,a,y}(z+ikh)}{\Xi }_{a,y}(z+ikh)\hfill \\ & +{\Xi }_{a,y}(z+ikh)\overline{{\Xi }_{a,y}(z+ikh)},\hfill \end{array}$$

(8)

where $\overline{z}$ denotes the complex conjugate of $z\in \mathbb{C}$. By the definition of ${\mathcal{Z}}_{n,a,y}\left(s\right)$,

$$\begin{array}{cc}\hfill {\mathcal{Z}}_{n,a,y}& (z+ikh)\overline{{\mathcal{Z}}_{n,a,y}(z+ikh)}\hfill \\ & ={\left(\frac{a-1}{n}\right)}^2\underset{log({\xi }_{l_1}/{\xi }_{l_2})=2\pi r/h}{\sum _{l_1=1}^n\sum _{l_2=1}^n}g({\xi }_{l_1},y)g({\xi }_{l_2},y){\xi }_{l_1}^{-z}{\xi }_{l_2}^{-\overline{z}}\hfill \\ & +{\left(\frac{a-1}{n}\right)}^2\underset{log({\xi }_{l_1}/{\xi }_{l_2})\ne 2\pi r/h}{\sum _{l_1=1}^n\sum _{l_2=1}^n}g({\xi }_{l_1},y)g({\xi }_{l_2},y){\xi }_{l_1}^{-z}{\xi }_{l_2}^{-\overline{z}}{\left(\frac{{\xi }_{l_1}}{{\xi }_{l_2}}\right)}^{-ikh},\hfill \end{array}$$

where $r\in \mathbb{Z}$ is arbitrary. Therefore,

$$\begin{array}{cc}\hfill & \frac{1}{N+1}\sum _{k=0}^N{\mathcal{Z}}_{n,a,y}(z+ikh)\overline{{\mathcal{Z}}_{n,a,y}(z+ikh)}\hfill \\ & ={\left(\frac{a-1}{n}\right)}^2\underset{log({\xi }_{l_1}/{\xi }_{l_2})=2\pi r/h}{\sum _{l_1=1}^n\sum _{l_2=1}^n}g({\xi }_{l_1},y)g({\xi }_{l_2},y){\xi }_{l_1}^{-z}{\xi }_{l_2}^{-\overline{z}}\hfill \\ & +O\left({\left(\frac{a-1}{n}\right)}^2\frac{1}{N}\underset{log({\xi }_{l_1}/{\xi }_{l_2})\ne 2\pi r/h}{\sum _{l_1=1}^n\sum _{l_2=1}^n}g({\xi }_{l_1},y)g({\xi }_{l2},y){\xi }_{l_1}^{-\mathrm{Re}z}{\xi }_{l_2}^{-\mathrm{Re}z}{\left|1-{\left(\frac{{\xi }_{l_1}}{{\xi }_{l_2}}\right)}^{-ih}\right|}^{-1}\right).\hfill \end{array}$$

Since

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{\left(\frac{a-1}{n}\right)}^2& \underset{log({\xi }_{l_1}/{\xi }_{l_2})=2\pi r/h}{\sum _{l_1=1}^n\sum _{l_2=1}^n}g({\xi }_{l_1},y)g({\xi }_{l_2},y){\xi }_{l_1}^{-z}{\xi }_{l_2}^{-\overline{z}}\hfill \\ & =\underset{log(x_1/x_2)=2\pi r/h}{{\int }_1^a{\int }_1^a}g(x_1,y)g(x_2,y)x_1^{-z}{x}_2^{-\overline{z}}\mathrm{d}{x}_1\mathrm{d}{x}_2=0,\hfill \end{array}$$

from this we obtain that, for all $z\in L$,

$$\underset{n\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N{\mathcal{Z}}_{n,a,y}(z+ikh)\overline{{\mathcal{Z}}_{n,a,y}(z+ikh)}=0.$$

(9)

By the definition of ${\Xi }_{a,y}\left(s\right)$, for all $z\in L$, we have

$$\begin{array}{cc}\hfill & \frac{1}{N+1}\sum _{k=0}^N{\Xi }_{a,y}(z+ikh)\overline{{\Xi }_{a,y}(z+ikh)}\hfill \\ & =\frac{1}{N+1}\sum _{k=0}^N{\int }_1^a{\int }_1^{a}g(x_1,y)g(x_2,y)x_1^{-z-ikh}{x}_2^{-\overline{z}+ikh}\mathrm{d}{x}_1\mathrm{d}{x}_2\hfill \\ & =\frac{1}{N+1}\sum _{k=0}^N\left(\underset{log(x_1/x_2)=2\pi r/h}{{\int }_1^a{\int }_1^a}+\underset{log(x_1/x_2)\ne 2\pi r/h}{{\int }_1^a{\int }_1^a}\right)g(x_1,y)g(x_2,y)x_1^{-z-ikh}{x}_2^{-\overline{z}+ikh}\mathrm{d}{x}_1\mathrm{d}{x}_2\hfill \\ & =\frac{1}{N+1}\underset{log(x_1/x_2)\ne 2\pi r/h}{{\int }_1^a{\int }_1^a}g(x_1,y)g(x_2,y)x_1^{-z}{x}_2^{-\overline{z}}\left(1-{\left(\frac{x_1}{x_2}\right)}^{-i(N+1)h}\right)\hfill \\ & \times {\left(1-{\left(\frac{x_1}{x_2}\right)}^{-ih}\right)}^{-1}\frac{1}{i}\mathrm{d}{x}_1\mathrm{d}{x}_2,\hfill \end{array}$$

where $r\in \mathbb{Z}$. Therefore,

$$\underset{n\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N{\Xi }_{a,y}(z+ikh)\overline{{\Xi }_{a,y}(z+ikh)}=0.$$

(10)

Since the sum of the last two terms in (7) is estimated as

$$\ll {\left(\sum _{k=0}^N{\left|{\mathcal{Z}}_{n,a,y}(z+ikh)\right|}^2\sum _{k=0}^N{\left|{\Xi }_{a,y}(z+ikh)\right|}^2\right)}^{1/2},$$

equality (5) follows from (6)–(10). □

For $A\in \mathcal{B}\left(\mathcal{H}\right(G\left)\right)$, define

$$P_{N,a,y,h}\left(A\right)=\frac{1}{N+1}\#\left\{0⩽k⩽N:{\Xi }_{a,y}(s+ikh)\in A\right\}.$$

Lemma 5. 

For every fixed $h>0$, on $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$, there exists a probability measure $P_{a,y,h}$ such that $P_{N,a,y,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_{a,y,h}$.

Proof. 

Let ${\theta }_{N,h}$ be a random variable defined on a certain probability space with measure P, and having the distribution

$$P\left\{{\theta }_{N,h}=kh\right\}=\frac{1}{N+1},k=0,1,\cdots ,N.$$

$X_{n,a,y,h}$ denotes the $\mathcal{H}\left(G\right)$-valued random element with the distribution $P_{n,a,y,h}$, where $P_{n,a,y,h}$ is the measure from Lemma 2, and define the $\mathcal{H}\left(G\right)$-valued random element

$$X_{N,n,a,y,h}=X_{N,n,a,y,h}\left(s\right)={\mathcal{Z}}_{n,a,y}(s+i{\theta }_{N,h}).$$

Then, in view of Lemma 2, we have

$$X_{N,n,a,y,h}\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}{X}_{n,a,y,h}.$$

(11)

Consider the sequence $\{P_{n,a,y,h}:n\in \mathbb{N}\}$. Let $K_l$ be the sets from the definition of the metric $\rho$. Then, applying the integral Cauchy formula and (9), we find that

$$\underset{n\in \mathbb{N}}{sup}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\mathcal{Z}}_{n,a,y}(s+ikh)\right|⩽C_{l,a,y,h}<\infty .$$

Fix $\epsilon >0$ and define $V_l=V_{l,a,y,h}=2^l{\epsilon }^{-1}{C}_{l,a,y,h}$. Then, using (11),

$$\begin{array}{cc}\hfill P\left\{\underset{s\in K_l}{sup}\left|X_{n,a,y,h}\left(s\right)\right|⩾V_l\right\}& =\underset{N\to \infty }{lim\; sup}P\left\{\underset{s\in K_l}{sup}\left|X_{N,n,a,y,h}\left(s\right)\right|⩾V_l\right\}\hfill \\ & ⩽\underset{n\in \mathbb{N}}{sup}\underset{N\to \infty }{lim\; sup}\frac{1}{V_l(N+1)}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\mathcal{Z}}_{n,a,y}(s+ikh)\right|⩽\frac{\epsilon }{2^l}\hfill \end{array}$$

for all $n,l\in \mathbb{N}$. Hence, taking

$$K=K\left(\epsilon \right)=\left\{g\in \mathcal{H}\left(G\right):\underset{s\in K_l}{sup}\left|g\left(s\right)\right|⩽V_l,l\in \mathbb{N}\right\},$$

we have

$$P\left\{X_{n,a,y,h}\in K\right\}=1-P\left\{X_{n,a,y,h}\notin K\right\}>1-\epsilon \sum _{l=1}^{\infty }{2}^{-l}=1-\epsilon$$

for all $n\in \mathbb{N}$. Since the set K is compact in the space $\mathcal{H}\left(G\right)$, this shows that the sequence $\left\{P_{n,a,y,h}\right\}$ is tight. Therefore, by the Prokhorov theorem; see, for example, [10], the sequence $\left\{P_{n,a,y,h}\right\}$ is relatively compact. Thus, there exists a subsequence $\left\{P_{n_r,a,y,h}\right\}$ weakly convergent to a certain probability measure $P_{a,y,h}$ on $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$ as $r\to \infty$. In other words,

$$X_{n_r,a,y,h}\underset{r\to \infty }{\overset{\mathcal{D}}{\to }}{P}_{a,y,h}.$$

(12)

Define one more $\mathcal{H}\left(G\right)$-valued random element

$$Y_{N,a,y,h}=Y_{N,a,y,h}\left(s\right)={\Xi }_{a,y}(s+i{\theta }_{N,h}).$$

Then, Lemma 4 implies that, for every $\epsilon >0$,

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}& \underset{N\to \infty }{lim\; sup}P\left\{\rho \left(Y_{N,a,y,h},X_{n_r,a,y,h}\right)⩾\epsilon \right\}\hfill \\ & ⩽\underset{n\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{\epsilon (N+1)}\sum _{k=0}^N\rho \left({\mathcal{Z}}_{n,a,y}(s+ikh),{\Xi }_{a,y}(s+ikh)\right)=0.\hfill \end{array}$$

(13)

Now, in view of (11)–(13), we may apply Lemma 3 for the random elements $Y_{N,a,y,h}$, $X_{N,n_r,a,y,h}$ and $X_{n_r,a,y,h}$. Then, we have the relation

$$Y_{N,a,y,h}\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}{P}_{a,y,h},$$

i.e., $P_{N,a,y,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_{a,y,h}$. □

Now, we are ready to prove a discrete limit lemma for the function

$${\Xi }_y\left(s\right)={\int }_1^{\infty }g(x,y)x^{-s}\mathrm{d}x.$$

Since $\zeta (1/2+it)\ll t^{1/6}$, $t⩾1$, and $v(x,y)$ decreases exponentially, the integral for ${\Xi }_y\left(s\right)$ is absolutely convergent for $\sigma >{\sigma }_0$ with arbitrary finite ${\sigma }_0$.

For $A\in \mathcal{B}\left(\mathcal{H}\right(G\left)\right)$, define

$$P_{N,y,h}\left(A\right)=\frac{1}{N+1}\#\left\{0⩽k⩽N:{\Xi }_y(s+ikh)\in A\right\}.$$

Lemma 6. 

For every fixed $h>0$, on $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$, there exists a probability measure $P_{y,h}$ such that $P_{N,y,h}\underset{N\to \infty }{\overset{\mathrm{W}}{\to }}{P}_{y,h}$.

Proof. 

Let ${\theta }_{N,h}$ be the same as in the proof of Lemma 5. Define

$$Y_{N,y,h}=Y_{N,y,h}\left(s\right)={\Xi }_y(s+i{\theta }_{N,h}),$$

and $X_{a,y,h}$ denotes the $\mathcal{H}\left(G\right)$-valued random element with distribution $P_{a,y,h}$. Then, by Lemma 5,

$$Y_{N,a,y,h}\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}{X}_{a,y,h}.$$

(14)

The integral Cauchy formula and (10) lead to

$$\underset{a⩾1}{sup}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\Xi }_{a,y}(s+ikh)\right|⩽C_{l,y,h}<\infty .$$

Therefore, taking $V_l=V_{l,y,h}=2^l{\epsilon }^{-1}{C}_{l,y,h}$, we find by (14) that

$$P\left\{\underset{s\in K_l}{sup}\left|X_{a,y,h}\left(s\right)\right|⩾V_l\right\}<\underset{a⩾1}{sup}\frac{1}{V_l(N+1)}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\Xi }_{a,y}(s+ikh)\right|⩽\frac{\epsilon }{2^l}$$

for all $a⩾1$ and $l\in \mathbb{N}$. This shows that, for $a⩾1$,

$$P\left\{X_{a,y,h}\in K\right\}⩾1-\epsilon ,$$

where

$$K=\left\{g\in \mathcal{H}\left(G\right):\underset{s\in K_l}{sup}\left|g\left(s\right)\right|⩽V_{l,y,h},l\in \mathbb{N}\right\}.$$

This means that the family of probability measures $\{P_{a,y,h}:a⩾1\}$ is tight. Hence, there exists a sequence $\left\{P_{a_r,y,h}\right\}\subset \left\{P_{a,y,h}\right\}$ weakly convergent to a certain probability measure $P_{y,h}$ as $r\to \infty$. Thus,

$$X_{a_r,y,h}\underset{r\to \infty }{\overset{\mathcal{D}}{\to }}{P}_{y,h}.$$

(15)

It remains to show the nearestness in the mean of ${\Xi }_{a,y}\left(s\right)$ and ${\Xi }_y\left(s\right)$. We have that, for a compact set $K\subset D$ and fixed $y>0$, $h>0$,

$${\Xi }_y(s+ikh)-{\Xi }_{a,y}(s+ikh)={\int }_a^{\infty }g(x,y)x^{-s-ikh}\mathrm{d}x{\ll }_y{\int }_a^{\infty }g(x,y)x^{-1/2}\mathrm{d}x=o_y\left(1\right)$$

as $a\to \infty$. From this, we have

$$\underset{a\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|{\Xi }_y(s+ikh)-{\Xi }_{a,y}(s+ikh)\right|=0,$$

and

$$\underset{a\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\rho \left({\Xi }_y(s+ikh),{\Xi }_{a,y}(s+ikh)\right)=0.$$

The latter equality, relations (14) and (15) together with Lemma 3 prove the lemma. □

To obtain a limit theorem for the function $\Xi \left(s\right)$, we use the integral representation for the function ${\Xi }_y\left(s\right)$. Define

$$a_y\left(s\right)=\frac{s}{\theta }\Gamma \left(\frac{s}{\theta }\right)y^s,$$

where $\Gamma \left(s\right)$ is the Euler gamma-function, and $\theta$ is from the definition of $v(x,y)$.

Lemma 7. 

For $s\in D$, the integral representation

$${\Xi }_y\left(s\right)=\frac{1}{2\pi i}{\int }_{\theta -i\infty }^{\theta +i\infty }\Xi (s+z)a_y\left(z\right)\frac{\mathrm{d}z}{z}$$

is valid.

Proof. 

The lemma is Lemma 9 proved in [7]. □

In addition, we need a discrete mean square estimate for $\Xi \left(s\right)$.

Lemma 8. 

Suppose that σ, $1/2<\sigma <1$, and $h>0$ are fixed, and $\tau \in \mathbb{R}$. Then, for every ${\epsilon }_1>0$,

$$\sum _{k=0}^N{\left|\Xi (\sigma +ikh+i\tau )\right|}^2{\ll }_{\sigma ,h,{\epsilon }_1}\left(N\right(1+{\left|\tau \right|\left)\right)}^{2-2\sigma +{\epsilon }_1}.$$

Proof. 

It is well known [4] that, for fixed $1/2<\sigma <1$, and any ${\epsilon }_1>0$,

$${\int }_0^T{\left|\Xi (\sigma +it)\right|}^2\mathrm{d}t{\ll }_{\sigma ,{\epsilon }_1}{T}^{2-2\sigma +{\epsilon }_1}.$$

From this, we find

$$\begin{array}{cc}\hfill {\int }_0^T{\left|\Xi (\sigma +it+i\tau )\right|}^2\mathrm{d}t& ={\int }_{\tau }^{T+\tau }{\left|\Xi (\sigma +it)\right|}^2\mathrm{d}t⩽2{\int }_0^{T+\left|\tau \right|}{\left|\Xi (\sigma +it)\right|}^2\mathrm{d}t\hfill \\ & {\ll }_{\sigma ,{\epsilon }_1}{(T+|\tau \left|\right)}^{2-2\sigma +{\epsilon }_1}.\hfill \end{array}$$

(16)

The latter estimate together with integral Cauchy formula gives

$${\int }_0^T{\left|{\Xi }^{\prime }(\sigma +it+i\tau )\right|}^2\mathrm{d}t{\ll }_{\sigma ,{\epsilon }_1}{(T+|\tau \left|\right)}^{2-2\sigma +{\epsilon }_1}.$$

(17)

Now, we apply the Gallagher lemma; see, for example, Lemma 1.4 of [11], connecting continuous and discrete mean squares of certain functions. Thus, by (16) and (17),

$$\begin{array}{cc}\hfill \sum _{k=2}^N& {\left|\Xi (\sigma +ikh+i\tau )\right|}^2{\ll }_h{\int }_0^{Nh}{\left|\Xi (\sigma +it+i\tau )\right|}^2\mathrm{d}t\hfill \\ & +{\left({\int }_0^{Nh}{\left|\Xi (\sigma +it+i\tau )\right|}^2\mathrm{d}t{\int }_0^{Nh}{\left|{\Xi }^{\prime }(\sigma +it+i\tau )\right|}^2\mathrm{d}t\right)}^{1/2}{\ll }_{\sigma ,h,{\epsilon }_1}\left(N\right(1+{\left|\tau \right|\left)\right)}^{2-2\sigma +{\epsilon }_1}.\hfill \end{array}$$

(18)

Since [4]

$$\Xi (\sigma +it){\ll }_{{\epsilon }_1}{\left|t\right|}^{1-\sigma +{\epsilon }_1},$$

for $0⩽\sigma ⩽1$, $\left|t\right|⩾t_0$, and ${\epsilon }_1>0$,

$$\sum _{k=0}^1{\left|\Xi (\sigma +ikh+i\tau )\right|}^2{\ll }_{\sigma ,h,{\epsilon }_1}{(1+|\tau \left|\right)}^{2-2\sigma +{\epsilon }_1}.$$

Therefore, in view of (18),

$$\sum _{k=0}^N{\left|\Xi (\sigma +ikh+i\tau )\right|}^2{\ll }_{\sigma ,h,{\epsilon }_1}{\left(N(1+|\tau \left|\right)\right)}^{2-2\sigma +{\epsilon }_1}.$$

(19)

□

The next lemma gives an approximation of $\Xi \left(s\right)$ by ${\Xi }_y\left(s\right)$.

Lemma 9. 

The equality

$$\underset{y\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\rho \left(\Xi (s+ikh),{\Xi }_y(s+ikh)\right)=0$$

holds for all $h>0$.

Proof. 

It is suffice to show that, for compact sets $K\subset G$,

$$\underset{y\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|\Xi (s+ikh)-{\Xi }_y(s+ikh)\right|=0.$$

Let $K\subset G$ be an arbitrary fixed compact set. Fix $\epsilon >0$ such that, for all $s=\sigma +it\in K$, the inequalities $1/2+2\epsilon ⩽\sigma ⩽1-\epsilon$ would be satisfied. Then, for such $\sigma$,

$${\theta }_1\stackrel{def}{=}\sigma -\epsilon -\frac{1}{2}>0.$$

Let $\theta =1/2+\epsilon$ in Lemma 7. The point $z=1-s$ is a double pole, and $z=0$ is a simple pole of the function

$$\Xi (s+z)\widehat{a}\left(z\right),\widehat{a}\left(z\right)=\frac{a\left(z\right)}{z};$$

therefore, Lemma 7 and the residue theorem give

$${\Xi }_y\left(s\right)-\Xi \left(s\right)=\frac{1}{2\pi i}{\int }_{-{\theta }_2-i\infty }^{-{\theta }_2+i\infty }\Xi (s+z){\widehat{a}}_y\left(z\right)\mathrm{d}z+r_y\left(s\right)$$

(20)

where

$$r_y\left(s\right)=\underset{z=1-s}{\mathrm{Res}}\Xi (s+z)\widehat{a}\left(z\right).$$

It is known [4] that, for $\sigma >-3/4$,

$$\Xi \left(s\right)=\frac{1}{{(s-1)}^2}+\frac{a_1}{s-1}+E\left(1\right)\pi (s-1)+s(s+1)(s+2){\int }_1^{\infty }{G}_1\left(x\right)x^{-s-3}\mathrm{d}x,$$

where $a_1=2{\gamma }_0-log2\pi$, ${\gamma }_0$ is the Euler constant, $E\left(T\right)$ is defined by

$${\int }_0^T{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^2\mathrm{d}t=Tlog\frac{T}{2\pi }+(2{\gamma }_0-1)T+E\left(T\right),$$

$$G_1\left(T\right)={\int }_1^{T}G\left(T\right)\mathrm{d}t,G\left(T\right)={\int }_1^{T}E\left(T\right)\mathrm{d}t-\pi T.$$

Therefore,

$$r_y\left(s\right)={\left(\widehat{a}\left(z\right)\right)}^{\prime }{|}_{z=1-s}+a_1\widehat{a}(1-s).$$

(21)

Equality (20), for all $s\in K$ and $h>0$, gives

$$\begin{array}{cc}\hfill {\Xi }_y(s+ikh)& -\Xi (s+ikh)\hfill \\ \hfill =& \frac{1}{2\pi i}{\int }_{-\infty }^{\infty }\Xi \left(\sigma +it-\sigma +\frac{1}{2}+\epsilon +ikh+i\tau \right)\widehat{a}\left(\frac{1}{2}+\epsilon -\sigma +i\tau \right)\mathrm{d}\tau +r_y(s+ikh)\hfill \\ \hfill =& \frac{1}{2\pi i}{\int }_{-\infty }^{\infty }\Xi \left(\frac{1}{2}+\epsilon +ikh+i\tau \right)\widehat{a}\left(\frac{1}{2}+\epsilon -s+i\tau \right)\mathrm{d}\tau +r_y(s+ikh)\hfill \\ \hfill \ll & {\int }_{-\infty }^{\infty }\left|\Xi \left(\frac{1}{2}+\epsilon +ikh+i\tau \right)\right|\underset{s\in K}{sup}\left|\widehat{a}\left(\frac{1}{2}+\epsilon -s+i\tau \right)\right|\mathrm{d}\tau +\underset{s\in K}{sup}\left|r_y(s+ikh)\right|\hfill \end{array}$$

after writing $\tau$ in place of $t+\tau$. Hence,

$$\begin{array}{cc}\hfill \frac{1}{N+1}\sum _{k=0}^N& \underset{s\in K}{sup}\left|\Xi (s+ikh)-{\Xi }_y(s+ikh)\right|\hfill \\ \hfill \ll & {\int }_{-\infty }^{\infty }\left(\frac{1}{N+1}\sum _{k=0}^N\Xi \left(\frac{1}{2}+\epsilon +ikh+i\tau \right)\right)\underset{s\in K}{sup}\left|\widehat{a}\left(\frac{1}{2}+\epsilon -s+i\tau \right)\right|\mathrm{d}\tau \hfill \\ & +\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|r_y(s+ikh)\right|\stackrel{def}{=}{I}_1+I_2.\hfill \end{array}$$

(22)

The classical estimate

$$\Gamma (\sigma +it)\ll exp\{-c|t\left|\right\},c>0,$$

(23)

which is uniform in any fixed strip ${\sigma }_1⩽\sigma ⩽{\sigma }_2$ is well-known. Thus, for $s=\sigma +it\in K$,the definition of ${\widehat{a}}_y\left(s\right)$ implies

$$\begin{array}{cc}\hfill \widehat{a}\left(\frac{1}{2}+\epsilon -s+i\tau \right)& {\ll }_{\theta }{y}^{1/2+\epsilon -\sigma }\left|\Gamma \left(\frac{1}{\theta }\left(\frac{1}{2}+\epsilon -\sigma -it+i\tau \right)\right)\right|{\ll }_{\theta }{y}^{-\epsilon }exp\left\{-\frac{c}{\theta }|t-\tau |\right\}\hfill \\ & {\ll }_{\theta }{y}^{-\epsilon }exp\left\{\frac{c}{\theta }\left|t\right|\right\}exp\left\{-\frac{c}{\theta }\left|\tau \right|\right\}{\ll }_{\theta ,K}{y}^{-\epsilon }exp\{-c_1\left|\tau \right|\},c_1>0.\hfill \end{array}$$

Therefore, using Lemma 8, we obtain with ${\epsilon }_1=2\epsilon$

$$I_1{\ll }_{\theta ,K,h,\epsilon ,{\epsilon }_1}{y}^{-\epsilon }{N}^{-\epsilon +{\epsilon }_1/2}{\int }_{-\infty }^{\infty }exp\{-c_1\left|\tau \right|\left\}\right(1+{\left|\tau \right|)}^{(2-2\sigma +{\epsilon }_1)/2}\mathrm{d}\tau {\ll }_{\theta ,K,h,\epsilon }{y}^{-\epsilon }.$$

(24)

To estimate $I_2$, first we evaluate $r_y\left(s\right)$. By (21),

$$r_y\left(s\right)=\frac{y^{1-s}}{\theta }\Gamma \left(\frac{1-s}{\theta }\right)\left(\frac{1}{\theta }\frac{{\Gamma }^{\prime }((1-s)/\theta )}{\Gamma \left(\right(1-s)/\theta )}+logy+a_1\right).$$

Hence, in virtue of (23) and the estimate ${\Gamma }^{\prime }\left(s\right)/\Gamma \left(s\right)\ll log\left|s\right|$,

$$\begin{array}{cc}\hfill r_y\left(s\right)& {\ll }_{\theta }{y}^{1-\sigma }\left|\Gamma \left(\frac{1-\sigma }{\theta }+i\frac{t+kh}{\theta }\right)\right|\left(\frac{1}{\theta }\left|\frac{{\Gamma }^{\prime }((1-\sigma )/\theta -i(t+kh)/\theta )}{\Gamma \left(\right(1-\sigma )/\theta -i(t+kh)/\theta )}\right|+logy+1\right)\hfill \\ & {\ll }_{\theta }{y}^{1-\sigma }exp\left\{-\frac{c}{\theta }|t+kh|\right\}\left(log\left|\frac{t+ikh}{\theta }\right|+logy+1\right)\hfill \\ & {\ll }_{\theta ,K,\epsilon }{y}^{1/2-\epsilon }exp\{-c_{2}kh\},c_2>0.\hfill \end{array}$$

This shows that

$$I_2{\ll }_{\theta ,K,\epsilon }\frac{y^{1/2-\epsilon }}{N}\sum _{k=0}^{N}exp\{-c_{2}kh\}{\ll }_{\theta ,K,\epsilon ,h}{y}^{1/2-\epsilon }{N}^{-1}logN.$$

Therefore, in view of (22) and (24),

$$\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|\Xi (s+ikh)-{\Xi }_y(s+ikh)\right|{\ll }_{\theta ,K,\epsilon ,h}{y}^{-\epsilon }+y^{1/2-\epsilon }{N}^{-1}logN.$$

From this, we find that

$$\underset{y\to \infty }{lim}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K}{sup}\left|\Xi (s+ikh)-{\Xi }_y(s+ikh)\right|=0,$$

and the lemma is proved. □

Recall that $P_{y,h}$ is the limit measure in Lemma 6.

Lemma 10. 

The family of probability measures $\{P_{y,h}:y⩾1\}$ is tight.

Proof. 

Let $K_l\subset G$ be a arbitrary compact set from the definition of metric in $\mathcal{H}\left(G\right)$. Then, for every fixed $h>0$,

$$\begin{array}{cc}\hfill \frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\Xi }_y(s+ikh)\right|⩽& \frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|\Xi (s+ikh)-{\Xi }_y(s+ikh)\right|\hfill \\ & +\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|\Xi (s+ikh)\right|.\hfill \end{array}$$

(25)

Estimate (19), for fixed $1/2<\sigma <1$, gives

$$\frac{1}{N+1}\sum _{k=0}^N{\left|\Xi (s+ikh)\right|}^2{\ll }_{\sigma ,{\epsilon }_1,h}{N}^{1-2\sigma +{\epsilon }_1}.$$

This and the integral Cauchy formula lead to

$$\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|\Xi (s+ikh)\right|⩽C_{l,h}<\infty .$$

Therefore, by (25) and the proof of Lemma 9,

$$\underset{y⩾1}{sup}\underset{N\to \infty }{lim\; sup}\frac{1}{N+1}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|{\Xi }_y(s+ikh)\right|⩽C_{l,h}<\infty .$$

Fix $\epsilon >0$ and take $V_l=V_{l,h}=2^l{\epsilon }^{-1}{C}_{l,h}$. Moreover, let $Y_{y,h}$ be the $\mathcal{H}\left(G\right)$-valued random element having the distribution $P_{y,h}$. Then, by Lemma 6,

$$\begin{array}{cc}\hfill P\left\{\underset{s\in K_l}{sup}\left|Y_{y,h}\left(s\right)\right|⩾V_l\right\}& =\underset{N\to \infty }{lim\; sup}P\left\{\underset{s\in K_l}{sup}\left|Y_{N,y,h}\left(s\right)\right|⩾V_l\right\}\hfill \\ & <\underset{y⩾1}{sup}\underset{N\to \infty }{lim\; sup}\frac{1}{(N+1)V_l}\sum _{k=0}^N\underset{s\in K_l}{sup}\left|\Xi (s+ikh)\right|⩽\frac{\epsilon }{2^l}.\hfill \end{array}$$

Hence, for all $y⩾1$,

$$P\left\{Y_{y,h}\in K_l\right\}⩾1-\epsilon ,$$

where

$$K=\left\{g\in \mathcal{H}\left(G\right):\underset{s\in K_l}{sup}\left|g\left(s\right)\right|⩽V_l,l\in \mathbb{N}\right\},$$

and the lemma is proved. □

3. Proofs of Theorems

Proof of Theorem 3. 

Lemma 10 and Prokhorov’s theorem imply the relative compactness of the family $\{P_{y,h}:y⩾1\}$. Thus, there exists a sequence $\left\{P_{y_r,h}\right\}\subset \left\{P_{y,h}\right\}$, such that $P_{y_r,h}\underset{r\to \infty }{\overset{\mathrm{W}}{\to }}{P}_h$, where $P_h$ is a certain probability measure on $\left(\mathcal{H}\right(G),\mathcal{B}(\mathcal{H}\left(G\right)\left)\right)$. Thus, in the above notation,

$$Y_{y_r,h}\underset{r\to \infty }{\overset{\mathcal{D}}{\to }}{P}_h.$$

(26)

Define the $\mathcal{H}\left(G\right)$-valued random element

$${\Xi }_{N,h}={\Xi }_{N,h}\left(s\right)=\Xi (s+i{\theta }_{N,h}).$$

Then, for every $\epsilon >0$ and $y⩾1$,

$$0⩽\underset{N\to \infty }{lim\; sup}P\left\{\rho \left({\Xi }_{N,h},{\Xi }_{N,y,h}\right)⩾\epsilon \right\}⩽\underset{N\to \infty }{lim\; sup}\frac{1}{(N+1)\epsilon }\sum _{k=0}^N\rho \left(\Xi (s+ikh),{\Xi }_y(s+ikh)\right).$$

Thus, Lemma 9 shows that

$$\underset{y\to \infty }{lim}\underset{N\to \infty }{lim\; sup}P\left\{\rho \left({\Xi }_{N,h},{\Xi }_{N,y,h}\right)⩾\epsilon \right\}=0.$$

This equality, (26) and Lemmas 6 and 3 prove that

$${\Xi }_{N,h}\underset{N\to \infty }{\overset{\mathcal{D}}{\to }}{P}_h.$$

The theorem is proved. □

Proof of Theorem 2. 

Let $F_h$ denote the support of the limit measure $P_h$ in Theorem 3, i.e., $F_h$ is the minimal closed subset of the space $\mathcal{H}\left(G\right)$ such that $P_h\left(F_h\right)=1$. For every element $f\in F_h$ and every open neighbourhood D of f, we have $P_h\left(D\right)>0$. Clearly, $F_h\ne ⌀$.

For $f\in F_h$, let

$$G_{\epsilon }=\left\{g\in \mathcal{H}\left(G\right):\underset{s\in K}{sup}|g\left(s\right)-f\left(s\right)<\epsilon \right\}.$$

Then, by the above mentioned property of the support,

$$P_h\left(G_{\epsilon }\right)>0.$$

(27)

Therefore, Theorem 3 and the equivalent of weak convergence in terms of open sets; see, for example, Theorem 2.1 of [10], give

$$\underset{N\to \infty }{lim\; inf}{P}_{N,h}\left(G_{\epsilon }\right)⩾P_h\left(G_{\epsilon }\right)>0.$$

This, the definitions of $P_{N,h}$ and $G_{\epsilon }$ prove the first inequality of theorem.

Since the boundary $\partial G_{\epsilon }$ of the set $G_{\epsilon }$ lies in the set

$$\left\{g\in \mathcal{H}\left(G\right):\underset{s\in K}{sup}|g\left(s\right)-f\left(s\right)=\epsilon \right\},$$

we have $\partial G_{{\epsilon }_1}\cap \partial G_{{\epsilon }_2}=⌀$ for different positive ${\epsilon }_1$ and ${\epsilon }_2$. Thus, $P_h(\partial G_{\epsilon })>0$ for all but at most countably many $\epsilon >0$, i.e., $G_{\epsilon }$ is a continuity set of the measure $P_h$ for all but at most countably many $\epsilon >0$. Therefore, Theorem 3 and the equivalent of weak convergence in terms of continuity sets [10] and (27) show that

$$\underset{N\to \infty }{lim}{P}_{N,h}\left(G_{\epsilon }\right)=P_h\left(G_{\epsilon }\right)>0$$

for all but at most countably many $\epsilon >0$, and the definitions of $P_{N,h}$ and $G_{\epsilon }$ prove the second inequality of the theorem. □