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

Abstract

This paper is devoted to the approximation of a certain class of analytic functions by shifts $\mathcal{Z}(s+i\tau )$, $\tau \in \mathbb{R}$, of the modified Mellin transform $\mathcal{Z}\left(s\right)$ of the square of the Riemann zeta-function $\zeta (1/2+it)$. More precisely, we prove the existence of a closed non-empty set F such that there are infinitely many shifts $\mathcal{Z}(s+i\tau )$, which approximate a given analytic function from F with a given accuracy. In the proof, the weak convergence of measures in the space of analytic functions is applied. Then, the set F coincides with the support of a limit measure.

1. Introduction

Recall that the Riemann zeta-function $\zeta \left(s\right)$, $s=\sigma +it$, is defined, for $\sigma >1$, by the Dirichlet series

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

and is analytically continuable to the whole complex plane, except for the point $s=1$, which is a simple pole with residue 1. It is well known that the function $\zeta \left(s\right)$ has good approximation properties, its shifts $\zeta (s+i\tau )$, $\tau \in \mathbb{R}$, approximate every non-vanishing analytic function defined on the strip $\{s\in \mathbb{C}:1/2<\sigma <1\}$. On the other hand, in the theory of the function $\zeta \left(s\right)$, there exists several important unsolved problems. One of them is the moment problem on the asymptotic behavior as $T\to \infty$ for

$$\underset{0}{\overset{T}{\int }}{\left|\zeta (\sigma +it)\right|}^{2k} \mathrm{d}t, \sigma ⩾\frac{1}{2}, k>0.$$

Y. Motohashi introduced [1], see also [2], the modified Mellin transforms

$${\mathcal{Z}}_k\left(s\right)=\underset{1}{\overset{\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2k}{x}^{-s} \mathrm{d}x, k\in \mathbb{N},$$

and applied them for investigation of the latter problem. He first considered the case $k=2$. The integral for ${\mathcal{Z}}_k\left(s\right)$ [3] is absolutely convergent for $\sigma >1$ if $0⩽k⩽2$, and for $\sigma >(k+2)/4$ if $2⩽k⩽6$. Hence, the function ${\mathcal{Z}}_k\left(s\right)$ is analytic in the corresponding half-planes. Later, the Mellin transforms ${\mathcal{Z}}_k\left(s\right)$ with applications were studied in [4,5,6].

We give one example from [4]. Define $E_2\left(T\right)$ by

$$\underset{0}{\overset{T}{\int }}{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^4 \mathrm{d}t=T{P}_4(logT)+E_2\left(T\right)$$

with

$$P_4\left(x\right)=\sum _{j=0}^4{a}_j{x}^j, a_4=\frac{1}{2{\pi }^2}.$$

There exists a problem to estimate $E_2\left(T\right)$. In [4], using the mean square estimates for $\mathcal{Z}\left(s\right)$, it was obtained that, for $\epsilon >0$,

$$E_2\left(T\right){\ll }_{\epsilon }{T}^{2/3+\epsilon },$$

and this estimate is the best up to $\epsilon$. The notation $a{\ll }_{\epsilon }b$, $a\in \mathbb{C}$, $b>0$, means that there exists a positive constant $c=c\left(\epsilon \right)$ such that $\left|a\right|⩽cb$.

In function theory, much attention is devoted to the approximation of analytic functions. We recall some results related to number theory. S.N. Mergelyan obtained [7] a very deep result connected to polynomials. Suppose that K is a compact set with a connected complement, and $f\left(s\right)$ a continuous function on K, which is analytic inside of K. Mergelyan proved [7] the existence of a polynomial sequence uniformly convergent on K to the function $f\left(s\right)$. From this, it follows that, for any $\epsilon >0$, we can find a polynomial $p_{f,\epsilon }\left(s\right)$ satisfying

$$\underset{s\in K}{sup}\left|f\left(s\right)-p_{f,\epsilon }\left(s\right)\right|<\epsilon .$$

Thus, a function satisfying the above hypotheses can be approximated by a polynomial.

In 1975, it turned out that there exist functions that approximate a whole class of analytic functions. The first example of such functions is the Riemann zeta-function. S.M. Voronin proved [8] that if $0<r<1/4$, the function $f\left(s\right)\ne 0$ is continuous on the disc $\left|s\right|⩽r$, and analytic inside that disc, then, for any $\epsilon >0$, there is a real number $\tau =\tau \left(\epsilon \right)$ satisfying the inequality

$$\underset{\left|s\right|⩽r}{max}\left|f\left(s\right)-\zeta \left(s+\frac{3}{4}+i\tau \right)\right|<\epsilon .$$

This shows that a set of non-vanishing analytic functions defined in the strip $\{s\in \mathbb{C}:1/2<\sigma <1\}$ is approximated by shifts $\zeta (s+i\tau )$ of one and the same function. In other words, $\zeta \left(s\right)$ is universal with respect to the approximation of analytic functions. The Voronin universality theorem was reinforced and extended for other zeta-functions. We recall its last form, see [9,10,11,12]. Suppose that $K\subset D$ is a compact set having a connected complement, $f\left(s\right)$ is continuous, having no zeros on K and analytic in inside of K function. Then, for any positive $\epsilon$,

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

Here, $\mu$ stands for the Lebesgue measure on the line $\mathbb{R}$.

The proof of the Voronin theorem in [8] is based on the rearrangement theorem for series in Hilbert space. B. Bagchi proposed [12] a new original probabilistic method that uses weak convergence of measures in the space of analytic functions. The Bagchi method was developed in [9,10]. Other results on the universality of zeta-functions are discussed in a survey paper [13]. We notice that an idea of application probabilistic methods in the theory of $\zeta \left(s\right)$ was proposed by H. Bohr and B. Jessen. In [14,15], they obtained the existence of the limit

$$\underset{T\to \infty }{lim}\frac{1}{T}\mathrm{J}\{t\in [0,T]:\zeta (\sigma +it)\in \mathrm{R}\}$$

for every rectangle $\mathrm{R}\subset \mathbb{C}$ with edges parallel to the axis and $\sigma >1/2$. Here, $\mathrm{J}$ denotes the Jordan measure on $\mathbb{R}$. A modern version of the Bohr–Jessen theorem in terms of weak convergence is presented in [9].

In general, for description value distribution of $\zeta \left(s\right)$, various methods and terms are used. For example, it was observed in [16] that the distribution of a-points of $\zeta \left(s\right)$, $a\ne 0$ (the solution of $\zeta \left(s\right)=a$), has a certain relation to a Julia line [17] with respect to the essential singularity of $\zeta \left(s\right)$ at infinity.

In the present paper, we are connected to a new problem—the approximation of analytic functions by the function $\mathcal{Z}\left(s\right)\stackrel{def}{=}{\mathcal{Z}}_1\left(s\right)$. We need some results from [3]. The function $\mathcal{Z}\left(s\right)$ is analytic in the half-plane $\sigma >-3/4$, except for a double pole at the point $s=1$, and has simple poles at the points $s=-(2k-1)$, $k\in \mathbb{N}$. Let ${\gamma }_0$ be the Euler constant, $E\left(T\right)$ be defined by

$$\underset{0}{\overset{T}{\int }}{\left|\zeta \left(\frac{1}{2}+it\right)\right|}^2 \mathrm{d}t=Tlog\frac{T}{2\pi }+(2{\gamma }_0-1)T+E\left(t\right),$$

and

$$G\left(T\right)=\underset{1}{\overset{T}{\int }}E\left(t\right) \mathrm{d}t-\pi T, G_1\left(T\right)=\underset{1}{\overset{T}{\int }}G\left(t\right) \mathrm{d}t.$$

Then, it was obtained that

$$\begin{array}{cc}\hfill \mathcal{Z}\left(s\right)=& \frac{1}{{(s-1)}^2}+\frac{2{\gamma }_0-log2\pi }{s-1}-E\left(1\right)+\pi (s+1)\hfill \\ & +s(s+1)(s+2)\underset{1}{\overset{\infty }{\int }}{G}_1\left(x\right)x^{-s-3} \mathrm{d}x, \sigma >-\frac{3}{4}.\hfill \end{array}$$

(1)

Moreover, for $0⩽\sigma ⩽1$, $t⩾t_0>0$, and fixed $\epsilon >0$,

$$\mathcal{Z}(\sigma +it){\ll }_{\epsilon }{t}^{1-\sigma +\epsilon },$$

and

$$\underset{1}{\overset{T}{\int }}{\left|\mathcal{Z}(\sigma +it)\right|}^2 \mathrm{d}t{\ll }_{\epsilon }\left\{\begin{array}{cc}{T}^{3-4\sigma +\epsilon }\hfill & \mathrm{if} 0⩽\sigma ⩽\frac{1}{2},\hfill \\ T^{2-2\sigma +\epsilon }\hfill & \mathrm{if} \frac{1}{2}⩽\sigma ⩽1.\hfill \end{array}\right.$$

(2)

Let $D=\{s\in \mathbb{C}:1/2<\sigma <1\}$, and $H\left(D\right)$ be the space of analytic on D functions equipped with the topology of uniform convergence on compacts. The main result of the paper is the following theorem.

Theorem 1.

There is a non-empty closed subset $F\subset H\left(D\right)$ such that, for arbitrary compact set $K\subset D$, $f\left(s\right)\in F$, and every $\epsilon >0$,

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

Moreover, except for at most a countably set of values of $\epsilon >0$, “lim inf” can be replaced by “lim”.

Theorem 1 implies that there are infinitely many shifts $\mathcal{Z}(s+i\tau )$ approximating a function from the set F. Theorem 1 is a certain version of the modern form of the Voronin universality theorem [8] for the Riemann zeta-function, see, for example, [9,10]. In the case of $\zeta \left(s\right)$, $F=\{g\in H(D):g(s)\ne 0 \mathrm{or} g(s)\equiv 0\}$.

Unfortunately, in the case of Theorem 1, the set F cannot be explicit described. Let $\mathcal{B}\left(\mathcal{X}\right)$ stand for Borel $\sigma$-field of the space $\mathcal{X}$. We will show that F is the support of a probability measure on $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$. Theorem 1 is a corollary of a limit theorem for weakly convergent measures in the space $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$. For $A\in \mathcal{B}\left(H\right(D\left)\right)$, set

$$P_T\left(A\right)=\frac{1}{T} \mu \{\tau \in [0,T]:\mathcal{Z}(s+i\tau )\in A\}.$$

Denote by $\underset{ }{\overset{W}{\to }}$ the weak convergence.

Theorem 2.

On the space $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$, there is a probability measure P such that $P_T\underset{T\to \infty }{\overset{W}{\to }}P$.

For the proof of Theorem 2, the auxiliary function

$${\mathcal{Z}}_y\left(s\right)=\underset{1}{\overset{\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2}v(x,y)x^{-s} \mathrm{d}x,$$

where

$$v(x,y)=exp\left\{-{\left(\frac{x}{y}\right)}^{{\sigma }_0}\right\}, x,y\in (1,\infty ),$$

with a fixed ${\sigma }_0>0$, will be useful.

2. Case of Finite Interval

Let $a>1$, and

$${\mathcal{Z}}_{a,y}\left(s\right)=\underset{1}{\overset{a}{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2}v(x,y)x^{-s} \mathrm{d}x,$$

For $A\in \mathcal{B}\left(H\right(D\left)\right)$, set

$$P_{T,a,y}\left(A\right)=\frac{1}{T} \mu \left\{\tau \in [0,T]:{\mathcal{Z}}_{a,y}(s+i\tau )\in A\right\}.$$

Lemma 1.

On the space $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$, there is a probability measure $P_{a,y}$ such that $P_{T,a,y}\underset{T\to \infty }{\overset{W}{\to }}{P}_{a,y}$.

The proof of Lemma 1 is divided into parts. In the first part, we will deal with weak convergence on a certain Cartesian product. Let $\gamma$ be the unit circle on $\mathbb{C}$, and

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

In virtue of the Tikhonov theorem, the set ${\Omega }_a$ with the product topology is a compact topological Abeliam group. For $A\in \mathcal{B}\left({\Omega }_a\right)$, set

$$Q_{T,a}\left(A\right)=\frac{1}{T} \mu \left\{\tau \in [0,T]:\left(u^{-i\tau }:u\in [1,a]\right)\in A\right\}.$$

Lemma 2.

On the space $({\Omega }_a,\mathcal{B}\left({\Omega }_a\right))$, there is a probability measure $Q_a$ such that $Q_{T,a}\underset{T\to \infty }{\overset{W}{\to }}{Q}_a$.

Proof. 

The character group of ${\Omega }_a$ is isomorphic to $\underset{u\in [1,a]}{\oplus }{\mathbb{Z}}_u$, where ${\mathbb{Z}}_u=\mathbb{Z}$ for all $u\in [1,a]$. Therefore, the Fourier transform $g_{T,a}(k_u:u\in [1,a])$ of $Q_{T,a}$ is given by

$$g_{T,a}(k_u:u\in [1,a])=\underset{{\Omega }_a}{\int }\prod _{u\in [1,a]}{x}_u^{k_u} \mathrm{d}{Q}_{T,a},$$

where $x_u\in \gamma$, $k_u\in \mathbb{Z}$, and only a finite number of $k_u$ are not zeros. Thus,

$$\begin{array}{cc}\hfill g_{T,a}(k_u:u\in [1,a])& =\frac{1}{T}\underset{0}{\overset{T}{\int }}exp\left\{-i\tau \sum _{u\in [1,a]}{k}_{u}logu\right\} \mathrm{d}\tau \hfill \\ & =\left\{\begin{array}{cc}1\hfill & \mathrm{if} {\displaystyle \sum _{u\in [1,a]}}{k}_{u}logu=0,\hfill \\ \frac{1-exp\left\{-i\tau \sum _{u\in [1,a]}{k}_{u}logu\right\}}{iT{\sum }_{u\in [1,a]}{k}_{u}logu}\hfill & \mathrm{if} {\displaystyle \sum _{u\in [1,a]}}{k}_{u}logu\ne 0.\hfill \end{array}\right.\hfill \end{array}$$

Therefore, we have

$$\underset{T\to \infty }{lim}{g}_{T,a}(k_u:u\in [1,a])\stackrel{def}{=}{g}_a(k_u:u\in [1,a])=\left\{\begin{array}{cc}1\hfill & \mathrm{if} {\displaystyle \sum _{u\in [1,a]}}{k}_{u}logu=0,\hfill \\ 0\hfill & \mathrm{if} {\displaystyle \sum _{u\in [1,a]}}{k}_{u}logu\ne 0.\hfill \end{array}\right.$$

This shows that $Q_{T,a}\underset{T\to \infty }{\overset{W}{\to }}{Q}_a$ with $Q_a$ on $({\Omega }_a,\mathcal{B}\left({\Omega }_a\right))$ with the Fourier transform $g_a(k_u:u\in [1,a])$. □

Lemma 2 implies a certain limit lemma in the space $H\left(D\right)$. We recall that if $h:\mathcal{X}\to {\mathcal{X}}_1$ is a $(\mathcal{B}\left(\mathcal{X}\right),\mathcal{B}\left({\mathcal{X}}_1\right))$-measurable mapping, then a probability measure P on $(\mathcal{X},\mathcal{B}(\mathcal{X}\left)\right)$ defines the unique probability measure $P{h}^{-1}$ on $({\mathcal{X}}_1,\mathcal{B}\left({\mathcal{X}}_1\right))$ defined by $P{h}^{-1}\left(A\right)=P\left(h^{-1}A\right)$, $A\in {\mathcal{X}}_1$. Moreover, if the mapping h is continuous, then the weak convergence is preserved, i.e., if $P_n\underset{n\to \infty }{\overset{W}{\to }}P$ in $\mathcal{X}$, then also $P_n{h}^{-1}\underset{n\to \infty }{\overset{W}{\to }}P{h}^{-1}$ in ${\mathcal{X}}_1$. The latter remark is sometimes very useful.

Let

$$S_{n,a,y}\left(s\right)=\frac{a-1}{n}\sum _{k=1}^n{\left|\zeta \left(\frac{1}{2}+i{\xi }_k\right)\right|}^{2}v({\xi }_k,y){\xi }_k^{-s},$$

where ${\xi }_k\in [x_{k-1},x_k]$ and $x_k=1+((a-1)/n)k$. For $A\in \mathcal{B}\left(H\right(D\left)\right)$, let

$$P_{T,n,a,y}\left(A\right)=\frac{1}{T} \mu \left\{\tau \in [0,T]:S_{n,a,y}(s+i\tau )\in A\right\}.$$

Lemma 3.

On the space $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$, there is a probability measure $P_{n,a,y}$ such that $P_{T,n,a,y}\underset{T\to \infty }{\overset{W}{\to }}{P}_{n,a,y}$.

Proof. 

Let the mapping $h_{n,a}:{\Omega }_a\to H\left(D\right)$ be given by the formula

$$h_{n,a}\left(\underline{y}\right)=\frac{a-1}{n}\sum _{k=1}^n{\left|\zeta \left(\frac{1}{2}+i{\xi }_k\right)\right|}^{2}v({\xi }_k,y){\xi }_k^{-s}{y}_{{\xi }_k}, \underline{y}=\{y_u\in \gamma :u\in [1,a]\}.$$

Then, $h_{n,a}$ is a continuous in the product topology, and $h_{n,a}\left(\{u^{-i\tau }:u\in [1,a]\}\right)=S_{n,a,y}(s+i\tau )$. Thus, $P_{T,n,a,y}=Q_{T,a}{h}_{n,a}^{-1}$, where $Q_{T,a}$ comes from Lemma 2. This equality, the continuity of $h_{n,a}$, Lemma 2 and the above remark on the preservation of weak convergence show that $P_{T,n,a,y}$ converges weakly to $P_{n,a,y}=Q_a{h}_{n,a}^{-1}$ as $T\to \infty$ with $Q_a$ defined in Lemma 2. □

In the sequel, we will use one lemma on the convergence in distribution ($\underset{ }{\overset{\mathcal{D}}{\to }}$). Recall that the random element $X_n$ converges in distribution to X as $n\to \infty$, if the distribution $P_n$ of $X_n$ converges weakly to that P of X as $n\to \infty$. In this case, we use the notation $X_n\underset{n\to \infty }{\overset{\mathcal{D}}{\to }}P$ as well.

Suppose that the metric space $(\mathcal{X},d)$ is separable and the $\mathcal{X}$-valued random elements X, $Y_n$ and $X_{nk}$ are defined on the same probability space with measure $\mathbb{P}$.

Lemma 4.

For $k\in \mathbb{N}$, let

$$X_{nk}\underset{n\to \infty }{\overset{\mathcal{D}}{\to }}{X}_k$$

and

$$X_k\underset{k\to \infty }{\overset{\mathcal{D}}{\to }}X.$$

If, for any $\epsilon >0$,

$$\underset{k\to \infty }{lim}\underset{n\to \infty }{lim\; sup}\mathbb{P}\left\{d(X_{nk},Y_n)⩾\epsilon \right\}=0,$$

then $Y_n\underset{n\to \infty }{\overset{\mathcal{D}}{\to }}X$.

The lemma is proved, for example, in [18], Theorem 4.2.

We note that the space $H\left(D\right)$ is separable and metrizable. It is known that there is a sequence $\{K_l:l\in \mathbb{N}\}\subset D$ of compact embedded sets such that D is the union of $K_l$, and every set $K\subset D$ lies in some set $K_l$. Then,

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

is a metric in $H\left(D\right)$ which induces its topology of uniform convergence on compacts.

Lemma 5.

The equality

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\rho \left(S_{n,a,y}(s+i\tau ),{\mathcal{Z}}_{a,y}(s+i\tau )\right) \mathrm{d}\tau =0$$

holds.

Proof. 

The definition of the metric $\rho$ implies that it suffices to show that

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|S_{n,a,y}(s+i\tau )-{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau =0$$

(3)

for every compact set $K\subset D$. Let L be a simple closed contour lying in D and enclosing a compact set K; suppose also that ${inf}_{s\in K}{inf}_{z\in L}|s-z|⩾c\left(L\right)>0$. Then, by the integral Cauchy formula, we have

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

Therefore,

$$\begin{array}{cc}\hfill \frac{1}{T}& \underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|S_{n,a,y}(s+i\tau )-{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau \hfill \\ & {\ll }_L\underset{L}{\int }\left|\mathrm{d}z\right|\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}\left|S_{n,a,y}(z+i\tau )-{\mathcal{Z}}_{a,y}(z+i\tau )\right|\mathrm{d}\tau \right).\hfill \end{array}$$

(4)

Clearly,

$$\begin{array}{cc}\hfill \frac{1}{T}& \underset{0}{\overset{T}{\int }}\left|S_{n,a,y}(z+i\tau )-{\mathcal{Z}}_{a,y}(z+i\tau )\right| \mathrm{d}\tau \hfill \\ & ⩽{\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left|S_{n,a,y}(z+i\tau )-{\mathcal{Z}}_{a,y}(z+i\tau )\right|}^2 \mathrm{d}\tau \right)}^{1/2}\hfill \end{array}$$

(5)

We have

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

(6)

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

$$\begin{array}{cc}\hfill S_{n,a,y}& (z+i\tau )\overline{S_{n,a,y}(z+i\tau )}={\left(\frac{a-1}{n}\right)}^2\sum _{k=1}^n{\left|\zeta \left(\frac{1}{2}+i{\xi }_k\right)\right|}^4{v}^2({\xi }_k,y){\xi }_k^{-2\mathrm{Re}z}\hfill \\ & +{\left(\frac{a-1}{n}\right)}^2\underset{k_1\ne k_2}{\sum _{k_1=1}^n\sum _{k_2=1}^n}{\left|\zeta \left(\frac{1}{2}+i{\xi }_{k_1}\right)\right|}^2{\left|\zeta \left(\frac{1}{2}+i{\xi }_{k_2}\right)\right|}^2\hfill \\ & \times v({\xi }_{k_1},y)v({\xi }_{k_2},y){\xi }_{k_1}^{-z}{\xi }_{k_2}^{-\overline{z}}{\left(\frac{{\xi }_{k_1}}{{\xi }_{k_2}}\right)}^{-i\tau }.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}\hfill \frac{1}{T}\underset{0}{\overset{T}{\int }}& S_{n,a,y}(z+i\tau )\overline{S_{n,a,y}(z+i\tau )} \mathrm{d}\tau \hfill \\ & ={\left(\frac{a-1}{n}\right)}^2\sum _{k=1}^n{\left|\zeta \left(\frac{1}{2}+i{\xi }_k\right)\right|}^4{v}^2({\xi }_k,y){\xi }_k^{-2\mathrm{Re}z}\hfill \\ & +O({\left(\frac{a-1}{n}\right)}^2\frac{1}{T}\underset{k_1\ne k_2}{\sum _{k_1=1}^n\sum _{k_2=1}^n}{\left|\zeta \left(\frac{1}{2}+i{\xi }_{k_1}\right)\right|}^2{\left|\zeta \left(\frac{1}{2}+i{\xi }_{k_2}\right)\right|}^2\hfill \\ & \times v({\xi }_{k_1},y)v({\xi }_{k_2},y){\xi }_{k_1}^{-\mathrm{Re}z}{\xi }_{k_2}^{\mathrm{Re}z}{\left|log\frac{{\xi }_{k_1}}{{\xi }_{k_2}}\right|}^{-1}).\hfill \end{array}$$

Since

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}\frac{a-1}{n}\sum _{k=1}^n{\left|\zeta \left(\frac{1}{2}+i{\xi }_k\right)\right|}^4{v}^2({\xi }_k,y){\xi }_k^{-2\mathrm{Re}z}=\underset{1}{\overset{a}{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^4{v}^2(x,y)x^{-2\mathrm{Re}z} \mathrm{d}x,\end{array}$$

hence we obtain, for all $z\in L$,

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}{S}_{n,a,y}(z+i\tau )\overline{S_{n,a,y}(z+i\tau )} \mathrm{d}\tau =0.$$

(7)

Similarly, by the definition of ${\mathcal{Z}}_{a,y}\left(s\right)$, for all $z\in L$,

$$\begin{array}{cc}\hfill & \underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}{\mathcal{Z}}_{a,y}(z+i\tau )\overline{{\mathcal{Z}}_{a,y}(z+i\tau )} \mathrm{d}\tau \hfill \\ & =\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}(\underset{1}{\overset{a}{\int }}\underset{1}{\overset{a}{\int }}{\left|\zeta \left(\frac{1}{2}+i{x}_1\right)\right|}^2{\left|\zeta \left(\frac{1}{2}+i{x}_2\right)\right|}^2\hfill \\ & \times v(x_1,y)v(x_2,y)x_1^{-z-i\tau }{x}_2^{-\overline{z}+i\tau } \mathrm{d}{x}_1 \mathrm{d}{x}_2) \mathrm{d}\tau \hfill \\ & =\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{x_1\ne x_2}{\underset{1}{\overset{a}{\int }}\underset{1}{\overset{a}{\int }}}{\left|\zeta \left(\frac{1}{2}+i{x}_1\right)\right|}^2{\left|\zeta \left(\frac{1}{2}+i{x}_2\right)\right|}^2\hfill \\ & \times v(x_1,y)v(x_2,y)x_1^{-z}{x}_2^{-\overline{z}}\left(e^{-iTlog(x_1/x_2)}-1\right){\left(log\frac{x_1}{x_2}\right)}^{-1} \mathrm{d}{x}_1 \mathrm{d}{x}_2=0.\hfill \end{array}$$

(8)

The latter equality suggests that the set of values of the function ${\mathcal{Z}}_{a,y}\left(s\right)$ is not dense. On the other hand, this case is convenient for our investigations. Moreover,

$$\begin{array}{cc}\hfill & \left|\frac{1}{T}\underset{0}{\overset{T}{\int }}{S}_{n,a,y}(z+i\tau )\overline{{\mathcal{Z}}_{a,y}(z+i\tau )} \mathrm{d}\tau \right|\hfill \\ & ⩽{\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left|S_{n,a,y}(z+i\tau )\right|}^2 \mathrm{d}\tau \right)}^{1/2}{\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left|{\mathcal{Z}}_{a,y}(z+i\tau )\right|}^2 \mathrm{d}\tau \right)}^{1/2},\hfill \end{array}$$

and this is true for the integral of $\overline{S_{n,a,y}(z+i\tau )}{\mathcal{Z}}_{a,y}(z+i\tau )$. This and (4)–(8) show that

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|S_{n,a,y}(s+i\tau )-{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau =0.$$

□

Proof of Lemma 1.

Suppose that ${\theta }_T$ is a random variable uniformly distributed on $[0,T]$ and defined on a certain probability space with measure $\mathbb{P}$. Define the $H\left(D\right)$-valued random element

$$X_{T,n,a,y}\left(s\right)=S_{n,a,y}(s+i{\theta }_T).$$

In view of Lemma 3, we have

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

(9)

where $X_{n,a,y}$ is the $H\left(D\right)$-valued random element with the distribution $P_{n,a,y}$.

Now, we will prove that the sequence $\{P_{n,a,y}:n\in \mathbb{N}\}$ is tight, i.e., that, for every $\epsilon >0$, there exists a compact set $K=K\left(\epsilon \right)\subset H\left(D\right)$ such that

$$P_{n,a,y}\left(K\right)>1-\epsilon$$

for all $n\in \mathbb{N}$. Let $K_l$ be a compact set in the definition of the metric $\rho$. Then, (7) and the integral Cauchy formula imply

$$\underset{n\in \mathbb{N}}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K_l}{sup}\left|S_{n,a,y}(s+i\tau )\right| \mathrm{d}\tau ⩽R_{l,a,y}<\infty .$$

Let $\epsilon >0$ be a fixed, and $M_l=M_{l,a,y}=R_{l,a,y}{2}^l{\epsilon }^{-1}$. Then, in view of (9),

$$\mathbb{P}\left(\underset{s\in K_l}{sup}\left|X_{n,a,y}\left(s\right)\right|>M_l\right)⩽\underset{n\in \mathbb{N}}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{M_{l}T}\underset{0}{\overset{T}{\int }}\underset{s\in K_l}{sup}\left|S_{n,a,y}(s+i\tau )\right| \mathrm{d}\tau ⩽\frac{\epsilon }{2^l}$$

(10)

for all n and $l\in \mathbb{N}$. Let

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

Then, the set K is compact in the space $H\left(D\right)$, and, by (10),

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

for all $n\in \mathbb{N}$. This and the definition of $P_{n,a,y}$ prove the tightness of the sequence $\{P_{n,a,y}:n\in \mathbb{N}\}$. In the theory of weak convergence of probability measures, the Prokhorov theorem, see, for example, [18], occupies an important place. Let $\left\{P\right\}$ be a family of probability measures on $(\mathcal{X},\mathcal{B}(\mathcal{X}\left)\right)$. The Prokhorov theorem connects the tightness and relative compactness of $\left\{P\right\}$; namely, if the family $\left\{P\right\}$ is tight, then it is relatively compact.

Since the sequence $\left\{P_{n,a,y}\right\}$ is tight, by the Prokhorov theorem [18], it is relatively compact, i.e., every subsequence $\left\{P_{n_k,a,y}\right\}$ contains a subsequence weakly convergent to a certain probability measure on $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$. Thus, there exists a probability measure $P_{a,y}$ on $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$ and a sequence $\left\{P_{n_r,a,y}\right\}$ such that $P_{n_r,a,y}$ converges weakly to $P_{a,y}$ as $r\to \infty$. In other words,

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

(11)

Now, we are in position to apply Lemma 4 for the random elements

$$Y_{T,a,y}\left(s\right)={\mathcal{Z}}_{a,y}(s+i{\theta }_T),$$

$X_{n_r,a,y}$ and $X_{a,y}$, where $X_{a,y}$ has the distribution $P_{a,y}$. By Lemma 5, we have, for every $\epsilon >0$,

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}& \underset{T\to \infty }{lim\; sup}\mathbb{P}\left(\rho (Y_{T,a,y},X_{n_r,a,y})⩾\epsilon \right)\hfill \\ & ⩽\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T\epsilon }\underset{0}{\overset{T}{\int }}\rho \left(S_{n_r,a,y}(s+i\tau ),{\mathcal{Z}}_{a,y}(s+i\tau )\right) \mathrm{d}\tau =0.\hfill \end{array}$$

This, (9) and (11) together with Lemma 5 prove Lemma 1, i.e., $P_{T,a,y}$ converges weakly to $P_{a,y}$ as $T\to \infty$. □

3. Case of Infinite Interval

In this section, we will prove a limit theorem for the function ${\mathcal{Z}}_y\left(s\right)$. Since $\zeta (1/2+it)\ll t^{1/6}$ as $t\to \infty$, and $v(x,y)$ with respect to x is decreasing exponentially, the integral for ${\mathcal{Z}}_y\left(s\right)$ is absolutely convergent for $\sigma >{\sigma }_0$ with every fixed ${\sigma }_0$ and $y>0$.

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

$$P_{T,y}\left(A\right)=\frac{1}{T} \mu \left\{\tau \in [0,T]:{\mathcal{Z}}_y(s+i\tau )\in A\right\}.$$

Lemma 6.

On $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$, there exists a probability measure $P_y$ such that $P_{T,y}\underset{T\to \infty }{\overset{W}{\to }}{P}_y$.

Proof. 

First, we observe that the equality

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\rho \left({\mathcal{Z}}_y(s+i\tau ),{\mathcal{Z}}_{a,y}(s+i\tau )\right) \mathrm{d}\tau =0$$

(12)

holds. As in the case of Lemma 5, it suffices to show that, for every compact set $K\subset D$,

$$\underset{n\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )-{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau =0.$$

(13)

It is easily seen that, for every fixed $y>0$ and $s\in K$,

$$\begin{array}{cc}\hfill {\mathcal{Z}}_y(s+i\tau )-{\mathcal{Z}}_{a,y}(s+i\tau )& =\underset{a}{\overset{\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2}v(x,y)x^{-s-i\tau } \mathrm{d}x\hfill \\ & {\ll }_y\underset{a}{\overset{\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^{2}v(x,y)x^{-1/2} \mathrm{d}x=o_y\left(1\right)\hfill \end{array}$$

as $a\to \infty$ in view of convergence of the integral. From this, equality (13) follows.

Let ${\theta }_T$ be the same random variable as in the proof of Lemma 1. Define

$$Y_{T,y}\left(s\right)={\mathcal{Z}}_y(s+i{\theta }_T),$$

and denote by $Y_{a,y}$ the $H\left(D\right)$-valued random element with the distribution $P_{a,y}$. Then, Lemma 1 implies the relation

$$Y_{T,a,y}\underset{T\to \infty }{\overset{\mathcal{D}}{\to }}{Y}_{a,y}.$$

(14)

Let $K_l$, $l\in \mathbb{N}$ be a compact set from the definition of metric $\rho$. Then, (8) and the integral Cauchy formula give

$$\underset{a⩾1}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K_l}{sup}\left|{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau ⩽R_{l,y}<\infty .$$

Thus, taking ${\widehat{M}}_l={\widehat{M}}_{l,y}=R_{l,y}{2}^l{\epsilon }^{-1}$, we find by (14)

$$\mathbb{P}\left(\underset{s\in K_l}{sup}\left|Y_{a,y}\left(s\right)\right|>{\widehat{M}}_l\right)⩽\underset{a⩾1}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{{\widehat{M}}_{l}T}\underset{0}{\overset{T}{\int }}\underset{s\in K_l}{sup}\left|{\mathcal{Z}}_{a,y}(s+i\tau )\right| \mathrm{d}\tau ⩽\frac{\epsilon }{2^l}.$$

This shows that

$$\mathbb{P}\left(Y_{a,y}\in K\right)>1-\epsilon ,$$

for all $a⩾1$, where $K=\{g\in H\left(D\right):{sup}_{s\in K_l}|g\left(s\right)|⩽{\widehat{M}}_l, l\in \mathbb{N}\}$. Therefore, the family of probability measures $\{P_{a,y}:a⩾1\}$ is tight. Thus, there exists a sequence $P_{a_r,y}$ weakly convergent to a certain probability measure $P_y$ as $r\to \infty$, i.e.,

$$Y_{a_r,y}\underset{r\to \infty }{\overset{\mathcal{D}}{\to }}{P}_y.$$

This, (12), (14) and Lemma 4 prove that

$$Y_{T,y}\underset{T\to \infty }{\overset{\mathcal{D}}{\to }}{P}_y,$$

(15)

and the lemma is proved. □

4. Formula for ${\mathcal{Z}}_{\mathbf{y}}\left(\mathbf{s}\right)$

As usual, denote by $\Gamma \left(s\right)$ the Euler gamma-function, and define

$$l_y\left(s\right)=\frac{s}{{\sigma }_0}\Gamma \left(\frac{s}{{\sigma }_0}\right)y^s,$$

where ${\sigma }_0$ is from definition of $v(x,y)$.

Lemma 7.

The integral representation, for $s\in D$,

$${\mathcal{Z}}_y\left(s\right)=\frac{1}{2\pi i}\underset{{\sigma }_0-i\infty }{\overset{{\sigma }_0+i\infty }{\int }}\mathcal{Z}(s+z)l_y\left(z\right)\frac{\mathrm{d}z}{z}$$

is valid.

Proof. 

We will apply the classical Mellin formula

$$\frac{1}{2\pi i}\underset{a-i\infty }{\overset{a+i\infty }{\int }}\Gamma \left(s\right)b^{-s} \mathrm{d}s=e^{-b}, a,b>0,$$

and obtain

$$\begin{array}{cc}\hfill \frac{1}{2\pi i}\underset{{\sigma }_0-i\infty }{\overset{{\sigma }_0+i\infty }{\int }}{l}_y\left(z\right)x^{-z}\frac{\mathrm{d}z}{z}& =\frac{1}{2\pi i}\underset{{\sigma }_0-i\infty }{\overset{{\sigma }_0+i\infty }{\int }}\frac{1}{{\sigma }_0}\Gamma \left(\frac{z}{{\sigma }_0}\right){\left(\frac{x}{y}\right)}^{-z} \mathrm{d}z\hfill \\ & =\frac{1}{2\pi i}\underset{1-i\infty }{\overset{1+i\infty }{\int }}\Gamma \left(z\right){\left(\frac{x}{y}\right)}^{-{\sigma }_{0}z}\mathrm{d}z\hfill \\ & =exp\left\{-{\left(\frac{x}{y}\right)}^{{\sigma }_0}\right\}=v(x,y).\hfill \end{array}$$

(16)

Setting, for brevity,

$$f(x,t)=\frac{1}{2\pi i}\frac{l_y({\sigma }_0+it)}{{\sigma }_0+it}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2{x}^{-s-{\sigma }_0-it}$$

and applying theorem from ([19], §1.84), we obtain

$$\underset{-T}{\overset{T}{\int }} \mathrm{d}t\underset{1}{\overset{X}{\int }}f(x,t) \mathrm{d}x=\underset{1}{\overset{X}{\int }} \mathrm{d}x\underset{-T}{\overset{T}{\int }}f(x,t) \mathrm{d}t,$$

(17)

for any $X,T>1$. Next, the well-known estimate

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

which is uniform in any fixed strip ${\sigma }_1<\sigma <{\sigma }_2$, together with the inequality

$$\underset{1}{\overset{X}{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2 \mathrm{d}x\ll X(logX)$$

imply

$$\underset{T}{\overset{+\infty }{\int }} \mathrm{d}t\underset{1}{\overset{X}{\int }}\left(\left|f\right(x,t\left)\right|+\left|f\right(x,-t\left)\right|\right) \mathrm{d}x, \underset{1}{\overset{X}{\int }} \mathrm{d}x\underset{T}{\overset{+\infty }{\int }}\left(\left|f\right(x,t\left)\right|+\left|f\right(x,-t\left)\right|\right) \mathrm{d}t\ll R,$$

where

$$\begin{array}{cc}\hfill R=R(X,T)& =y^{{\sigma }_0}\underset{T}{\overset{+\infty }{\int }}{e}^{-ct/{\sigma }_0} \mathrm{d}t\underset{1}{\overset{X}{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2{x}^{-\sigma -{\sigma }_0} \mathrm{d}x\hfill \\ & \ll y^{{\sigma }_0}exp\left\{-\frac{c}{{\sigma }_0}T\right\}\left(1+X^{1-\sigma -{\sigma }_0}\right){(logX)}^2.\hfill \end{array}$$

(18)

Hence, using (17) and (18), we find that

$$\begin{array}{cc}\hfill \underset{-\infty }{\overset{+\infty }{\int }} \mathrm{d}t\underset{1}{\overset{X}{\int }}f(x,t) \mathrm{d}x& =\left(\underset{-T}{\overset{T}{\int }}+\underset{T}{\overset{+\infty }{\int }}+\underset{-\infty }{\overset{-T}{\int }}\right) \mathrm{d}t\underset{1}{\overset{X}{\int }}f(x,t) \mathrm{d}x\hfill \\ & =\underset{1}{\overset{X}{\int }} \mathrm{d}x\underset{-T}{\overset{T}{\int }}f(x,t) \mathrm{d}t+O\left(R\right)\hfill \\ & =\underset{1}{\overset{X}{\int }} \mathrm{d}x\left(\underset{-\infty }{\overset{+\infty }{\int }}-\underset{T}{\overset{+\infty }{\int }}-\underset{-\infty }{\overset{-T}{\int }}\right)f(x,t) \mathrm{d}t+O\left(R\right)\hfill \\ & =\underset{1}{\overset{X}{\int }} \mathrm{d}x\underset{-\infty }{\overset{+\infty }{\int }}f(x,t) \mathrm{d}t+O\left(R\right).\hfill \end{array}$$

Tending $T\to \infty$, we obtain

$$\underset{-\infty }{\overset{+\infty }{\int }} \mathrm{d}t\underset{1}{\overset{X}{\int }}f(x,t) \mathrm{d}x=\underset{1}{\overset{X}{\int }} \mathrm{d}x\underset{-\infty }{\overset{+\infty }{\int }}f(x,t) \mathrm{d}t$$

for any $X>1$. Therefore, the application of a theorem from ([19], §1.84) together with (16) yields

$$\begin{array}{cc}\hfill \frac{1}{2\pi i}& \underset{{\sigma }_0-i\infty }{\overset{{\sigma }_0+i\infty }{\int }}\mathcal{Z}(s+z)l_y\left(z\right)\frac{\mathrm{d}z}{z}\hfill \\ & =\underset{-\infty }{\overset{+\infty }{\int }}\underset{1}{\overset{+\infty }{\int }}\frac{1}{2\pi i}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2\frac{l_y({\sigma }_0+it)}{{\sigma }_0+it} x^{-s-{\sigma }_0-it}\mathrm{d}x\hfill \\ & =\underset{-\infty }{\overset{+\infty }{\int }} \mathrm{d}t\underset{1}{\overset{+\infty }{\int }}f(x,t) \mathrm{d}x=\underset{1}{\overset{+\infty }{\int }} \mathrm{d}x\underset{-\infty }{\overset{+\infty }{\int }}f(x,t)\mathrm{d}t\hfill \\ & =\underset{1}{\overset{+\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2{x}^{-s}\frac{1}{2\pi i}\underset{{\sigma }_0-i\infty }{\overset{{\sigma }_0+i\infty }{\int }}{l}_y\left(z\right)x^{-z}\frac{\mathrm{d}z}{z} \mathrm{d}x\hfill \\ & =\underset{1}{\overset{+\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2{x}^{-s}v(x,y) \mathrm{d}x={\mathcal{Z}}_y\left(s\right).\hfill \end{array}$$

□

5. Approximation of $\mathcal{Z}\left(\mathit{s}\right)$ by ${\mathcal{Z}}_{\mathit{y}}\left(\mathit{s}\right)$

Lemma 8.

The equality

$$\underset{y\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\rho \left(\mathcal{Z}(s+i\tau ),{\mathcal{Z}}_y(s+i\tau )\right) \mathrm{d}\tau =0$$

holds.

Proof. 

Let K be an arbitrary fixed compact set of the strip D. Then, there exists a number $\epsilon >0$ such that, for all $s=\sigma +it\in K$, $1/2+2\epsilon ⩽\sigma ⩽1-\epsilon$. Denote

$${\sigma }_1=\sigma -\epsilon -\frac{1}{2}, {\sigma }_0=\frac{1}{2}+\epsilon .$$

Then, ${\sigma }_1>0$ for all $s\in K$. Since the point $z=1-s$ is a double pole, and $z=0$ is a simple pole of the function

$$\mathcal{Z}(s+z)\frac{l_y\left(z\right)}{z},$$

Lemma 7 and the residue theorem imply

$${\mathcal{Z}}_y\left(s\right)-\mathcal{Z}\left(s\right)=\frac{1}{2\pi i}\underset{-{\sigma }_1-i\infty }{\overset{-{\sigma }_1+i\infty }{\int }}\mathcal{Z}(s+z)l_y\left(z\right)\frac{\mathrm{d}z}{z}+R_y\left(s\right),$$

(19)

where

$$R_y\left(s\right)=\underset{z=1-s}{\mathrm{Res}}\mathcal{Z}(s+z)\frac{l_y\left(z\right)}{z}.$$

Let $a_0=2{\gamma }_0-log2\pi$. Then, in view of (1),

$$R_y\left(s\right)={\left(\frac{l_y\left(z\right)}{z}\right)}^{\prime }{|}_{z=1-s}+a_0\frac{l_y(1-s)}{1-s}.$$

(20)

By (19), for all $s\in K$, we have

$$\begin{array}{cc}\hfill {\mathcal{Z}}_y(s+i\tau )-\mathcal{Z}(s+i\tau )=& \frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\mathcal{Z}\left(\sigma +it-\sigma +\frac{1}{2}+\epsilon +i\tau +iv\right)\hfill \\ & \times \frac{l_y(1/2+\epsilon -\sigma +iv)}{1/2+\epsilon -\sigma +iv} \mathrm{d}v+R_y(s+i\tau ).\hfill \end{array}$$

Hence, writing v in place of $t+v$, gives, for $s\in K$,

$$\begin{array}{cc}\hfill {\mathcal{Z}}_y(s+i\tau )-\mathcal{Z}(s+i\tau )=& \frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\frac{l_y(1/2+\epsilon -s+iv)}{1/2+\epsilon -s+iv} \mathrm{d}v\hfill \\ & +R_y(s+i\tau )\hfill \\ \hfill \ll & \underset{-\infty }{\overset{\infty }{\int }}\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\right|\underset{s\in K}{sup}\left|\frac{l_y(1/2+\epsilon -s+iv)}{1/2+\epsilon -s+iv}\right| \mathrm{d}v\hfill \\ & +\underset{s\in K}{sup}\left|R_y(s+i\tau )\right|.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill \frac{1}{T}& \underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )-\mathcal{Z}(s+i\tau )\right| \mathrm{d}\tau \hfill \\ & \ll \underset{-\infty }{\overset{\infty }{\int }}\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\right| \mathrm{d}\tau \right)\underset{s\in K}{sup}\left|\frac{l_y(1/2+\epsilon -s+iv)}{1/2+\epsilon -s+iv}\right| \mathrm{d}v\hfill \\ & +\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|R_y(s+i\tau )\right| \mathrm{d}\tau \stackrel{def}{=}{I}_1+I_2.\hfill \end{array}$$

(21)

First we estimate $I_1$. The definition of $l_y\left(s\right)$ and (17) imply that, for $s=\sigma +it\in K$,

$$\begin{array}{cc}\hfill \frac{l_y(1/2+\epsilon -s+iv)}{1/2+\epsilon -s+iv}& {\ll }_{{\sigma }_0}{y}^{1/2+\epsilon -\sigma }\left|\Gamma \left(\frac{1}{{\sigma }_0}\left(\frac{1}{2}+\epsilon -\sigma -it+iv\right)\right)\right|\hfill \\ & {\ll }_{{\sigma }_0}{y}^{1/2+\epsilon -\sigma }exp\left\{-\frac{c}{{\sigma }_0}|t-v|\right\}\hfill \\ & {\ll }_{{\sigma }_0}{y}^{1/2+\epsilon -\sigma }exp\left\{-\frac{c}{{\sigma }_0}|v-t|\right\}\hfill \\ & {\ll }_{{\sigma }_0}{y}^{1/2+\epsilon -\sigma }exp\left\{\frac{c}{{\sigma }_0}\left|t\right|\right\}exp\left\{-\frac{c}{{\sigma }_0}\left|v\right|\right\}.\hfill \end{array}$$

(22)

Since $\sigma ⩾1/2+2\epsilon$, then $1/2+\epsilon -\sigma ⩽-\epsilon$ for any $s\in K$. Therefore,

$$\underset{s\in K}{sup}\left|\frac{l_y(1/2+\epsilon +iv-s)}{1/2+\epsilon +iv-s}\right|{\ll }_{{\sigma }_0,K}{y}^{-\epsilon }exp\{-c_1\left|v\right|\}.$$

Thus, we obtain

$$I_1{\ll }_{{\sigma }_0,K}{y}^{-\epsilon }\underset{-\infty }{\overset{+\infty }{\int }}exp\left\{-\frac{c}{{\sigma }_0}\left|v\right|\right\} \frac{1}{T}\underset{0}{\overset{T}{\int }}\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\right| \mathrm{d}\tau \mathrm{d}v.$$

Given $0<{\epsilon }_1<\epsilon$, Cauchy inequality together with (2) imply

$$\begin{array}{cc}\hfill \frac{1}{T}\underset{0}{\overset{T}{\int }}\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\right| \mathrm{d}\tau & ⩽{\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau +iv\right)\right|}^2 \mathrm{d}\tau \right)}^{1/2}\hfill \\ & ={\left(\frac{1}{T}\underset{v}{\overset{T+v}{\int }}{\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau \right)\right|}^2 \mathrm{d}\tau \right)}^{1/2}\hfill \\ & ⩽{\left(\frac{2}{T}\underset{0}{\overset{T+\left|v\right|}{\int }}{\left|\mathcal{Z}\left(\frac{1}{2}+\epsilon +i\tau \right)\right|}^2 \mathrm{d}\tau \right)}^{1/2}\hfill \\ & {\ll }_{\epsilon }{\left(\frac{1}{T}{(T+|v\left|\right)}^{2-2(1/2+\epsilon )+{\epsilon }_1}\right)}^{1/2}\hfill \\ & {\ll }_{\epsilon }{T}^{-\epsilon +{\epsilon }_1/2}+\frac{1}{\sqrt{T}}{\left|v\right|}^{1/2-\epsilon +{\epsilon }_1/2}\hfill \\ & {\ll }_{\epsilon }{T}^{-\epsilon /2}+\frac{{\left|v\right|}^{1/2}}{\sqrt{T}}.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill I_1& {\ll }_{{\sigma }_0,K,\epsilon }{y}^{-\epsilon }\underset{-\infty }{\overset{+\infty }{\int }}{e}^{c_1\left|v\right|}\left(T^{-\epsilon /2}+\frac{{\left|v\right|}^{1/2}}{\sqrt{T}}\right) \mathrm{d}v{\ll }_{{\sigma }_0,K,\epsilon }{y}^{-\epsilon }\left(T^{-\epsilon /2}+T^{-1/2}\right)\hfill \\ & {\ll }_{{\sigma }_0,K,\epsilon }{y}^{-\epsilon }{T}^{-\epsilon /2}.\hfill \end{array}$$

Passing to the estimate of $I_2$, and setting $g\left(z\right)=l_y\left(z\right)/z$, we obtain

$$g^{\prime }\left(z\right)=\frac{y^z}{{\sigma }_0}\Gamma \left(\frac{z}{{\sigma }_0}\right)\left(\frac{1}{{\sigma }_0}\psi \left(\frac{z}{{\sigma }_0}\right)+logy\right)$$

where $\psi \left(x\right)={\Gamma }^{\prime }\left(x\right)/\Gamma \left(x\right)$ denotes the digamma-function. Hence,

$$\begin{array}{cc}\hfill R_y\left(s\right)& =g^{\prime }(1-s)+(2{\gamma }_0-log2\pi )g(1-s)\hfill \\ & =\frac{y^{1-s}}{{\sigma }_0} \Gamma \left(\frac{1-s}{{\sigma }_0}\right)\left(\frac{1}{{\sigma }_0}\psi \left(\frac{1-s}{{\sigma }_0}\right)+logy+2{\gamma }_0-log2\pi \right).\hfill \end{array}$$

Thus, we find

$$\begin{array}{cc}\hfill R_y(s+i\tau )& \ll y^{1-\sigma }\left|\Gamma \left(\frac{1-\sigma }{{\sigma }_0}+i\frac{t+\tau }{{\sigma }_0}\right)\right|\left(\left|\psi \left(\frac{1-\sigma }{{\sigma }_0}+i\frac{t+\tau }{{\sigma }_0}\right)\right|+logy+1\right)\hfill \\ & {\ll }_{{\sigma }_0}{y}^{1-\sigma }exp\left\{-\frac{c}{{\sigma }_0}|t+\tau |\right\}\left(log\left|\frac{t+\tau }{{\sigma }_0}\right|+logy+1\right)\hfill \end{array}$$

and therefore

$$\begin{array}{cc}\hfill \underset{s\in K}{sup}\left|R_y(s+i\tau )\right|& {\ll }_{{\sigma }_0,K}{y}^{1-(1/2+2\epsilon )}exp\left\{-\frac{c}{{\sigma }_0}\tau \right\}\left(log(\tau +2)+logy\right)\hfill \\ & {\ll }_{{\sigma }_0,K,\epsilon }{y}^{1/2-\epsilon }exp\{-c_2\left|\tau \right|\}.\hfill \end{array}$$

Thus, we obtain

$$I_2=\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|R_y(s+i\tau )\right| \mathrm{d}\tau {\ll }_{{\sigma }_0,K,\epsilon }\frac{y^{1/2-\epsilon }}{T}\underset{0}{\overset{T}{\int }}exp\{-c_2\tau \} \mathrm{d}\tau {\ll }_{{\sigma }_0,K,\epsilon }\frac{y^{1/2-\epsilon }}{T}.$$

Consequently,

$$\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )-\mathcal{Z}(s+i\tau )\right| \mathrm{d}\tau =I_1+I_2{\ll }_{{\sigma }_0,K,\epsilon }{y}^{-\epsilon }{T}^{-\epsilon /2}+\frac{y^{1/2-\epsilon }}{T}.$$

Tending $T\to \infty$, we find

$$\underset{T\to \infty }{lim}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )-\mathcal{Z}(s+i\tau )\right| \mathrm{d}\tau =0.$$

Then, the desired assertion follows. □

6. Proof of Theorem 2

We return to the limit measure $P_y$ in Lemma 6. We recall that $P_{T,y}\underset{T\to \infty }{\overset{W}{\to }}{P}_y$.

Lemma 9.

The family of probability measures $\{P_y:y>1\}$ is tight.

Proof. 

Let $K\subset D$ be a compact set. Then, we have

$$\begin{array}{cc}\hfill \frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )\right| \mathrm{d}\tau ⩽& \frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|\mathcal{Z}(s+i\tau )-{\mathcal{Z}}_y(s+i\tau )\right| \mathrm{d}t\hfill \\ & +\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|\mathcal{Z}(s+i\tau )\right| \mathrm{d}\tau .\hfill \end{array}$$

(23)

Let L be a simple closed contour lying in D, enclosing a compact K and such that

$$\underset{z\in L}{inf}\underset{s\in K}{inf}|z-s|⩾c\left(L\right)>0.$$

Then, the application of the integral Cauchy formula gives

$$\left|\mathcal{Z}(s+i\tau )\right|⩽\frac{1}{2\pi }\underset{L}{\int }\frac{\left|\mathcal{Z}\right(z+i\tau \left)\right|}{|z-s|} \left|\mathrm{d}z\right|,$$

and

$$\underset{s\in K}{sup}\left|\mathcal{Z}(s+i\tau )\right|⩽\frac{1}{2\pi c}\underset{L}{\int }\left|\mathcal{Z}(z+i\tau )\right| \left|\mathrm{d}z\right|{\ll }_L\underset{L}{\int }\left|\mathcal{Z}(z+i\tau )\right| \left|\mathrm{d}z\right|.$$

Setting $z=\sigma +it$ for $z\in L$, and using the inequality (2), we find

$$\begin{array}{cc}\hfill \frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|\mathcal{Z}(s+i\tau )\right| \mathrm{d}\tau & {\ll }_L\underset{L}{\int }\frac{1}{T}\underset{0}{\overset{T}{\int }}\left|\mathcal{Z}(z+i\tau )\right| \mathrm{d}\tau \left|\mathrm{d}z\right|\hfill \\ & {\ll }_L\underset{L}{\int }{\left(\frac{1}{T}\underset{0}{\overset{T}{\int }}{\left|\mathcal{Z}(z+i\tau )\right|}^2 \mathrm{d}\tau \right)}^{1/2} \left|\mathrm{d}z\right|\hfill \\ & {\ll }_L\underset{L}{\int }{\left(\frac{1}{T}\underset{0}{\overset{T+\left|t\right|+1}{\int }}{\left|\mathcal{Z}(\sigma +i\tau )\right|}^2 \mathrm{d}\tau \right)}^{1/2} \left|\mathrm{d}z\right|\hfill \\ & {\ll }_L\underset{L}{\int }{\left(\frac{1}{T}{(T+|t\left|\right)}^{2-2\sigma +{\epsilon }_1}\right)}^{1/2} \left|\mathrm{d}z\right|{\ll }_L{T}^{-\epsilon /2}.\hfill \end{array}$$

Thus,

$$\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|\mathcal{Z}(s+i\tau )\right| \mathrm{d}t⩽C<\infty .$$

Therefore, by (23) and Lemma 8,

$$\underset{y⩾1}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{T}\underset{0}{\overset{T}{\int }}\underset{s\in K}{sup}\left|{\mathcal{Z}}_y(s+i\tau )\right| \mathrm{d}t⩽R<\infty .$$

Fix $\epsilon >0$ and put $M_l=R{2}^l{\epsilon }^{-1}$, $l\in \mathbb{N}$. Let $Y_y\left(s\right)$ be a $H\left(D\right)$-valued random element with the distribution $P_y$. Then,

$$\mathbb{P}\left(\underset{s\in K_l}{sup}\left|Y_y\left(s\right)\right|>M_l\right)⩽\underset{y⩾1}{sup}\underset{T\to \infty }{lim\; sup}\frac{1}{T{M}_l}\underset{0}{\overset{T}{\int }}\underset{s\in K_l}{sup}\left|{\mathcal{Z}}_y(s+i\tau )\right| \mathrm{d}\tau ⩽\frac{\epsilon }{2^l}.$$

Hence, we have

$$\mathbb{P}(Y_y\left(s\right)\in K)⩾1-\epsilon$$

for all $y⩾1$, where

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

and the lemma is proved. □

Proof of Theorem 2.

By Lemma 9 and the Prokhorov theorem, the family $\{P_y:y⩾1\}$ is relatively compact. Therefore, there exists a sequence $\left\{y_k\right\}$, $y_k\to \infty$ as $k\to \infty$ such that $\left\{P_{y_k}\right\}$ converges weakly to a certain probability measure P on $\left(H\right(D),\mathcal{B}(H\left(D\right)\left)\right)$ as $k\to \infty$. Since the distribution of $Y_{y_k}$ is $P_{y_k}$, we have

$$Y_{y_k}\underset{k\to \infty }{\overset{\mathcal{D}}{\to }}P.$$

(24)

Define

$$Z_T\left(s\right)=\mathcal{Z}(s+i{\theta }_T).$$

Then, in view of Lemma 8, for every $\epsilon >0$,

$$\begin{array}{cc}\hfill \underset{y\to \infty }{lim}& \underset{T\to \infty }{lim\; sup}\mathbb{P}\left(\rho (Z\left(s\right),Y_{T,Y}\left(s\right))⩾\epsilon \right)\hfill \\ & ⩽\underset{y\to \infty }{lim}\underset{T\to \infty }{lim\; sup}\frac{1}{T\epsilon }\underset{0}{\overset{T}{\int }}\rho (\mathcal{Z}(s+i\tau ),{\mathcal{Z}}_y(s+i\tau )) \mathrm{d}\tau =0.\hfill \end{array}$$

The latter equality, relations (15) and (24) show that all hypotheses of Lemma 4 are satisfied. Therefore, we obtain the relation

$$Z_T\underset{T\to \infty }{\overset{\mathcal{D}}{\to }}P,$$

and this is equivalent to the assertion of the theorem. □

7. Proof of Theorem 1

We derive Theorem 1 from Theorem 2 by applying properties of weak convergence of probability measures. We will approximate functions from the support of the limit measure P of $P_T$ in Theorem 2. The application of Theorem 2 is based on the following equivalents of weak convergence, see, for example, [18].

Lemma 10.

Let $P_n$, $n\in \mathbb{N}$, and P be probability measures on $(\mathcal{X},\mathcal{B}(\mathcal{X}\left)\right)$. Then, the following statements are equivalent.

1°

$P_n\underset{n\to \infty }{\overset{W}{\to }}P$.

2°

For every open set $G\subset \mathcal{X}$,

$$\underset{n\to \infty }{lim\; inf}{P}_n\left(G\right)⩾P\left(G\right).$$

3°

For every continuity set A of the measure P (A is a continuity set of P if $P(\partial A)=0$, where $\partial A$ is a boundary of A),

$$\underset{n\to \infty }{lim}{P}_n\left(A\right)=P\left(A\right).$$

Proof of Theorem 1.

Denote by F the support of the limit measure P in Theorem 2. The set F is a minimal closed subset of $H\left(D\right)$ such that $P\left(F\right)=1$. The set F consists of all elements $f\in H\left(D\right)$ such that, for every open neighborhood G of f, the inequality $P\left(G\right)>0$ is satisfied. Obviously, $F\ne ⌀$.

For $f\in F$, define

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

Then, $G_{\epsilon }$ is an open neighborhood of the element f of the support of P. Therefore,

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

(25)

Thus, by Theorem 2, and $1^{\circ }$ and $2^{\circ }$ of Lemma 10,

$$\underset{T\to \infty }{lim\; inf}\frac{1}{T} \mu \left\{\tau \in [0,T]:\underset{s\in K}{sup}|\mathcal{Z}(s+i\tau )-f\left(s\right)|<\epsilon \right\}⩾P\left(G_{\epsilon }\right)>0.$$

To prove the second assertion of the theorem, we notice that the boundary $\partial G_{\epsilon }$ lies in

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

Hence, $\partial G_{{\epsilon }_1}\cap \partial G_{{\epsilon }_2}=⌀$ for ${\epsilon }_1\ne {\epsilon }_2$. Thus, $P(\partial G_{\epsilon })$ can be positive for at most countably many positive $\epsilon$, in other words, $G_{\epsilon }$ is a continuity set of P, except for all but at most a countable set of values $\epsilon >0$. Therefore, Theorem 2, $1^{\circ }$ and $3^{\circ }$ of Lemma 10, and (25) show that the limit

$$\underset{T\to \infty }{lim}\frac{1}{T} \mu \left\{\tau \in [0,T]:\underset{s\in K}{sup}|\mathcal{Z}(s+i\tau )-f\left(s\right)|<\epsilon \right\}=P\left(G_{\epsilon }\right)>0.$$

exists for all but at most countably many $\epsilon >0$. The theorem is proved. □

8. Conclusions

In this paper, we found that the shifts of the Mellin transform of the square of the Riemann zeta-function $\zeta \left(s\right)$

$$\mathcal{Z}\left(s\right)=\underset{1}{\overset{\infty }{\int }}{\left|\zeta \left(\frac{1}{2}+ix\right)\right|}^2{x}^{-s} \mathrm{d}x$$

approximate a certain class F of analytic functions defined in the strip $\{s\in \mathbb{C}:1/2<\sigma <1\}$. The main ingredient of the proof of the above result is a limit theorem for the function $\mathcal{Z}\left(s\right)$ in the space of analytic functions. Note that a problem of approximation of analytic functions by shifts of the function $\mathcal{Z}\left(s\right)$ is new, and it is discussed for the first time. The main result and method are inspired by universality theorems for $\zeta \left(s\right)$. Unfortunately, the set F is not explicitly given. This is a complicated future problem. Additionally, we are planning to extend the results of the paper for Mellin transforms of other powers of $\zeta \left(s\right)$.