Approximation of analytic functions by an absolutely convergent Dirichlet series

Abstract

In the paper, an absolutely convergent Dirichlet series whose shifts approximate a wide class of analytic functions is constructed. This series is close in the mean to the Riemann zeta-function.

1 Introduction

Let $\zeta (s)$, $s=\sigma +it$, be the Riemann zeta-function, i.e., for $\sigma >1$,

$$\begin{aligned} \zeta (s)=\sum _{m=1}^\infty \frac{1}{m^s}. \end{aligned}$$

The function $\zeta (s)$ has an analytic continuation to the whole complex plane, except for a simple pole at the point $s=1$ with residue 1. It is well known that the function $\zeta (s)$ has a universality property discovered by Voronin [13] on the approximation of a wide class of analytic functions by shifts $\zeta (s+i\tau )$, $\tau \in {\mathbb {R}}$. Let $D=\{ s\in {\mathbb {C}}: 1/2<\sigma <1\}$. Denote by ${\mathcal {K}}$ the class of compact subsets of the strip D with connected complements, and by $H_0(K)$ with $K\in {\mathcal {K}}$ the class of continuous non-vanishing functions on K that are analytic in the interior of K. Then the last version of the Voronin theorem, see, for example, [5], says that, for every $K\in {\mathcal {K}}$, $f(s)\in H_0(K)$, and $\varepsilon >0$,

$$\begin{aligned} \liminf _{T\rightarrow \infty } \frac{1}{T} \mathrm {meas}\left\{ \tau \in [0,T]: \sup _{s\in K} |\zeta (s+i\tau )-f(s)|<\varepsilon \right\} >0. \end{aligned}$$

(1.1)

The latter inequality shows that there are infinitely many shifts $\zeta (s+i\tau )$ that approximate uniformly on K a given function $f(s)\in H_0(K)$ with accuracy $\varepsilon >0$. On the other hand, no concrete such shift is known. In [7] and [9], it was obtained that “$\liminf$” in (1.1) can be replaced by “$\lim$” for all but at most countably many $\varepsilon >0$.

Some other zeta-functions and their classes are universal in the above sense as well, see [1, 3, 6, 11] and the very informative survey paper [8].

The Riemann zeta-function $\zeta (s)$, for $s\in D$, cannot be written as a convergent Dirichlet series, and is defined by analytic continuation. The aim of this note is to present a certain absolutely convergent Dirichlet series in D having an approximation property similar to that of the function $\zeta (s)$. Obviously, an absolutely convergent Dirichlet series cannot be universal. Therefore, such a series must be close in a certain sense to $\zeta (s)$.

For $u>0$, let $\theta >1/2$ be a fixed number and, for $m\in {\mathbb {N}}$,

$$\begin{aligned} v_u(m)=\exp \left\{ -\left( \frac{m}{u}\right) ^{\theta }\right\} , \end{aligned}$$

where $\exp \{a\}=\mathrm {e}^a$. Define the series

$$\begin{aligned} \zeta _u(s)=\sum _{m=1}^\infty \frac{v_u(m)}{m^s}. \end{aligned}$$

It will be proved below that the latter series is absolutely convergent in the half plane $\sigma >1/2$, thus, for $s\in D$ as well.

For the precise statement of the approximation theorem for the function $\zeta _u(s)$, we need one particular topological group. Let $\gamma =\{s\in {\mathbb {C}}: |s|=1\}$, ${\mathbb {P}}$ denote the set of all prime numbers, and

$$\begin{aligned} \Omega = \prod _{p\in {\mathbb {P}}} \gamma _p, \end{aligned}$$

where $\gamma _p=\gamma$ for all $p\in {\mathbb {P}}$. By the Tikhonov theorem, the infinite-dimensional torus $\Omega$ with pointwise multiplication and the product topology is a compact topological Abelian group. Therefore, on $(\Omega , {\mathcal {B}}(\Omega ))$ (${\mathcal {B}}({\mathbb {X}})$ is the Borel $\sigma$-field of the space ${\mathbb {X}}$), the probability Haar measure $m_H$ can be defined. This leads to the probability space $(\Omega , {\mathcal {B}}(\Omega ), m_H)$. Denote by H(D) the space of analytic functions on D endowed with the topology of uniform convergence on compacta, by $\omega (p)$ the pth component of an element $\omega \in \Omega$, $p\in {\mathbb {P}}$, and on the probability space $(\Omega , {\mathcal {B}}(\Omega ), m_H)$, define the H(D)-valued random element

$$\begin{aligned} \zeta (s, \omega )= \prod _{p\in {\mathbb {P}}} \left( 1-\frac{\omega (p)}{p^s}\right) ^{-1}. \end{aligned}$$

Note that the latter infinite product converges uniformly on compact subsets of the strip D for almost all $\omega \in \Omega$, see [1] or [5].

The main result of the paper is the following theorem.

Theorem 1.1

Suppose that $n_T\rightarrow \infty$ and $n_T\ll T^2$ as $T\rightarrow \infty$. Let $K\in {\mathcal {K}}$ and $f(s)\in H_0(K)$. Then the limit

$$\begin{aligned}&\lim _{T\rightarrow \infty } \frac{1}{T} \mathrm {meas} \left\{ \tau \in [0,T]: \sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|<\varepsilon \right\} \\&\quad = m_H\left\{ \omega \in \Omega : \sup _{s\in K} |\zeta (s, \omega )-f(s)|<\varepsilon \right\} >0 \end{aligned}$$

exists for all but at most countably many $\varepsilon >0$.

Theorem 1.1 implies that there exists $T_0=T_0(f,K, \varepsilon )>0$ such that, for every $T\ge T_0$, there are infinitely many shifts $\zeta _{n_T}(s+i\tau )$ approximating a given function f(s).

2 The function $\zeta _u(s)$

Denote by $\Gamma (s)$ the Euler gamma-function, and define

$$\begin{aligned} l_u(s)=\frac{s}{\theta }\Gamma \left( \frac{s}{\theta }\right) u^s, \end{aligned}$$

where $\theta >1/2$ is from the definition of $v_u(m)$. For convenience, recall some properties of $\Gamma (s)$.

Lemma 2.1

For arbitrary $\sigma _1<\sigma _2$, there exists $c>0$ such that the estimate

$$\begin{aligned} \Gamma (\sigma +it)\ll \exp \{-c|t|\} \end{aligned}$$

holds uniformly in $\sigma$, $\sigma _1\le \sigma \le \sigma _2$.

Proof of the lemma can be found, for example, in [4].

Lemma 2.2

For positive a and b,

$$\begin{aligned} \frac{1}{2\pi i} \int \limits _{b-i\infty }^{b+i\infty } \Gamma (s)a^{-s}\, \mathrm {d}s=\mathrm {e}^{-a}. \end{aligned}$$

The equality of the lemma is the classical Mellin formula, for the proof, see, for example, [12].

Lemma 2.3

The series for $\zeta _u(s)$ is absolutely convergent for $\sigma >1/2$. Moreover, the equality

$$\begin{aligned} \zeta _u(s)= \frac{1}{2\pi i} \int \limits _{\theta -i\infty }^{\theta +i\infty } \zeta (s+z) l_u(z) \frac{\mathrm {d} z}{z} \end{aligned}$$

is valid.

Proof

In view of Lemma 2.2,

$$\begin{aligned}&\frac{1}{2\pi i} \int \limits _{\theta -i\infty }^{\theta +i\infty } \frac{1}{\theta } \Gamma \left( \frac{s}{\theta }\right) \left( \frac{u}{m}\right) ^s\, \mathrm {d} s \nonumber \\&\quad = \frac{1}{2\pi i} \int \limits _{1-i\infty }^{1+i\infty } \Gamma (s) \left( \left( \frac{m}{u}\right) ^\theta \right) ^{-s}\, \mathrm {d} s\nonumber \\&\quad = \exp \left\{ -\left( \frac{m}{u}\right) ^\theta \right\} . \end{aligned}$$

(2.1)

Therefore, by Lemma 2.1,

$$\begin{aligned} \exp \left\{ -\left( \frac{m}{u}\right) ^\theta \right\} \ll _\theta \int \limits _{-\infty }^\infty \left| \Gamma \left( 1+\frac{it}{\theta }\right) \right| \frac{u^\theta }{m^{\theta }}\, \mathrm {d} t\ll _{\theta ,u} m^{-\theta }. \end{aligned}$$

Since $\theta > 1/2$, this shows the absolute convergence in the half-plane $\sigma >1/2$ of the series for $\zeta _u(s)$.

Now, using (2.1) and the definition of $l_u(s)$, we find

$$\begin{aligned} \zeta _u(s)&= \frac{1}{2 \pi i} \sum _{m=1}^\infty \frac{1}{m^s} \int \limits _{\theta -i\infty }^{\theta +i\infty } \frac{z}{\theta } \Gamma \left( \frac{z}{\theta }\right) \left( \frac{u}{m}\right) ^z \frac{\mathrm {d}z}{z}\\&= \frac{1}{2 \pi i} \int \limits _{\theta -i\infty }^{\theta +i\infty } \frac{l_u(z)}{z} \sum _{m=1}^\infty \frac{1}{m^{s+z}}\, \mathrm {d} z= \frac{1}{2 \pi i} \int \limits _{\theta -i\infty }^{\theta +i\infty } \zeta (s+z) \frac{l_u(z) \mathrm {d} z}{z}. \end{aligned}$$

$\square$

3 Mean distance between $\zeta _{n_T}(s)$ and $\zeta (s)$

For $n_T$ sufficiently large, the coefficients of the series for $\zeta _{n_T}(s)$ are close to 1. This suggests that the function $\zeta _{n_T}(s)$ is close to $\zeta (s)$ even in the strip D. Actually, those two functions are close in the mean.

Lemma 3.1

Suppose that $n_T\rightarrow \infty$ and $n_T\ll T^2$ as $T\rightarrow \infty$. Then, for every compact set $K\subset D$,

$$\begin{aligned} \lim _{T\rightarrow \infty } \frac{1}{T} \int \limits _0^T \sup _{s\in K} |\zeta (s+i\tau )-\zeta _{n_T}(s+i\tau )|\, \mathrm {d} \tau =0. \end{aligned}$$

Proof

By Lemma 2.3, for $s\in D$, we have

$$\begin{aligned} \zeta _{n_T}(s)=\frac{1}{2\pi i} \int \limits _{\theta -i\infty }^{\theta +i\infty } \zeta (s+z)l_{n_T}(z)\frac{\mathrm {d}z}{z}. \end{aligned}$$

(3.1)

Let $0<{\hat{\theta }}<2\theta$. Then (3.1), properties of the function $\zeta (s)$, and the residue theorem imply

$$\begin{aligned} \zeta _{n_T}(s)-\zeta (s)= \frac{1}{2\pi i} \int \limits _{-{\hat{\theta }}-i\infty }^{-{\hat{\theta }}+i\infty } \zeta (s+z) l_{n_T}(z)\frac{\mathrm {d}z}{z} +\frac{l_{n_T}(1-s)}{1-s}. \end{aligned}$$

(3.2)

Denote the points of the set K by $s=\sigma +iv$. Then there exists $\varepsilon >0$ such that $1/2+2\varepsilon \le \sigma \le 1-\varepsilon$ for all $s\in K$. Taking

$$\begin{aligned} {\hat{\theta }}= \sigma -\frac{1}{2}-\varepsilon >0, \end{aligned}$$

we obtain from (3.2) that, for all $s\in K$,

$$\begin{aligned}&\zeta _{n_T}(s+i\tau )-\zeta (s+i\tau )= \frac{1}{2 \pi i} \int \limits _{-\infty }^\infty \zeta \left( \sigma +iv +i\tau -\sigma +\frac{1}{2}+it+\varepsilon \right) \\&\quad \times l_{n_T} \left( \frac{1}{2}+\varepsilon -s+i(v+t)\right) \frac{\mathrm {d}t}{1/2+\varepsilon -s+i(v+t)} +\frac{l_{n_T}(1-s-i\tau )}{1-s-i\tau }. \end{aligned}$$

Hence, putting t in place of $v+t$ gives

$$\begin{aligned} \zeta _{n_T}(s+i\tau )-\zeta (s+i\tau )=&\frac{1}{2 \pi i} \int \limits _{-\infty }^\infty \zeta \left( \frac{1}{2}+\varepsilon +i\tau +it\right) \\&\times \frac{l_{n_T}(1/2+\varepsilon -s +it)}{1/2+\varepsilon -s +it}\, \mathrm {d} t+\frac{l_{n_T}(1-s-i\tau )}{1-s-i\tau }\\ \ll&\int \limits _{-\infty }^\infty \left| \zeta \left( \frac{1}{2}+\varepsilon +i\tau +it\right) \right| \\ {}&\times \sup _{s\in K}\frac{|l_{n_T}(1/2+\varepsilon -s +it)|}{|1/2+\varepsilon -s +it|}\, \mathrm {d} t \\ {}&\quad + \sup _{s\in K}\frac{|l_{n_T}(1-s-i\tau )|}{|1-s-i\tau |}. \end{aligned}$$

Therefore, in view of Lemma 2.1,

$$\begin{aligned} \frac{1}{T} \int \limits _0^T \sup _{s\in K} \left| \zeta _{n_T}(s+i\tau )-\zeta (s+i\tau )\right| \, \mathrm {d}\tau \ll I_{1T}+ I_{2T}, \end{aligned}$$

(3.3)

where

$$\begin{aligned} I_{1T}= \int \limits _{-\infty }^\infty \left( \frac{1}{T} \int \limits _0^T\left| \zeta \left( \frac{1}{2}+\varepsilon +i\tau +it\right) \right| \, \mathrm {d}\tau \right) \sup _{s\in K}\frac{|l_{n_T}(1/2+\varepsilon -s +it)|}{|1/2+\varepsilon -s +it|}\, \mathrm {d} t \end{aligned}$$

and

$$\begin{aligned} I_{2T}= \frac{1}{T} \int \limits _0^T \sup _{s\in K} \left| \frac{l_{n_T}(1-s-i\tau )}{1-s-i\tau }\right| \, \mathrm {d}\tau . \end{aligned}$$

It is well known that, for fixed $\sigma$, $1/2<\sigma <1$,

$$\begin{aligned} \int \limits _{-T}^T|\zeta (\sigma +it)|^2\, \mathrm {d} t \ll _{\sigma } T. \end{aligned}$$

Therefore, for all $t\in {\mathbb {R}}$ and $T\ge 1$,

$$\begin{aligned} \begin{aligned} \frac{1}{T} \int \limits _{0}^T \left| \zeta \left( \frac{1}{2}+\varepsilon +i\tau +it\right) \right| \, \mathrm {d}\tau&\le \left( \frac{1}{T} \int \limits _{0}^T \left| \zeta \left( \frac{1}{2}+\varepsilon +i\tau +it\right) \right| ^2\, \mathrm {d}\tau \right) ^{1/2}\\&\ll \left( \frac{1}{T} \int \limits _{-|t|}^{T+|t|} \left| \zeta \left( \frac{1}{2}+\varepsilon +i\tau \right) \right| ^2\, \mathrm {d}\tau \right) ^{1/2}\\&\ll _\varepsilon \left( \frac{T+|t|}{T}\right) ^{1/2} \ll _\varepsilon 1+|t|. \end{aligned} \end{aligned}$$

(3.4)

Moreover, taking into account Lemma 2.1, we find that, for all $s\in K$,

$$\begin{aligned} \frac{l_{n_T}(1/2+\varepsilon -s+it)}{1/2+\varepsilon -s+it}&\ll _{\theta } n_T^{1/2+\varepsilon -\sigma } \left| \Gamma \left( \frac{1}{\theta } \left( \frac{1}{2} +\varepsilon -\sigma -iv+it\right) \right) \right| \\&\ll _\theta n_T^{-\varepsilon } \exp \left\{ -\frac{c}{\theta }|t-v|\right\} \\&\ll _{\theta ,K} n_T^{-\varepsilon } \exp \{-c_1|t|\}, \quad c_1>0. \end{aligned}$$

This and (3.4) show that

$$\begin{aligned} I_{1T}\ll _{\varepsilon , \theta , K} n_T^{-\varepsilon }\int \limits _{-\infty }^\infty (1+|t|)\exp \{-c_1|t|\}\, \mathrm {d} t\ll _{\varepsilon , \theta , K} n_T^{-\varepsilon }. \end{aligned}$$

(3.5)

Similarly, we obtain that, for $s\in K$,

$$\begin{aligned} \frac{l_{n_T}(1-s-i\tau )}{1-s-i\tau }&\ll _\theta n_T^{1-\sigma } \left| \Gamma \left( \frac{1}{\theta } (1-\sigma -iv-i\tau ) \right) \right| \\&\ll _{\theta } n_T^{1/2-2\varepsilon }\exp \left\{ -\frac{c}{\theta }|\tau +v|\right\} \ll _{\theta , K} n_T^{1/2-2\varepsilon }\exp \{-c_1|\tau |\}. \end{aligned}$$

Therefore,

$$\begin{aligned} I_{2T}&\ll _{\theta ,K} n_T^{1/2-2\varepsilon } \frac{1}{T} \int \limits _0^T \exp \{-c_1|\tau |\}\, \mathrm {d}\tau \ll _{\theta ,K} \frac{n_T^{1/2-2\varepsilon }}{T} \int \limits _0^\infty \exp \{-c_1\tau \}\, \mathrm {d}\tau \\&\ll _{\theta ,K} \frac{n_T^{1/2-2\varepsilon }}{T}. \end{aligned}$$

Since $n_T\ll T^2$, this, (3.3), and (3.5) prove the lemma. $\square$

4 Limit theorem

The inequality (1.1) and its modification with “$\lim$” are derived in [1, 5, 7] from a probabilistic limit theorem for measures in the space of analytic functions H(D). For $A\in {\mathcal {B}}(H(D))$, define

$$\begin{aligned} P_T(A)=\frac{1}{T} \mathrm {meas} \left\{ \tau \in [0,T]: \zeta (s+i\tau )\in A\right\} , \end{aligned}$$

and denote by $P_\zeta$ the distribution of the H(D)-valued random element $\zeta (s, \omega )$, i.e.,

$$\begin{aligned} P_\zeta (A)= m_H\{ \omega \in \Omega : \zeta (s, \omega )\in A\}. \end{aligned}$$

Then the following statement is true [1, 5].

Lemma 4.1

$P_T$ converges weakly to $P_\zeta$ as $T\rightarrow \infty$. Moreover, the support of the measure $P_\zeta$ is the set $S=\{g\in H(D): g(s)\ne 0\; or\; g(s)\equiv 0\}$.

Several equivalents of weak convergence of probability measures are known, see, for example, [2]. For us, the equivalent in terms of continuity sets is useful. Recall that a set $A\in {\mathcal {B}}({\mathbb {X}})$ is called a continuity set of a measure P on $({\mathbb {X}}, {\mathcal {B}}({\mathbb {X}}))$ if $P(\partial A)=0$, where $\partial A$ denotes the boundary of the set A.

Lemma 4.2

Suppose that $P_n$, $n\in {\mathbb {N}}$, and P are probability measures on $({\mathbb {X}}, {\mathcal {B}}({\mathbb {X}}))$. Then $P_n$ converges weakly to P as $n\rightarrow \infty$ if and only if, for every continuity set A of P,

$$\begin{aligned} \lim _{n\rightarrow \infty } P_n(A)=P(A). \end{aligned}$$

Proof of the lemma can be found, for example, in [2].

5 Proof of Theorem 1.1

We will apply a method of characteristic functions which is used in the theory of weak convergence of probability measures on $({\mathbb {R}}, {\mathcal {B}}({\mathbb {R}}))$. We recall that every left continuous non-decreasing function F(x) on ${\mathbb {R}}$ such that $F(+\infty )=1$ and $F(-\infty )=0$ coincides with a certain distribution function. Note that the left continuity of distribution functions can be replaced by right continuity, however, for our aims the left continuity is more convenient. We say that the distribution function $F_n(x)$, $n\in {\mathbb {N}}$, converges weakly to a distribution function F as $n\rightarrow \infty$ if $\lim _{n\rightarrow \infty } F_n(x)= F(x)$ for every continuity point x of F(x).

Every distribution function F(x) is uniquely defined by its characteristic function

$$\begin{aligned} \int \limits _{-\infty }^\infty \mathrm {e}^{iu x}\, \mathrm {d} F(x), \quad u\in {\mathbb {R}}. \end{aligned}$$

Moreover, the following classical continuity theorem is valid.

Lemma 5.1

Suppose that $F_n(x)$, $n\in {\mathbb {N}}$, and F(x) are distribution functions, and $g_n(u)$ and g(u) are the corresponding characteristic functions. If $F_n(x)$, as $n\rightarrow \infty$, converges weakly to F(x), then $\lim _{n\rightarrow \infty } g_n(u)= g(u)$, $u\in {\mathbb {R}}$. This convergence is uniform in every finite interval. If $\lim _{n\rightarrow \infty }g_n(u)=g(u)$, $u\in {\mathbb {R}}$, where g(u) is a continuous function at $u=0$, then there exists a distribution function F(x) such that $F_n(x)$ converges weakly to F(x) as $n\rightarrow \infty$. In this case, g(u) is the characteristic function of F(x).

For the proof of Theorem 1.1, the Mergelyan theorem on the approximation of analytic functions by polynomials [10] is also needed, see the next lemma.

Lemma 5.2

Suppose that $K\subset {\mathbb {C}}$ is a compact set with connected complement, and g(s) is a continuous function in K and analytic in the interior of K. Then, for every $\varepsilon >0$, there exists a polynomial p(s) such that

$$\begin{aligned} \sup _{s\in K} |g(s)-p(s)|<\varepsilon . \end{aligned}$$

Proof of Theorem 1.1

By Lemma 4.1, $P_T$ converges weakly to $P_\zeta$ as $T\rightarrow \infty$. Define the set

$$\begin{aligned} A_\varepsilon =\left\{ g\in H(D): \sup _{s\in K} |g(s)-f(s)|<\varepsilon \right\} . \end{aligned}$$

The boundaries $\partial A_{\varepsilon _1}$ and $\partial A_{\varepsilon _2}$ do not intersect for different positive $\varepsilon _1$ and $\varepsilon _2$. Therefore, the set $A_\varepsilon$ is a continuity set of the measure $P_\zeta$ for all but at most countably many $\varepsilon >0$. This, the weak convergence of $P_T$, and Lemma 4.2 imply the relation

$$\begin{aligned} \lim _{T\rightarrow \infty } P_T(A_\varepsilon )= P_\zeta (A_\varepsilon ) \end{aligned}$$

(5.1)

for all but at most countably many $\varepsilon >0$. Moreover, since by Lemma 4.1, the support of $P_\zeta$ is the set S, the inequality

$$\begin{aligned} P_\zeta ({\hat{A}}_\varepsilon )>0, \end{aligned}$$

(5.2)

where p(s) is a polynomial and

$$\begin{aligned} {\hat{A}}_\varepsilon =\left\{ g\in H(D): \sup _{s\in K} \left| g(s)-\mathrm {e}^{p(s)}\right| <\frac{\varepsilon }{2}\right\} , \end{aligned}$$

is true. Lemma 5.2 ensures the choice of the polynomial p(s) satisfying

$$\begin{aligned} \sup _{s\in K} \left| f(s)-\mathrm {e}^{p(s)}\right| <\frac{\varepsilon }{2}. \end{aligned}$$

The latter inequality together with the definitions of the sets $A_\varepsilon$ and ${\hat{A}}_\varepsilon$ implies the inclusion ${\hat{A}}_\varepsilon \subset A_\varepsilon$. Therefore, in view of (5.2),

$$\begin{aligned} P_\zeta (A_\varepsilon )>0. \end{aligned}$$

By the definitions of $P_T$, $P_\zeta$, and $A_\varepsilon$,

$$\begin{aligned} P_T(A_\varepsilon )=\frac{1}{T} \mathrm {meas} \left\{ \tau \in [0,T]: \sup _{s\in K} |\zeta (s+i\tau )-f(s)|<\varepsilon \right\} \end{aligned}$$

and

$$\begin{aligned} P_\zeta (A_\varepsilon )=m_H \left\{ \omega \in \Omega : \sup _{s\in K} |\zeta (s, \omega )-f(s)|<\varepsilon \right\} . \end{aligned}$$

Thus, $F_T(\varepsilon )\mathop {=}\limits ^{def} P_T(A_\varepsilon )$ and $F_\zeta (\varepsilon )\mathop {=}\limits ^{def} P_\zeta (A_\varepsilon )$ with respect to $\varepsilon$ are distribution functions. Moreover,

$$\begin{aligned} P_\zeta (B_\varepsilon )- P_\zeta (A_\varepsilon )= P_\zeta (\partial A_\varepsilon ), \end{aligned}$$

where

$$\begin{aligned} B_\varepsilon =\left\{ g\in H(D): \sup _{s\in K} |g(s)-f(s)|\le \varepsilon \right\} . \end{aligned}$$

Thus,

$$\begin{aligned} P_\zeta (\partial A_\varepsilon )= F_\zeta (\varepsilon +0)-F_\zeta (\varepsilon ). \end{aligned}$$

This shows that $F_\zeta (\varepsilon )$ is right continuous, thus continuous, only if the set $A_\varepsilon$ is a continuity set of the measure $P_\zeta$. Therefore, in view of (5.1), the distribution function $F_T$ converges weakly to $F_\zeta$ as $T\rightarrow \infty$. Hence, by the first part of Lemma 5.1,

$$\begin{aligned} \lim _{T\rightarrow \infty } g_T(u)= g_\zeta (u) \end{aligned}$$

(5.3)

uniformly in u in every finite interval, where $g_T(u)$ and $g_\zeta (u)$ are the characteristic functions of the distribution functions $F_T$ and $F_\zeta$, respectively.

Denote by ${\hat{g}}_T(u)$ the characteristic function of the distribution function

$$\begin{aligned} {\hat{F}}_T(\varepsilon ) \mathop {=}\limits ^{def} \frac{1}{T} \mathrm {meas} \left\{ \tau \in [0,T]: \sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|< \varepsilon \right\} . \end{aligned}$$

Using the inequality $|\mathrm {e}^{iu}-1|\le |u|$, $u\in {\mathbb {R}}$, and the triangle inequality

$$\begin{aligned} \left| \sup _{s\in K} \left| \zeta _{n_T}(s+i\tau )-f(s)\right| -\sup _{s\in K} \left| \zeta (s+i\tau )-f(s)\right| \right| \\ \le \sup _{s\in K} \left| \zeta (s+i\tau )-\zeta _{n_T}(s+i\tau )\right| , \end{aligned}$$

we obtain by (5.3) that

$$\begin{aligned} {\hat{g}}_T(u)=&\int \limits _{-\infty }^\infty \mathrm {e}^{iu\varepsilon }\, \mathrm {d}{\hat{F}}_T(\varepsilon )= \frac{1}{T} \int \limits _0^T \exp \left\{ iu \sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|\right\} \, \mathrm {d} \tau \\=&\frac{1}{T} \int \limits _0^T \exp \left\{ iu\left( \sup _{s\in K} |\zeta (s+i\tau )-f(s)| +\sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|\right. \right. \\ {}&-\left. \left. \sup _{s\in K} |\zeta (s+i\tau )-f(s)|\right) \right\} \, \mathrm {d}\tau \\ =&\frac{1}{T} \int \limits _0^T \exp \left\{ iu\sup _{s\in K} |\zeta (s+i\tau )-f(s)|\right\} \, \mathrm {d}\tau \\ {}&+O\left( \frac{|u|}{T} \int \limits _0^T\left| \sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|\right. \right. \\ {}&\quad -\left. \left. \sup _{s\in K} |\zeta (s+i\tau )-f(s)|\right| \, \mathrm {d}\tau \right) \\=&g_\zeta (u)+o(1)+ O\left( \frac{|u|}{T} \int \limits _0^T \sup _{s\in K}|\zeta (s+i\tau )-\zeta _{n_T}(s+i\tau )|\, \mathrm {d}\tau \right) \end{aligned}$$

as $T\rightarrow \infty$. Therefore, in view of Lemma 3.1, we have

$$\begin{aligned} {\hat{g}}_T(u)=g_\zeta (u)+o(1) \end{aligned}$$

as $T\rightarrow \infty$, uniformly in u in every finite interval. The function $g_\zeta (u)$, $|u|\le C$ with every $C>0$, as a characteristic function is continuous at the point $u=0$. This and the second part of Lemma 5.1 show that ${\hat{F}}_T$, as $F_T$, converges weakly to $F_\zeta$ as $T\rightarrow \infty$, and we have

$$\begin{aligned} \lim _{T\rightarrow \infty } \frac{1}{T}&\mathrm {meas} \left\{ \tau \in [0,T]: \sup _{s\in K} |\zeta _{n_T}(s+i\tau )-f(s)|<\varepsilon \right\} \\ {}&= m_H\left\{ \omega \in \Omega : \sup _{s\in K} |\zeta (s, \omega )-f(s)|<\varepsilon \right\} >0 \end{aligned}$$

for all but at most countably many $\varepsilon >0$. $\square$