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

In the paper, it is obtained that there are infinite discrete shifts Ξ(s + ikh), h > 0, k ∈ N0 of the Mellin transform Ξ(s) 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.


Introduction
As usual, ζ(s) is denoted by s = σ + it, the Riemann zeta-function, which, for σ > 1, is defined by 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 ζ(s), the modified Mellin transforms Ξ k (s) play an important role.For k 0 and σ > σ(k) > 1, the functions Ξ k (s) are defined by x −s dx.
The functions Ξ k (s) were introduced in [1,2] and are applied for the investigation of the moments In general, Ξ k (s) are attractive analytic functions and are widely studied; see, for example, [3][4][5][6].
In [7], the approximation properties of the function Ξ 1 (s) were studied.Let G = {s ∈ C : 1/2 < σ < 1}.H(G) is denoted by the space of analytic functions on G endowed with the topology of uniform convergence on compacta, and by measA the Lebesgue measure of a measurable set A ⊂ R.Then, in [7], the following theorem is proven.Theorem 1.There exists a closed, non-empty set F ⊂ H(G), such that, for every compact set K ⊂ G, function f (s) ∈ F, and ε > 0, lim inf Moreover, the limit exists and is positive for all but, at most, is a countable number ε > 0.
Theorem 1 is of continuous type, τ in the shifts Ξ 1 (s + iτ) takes arbitrary real values.The aim of this paper is to obtain a discrete version of Theorem 1 with shifts Ξ 1 (s + ikh), where h > 0 is a fixed number and k ∈ N ∪ {0} de f = N 0 .#A denotes the cardinality of a set A ⊂ R. For brevity, we write Ξ(s) in place of Ξ 1 (s).
Let N run over the set N 0 .
Theorem 2. For every h > 0, there exists a closed non-empty set F h ⊂ H(G) such that, for every compact set K ⊂ G, function f (s) ∈ F h , and ε > 0, lim inf Moreover, the limit exists and is positive for all but at most countably many ε > 0.
Theorem 2 shows that the set of discrete shifts Ξ(s + ikh) approximating with a given accuracy the function f (s) ∈ 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 Ξ(s).In some sense, Theorems 1 and 2 recall universality theorems for the function ζ(s).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 H(G)-valued random element defined in terms of Ξ(s).The distribution of that random element is the limit measure in a probabilistic discrete limit theorem for the function Ξ(s).B(X ) denotes the Borel σ-field of the space X , by W −→ the weak convergence of probability measures, and, for Theorem 3.For every fixed h > 0, on (H(G), B(H(G))), there exists a probability measure P h such that P N,h

Some Lemmas
Let a > 1 be a fixed number.Define the set where As a Cartesian product of compact sets, the torus Ω a is a compact topological Abelian group.Let ω = {ω u : u ∈ [1, a]} be elements of Ω a .
For A ∈ B(Ω a ) and h > 0, define Lemma 1.On (Ω a , B(Ω a )), there exists a probability measure Q a,h such that Q N,a,h Proof.We apply the Fourier transform method.Let ), be the Fourier transform of Q N,a,h , i.e., where " * " shows that only a finite number of integers k u are non-zero.Thus, by the definition of Q N,a,h , If then If k = (k u : u ∈ [1, a]) does not satisfy (2), then using the formula of geometric progression gives where This shows that Q N,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 ∈ [1, ∞] and fixed θ > 1/2, define Z n,a,y (s) denotes the integral sum of the function g(x, y)x −s over the interval [1, a], i.e., where , there exists a probability measure P n,a,y,h such that Proof.We apply the following simple remark on the preservation of weak convergence under continuous mappings.Let w : X 1 → X 2 be a (B(X 1 ), B(X 2 ))-measurable mapping.Then, every probability measure P on (X 1 , B(X 1 )) induces the unique probability measure Pw −1 in the space X 2 as well [10].
Define the mapping w n,a,y : Ω a → H(G) by the formula Since the above sum is finite, the mapping w n,a is continuous in the product topology.Moreover, Hence, P N,n,a,y,h = Q N,a,h w −1 n,a,y .Therefore, the above remark, continuity of w n,a,y and Lemma 1 show that P N,n,a,y,h The next step consists of the passage from Z n,a,y (s) to Ξ a,y (s) in Lemma 2. For this, one statement on convergence in distribution ( D −→) of H(G)-valued random elements is useful, and we recall it.There exists a sequence {K l : l ∈ N} ⊂ G of compact embedded sets such that G is union of sets K l , and every compact K ⊂ G lies in some set K l .Then is a metric in H(G) which induces the topology of uniform convergence on compacta.
Lemma 3. Suppose that X, Y N and Y Nl are H(G)-valued random elements defined on the same probability space with measure P such that, for l ∈ N, Moreover, let, for every ε > 0, Proof.Since the space H(G) 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 Proof.In view of the definition of the metric ρ, it is suffice to show that, for arbitrary compact set K ⊂ G, Let L be a simple closed contour lying in G and enclosing a compact set K ⊂ G.Then, by the integral Cauchy formula, where a ξ b, b > 0, means that there exists a constant c = c(ξ) > 0 such that |a| cb.Hence, By the Cauchy-Schwarz inequality, Obviously, where z denotes the complex conjugate of z ∈ C. By the definition of Z n,a,y (s), where r ∈ Z is arbitrary.Therefore, from this we obtain that, for all z ∈ L, By the definition of Ξ a,y (s), for all z ∈ L, we have where r ∈ Z.Therefore, Since the sum of the last two terms in ( 7) is estimated as , equality (5) follows from ( 6)- (10).
For A ∈ B(H(G)), define Lemma 5.For every fixed h > 0, on (H(G), B(H(G))), there exists a probability measure P a,y,h such that P N,a,y,h Proof.Let θ N,h be a random variable defined on a certain probability space with measure P, and having the distribution X n,a,y,h denotes the H(G)-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 H(G)-valued random element X N,n,a,y,h = X N,n,a,y,h (s) = Z n,a,y (s + iθ N,h ).
Then, in view of Lemma 2, we have Consider the sequence {P n,a,y,h : n ∈ N}.Let K l be the sets from the definition of the metric ρ.Then, applying the integral Cauchy formula and ( 9), we find that sup n∈N lim sup Fix ε > 0 and define V l = V l,a,y,h = 2 l ε −1 C l,a,y,h .Then, using (11), for all n, l ∈ N. Hence, taking Since the set K is compact in the space H(G), this shows that the sequence {P n,a,y,h } is tight.Therefore, by the Prokhorov theorem; see, for example, [10], the sequence {P n,a,y,h } is relatively compact.Thus, there exists a subsequence {P n r ,a,y,h } weakly convergent to a certain probability measure P a,y,h on (H(G), B(H(G))) as r → ∞.In other words, Define one more H(G)-valued random element Then, Lemma 4 implies that, for every ε > 0, 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 Now, we are ready to prove a discrete limit lemma for the function Since ζ(1/2 + it) t 1/6 , t 1, and v(x, y) decreases exponentially, the integral for Ξ y (s) is absolutely convergent for σ > σ 0 with arbitrary finite σ 0 .
For A ∈ B(H(G)), define Lemma 6.For every fixed h > 0, on (H(G), B(H(G))), there exists a probability measure P y,h such that P N,y,h Proof.Let θ N,h be the same as in the proof of Lemma 5. Define and X a,y,h denotes the H(G)-valued random element with distribution P a,y,h .Then, by Lemma 5, Y N,a,y,h The integral Cauchy formula and (10) lead to sup Therefore, taking V l = V l,y,h = 2 l ε −1 C l,y,h , we find by ( 14) that for all a 1 and l ∈ N.This shows that, for a 1, where This means that the family of probability measures {P a,y,h : a 1} is tight.Hence, there exists a sequence {P a r ,y,h } ⊂ {P a,y,h } weakly convergent to a certain probability measure P y,h as r → ∞.Thus, It remains to show the nearestness in the mean of Ξ a,y (s) and Ξ y (s).We have that, for a compact set K ⊂ D and fixed y > 0, h > 0, as a → ∞.From this, we have Proof.The lemma is Lemma 9 proved in [7].
In addition, we need a discrete mean square estimate for Ξ(s).
Lemma 9.The equality Let K ⊂ G be an arbitrary fixed compact set.Fix ε > 0 such that, for all s = σ + it ∈ K, the inequalities 1/2 + 2ε σ 1 − ε would be satisfied.Then, for such σ, Let θ = 1/2 + ε in Lemma 7. The point z = 1 − s is a double pole, and z = 0 is a simple pole of the function therefore, Lemma 7 and the residue theorem give where r y (s) = Res z=1−s Ξ(s + z) a(z).

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 {P y r ,h } ⊂ {P y,h }, such that 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 H(G) such that P h (F h ) =