On the Voronin s universality theorem for the Riemann zeta-function

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1 On the Voronin s universality theorem for the Riemann zeta-function Ramūnas Garunštis Abstract. We present a slight modification of the Voronin s proof for the universality of the Riemann zeta-function. The difference (and simplification) is that we do not use the rearrangement of terms in functional series. 1 Introduction In [8] Voronin proved the well nown universality theorem, which states: let 0 < r < 1/4. Suppose that f(s) is an analytic function in the interior of the disc s r and is continuous up to the boundary of this disc. If f(s) 0 for s < r, then for every ε > 0 there exists T = T (ε) such that max g(s) ζ(s + 3/4 + it ) < ε. s r The Voronin s proof is ineffective, that means that from it an upper bound for T = T (ε) can not be derived. One of the reasons is that Voronin uses the theorem of Pechersii on the rearrangement of terms in functional series. In some partial cases the effective proof was obtained in Garunštis [3]. The mentioned paper, wors of Gone [4] and Good [5] suggest some simplification of Voronin s proof. In the next chapter (see Proposition 1) we give a slight modification of Main Lemma from Voronin [8] (or Lemma 1 from [6], 7.1). In the proof of Proposition 1 we avoid the rearrangements of series. The remaining proof of the universality theorem for the Riemann zeta-function is the same as in 7.1 of [6]. For another way to prove universality theorems, using the method of limit theorems, see Bagchi [1], Laurinčias [7]. In this case the proof is also ineffective. Partially supported by Grant from Lithuanian Foundation of Studies and Science. 1

2 2 Proof For a complex s and for a real vector θ = (θ 1,..., θ m ) we define ζ m (s, θ) := ( ) 1 1 e 2πiθp. p s p m Proposition 1 Let 0 < r < 1/4. Suppose that g(s) is analytic for s < r and continuous for s r. Then for any ε > 0 and y > 0 there exist integer m y and a vector θ = (θ 1,..., θ n ) with θ j {0, 1/4, 1/2, 3/4}, j = 1,..., n such that where max s r g(s) log ζ m (s + 3 ) 4, θ < ε, log ζ m (s, θ) = log p m ( 1 e 2πiθ p p s To proof the proposition we need few lemmas. ) = p m ( e 2πiθ p p s e 2 2πiθ ) p p 2s Lemma 2 Let X be a linear normed space and let D X be a convex set, closed in view of the norm of X. Then for any vector s X \ D there are ε > 0 and linear functional f X, such that for all x D. Rf(x) Rf(s) ɛ For a proof of Lemma 2 see ([2], 5.2). From Lemma 2 we obtain: Lemma 3 Let H be a real Hilbert space and let D X be a convex set, closed in view of the norm of X. If D H then there is a vector e H, e = 1, such that sup(x, e) +. x D Lemma 4 Let u n, n IN be vectors of a real Hilbert space H and let the series u n satisfies the condition u n 2 <. Let for any e H, e = 1 the sequence { M (u n, e) : M IN } is unbounded. Then for any number y, any s H and any ε > 0 there are integer N y and numbers α 1,..., α n equal to 1 or 1 such that α n u n < ε. 2

3 Proof. Let s H, and let ε > 0 be arbitrary. Choose m so that m y and u n 2 < ε 2 /36. Let P m denote the set of all vectors x H of the form N x = λ n u n, where λ n [ 1, 1], n = m, m + 1,..., N, and N = m, m + 1, m + 2,.... The set P m is convex. Let P m be the closure of P m in the norm of H. Then P m is a closed convex set. By assumption and Lemma 3 we see that P m = H. Consequently, there exist N = N(s) m and λ n 1, n = m, m + 1,..., N such that λ n u n < ε/3. Using induction and the property x = (x, x), it is easy to construct numbers α m,..., α N equal to -1 or 1 such that Then N N 2 N λ n u n α n u n 4 u n 2 < ε 2 /9. α n u n < ε. Now Lemma 4 follows, if we apply the above to s m 1 u n instead of s. Proof of Proposition 1. γ 2 r < 1/4 and By the continuity of g(s), there exists γ > 1 such that max g(s) s r g(s/γ2 ) < ε. The function g(s/γ 2 ) belongs to the Hardy space H (γr) 2. We remind that the Hardy space H (R) 2 is the set of functions f(s) analytic in s < R with the norm f = lim f(s) dσdt <. r R s <r 3

4 Let s define a scalar product in H (R) 2 by the formula (ϕ 1 (s), ϕ 2 (s)) = R ϕ 1 (s)ϕ 2 (s)dσdt. s R This maes H (R) 2 into a real Hilbert space. To prove the proposition we apply Lemma 4. Since the series ( ) log 1 e 2πi/4 =1 p s+3/4 differs by an absolutely convergent series from the series e 2πi/4 η p (s) := =1 =1 p s+3/4 it suffices to verify the conditions of Lemma 4 for the last series. We have η p (s) <, =1 =1 p 3/4 R since 0 < R < 1/4. Next, let ϕ(s) H (R) 2 with ϕ(s) = 1. Then Voronin [8] (or see [6], 7.1) proved that there exists a subseries of (η (s), ϕ(s)) =1 diverging to. Proposition 1 is proved. References [1] B. Bagchi, The statistical behaviour and universality properties of the Riemann zeta-function and other alied Dirichlet series, Ph. D. Thesis, Calcutta, Indian Statistical Institute, [2] N. Dunford and J. T. Schwartz, Linear operators. II: General theory, Intercience publishers, New Yor, London [3] R. Garunštis, The effective universality theorem for the Riemann zeta function, submited [4] S. M. Gone, Analytic properties of zeta and L-functions, Ph. D. Thesis, University of Michigan, (1979). 4

5 [5] A. Good, On the distribution of the values of Riemann s Zeta-function, Acta Arith. 38 (1981), [6] Karatsuba and Voronin, The Riemann zeta-function, De Gruyter Expositions in Mathematics. 5. Berlin etc.: W. de Gruyter. xii, (1992). [7] A. Laurinčias, Limit theorems for the Riemann zeta-function, Kluwer Academic Publishers, Dordrecht, (1996). [8] S.M. Voronin, Theorem on the universality of the Riemann zeta-function, Izv. Aad. Nau SSSR, Ser. Matem., 39 (1975) (in Russian); Math. USSR Izv. 9 (1975), Ramūnas Garunštis Department of Mathematics and Informatics Vilnius University Naugarduo Vilnius Lithuania ramunas.garunstis@maf.vu.lt 5