Discrete uniform limit law for additive functions on shifted primes
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1 Nonlinear Analysis: Modelling and Control, Vol. 2, No. 4, ISSN Discrete uniform it law for additive functions on shifted primes Gediminas Stepanauskas, Laura Žvinytė Faculty of Mathematics and Informatics, Vilnius University, Naugarduko str. 24, LT Vilnius, Lithuania Received: September 8, 205 / Revised: January 5, 206 / Published online: March 7, 206 Abstract. The sufficient and necessary conditions for a weak convergence of distributions of a set of strongly additive functions f x, x 2, the arguments of which run through shifted primes, to the discrete uniform law are obtained. The case when f xp) {0, } for every prime p is considered. Keywords: additive function, discrete uniform law, frequency, weak convergence. Introduction Let f x, x 2, be a set of strongly additive functions such that f x p) {0, } for all primes p and all x 2. It follows from the strong additivity that for every positive integer n f x n) = f x p) = p n p n f xp)=. The problem of a weak convergence of distributions n x, fx n) < u ) := [x] n x f xn)<u as x is of key importance in probabilistic number theory. There are interesting general conditions of convergence, classes of possible it distributions, conditions of convergence to particular distributions. A detailed account of particular and general results you can find in the monographs [, 2, 5]. c Vilnius University, 206
2 438 G. Stepanauskas, L. Žvinytė In the articles [7, 8, 9], the case of the Poisson distribution as a it law was considered. It was shown there that the Poisson law can occur as a it one for the distributions: p x, fx p + ) < u ), n x, fx n) + g x n + ) < u ), p x, fx p + ) + g x p + 2) < u ). The Bernoulli, geometrical, binomial, discrete uniform distributions as it ones for f x n) < u) were investigated in [0,, 2]. Several results general enough) can be found in [, 4, 3, 4]. In this work, we consider the weak convergence of distribution functions fx p + ) < u ) = p x, fx p + ) < u ) := πx) #{ p x: f x p + ) < u, p prime } ) to the discrete uniform law Uu, L) := k=0,,...,l k<u L, 2) where the parameter L N, L 2. Similarly as in [9], we use in the proofs the method of factorial moments and we have to restrict the behaviour of additive functions on large primes see condition H)). But the authors think that this condition is not necessary for the weak convergence to the discrete uniform distribution. Maybe, the problem could be solved applying the Kubilius model of probability spaces [, 2, 5]. But the large primes have to be overcome there as well. Throughout the paper, we keep the following notation. The values of p, p, p 2,... mean prime numbers; c is an absolute positive constant not always the same. By the symbol εx) we denote values vanishing as x. The notation a b is equivalent to the inequality a cb. If the constant c, the constant included in, or the vanishing function εx) depend on a parameter a, we write c a, a, ε a x). The notation F x u) F u) means that the distribution functions F x u) converge weakly to the distribution function F u) as x. The superscript at the signs of sum or maximum,, max, means that the summation or maximum is expanded over the primes p for which f x p) =. The other notation is generally accepted or is later discussed in the text. 2 Main result and auxiliary lemmas Theorem. Let f x, x 2, be a set of strongly additive functions. Assume that f x p) {0, } for all prime numbers p and log x x γ < p = 0 H)
3 Uniform law for additive functions 439 for all γ 0, ). The distributions f x p + ) < u) converge weakly to the it discrete uniform law Uu, L) as x if and only if L = 2 and f x 3) =, 3< = 0. 3) p The proof of this main theorem is based on the following three lemmas on the it behaviour of factorial moments of the distribution f x p + ) < u). Lemma. Let f x, x 2, be a set of strongly additive functions such that f x p) {0, } for all primes p. If distributions ) converge weakly to some distribution function F u) with a jump at the point u = 0 as x, then the quantities βl, x) := πx) f x p + ) f x p + ) ) f x p + ) l + ), l =, 2,..., have finite its where g l is the lth factorial moment of the it law. βl, x) = g l, 4) Lemma 2. See [9, Lemma 2].) If a set of strongly additive functions f x satisfies the conditions of Theorem and, 5) p then l =, 2,.... βl, x) = p,p 2,...,p l x p )p 2 ) p l ) + ε lx), According to this statement and equality 4) in the case of convergence of f x p + ) < u), we have that for each l {, 2,... }. p,p 2,...,p l x p )p 2 ) p l ) = g l 6) Lemma 3. Let f x, x 2, be a set of strongly additive functions such that f x p) {0, } for all primes p and condition H) hold. If distributions ) converge weakly to the distribution F ξ of the random variable ξ with a finite support {0,,..., L }, L 2, then there exists some constant D 2 such that sup # { p D: f x p) = } L, 7) Nonlinear Anal. Model. Control, 24):
4 440 G. Stepanauskas, L. Žvinytė p,p 2,...,p l D = 0, p D< /L 8) p )p 2 ) p l ) = g l, 9) l =, 2,..., L. Moreover, the characteristic function of the it distribution F ξ is equal to L g l + e it ) l. l! l= From Theorem we get the following example. Example. Let if p = 3, f x p) = if x α p < x α+εx, 0 otherwise, where ε x > 0 and ε x log x 0 as x. Then fx p + ) < u ) Uu, 2) as x. 3 Proofs of lemmas Proof of Lemma. Suppose that distribution functions ) converge weakly to the it distribution F u) with the jump at the point u = 0. From the weak convergence we have that ν x fx p + ) = 0 ) = F 0+) F 0) c. Using this estimate, it is proved in [9] see inequality 0)) that According to this, πx) f xp+)=k βl, x) l, l. 0) = πx) f xp+)=k kk ) k l ) kk ) k l ) βl + 2, x) kk ) k l ) l kk ) k l )
5 Uniform law for additive functions 44 for every k l + 2. Therefore, F k+) F k) = ν x fx p + ) = k ) l kk ) k l ) for every k l + 2. Let us fix l N and choose K > l + 2. Using estimate 0), analogously as in [9], we get βl, x) = πx) = + πx) f xp+) K f xp+)>k K kk ) k l + ) k=l f x p + ) f x p + ) ) f x p + ) l + ) f x p + ) f x p + ) ) f x p + ) l + ) f x p + ) l f x p + ) l πx) f xp+)=k ) βl +, x) + O K l K = kk ) k l + ) F k+) F k) ) + ε K,l x) + O l K l k=l = g l kk ) k l + ) F k+) F k) ) + ε K,l x) + O l k=k+ = g l + ε K,l x) + O l = g l + ε K,l x) + O l K k=k+ ). ) ) + O l k l)k l ) K ) K l Taking the it in the last equality as x tends to infinity and then as K tends to infinity, we obtain relation 4). Lemma is proved. Proof of Lemma 3. The proof of the lemma almost coincides with the proof of the necessity part of Corollary 4 from [6]. From the conditions of the lemma we have that there is some k {0,,..., L } such that Thus, from the inequality see [3]) it follows inequality 5). fx p + ) = k ) = F ξ k+) F ξ k) c. fx p + ) = k ) 4 + ) /2 ) p ) Nonlinear Anal. Model. Control, 24):
6 442 G. Stepanauskas, L. Žvinytė Now according to Lemma 2, equality 6) holds. Let d 2 be a temporarily fixed positive integer. If x/d L > d, we have from equality 6) that g L sup p,p 2,...,p L d p i p j,i j p )p 2 ) p L ). Assuming that sup # { p d: f x p) 0 } L, we get a contradiction to the condition g L = 0. Thus, sup # { p d: f x p) 0 } L for each fixed positive integer d. Put a d = sup # { p d: f x p) 0 }. The sequence a d d 2) is integer-valued, non-decreasing, and bounded. Thus, there exists a positive integer D 2 such that sup # { p D: f x p) 0 } = a d for d D. Since a d L for all positive integers d, we obtain condition 7). On the other hand, from the reasoning above it follows that f xp) = 0 for each fixed prime p > D. Consequently, we obtain that Since for every pair i, j, i < j L, equality 6) shows that D<p,...,p L x /L p i=p j max D< max D< p = 0. p ) p L ) p ) L 0 x ), p ) L g L sup. p D< /L Thus, the condition g L = 0 implies 8). The last condition 9) of the lemma now follows from 6), equality 8), and condition H). Lemma 3 is proved.
7 Uniform law for additive functions Proof of Theorem Necessity. Suppose that fx p + ) < u ) Uu, L) as x 2) with parameter L 2. From Lemma we obtain that βx, l) = L )! L l)! l + ) for l =, 2,..., L. The values of g l are the factorial moments of the it distribution. In the case of the uniform distribution, we get that g = L, g 2 = 2 g k = L )L 2), 3 L )L 2) L k), k = 3, 4,..., L, k + g k = 0, k = L, L +,.... From 2) we have that ν x fx p + ) = 0 ) = U0+, L) U0, L) = L > 0. Thus, inequality 5) follows from inequality ) with k = 0. We apply now Lemma 2. The values of g l are the factorial moments of the it distribution. It is clear that g l g l k g k for all l = 2, 3,... and all k =, 2,..., l. In the particular case, Therefore, g 2 g 2. L )L 2) 3 ) 2 L, 2 which implies L 5. Further, we examine separately the cases L = 2, 3, 4, 5. Let L = 2. In this case, g = 2, g 2 = g 3 = = 0. Using Lemma 3, we have that, for some D 2, sup # { p D: f x p) = } = κ, Nonlinear Anal. Model. Control, 24):
8 444 G. Stepanauskas, L. Žvinytė D< /2 p = 0, p D p = 2. 3) If κ =, then we obtain from relations above and condition H) that there is only one case f x 3) = for large x and 3< p = 0. If κ = 0, then f x p) = 0 for every fixed p and sufficiently large x. In this case, equality 3) cannot be satisfied. It follows that the case κ = 0 cannot occur. Let L = 3. Then g =, g 2 = 2 3, g 3 = g 4 = = 0. According to Lemma 3, we have that, for some fixed D 2, the following conditions hold: sup # { p D: f x p) = } = κ 2, D< /3 p D p,p 2 D p p 2 p = 0, =, 4) p p )p 2 ) = ) First, we suppose that κ = 2. From 4) and condition H) we have that for large x and f x p ) = f x p 2 ) = p + p 2 = for some fixed primes p < p 2. Since the last equality is impossible for any pair of different primes p, p 2, then the case κ = 2 cannot occur. From equality 5) it follows that the cases κ =, κ = 0 cannot occur as well.
9 Uniform law for additive functions 445 Let L = 4. Then g = 3 2, g 2 = 2, g 3 = 3 2, g 4 = g 5 = = 0. According to Lemma 3, there exists some constant D 2 for which sup # { p D: f x p) = } = κ 3, 6) D< /4 p D p,p 2 D p p 2 p,p 2,p 3 D p = 0, p = 3 2, 7) p )p 2 ) = 2, p )p 2 )p 3 ) = ) It follows from 8) that κ cannot be 0,, and 2. Suppose κ = 3. Then equalities 6) and 7) imply that there exist fixed primes p < p 2 < p 3 for which f x p ) = f x p 2 ) = f x p 3 ) = for large x and p + p 2 + p 3 = ) But there are no primes satisfying equality 9). So, the case κ = 3 is impossible as well. Let L = 5. Then g = 2, g 2 = 4, g 3 = 6, g 4 = 24 5, g 5 = g 6 = = 0. According to Lemma 3, there exists some D 2 for which sup # { p D: f x p) = } = κ 4, p D p,p 2 D p p 2 D< /5 p = 0, = 2, 20) p p )p 2 ) = 4, Nonlinear Anal. Model. Control, 24):
10 446 G. Stepanauskas, L. Žvinytė p,p 2,p 3 D p,p 2,p 3,p 4 D p )p 2 )p 3 ) = 6, p )p 2 )p 3 )p 4 ) = ) It follows from equality 2) that κ cannot be equal to 0,, 2, and 3. Suppose κ = 4. Then from 20) we deduce that there exist primes p < p 2 < p 3 < p 4 such that f x p ) = f x p 2 ) = f x p 3 ) = f x p 4 ) = for large x and p + p 2 + p 3 + p 4 = 2. But p + p 2 + p 3 + p < 2. So, the case κ = 4 is impossible as well. Sufficiency. Assume that conditions 3) of the theorem together with additional condition H) are satisfied. Estimate 5) follows from condition 3). Now using Lemma 2, we obtain that if l > and Put βl, x) = for x 2 and t R. Since for all r {0} N, we have p,p 2,...,p l x p )p 2 ) p l ) = 0 β, x) = 2. ψ x t) = πx) e itfxp+) e itr r e it ) rr ) e it 2 2 ψ x t) = + β, x) e it ) + O β2, x) ). Taking the it in the last equality, we conclude that ψ xt) = e it + ). 2 But this is the characteristic function of the uniform distribution Uu, 2). So, the sufficiency of the theorem follows. Theorem is now proved.
11 Uniform law for additive functions 447 References. P.D.T.A. Elliott, Probabilistic Number Theory. I: Mean Value Theorems, Springer, New York, Heidelberg, Berlin, P.D.T.A. Elliott, Probabilistic Number Theory. II: Central Limit Theorems, Springer, New York, Heidelberg, Berlin, P.D.T.A. Elliott, The concentration function of additive functions on shifted primes, Acta Math., 73: 35, I. Kátai, On the distribution of arithmetical functions on the set of primes plus one, Compos. Math., 9: , J. Kubilius, Probabilistic Methods in the Theory of Numbers, Transl. Math. Monogr., Vol., Amer. Math. Soc., Providence, RI, J. Šiaulys, Factorial moments of distributions of additive functions, Lith. Math. J., 404):389 40, J. Šiaulys, G. Stepanauskas, Poisson distribution for a sum of additive functions, Acta Appl. Math., 97: , J. Šiaulys, G. Stepanauskas, Poisson distribution for a sum of additive functions on shifted primes, Acta Arith., 304):403 44, J. Šiaulys, G. Stepanauskas, The Poisson law for additive functions on shifted primes, in E. Manstavičius, A. Laurinčikas Eds.), Analytic and Probabilistic Methods in Number Theory. Proceedings of the Fourth International Conference in Honour of J. Kubilius, Palanga, Lithuania, September 24 30, 2006, TEV, Vilnius, 2007, pp J. Šiaulys, G. Stepanauskas, Some it laws for strongly additive prime number indicators, Šiauliai Math. Semin., 3): , J. Šiaulys, G. Stepanauskas, Binomial it law for additive prime number indicators, Lith. Math. J., 54): , J. Šiaulys, G. Stepanauskas, Discrete uniform it law for additive prime number indicators, in A. Laurinčikas, E. Manstavičius, G. Stepanauskas Eds.), Analytic and Probabilistic Methods in Number Theory. Proceedings of the Fifth International Conference in Honour of J.Kubilius, Palanga, Lithuania, September 25 29, 20, TEV, Vilnius, 202, pp N.N. Timofeev, The distribution of values of additive functions on the sequence {p + }, Mat. Zametki, 336): , 983 in Russian). 4. N.N. Timofeev, Arithmetic functions on the set of shifted primes, Proc. Steklov Inst. Math., 6:3 37, 995. Nonlinear Anal. Model. Control, 24):
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