CHARACTERIZATIONS OF THE LOGARITHM AS AN ADDITIVE FUNCTION

Transcription

1 Annales Univ. Si. Budapest., Set. Comp. 6 (2004) CHARACTERIZATIONS OF THE LOGARITHM AS AN ADDITIVE FUNCTION Bui Minh Phong (Budapest, Hungary) Dediated to Professor János Balázs on the oasion of his 75th birthday. Introdution An arithmeti funtion f(n) is said to be additive if (n, m) = implies that f(nm) = f(n) + f(m), it is ompletely additive if the above equality holds for all positive integers n m. Let A A denote the set of all omplex-valued additive ompletely additive funtions, respetively. The problem onerning the haraterization of funtions f(n) = U log n as additive arithmeti funtions was studied by several authors. It is lear that f(n) = U log n belongs to A. Normally log is onsidered as a mapping IR IR in this ontext it is well known that ontinuity along with the funtional equation f(xy) = f(x) + f(y) haraterizes the logarithm up to a onstant fator. Restriting the domain from R to N reates an interesting question: What further properties along with (omplete) additivity will ensure that an arithmeti funtion f is in fat U log n? Most of the suffiient onditions that are known an be formulated in terms of their differenes. The first suh haraterization is apparently that of P.Erdős. Among other results on additive funtions, P.Erdős [7] proved in 946 that if a real-valued additive funtion f satisfies f(n + ) f(n) 0 (n =, 2,...) or f(n + ) f(n) = o() as n,

2 46 Bui Minh Phong then f(n) is a onstant multiple of log n. He stated several onjetures onerning possible improvements generalizations of his results. In addition P.Erdős onjetured that the last ondition ould be weakened to () f(n + ) f(n) = o(x) as x. This was later proved by I.Kátai [0]. An alternative proof of a slightly stronger form of this onjeture was given by E.Wirsing [6]. Another one of Erdős onjetures was that if an additive funtion f satisfies f(n + ) f(n) = O() as n, then f = U log for ompletely additive f f = U log +O() for additive f. Any additive funtion f with f(n) = 0 for odd n bounded f(2 k ) shows that the term O() must not be dropped. This onjeture was proved by E.Wirsing in [5]. P.Erdős also onjetured that the ondition () ould be replaed by f(n + ) f(n) = o() (n through a set of density ). This onjeture has been proved only very reently by A.Hildebr [5] as a orollary to a more general result on the limit distribution for f(n + ) f(n). Sine the appearane of Erdős paper several new haraterizations of the logarithm have been found that generalize or sharpen Erdős original results in a variety of ways. I.Kátai [8], [9] proposed the problem to obtain similar haraterizations when n n + are replaed by two linear forms an + b An + B. Speifially, I.Kátai asked for a haraterization of those real-valued additive funtions f whih satisfy f(an + B) f(an + b) C as n for some integers A > 0, B, a > 0, b with Ab ab 0 for a real number C. I.Kátai onsidered this problem with b = 0 small values of A B in [8], [9]. The general ase has been treated ompletely solved by P.D.T.A.Elliott [], [2], [3]. Namely, P.D.T.A.Elliott [3] showed that if a realvalued additive funtion f satisfies the above ondition, then f(n) = U log n holds for all positive integers n whih are prime to Aa(Ab ab). On the other h, the ondition () was weakened by I.Kátai in 978. He proved in [] that a funtion f A satisfies lim inf x log x f(n + ) f(n) = 0 n

3 Charaterizations of the logarithm as additive funtion 47 must be of the form f = U log for some omplex onstant U. For further results generalizations of these questions related open problems see the exellent book of P.D.T.A.Elliott [3] the survey papers of I.Kátai [2], Hildebr [6]. In [4] we obtained a omplete haraterization of those funtions f A f 2 A for whih the relation f (an + b) f 2 (n) d = o(x) as x holds for some fixed positive integers a, b for a omplex onstant d. We dedued from the above relation that there are a omplex onstant U funtions F A, F 2 A suh that f (n) = U log n + F (n), f 2 (n) = U log n + F 2 (n) F (an + b) F 2 (n) d + U log a = 0 hold for all positive integers n. We note that this result an be derived from a reent result due to P.D.T.A.Elliott [4], whih was obtained with analyti methods. Our proof in [4] is elementary, it was used in [3]. Our purpose in this paper is to improve some results mentioned above. We shall prove the following Theorem. Let a, b, be positive integers let d be a omplex onstant. Then f A f 2 A satisfy the ondition (2) n f (an + b) f 2 (n) d = o(log x) as x if only if there are a omplex onstant U funtions F A, F 2 A suh that f (n) = U log n + F (n), f 2 (n) = U log n + F 2 (n) ( a ) F (an + b) F 2 (n) d + U log = 0 hold for all positive integers n. Remark. From our proof it will follow that F 2 (n) = F 2 [ (n, b 2 N 2 ) ] (n =, 2,...)

4 48 Bui Minh Phong F (n) = 0 if (n, abn 2 ) =, where N 2 {, 2} satisfying (2, an 2 + ) =. We shall prove Theorem by using the similar result onerning the ase when f = f 2. Theorem 2. Assume that f A satisfies the ondition (3) f(an + B) f(cn) D = o(log x) as x n for some positive integers A, B, C for a omplex onstant D. Then there are a omplex onstant U a funtion F A suh that f(n) = U log n + F (n) F (n) = F [(n, BCC A )] hold for all positive integers n, where C A denotes the produt of all prime divisors of C whih are prime to A. Theorem 3. Let a, b be positive integers. If f A f 2 A satisfy the ondition f (an + b) f 2 (n) = O() as n, then there are a omplex onstant U a funtion F A suh that f (n) = U log n + F (n), f 2 (n) = U log n + O() F (an + b) = O() hold for all positive integers n. In partiular, we have F (n) = O() for all (n, a) =. We note that Theorem 3 is known result, namely it is a onsequene of the theorem of P.D.T.A.Elliott [2]. But our proof is elementary it will use only the result of E.Wirsing [5] onerning the ase a = b = =.

5 Charaterizations of the logarithm as additive funtion Auxiliary lemmas In this setion we introdue some notations prove two lemmas whih will be used at the proofs of our theorems. Let A, B C be fixed positive integers. We shall denote by C A the produt of all distint prime divisors of C whih are prime to A. For an arbitrary positive integer n let E(n) = E B (n) be the produt of all prime power fators of B omposed from the prime divisors of n, i.e. E(n) B, (n, B/E(n)) = = every prime divisor of E(n) is a divisor of n. Lemma. Assume that f A satisfies the ondition (3) for some positive integers A, B, C for a omplex onstant D. Then for eah positive integer k Q we have (4) f ( BCC A Q k) = kf(bcc A Q) (k )f(bcc A ) (5) f ( ACC 2 AE(C) ) = 2f (CC A E(C)) f (E(C)) + D. Proof. For eah positive integer Q we define the sequene R = R(AC A Q) = {R k (AC A Q)} k= by the initial term R (AC A Q) = by the formula (6) R k (AC A Q) = + AC A Q (AC A Q) k for all integers k 2. Moreover, let (7) T k (n, Q) = (AC A Q) k E(CQ)n + BR k (AC A Q). By using (6) (7), we have (8) T k+ (n, Q) = AC A QT k (n, Q) + B (9) (CC A QE(CQ), T k (n, Q)/E(CQ)) =

6 50 Bui Minh Phong for all positive integers k. Thus, using (3), (7), (8), (9) the additivity of f, we have n f (T (n, Q)) f (CC A QE(CQ)n) D = o(log x) as x n f (T k(n, Q)) f (T k (n, Q)) H(Q) = o(log x) as x for eah integer k 2, where H(Q) := f (CC A QE(CQ)) f (E(CQ)) + D. These imply that (0) n f (T k(n, Q)) f (CC A QE(CQ)n) (k )H(Q) D = o(log x) holds for eah positive integer k. We shall dedue from (0) that () f ( A k CC k AQ k P E(CQ) ) = (k )H(Q) + f (CC A QP E(CQ)) holds for every positive integer k, Q P. Let k, Q P be positive integers. Let R k = R k (AC A Q). Considering (2) n := P R k {AP CQR k m + } taking into aount (0), one an dedue that () holds if k, Q P satisfy the ondition (3) (P, R k ) = (P E(CQ) + B, R k ) =. It is obvious that (3) is satisfied in the following ases: P =, Q = 2B P =, Q = 2pB,

7 Charaterizations of the logarithm as additive funtion 5 where p is a prime number. Thus, we get from (), using the fat E(2BC) = = E(2pBC) = B, that beause f ( p k) = kf(p) if (p, 2ABC) =, f ( A k CC k A(2B) k B ) = (k )H(2B) + f ( CC A 2B 2), f ( A k CC k A(2pB) k B ) = (k )H(2pB) + f ( CC A 2pB 2) H(2pB) = f(p) + H(2B) if (p, 2BC) =. Therefore, by using the additivity of f, we have (4) f(nm) = f(n) + f(m) if (n, m, 2ABC) =. Thus, using (0), (2) (4), we see that () also holds if we relax the ondition (3) to (5) (P, R k, 2B) = (P E(CQ) + B, R k, 2) =. Assume that (2, ABC) = k is an odd positive integer. In this ase one an hek that the ondition (5) is satisfied for P = Q = P =, Q = 2. Thus, () holds in these ases, so we an dedue in the same way as above that (6) f ( 2 k) = kf(2) for all odd positive integers k. On the other h, (5) also holds for P = 2 ν, Q = 2 k = 2, where ν 0 is an integer. From () we have (7) f ( ACC 2 A2 ν+2 E(C) ) = H(2) + f ( CC A 2 ν+ E(C) ). Thus, we get from (7) that f ( 2 k) = kf(2) + (k ) [ H() + f (CC A E(C)) f ( ACC 2 AE(C) )] holds for every positive integer k, whih with (6) shows that H() + f(cc A E(C)) f ( ACC 2 AE(C) ) = 0.

8 52 Bui Minh Phong Thus, the last two relations imply so by (4) we have f ( 2 k) = kf(2) (k =, 2,...), (8) f(nm) = f(n) + f(m) if (n, m, ABC) =. Similarly as above, by using (0), (2) (8), we see that () holds if k, Q P satisfy (9) (P, R k, B) =. Finally, let P = P P 2, where (P, P 2 ) = (P, AC A Q) = every prime divisor of P 2 is a divisor of AC A Q. We have (P 2, R k, B) =, therefore by () (9) it follows that f ( A k CC k AQ k P 2 E(CQ) ) = (k )H(Q) + f (CC A QP 2 E(CQ)). Sine (P, AC A QP 2 ) =, by using the additivity of f we get f ( A k CC k AQ k P E(CQ) ) = f ( A k CC k AQ k P 2 E(CQ) ) + f(p ) = whih proves (). = (k )H(Q) + f (CC A QP E(CQ)), Applying () in the ase Q =, we obtain that f ( A k CC k AP E(C) ) = (k )H() + f (CC A P E(C)) holds for every positive integer k P. Therefore, applying this relation with P = Q k E(CQ), we infer that E(C) (20) f ( A k CC k AQ k E(CQ) ) = (k )H() + f ( CC A Q k E(CQ) ) holds for all positive integers k Q. On the other h, () with P = implies f ( A k CC k AQ k E(CQ) ) = (k )H(Q) + f (CC A QE(CQ)),

9 Charaterizations of the logarithm as additive funtion 53 whih with (20) gives f ( CC A Q k E(CQ ) = (k ) [H(Q) H()] + f (CC A QE(CQ)). This, using the fat (CQE(CQ), B/E(CQ)) = the additivity of f, shows that f ( BCC A Q k) = kf(bcc A Q) (k )f(bcc A ). So, we have proved Lemma, beause (5) follows from () in the ase k = 2 P = Q =. Lemma 2. Let A, B be positive integers let D be a omplex onstant. If f A satisfies the ondition (2) f(an + B) f(n) D = o(log x) as x, n then there is a omplex onstant U suh that f(n) = U log n (n =, 2,...). Proof. Assume that f A satisfies the ondition (2) for some positive integers A, B a omplex onstant D. We first note that, by using (5) of Lemma the fat C =, (2) implies that (22) f(a) = D. We denote by I f those pairs (q, r) of positive integers for whih f(qn + r) f(qn) = o(log x) as x. n Thus, it follows from (2) (22) that (A, B) I f, therefore by using the omplete additivity of f, we have (A, ) I f. We shall prove that (23) (q, r) I f if 0 < r < q. First we show the following assertions: (a) (q, ) I f if (k, ) I f q k; (b) (q, r) I f if (k, ) I f, k 2 0 < r < q/(k );

10 54 Bui Minh Phong () (h, ) I f if (h +, ) I f h 2. Assume that (k, ) I f. By using the omplete additivity of f, we have f((k + )n + ) f((k + )n) = = [f(kn + ) f(kn)] [f(k((k + )n + ) + ) f(k((k + )n + ))], so, by using the fat (k, ) I f, we dedue that (k +, ) I f. By using indution, we have proved that (a) holds. Assume again that (k, ) I f k 2. We shall prove (b) by indution on r. From (a) it is lear that (b) is satisfied for r =. Assume that (q, r) I f holds for all integers q r satisfying 0 < r < q/(k ) r < r 0, where r 0 is an integer. Let q 0 be an integer suh that (24) 0 < r 0 < q 0 k. In order to show (b) it suffies to prove that (q 0, r 0 ) I f. generality we may assume that (q 0, r 0 ) =. Let q r be positive integers suh that Without loss of (25) r 0 q = q 0 r + r < r 0. It follows by (24) (25) that 0 < r < (q 0 r + )/q 0 = r 0 q/q 0 < q/(k ). Thus, by using our assumption the fat r < r 0, we have (q, r) I f. On the other h, by (25), we get f(q 0 n + r 0 ) f(q 0 n) = [f(q 0 (qn + r) + ) f(q 0 (qn + r))] + [f(qn + r) f(qn)], therefore, by using the fat (q 0, ) I f (q, r) I f, we have (q 0, r 0 ) I f. Thus, we have proved (b). Finally, we prove (). Assume that (h +, ) I f h 2. Let S(x) := f(hn + ) f(hn). n For eah integer d with 0 d h we an hoose positive integers q = q(d) r = r(d) suh that (26) (hd + )q = h 2 r +.

11 Charaterizations of the logarithm as additive funtion 55 We have S(x) = f(hn + ) f(hn) = n (27) = h d=0 hm+d x h d=0 hm+d x h + d=0 hm+d x hm + d f(h2 m + hd + ) f(h(hm + d)) = hm + d f(h2 (qm + r) + ) f(h 2 (qm + r)) + f(q((hm + d) + hr qd) f(q(hm + d)), hm + d so S(x) = o(log x) if hr qd = 0, beause, by using (a), (h +, ) I f h 2 implies (h 2, ) I f. If hr qd 0, then we get from (26) that 0 < hr qd = (q ) h < q h, whih, by applying (b) with k = h +, implies that (q, hr qd) I f. This, with (h 2, ) I f shows that S(x) = o(log x). This ompletes the proof of (). Now we prove (23). As we have seen above, (A, ) I f. If A =, then (23) holds. If A 2, then by using () one an dedue that (2, ) I f, so by applying (b) with k = 2, it follows that (q, r) I f for all positive integers q r whih satisfy 0 < r < q. This ompletes the proof of (23). We now prove Lemma 2. Let q be a fixed positive integer. Let T (x) := f(n) n. Then, by using the omplete additivity of f, we have (28) n 0(modq) f(n) n = m x/q f(q) + f(m) qm = f(q) q log x + q T ( x q ) + O().

12 56 Bui Minh Phong Let r be an integer for whih 0 < r < q. Then, by (23), we have (q, r) I f, so (29) n r(modq) f(n) n = qm+r x = f(q) q f(qm + r) f(qm) qm + r log x + q T ( x q + ) + o(log x). qm+r x f(qm) qm + r = By summing (28) (29) over the range 0 r q, we infer that (30) T (x) = f(q) log x + T ( ) x + o(log x) as x. q Let k = k(x) be a positive integer suh that q k x < q k+. Then it is lear that k = log x log q + O(). From the asymptoti formula (30) of T (x) we get T (x) = f(q){log x + log(x/q) log(x/q k )} + o((k + ) log x) = Thus, we have = (k + )f(q) log x [ k]f(q) log q + o((k + ) log x) = = f(q) 2 log q log2 x + o(log 2 x). f(q) log q = lim 2T (x) x log 2 := U. x This holds for eah positive integer q, so it also holds for all positive integers q. Thus, the proof of Lemma 2 is finished. 3. Proofs of Theorem Theorem 2 Proof of Theorem 2. Assume that f A satisfies the ondition (3). Then from Lemma (3) f(bcc A Q k ) = kf(bcc A Q) (k )f(bcc A )

13 Charaterizations of the logarithm as additive funtion 57 holds for every integer k Q, where C A denotes the produt of all prime divisors of C whih are prime to A. For eah prime p let e = e(p) be a non-negative integer for whih p e BCC A. Then for all integers β e we dedue from (3) that (32) f(p β+ ) f(p β ) = f(p e+ ) f(p e ). Now we write f(n) = f (n) + F (n), where f is a ompletely additive funtion defined as follows: (33) f (p) := f(p e+ ) f(p e ), e = e(p). Then, from (32) (33) it follows that F (p β+ ) = F (p β ) (β = e, e +,...), whih implies Thus, we have F (p k ) = F [(p k, BCC A )] (k = 0,,...). (34) F (n) = F [(n, BCC A )] (n =, 2,...). We shall prove that f = U log for some onstant U. We note that, by onsidering n = BCC A m taking into aount (3), we have (35) n f(abcc Am + B) f(bc 2 C A m) D = o(log x) as x. Sine f = f + F, from (34) we get f(abcc A m + B) f(bc 2 C A m) D = f (ABCC A m + B) f (BC 2 C A m) + F (ABCC A m + B) F (BC 2 C A m) D = = f (ACC A m + ) f (m) {f (C 2 C A ) F (B) + F (BCC A ) + D}

14 58 Bui Minh Phong so, by (35) Lemma 2, there is a omplex onstant U suh that f = = U log. This ompletes the proof of Theorem 2. Proof of Theorem. It is lear to show that if there are a omplex onstant U funtions F i A (i =, 2) suh that ( a ) F (an + b) F 2 (n) d + U log = 0 holds for all positive integers n, then the funtions f i (n) = U log n + F i (n) (i =, 2) are additive they satisfy the ondition (2). suffiieny part of Theorem. Thus, we have proved the In the following we shall prove the neessity part. Assume that f A f 2 A satisfy the ondition (2) for some positive integers a, b, for a omplex onstant d, i.e. (36) n f (an + b) f 2 (n) d = o(log x) as x. We shall prove that there are a omplex onstant U funtions F i (i =, 2) suh that A f i (n) = U log n + F i (n) (i =, 2) ( a ) F (an + b) F 2 (n) d + U log = 0 hold for all positive integers n. Let I(a, b) denote the set of those positive integers N for whih (an +, b) =. Then for eah positive integer N I(a, b) we have (an +, a(an + )n + b) = (an + )(a(an + )n + b) = a[(an + ) 2 n + bn] + b

15 Charaterizations of the logarithm as additive funtion 59 for every positive integer n. Thus, by using the additivity of f, we get f 2 [(an + ) 2 n + bn] f 2 [(an + )n] f (an + ) = = {f [(an + )(a(an + )n + b)] f 2 [(an + ) 2 n + bn] d}+ + {f [a(an + )n + b] f 2 [(an + )n] d}, whih with (36) implies that (37) n f 2[(aN + ) 2 n + bn] f 2 [(an + )n] f (an + ) = o(log x) holds for eah N I(a, b). Applying Lemma with A = (an + ) 2, B = bn C = (an + ), it follows from (37) that for eah positive integer N I(a, b) (38) f 2 [b 2 (an + )NQ k ] = kf 2 [b 2 (an + )NQ] (k )f 2 [b 2 (an + )N] holds for every positive integer k Q. Sine (38) holds for eah fixed positive integer N I(a, b), so (38) also holds for every positive integer N I(a, b). Let N 2 {, 2} satisfying (2, an 2 + ) =. For eah prime p, let M p be the smallest positive integer for whih (pb, am p + ) = N 2 M p, (p, M p ) = for p 2; ( p, M ) p = for p = 2. N 2 We note that for eah prime p suh M p exists as well. Indeed, in ase p 2, one an show that there is a positive integer r with (r, p) = (ar +, p) =, therefore for some positive integer t (pb, a(pt + r) + ) =. This shows that M p exists in this ase. For p = 2 we an find M 2 in the form N 2 (2t + ). Thus, M p exists in eah ase, furthermore M p I(a, b). Now we define the ompletely additive funtion f2 for eah prime p as follows: (39) f 2 (p) := f 2 (b 2 (am p + )M p p) f 2 (b 2 (am p + )M p ).

16 60 Bui Minh Phong Let (40) f 2 (n) := f 2 (n) + F 2 (n) (n =, 2,...). Sine M p I(a, b), we apply (38) with Q = p N = M p, by using (38)-(40) the definition of M p, we have (4) F 2 (b 2 p k ) = F 2 (b 2 ) F 2 (b 2 N 2 2 k ) = F 2 (b 2 N 2 ) for all primes p positive integers k. Sine (p, N 2 ) = for p 2, one an hek from (4) that (42) F 2 (n) = F 2 [(n, b 2 N 2 )] (n =, 2,...). Thus, (40) (42) imply that (43) f 2 [(abn 2 M + ) 2 bn 2 m + b 2 2 N 2 M)] f 2 [(abn 2 M + )bn 2 m] f (abn 2 M + ) = f 2 [(abn 2 M + ) 2 m + bm] f 2 (m) D, where beause by (42) D = f (abn 2 M + ) + f 2 (abn 2 M + ), F 2 [(abn 2 M + ) 2 bn 2 m + b 2 2 N 2 M)] F 2 [(abn 2 M + )bn 2 m] = 0 for all positive integers m. Applying (37) with n = bn 2 m N = bn 2 M, by using (43) the fat N I(a, b) in this ase, we get m x m f 2 [(abn 2 M + ) 2 m + bm] f 2 (m) D = o(log x) whih, by using (5) Lemma 2, implies (44) f 2 = U log for some onstant U f (abn 2 M + ) = f 2 (abn 2 M + ) = U log(abn 2 M + ). The last relation holds for every positive integer M, onsequently f (m) = U log m

17 Charaterizations of the logarithm as additive funtion 6 holds for all positive integers m whih are prime to abn 2. Let (45) f (m) := F (m) + U log m (m =, 2,...). Then, we have (46) F (m) = 0 if (m, abn 2 ) =. Finally, we shall prove that ( a ) F (an + b) F 2 (n) d + U log = 0 (n =, 2,...), whih with (40), (44) (45) ompletes the proof of Theorem. Sine ( a ) F (an + b) F 2 (n) d + U log = [f (an + b) f 2 (n) d] we obtain from (36) that (47) = [ U log(an + b) U log(n) U log ( a ) F (an + b) F 2 (n) d + U log = o(log x). n ( a )], Let K be a positive integer. By (42) (46), we have F (abn 2 t + ) = 0 F 2 [ (ak + b)b 2 N 2 t + K ] = F 2 (K) hold for all positive integers t, onsequently ( a ) F (ak + b) F 2 (K) d + U log = ( a ) (48) = F (ak + b) + F (abn 2 t + ) F 2 (K) d + U log = ( a ) = F [a((ak + b)bn 2 t + K) + b] F 2 [(ak + b)b 2 N 2 t + K] d + U log

18 62 Bui Minh Phong holds for every positive integer t. Thus, by applying (47) with using (48), we have (49) t x n = (ak + b)bn 2 t + K ( a ) F (ak + b) F 2 (K) d + U log = o(log x), t whih implies ( a ) F (ak + b) F 2 (K) d + U log = 0 for eah positive integer K, i.e. (49) holds for every positive integer K. This ompletes the proof of Theorem. 4. Proof of Theorem 3 We first onsider the ase when f = f 2. Assume that a funtion f A satisfies the ondition (50) f(an + B) f(cn) = O() as n for some positive integers A, B C. For eah positive integer Q, as in the proof of Lemma, we define the sequene R = R(AC A Q) = {R k (AC A Q)} k= by the initial term R (ACC A ) = by the formula R k (AC A Q) = + AC A Q (AC A Q) k for all integers k 2 let T k (n, Q) = (AC A Q) k E(CQ)n + BR k (AC A Q). By using (8) (9), one an dedue from (50) that f(t (n, Q)) f(cc A QE(CQ)n) = O() as n

19 Charaterizations of the logarithm as additive funtion 63 f(t k (n, Q)) f(t k (n, Q)) f(bcc A Q) = O() as n for eah integer k 2. Here in the remainder of this paper the implied onstants O() depend at most on the given initial integers given funtions, but they do not depend on n, Q, k. The last relations imply that (5) f(t k (n, Q)) f(cc A QE(CQ)n) (k )f(bcc A Q) = O(k) holds for all positive integers k Q. We shall dedue from (5) that (52) f(bcc A Q k ) kf(bcc A Q) = O(k) (53) f(a k CC k AQ k P E(CQ)) = = (k )f(bcc A Q) + f(cc A QP E(CQ)) + O(k) hold for all positive integers Q, k, P. Let k, Q P be positive integers. Let R k = R k (AC A Q). Considering n := P R k {AP CQR k m + } taking into aount (5), one an see that (53) holds if k, Q P satisfy the relation (54) (P, R k ) = (P E(CQ) + B, R k ) =. It is lear that (54) is satisfied if P = 2B Q. Thus, we an apply (53) in the following ases: P =, Q = 2B P =, Q = 2hB, where h is a positive integer number. Thus, we get from (53), using the fat E(2BC) = E(2kBC) = B, that (55) f(h k ) = kf(h) + O(k) if (h, 2ABC) =. Let h be a positive integer satisfying (h, 2ABC) =. By applying (55) with k = uv in the two possible way, we have f(h uv ) = uf(h v ) + O(u) = vf(h u ) + O(v),

20 64 Bui Minh Phong onsequently f(h u ) u f(hv ) v = O ( ) ( ) + O. u v By using Cauhy s riterion, it follows from the above relation that f(h k ) lim := f (h) k k exists. This with (55) shows that f(h) = f (h) + O() the funtion f satisfies f (nm) = f (n) + f (m) if (n, m, 2ABC) =. Thus, we an assume that (50) holds for a funtion f satisfying (56) f(nm) = f(n) + f(m) if (n, m, 2ABC) =. Therefore, by using (50) (56), one an prove in same way as above that (53) also holds if k, Q, P satisfy (57) (P, R k, 2B) = (P E(CQ) + B, R k, 2) =. Assume that (2, ABC) =. Then (57) holds for P = 2 ν, Q = 2 k = 2, where ν 0 is an integer. From (53), we have whih implies f(2 ν+2 ) = f(2) + f(2 ν+ ) + O(), f(2 k ) = kf(2) + O(k) (k =, 2,...). Similarly as above, the last relation with (56) shows that (53) holds if (P, R k, B) =, so by the additivity of f, the proof of (53) is finished. Applying (53) in the ase Q =, we obtain that f(a k CC k AP E(C)) = f(cc A P E(C)) + O(k) holds for every positive integer k P. Therefore, applying this relation with P = Q k E(CQ), we infer that E(C) f(a k CC k AQ k E(CQ)) = f(cc A Q k E(CQ)) + O(k), whih with (53) onerning P = proves (52). So, (52) is proved.

21 Charaterizations of the logarithm as additive funtion 65 Sine (52) holds for all positive integers k Q, as we have seen above, by (52) it is easy to show in the same way as above that exists f(bcc A Q k ) lim := f (Q) k k f(bcc A Q) = f (Q) + O(), where f A. Let f = f + F. Then the last relation implies that F (BCC A Q) = O() for all positive integers Q, from whih F (n) = O() for all n. Thus, f(n) = f (n) + O(), whih with (50) implies (58) f (An + B) f (Cn) = O() as n. In order to prove Theorem 3 in the ase f = f 2 it suffies to dedue from (58) that f = Ulog for some onstant U. To show this, we shall prove that for eah positive integer t we have (59) f (An + t) f (An) = O(t) as n. From (58) it follows that (59) holds for t =. Assume that (59) holds for t. By using the omplete additivity of f, we have f (An + t + ) f (An) = {f (An + t) f (An)} {f [An(An + t + ) + t] f [An(An + t + )]}+ + {f (An + ) f (An)}, whih with the assumption of indution shows that (59) holds. Finally, by applying (59) with t = A, we get f (n + ) f (n) = O(). This with the result of Wirsing [6] implies f = U log so the proof of Theorem 3 in ase f = f 2 is finished. Now we prove Theorem 3. Assume that f A f 2 A satisfy the ondition (60) f (an + b) f 2 (n) = O() as n for some positive integers a, b.

22 66 Bui Minh Phong We use an argument similar to the proof of Theorem, one an dedue by (60) the fat (ab +, a(ab + )n + b) = that f 2 [(ab + ) 2 n + b 2 ] f 2 [(ab + )n] = O() as n. As we have proved above, the last relation implies that there is a onstant U suh that (6) f 2 (n) = U log n + O(). Let (62) f (n) = U log n + F (n). From (60) (6) it follows that F (an + b) = O(), onsequently F (n) = O() for all positive integers n whih are prime to a. This with (6) (62) ompletes the proof of Theorem 3. Referenes [] Elliott P.D.T.A., On sums of an additive arithmeti funtion with shifted arguments, J.London Math. So., 22 (980), [2] Elliott P.D.T.A., On additive arithmeti funtion f(n) for whih f(an+ +b) f(n + d) is bounded, J.Number Theory, 6 (983), [3] Elliott P.D.T.A., Arithmeti funtions integer produts, Grund.der Math.Wiss. 272, Springer, 979. [4] Elliott P.D.T.A., The value distribution of differenes of additive arithmeti funtions, J.Number Theory, 32 (989), [5] Hildebr A., An Erdős-Wintner theorem for differenes of additive funtions, Trans. Amer. Math. So., 30 (988), [6] Hildebr A., Charaterizations of the logarithm as an additive arithmeti funtion, Pro. st Conf. Can. Number Theory Asso., 988, [7] Erdős P., On the distribution funtion of additive funtions, Ann.Math., 47 (946), -20.

23 Charaterizations of the logarithm as additive funtion 67 [8] Kátai I., Some results problems in the theory of additive funtions, Ata Si. Math. Szeged, 30 (969), [9] Kátai I., On number-theoretial funtions, Coll. Math. So. János Bolyai 2, 970, [0] Kátai I., On a problem of P.Erdős, J.Number Theory, 2 (970), -6. [] Kátai I., On additive funtions, Publ. Math. Debreen, 25 (978), [2] Kátai I., Charaterization of arithmetial funtions, problems results, Pro. st Conf. Can. Number Theory Asso., 988, [3] Phong B.M., A haraterization of some arithmetial multipliative funtions, Ata Math. Hungar., 63 () (994), [4] Phong B.M., On a theorem of Kátai-Wirsing, Ata Si. Math. Szeged, 56 (992), [5] Wirsing E., A haraterization of log n as an additive funtion, Symposia Math., IV (970), [6] Wirsing E., Charaterization of the logarithm as an additive funtion, Sympos. Pure Math., 20, Amer. Math. So., Providene, 97, Bui Minh Phong Department of Computer Algebra Eötvös Loránd University VIII. Múzeum krt H-088 Budapest, Hungary