aa r X i v : . [ m a t h . N T ] A ug ON THE MULTIPLICATIVE ERD ˝OS DISCREPANCY PROBLEM
MICHAEL COONS
Abstract.
As early as the 1930s, P´al Erd˝os conjectured that: for any mul-tiplicative function f : N → {− , } , the partial sums P n x f ( n ) are un-bounded. In this paper, after providing a counterexample to this conjecture,we consider completely multiplicative functions f : N → {− , } as well as aclass of similar multiplicative functions f satisfying X p x f ( p ) = c · x log x (1 + o (1)) . We prove that if c > f are unbounded, and if c < µf are unbounded. Extensions of this result are alsogiven. Introduction
Erd˝os [2] asked the following question, sometimes known as the Erd˝os Discrep-ancy Problem. “Let f ( n ) = ± be an arbitrary number theoretic function. Is ittrue that to every c there is a d and an m for which (1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) n X k =1 f ( kd ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > c ? Inequality (1) is one of my oldest conjectures.” (This particular quote is taken froma restatement of the conjecture in [3, p.78]. See also [4] and [5].) Erd˝os offered 500dollars for a proof of this conjecture. Erd˝os [2, p.293] wrote in 1957 that thisconjecture is twenty-five years old, placing its origin at least as far back as theearly 1930s. In [2, 3, 4], Erd˝os also stated a multiplicative form of his conjecture.
Conjecture 1.1 (Erd˝os) . Let f ( n ) = ± be a multiplicative function, (i.e., f ( ab ) = f ( a ) f ( b ) , when gcd( a, b ) = 1 ). Then (2) lim sup x →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n x f ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ∞ ; that is, the partial sums of f are unbounded. Erd˝os added in [3] that “clearly (2) would follow from (1) but as far as I know (2) has never been proved. Incidentally (2) was also conjectured by Tchudakoff.”
Conjecture 1.1 as stated is not true, and while this may be known to others inthis field, there seems to be no account of it in the literature.
Date : August 13, 2018.2010
Mathematics Subject Classification.
Primary 11N37; 11N56 Secondary 11A25.
Key words and phrases.
Multiplicative functions, partial sums, mean values.The research of M. Coons is supported by a Fields-Ontario Fellowship and NSERC..
For a counterexample, consider the multiplicative function g defined by g (1) = 1,and on prime powers by(3) g ( p k ) = ( − p = 2 and k ≥
11 if p = 2 and k ≥ . Then g is periodic with period 2 and for all n ≥ g (2 n ) = − g (2 n −
1) = 1. Thus X n ≤ x g ( n ) = ( x ] is odd0 if [ x ] is even , and so lim sup x →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n x g ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = 1 . It may very well be the case that the function g defined above is the only coun-terexample to Conjecture 1.1, but at least at this point, we can say that this is theonly known counterexample.Along with Conjecture 1.1, Erd˝os [2] conjectured a result on the mean valuesof multiplicative functions. A number–theoretic function f : N → C has a meanvalue , denoted M ( f ), provided the limit(4) M ( f ) := lim x →∞ x X n x f ( n )exists. Erd˝os [2, 3] (among others) conjectured that any multiplicative functiontaking the values ± f satisfying | f | Theorem 1.2 (Delange, Wirsing, Hal´asz) . Let f : N → {− , } be a multiplicativefunction. If (5) X p x − f ( p ) p is bounded then M ( f ) exists and is positive, and if (5) is unbounded then M ( f ) = 0 . We note that the ideas of Theorem 1.2 have been generalized by many authors,including Granville and Soundararajan [7, 8] and Goldmakher [6]. In these worksthe authors use properties of a generalization of (5) to give some new results con-cerning sums of certain types of Dirichlet characters. The generalization of (5) isusually made by considering a special multiplicative function g (e.g., a Dirichletcharacter) and comparing it to the multiplicative function of interest f (e.g., aDirichlet character) by means of investigating the asymptotics of X p x − ℜ ( f g ( p )) p . This sum can be thought of as a metric [7], and in some sense measures how g mimics f ; this terminology was introduced in [6]. N THE MULTIPLICATIVE ERD ˝OS DISCREPANCY PROBLEM 3
In contrast to this “mimicry metric,” we consider the asymptotics of X p x c − f ( p ) p for c not necessarily equal to 1. By considering sums like like this, we are able togive the following result toward Conjecture 1.1. Theorem 1.3.
Let f : N → {− , } be a multiplicative function such that there issome k ≥ with f (2 k ) = 1 . Suppose that for some c ∈ [ − , we have X p x f ( p ) = c · x log x (1 + o (1)) . If c > then the partial sums of f are unbounded, and if c < the partial sums of µf are unbounded. Some extensions of this theorem are given in Section 4, including some instancesof the case c = 0. In Section 2, we show that this theorem is true for completelymultiplicative functions without the assumption that there is some k ≥ f (2 k ) = 1. 2. Completely multiplicative functions
If a multiplicative function f : N → {− , } has positive mean value, then clearlythe partial sums of f are unbounded; they are asymptotic to M ( f ) · x . The trivialityleaves when we consider functions with M ( f ) = 0. Theorem 2.1.
Let f : N → {− , } be a completely multiplicative function (i.e., f ( ab ) = f ( a ) f ( b ) for all a, b ∈ N ) and suppose that c ∈ [ − , . If X p c − f ( p ) p < ∞ , then the mean value of f exists and is equal to .Proof. This follows from Theorem 1.2 in a very straightforward way. We need onlynote that X p x − f ( p ) p = X n x − c + c − f ( p ) p = (1 − c ) X n x p + X n x c − f ( p ) p = (1 − c ) log log x + O (1) . (cid:3) To prove Theorem 1.3, we will first prove the result for completely multiplicativefunctions f : N → {− , } . The bulk of the work is taken up by the followinglemma. Lemma 2.2.
Let f : N → {− , } be a completely multiplicative function. Supposethat c ∈ [ − , is nonzero and X p c − f ( p ) p < ∞ . If c > then the partial sums of f are unbounded, and if c < the partial sums of µf are unbounded. MICHAEL COONS
Proof.
Suppose firstly that c >
0. To give the desired result, it is enough to showthat lim x →∞ X n x f ( n ) n = ∞ . To this end, note that for σ > F ( σ ) = log X n > f ( n ) n σ = − X p log (cid:18) − f ( p ) p (cid:19) = X p X k > f ( p ) k kp kσ = X p f ( p ) p σ + X p X k > f ( p ) k kp kσ = X p f ( p ) p σ + O (1) , where the O (1) term is valid for σ > /
2. Since X p c − f ( p ) p < ∞ , we have that(7) X p x f ( p ) p = c log log x + O (1) . The condition that c > s → + X p f ( p ) p σ = ∞ , and so the divergence of log F ( σ ) at σ = 1 occurs because lim σ → + F ( σ ) = ∞ .In the light of (6) it must be the case that(8) lim x →∞ X n x f ( n ) n = ∞ . Thus we have that lim sup x →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n x f ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ∞ . For if not, there is a real number
M > (cid:12)(cid:12)(cid:12)P n x f ( n ) (cid:12)(cid:12)(cid:12) < M, and bypartial summation, we would then have that X n x f ( n ) n = 1 x X n x f ( n ) + Z x X n t f ( n ) dtt = O (cid:18)Z x dtt (cid:19) = O (1) , which contradicts (8).Now suppose that c <
0. In this case, instead of F ( σ ), we consider the function1 /F ( σ ). Running through the above argument gives(9) − log F ( σ ) = − X p f ( p ) p σ + O (1) , N THE MULTIPLICATIVE ERD ˝OS DISCREPANCY PROBLEM 5 where again the O (1) term is valid for σ > / . Similar to the above, using theassumption of the lemma, we have that(10) − X p x f ( p ) p = | c | log log x + O (1) , which in turn gives, due to (9) thatlim σ → + F ( σ ) = ∞ . This implies that lim x →∞ X n x µ ( n ) f ( n ) n = ∞ , which using a similar argument as the case c >
0, give thatlim sup x →∞ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X n x µ ( n ) f ( n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ∞ . This completes the proof of the lemma. (cid:3)
Our proof of the main theorem follows from the similar result for completelymultiplicative functions. Using partial summation we have the following theorem.
Theorem 2.3.
Let f : N → {− , } be a completely multiplicative function. Sup-pose that for some c ∈ [ − , we have X p x f ( p ) = c · x log x (1 + o (1)) . If c > then the partial sums of f are unbounded, and if c < the partial sums of µf are unbounded.Proof. This follows directly from Lemma 2.2. The condition X p x f ( p ) = c · x log x (1 + o (1))gives by partial summation that X p x f ( p ) p = 1 x X p x f ( p ) + Z x X p t f ( p ) dtt = c · x (1 + o (1)) + c Z x t log t (1 + o (1)) dt = c log log x (1 + o (1)) . (11)Note that the proof of the lemma follows from the divergent behavior of P p x f ( p ) p in both (7) and (10), and that this divergence is satisfied by (11). Thus using (11)in the place of (7) and (10) is enough to prove Lemma 2.2, and thus the condition(11) implies the result of the theorem. (cid:3) MICHAEL COONS Extension to multiplicative functions
The results of the previous section are extendable to multiplicative functions f : N → {− , } with the added condition that there is some k ≥ f (2 k ) = 1.In this section, by relating a multiplicative function f : N → {− , } to a relatedcompletely multiplicative function, we are able to deduce Theorem 1.3 as a corollaryto Theorem 2.3. This is obtained via the following lemma. Lemma 3.1.
Let f : N → {− , } be a multiplicative function such that there issome k ≥ with f (2 k ) = 1 . Then F ( σ ) = X n > f ( n ) n σ = Π( σ ) · Y p (cid:18) − f ( p ) p σ (cid:19) − ( σ > , where Π( σ ) = Y p X k > f ( p k ) − f ( p k − ) f ( p ) p kσ . Moreover, there is a σ ( f ) ∈ (0 , such that Π( σ ) is absolutely convergent for σ > σ ( f ) . Proof.
Note that if f is multiplicative, then for σ > F ( σ ) := X n > f ( n ) n σ = Y p (cid:18) f ( p ) p σ + f ( p ) p σ + · · · (cid:19) = Y p (cid:18) − f ( p ) p σ (cid:19) − · Y p X k > f ( p k ) − f ( p k − ) f ( p ) p kσ . It remains to show that Π( σ ) is absolutely convergent for σ > log ϕ log 2 . Firstly, notethat for each prime p we have(12) 1 + X k > f ( p k ) − f ( p k − ) f ( p ) p kσ ≥ max X k > f (2 k ) − f (2 k − ) f (2)2 kσ , − σ (3 σ − . To ensure that none of the factors of the product Π( σ ) is zero, we will show thatthe right–hand side of (12) is greater than zero for σ > log ϕ log 2 . Now if f (2 k ) = 1 forall k ≥
1, then 1 + X k > f (2 k ) − f (2 k − ) f (2)2 kσ = 1 > , regardless of the range of σ . Thus we may suppose that f (2 k ) = 1 identically.Then, using our assumption, since f (2 k ) = 1 for at least one k ≥
1, we have forsome k ≥ f (2) = f (2 k ). Rephrased, this means that there is a k ≥ f (2 k − ) f (2) = −
1. Denote k := min { k ≥ f (2 k − ) f (2) = − } . N THE MULTIPLICATIVE ERD ˝OS DISCREPANCY PROBLEM 7
Then1 + X k > f (2 k ) − f (2 k − ) f (2)2 kσ ≥ − X k ≥ kσ + 22 k σ = 1 − σ (2 σ −
1) + 22 k σ . Note that for σ > k ≥
2, the function1 − σ (2 σ −
1) + 22 k σ is continuous and increasing. Also at σ = 1 we have1 − (2 −
1) + 22 k = 22 k > , so that by continuity and the fact that k ≥
3, there is some minimal α := α ( k ) ∈ (0 ,
1) such that for σ > α we have1 − σ (2 σ −
1) + 22 k σ > . Also, we have that 3 σ − σ − > σ > log 2log 3 by the quadratic formula. Since 3 σ − σ − > − σ (3 σ − >
0, combining this with the above, we have that each of the terms ofthe product Π( σ ) is positive for all σ > σ ( f ) := max (cid:26) α, log 2log 3 (cid:27) . Since this maximum is strictly less that one, the only thing left to show is that thesum P p P k > f ( p k ) − f ( p k − ) f ( p ) p kσ is absolutely convergent. We have that X p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)X k > f ( p k ) − f ( p k − ) f ( p ) p kσ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X p X k > p kσ = 2 X p p σ · p σ − , which is convergent when σ > /
2, proving the lemma. (cid:3)
It is worth remarking that assuming that f (2 k ) = 1 for some k ≥ g defined in (3).We now give the proof of Theorem 1.3 as a corollary to Theorem 2.3. Proof of Theorem 1.3.
Let f : N → {− , } be a multiplicative function such that f (2 k ) = 1 for some k ≥
1, and denote F ( σ ) = P n > f ( n ) n σ . By Lemma 3.1, we havethat F ( σ ) = Π( σ ) F c ( σ ) , where Π( σ ) is defined by as in Lemma 3.1 and F c ( σ ) is the generating Dirichlet seriesfor the completely multiplicative function f c : N → {− , } defined by f c ( p ) = f ( p )for all primes p . Similar to (6), we have thatlog F ( σ ) = log Π( σ ) + log F c ( σ ) = X p f ( p ) p σ + O (1) , since Π( σ ) > σ ≥
1. If c > , then considering the proof of Theorem 2.3 for F c ( σ ) gives that lim σ → + F c ( σ ) = ∞ , MICHAEL COONS and so lim σ → + F ( σ ) = ∞ , which in turn gives that the partial sums P n x f ( n ) are unbounded.If c < /F ( σ ), anduse the equation log 1 F ( σ ) = − log F ( σ ) = − log Π( σ ) − log F c ( σ )to yield the result. (cid:3) Weakening of hypotheses and further extensions
In Theorem 2.3 we can replace the condition(13) X p c − f ( p ) p < ∞ with something considerably weaker.Note that assumption (13) is given so that we may use an asymptotic of the form X p x f ( p ) p = c log log x + O (1) , for nonzero c ∈ [ − , c we can weaken the condition to(14) lim x →∞ X p x f ( p ) p = ∞ , and in the case of negative c we can weaken the condition to(15) lim x →∞ X p x f ( p ) p = −∞ . Then if (14) holds we have that P n x f ( n ) is unbounded, and if (15) holds wehave that P n x µ ( n ) f ( n ) is unbounded. As far as “density conditions” the abovelimits are satisfied when we take X p x f ( p ) = c · x log x log x · · · log k x (1 + o (1)) , where log j x denotes log log · · · log x with “log” written j times, k is any nonnegativeinteger, and c = 0 is taken to be positive or negative depending on the desired case;this is easily seen via partial summation.We can do a little in the case that c = 0. Indeed, all we really need is to havefor some σ > / x →∞ X p x f ( p ) p σ = ∞ or lim x →∞ X p x f ( p ) p σ = −∞ . In this case, the same method gives the following theorem, though consideration ofthe limits in (16) directly gives a more exact result.
N THE MULTIPLICATIVE ERD ˝OS DISCREPANCY PROBLEM 9
Theorem 4.1.
Let f : N → {− , } be a completely multiplicative function, σ > / , k a nonnegative integer, and suppose that X p x f ( p ) = c · x σ log x log x · · · log k x (1 + o (1)) . If c > then the partial sums of f are unbounded, and if c < then the partialsums of µf are unbounded. As discussed above, the proof of Theorem 4.1 follows exactly the same as thatof Theorem 2.3, and as such we omit it for fear of sounding redundant. Theorem1.3 can be generalized similarly, but with the added assumptions that both f = g for g as defined in (3), and that σ > σ ( f ) as defined in the proof of Lemma 3.1.5. Concluding remarks
Functions satisfying (13) for positive c are, in some sense, large. In fact, since F ( σ ) are divergent at σ = 1, we have, using an obvious abuse of notation, at leastthat X n x f ( n ) ≫ x − ε for any ε >
0. Probably this can be improved, but our original purpose was tojust prove the unboundedness of partial sums. Indeed, using the terminology ofGoldmakher [6], we should have that the function f ( n ) mimics the function c Ω( n ) and the partial sums of this function are quite large; we have X n x c Ω( n ) > X p x c = c · π ( x ) . As an extension of the results for negative c , it would be nice if one could showthat since P n x µ ( n ) f ( n ) is unbounded, so is P n x f ( n ). We suspect that one mayhave to consider cases whether or not the Riemann hypothesis holds. Nonetheless,since we have X n x µ ( n ) f ( n ) ≫ x − ε for any ε > µ are not too small, P n x µ ( n ) = O ( x / ) , something may be able to be done in this case. Indeed, we conjecture that in thiscase one should have at least that X n x f ( n ) ≫ x / − ε for any ε >
0. Towards something like this we have tried to factor F ( s ) in anenlightening way (to find a singularity at s = 1 /
2, but to no avail. We note thatone has for any such series F ( s ) and c ∈ [ − , F ( s ) = (cid:18) ζ (2 s ) ζ ( s ) (cid:19) | c | e P (2 s )2 D ( s ) ζ (2 s ) | c | · exp " − X p c − f ( p ) p s , where P ( s ) is the prime zeta function and the function D ( s ) is absolutely convergentfor ℜ ( s ) > / References
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