Bounded discrepancy multiplicative functions do not correlate with characters

Theorem Let [math]\displaystyle{ g: {\Bbb N} \to S^1 }[/math] be a completely multiplicative function of bounded discrepancy. Let [math]\displaystyle{ \chi }[/math] be a non-principal Dirichlet character of some conductor q. Then [math]\displaystyle{ \sum_{p} \frac{|1-g(p)\overline{\chi(p)}|}{p} = \infty }[/math].

Remark By a theorem of Wintner on mean values of completely multiplicative functions, this implies that [math]\displaystyle{ \sum_{n \leq N} g(n) \overline{\chi(n)} = o(N) }[/math], thus g asymptotically does not correlate with any Dirichlet character. Added later Actually, it may be that Wintner's theorem only applies in the real case. In the complex case one also has to deal with functions like [math]\displaystyle{ \chi(n) n^{it} }[/math].

Proof By deleting the exceptional prime 2 from the character, we may assume that the conductor q is even. Suppose for contradiction that the sum was finite. We let the implied constants in the O() notation depend on q, the discrepancy of g, and on the sum, thus q = O(1),

[math]\displaystyle{ g(1)+\ldots+g(n) = O(1) }[/math], (0)

and

[math]\displaystyle{ \sum_{p} \frac{|1-g(p)\overline{\chi(p)}|}{p} = O(1) }[/math]. (1)

Let [math]\displaystyle{ \tilde \chi }[/math] be the completely multiplicative function which equals [math]\displaystyle{ \chi }[/math] for numbers coprime to q, and such that [math]\displaystyle{ \tilde \chi(p) = g(p) }[/math] for p dividing q. Let h be the completely multiplicative function with [math]\displaystyle{ h(p) = g(p) \overline{\chi(p)} }[/math] for p coprime to q, and h(p)=1 otherwise. Then we have the factorisation

[math]\displaystyle{ g(n) = h(n) \tilde \chi(n) }[/math]. (2)

By hypothesis, we have

[math]\displaystyle{ \sum_p \frac{|1-h(p)|}{p} = O(1) }[/math]. (3)

As for [math]\displaystyle{ \tilde \chi }[/math], we observe the following limited amount of periodicity:

Lemma Let k be a positive integer, and let Q be a sufficiently large power of q (depending on k). Then for every [math]\displaystyle{ 1 \leq i \leq q^k }[/math], one has [math]\displaystyle{ \tilde \chi(Qn+i) = \tilde \chi(i) }[/math] for all n.

Proof Let r be the lcm of Q and i, then it suffices by multiplicativity to show that [math]\displaystyle{ \tilde \chi(Qn/r + i/r) = \tilde \chi(i/r) }[/math]. But by construction of i and Q, i/r is coprime to q, and Q/r is a multiple of q. We then have

[math]\displaystyle{ \tilde \chi(Qn/r + i/r) = \chi(Qn/r + i/r) = \chi(i/r) = \tilde \chi(i/r) }[/math]

as desired. [math]\displaystyle{ \Box }[/math]

We're going to use the Maier matrix method. Let k be a large integer, and let Q be as in the lemma. Let M be a large number (depending on k, Q) to be chosen later, and let N be an enormous number (depending on k, Q, M). It may be helpful to keep the hierarchy

[math]\displaystyle{ k \ll Q \ll M \ll N }[/math]

in mind.

For any n,

[math]\displaystyle{ \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} g( Q n + i )|^2 = O(1) }[/math]

by bounded discrepancy (0). We average this over n in an interval [n,n+M] to get

[math]\displaystyle{ \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \frac{1}{M} \sum_{m=1}^M g( Q(n+m) + i )|^2 = O(1). }[/math]

Pick n at random from 1 to N. By the lemma, we have [math]\displaystyle{ \tilde \chi(Q(n+m)+i) = \tilde \chi(i) }[/math]. Thus by (2) we have

[math]\displaystyle{ \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i) \frac{1}{M} \sum_{m=1}^M h( Q(n+m) + i )|^2 = O(1). }[/math]

Now we perform the m summation. By construction, Q is coprime to all the primes for which h(p) is different from 1. An application of the second moment method shows that

[math]\displaystyle{ \frac{1}{M} \sum_{m=1}^M h(Q(n+m)+i) = c + o(1) }[/math]

with probability 1-o(1) for each i, where c is the singular series

[math]\displaystyle{ c = \prod_{p \in {\mathcal P}} \frac{1-1/p}{1-h(p)/p} }[/math]

and o(1) denotes a quantity that goes to zero if M is large enough (and N is huge enough depending on M), for fixed k and Q. Thus, with probability 1-o(1), one has

[math]\displaystyle{ c \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i)|^2 = O(1). }[/math]

However, by (3), c is non-zero (this is why we needed to eliminate the exceptional prime 2), thus

[math]\displaystyle{ \frac{1}{q^k} \sum_{a=1}^{q^k} |\sum_{i=1}^{a} \tilde \chi(i)|^2 = O(1). }[/math]

In other words, if a is chosen at random from 1 to q^k, then [math]\displaystyle{ \sum_{i=1}^{a} \tilde \chi(i) }[/math] has bounded variance. But in fact this variance grows linearly in k, which gives a contradiction by setting k large enough.

It's easiest to prove this linear growth in the case when q is prime. Then we can split this random sum as the sum of k terms

[math]\displaystyle{ \tilde \chi(q)^l \sum_{1 \leq i \leq a/q^l: (a,q)=1} \chi(i) }[/math]

for [math]\displaystyle{ l=0,\ldots,k-1 }[/math] (plus a O(1) error), but each such term has a variance comparable to 1 and depends only on the l^th digit in the base q expansion of a, so these terms have negligible correlation with respect to each other and their sum has a variance growing linearly in k. I'm pretty sure one can adapt this argument to composite Q.