A converse to Halasz's theorem - IAS

12 MAKSYM RADZIWI̷L̷L
Lemma 7. Let Ψ(·) be a distribution function. Let f be a positive strongly additive function.
Suppose that 0 f(p) O(1) for all primes p. Let
K f (x; t) :=
1
B 2 (f; x)
∑
p x
f(p) t
If K f (x; t) − Ψ(t) ≪ 1/B 2 (f; x) uniformly in t ∈ R, then
f(p) 2
ρ Ψ (B(f; x); ∆) − η f (x; ∆) = o(1/B 2 (f; x))
uniformly in 1 ∆ o(B(f; x)). The symbols ρ Ψ (B(x); ∆) and η f (x; ∆) are defined in
lemma 6 and lemma 5 respectively.
Proof. Let η = η f (x; ∆). Recall that by lemma 5, η = o(1) in the range 1 ∆ o(B(f; x)).
This will justify the numerous Taylor expansions involving the parameter η. With K f (x; t)
defined as in the statement of the lemma, we have
(15)
∑
px
f(p)e ηf(p)
p
− µ(f; x) = ∑ f(p)(e ηf(p) − 1)
p
px
∫
= B 2 e ηt − 1
(f; x) dK f (x; t)
R t
Let M > 0 be a real number such that 0 f(p) M for all p. Since the f(p) are bounded,
for each x > 0 the distribution function K f (x; t) is supported on [0; M]. Furthermore since
K f (x; t) → Ψ(t) the distribution function Ψ(t) is supported on exactly the same interval.
From these considerations, it follows that
∫
e ηt ∫
− 1
M
e ηt − 1
dK f (x; t) =
dK f (x; t)
(16)
=
R
∫ M
0
t
e ηt − 1
dΨ(t) +
t
∫ M
0
p
0 t
(
Kf (x; t) − Ψ(t) ) [ ]
e ηt − ηt · e ηt − 1
· dt
t 2
By a simple Taylor expansion e ηt − ηt · e ηt − 1 = O(η 2 t 2 ). Therefore the integral on the right
hand side is bounded by O(η 2 /B 2 (f; x)). We conclude from (15) and (16) that
∑ f(p)e ηf(p)
∫
(17)
− µ(f; x) = B 2 e ηt − 1
(f; x) dΨ(t) + O ( η 2)
p
t
px
By definition of ρ Ψ and η f ,
∫
(18) B 2 e ρΨ(B;∆)t − 1
(f; x)
dΨ(t) = ∆B(f; x) = ∑ f(p)e η f (x;∆)f(p)
− µ(f; x)
R t
p
px
From (17) and (18) it follows that
∫
(19) B 2 e ρΨ(B;∆)t − e η f (x;∆)t
(f; x)
dΨ(t) = O ( η 2
t
f(x; ∆) )
R
Since Ψ(t) is supported on [0; M] we can restrict the above integral to [0; M]. By lemma 5,
we have η f (x; ∆) ∼ ∆/B(f; x) = o(1) in the range ∆ o(B(f; x)). Also 0 ρ Ψ (x; ∆)
R