Author:
(1) Yitang Zhang.
Appendix A. Some Euler products
Appendix B. Some arithmetic sums
In this section we prove Proposition 2.2. We henceforth assume that ψ(mod p) ∈ Ψ1. This assumption will not be repeated in the statements of Lemma 4.1-4.8.
We begin by proving some consequences of the inequalities (3.4)-(3.6).
Lemma 4.1. Let
Proof. By the Stieltjes integral we may write
Hence, by partial integration,
For G(s, ψ) an entirely analogous bound is valid. The result now follows by (3.4).
Lemma 4.2. If s ∈ Ω1, then
Proof. We have
Thus, similar to (4.2), by partial integration we obtain
Lemma 4.3. Let
We proceed to establish an approximate formula for L(s, ψ)L(s, χψ). For this purpose we first introduce a weight g(x) that will find application at various places. Let
with
We may write
Since
it follows, by changing the order of integration, that
Thus the function g(x) is increasing and it satisfies 0 < g(x) < 1. Further we have
Note that χψ is a primitive character (modDp). Write
so that
By (2.4) with θ = ψ and θ = χψ we have
This yields, by Stirling’s formula,
and
Lemma 4.4. Let
If s ∈ Ω3, then
Proof. By the residue theorem,
By (4.2) and (4.3),
By (3.5) and partial summation, the second sum on the right side above is
On the other hand, by the functional equation, for u = −σ − 1/2,
and
To prove (4.7) we move the contour of integration to the vertical segments
and to the two connecting horizontal segments
By a trivial bound for ω1(w), (4.5) and the residue theorem we obtain (4.7).
To prove (4.8) we move the contour of integration to the vertical segments
and to the two connecting horizontal segments
By a trivial bound for ω1(w) and (4.5) we see that the left side of (4.8) is
with s ∗ = 1 + α − s¯. By partial integration,
The estimate (4.9) follows by moving the contour of integration to the vertical segments
and to the two connecting horizontal segments
and applying (4.5) and trivial bounds for ω1(w) and the involved sum.
In order to prove Proposition 2.2, it is appropriate to deal with the function
By Lemma 4.2, A(s, ψ) is analytic and it has the same zeros as L(s, ψ)L(s, ψχ) in Ω1. Further, for s ∈ Ω1, we have
by Lemma 4.1 and 4.2. This together with Lemma 4.4 implies that
for s ∈ Ω3, where
The proof of Proposition 2.2 is reduced to proving three lemmas as follows.
Lemma 4.5. If
then
Proof. We discuss in two cases.
By Lemma 4.2 and trivial estimation,
Hence, by (4.5),
The result now follows by (4.10).
Since |B(1/2 + it, ψ)| = 1, it follows that
Hence, by (4.10),
Lemma 4.6. Suppose ρ = β + iγ is a zero of A(s, ψ) satisfying
Proof. It suffices to show that the function A(1/2 + iγ + w, ψ) has exactly one zero inside the circle |w| = α(1 − c ′αL), counted with multiplicity. By the Rouch´e theorem, this can be reduced to proving that
In either case (4.16) holds.
Lemma 4.5 and 4.6 together imply the assertions (i) and (ii) of Proposition 2.2. It is also proved that the gap between any distinct zeros of A(s, ψ) in Ω is > α(1 − c ′αL). To complete the proof of the gap assertion (iii), it now suffices to prove
Proof. In a way similar to the proof of Lemma 4.6, it is direct to verify that
We conclude this section by giving a result which is implied in the proof of Proposition 2.2.
Lemma 4.8. Assume that ρ is a zero of L(s, ψ)L(s, χψ) in Ω. Then we have
Proof. It follows from Lemma 4.4 that