Download Zero-Free Regions for Riemann Zeta Function: Proof of No Vanishing at σ = 1 and more Study notes Number Theory in PDF only on Docsity!
Math 259: Introduction to Analytic Number Theory A zero-free region for ζ(s)
We first show, as promised, that ζ(s) does not vanish on σ = 1. As usual nowa- days, we give Mertens’ elegant version of the original arguments of Hadamard and (independently) de la Vall´ee Poussin. Recall that
ζ′(s) ζ(s)
∑^ ∞
n=
Λ(n) ns
has a simple pole at s = 1 with residue +1. If ζ(s) were to vanish at some 1 + it then −ζ′/ζ would have a simple pole with residue −1 (or − 2 , − 3 ,.. .) there. The idea is that
n Λ(n)/n s (^) converges for σ > 1, and as s approaches 1
from the right all the terms contribute towards the positive-residue pole. As σ → 1 + it from the right, the corresponding terms have the same magnitude but are multiplied by n−it, so a pole with residue −1 would force “almost all” the phases n−it^ to be near −1. But then near 1 + 2it the phases n−^2 it^ would again approximate (−1)^2 = +1, yielding a pole of positive residue, which is not possible because then ζ would have another pole besides s = 1.
To make precise the idea that if n−it^ ≈ −1 then n−^2 it^ ≈ +1, we use the identity
2(1 + cos θ)^2 = 3 + 4 cos θ + cos 2θ,
from which it follows that the right-hand side is positive. Thus if θ = t log n we have 3 + 4 Re(n−it) + Re(n−^2 it) ≥ 0.
Multiplying by Λ(n)/nσ^ and summing over n we find
3
[
ζ′ ζ
(σ)
]
[
ζ′ ζ
(σ + it)
]
[
ζ′ ζ
(σ + 2it)
]
for all σ > 1 and t ∈ R. Fix t 6 = 0. As σ→1+, the first term in the LHS of this inequality is 3/(σ − 1) + O(1), and the remaining terms are bounded below. If ζ had a zero of order r > 0 at 1+it, the second term would be − 4 r/(σ −1)+O(1). Thus the inequality yields 4r ≤ 3. Since r is an integer, this is impossible, and the proof is complete.
We next use (1), together with the partial-fraction formula
ζ′ ζ
(s) =
s − 1
+ B 1 +
s 2
ρ
s − ρ
ρ
to show that even the existence of a zero close to 1 + it is not possible. How close depends on t; specifically, we show:^1 (^1) See for instance Chapter 13 of Davenport’s book [Davenport 1967] cited earlier. This
classical bound has been improved; the current record of 1 − σ log−^2 /^3 −^ |t|, due to Korobov and perhaps Vinogradov, has stood for 40 years. See [Walfisz 1963] or [Montgomery 1971, Chapter 11].
Theorem. There is a constant c > 0 such that if |t| > 2 and ζ(σ + it) = 0 then
σ < 1 − c log |t|
Proof : Let σ ∈ [1, 2] and^2 |t| ≥ 2 in the partial-fraction formula. Then the B 1 and Γ′/Γ terms are O(log |t|), and each of the terms 1/(s − ρ), 1/ρ has positive real part as noted in connection with von Mangoldt’s theorem on N (T ). Therefore^3
− Re ζ′ ζ
(σ + 2it) < O(log |t|),
and if some ρ = 1 − δ + it then
− Re
ζ′ ζ
(σ + 2it) < O(log |t|) −
σ + δ − 1
Thus (1) yields 4 σ + δ − 1
σ − 1
In particular, taking^4 σ = 1+4δ yields 1/ 20 δ < O(log |t|). Hence δ (log |t|)−^1 , and our claim (2) follows.
Once we obtain the functional equation and partial-fraction decomposition for Dirichlet L-functions L(s, χ), the same argument will show that (2) also gives a zero-free region for L(s, χ), though with the implied constant depending on χ.
Remarks
The only properties of Λ(n) that we used in the proof of ζ(1 + it) 6 = 0 are that facts that Λ(n) ≥ 0 for all n and that
n Λ(n)/n
s (^) has an analytic continuation
with a simple pole at s = 1 and no other poles of real part ≥ 1. Thus the same argument exactly will show that
χ mod q L(s, χ), and thus each of the factors L(s, χ), has no zero on the line σ = 1.
The 3 + 4 cos θ + cos 2θ trick is worth remembering, since it has been adapted to other uses. For instance we shall revisit and generalize it when we develop the Drinfeld-Vl˘adut¸ upper bounds on points of a curve over a finite field and the Odlyzko-Stark lower bounds on discriminants of number fields. See also the following Exercises. (^2) Any lower bound > 1 would do — and the only reason we cannot go lower is that our bounds are in terms of log |t| so we do not want to allow log |t| = 0. (^3) Note that we write < O(log |t|), not = O(log |t|), to allow the possibility of an arbitrarily large negative multiple of | log t|. (^4) 1 + αδ will do for any α > 3. This requires that αδ ≤ 1, for instance δ ≤ 1 /4 for our choice of α = 4, else σ > 2; but we’re concerned only with δ near zero, so this does not matter.
