13 Zero free regions

A zero of the Riemann zeta function is a complex number \(\rho = \beta + i\gamma \) for which \(\zeta (\rho ) = 0\).

There are infinitely many “trivial" zeros of the form \(\rho = -2n\) for integer \(n \ge 1\); these zeros are well understood.
There are a countably infinite number of “non-trivial zeros lying inside the critical strip \(0 {\lt} \Re z {\lt} 1\).

Definition 13.1 Zero free region of \(\zeta (s)\)

A zero-free region of the Riemann zeta function is a set \(D \subset \mathbb {C}\) for which \(\zeta (s) \ne 0\) for all \(s \in D\).

Lemma 13.2 Basic properties of zero free regions

The following properties hold:

(Symmetry about the real axis) If \(\zeta (\sigma + it) \ne 0\) then \(\zeta (\sigma - it) \ne 0\).
(Symmetry about the critical line \(\Re s = 1/2\)) For \(0 \le \sigma \le 1\), if \(\zeta (\sigma + it) \ne 0\) then \(\zeta (1 - \sigma + it) \ne 0\).
(Non vanishing for \(\Re s {\gt} 1\)) If \(\Re s {\gt} 1\) then \(\zeta (s) \ne 0\).

Proof ▶

Claim (i) follows directly from the property \(\overline{\zeta (s)} = \zeta (\overline{s})\). Claim (ii) follows from the functional equation

\[ \zeta (s) = 2^s \pi ^{s - 1}\sin (\pi s/2) \Gamma (1 -s)\zeta (1-s) \]

and claim (iii) follows from the Euler product formula

\[ \zeta (s) = \prod _{p}(1 - p^{-s})^{-1},\qquad (\Re s {\gt} 1). \]

A well-known conjecture regarding the non-trivial zeroes of \(\zeta (s)\) is the Riemann hypothesis.

Heuristic 13.3 Riemann hypothesis

If \(\rho \) is a non-trivial zero of the Riemann zeta function, then \(\Re \rho = 1/2\).

In light of Lemma 13.2, for the rest of the chapter we will focus on the quadrant

\[ D\subseteq \{ z \in \mathbb {C}: \Re z {\gt} 1/2, \Im z {\gt} 0\} . \]

The first non-trivial zero-free region was due to de la Vallée Poussin [ 56 ] and Hadamard [ 89 ] , who showed independently that:

Theorem 13.4 Non-vanishing on the 1-line

One has \(\zeta (1 + it) \ne 0\) for any real \(t\).

Proof ▶

For \(\Re s {\gt} 1\), one has

\[ \Re \log \zeta (s) = \sum _{p}\sum _{m = 1}^{\infty }\frac{\cos (t \log p^m)}{mp^{m\sigma }} \]

where the outer sum runs through all primes. Applying this formula at \(s = \sigma , \sigma + it\) and \(\sigma + 2it\) (\(t \ne 0\)), and since \(3 + 4\cos \theta + \cos 2\theta = 2(1 + \cos \theta )^2 \ge 0\), one has

\[ 3\Re \log \zeta (\sigma ) + 4\Re \log \zeta (\sigma + it) + \Re \log \zeta (\sigma + 2it) = \sum _{p}\sum _{m = 1}^{\infty }\frac{3 + \cos (t \log p^m) + \cos (2t\log p^m)}{mp^{m\sigma }} \ge 0. \]

It follows that \(|\zeta (\sigma )^3 \zeta (\sigma + it)^4 \zeta (\sigma + 2it)| \ge 1\). Now as \(\sigma \to 1\) from above (and \(t\) remains fixed), one has

\[ \zeta (\sigma ) \ll \frac{1}{\sigma - 1},\qquad \zeta (\sigma + 2it) \ll 1, \]

since \(\zeta \) has a simple pole at \(s = 1\) and no pole at \(\sigma + 2it\). If \(\zeta (1 + it) = 0\), then \(\zeta (\sigma + it) \ll \sigma - 1\) so that \(|\zeta (\sigma )^3 \zeta (\sigma + it)^4 \zeta (\sigma + 2it)| \ll \sigma - 1\), a contradiction.

This was used to prove the prime number theorem \(\pi (x) \sim x/\log x\) as \(x \to \infty \) (in fact the two statements are equivalent).

13.1 Relation to growth rates of zeta

Using estimates of \(\zeta (\sigma + it)\) close to the line \(\sigma = 1\), one can extend the zero free region slightly inside the critical strip.

Lemma 13.5 Relation to growth exponents of zeta

Suppose and \(0 {\lt} g(t) \le 1 {\lt} f(t)\) are real-valued functions for \(t \ge 0\), with \(f(t)\) non-decreasing and tending to infinity with \(t\), and \(g(t)\) non-increasing. Suppose further that \(e^{f(t)}/g(t) = o(f(t))\). If

\[ \zeta (\sigma + it) \ll f(t)\qquad (1 - g(t) \le \sigma \le 2, t \ge 0) \]

then \(\zeta (\sigma + it) \ne 0\) for

\[ \sigma \ge 1 - A\frac{g(2t + 1)}{\log f(2t + 1)} \]

where \(A {\gt} 0\) is an absolute constant.

Proof ▶

See [ 277 , Theorem 3.10 ] .

Theorem 13.6 Classical zero free region

One has \(\zeta (\sigma + it) \ne 0\) if

\[ \sigma \ge 1 - \frac{A}{\log t}. \]

for an absolute constant \(A {\gt} 0\) and \(t\) sufficiently large.

Proof ▶

Thanks to the convexity bound \(\mu (\sigma ) \le (1-\sigma )/2\), one may take \(g(t) = 1/2\), \(f(t) = t^{1/4 + o(1)}\) in Lemma 13.5. The result follows.

This classical result has been improved in a number of works, most of which make crucial use of non-trivial estimates of certain types of exponential sums.

Theorem 13.7 Littlewood zero free region

One has \(\zeta (\sigma + it) \ne 0\) if

\[ \sigma \ge 1 - \frac{A \log \log t}{\log t} \]

for an absolute constant \(A {\gt} 0\) and \(t\) sufficiently large.

Proof ▶

Follows from the zeta bound corresponding to

\[ \mu \left(1 - \frac{k}{2^k - 2}\right) \le \frac{1}{2^k - 2} \]

for integer \(k\ge 3\), which is generated by the van der Corput exponent pair \(A^{k - 2}B(0, 1) = (\frac{1}{2^k - 2}, 1 - \frac{k - 1}{2^k - 2})\). However, one needs to make explicit the \(o(1)\) term in the bound \(\zeta (\sigma + it) \ll t^{\mu (\sigma ) + o(1)}\). In particular, by [ 277 , Theorem 5.14 ] , one has

\[ \zeta (1 - \frac{k}{2^k - 2} + it) \ll t^{1/(2^k - 2)}\log t. \]

Taking

\[ k = \left\lfloor \frac{1}{\log 2}\log \left(\frac{\log t}{\log \log t}\right)\right\rfloor \]

and using the Phragmén Lindelöf principle, one has

\[ \zeta (\sigma + it) \ll (\log t)^5,\qquad (\sigma \ge 1 - \frac{(\log \log t)^2}{\log t}), \]

so we may take \(f(t) = (\log t)^5\) and \(g(t) = (\log \log t)^2/\log t\) in Lemma 13.5.

Theorem 13.8 Chudakov zero free region

One has \(\zeta (\sigma + it) \ne 0\) if

\[ \sigma \ge 1 - \frac{1}{(\log t)^{3/4 + o(1)}} \]

for \(t\) sufficiently large.

Theorem 13.9 Korobov-Vinogradov zero free region

One has \(\zeta (\sigma + it) \ne 0\) if

\[ \sigma \ge 1 - \frac{A}{(\log t)^{2/3}(\log \log t)^{1/3}} \]

for an absolute constant \(A {\gt} 0\) and \(t\) sufficiently large.

Proof ▶

Via estimates of Vinogradov’s integral, one may obtain an estimate of the form (see e.g. Richert [ 248 ] )

\[ \zeta (\sigma + it) \ll t^{B(1 - \sigma )^{3/2}}(\log t)^{O(1)}\qquad (1/2 \le \sigma \le 1) \]

where \(B {\gt} 0\) is a constant and \(t\) sufficiently large. Take

\[ g(t) = \left(\frac{\log \log t}{\log t}\right)^{2/3} \]

so that

\[ \zeta (\sigma + it) \ll f(t) = (\log t)^{O(1)}\qquad (\sigma \ge 1 - g(t)). \]

The result follows from applying Lemma 13.5.